Chemistry In action
Foundations of
College Chemistry
Foundations of
College Chemistry
fourteenth Edition
Morris Hein
Mount San Antonio College
Susan Arena
University of Illinois, Urbana-Champaign
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10 9 8 7 6 5 4 3 2 1
About the Authors
Morris Hein earned a BS and MS in Chemistry at the University of Denver and his
PhD at the University of Colorado, Boulder. He is Professor Emeritus of Chemistry
at Mt. San Antonio College, where he regularly taught the preparatory chemistry
course and organic chemistry. He is the original author of Foundations of College
Chemistry, and his name has become synonymous with clarity, meticulous accuracy,
and a step-by-step approach that students can follow. Over the years, more than
three million students have learned chemistry using a text by Morris Hein. In addition to Foundations of College Chemistry, Fourteenth Edition, he is co-author of
Introduction to General, Organic, and Biochemistry, Tenth Edition, and Introduction
to Organic and Biochemistry. He is also co-author of Foundations of Chemistry in
the Laboratory, Fourteenth Edition, and Introduction to General, Organic and Biochemistry in the Laboratory, Tenth Edition.
Susan Arena earned a BS and MA in Chemistry at California State University-Fullerton.
She has taught science and mathematics at all levels, including middle school, high
school, community college, and university. At the University of Illinois she developed
a program for increasing the retention of minorities and women in science and engineering. This program focused on using active learning and peer teaching to encourage students to excel in the sciences. She has coordinated and led workshops and
progrms for science teachers from elementary through college levels that encourage
and support active learning and creative science teaching techniques. For several
years she was director of an Institute for Chemical Education (ICE) field center in
Southern California. In addition to Foundations of College Chemistry, Fourteenth
edition, she is co-author of Introduction to General, Organic and Biochemistry, Tenth
edition. Susan enjoys reading, knitting, traveling, classic cars, and gardening in her
spare time when she is not playing with her grandchildren.
v
Brief Contents
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
An Introduction to Chemistry
1
Standards for Measurement
13
Elements and Compounds
44
Properties of Matter
62
Early Atomic Theory and Structure
82
Nomenclature of Inorganic Compounds
98
Quantitative Composition of Compounds
121
Chemical Equations
143
Calculations from Chemical Equations
167
M
odern Atomic Theory and the Periodic Table
191
C
hemical Bonds: The Formation of Compounds from Atoms
212
The Gaseous State of Matter
248
Liquids
282
Solutions
305
Acids, Bases, and Salts
337
Chemical Equilibrium
363
Oxidation–Reduction
390
Nuclear Chemistry
417
Introduction to Organic Chemistry
441
Introduction to Biochemistry
483
Appendices
A-1
Glossary
G-1
Index
I-1
vii
contents
1 An Introduction to Chemistry
1
1.1 The Nature of Chemistry
2
Thinking Like a Chemist 2
1.2 A Scientific Approach to Problem Solving
Chemistry in action
Egyptians, the First Medicinal Chemists
3
4
The Scientific Method 4
1.3 The Particulate Nature of Matter
5
3 Elements and Compounds
7
3.1 Elements
Physical States of Matter 6
1.4 Classifying Matter
2 Standards for Measurement
2.1 Scientific Notation
2.2 Measurement and Uncertainty
2.3 Significant Figures
9
10
11
12
13
18
Multiplication or Division 18
Addition or Subtraction 19
2.5 The Metric System
21
Measurement of Length 22
Unit Conversions 23
Measurement of Mass 24
Chemistry in action
Keeping Track of Units
25
Measurement of Volume 26
2.6 D
imensional Analysis: A ProblemSolving Method
2.7 Measurement of Temperature
27
30
Chemistry in action
Setting Standards
32
Chemistry in action
Taking the Temperature of Old Faithful
33
43
44
45
46
47
Symbols of the Elements
47
Chemistry in action
Naming Elements
48
3.2 Introduction to the Periodic Table
Metals, Nonmetals, and Metalloids
14
15
16
Rounding Off Numbers 17
2.4 Significant Figures in Calculations
Distribution of Elements
Names of the Elements
34
37
38
39
41
45
Natural States of the Elements
Distinguishing Mixtures from Pure Substances 8
Review
Review Questions
Paired Exercises, Additional Exercises
Answers to Practice Exercises
2.8 Density
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to Practice
Exercises
49
50
Diatomic Elements 51
Chemistry in action
Atomic Oxygen, Friend or Foe?
3.3 Compounds and Formulas
52
Molecular and Ionic Compounds
52
Writing Formulas of Compounds
54
Composition of Compounds
52
55
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
57
58
59
60
61
4 Properties of Matter
62
4.1 Properties of Substances
63
Chemistry in action
Making Money
64
4.2 Physical and Chemical Changes
4.3 Learning to Solve Problems
65
68
x
Contents
4.4 Energy
Energy in Chemical Changes 69
Binary Ionic Compounds Containing a Metal
That Can Form Two or More Types of Cations
Conservation of Energy 70
Binary Compounds Containing Two Nonmetals
68
4.5 Heat: Quantitative Measurement
4.6 Energy in the Real World
Chemistry in action
Popping Popcorn
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
70
72
73
74
75
76
77
78
Putting It Together
C hapters
1–4 review
79
82
5.1 Dalton’s Model of the Atom
5.2 Electric Charge
83
84
Discovery of Ions 84
85
87
General Arrangement of Subatomic Particles 88
Atomic Numbers of the Elements 89
5.5 Isotopes of the Elements
Chemistry in action
Isotope Detectives
5.6 Atomic Mass
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercise, Answers to
Practice Exercises
6 Nomenclature of Inorganic
Compounds
6.1 Common and Systematic Names
6.2 Elements and Ions
Chemistry in action
What’s in a Name?
6.3 W
riting Formulas from Names of
Ionic Compounds
6.4 Naming Binary Compounds
Binary Ionic Compounds Containing a Metal
Forming Only One Type of Cation 105
Binary Acids
108
109
111
111
Naming Oxy-Acids
112
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercise, Answers to
Practice Exercises
114
115
116
117
118
Putting It Together
5 Early Atomic Theory and Structure
5.3 Subatomic Parts of the Atom
5.4 The Nuclear Atom
6.5 N
aming Compounds Containing
Polyatomic Ions
6.6 Acids
106
89
91
92
93
94
95
96
97
98
99
100
101
103
105
C hapters
5–6 review
7 Quantitative Composition of
Compounds
7.1 The Mole
7.2 Molar Mass of Compounds
7.3 Percent Composition of Compounds
119
121
122
126
129
Percent Composition from Formula 130
Chemistry in action
Vanishing Coins?
132
7.4 Calculating Empirical Formulas
7.5 Calculating the Molecular Formula from
the Empirical Formula
Review
Review Questions, Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
133
135
138
139
141
142
8 Chemical Equations
143
8.1 The Chemical Equation
144
Conservation of Mass
145
8.2 Writing and Balancing Chemical Equations
145
Information in a Chemical Equation 149
8.3 Types of Chemical Equations
Combination Reaction
150
150
Chemistry in Action
CO Poisoning—A Silent Killer
Decomposition Reaction
151
151
Single-Displacement Reaction
Double-Displacement Reaction
152
153
Contents
8.4 Heat in Chemical Reactions
8.5 Global Warming: The Greenhouse Effect
Chemistry in action
Decreasing Carbon Footprints
Review
Review Questions, Paired Exercises
Additional Exercises
Challenge Exercise, Answers to
Practice Exercises
9 Calculations from Chemical
Equations
9.1 Introduction to Stoichiometry
156
159
160
161
163
165
166
167
168
Mole–Mole Calculations
Mole–Mass Calculations
Mass–Mass Calculations
Limiting Reactant and Yield Calculations
Chemistry in action
A Shrinking Technology
170
173
174
176
177
Review
Review Questions, Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
182
183
185
187
Putting It Together
C hapters
7–9 review
10 Modern Atomic Theory and
the Periodic Table
10.1 Electromagnetic Radiation
Electromagnetic Radiation
10.2 The Bohr Atom
10.3 Energy Levels of Electrons
Chemistry in action
Atomic Clocks
10.4 A
tomic Structures of the
First 18 Elements
10.5 Electron Structures and the
Periodic Table
211
of Compounds from Atoms
212
11.1 Periodic Trends in Atomic Properties
213
Atomic Radius
213
214
Ionization Energy
214
11.2 Lewis Structures of Atoms
11.3 The Ionic Bond: Transfer of Electrons
from One Atom to Another
11.4 Predicting Formulas of Ionic Compounds
11.5 The Covalent Bond: Sharing Electrons
11.6 Electronegativity
Chemistry in Action
Trans-forming Fats
191
192
193
193
195
216
217
222
224
226
228
11.7 Lewis Structures of Compounds
229
Chemistry in Action
Strong Enough to Stop a Bullet?
232
11.8 Complex Lewis Structures
11.9 Compounds Containing Polyatomic Ions
Chemistry in Action
Chemistry or Art?
232
234
235
11.10 Molecular Shape
192
Chemistry in action
You Light Up My Life
Chemistry in action
Collecting the Elements
188
206
207
208
210
11 Chemical Bonds: The Formation
Metals and Nonmetals
A Short Review 168
9.2
9.3
9.4
9.5
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
235
The Valence Shell Electron Pair Repulsion
(VSEPR) Model 235
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises
Answers to Practice Exercises
239
240
241
243
244
245
Putting It Together
C hapters
10–11 review
246
197
198
201
202
12 The Gaseous State of Matter
12.1 Properties of Gases
Measuring the Pressure of a Gas
248
249
249
Pressure Dependence on the Number
of Molecules and the Temperature 251
xi
Contents
xii
13.7 Water, a Unique Liquid
Chemistry in action
What the Nose Knows
12.2
12.3
12.4
12.5
252
Boyle’s Law
Charles’ Law
Avogadro’s Law
Combined Gas Laws
252
256
259
260
Mole–Mass–Volume Relationships
of Gases 262
12.6 Ideal Gas Law
264
The Kinetic-Molecular Theory 266
Real Gases
266
Chemistry in action
Air Quality
267
12.7 Dalton’s Law of Partial Pressures
Chemistry in action
Getting High to Lose Weight?
12.8 Density of Gases
12.9 Gas Stoichiometry
Chemistry in Action
Reverse Osmosis?
298
Structure of the Water Molecule 298
Sources of Water for a Thirsty World 299
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
14 Solutions
268
14.1 General Properties of Solutions
14.2 Solubility
Mole–Volume and Mass–Volume
Calculations 270
300
301
302
303
304
305
306
307
The Nature of the Solute and Solvent
274
276
277
279
309
310
Saturated, Unsaturated, and Supersaturated
Solutions 310
14.3 Rate of Dissolving Solids
14.4 Concentration of Solutions
311
312
Dilute and Concentrated Solutions
Mass Percent Solution
281
308
The Effect of Temperature on Solubility
The Effect of Pressure on Solubility
272
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
297
297
267
270
270
Volume–Volume Calculations
Physical Properties of Water
313
Mass/Volume Percent (m/v)
Volume Percent
313
315
315
Molarity 315
Dilution Problems
13 Liquids
282
13.1 States of Matter: A Review
13.2 Properties of Liquids
Surface Tension
Evaporation
283
283
283
284
Vapor Pressure
285
13.3 Boiling Point and Melting Point
286
Chemistry in Action
Chemical Eye Candy
288
13.4 Changes of State
13.5 Intermolecular Forces
288
290
Dipole–Dipole Attractions
14.6 Osmosis and Osmotic Pressure
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
324
325
327
328
329
332
333
Putting It Together
C hapters
12–14 review
15 Acids, Bases, and Salts
Chemistry in Action
How Sweet It Is!
13.6 Hydrates
Chemistry in action
The Scoop on Ice Cream
320
334
290
The Hydrogen Bond 291
London Dispersion Forces
319
14.5 Colligative Properties of Solutions
293
294
295
15.1 Acids and Bases
Chemistry in action
Drug Delivery: An Acid–Base Problem
337
338
341
Contents
15.2 Reactions of Acids and Bases
Acid Reactions
342
Base Reactions
343
342
15.3 Salts
343
Chemistry in action
A Cool Fizz
344
15.4 Electrolytes and Nonelectrolytes
17 Oxidation–Reduction
346
17.1 Oxidation Number
Colligative Properties of Electrolyte
Solutions 348
Ionization of Water
Oxidation–Reduction
348
15.5 Introduction to pH
349
Chemistry in action
Ocean Corals Threatened by Increasing
Atmospheric CO 2 Levels
15.6 Neutralization
15.7 Writing Net Ionic Equations
15.8 Acid Rain
Review
Review Questions, Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
351
362
363
364
365
365
16.3 Le Châtelier’s Principle
366
Chemistry in Action
New Ways in Fighting Cavities and
Avoiding the Drill
393
17.2 Balancing Oxidation–Reduction Equations
17.3 Balancing Ionic Redox Equations
395
398
400
17.4 Activity Series of Metals
17.5 Electrolytic and Voltaic Cells
Chemistry in Action
Superbattery Uses Hungry Iron Ions
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
C hapters
15–17 review
401
403
407
407
409
410
412
413
417
18.1 Discovery of Radioactivity
418
Effect of Temperature on Equilibrium
Effect of Catalysts on Equilibrium
373
374
376
378
381
422
Gamma Rays
422
Transmutation of Elements
382
424
425
Artificial Radioactivity
425
Transuranium Elements
426
18.4 Measurement of Radioactivity
18.5 Nuclear energy
Nuclear Fission
421
421
Beta Particles
18.3 Radioactive Disintegration Series
372
Equilibrium Constants
Ion Product Constant for Water
Ionization Constants
Solubility Product Constant
Buffer Solutions: The Control of pH
Chemistry in Action
Exchange of Oxygen and Carbon
Dioxide in the Blood
371
419
18.2 A
lpha Particles, Beta Particles, and
Gamma Rays
Alpha Particles
370
414
18 Nuclear Chemistry
Natural Radioactivity
367
Effect of Concentration on
Equilibrium 368
16.4
16.5
16.6
16.7
16.8
390
Putting It Together
16.1 Rates of Reaction
16.2 Chemical Equilibrium
Effect of Volume on Equilibrium
389
391
Chemistry in Action
Sensitive Sunglasses
352
354
356
357
359
361
16 Chemical Equilibrium
Reversible Reactions
383
384
385
387
344
Dissociation and Ionization of
Electrolytes 345
Strong and Weak Electrolytes
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
426
427
427
Chemistry in Action
Does Your Food Glow in the Dark?
428
xiii
xiv
Contents
19.10 Carboxylic Acids
19.11 Esters
Nuclear Power 430
431
The Atomic Bomb
Nuclear Fusion
432
18.6 M
ass–Energy Relationship in
Nuclear Reactions
18.7 Biological Effects of Radiation
Acute Radiation Damage
433
434
434
Long-Term Radiation Damage 434
Genetic Effects
434
Chemistry in Action
A Window into Living Organisms
436
437
438
439
441
Chemistry in action
Biodiesel: Today’s Alternative Fuel
442
442
444
483
20.1 Chemistry in Living Organisms
20.2 Carbohydrates
Monosaccharides
486
488
Chemistry in action
The Taste of Umami
20.5 Enzymes
20.6 Nucleic Acids, DNA, and Genetics
DNA and Genetics
Structural Formulas and Isomerism
446
448
19.4 Alkenes and Alkynes
452
Naming Alkenes and Alkynes
Reactions of Alkenes
496
497
499
504
505
507
Putting It Together
C hapters
453
488
492
502
Review
Review Questions and Exercises
Answers to Practice Exercises
445
484
484
484
20.3 Lipids
20.4 Amino Acids and Proteins
444
19.3 Alkanes
18–20 review
508
455
455
19.5 Aromatic Hydrocarbons
456
Naming Aromatic Compounds
Monosubstituted Benzenes
Disubstituted Benzenes
457
Tri- and Polysubstituted Benzenes
Alkyl Halides
Appendices
457
457
19.6 Hydrocarbon Derivatives
458
459
460
19.7 Alcohols
Ethanol
20 Introduction to Biochemistry
Disaccharides
19.1 The Beginnings of Organic Chemistry
19.2 Why Carbon?
Methanol
473
474
477
481
482
Polysaccharides
Chemistry
Naming Alkanes
472
19.12 Polymers—Macromolecules
Review
Review Questions, Paired Exercises
Additional Exercises
Answers to Practice Exercises
440
19 Introduction to Organic
Addition
Chemistry in action
Getting Clothes CO 2 Clean!
435
Review
Review Questions
Paired Exercises
Additional Exercises
Challenge Exercises, Answers to
Practice Exercises
Hydrocarbons
469
471
461
462
463
Naming Alcohols 463
19.8 Ethers
Naming Ethers
Mathematical and Review
Using a Scientific Calculator
Units of Measurement
Vapor Pressure of Water at Various
Temperatures
V. Solubility Table
VI. Answers to Selected Exercises
VII. Answers to Putting It Together Review
Exercises
A-1
A-10
A-14
A-15
A-16
A-17
A-33
465
466
19.9 Aldehydes and Ketones
Naming Aldehydes
Naming Ketones
I.
II.
III.
IV.
467
Glossary
G-1
467
468
index
I-1
PREFACE
This new Fourteenth Edition of Foundations of College Chemistry presents chemistry as a
modern, vital subject and is designed to make introductory chemistry accessible to all beginning students. The central focus is the same as it has been from the first edition: to make
chemistry interesting and understandable to students and teach them the problem-solving
skills they will need. In preparing this new edition, we considered the comments and suggestions of students and instructors to design a revision that builds on the strengths of previous
editions. We have especially tried to relate chemistry to the real lives of our students as we
develop the principles that form the foundation for the further study of chemistry, and to provide them with problem-solving skills and practice needed in their future studies.
Foundations of College Chemistry, 14th Edition, is intended for students who have never
taken a chemistry course or those who have had a significant interruption in their studies
but plan to continue with the general chemistry sequence. Since its inception, this book has
helped define the preparatory chemistry course and has developed a much wider audience.
In addition to preparatory chemistry, our text is used extensively in one-semester general
purpose courses (such as those for applied health fields) and in courses for nonscience majors.
72
chAPTer 4
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Customization and Flexible Options to Meet
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145
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balanced
significant
figures.
We contains
have added
inequation
the text
to
highlight
(212 g)(specific heat metal)(90.8°C) = 1.60 * 104 J
each side of the equation. The balanced equation therefore obeys the law of conservation
To
determine
specific heat of the metal, we rearrange the equation
the
steps
needed
to
solve
chemistry
problems.
Painstaking
care
has
been
taken
to showtheeach
of mass.
heat)(t) = heat
Every
student must learn to balance
equations.
equations
are balanced
by solving example (mass)(specific
stepchemistry
in the problem-solving
process
andMany
to use
these
steps in
problems. We
solving for the specific heat of the metal
trial and error, but care and attention to detail are still required. The way to balance an equation
continue to use four significant figures for atomic and molar masses for consistency and for 1.60 * 104 J
is to adjust the number of atoms of each element so that they are the same on each side of the
specific
heat metal =
= 0.831 J>g°C
rounding
offformula
answers
appropriately.
haveanbeen
meticulous
providing
answers,
correctly
equation,
but a correct
is never
changed in orderWe
to balance
equation.
The general incorrect
formulas
are not
changed
(212 g)(90.8°C)
an equation.
procedure
for balancing
equations is who
as follows:
rounded,
for students
have difficulty with mathematics. to balance
Check •
Note the units in the answer agree with units for specific heat.
prObleM-SOlving Strategy: For Writing and balancing a Chemical equation
1. Identify the reaction. Write a description or word equation for the reaction. For example,
let’s consider mercury(II) oxide decomposing into mercury and oxygen.
mercury(II) oxide ¡ mercury + oxygen
2. Write the unbalanced (skeleton) equation. Make sure that the formula for each substance
is correct and that reactants are written to the left and products to the right of the arrow. For
our example,


HgO ¡
Hg + O2
The correct formulas must be known or determined from the periodic table, lists of ions, or
experimental data. Strategy
Problem-Solving
learning ObjeCtive
3. Balance the equation. Use the following process as necessary:
(a) Count and compare the number of atoms of each element on each side of the equation
and determine those that must be balanced:
Hg is balanced
(1 on each side)
O needs to be balanced
(1 on reactant side, 2 on product side)
PraCtiCe 4.3
Calculate the quantity of energy needed to heat 8.0 g of water from 42.0°C to 45.0°C.
study this procedure carefully
and refer
it when
you4work
P r to
aC
tiCe
.4
examples.
A 110.0-g sample of metal at 55.5°C raises the temperature of 150.0 g of water from
23.0°C to 25.5°C. Determine the specific heat of the metal in J>g°C.
Practice Problems
4.6 energy in The reAl World
xv
Define a hydrocarbon compound and explain its role in the world’s energy supply.
Coal, petroleum, natural gas, and woody plants provide us with a vast resource of energy, all
of which is derived from the sun. Plants use the process of photosynthesis to store the sun’s
energy, and we harvest that energy by burning the plants or using the decay products of the
plants. These decay products have been converted over millions of years to fossil fuels. As the
plants died and decayed, natural processes changed them into petroleum deposits that we now
xvi
Preface
Fostering Student Skills Attitude plays a critical role in problem solving. We encour-
age students to learn that a systematic approach to solving problems is better than simple
memorization. Throughout the book we emphasize the use of our approach to problem solving
to encourage students to think through each problem. Once we have laid the foundations of
concepts, we highlight the steps so students can locate them easily. Important rules and equations are highlighted for emphasis and ready reference.
Student Practice Practice problems follow the examples in the text, with answers pro-
vided at the end of the chapter. The end of each chapter begins with a Chapter Review and Review Questions section, which help students review key terms and concepts, as well as material
presented in tables and figures. This is followed by Paired Exercises, covering concepts and
numerical exercises, where two similar exercises are presented side by side. The final section,
Additional Exercises, includes further practice problems presented in a more random order. In
our new edition we have changed a significant number of exercises per chapter, added many
new practice problems and Examples throughout the text and added many new exercises which
emphasize practical and real-life applications of the chemical principles discussed in the text.
In addition, for many of these examples enhanced versions have been developed for use with
either the online book or WileyPlus.
Organization
We continue to emphasize the less theoretical aspects of chemistry early in the book, leaving
the more abstract theory for later. This sequence seems especially appropriate in a course where
students are encountering chemistry for the very first time. Atoms, molecules, and reactions
are all an integral part of the chemical nature of matter. A sound understanding of these topics
allows the student to develop a basic understanding of chemical properties and vocabulary.
Chapters 1 through 3 present the basic mathematics and the language of chemistry, including an explanation of the metric system and significant figures. In Chapter 4 we present chemical properties—the ability of a substance to form new substances. Then, in Chapter 5, students
encounter the history and language of basic atomic theory.
We continue to present new material at a level appropriate for the beginning student by
emphasizing nomenclature, composition of compounds, and reactions in Chapters 6 through 9
before moving into the details of modern atomic theory. Some applications of the Periodic Table
are shown in early chapters and discussed in detail in Chapters 10 and 11. Students gain confidence in their own ability to identify and work with chemicals in the laboratory before tackling
the molecular models of matter. As practicing chemists we have little difficulty ­connecting
molecular models and chemical properties. Students, especially those with no prior chemistry
background, may not share this ability to connect the molecular models and the macroscopic
properties of matter. Those instructors who feel it is essential to teach atomic theory and bonding early in the course can cover Chapters 10 and 11 immediately following Chapter 5.
New to This Edition
In Fourteenth Edition we have tried to build on the strengths of the previous editions. We
have added a new contributor to our author team, Cary Willard from Grossmont ­College
in California. Cary has added a new perspective on problem-solving and technology which
has substantially added new content to our instructional package. Cary revised the endof-chapter materials and added new real-world exercises, including many applications in
fields of interest for our students. She designed and authored the new Enhanced Examples
and worked to correlate the end-of-chapter material to the concept modules and learning
objectives as well as updating the solutions for this material. We are delighted to have her
join our team.
We continually strive to keep the material at the same level so that students can easily
read and use the text and supplemental material to learn chemistry. With a focus on problem
solving, student engagement, and clarity, some of the specific changes are highlighted below:
•
Concept Modules. Each chapter was reviewed and reorganized into concept modules
to facilitate student learning. Each module is structured to ensure that students are focused
on learning the key concepts and skills presented within the module. While the chapters
Preface
may look different in their section titles, the content remains essentially the same; it has
simply been repackaged to improve the flow and place conceptual material together.
• Each module begins with a Learning Objective which states the goal for the section
and prompts students to recognize the important concepts, providing the scaffolding for
understanding.
• New Examples model problem solving for students, and Practice Problems have been
added to provide students with opportunities to apply their knowledge. Key Terms in
each section are highlighted in bold and listed at the beginning of each section as well
as in the chapter Review. Each Key Term is also defined in the Glossary.
•
•
•
•
•
•
12.1
xvii
• properties of Gases
251
it is necessary to make conversions among the various pressure units. Since atm, torr, and mm
Hg are common
pressurethe
units,examples
we give examples
involving all three of these units:
New Enhanced Examples within WileyPLUS
extend
presented
1 atm = interactive
760 torr = 760 mm
within the Concept Modules and are designed to provide more detailed,
ex-Hg
amples and practice for students.
example 12.1
• Interactive concept questions help the stuThe average atmospheric pressure at Walnut, California, is 740. mm Hg. Calculate this
dent understand the underlying concepts
pressure in (a) torr and (b) atmospheres.
SOlutiOn
presented in the text example. The student
Let’s use conversion factors that relate one unit of pressure to another.
then has the opportunity to use the concepts
(a) To convert mm Hg to torr, use the conversion factor
to solve interactive problems closely relat760 torr>760 mm Hg (1 torr>1 mm Hg):
ed to the text examples.
1 torr
(740. mm Hg)a
b = 740. torr
• Question assistance If the student gets a
1 mm Hg
problem incorrect, immediate feedback
(b) To convert mm Hg to atm, use the conversion factor
is given to assist the student in working
1 atm>760. mm Hg:
through the problem.
1 atm
(740. mm Hg)a
b = 0.974 atm
760. mm Hg
• GOTutorials In more complex examples
the students are also given a GOTutorial to
pracTice 12.1
aid them by providing step-by-step guided
A barometer reads 1.12 atm. Calculate the corresponding pressure in (a) torr, (b) mm Hg,
and (c) kilopascals.
questions to work toward the correct solution to the problem.
Enhanced Examples have been clustered together to reinforce the connection between
pressure Dependence on the number of Molecules
similar problems and to help students recognize how
a single idea or relationship can be
and the temperature
utilized to solve a variety of problems. The problems within the Enhanced Examples are
Pressure is produced by gas molecules colliding with the walls of a container. At a specific
randomized to provide multiple opportunities for atemperature
student to
variousdepends
typeson the number of gas molecules
andpractice
volume, the solving
number of collisions
present.
The number
of collisions
can be increased by increasing the number of gas molecules
of problems. Enhanced Examples are designated by
a margin
note
and icon.
ENHANCED EXAMPLE
present. If we double the number of molecules, the frequency of collisions and the pressure
should double. We find, for an ideal gas, that this doubling is actually what happens. When the
Chemistry In Action boxes have been updated and
new boxes have been added to include
temperature and mass are kept constant, the pressure is directly proportional to the number of
different applications of the concepts in the text. moles or molecules of gas present. Figure 12.5 illustrates this concept.
A good example of this molecule–pressure relationship may be observed in an ordinary
cylinder
gas equipped with
a pressure
gauge. When the valve is opened, gas
A new concept module has been added to Chapter
13 ofoncompressed
Intermolecular
Forces
which
escapes from the cylinder. The volume of the cylinder is constant, and the decrease in quantity
includes dipole-dipole forces, hydrogen bonding, (moles)
and London
dispersion
of gas is registered
by a dropforces.
in pressure indicated on the gauge.
Some of the older industrial chemistry applications have been removed and newer applications added as appropriate throughout the text.
Figure 12.5
The pressure exerted by a gas
is directly proportional to the
number of molecules present.
In each case shown, the volume
is 22.4 L and the temperature
is 0°C.
New, modern design. The entire text has been redesigned to foster greater accessibility
and increase student engagement.
The Illustration Program has been improved to include new photos and updated art,
with an emphasis on molecular art to illustrate for the student what happens at the mi2 mol H 2
1 mol H 2
croscopic level.
P = 2 atm
P = 1 atm
0.5 mol H 2
P = 0.5 atm
Learning Aids
To help the beginning student gain the confidence necessary to master technical material, we
have refined and enhanced a series of learning aids:
Hein_14ed_c12_248-281hr.indd 251
•
•
Learning Objectives highlight the concept being taught in each section. These objectives
are tied to Example, Practice Problems, Review Exercises, and Exercises to assist the
student in mastering each concept module and objective.
Important terms are set off in bold type where they are defined, and are listed in grey at the
beginning of each section. All Key Terms listed in the Chapter Review are also defined
in the Glossary.
9/3/12 4:02 PM
xviii
Preface
•
•
•
Worked examples show students the how of problem solving using Problem-Solving
Strategies and Solution Maps before they are asked to tackle problems on their own.
Practice problems permit immediate reinforcement of a skill shown in the example problems. Answers are provided at the end of the chapter to encourage students to check their
problem solving immediately.
Marginal notations help students understand basic concepts and problem-solving techniques. These are printed in blue to clearly distinguish them from text and vocabulary terms.
Learning Aids: Math Skills For students who may need help with the mathematical
aspects of chemistry, the following learning aids are available:
A Review of Mathematics, covering the basic functions, is provided in Appendix I.
Math Survival Guide: Tips and Tricks for Science Students, 2nd Edition, by Jeffrey
R. Appling and Jean C. Richardson, a brief paperback summary of basic skills that can be
packaged with the text, provides an excellent resource for students who need help with the
mathematical aspects of chemistry.
•
•
Supplements Package
For the Student Study Guide by Rachael Henriques Porter is a self-study guide for
students. For each chapter, the Study Guide includes a self-evaluation section with student
exercises, a summary of chapter concepts, one or more “challenge problems,” and answers and
solutions to all Study Guide exercises.
Solutions Manual by Morris Hein, Cary Willard, Susan Arena, and Kathy Mitchell
­includes answers and solutions to all end-of-chapter questions and exercises.
Math Survival Guide: Tips and Tricks for Science Students, 2nd Edition, by Jeffrey
Appling and Jean Richardson, is a paperback summary of basic skills with practice exercises
in every chapter.
Foundations of Chemistry in the Laboratory, 14th Edition, by Morris Hein, Judith N.
Peisen, and Robert L. Miner includes 28 experiments for a laboratory program that may accompany the lecture course. Featuring updated information on waste disposal and emphasizing
safe laboratory procedures, the lab manual also includes study aids and exercises.
For the Instructor Test Bank, by Raymond Sadeghi, includes chapter tests with ad-
ditional test questions and answers to all test questions.
Computerized Test Bank. The test bank contains true-false, multiple-choice, and openended questions, and is available in two formats.
Alternate Editions For the convenience of instructors and to accommodate the various
lengths of academic terms, two versions of this book are available. Foundations of College
Chemistry, 14th Edition, includes 20 chapters and is our main text. Foundations of College
Chemistry, Alternate 14th Edition, provides a shorter, 17-chapter text in paperback with the
same material, but without the nuclear, organic, and biochemistry chapters.
WileyPLUS is an innovative, research-based online environment for effective teaching and
learning. WileyPLUS builds students’ confidence because it takes the guesswork out of studying by providing students with a clear roadmap: what to do, how to do it, if they did it right.
This interactive approach focuses on:
CONFIDENCE: Research shows that students experience a great deal of anxiety over studying.
That’s why we provide a structured learning environment that helps students focus on what to
do, along with the support of immediate resources.
MOTIVATION: To increase and sustain motivation throughout the semester, WileyPLUS helps
students learn how to do it at a pace that’s right for them. Our integrated resources—available
24/7—function like a personal tutor, directly addressing each student’s demonstrated needs
with specific problem-solving techniques.
Preface
SUCCESS: WileyPLUS helps to assure that each study session has a positive outcome by putting students in control. Through instant feedback and study objective reports, students know
if they did it right, and where to focus next, so they achieve the strongest results.
With WileyPLUS, our efficacy research shows that students improve their outcomes by as
much as one letter grade. WileyPLUS helps students take more initiative, so you’ll have greater
impact on their achievement in the classroom and beyond.
What Do Students Receive with WileyPLUS?
•
•
•
The complete digital textbook, saving students up to 60% off the cost of a printed text.
Question assistance, including links to relevant sections in the online digital textbook.
Immediate feedback and proof of progress, 24/7.
WileyPLUS addresses different learning styles, different levels of proficiency, and different
levels of preparation—each of your students is unique. WileyPLUS empowers them to take
advantage of their individual strengths:
•
•
•
•
Students receive timely access to resources that address their demonstrated needs, and get
immediate feedback and remediation when needed.
Integrated, multimedia resources—including audio and visual exhibits, demonstration
problems, and much more—provide multiple study paths to fit each student’s learning
preferences and encourage more active learning.
WileyPLUS includes many opportunities for self-assessment linked to the relevant portions
of the text. Students can take control of their own learning and practice until they master
the material.
Complete online version of the textbook, including relevant student study tools and learning resources to ensure positive learning outcomes, including:
• Enhanced Examples—these interactive examples are designed to promote conceptual
understanding and practice quantitative problem-solving skills. They are designated
with this icon
in the text.
ENHANCED EXAMPLE
• MathSkills module provides remedial review.
• Interactive Periodic Table
What Do Instructors Receive with WileyPLUS?
•
•
Reliable resources that reinforce course goals inside and outside of the classroom.
The ability to easily identify those students who are falling behind.
• WileyPLUS Quickstart assignments and presentations come preloaded for every chapter. Use as is, edit, or start from scratch.
• Lecture Notes PowerPoint™ Slides to summarize chapter
content.
• Image Gallery includes all line art, tables, and photos in jpeg
format. Leader lines and labels have been enhanced for projection display.
• Test Bank questions for use during in-class exams or as
online quizzes—Student Area
• Classroom Response System (Clickers) questions
• Instructor’s Solutions Manual
• All content is tagged to Learning Objectives
• New visualizations of key concepts
xix
xx
Preface
•
•
•
•
•
•
WileyPLUS Includes a robust assessment package with options created specifically to support undergraduate education in Chemistry.
All end-of-chapter questions are available for assignment and automatic grading in
­WileyPLUS.
Every end-of-chapter question has several forms of assistance that are released to students
at the instructor’s discretion. This assistance can include:
• Hints
• Step-by-step tutorials
• Wrong answer feedback
• Links to the textbook or other media
•
•
Contextual help as they are working
Test Bank
Select end-of-chapter questions which may include review questions, problems, and additional exercises
CATALYST: true concept mastery assignments for over 150 topics in the course
GO (Guided Online) Tutorial problems
•
Prelecture Checkpoint questions
Study Objective reporting delivers an individualized report on how the student has
been doing against each study objective that’s been covered within a specific assignment. It
enables students to determine where they should focus their time.
Instructors can see students’ performance on assignments and identify problem areas at a glance.
Learn More.
www.wileyplus.com
All Instructor Resources are available within WileyPLUS, or they can be accessed by contacting your local Wiley Sales Representative. Many of the assets are located on the book companion site, www.wiley.com/college/hein
Preface
For the Laboratory Foundations of Chemistry in the Laboratory, 14th Edition, by
Morris Hein, Judith Peisen, and Robert Miner, has been completely updated and revised to
reflect the most current terminology and environmental standards. Instructors can customize
their own lab manual to meet the distinct needs of their laboratory by selecting from any of the
28 experiments, adding their own experiments or exercises.
Instructor’s Companion Web Site Instructors have access to all:
•
•
•
Digital Image Library: Images from the text are available online in JPEG format. Instructors may use these to customize their presentations and to provide additional visual support
for quizzes and exams.
Test Bank: true-false, multiple-choice, and free-response questions.
Power Point Lecture Slides: Created by Chris Bradley, these slides contain lecture outlines and key topics from each chapter of the text, along with supporting artwork and
figures from the text.
Acknowledgments
Books are the result of a collaborative effort of many talented and dedicated people. We particularly want to thank our editor, Nick Ferrari, who guided the project through a complex
revision. We are grateful to Mary Ann Price for finding new and interesting photos which add
so much to the pages of our text. We also want to thank Elizabeth Swain, our production editor who kept us on track and kept all of the pieces of production running smoothly. Assunta
Petrone, of cMPreparé in Italy was amazing as our compositor and turned our revisions into
the beautiful text in our 14th edition. We also especially appreciate the work of Geraldine Osnato, Senior Product Designer who helped us figure out how to bring our text into the world of
WileyPLUS and provided so much support in the process of adding enhanced examples and
improving our end-of-chapter online materials. Thanks to others who provided valuable assistance on this revision including Marketing Manager Kristine Ruff, Senior Designer Maureen
Eide, Media Editor Evelyn Brigandi, Content Editor Alyson Rentrop, and Editorial Assistant
Ashley Gayle. We are grateful for the many helpful comments from colleagues and students
who, over the years, have made this book possible. We hope they will continue to share their
ideas for change with us, either directly or through our publisher.
We are especially thankful for the help and support of Tom Martin, our developmental editor. His positive attitude, attention to detail, efficiency, good humor, and willingness to help
in any way were indispensable in this revision. Special thanks as well to Cary for all her hard
work and quick mind. She has added to our team immeasurably.
Our sincere appreciation goes to the following reviewers who were kind enough to read and
give their professional comments.
Reviewers
For the 13th Edition:
Madeline Adamczeski
San Jose City College
Edward L. Barnes, Jr.
Fayetteville Technical
­Community College
Sean Birke
Jefferson College
Jing-Yi Chin
Suffolk Community
College
Joe David Escobar
Jr., Oxnard College
Theodore E. Fickel
Los Angeles Valley College
Melodie Graber
Oakton Community College
Dawn Richardson
Collin College
Lydia Martinez Rivera
The University of Texas—
San Antonio
Karen Sanchez
Florida Community
­College—Jacksonville
Ali O. Sezer
California University of
Pennsylvania
David B. Shaw
Madison Area Technical
College
Joy Walker
Truman College
xxi
xxii
Preface
For the 14th Edition:
Jeffrey Allison
Austin Community College
Tamara Hanna
Texas Tech Lubbock
David Peitz
Wayne State College
Jeanne Arquette
Phoenix College
Chris Hamaker
Illinois State University
Sharadha Sambasivan
Suffolk Community College
Rebecca Broyer
University of Southern
California
Claudia Hein
Diablo Valley College
Hussein Samha
Southern Utah University
Donna Iannotti
Brevard Community College
Mary Shoemaker
Pennsylvania State
­University—University Park
Michael Byler
­Community College of
­Philadelphia
Kevin Cannon
Penn State Abington
Rong Cao
­Community College of
­Allegheny City
Ken Capps
College of Central Florida
Charles Carraher
Florida Atlantic University
Loretta Dorn
Fort Hays State University
Robert Eves
Southern Utah University
Mitchel Fedak
Duquesne University
Paul Fox
Danville Community
College
Erick Fuoco
Richard J. Daley College
Amy Grant
El Camino College
Crystal Jenkins
Santa Ana College
Lee Silverberg
Penn State—Schuylkill
Jodi Kreiling
University of Nebraska at
Omaha
Gabriela Smeureanu
Hunter College
Julie Larson
Bemidji State University
Sunanda Sukumar
Albany College of Pharmacy
Anne Lerner
Santa Fe College
Paris Svoronos
QCC of Cuny
Lauren McMills
Ohio University—Main
­Campus
Susan Thomas
University of Texas—San
Antonio
Mitchel Millan
Casper College
Timothy Minger
Mesa Community College
Franklin Ow
East LA City College
Sergey Trusov
Oxnard College
Elaine Vickers
Southern Utah University
Loretta Vogel
Ocean County College
Ethel (April) Owusu
Santa Fe College
Liwen Yu
Inver Hills Community
College
Fumin Pan
Mohawk Valley Community
College
Karl Wallace
The University of Southern
Mississippi
Morris Hein and Susan Arena
Foundations of
College Chemistry
This page intentionally left blank
C hapt e r
1
An Introduction
to Chemistry
D
o you know how the beautiful, intricate fireworks
displays are created? Have you ever wondered how a
tiny seedling can grow into a cornstalk taller than you
in just one season? Perhaps you have been mesmerized by the
flames in your fireplace on a romantic evening as they change
color and form. And think of your relief when you dropped a
container and found that it was plastic, not glass. These phenomena are the result of chemistry that occurs all around us, all
the time. Chemical changes bring us beautiful colors, warmth,
light, and products to make our lives function more smoothly.
Understanding, explaining, and using the diversity of materials
we find around us is what chemistry is all about.
Chapter Outline
1.1 The Nature of Chemistry
1.2 A Scientific Approach to
Problem Solving
1.3 The Particulate Nature
of Matter
1.4 Classifying Matter
© Stocktrek Images/SuperStock
The spectacular colors of the
aurora borealis are the result of
chemistry in our atmosphere.­
2 chapter 1
• An Introduction to Chemistry
1.1 The Nature of Chemistry
L earning obje c tive
State the definition of chemistry and why the study of chemistry is important.
key term
A knowledge of chemistry is useful to virtually everyone—we see chemistry occurring around
us every day. An understanding of chemistry is useful to engineers, teachers, health care professionals, attorneys, homemakers, business people, firefighters, and environmentalists, just to
name a few. Even if you’re not planning to work in any of these fields, chemistry is important
and is used by us every day. Learning about the benefits and risks associated with chemicals will
help you to be an informed citizen, able to make intelligent choices concerning the world around
you. Studying chemistry teaches you to solve problems and communicate with others in an organized and logical manner. These skills will be helpful in college and throughout your career.
What is chemistry? One dictionary gives this definition: “Chemistry is the science of the
composition, structure, properties, and reactions of matter, especially of atomic and molecular
systems.” Another, somewhat simpler definition is “Chemistry is the science dealing with
the composition of matter and the changes in composition that matter undergoes.” Neither of
these definitions is entirely adequate. Chemistry and physics form a fundamental branch of
knowledge. Chemistry is also closely related to biology, not only because living organisms are
made of material substances but also because life itself is essentially a complicated system of
interrelated chemical processes.
The scope of chemistry is extremely broad. It includes the whole universe and everything,
animate and inanimate, in it. Chemistry is concerned with the composition and changes in the
composition of matter and also with the energy and energy changes associated with matter.
Through chemistry we seek to learn and to understand the general principles that govern the
behavior of all matter.
The chemist, like other scientists, observes nature and attempts to understand its secrets:
What makes a tulip red? Why is sugar sweet? What is occurring when iron rusts? Why is carbon monoxide poisonous? Problems such as these—some of which have been solved, some of
which are still to be solved—are all part of what we call chemistry.
A chemist may interpret natural phenomena, devise experiments that reveal the composition and structure of complex substances, study methods for improving natural processes, or
synthesize substances. Ultimately, the efforts of successful chemists advance the frontiers of
knowledge and at the same time contribute to the well-being of humanity.
chemistry
iStockphoto
Key terms are highlighted in bold
to alert you to new terms defined
in the text.
Image Source/SuperStock
A health care professional needs
to understand chemistry in order
to administer the correct dose of
medication.
Thinking Like a Chemist
Chemists take a special view of things in order to understand the nature of the chemical ­changes
taking place. Chemists “look inside” everyday objects to see how the basic components are
behaving. To understand this approach, let’s consider a lake. When we view the lake from a distance, we get an overall picture of the water and shoreline.
This overall view is called the macroscopic picture.
As we approach the lake we begin to see more
­details—rocks, sandy beach, plants submerged in the
water, and aquatic life. We get more and more curious.
What makes the rocks and sand? What kind of organisms
live in the water? How do plants survive underwater?
What lies hidden in the water? We can use a microscope to learn the answers to some of these questions. Within the water and the plants, we can
see single cells and inside them organelles
working to keep the organisms alive. For
answers to other questions, we need to
go even further inside the lake. A drop of
lake water can itself become a mysterious
and fascinating microscopic picture full of
Figure 1.1
molecules and motion. (See Figure 1.1.) A
chemist looks into the world of atoms and
Inside a drop of lake water we find water molecules,
molecules and their motions. Chemistry makes
some ­dissolved substances.
1.2
• A Scientific Approach to Problem Solving 3
the connection between the microscopic world of molecules and the macroscopic world of
everyday objects.
Think about the water in the lake. On the surface it has beauty and colors, and it gently
laps the shore of the lake. What is the microscopic nature of water? It is composed of tiny
molecules represented as
H
H
O
H
≈ ≈
O
H
In this case H represents a hydrogen atom and O an oxygen atom. The water molecule is
represented by H2O since it is made up of two hydrogen atoms and one oxygen atom.
example 1.1
You are given eight oxygen atoms and fifteen hydrogen atoms. How many water molecules can you make from them?
ENHANCED EXAMPLE
Solution
From the model shown above for a water molecule you can see that one molecule of water
contains one atom of oxygen and two atoms of hydrogen. Using this model as reference,
you can make eight water molecules from eight oxygen atoms. But you can make only
seven water molecules from fifteen hydrogen atoms with one H atom and one O atom left
over. The answer is seven water molecules.
Practice 1.1
You are given ten hydrogen atoms and eight oxygen atoms. How many water molecules
can you make from them?
1.2A Scientific Approach
to Problem Solving
Describe the steps involved in the scientific method.
Learning objective
One of the most common and important things we do every day is to solve problems.
For example,
Key Terms
• You have two exams and a laboratory report due on Monday. How should you ­divide
your time?
• You leave for school and learn from the radio that there is a big accident on the freeway.
What is your fastest alternate route to avoid the traffic problem?
• You need to buy groceries, mail some packages, attend your child’s soccer game, and
pick up the dry cleaning. What is the most efficient sequence of events?
We all face these kinds of problems and decisions. A logical approach can be useful for solving daily problems:
1. Define the problem. We first need to recognize we have a problem and state it clearly,
including all the known information. When we do this in science, we call it making an
observation.
2. Propose possible solutions to the problem. In science this is called making a hypothesis.
3. Decide which is the best way to proceed or solve the problem. In daily life we use our
memory of past experiences to help us. In the world of science we perform an experiment.
Using a scientific approach to problem solving is worthwhile. It helps in all parts of your life
whether you plan to be a scientist, doctor, business person, or writer.
scientific method
hypothesis
theory
scientific laws
4 chapter 1
• An Introduction to Chemistry
>Chemistry in action
Look at any images of the ancient Egyptians and notice
the black eyeliner commonly worn at that time. As chemists analyzed the composition of a sample of this eyeliner
in the antiquities collection at the Louvre Museum in Paris,
they were appalled to discover the high concentration
of lead in the samples. Today lead is
routinely removed from most consumer
products because it is very toxic even in
low concentrations. It is toxic to many
organs and can cause symptoms such
as abdominal pain, dementia, anemia,
seizures, and even death. It turns out the
lead compounds found in the Egyptian
eyeliner are not found in nature but
must be synthesized. The synthesis of
these lead salts is complicated and the
products are not lustrous. This led chemists to question why the Egyptians would
add these compounds to their eyeliner.
The answer was revealed after reading
some of the ancient manuscripts from
that time. Lead salts were synthesized for use in treating
eye ailments, scars, and discolorations. So even if the lead
salts were not the best ingredients for beauty, they were
added for the perceived health benefits.
Since we now know that lead compounds are very toxic,
Christian Amatore, an analytical chemist at the Ecole
Normale Supérieure in Paris, wondered if
the lead compounds in Egyptian eyeliner
could have actually conferred any health
benefits. He introduced lead salts into
samples of human tissue growing in the
laboratory and observed that the cells
began forming compounds that trigger
an immune response. Perhaps the ancient
Egyptians did know something about
medicinal chemistry after all. So should
we follow the Egyptians example and add
lead to our cosmetics? This is probably not
a good idea because the risks associated
with prolonged lead exposure outweigh
the benefits.
age fotostock/SuperStock
Egyptians, the First Medicinal Chemists
Terra cotta sculpture of Nefertiti
The Scientific Method
Chemists work together and also with other scientists to solve problems. As scientists conduct
studies they ask many questions, and their questions often lead in directions that are not part
of the original problem. The amazing developments from chemistry and technology usually
involve what we call the ­scientific method, which can generally be described as follows:
Scott Bauer/Courtesy USDA
1. Collect the facts or data that are relevant to the problem or question at hand. This is
usually done by planned experimentation. The data are then analyzed to find trends or
regularities that are pertinent to the problem.
2. Formulate a hypothesis that will account for the data and that can be tested by further
experimentation.
3. Plan and do additional experiments to test the hypothesis.
4. Modify the hypothesis as necessary so that it is compatible with all the pertinent data.
Scientists employ the scientific­
method every day in their
laboratory­work.
Confusion sometimes arises regarding the exact meanings of the words hypothesis, theory, and law. A hypothesis is a tentative explanation of certain facts that provides a basis
for further experimentation. A well-established ­hypothesis is often called a theory or model.
Thus, a theory is an explanation of the general principles of certain phenomena with considerable evidence or facts to support it. Hypotheses and theories explain natural phenomena,
whereas scientific laws are simple statements of natural phenomena to which no exceptions
are known under the given conditions.
These four steps are a broad outline of the general procedure that is followed in most scientific work, but they are not a “recipe” for doing chemistry or any other science (Figure 1.2).
Chemistry is an experimental science, however, and much of its progress has been due to
application­of the scientific method through systematic ­research.
1.3
Laws
Observations
(analysis)
• The Particulate Nature of Matter 5
(explanation)
Hypothesis
Experiment
(analysis)
Theory
(model)
Figure 1.2
The scientific method.
We study many theories and laws in chemistry; this makes our task as ­students easier
because theories and laws summarize important aspects of the sciences. Certain theories
and models advanced by great scientists in the past have since been substantially altered
and modified­. Such changes do not mean that the discoveries of the past are any less
significant. Modification of existing theories and models in the light of new experimental ­evidence is essential to the growth and evolution of scientific knowledge. Science is
dynamic.
1.3 The Particulate Nature of Matter
Describe the characteristics of matter, including the states of matter.
Learning objectiv e
The entire universe consists of matter and energy. Every day we come into contact with countless kinds of matter. Air, food, water, rocks, soil, glass, and this book are all different types of
matter. Broadly defined, matter is anything that has mass and occupies space.
Matter may be quite invisible. For example, if an apparently empty test tube is submerged
mouth downward in a beaker of water, the water rises only slightly into the tube. The water
cannot rise further because the tube is filled with invisible matter: air (see Figure 1.3).
To the macroscopic eye, matter appears to be continuous and unbroken. We are impressed
by the great diversity of matter. Given its many forms, it is difficult to believe that on a microscopic level all of matter is composed of discrete, tiny, fundamental particles called atoms
(Figure 1.4). It is truly amazing to understand that the fundamental particles in ice cream
are very similar to the particles in air that we breathe. Matter is actually discontinuous and is
composed of discrete, tiny particles called atoms.
key terms
matter
solid
amorphous
liquid
gas
Figure 1.3
An apparently empty test tube is
submerged, mouth downward,
in water. Only a small volume of
water rises into the tube, which is
actually filled with invisible
matter–air.
Figure 1.4
© Andrew Dunn/Alamy
Silicon atoms on a silicon chip produced this
image using a scanning tunneling microscope.
• An Introduction to Chemistry
The three states of matter.
(a) Solid—water molecules are
held together rigidly and are
very close to each other.
(b) Liquid—water molecules are
close together but are free to
move around and slide over each
other. (c) Gas—water molecules
are far apart and move freely and
randomly.
Solid (Ice)
(a)
Liquid (Water)
(b)
© Can Balcioglu/iStockphoto
Figure 1.5
© Carlos Alvarez/iStockphoto
chapter 1
© Diane Diederich/iStockphoto
6 Gas (Steam)
(c)
Physical States of Matter
Charles D. Winters/Photo Researchers, Inc.
Matter exists in three physical states: solid, liquid, and gas (see Figure 1.5). A solid has a
definite shape and volume, with particles that cling rigidly to one another. The shape of a solid
can be independent of its container. In Figure 1.5a we see water in its solid form. Another
example, a crystal of sulfur, has the same shape and volume whether it is placed in a beaker
or simply laid on a glass plate.
Most commonly occurring solids, such as salt, sugar, quartz, and metals, are crystalline.
Crystalline solid
The particles that form crystalline materials exist in regular, repeating, three-dimensional,
geometric patterns (see Figure 1.6). Some solids such as plastics, glass, and gels do not have
any regular, internal geometric pattern. Such solids are called amorphous solids. (Amorphous
means “without shape or form.”)
A liquid has a definite volume but not a definite shape, with particles that stick firmly but
not rigidly. Although the particles are held together by strong attractive forces and are in close
contact with one another, they are able to move freely. Particle mobility gives a liquid fluidity
and causes it to take the shape of the container in which it is stored. Note how water looks as
a liquid in Figure 1.5b.
Amorphous solid
A gas has indefinite volume and no fixed shape, with particles that move independently of one another. Particles in the gaseous state have gained enough energy to overcome the attractive forces that held them together as liquids or solids.
A gas presses continuously in all directions on the walls of any container.
Because of this quality, a gas completely fills a container. The particles
of a gas are relatively far apart compared with those of solids and liquids. The actual volume of the gas particles is very small compared
with the volume of the space occupied by the gas. Observe the large
space between the water molecules in Figure 1.5c compared to ice and
liquid water. A gas therefore may be compressed into a
very small volume or expanded almost indefinitely. Liquids cannot be compressed to
any great extent, and solids are even
less compressible than liquids.
If a bottle of ammonia solution
is opened in one corner of the laboratory, we can soon smell its familiar
Figure 1.6
odor in all parts of the room. The amA large crystal of table salt. A salt crystal is
monia gas escaping from the solution
composed of a three-dimensional array of particles.
demonstrates that gaseous particles
move freely and rapidly and tend to
Na + –
permeate the entire area into which they
Cl
are released.
1.4
• Classifying Matter 7
Although matter is discontinuous, attractive forces exist that hold the particles together
and give matter its appearance of continuity. These attractive forces are strongest in solids, giving them rigidity; they are weaker in liquids but still strong enough to hold liquids to definite
volumes. In gases, the attractive forces are so weak that the particles of a gas are practically
independent of one another. Table 1.1 lists common materials that exist as solids, liquids, and
gases. Table 1.2 compares the properties of solids, liquids, and gases.
Table 1.1
Common Materials in the Solid, Liquid, and Gaseous States of Matter
Solids
Liquids
Gases
Aluminum
Copper
Gold
Polyethylene
Salt
Sand
Steel
Sulfur
Alcohol
Blood
Gasoline
Honey
Mercury
Oil
Vinegar
Water
Acetylene
Air
Butane
Carbon dioxide
Chlorine
Helium
Methane
Oxygen
Table 1.2
Physical Properties of Solids, Liquids, and Gases
State
Shape
Volume
Particles
Solid
Definite
Definite
Rigidly clinging;
tightly packed
Liquid
Indefinite
Definite
Mobile; adhering
Indefinite
Indefinite
Independent
Gas
of each other
and relatively
far apart
Compressibility
Very slight
Slight
High
1.4 Classifying Matter
Distinguish among a pure substance, a homogeneous mixture, and a heterogeneous
mixture.
Learning objective
The term matter refers to all materials that make up the universe. Many thousands of distinct
kinds of matter exist. A substance is a particular kind of matter with a definite, fixed composition. Sometimes known as pure substances, substances are either elements or compounds.
Familiar examples of elements are copper, gold, and oxygen. Familiar compounds are salt,
sugar, and water. We’ll discuss elements and compounds in more detail in Chapter 3.
We classify a sample of matter as either homogeneous or heterogeneous by ­examining it.
Homogeneous matter is uniform in appearance and has the same properties throughout. Matter
consisting of two or more physically distinct phases is heterogeneous. A phase is a homogeneous part of a system separated from other parts by physical boundaries. A system is simply
the body of matter under consideration. Whenever we have a system in which visible boundaries
exist between the parts or components, that system has more than one phase and is heterogeneous. It does not matter whether these components are in the solid, liquid, or gaseous states.
A pure substance may exist as different phases in a heterogeneous system. Ice floating in
water, for example, is a two-phase system made up of solid water and liquid water. The water
in each phase is homogeneous in composition, but because two phases are present, the system
is heterogeneous.
A mixture is a material containing two or more substances and can be either heterogeneous
or homogeneous. Mixtures are variable in composition. If we add a spoonful of sugar to a glass of
water, a heterogeneous mixture is formed immediately. The two phases are a solid (sugar) and a
liquid (water). But upon stirring, the sugar dissolves to form a homogeneous ­mixture or solution.
key terms
substance
homogeneous
heterogeneous
phase
system
mixture
chapter 1
• An Introduction to Chemistry
Richard Megna/Fundamental Photographs
(a) Water is the liquid in the
beaker, and the white solid in the
spoon is sugar. (b) Sugar can be
dissolved in the water to produce
a solution.
(a)
Flowcharts can help you to
visualize the connections
between­concepts.
Pure substances
(homogeneous composition)
Elements
Compounds
Richard Megna/Fundamental Photographs
8 (b)
Both substances are still present: All parts of the solution are sweet and wet. The proportions of
sugar and water can be varied simply by adding more sugar and stirring to dissolve. Solutions
do not have to be liquid. For example, air is a homogeneous mixture of gases. Solid solutions
also exist. Brass is a homogeneous solution of copper and zinc.
Many substances do not form homogeneous
Matter
mixtures. If we mix sugar and fine white sand, a
heterogeneous mixture is formed. Careful exami­
nation may be needed to decide that the mixture is
Mixtures of two
heterogeneous because the two phases (sugar and
or more substances
sand) are both white solids. Ordinary matter ex­
ists mostly as mixtures. If we examine soil, gran­
ite, iron ore, or other naturally occurring mineral
Solutions
Heterogeneous
(homogeneous
mixtures
deposits, we find them to be heterogeneous mix­
composition—
(two or more
tures. Figure 1.7 illustrates the relationships of
one phase)
phases)
­substances and mixtures.
Figure 1.7
Classification of matter. A
pure substance is always
homogeneous in composition,
whereas a mixture always
contains two or more substances
and may be either homogeneous
or heterogeneous.
Distinguishing Mixtures from Pure Substances
Single substances—elements or compounds—seldom occur naturally in a pure state. Air is a
mixture of gases; seawater is a mixture of a variety of dissolved minerals; ordinary soil is a
complex mixture of minerals and various organic materials.
Mixture
Pure Substance
1.A mixture always contains two or more
substances that can be present in vary­
ing amounts.
1.A pure substance (element or compound)
always has a definite composition by
mass.
2.The components of a mixture do not
lose their identities and may be sepa­
rated by physical means.
2.The elements in a compound lose their
identities and may be separated only by
chemical means.
Practice 1. 2
Which of the following is a mixture and which is a pure substance?
Explain your answer.
(a) vinegar (4% acetic acid and 96% water)
(b) sodium chloride (salt) solution
(c) gold
(d) milk
Review 9
(a)
Ken Karp
Ken Karp
How is a mixture distinguished from a pure substance? A mixture always contains
two or more substances that can be present in varying concentrations. Let’s consider two
examples.
Homogeneous Mixture Homogeneous mixtures (solutions) containing either 5% or
10% salt in water can be prepared simply by mixing the correct amounts of salt and water.
These mixtures can be separated by boiling away the water, leaving the salt as a residue.
H
eterogeneous Mixture The composition of a heterogeneous mixture of sulfur crystals
and iron filings can be varied by merely blending in either more sulfur or more iron filings.
This mixture can be separated physically by using a magnet to attract the iron.
(b)
C hapt e r
(a) When iron and sulfur exist as
pure substances, only the iron
is attracted to a magnet. (b) A
mixture of iron and sulfur can be
separated by using the difference
in magnetic attraction.
1 review
1.1 The Nature of Chemistry
• Chemistry is important to everyone because chemistry occurs all around us in our daily lives.
• Chemistry is the science dealing with matter and the changes in composition that matter undergoes.
• Chemists seek to understand the general principles governing the behavior of all matter.
• Chemistry “looks inside” ordinary objects to study how their components behave.
• Chemistry connects the macroscopic and microscopic worlds.
Key Term
chemistry
1.2 A Scientific Approach to Problem Solving
• Scientific thinking helps us solve problems in our daily lives.
• General steps for solving problems include:
• Defining the problem
• Proposing possible solutions
• Solving the problem
• The scientific method is a procedure for processing information in which we:
• Collect the facts
• Formulate a hypothesis
• Plan and do experiments
• Modify the hypothesis if necessary
Laws
Observations
(analysis)
(explanation)
Hypothesis
Experiment
(analysis)
Theory
(model)
Key Terms
scientific method
hypothesis
theory
scientific laws
10 chapter 1
• An Introduction to Chemistry
1.3 The Particulate Nature of Matter
Key Terms
matter
solid
amorphous
liquid
gas
• Matter is anything with the following two characteristics:
• Has mass
• Occupies space
• On the macroscopic level matter appears continuous.
• On the microscopic level matter is discontinuous and composed of atoms.
• Solid—rigid substance with a definite shape
• Liquid—fluid substance with a definite volume that takes the shape of its container
• Gas—takes the shape and volume of its container
1.4 Classifying Matter
Key Terms
substance
homogeneous
heterogeneous
phase
system
mixture
• Matter can be classified as a pure substance or a mixture.
• A mixture has variable composition:
• Homogeneous mixtures have the same properties throughout.
• Heterogeneous mixtures have different properties in different parts of the system.
• A pure substance always has the same composition. There are two types of pure substances:
• Elements
• Compounds
Matter
Pure substances
(homogeneous composition)
Elements
Compounds
Mixtures of two
or more substances
Solutions
(homogeneous
composition—
one phase)
Heterogeneous
mixtures
(two or more
phases)
review Questions
5. Define a phase.
6. How many phases are present in the graduated cylinder?
7. What is another name for a homogeneous
mixture?
8. Which liquids listed in Table 1.1 are not
mixtures?
9. Which of the gases listed in Table 1.1 are not
pure substances?
10. When the stopper is removed from a partly filled bottle containing
solid and liquid acetic acid at 16.7°C, a strong vinegar-like odor
is noticeable immediately. How many acetic acid phases must be
present in the bottle? Explain.
11. Is the system enclosed in the bottle in Question 10 homogeneous or
heterogeneous? Explain.
12. Is a system that contains only one substance necessarily
­homogeneous? Explain.
13. Is a system that contains two or more substances necessarily
­heterogeneous? Explain.
14. Distinguish between homogeneous and heterogeneous mixtures.
15. Which of the following are pure substances?
(a) sugar
(d) maple syrup
(b) sand
(e) eggs
(c) gold
16. Use the steps of the scientific method to help determine the reason
that your cell phone has suddenly stopped working:
(a) observation
(c) experiment
(b) hypothesis
(d) theory
Richard Megna/
Fundamental Photographs
1. Explain the difference between
(a) a hypothesis and a theory
(b) a theory and a scientific law
2. Consider each of the following statements and determine whether
it represents an observation, a hypothesis, a theory, or a scientific
law:
(a) The battery in my watch must be dead since it is no longer
keeping time.
(b) My computer must have a virus since it is not working properly.
(c) The air feels cool.
(d) The candle burns more brightly in pure oxygen than in air because oxygen supports combustion.
(e) My sister wears red quite often.
(f) A pure substance has a definite, fixed composition.
3. Determine whether each of the following statements refers to a
solid, a liquid, or a gas:
(a) It has a definite volume but not a definite shape.
(b) It has an indefinite volume and high compressibility.
(c) It has a definite shape.
(d) It has an indefinite shape and slight compressibility.
4. Some solids have a crystalline structure, while others have an amorphous structure. For each of the following 5 descriptions, determine
whether it refers to a crystalline solid or an amorphous solid:
(a) has a regular repeating pattern
(b) plastic
(c) has no regular repeating pattern
(d) glass
(e) gold
Additional Exercises 11
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com)
All exercises with blue numbers have answers in Appendix VI.
PAIRE D E x e r c i s e s
3.Look at the photo and determine whether it represents a homogeneous or heterogeneous mixture.
4.Look at the maple leaf below and determine whether it represents a
homogeneous or heterogeneous mixture.
5.For each of the following mixtures, state whether it is homogeneous
or heterogeneous:
(a) tap water
(b) carbonated beverage
(c) oil and vinegar salad dressing
(d) people in a football stadium
© Justin Horrocks/iStockphoto
2.Refer to the illustration and determine which states(s) of matter
are present.
Richard Megna/Fundamental
Photographs
1. Refer to the illustration and determine which state(s) of matter
are present.
6.For each of the following mixtures, state whether it is homogeneous
or heterogeneous:
(a) stainless steel
(b) motor oil
(c) soil
(d) a tree
A D D ITIONAL EXERCISES
7. A
t home, check your kitchen and bathroom cabinets for five
different substances; then read the labels and list the first ingredient of each.
8.During the first week of a new semester, consider that you have enrolled
in five different classes, each of which meets for 3 hours per week. For
every 1 hour that is spent in class, a minimum of 1 hour is required
outside of class to complete assignments and study for exams. You also
work 20 hours per week, and it takes you 1 hour to drive to the job site
and back home. On Friday nights, you socialize with your friends. You
are fairly certain that you will be able to successfully complete the semester with good grades. Show how the steps in the scientific method
can help you predict the outcome of the semester.
12 chapter 1
• An Introduction to Chemistry
Use the following food label to answer Exercises 9–10.
Nutrition Amount/serving
%DV*
Facts
18%
Total Fat 12g
Serv. Size 1 cup
30%
Sat. Fat 6g
(249g)
Servings About 3 Polyunsat. Fat 1.5g
Calories 250
Monounsat. Fat 2.5g
Fat Cal. 110
20%
*Percent Daily Values Cholest. 60mg
(DV) are based on a
Vitamin A 0% Vitamin C 0%
2,000 calorie diet.
Amount/serving %DV*
Sodium 940mg 39%
Total Carb. 24g 8%
Dietary Fiber 1g 4%
Sugars 1g
Protein 10g 20%
Calcium 6% Iron 8%
12. Various chemical elements are pictured. For each one a microscopic
view is also shown. Determine the number of phases for each substance below and identify them.
(a) iodine
INGREDIENTS: WATER, CHICKEN STOCK, ENRICHED PASTA (SEMOLINA
WHEAT FLOUR, EGG WHITE SOLIDS, NIACIN, IRON, THIAMINE MONONITRATE
[VITAMIN B1], RIBOFLAVIN [VITAMIN B2] AND FOLIC ACID), CREAM (DERIVED
FROM MILK), CHICKEN, CONTAINS LESS THAN 2% OF: CHEESES (GRANULAR,
PARMESAN AND ROMANO PASTE [PASTERURIZED COW’S MILK, CULTURES,
SALT, ENZYMES], WATER, SALT, LACTIC ACID, CITRIC ACID AND DISODIUM
PHOSPHATE), BUTTER (PASTERURIZED SWEET CREAM [DERIVED FROM MILK]
AND SALT), MODIFIED CORN STARCH, SALT, WHOLE EGG SOLIDS, SUGAR,
RICE STARCH, GARLIC, SPICE, XANTHAM GUM, MUSTARD FLOUR, ISOLATED
SOY PROTEIN AND SODIUM PHOSPHATE.
© 2001 Richard Megna/
Fundamental Photographs
(b) bromine
9.Identify the following ingredients as either a pure substance or a
­mixture.
(a) water
(b) chicken stock (the liquid that remains in the pot after cooking the
chicken)
(c) salt
(d) mustard flour
10.Using the food label, make a hypothesis regarding the nutritional
quality of this food. Propose a way to determine whether your hypothesis is valid.
© 1995 Richard Megna/
Fundamental Photographs
(c) sulfur
11. Read the following passage (from Science News) and identify the
observation and hypothesis.
Could an invisibility cloak, like the one used by Harry Potter,
really exist? For scientists in Cambridge, Massachusetts, it may
be a reality. Researchers at MIT have developed an invisibility
cloak for small objects such as an ant or a grain of sand. Calcite
crystals have the ability to reflect light around an object, rendering it invisible, or at least “unseeable.” If a larger crystal were to
be used, it should be able to hide larger objects. The invisibility
cloak works only with laser light aimed directly at the crystal.
Future work will improve the effectiveness of these invisibility
cloaks.
Stocktrek Images/Richard Roscoe/Getty
Images, Inc.
Answers to Practice Exercises
1.1
five water molecules: H2O, H2O, H2O, H2O, H2O
1.2
( a) mixture; concentration can be changed by adding more acetic
acid or more water. (b) mixture; concentration can be changed by
adding more or less salt. (c) pure substance; gold is 100% gold.
(d) mixture; milk contains several substances.
Chapter
2
Standards for
Measurement
D
oing an experiment in chemistry is very much like cooking a
meal in the kitchen. It’s important to know the ingredients
and the amounts of each in order to have a tasty product.
Working on your car requires specific tools in exact sizes. Buying
new carpeting or draperies is an exercise in precise and ­accurate
measurement for a good fit. A small difference in the concentration
or amount of medication a pharmacist gives you may have significant effects on your well-being. As we saw in Chapter 1, observation
is an important part of the scientific process. ­Observations can be
qualitative (the substance is a blue solid) or quantitative (the mass
of the substance is 4.7 grams). A quantitative observation is called
a measurement. Both a number and a unit are ­required for a measurement. For example, at home in your kitchen measuring 3 flour is
not ­possible. We need to know the unit on the 3. Is it cups, grams, or
tablespoons? A measurement of 3 cups tells us both the amount and
the size of the measurement. In this chapter we will discuss measurements and the rules for calculations using these measurements.
Chapter Outline
2.1 Scientific Notation
2.2 Measurement and Uncertainty
2.3 Significant Figures
2.4 Significant Figures in Calculations
2.5 The Metric System
2.6 Dimensional Analysis:
A Problem-Solving Method
2.7 Measurement of Temperature
2.8 Density
Blend Images/SuperStock
Careful and accurate measurements for each ingredient are essential when baking or cooking as
well as in the chemistry laboratory.
14 chapter 2
• Standards for Measurement
2.1 Scientific Notation
Learning objective
Write decimal numbers in scientific notation.
key terms
Scientists often use numbers that are very large or very small in measurements. For example,
the Earth’s age is estimated to be about 4,500,000 (4.5 billion) years. Numbers like these are
bulky to write, so to make them more compact scientists use powers of 10. Writing a number
as the product of a number ­between 1 and 10 multiplied by 10 raised to some power is called
scientific notation.
To learn how to write a number in scientific notation, let’s consider the number 2468. To
write this number in scientific notation:
measurement
scientific notation
1. Move the decimal point in the original number so that it is located after the first nonzero
digit.
2468
2.468 (decimal moves three places to the left)
2. Multiply this new number by 10 raised to the proper exponent (power). The proper exponent is equal to the number of places that the decimal point was moved.
Examples show you ­
problem-solving techniques
in a step-by-step form. Study
each one and then try the
­Practice Exercises.
2.468 * 103
3. The sign on the exponent indicates the direction the decimal was moved.
moved right S negative exponent
moved left S positive exponent
Example 2.1
ENHANCED EXAMPLE
Write 5283 in scientific notation.
SOLUTION
5283. Place the decimal between the 5 and the 2. Since the decimal
3
was moved three places to the left, the power of 10 will be 3
and the number 5.283 is multiplied by 103.
5.283 * 103
Example 2.2
Write 4,500,000,000 in scientific notation (two digits).
SOLUTION
4 500 000 000. Place the decimal between the 4 and the 5. Since the
decimal was moved nine places to the left, the power of
9
10 will be 9 and the number 4.5 is multiplied by 109.
NASA
Scientific notation is a useful
way to write very large numbers,
such as the distance between the
Earth and the moon, or very small
numbers, such as the length of
these E. coli bacteria (shown here
as a colored scanning electron
micrograph * 14,000).
Dr. Gopal Murti/Photo Researchers, Inc.
4.5 * 109
2.2
• Measurement and Uncertainty 15
Example 2.3
Write 0.000123 in scientific notation.
SOLUTION
0.000123 Place the decimal between the 1 and the 2. Since the deci4
mal was moved four places to the right, the power of 10
will be -4 and the number 1.23 is multiplied by 10-4.
1.23 * 10-4
Answers to Practice Exercises are
found at the end of each chapter.
Practice 2.1
Write the following numbers in scientific notation:
(a) 1200 (four digits)
(c) 0.0468
(b) 6,600,000 (two digits)
(d) 0.00003
2.2 Measurement and Uncertainty
Explain the significance of uncertainty in measurement in chemistry and how
­significant figures are used to indicate a measurement’s certainty.
Learning objective
To understand chemistry, it is necessary to set up and solve problems. Problem solving requires
an understanding of the mathematical operations used to ­manipulate numbers. Measurements
are made in an experiment, and chemists use these data to calculate the extent of the physical and chemical changes ­occurring in the substances that are being studied. By appropriate
calculations, an experiment’s results can be compared with those of other experiments and
summarized in ways that are meaningful.
A measurement is expressed by a numerical value together with a unit of that measurement.
For example,
key term
significant figures
A measurement always
requires a unit.
numericalvalue
value
numerical
70.0 kilograms �154 pounds
unit
Whenever a measurement is made with an instrument
such as a thermo­meter or ruler, an estimate is required. We
can illustrate this by measuring temperature. Suppose we
measure temperature on a thermometer cali­brated in degrees
and observe that the mercury stops between 21 and 22 (see
Figure 2.1a). We then know that the temperature is at least
21 degrees and is less than 22 degrees. To express the temperature with one more digit, we estimate that the mercury
is about two-tenths the distance between 21 and 22. The
temperature is the­re­fore 21.2 degrees. The last digit (2) has
some ­uncertainty because it is an estimated value. Because
the last digit has some uncertainty (we made a visual estimate), it may be different when another ­person makes the
same measurement. If three more people make this same
reading, the results might be
Person
Measurement
1
21.2
2
21.3
3
21.1
4
21.2
(a)
(b)
(c)
30
30
30
20
20
20
21.2°C
°
precise to
o 0.1°C
22.0°C
precise to 0.1°C
22.11°C
precise to 0.01°C
Figure 2.1
Measuring temperature (°C) with various degrees of precision.
16 chapter 2
• Standards for Measurement
Notice that the first two digits of the measurements did not change (they are certain). The last
digit in these measurements is uncertain. The custom in science is to record all of the certain
digits and the first uncertain digit.
Numbers obtained from a measurement are never exact values. They always have some
degree of uncertainty due to the limitations of the measuring ­instrument and the skill of the
individual making the measurement. It is customary when recording a measurement to include
all the digits that are known plus one digit that is estimated. This last ­estimated digit introduces
some uncertainty. Because of this uncertainty, every number that expresses a measurement can
have only a limited number of digits. These digits, used to express a measured quantity, are
known as significant figures.
Now let’s return to our temperature measurements. In Figure 2.1a the temperature is
recorded as 21.2 degrees and is said to have three significant figures. If the mercury stopped
exactly on the 22 (Figure 2.1b), the temperature would be recorded as 22.0 degrees. The zero is
used to indicate that the temperature was estimated to a precision of one-tenth degree. Finally,
look at Figure 2.1c. On this thermometer, the temperature is recorded as 22.11°C (four significant figures). Since the thermometer is calibrated to tenths of a degree, the first ­estimated
digit is the hundredths.
Example 2.4
The length of a ballpoint pen was recorded by three students as 14.30 cm, 14.33 cm, and
14.34 cm.
(a) What are the estimated digits in these measurements?
(b) How many figures are certain?
Solution
(a) The estimated digits are 0, 3, and 4, respectively.
(b) Three figures are certain in each measurement
Practice 2.2
Three measurements for the boiling point of water are 100.4°C, 100.1°C, and 100.0°C.
The accepted boiling point is 100.0°C.
(a) What are the estimated digits?
(b) How many figures are certain?
2.3 Significant Figures
Learning objective
Determine the number of significant figures in a given measurement and round measurements to a specific number of significant figures.
key term
Because all measurements involve uncertainty, we must use the proper number of significant figures in each measurement. In chemistry we frequently do calculations involving measurements,
so we must understand what happens when we do arithmetic on numbers containing uncertainties.
We’ll learn several rules for doing these calculations and figuring out how many significant figures
to have in the result. You will need to follow these rules throughout the calculations in this text.
The first thing we need to learn is how to determine how many significant figures are in
a number.
rounding off numbers
Rules for counting significant figures
1. Nonzero digits. All nonzero digits are significant.
2. Exact numbers. Some numbers are exact and have an infinite number of significant figures. Exact numbers occur in simple counting operations; when
you count 25 dollars, you have exactly 25 dollars. Defined numbers, such as
2.3
12 inches in 1 foot, 60 minutes in 1 hour, and 100 centimeters in 1 meter, are
also considered to be exact numbers. Exact numbers have no uncertainty.
3. Zeros. A zero is significant when it is
• between nonzero digits:
205 has three significant figures (2, 0, 5)
2.05 has three significant figures (2, 0, 5)
61.09 has four significant figures (6, 1, 0, 9)
• at the end of a number that includes a decimal point:
0.500 has three significant figures (5, 0, 0)
25.160 has five significant figures (2, 5, 1, 6, 0)
3.00 has three significant figures (3, 0, 0)
20. has two significant figures (2, 0)
A zero is not significant when it is
• before the first nonzero digit. These zeros are used to locate a decimal point:
0.0025 has two significant figures (2, 5)
0.0108 has three significant figures (1, 0, 8)
• Significant Figures 17
Rules for significant figures
should be memorized for use
throughout the text.
• at the end of a number without a decimal point:
1000 has one significant figure (1)
590 has two significant figures (5, 9)
One way of indicating that these zeros are significant is to write the number using scientific
notation. Thus if the value 1000 has been determined to four significant figures, it is written as
1.000 * 103. If 590 has only two significant figures, it is written as 5.9 * 102.
Example 2.5
How many significant figures are in each of the following measurements?
(a) 45 apples
(b) 0.02050 cm
(c) 3500 ft
Solution
(a) The number is exact and has no uncertainty associated with it (Rule 2).
(b) Four significant figures. The zeroes before the 2 are not significant, the 2 is significant, the
zero between the 2 and the 5 is significant, the 5 is significant, and the final zero is significant.
(c) Two significant figures. The zeroes at the end of a number without a decimal point are
not significant.
Practice 2.3
How many significant figures are in each of these measurements?
(a) 4.5 inches
(e) 25.0 grams
(b) 3.025 feet
(f) 12.20 liters
(c) 125.0 meters
(g) 100,000 people
(d) 0.001 mile
(h) 205 birds
Rounding Off Numbers
When we do calculations on a calculator, we often obtain answers that have more digits than
are justified. It is therefore necessary to drop the excess digits in order to express the answer
with the proper number of significant figures. When digits are dropped from a number, the
value of the last digit retained is determined by a process known as rounding off numbers.
Two rules will be used in this book for rounding off numbers.
ENHANCED EXAMPLE
18 chapter 2
• Standards for Measurement
Rules for rounding off
Not all schools use the same rules
for rounding. Check with your
instructor for variations in these
rules.
1. When the first digit after those you want to retain is 4 or less, that digit and all
others to its right are dropped. The last digit retained is not changed. The following examples are rounded off to four digits:
74.693 � 74.69
1.00629 � 1.006
This digit is dropped.
These two digits are dropped.
2. When the first digit after those you want to retain is 5 or greater, that digit and
all others to the right are dropped and the last digit retained is increased by
one. These examples are rounded off to four digits:
1.026868 � 1.027
18.02500 � 18.03
These three digits are dropped.
This digit is changed to 7.
These three digits are dropped.
This digit is changed to 3.
12.899 � 12.90
This digit is dropped.
These two digits are changed to 90.
Practice 2.4
Round off these numbers to the number of significant figures indicated:
(a) 42.246 (four)
(d) 0.08965 (two)
(b) 88.015 (three)
(e) 225.3 (three)
(c) 0.08965 (three)
(f) 14.150 (three)
2.4 Significant Figures in Calculations
Learning objective
Apply the rules for significant figures, in calculations involving addition, subtraction,
multiplication, and division.
The results of a calculation based on measurements cannot be more precise than the least
precise measurement.
Multiplication or Division
In calculations involving multiplication or division, the answer must contain the same number
of significant figures as in the measurement that has the least number of significant figures.
Consider the following examples:
Use your calculator to check your
work in the examples. ­Compare
your results to be sure you
­understand the mathematics.
ENHANCED EXAMPLE
Example 2.6
(190.6)(2.3) = 438.38
SOLUTION
The value 438.38 was obtained with a calculator. The answer should have two significant
figures because 2.3, the number with the fewest significant figures, has only two significant figures.
Round off this digit to 4.
Drop these three digits.
438.38
Move the decimal two places to the left to express in scientific notation.
The correct answer is 440 or 4.4 * 102.
2.4
• Significant Figures in Calculations 19
Example 2.7
(13.59)(6.3)
= 7.13475
12
SOLUTION
The value 7.13475 was obtained with a calculator. The answer should contain two significant figures because 6.3 and 12 each have only two significant ­figures.
Drop these four digits.
7.13475
This digit remains the same.
The correct answer is 7.1.
Practice 2.5
(a) (134 in.)(25 in.) = ?
(d) 0.0321 * 42 = ?
(b)
213 miles
= ?
4.20 hours
(e)
0.0450
= ?
0.00220
(c)
(2.2)(273)
= ?
760
(f)
1.280
= ?
0.345
Addition or Subtraction
The results of an addition or a subtraction must be expressed to the same precision as the least
precise measurement. This means the result must be rounded to the same number of decimal
places as the value with the fewest decimal places (blue line in examples).
Example 2.8
Add 125.17, 129, and 52.2.
SOLUTION
125. 17
129
52. 2
306. 37
he number with the least precision is 129. Therefore the answer is rounded off to the
T
­nearest unit: 306.
Example 2.9
Subtract 14.1 from 132.56.
SOLUTION
132. 5 6
- 14. 1
118. 4 6
The number with the least precision is 14.1. Therefore the answer is rounded off to the
nearest tenth: 118.5.
20 chapter 2
• Standards for Measurement
Example 2.10
Subtract 120 from 1587.
SOLUTION
158 7
- 12 0
146 7
The number with the least precision is 120. The zero is not considered significant; therefore the answer must be rounded to the nearest ten: 1470 or 1.47 * 103.
Example 2.11
Add 5672 and 0.00063.
SOLUTION
Note: When a very small number is
added to a large number, the result
is simply the original number.
5672
+
0. 00063
5672. 00063
The number with the least precision is 5672. So the answer is rounded off to the nearest
unit: 5672.
Example 2.12
1.039 - 1.020
= ?
1.039
SOLUTION
1.039 - 1.020
= 0.018286814
1.039
The value 0.018286814 was obtained with a calculator. When the subtraction in the
­numerator is done,
1.039 - 1.020 = 0.019
the number of significant figures changes from four to two. Therefore the ­answer should
contain two significant figures after the division is carried out:
Drop these six digits.
0.018286814
This digit remains the same.
The correct answer is 0.018, or 1.8 * 10-2.
Practice 2.6
If you need to brush up on
your math skills, refer to the
“Mathematical Review” in
Appendix I.
How many significant figures should the answer in each of these calculations contain?
(a) (14.0)(5.2)
(e) 119.1 - 3.44
(b) (0.1682)(8.2)
(f)
(c)
(160)(33)
4
(d) 8.2 + 0.125
94.5
1.2
(g) 1200 + 6.34
(h) 1.6 + 23 - 0.005
2.5
• The Metric System 21
Rules for significant figures in calculations
Multiplication or Division
The answer contains the same number of significant figures as the measurement
with the least number of significant figures.
Addition or Subtraction
The answer contains the same number of significant figures as the least precise
measurement.
Additional material on mathematical operations is given in Appendix I, “Mathematical
Review.” Study any portions that are not familiar to you. You may need to do this at various
times during the course when additional knowledge of mathematical operations is required.
2.5 The Metric System
Name the units for mass, length, and volume in the metric system and convert from
one unit to another.
Lea rni ng objecti ve
The metric system, or International System (SI, from Système International), is a decimal system of units for measurements of mass, length, time, and other physical quantities. Built around
a set of standard units, the metric system uses factors of 10 to express larger or smaller numbers
of these units. To express quantities that are larger or smaller than the standard units, prefixes
are added to the names of the units. These prefixes represent multiples of 10, making the metric
system a decimal system of measurements. Table 2.1 shows the names, symbols, and numerical
values of the common prefixes. Some examples of the more commonly used prefixes are
key terms
1 kilometer = 1000 meters
1 kilogram = 1000 grams
metric system or International
System (SI)
meter (m)
conversion factor
solution map
mass
weight
kilogram (kg)
volume
liter (L)
1 millimeter = 0.001 meter
1 microsecond = 0.000001 second
The common standard units in the International System, their abbreviations, and the quantities they measure are given in Table 2.2. Other units are derived from these units. The metric
system, or International System, is currently used by most of the countries in the world, not
only in scientific and technical work but also in commerce and industry.
Table 2.1 Common Prefixes and Numerical Values for SI Units
Prefix
Symbol
Numerical value
giga
mega
kilo
hecto
deka
—
deci
centi
milli
micro
nano
pico
femto
G
M
k
h
da
—
d
c
m

n
p
f
1,000,000,000
1,000,000
1,000
100
10
1
0.1
0.01
0.001
0.000001
0.000000001
0.000000000001
0.000000000000001
Power of 10
equivalent
109
106
103
102
101
100
10-1
10-2
10-3
10-6
10-9
10-12
10-15
The prefixes most commonly
used in chemistry are shown in
bold in Table 2.1.
22 chapter 2
• Standards for Measurement
Table 2.2
International System’s Standard Units of Measurement
Quantity
Name of unit
Abbreviation
meter
kilogram
kelvin
second
mole
ampere
candela
m
kg
K
s
mol
A
cd
Kathleen Pepper/Copyright John Wiley & Sons, Inc.
Length
Mass
Temperature
Time
Amount of substance
Electric current
Luminous intensity
Measurement of Length
Common length relationships:
1m
1 cm
1 in.
1 mile
=
=
=
=
=
106 m = 1010 Å
100 cm = 1000 mm
10 mm = 0.01 m
2.54 cm
1.609 km
See inside back cover for a table
of conversions.
© tomograf/iStockphoto
Most products today list both
systems of measurement on their
labels.
The standard unit of length in the metric system is the meter (m). When the metric system
was first introduced in the 1790s, the meter was defined as one ten-millionth of the distance
from the equator to the North Pole, measured along the meridian passing through
Dunkirk, France. The latest definition describes a meter as the distance that light
travels in a vacuum during 1/299,792,458
of a second.
A meter is 39.37 inches, a little longer
than 1 yard. One meter equals 10 dec­imeters,
100 centimeters, or 1000 millimeters (see
Figure 2.2). A kilometer contains 1000 meters. Table 2.3 shows the relationships of
these units.
The nanometer (10-9 m) is used extensively in expressing the wavelength of
light, as well as in atomic dimensions. See
the inside back cover for a complete table of
common conversions.
1 in.
Inches
1
1
2
2
3
2.54 cm
25.4 mm
Table 2.3
4
5
3
6
Centimeters
7
8
Figure 2.2
Comparison of the metric and
American systems of length
measurement: 2.54 cm = 1 in.
A playing card measures 2 1/2 in.
or 6.3 cm.
Metric Units of Length
Unit
Abbreviation
Meter equivalent
kilometer
meter
decimeter
centimeter
millimeter
micrometer
nanometer
angstrom
1000 m
1m
0.1 m
0.01 m
0.001 m
0.000001 m
0.000000001 m
0.0000000001 m
km
m
dm
cm
mm
m
nm
Å
Exponential
equivalent
103 m
100 m
10-1 m
10-2 m
10-3 m
10-6 m
10-9 m
10-10 m
2.5
• The Metric System 23
Unit Conversions
One of the benefits of using the metric system is the ease with which we can convert from one
unit to another. We do this by using the relationships between units—math equations called
equivalent units. Table 2.4 shows common equivalent units for length in the metric system.
To convert from one unit to another we must use a conversion factor. A conversion factor is a ratio of equivalent quantities. Where do we find these conversion factors? They are
formed from equivalent units. For example, since 1 m = 100 cm (Table 2.4) we could write
1m
100 cm
two ­ratios:
or
. In both cases the ratio is 1 since the units are equal.
100 cm
1m
We can convert one metric unit to another by using conversion factors.
unit1 : conversion factor  unit2
Table 2.4
Length Equivalent Units
1m
1m
1m
1 kg
1 cm
=
=
=
=
=
10 dm
100 cm
1000 mm
1000 g
10 mm
Important equations and statements are boxed or highlighted
in color.
If you want to know how many millimeters are in 2.5 meters, you need to convert meters (m)
to millimeters (mm). Start by writing
m * conversion factor = mm
This conversion factor must accomplish two things: It must cancel (or eliminate) meters, and
it must introduce millimeters—the unit wanted in the answer. Such a conversion factor will
be in fractional form and have meters in the denominator and millimeters in the numerator:
mm
= mm
m *
m
We know that 1 m = 1000 mm. From this relationship we can write two conversion factors.
1m
1000 mm
Choosing the conversion factor
2.5 m to millimeters:
and
1000 mm
1m
1000 mm
, we can set up the calculation for the conversion of
1m
We know that multiplying a measurement by 1 does not change
its value. Since our conversion
factors both equal 1, we can
multiply the measurement by the
appropriate one to convert units.
1000 mm
b = 2500 mm or 2.5 * 103 mm (two significant figures)
1m
Note that, in making this calculation, units are treated as numbers; meters in the ­numerator are
canceled by meters in the denominator.
(2.5 m )a
Example 2.13
Change 215 centimeters to meters.
Solution
We will use a solution map to outline our path for the conversion for this example:
Solution map: cm S m
We start with
cm * conversion factor = m
The conversion factor must have centimeters in the denominator and meters in the numerator:
cm *
m
= m
cm
From the relationship 100 cm = 1 m, we can write a factor that will accomplish this conversion:
1m
100 cm
ENHANCED EXAMPLE
24 chapter 2
• Standards for Measurement
Now set up the calculation using all the data given:
(215 cm )a
1m
b = 2.15 m
100 cm
Practice 2.7
Convert the following measurements:
(a) 567 mm to m
(b) 68 cm to mm
(c) 125 m to km
Measurement of Mass
Although we often use mass and weight interchangeably in our everyday lives, they have quite
different meanings in chemistry. In science we define the mass of an object as the amount of
matter in the object. Mass is measured on an instrument called a balance. The weight of an
object is a measure of the effect of gravity on the object. Weight is determined by using an
instrument called a scale, which measures force against a spring. This means mass is independent of the location of an object, but weight is not. In this text we will use the term mass
for all of our metric mass measurements. The gram is a unit of mass measurement, but it is a
tiny amount of mass; for instance, a U.S. nickel has a mass of about 5 grams. Therefore the
standard unit of mass in the SI system is the kilogram (kg) (equal to 1000 g). The amount
of mass in a kilogram is defined by international agreement as exactly equal to the mass of a
platinum-iridium cylinder (international prototype kilogram) kept in a vault at Sèvres, France.
Comparing this unit of mass to 1 lb (16 oz), we find that 1 kg is equal to 2.205 lb. A pound
is equal to 453.6 g (0.4536 kg). The same prefixes used in length measurement are used to
indicate larger and smaller gram units (see Table 2.5). A balance is used to measure mass. Two
examples of balances are shown in Figure 2.3.
Table 2.5
Metric Units of Mass
Unit
Abbreviation
Gram equivalent
kilogram
gram
decigram
centigram
milligram
microgram
kg
g
dg
cg
mg
g
Exponential
equivalent
103 g
100 g
10-1 g
10-2 g
10-3 g
10-6 g
1000 g
1g
0.1 g
0.01 g
0.001 g
0.000001 g
Figure 2.3
© Sciencephotos/Alamy
© 2005 Richard Megna/Fundamental Photographs
(a) Digital electronic, top-loading
balance with a precision of 0.001 g
(1 mg); and (b) a digital electronic
analytical balance with a precision
of 0.0001 g (0.1 mg).
(a)
(b)
2.5
>Chemistry in action
Keeping Track of Units
Why do scientists worry about units? The National
Aeronautics and Space Administration (NASA) recently
was reminded of just why keeping track of units is so
important. In 1999 a $125
million satellite was lost in the
atmosphere of Mars because
scientists made some improper assumptions about
units. NASA’s scientists at the
Jet Propulsion Lab (JPL) in
Pasadena, ­California, received
thrust data from the satellite’s
manufacturer, Lockheed Martin
Aeronautics in Denver, Colorado. Unfortunately, the Denver
scientists used American units
in their measurements and
the JPL scientists ­assumed the
units were metric. This ­mistake
caused the satellite to fall
100 km lower into the Mars ­atmosphere than planned.
The spacecraft burned up from the friction with the
atmosphere.
Measuring and using units correctly is very important. In
fact, it can be critical, as we have
just seen. For example, a Canadian jet almost crashed when
the tanks were filled with 22,300
pounds (instead of kilograms)
of fuel. Calculations for distance
were based on kilograms, and
the jet almost ran out of fuel before landing at the destination.
Correct units are also important
to ensure perfect fits for household purchases such as drapes,
carpet, or appliances. Be sure to
pay attention to units in both
your chemistry problems and in
everyday life!
JPL/NASA
­
• The Metric System 25
Mars climate orbiter
Example 2.14
Change 25 grams to milligrams.
Common mass relationships:
1 g = 1000 mg
1 kg = 1000 g
1 kg = 2.205 lb
1 lb = 453.6 g
Solution
We use the conversion factor 1000 mg>g.
Solution map: g S mg
(25 g )a
1000 mg
b = 25,000 mg
1g
(2.5 * 104 mg)
ote that multiplying a number by 1000 is the same as multiplying the number by 103
N
and can be done simply by moving the decimal point three places to the right:
(6.428)(1000) = 6428 (6.428)
3
Practice 2.8
Convert the following measurements:
(a) 560 mg to g
(b) 525 g to kg
(c) 175 g to mg
To change milligrams to grams, we use the conversion factor 1 g>1000 mg. For example,
we convert 155 mg to grams as follows:
(155 mg )a
1g
b = 0.155 g
1000 mg
26 chapter 2
• Standards for Measurement
Measurement of Volume
Volume, as used here, is the amount of space occupied by matter. The SI unit of volume is the
Common volume relationships:
1 L = 1000 mL = 1000 cm3
1 mL = 1 cm3
1 L = 1.057 qt
946.1 mL = 1 qt
cubic meter (m3). However, the liter (pronounced lee-ter and ­abbreviated L) and the milliliter
(abbreviated mL) are the standard units of ­volume used in most chemical laboratories. A liter
(L) is usually defined as 1 cubic decimeter (1 kg) of water at 4°C.
The most common instruments or equipment for measuring liquids are the graduated
cylinder, volumetric flask, buret, pipet, and syringe, which are illustrated in Figure 2.4.
These pieces are usually made of glass and are available in various sizes. Let’s try some
examples.
mL
0
ml
100
mL
5
5
90
80
4
10
3
70
60
2
15
50
20
40
mL
10
1
0
30
25
20
10
100 mL
Graduated
cylinder
Volumetric
flask
Buret
Pipet
Syringe
Figure 2.4
Calibrated glassware for measuring the volume of liquids.
Example 2.15
How many milliliters are contained in 3.5 liters?
SOLUTION
The conversion factor to change liters to milliliters is 1000 mL>L:
Solution map: L S mL
(3.5 L )a
1000 mL
b = 3500 mL
L
(3.5 * 103 mL)
iters may be changed to milliliters by moving the decimal point three places to the right
L
and changing the units to milliliters:
1.500 L � 1500. mL
3
Practice 2.9
Convert the following measurements:
(a) 25 mL to L
(b) 1.55 L to mL
(c) 4.5 L to dL
2.6
• Dimensional Analysis: A Problem-Solving Method 27
2.6 Dimensional Analysis:
a problem-solving method
Use dimensional analysis to solve problems involving unit conversions.
Many chemical principles can be illustrated mathematically. Learning how to set up and solve
numerical problems in a systematic fashion is essential in the study of chemistry.
Usually a problem can be solved by several methods. But in all methods it is best to use
a systematic, orderly approach. The dimensional analysis method is emphasized in this book
because it
• provides a systematic, straightforward way to set up problems.
• gives a clear understanding of the principles involved.
• trains you to organize and evaluate data.
• helps to identify errors, since unwanted units are not eliminated if the setup of the problem
is incorrect.
Learning objective
See Appendix II for help in using
a scientific calculator.
Dimensional analysis converts one unit to another unit by using conversion factors.
Label all factors with the proper units.
The following examples show the conversion between American units and metric units.
Mass conversions from American to metric units are shown in Examples 2.16 and 2.21.
Example 2.16
A 1.50-lb package contains how many grams of baking soda?
ENHANCED EXAMPLE
SOLUTION
We are solving for the number of grams equivalent to 1.50 lb. Since 1 lb = 453.6 g, the
conversion factor is 453.6 g>lb:
Solution map: lb S g
(1.50 lb )a
453.6 g
b = 680. g
1 lb
Practice 2.10
A tennis ball has a mass of 65 g. Determine the American equivalent in pounds.
The volume of a cubic or rectangular container can be determined by ­multiplying
its length * width * height. Thus a box 10 cm on each side has a volume of (10 cm)
(10 cm) (10 cm) = 1000 cm3.
Example 2.17
How many cubic centimeters are in a cube that is 11.1 inches on a side?
SOLUTION
First we change inches to centimeters; our conversion factor is 2.54 cm>in.:
Solution map: in. S cm
2.54 cm
(11.1 in. )a
b = 28.2 cm on a side
1 in.
Then we determine volume (length * width * height):
(28.2 cm)(28.2 cm)(28.2 cm) = 22,426 cm3
(2.24 * 104 cm3)
When doing problems with
multiple steps, you should round
only at the end of the problem.
We are rounding at the end of
each step in example problems
to illustrate the proper significant
figures.
28 chapter 2
• Standards for Measurement
Example 2.18
How many cubic centimeters (cm3) are in a box that measures 2.20 in. by 4.00 in. by
6.00 in.?
SOLUTION
First we need to determine the volume of the box in cubic inches (in.3) by ­multiplying the
length * width * height:
(2.20 in.)(4.00 in.)(6.00 in.) = 52.8 in.3
Now we convert in.3 to cm3 by using the inches and centimeters relationship
(1 in. = 2.54 cm) three times:
Solution map: (in. S cm)3
in.3 *
cm
cm
cm
*
*
= cm3
in.
in.
in.
(52.8 in.3 )a
Units are emphasized in problems
by using color and solution maps
to help visualize the steps in the
process.
2.54 cm 2.54 cm 2.54 cm
ba
ba
b = 865 cm3
1 in.
1 in.
1 in.
Some problems require a series of conversions to reach the correct units in the answer. For
example, suppose we want to know the number of seconds in 1 day. We need to convert from
the unit of days to seconds using this solution map:
Solution map: day S hours S minutes S seconds
Each arrow in the solution map represents a conversion factor. This sequence requires three
conversion factors, one for each step. We convert days to hours (hr), hours to minutes (min),
and minutes to seconds (s). The conversions can be done individually or in a continuous
sequence:
day *
hr
min
s
¡ hr *
¡ min *
= s
day
hr
min
day *
hr
min
s
*
*
= s
day
hr
min
Inserting the proper factors, we calculate the number of seconds in 1 day to be
(1 day )a
24 hr
60 min
60 s
ba
ba
b = 86,400. s
1 day
1 hr
1 min
All five digits in 86,400. are significant, since all the factors in the calculation are exact
­numbers.
Example 2.19
How many centimeters are in 2.00 ft?
SOLUTION
The stepwise conversion of units from feet to centimeters may be done in this sequence.
Convert feet to inches; then convert inches to centimeters:
Solution map: ft S in. S cm
The conversion factors needed are
12 in.
1 ft
and
2.54 cm
1 in.
2.6
(2.00 ft )a
(24.0 in. )a
NOTE
• Dimensional Analysis: A Problem-Solving Method 29
12 in.
b = 24.0 in.
1 ft
2.54 cm
b = 61.0 cm
1 in.
Since 1 ft and 12 in. are exact numbers, the number of significant figures
­allowed in the answer is three, based on the number 2.00.
Example 2.20
How many meters are in a 100.-yd football field?
SOLUTION
The stepwise conversion of units from yards to meters may be done by this ­sequence, ­
using the proper conversion factors:
Solution map: yd S ft S in. S cm S m
(100. yd )a
(300. ft )a
(3600 in. )a
(9144 cm )a
3 ft
b = 300. ft
1 yd
12 in.
b = 3600 in.
1 ft
2.54 cm
b = 9144 cm
1 in.
1m
b = 91.4 m
100 cm
(three significant figures)
Practice 2.11
(a) How many meters are in 10.5 miles?
(b) What is the area of a 6.0-in. * 9.0-in. rectangle in square ­meters?
Examples 2.19 and 2.20 may be solved using a linear expression and ­writing down conversion
factors in succession. This method often saves one or two ­calcu­lation steps and allows numerical values to be reduced to simpler terms, leading to simpler calculations. The single linear
expressions for Examples 2.19 and 2.20 are
(2.00 ft )a
(100. yd )a
12 in.
2.54 cm
ba
b = 61.0 cm
1 ft
1 in.
3 ft
12 in.
2.54 cm
1m
ba
ba
ba
b = 91.4 m
1 yd
1 ft
1 in.
100 cm
Using the units alone (Example 2.20), we see that the stepwise cancellation proceeds in succession until the desired unit is reached:
yd *
ft
in.
cm
m
*
*
*
= m
cm
yd
ft
in.
Example 2.21
A driver of a car is obeying the speed limit of 55 miles per hour. How fast is the car
­traveling in kilometers per second?
SOLUTION
Several conversions are needed to solve this problem:
Solution map: mi. S km
hr S min S s
30 chapter 2
• Standards for Measurement
To convert mi S km,
a
55 mi
1.609 km
km
ba
b = 88
hr
1 mi
hr
a
88 km
1 hr
1 min
km
ba
ba
b = 0.024
s
hr
60 min
60 s
Next we must convert hr S min S s. Notice that hours is in the denominator of our
quantity, so the conversion factor must have hours in the numerator:
Practice 2.12
How many cubic meters of air are in a room measuring 8 ft * 10 ft * 12 ft?
Example 2.22
Suppose four ostrich feathers weigh 1.00 lb. Assuming that each feather is equal in mass,
how many milligrams does a single feather weigh?
SOLUTION
The unit conversion in this problem is from 1.00 lb>4 feathers to milligrams per feather.
Since the unit feathers occurs in the denominator of both the starting unit and the desired
unit, the unit conversions are
Solution map: lb S g S mg
a
453.6 g
1000 mg
113,400 mg
1.00 lb
ba
ba
b =
(1.13 * 105 mg>feather)
4 feathers
1 lb
1g
feather
Practice 2.13
You are traveling in Europe and wake up one morning to find your mass is 75.0 kg.
Determine the American equivalent (in pounds) to see whether you need to go on a diet
before you return home.
Practice 2.14
A bottle of excellent Chianti holds 750 mL. What is its volume in quarts?
Practice 2.15
Milk is often purchased by the half gallon. Determine the number of liters equal to this
amount.
2.7 Measurement of Temperature
Learning objective
Convert measurements among the Fahrenheit, Celsius, and Kelvin temperature scales.
key terms
Thermal energy is a form of energy associated with the motion of small particles of ­matter.
Depending on the amount of thermal energy present, a given system is said to be hot or cold.
Temperature is a measure of the intensity of thermal energy, or how hot a system is, ­regardless
of its size. The term heat refers to the flow of energy due to a temperature difference. Heat
always flows from a region of higher temperature to one of lower temperature. The SI unit of
temperature is the kelvin. The common laboratory instrument for measuring temperature is a
thermometer (see Figure 2.5).
thermal energy
temperature
heat
2.7
100
212
100
370
373
Boiling point
of water
273
Freezing point
of water
200
90
190
360
180
80
350
170
160
70
340
150
60
140
330
130
180
• Measurement of Temperature 31
100
120
50
100
320
110
40
100
310
90
30
300
80
70
20
290
60
50
10
280
40
30
32
0
0
270
20
10
10
260
0
Figure 2.5
Fahrenheit, °F
Celsius, °C
Kelvin, K
Comparison of Celsius, Kelvin,
and Fahrenheit temperature
scales.
The temperature of a system can be expressed by several different scales. Three commonly used temperature scales are Celsius (pronounced sell-see-us), Kelvin (also called absolute), and Fahrenheit. The unit of temperature on the Celsius and ­Fahrenheit scales is called a
degree, but the size of the Celsius and the Fahrenheit degree is not the same. The symbol for
the Celsius and Fahrenheit degrees is °, and it is placed as a superscript after the number and
before the symbol for the scale. Thus, 10.0°C means 10.0 degrees Celsius. The degree sign is
not used with Kelvin temperatures:
degrees Celsius = °C
Kelvin (absolute) = K
degrees Fahrenheit = °F
On the Celsius scale the interval between the freezing and boiling temperatures of
water is divided into 100 equal parts, or degrees. The freezing point of water is assigned
a temperature of 0°C and the boiling point of water a ­temperature of 100°C. The Kelvin
temperature scale is known as the absolute temperature scale because 0 K is the lowest
temperature theoretically attainable. The Kelvin zero is 273.15 kelvins below the Celsius
zero. A kelvin is equal in size to a Celsius degree. The freezing point of water on the
Kelvin scale is 273.15 K. The Fahrenheit scale has 180 degrees between the freezing and
boiling temperatures of water. On this scale the freezing point of water is 32°F and the
boiling point is 212°F:
0°C  273 K  32°F
100°C  373 K  212°F
The three scales are compared in Figure 2.5. Although absolute zero (0 K) is the lower
limit of temperature on these scales, temperature has no upper limit. Temperatures of several
million degrees are known to exist in the sun and in other stars.
By examining Figure 2.5, we can see that there are 100 Celsius degrees and 100 kelvins
between the freezing and boiling points of water, but there are 180 Fahrenheit degrees between
these two temperatures. Hence, the size of the Celsius degree and the kelvin are the same, but
1 Celsius degree is equal to 1.8 Fahrenheit degrees.
Hein_14ed_c02_013-043hr.indd 31
180
= 1.8
100
7/2/12 1:09 PM
32 chapter 2
• Standards for Measurement
>Chemistry in action
The kilogram is the standard base unit for mass in the SI
system. But who decides just what a kilogram is? Before
1880, a kilogram was defined as the mass of a cubic decimeter of water. But this is difficult to reproduce very accurately since impurities and air are dissolved in water. The
density of water also changes with temperature, leading
to inaccuracy. So in 1885 a Pt-Ir cylinder was made with a
mass of exactly 1 kilogram. This cylinder is kept in a vault at
Sèvres, France, outside Paris, along with six copies. In 1889
the kilogram was determined to be the base unit of mass
for the metric system and has been weighed against its
copies three times in the past 100 years (1890, 1948, 1992)
to make sure that it is accurate. The trouble with all this is
that the cylinder itself can vary. Experts in the science of
measurement want to change the definition of the kilogram
to link it to a property of matter that does not change.
How else could we define a kilogram? Scientists have several ideas. They propose to fix the kilogram to a set value
of a physical constant (which does not change). These
constants connect experiment to theory. Right now they
are continuing to struggle to find a new way to determine
the mass of the kilogram.
Official contribution of the National Institute of Standards
and Technology, not subject to copyright in the United States.
Sissy Riley, Information Services division/NIST
Setting Standards
The kilogram is the only one of seven SI base units defined in terms of a physical object instead of a property
of matter. The meter (redefined in 1983) is defined in
terms of the speed of light (1 m = distance light travels
1
in 299,792,458
s). The second is defined in terms of the natural vibration of the cesium atom. Unfortunately, three of
the SI base units (mole, ampere, and candela) depend on
the definition of the kilogram. If the mass of the kilogram
is uncertain, then all of these units are uncertain.
Kilogram artifact in
Sèvres, France.
From these data, mathematical formulas have been derived to convert a temperature on
one scale to the corresponding temperature on another scale:
K = °C + 273.15
°F = (1.8 * °C) + 32
°C =
°F - 32
1.8
Example 2.23
ENHANCED EXAMPLE
The temperature at which table salt (sodium chloride) melts is 800.°C. What is this
temperature on the Kelvin and Fahrenheit scales?
SOLUTION
To calculate K from °C, we use the formula
K = °C + 273.15
K = 800.°C + 273.15 = 1073 K
To calculate °F from °C, we use the formula
°F = (1.8 * °C) + 32
°F = (1.8)(800.°C) + 32
°F = 1440 + 32 = 1472°F
Summarizing our calculations, we see that
800.°C = 1073 K = 1472°F
Remember that the original measurement of 800.°C was to the units place, so the
­converted temperature is also to the units place.
2.7
>Chemistry in action
Taking the Temperature of Old Faithful
During the first 20–30 seconds of an
eruption, steam and boiling water
shoot through the narrowest part of
the vent at near the speed of sound.
The narrow tube limits the rate at
which the water can shoot from the
geyser. When the pressure falls below a critical value, the process slows
and Old Faithful begins to quiet
down again.
The frequency of Old Faithful’s eruptions is not on a precise schedule,
but varies from 45 to 105 minutes—
the average is about 79 minutes. The
variations in time between eruptions
depend on the amount of boiling water left in the fissure. Westphal says,
“There’s no real pattern except that a
short eruption is always followed by a
long one.” Measurements of
temperatures inside Old Faithful have
given scientists a better understanding of what causes a geyser
to erupt.
Poterfield/Chickoring/Photo Researchers, Inc.
If you have ever struggled to take the
temperature of a sick child, imagine
the difficulty in taking the temperature of a geyser. Such are the tasks
scientists set for themselves! In 1984,
James A. Westphal and Susan W.
Keiffer, geologists from the California
Institute of Technology, measured
the temperature and pressure inside
Old Faithful during several eruptions
in order to learn more about how a
geyser functions. The measurements,
taken at eight depths along the upper part of the geyser were so varied
and complicated that the researchers returned to Yellowstone in 1992
to further investigate Old Faithful’s
structure and functioning. To see
what happened between eruptions,
Westphal and Keiffer lowered an
insulated 2-in. video camera into the
geyser. Keiffer had assumed that the
vent (opening in the ground) was a
uniform vertical tube, but this is not
the case. Instead, the geyser appears
to be an east–west crack in the Earth
that extends downward at least 14 m.
In some places it is over 1.8 m wide,
and in other places it narrows to less
than 15 cm. The walls of the vent contain many cracks, allowing water to
enter at several depths. The complicated nature of the temperature data
is explained by these cracks. Cool
water enters the vent at depths of
5.5 m and 7.5 m. Superheated water
and steam blast into the vent 14 m
underground. According to Westphal,
temperature increases of up to 130°C
at the beginning of an eruption suggest that water and steam also surge
into the vent from deeper geothermal
sources.
Old Faithful eruptions shoot into
the air an average of 130 feet.
• Measurement of Temperature 33
34 chapter 2
• Standards for Measurement
Example 2.24
The temperature on December 1 in Honolulu, Hawaii, was 110.°F, a new record. Convert
this temperature to °C.
SOLUTION
We use the formula
°C =
°F - 32
1.8
°C =
110. - 32
78
=
= 43°C
1.8
1.8
Example 2.25
What temperature on the Fahrenheit scale corresponds to -8.0°C? (Notice the negative
sign in this problem.)
SOLUTION
°F = (1.8 * °C) + 32
°F = (1.8)( -8.0) + 32 = -14 + 32
°F = 18°F
Temperatures used throughout this book are in degrees Celsius (°C) unless specified
otherwise. The temperature after conversion should be expressed to the same precision as
the original measurement.
Practice 2.16
Helium boils at 4 K. Convert this temperature to °C and then to °F.
Practice 2.17
“Normal” human body temperature is 98.6°F. Convert this to °C and K.
2.8 Density
Learning objective
Solve problems involving density
key terms
Density (d) is the ratio of the mass of a substance to the volume occupied by that mass; it is
the mass per unit of volume and is given by the equation
mass
d =
volume
density
specific gravity
Density is a physical characteristic of a substance and may be used as an aid to its identification. When the density of a solid or a liquid is given, the mass is usually expressed in grams
and the volume in milliliters or cubic centimeters.
d =
g
mass
=
volume
mL
or d =
g
cm3
Since the volume of a substance (especially liquids and gases) varies with temperature, it is
important to state the temperature along with the density. For example, the volume of 1.0000 g
of water at 4°C is 1.0000 mL; at 20°C, it is 1.0018 mL; and at 80°C, it is 1.0290 mL. Density
therefore also varies with temperature.
2.8
• Density 35
10.00 mL
Water
4.83 mL
0.518
mL
1.00 g
2.07 g
19.3 g
Sulfur
Gold
Water
Sulfur
Gold
Mass, 10.0 g
Volume, 1.00 cm3
Comparing volumes of equal masses
Comparing masses of equal volumes
Figure 2.6
Comparison of mass and volume.
The density of water at 4°C is 1.0000 g>mL, but at 80°C the density of water is
0.9718 g>mL:
1.0000 g
d 4°C =
= 1.0000 g>mL
1.0000 mL
1.0000 g
d 80°C =
= 0.97182 g>mL
1.0290 mL
The density of iron at 20°C is 7.86 g>mL.
7.86 g
d 20°C =
= 7.86 g>mL
1.00 mL
The densities of a variety of materials are compared in Figure 2.6.
Densities for liquids and solids are usually represented in terms of grams per milliliter
(g>mL) or grams per cubic centimeter (g>cm3). The density of gases, however, is expressed
in terms of grams per liter (g>L). Unless otherwise stated, gas densities are given for 0°C and
1 atmosphere pressure ­(discussed further in Chapter 13). Table 2.6 lists the densities of some
common materials.
Table 2.6
Densities of Some Selected Materials
Liquids and solids
Substance
Wood (Douglas fir)
Ethyl alcohol
Vegetable oil
Water (4°C)
Sugar
Glycerin
Karo syrup
Magnesium
Sulfuric acid
Sulfur
Salt
Aluminum
Silver
Lead
Mercury
Gold
Gases
Density
(g/mL at 20°C)
Substance
0.512
0.789
0.91
1.000*
1.59
1.26
1.37
1.74
1.84
2.07
2.16
2.70
10.5
11.34
13.55
19.3
Hydrogen
Helium
Methane
Ammonia
Neon
Carbon monoxide
Nitrogen
Air
Oxygen
Hydrogen chloride
Argon
Carbon dioxide
Chlorine
Density
(g/L at 0°C)
0.090
0.178
0.714
0.771
0.90
1.25
1.251
1.293*
1.429
1.63
1.78
1.963
3.17
*For comparing densities, the density of water is the reference for solids and liquids; air is the reference for gases.
36 chapter 2
• Standards for Measurement
When an insoluble solid object is dropped into water, it will sink or float, depending on
its density. If the object is less dense than water, it will float, displacing a mass of water equal
to the mass of the object. If the object is more dense than water, it will sink, displacing a volume
of water equal to the volume of the object. This information can be used to determine the
volume (and density) of irregularly shaped objects.
The specific gravity (sp gr) of a substance is the ratio of the density of that substance to
the density of another substance, usually water at 4°C. Specific gravity has no units because
the density units cancel. The specific gravity tells us how many times as heavy a liquid, a
solid, or a gas is as compared to the reference material. Since the density of water at 4°C is
1.00 g>mL, the specific gravity of a solid or liquid is the same as its density in g>mL without
the units.
sp gr =
density of a liquid or solid
density of water
Sample calculations of density problems follow.
Example 2.26
What is the density of a mineral if 427 g of the mineral occupy a volume of 35.0 mL?
SOLUTION
We need to solve for density, so we start by writing the formula for ­calculating density:
d =
mass
volume
Then we substitute the data given in the problem into the equation and solve:
mass = 427 g
d =
volume = 35.0 mL
427 g
mass
=
= 12.2 g>mL
volume
35.0 mL
Example 2.27
The density of gold is 19.3 g>mL. What is the mass of 25.0 mL of gold? Use density as a
conversion factor, converting
Solution map: mL S g
The conversion of units is
g
mL *
= g
mL
(25.0 mL )a
19.3 g
b = 483 g
mL
Example 2.28
Calculate the volume (in mL) of 100. g of ethyl alcohol.
SOLUTION
From Table 2.6 we see that the density of ethyl alcohol is 0.789 g>mL. This density also
means that 1 mL of the alcohol has a mass of 0.789 g (1 mL>0.789 g).
For a conversion factor, we can use ­either
g
mL
or
mL
g
Review 37
Solution map: g S mL,
(100. g) a
1 mL
b = 127 mL of ethyl alcohol
0.789 g
Example 2.29
The water level in a graduated cylinder stands at 20.0 mL before and at 26.2 mL
after a 16.74-g metal bolt is submerged in the water. (a) What is the volume of the bolt?
(b) What is the density of the bolt?
SOLUTION
(a) The bolt will displace a volume of water equal to the volume of the bolt. Thus, the
increase in volume is the volume of the bolt:
26.2 mL = volume of water plus bolt
-20.0 mL = volume of water
6.2 mL = volume of bolt
(b) d =
16.74 g
mass of bolt
=
= 2.7 g>mL
volume of bolt
6.2 mL
Practice 2.18
Pure silver has a density of 10.5 g>mL. A ring sold as pure silver has a mass of 18.7 g. When
it is placed in a graduated cylinder, the water level rises 2.0 mL. Determine whether the
ring is actually pure silver or whether the customer should contact the Better Business
Bureau.
Practice 2.19
The water level in a metric measuring cup is 0.75 L before the addition of 150. g of shortening. The water level after submerging the shortening is 0.92 L. Determine the density
of the shortening.
C h a p t e r
2 review
2.1 Scientific Notation
• Quantitative observations consist of a number and a unit and are called measurements.
• Very large and very small numbers can be represented compactly by using scientific notation:
• The number is represented as a decimal between 1 and 10 and is multiplied by 10 raised to the
appropriate exponent.
• The sign on the exponent is determined by the direction the decimal point is moved in the
original number.
key terms
measurement
scientific notation
2.2 Measurement and Uncertainty
• A
ll measurements reflect some amount of uncertainty, which is indicated by the number of significant figures in the measurement.
• The significant figures include all those known with certainty plus one estimated digit.
key term
significant figures
38 chapter 2
• Standards for Measurement
2.3 Significant Figures
key term
rounding off numbers
• Rules exist for counting significant figures in a ­measurement:
• Nonzero numbers are always significant.
• Exact numbers have an infinite number of significant figures.
• The significance of a zero in a measurement is ­determined by its position within the number.
• Rules exist for rounding off the result of a calculation to the correct number of significant figures.
• If the first number after the one you want to retain is 4 or less, that digit and all those after it are
dropped.
• If the first number after the one you want to retain is 5 or greater, that digit and all those after it
are dropped and the last digit retained is increased by one.
2.4 Significant Figures in Calculations
• The results of a calculation cannot be more precise than the least precise measurement.
• Rules exist for determining the correct number of significant figures in the result of a calculation.
2.5 The Metric System
key terms
metric system or
International System (SI)
meter (m)
conversion factor
solution map
mass
weight
kilogram (kg)
volume
liter (L)
• The metric system uses factors of 10 and a set of standard units for measurements.
• Length in the metric system is measured by the standard unit of the meter.
• The standard unit for mass in the metric system is the kilogram. In chemistry we often use the gram
instead, as we tend to work in smaller quantities.
• Volume is the amount of space occupied by matter.
• The standard unit for volume is the cubic meter. In chemistry we usually use the volume unit of the
liter or the milliliter.
2.6 Dimensional Analysis: A problem-solving method
• Dimensional analysis is a common method used to convert one unit to another.
• Conversion factors are used to convert one unit into another:
unit1 * conversion factor = unit2
• A solution map is used to outline the steps in the unit conversion.
2.7 Measurement of Temperature
key terms
thermal energy
temperature
heat
• There are three commonly used temperature scales: Fahrenheit, Celsius, and Kelvin.
• We can convert among the temperature scales by using mathematical formulas:
• K = °C + 273.15
• °F = (1.8 * °C) + 32
°F - 32
• °C =
1.8
2.8 Density
key terms
• The density of a substance is the amount of matter (mass) in a given volume of the substance:
density
specific gravity
d =
mass
volume
• Specific gravity is the ratio of the density of a substance to the density of another reference substance
(usually water).
review Questions
1. When a very large number is written in scientific notation, should
the exponent be a positive or a negative number?
2. To write 635.2 * 10-8 in the proper scientific notation form, should
the exponent increase or ­decrease?
3. Explain why the last digit in a measurement is uncertain.
4. How can the number 642,000 g be written to indicate that there are
four significant figures?
5. How are significant zeroes identified?
Paired Exercises 39
6. State the rules used in this text for rounding off numbers.
7. Is it possible for the calculated answer from a multiplication or
division problem to contain more significant figures than the measurements used in the calculation?
8. Is it possible for the calculated answer from an addition or subtraction problem to contain more significant figures than the measurements used in the calculation?
9. How many nanometers are in 1 cm? (Table 2.3)
10. How many milligrams are in 1 kg?
11. Why does an astronaut weigh more on Earth than in space when his
or her mass remains the same in both places?
12. What is the relationship between milliliters and cubic centimeters?
13. What is the metric equivalent of 3.5 in.?
14. Distinguish between heat and temperature.
15. Compare the number of degrees between the freezing point of
water and its boiling point on the Fahrenheit, Kelvin, and Celsius
temperature scales. (Figure 2.5)
16. Describe the order of the following substances (top to bottom) if
these three substances were placed in a 100-mL graduated cylinder:
25 mL glycerin, 25 mL mercury, and a cube of magnesium 2.0 cm
on an edge. (Table 2.6)
17. Arrange these materials in order of increasing density: salt, vegetable oil, lead, and ethyl alcohol. (Table 2.6)
18. Ice floats in vegetable oil and sinks in ethyl alcohol. The density of
ice must lie between what numerical values? (Table 2.6)
19. Distinguish between density and specific gravity.
20. Ice floats in water, yet ice is simply frozen water. If the density of
water is 1.0 g>mL, how is this possible?
21. If you collect a container of oxygen gas, should you store it with
the mouth up or down? Explain your answer. (Table 2.6)
22. Which substance has the greater volume; 25 g gold or 25 g silver?
(Table 2.6)
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
PAIR E D E x e r c i s e s
1. What is the numerical meaning of each of the following?
(a) kilogram
(d) millimeter
(b) centimeter
(e) deciliter
(c) microliter
2. What is the correct unit of measurement for each of the ­following?
(a) 1000 m
(d) 0.01 m
(b) 0.1 g
(e) 0.001 L
(c) 0.000001 L
3. State the abbreviation for each of the following units:
(a) gram
(d) micrometer
(b) microgram
(e) milliliter
(c) centimeter
(f) deciliter
4. State the abbreviation for each of the following units:
(a) milligram
(d) nanometer
(b) kilogram
(e) angstrom
(c) meter
(f) microliter
5. Determine whether the zeros in each number are significant:
(a) 2050
(d) 0.075
(b) 9.00 * 102
(e) 300.
(c) 0.0530
(f) 285.00
6. Determine whether the zeros in each number are significant:
(a) 0.005
(d) 10.000
(b) 1500
(e) 6.070 * 104
(c) 250.
(f) 0.2300
7. How many significant figures are in each of the following ­numbers?
(a) 0.025
(c) 0.0404
(b) 22.4
(d) 5.50 * 103
8. State the number of significant figures in each of the ­following
numbers:
(a) 40.0
(c) 129,042
(b) 0.081
(d) 4.090 * 10-3
9. Round each of the following numbers to three significant figures:
(a) 93.246
(c) 4.644
(b) 0.02857
(d) 34.250
10. Round each of the following numbers to three significant ­figures:
(a) 8.8726
(c) 129.509
(b) 21.25
(d) 1.995 * 106
11. Express each of the following numbers in exponential ­notation:
(a) 2,900,000
(c) 0.00840
(b) 0.587
(d) 0.0000055
12. Write each of the following numbers in exponential ­notation:
(a) 0.0456
(c) 40.30
(b) 4082.2
(d) 12,000,000
13. Solve the following problems, stating answers to the proper number
of significant figures:
14. Evaluate each of the following expressions. State the answer to the
proper number of significant figures:
(a) 12.62 + 1.5 + 0.25 = ?
3
4
(b) (2.25 * 10 )(4.80 * 10 ) = ?
(c)
(452)(6.2)
= ?
14.3
(d) (0.0394)(12.8) = ?
(e)
0.4278
= ?
59.6
(f) 10.4 + 3.75(1.5 * 104) = ?
(a) 15.2 - 2.75 + 15.67
(b) (4.68)(12.5)
(c)
182.6
4.6
(d) 1986 + 23.84 + 0.012
(e)
29.3
(284)(415)
(f) (2.92 * 10-3)(6.14 * 105)
40 chapter 2
• Standards for Measurement
15. Change these fractions into decimals. Express each ­answer to three
significant figures:
5
3
12
9
(a) (b) (c) (d)
6
7
16
18
16. Change each of the following decimals to fractions in lowest terms:
(a) 0.25 (b) 0.625 (c) 1.67 (d) 0.8888
17. Solve each of these equations for x:
0.525
(a) 3.42x = 6.5 (c)
= 0.25
x
x
(b)
= 7.05 12.3
18. Solve each equation for the variable:
212 - 32
(a) x =
(c) 72°F = 1.8x + 32
1.8
g
40.90 g
(b) 8.9
=
x
mL
Unit Conversions
19. Complete the following metric conversions using the correct number
of significant figures:
(a) 28.0 cm to m
(e) 6.8 * 104 mg to kg
(b) 1000. m to km
(f) 8.54 g to kg
(c) 9.28 cm to mm
(g) 25.0 mL to L
(d) 10.68 g to mg
(h) 22.4 L to L
20. Complete the following metric conversions using the correct number of significant figures:
(a) 4.5 cm to Å
(e) 0.65 kg to mg
(b) 12 nm to cm
(f) 5.5 kg to g
(c) 8.0 km to mm
(g) 0.468 L to mL
(d) 164 mg to g
(h) 9.0 L to mL
21. Complete the following American/metric conversions using the
correct number of significant figures:
(a) 42.2 in. to cm
(d) 42.8 kg to lb
(b) 0.64 m to in.
(e) 3.5 qt to mL
(c) 2.00 in.2 to cm2
(f) 20.0 L to gal
22. Make the following conversions using the correct number of
significant figures:
(a) 35.6 m to ft
(d) 95 lb to g
(b) 16.5 km to mi
(e) 20.0 gal to L
(c) 4.5 in.3 to mm3
(f) 4.5 * 104 ft3 to m3
23. After you have worked out at the gym on a stationary bike for
45 min, the distance gauge indicates that you have traveled 15.2 mi.
What was your rate in km/hr?
25. A pharmacy technician is asked to prepare an antibiotic IV solution
that will contain 500. mg of cephalosporin for every 100. mL of normal
saline solution. The total volume of saline solution will be 1 L. How
many grams of the cephalosporin will be needed for this IV solution?
24. A competitive college runner ran a 5-K (5.0-km) race in 15 min, 23 s.
What was her pace in miles per hour?
27. The maximum speed recorded for a giant tortoise is 0.11 m/sec.
How many miles could a gaint tortoise travel in 5.0 hr?
28. How many days would it take for a crepe myrtle tree to grow 1 cm
in height if it grows 3.38 feet per year? (Report your answer to three
significant figures.)
29. A personal trainer uses calipers on a client to determine his percent
body fat. After taking the necessary measurements, the personal
trainer determines that the client’s body contains 11.2% fat by mass
(11.2 lb of fat per 100 lb of body mass). If the client weighs 225 lb,
how many kg of fat does he have?
31. A competitive high school swimmer takes 52 s to swim 100. yards.
What is his rate in m/min?
30. The weight of a diamond is measured in carats. How many pounds
does a 5.75-carat diamond weigh? (1 carat = 200. mg)
33. In 2006, Christian Stengl climbed to the top of Mount Everest,
elevation 29,035 ft, from a starting point of 21,002 ft in a record
time of 16 hr, 42 min. Determine his average rate of climb in
(a) miles per minute
(b) meters per second
34. The Alvin, a submersible research vessel, can descend into the
ocean to a depth of approximately 4500 m in just over 5 hr. Determine its average rate of submersion in
(a) feet per minute
(b) kilometers per second
35. The world’s record for the largest cup of coffee was broken on
­October 15, 2010, with a 2010-gal cup of coffee in Las Vegas, Nevada.
If a cup of coffee contains 473 mL of coffee, how many cups of
coffee would be required to fill this coffee cup?
36. Tilapia is rapidly becoming an important source of fish around the
world. This is because it can be farmed easily and sustainably. In
2011, 475 million lb of tilapia were consumed by Americans. If
the average tilapia has a mass of 535 g, how many tilapia were
consumed in 2011?
26. An extra-strength aspirin tablet contains 0.500 g of the active ingredient,
acetylsalicylic acid. Aspirin strength used to be measured in grains.
If 1 grain = 60 mg, how many grains of the active ingredient are in
1 tablet? ­(Report your answer to three significant figures.)
PRNewsFoto/GourmetGift
Baskets.com/AP/Wide World
Photos
32. In 2005, Jarno Trulli was the pole winner of the U.S. Grand Prix
Race with a speed of 133 mi per hr. What was his speed in cm/s?
37. Assuming that there are 20. drops in 1.0 mL, how many drops are
in 1.0 gallon?
38. How many liters of oil are in a 42-gal barrel of oil?
Additional Exercises 41
39. Calculate the number of milliliters of water in a cubic foot of water.
40. Oil spreads in a thin layer on water called an “oil slick.” How much
area in m2 will 200 cm3 of oil cover if it forms a layer 0.5 nm thick?
41. A textbook is 27 cm long, 21 cm wide and 4.4 cm thick. What is the
volume in:
(a) cubic centimeters?
(c) cubic inches?
(b) liters?
42. An aquarium measures 16 in. * 8 in. * 10 in. How many liters of
water does it hold? How many gallons?
43. A toddler in Italy visits the family doctor. The nurse takes the
child’s temperature, which reads 38.8°C.
(a) Convert this temperature to °F.
(b) If 98.6°F is considered normal, does the child have a fever?
44. Driving to the grocery store, you notice the temperature is 45°C.
Determine what this temperature is on the Fahrenheit scale and
what season of the year it might be.
45. Make the following conversions and include an equation for each
one:
(a) 162°F to °C
(c) - 18°C to °F
(b) 0.0°F to K
(d) 212 K to °C
46. Make the following conversions and include an equation for each
one:
(a) 32°C to °F
(c) 273°C to K
(b) -8.6°F to °C
(d) 100 K to °F
47. At what temperature are the Fahrenheit and Celsius ­temperatures
exactly equal?
48. At what temperature are Fahrenheit and Celsius temperatures the
same in value but opposite in sign?
49. The average temperature on Venus is 460°C. What is this temperature
in °F?
50. The average temperature at the top of Jupiter’s clouds is - 244°F.
What is this temperature in °C?
51. What is the density of a sample of 65.0 mL of automobile oil having
a mass of 59.82 g?
52. A 25.2-mL sample of kerosene was determined to have a mass of
20.41 g. What is the density of kerosene?
53. A student weighed an empty graduated cylinder and found that it
had a mass of 25.23 g. When filled with 25.0 mL of an unknown
liquid, the total mass was 50.92 g.What is the density of the liquid?
54. A total of 32.95 g of mossy zinc was placed into a graduated cylinder containing 50.0 mL of water. The water level rose to 54.6 mL.
Determine the density of the zinc.
55. Linseed oil has a density of 0.929 g/mL. How many mL are in 15 g
of the oil?
56. Glycerol has a density of 1.20 g/mL. How many mL are in 75 g of
glycerol?
Additional Exercises
57. You calculate that you need 10.0123576 g of NaCl for an experiment. What amount should you measure out if the ­precision of the
balance is
(a) + or - 0.01 g?
(c) + or - 0.0001 g?
(b) + or - 0.001 g?
58. Often small objects are measured by mass in order to count them.
(a) If the mass of 1 Skittle is 1.134, what mass of Skittles should
be packaged in a bag containing 175 Skittles?
(b) The volume of exactly 6 Skittles is 5.3 mL as measured in a
10.0-mL graduated cylinder. What volume (in liters) of Skittles
should be packaged in a bag containing 175 Skittles?
(c) Determine the number of Skittles expected in 325.0-g bag of
Skittles.
(d) Determine the number of Skittles expected in a beaker containing 0.550 L of Skittles.
(e) Five bags were filled with 350.0 g of Skittles and 5 more bags
were filled with 0.325 L of Skittles. The Skittles in each bag
were then counted and the data are tabulated below:
Mass Skittles
Number of
Skittles
Volume Skittles
Number of
Skittles
350.0 g
310
0.325 L
392
350.0 g
313
0.325 L
378
350.0 g
308
0.325 L
401
350.0 g
309
0.325 L
369
350.0 g
312
0.325 L
382
Which measurement method is more accurate? More precise?
­Explain your ­reasoning.
59. Suppose you want to add 100 mL of solvent to a reaction flask.
Which piece of glassware shown in Figure 2.3 would be the best
choice for accomplishing this task and why?
60. A reaction requires 21.5 g of CHCl3. No balance is ­available, so it
will have to be measured by volume. How many mL of CHCl3 need
to be taken? (Density of CHCl3 is 1.484 g/mL.)
61. A 25.27-g sample of pure sodium was prepared for an
­experiment. How many mL of sodium is this? (Density of
­sodium is 0.97 g/mL.)­
62. In the United States, coffee consumption averages 4.00 * 108 cups
per day. If each cup of coffee contains 160 mg caffeine, how many
pounds of caffeine are being consumed each day?
63. In a role-playing video game (RPG) your character is a Human
Paladin that can carry 115 lb of gear. Your character is carrying 92
lb of gear and a vial of strength potion (which allows you to carry
an additional 50.0 lb of gear). If you find a cave filled with mass
potions (used for resisting strong winds), after using the strength
potion, how many vials can you collect if the vials each contain
50.0 mL of mass potion with a density of 193 g/mL? The vials have
negligible mass.
64. A cape is designed for Lady Gaga’s concert with 4560 sequins. If
a sequin has a volume of 0.0241 cm3 and the sequins have a density
of 41.6 g/cm3, what is the mass in lb and kg of sequins on Lady
Gaga’s cape?
65. Will a hollow cube with sides of length 0.50 m hold 8.5 L
of solution? Depending on your answer, how much additional solution would be required to fill the container or how
many times would the container need to be filled to measure
the 8.5 L?
• Standards for Measurement
66. The accepted toxic dose of mercury is 300 g/day. ­Dental offices
sometimes contain as much as 180 g of mercury per cubic meter
of air. If a nurse working in the office ingests 2 * 104 L of air per
day, is he or she at risk for mercury ­poisoning?
67. Hydrogen becomes a liquid at 20.27 K. What is this temperature in
(a) °C?
(b) °F?
68. Scientists led by Rob Eagle at CalTech have examined 150-millionyear-old fossilized teeth of sauropods, huge four-legged dinosaurs,
to determine their average body temperature. They analyzed the
type and number of carbon-oxygen bonds, leading them to conclude that the sauropods’ internal temperature was between 36 and
38°C. Determine this temperatue range in °F and compare the body
temperature to other modern animals. Which of the following animals do these dinosaurs most closely resemble?
Animal
Temperature
dogs
100.5–102.5°F
tropical fish
78°F
bottlenose dolphin
97–99°F
tortoises
78–82°F
humans
99°F
birds
105°F
cows
102°F
69. According to the National Heart, Lung, and Blood Institute,
LDL-cholesterol levels of less than 130 mg of LDL-cholesterol
per deciliter of blood are desirable for heart health in humans. On
the average, a human has 4.7 L of whole blood. What is the maximum number if grams of LDL-cholesterol that a human should
have?
70. You have been sent to buy gold from prospectors in the West.
You have no balance but you do have a flask full of mercury. You
have been told that many prospectors will bring you fool’s gold
or iron pyrite in the hopes that you cannot tell the difference.
Given the following densities for gold, mercury, and iron pyrite,
how could you determine whether the samples brought to you are
really gold?
gold
mercury
iron pyrite
76. A cube of aluminum has a mass of 500. g. What will be the mass
of a cube of gold of the same dimensions?
77. A 25.0-mL sample of water at 90°C has a mass of 24.12 g. ­Calculate
the density of water at this temperature.
78. The mass of an empty container is 88.25 g. The mass of the container when filled with a liquid (d = 1.25 g>mL) is 150.50 g. What
is the volume of the container?
79. Which liquid will occupy the greater volume, 50 g of water or 50 g
of ethyl alcohol? Explain.
80. The Sacagawea gold-colored dollar coin has a mass of 8.1 g and
contains 3.5% manganese. What is its mass in ounces (1 lb = 16 oz),
and how many ounces of Mn are in this coin?
81. The density of sulfuric acid is 1.84 g>mL. What volume of this acid
will weigh 100. g?
82. Dark-roasted coffee contains increased amonts of N-­methylpyridinium,
or NMP, a ringed compound that is not present in green unroasted
coffee beans. This compound seems to decrease the stomach acid
production normally associated with drinking coffee. In coffee made
from dark-roasted coffee, the concentration of NMP is 31.4 mg/L.
How many milligrams of NMP would you consume if you drank 2.00
large cups of dark-roasted coffee? (One coffee cup contains 10.0 fluid
ounces of liquid.)
83. The density of palladium at 20°C is 12.0 g>mL, and at 1550°C the
density is 11.0 g>mL. What is the change in volume (in mL) of
1.00 kg Pd in going from 20°C to 1550°C?
84. As a solid substance is heated, its volume increases, but its mass
remains the same. Sketch a graph of density ­versus temperature
showing the trend you expect. Briefly explain.
85. The first Apple computer had 5.0 Mbytes of storage space on its
hard drive and the cost of this computer was $9995. An Apple
iPad II has 64 Gbytes of storage for a cost of $699. Calculate the
cost per byte for each of these two Apple products. Which is a
better buy?
18.3 g/mL
13.6 g/mL
5.00 g/mL
71. The height of a horse is measured in hands (1 hand = exactly 4 in.).
How many meters is a horse that ­measures 14.2 hands?
72. Camels have been reported to drink as much as 22.5 gal of water in
12 hr. How many liters can they drink in 30. days?
73. You are given three cubes, A, B, and C; one is magnesium, one is
aluminum, and the third is silver. All three cubes have the same
mass, but cube A has a volume of 25.9 mL, cube B has a volume of
16.7 mL, and cube C has a volume of 4.29 mL. Identify cubes A,
B, and C.
74. When a chunk of wood burns, much more than just smoke is produced. In addition nanotubes made of pure carbon are formed that
they have a structure resembling a roll of chicken wire. These nanotubes are stronger than steel, resistant to fire, and make good heat
conductors. If a typical carbon nanotube has a diameter of 1.3 nm,
how many nanotubes would need to be laid side by side to construct
a bridge 40.0 feet wide?
75. In the United States, land is measured in acres. There are 43,560 ft2
in each acre. How many km2 are in 125 acres?
© Christine Glade/iStockphoto
chapter 2
Rama & Musée
Bolo/Wikimedia/http://en.
wikipedia.org/wiki/File:
Apple-II.jpg
42 (a) Apple II computer
(b) iPad
86. A gold bullion dealer advertised a bar of pure gold for
sale. The gold bar had a mass of 3300 g and measured
2.00 cm * 15.0 cm * 6.00 cm. Was the bar pure gold? Show
evidence for your answer.
87. A 35.0-mL sample of ethyl alcohol (density 0.789 g/mL) is
added to a graduated cylinder that has a mass of 49.28 g. What
will be the mass of the cylinder plus the alcohol?
88. Several years ago, pharmacists used the Apothecary System of
Measurement. In this system, 1 scruple is equal to 20 grains
(gr). There are 480 gr in 1 oz, and in this system, there are 373 g
in 12 oz. How many scruples would be in 695 g?
Answers to Practice Exercises 43
Challenge Exercises
89. You have just purchased a 500-mL bottle of a decongestant
medication. The doctor prescribed 2 teaspoons 4 times a day for
10 days. Have you purchased enough medication? (1 teaspoon (tsp)
= 5 mL)
90. Your boss found a piece of metal in the lab and wants you to determine what the metal is. She is pretty sure that the metal is either lead, aluminum, or silver. The lab bench has a balance and a
100-mL graduated cylinder with 50 mL of water in it. You decide
to weigh the metal and find that it has a mass of 20.25 g. After
dropping the piece of metal into the graduated cylinder containing
water, you observe that the volume ­increased to 57.5 mL. Identify
the metal.
91. Forgetful Freddie placed 25.0 mL of a liquid in a graduated cylinder with a mass of 89.450 g when empty. When Freddie placed a
metal slug with a mass of 15.454 g into the cylinder, the volume
rose to 30.7 mL. Freddie was asked to calculate the density of the
liquid and of the metal slug from his data, but he forgot to obtain the mass of the liquid. He was told that if he found the mass
of the cylinder containing the liquid and the slug, he would have
enough data for the calculations. He did so and found its mass to be
125.934 g. Calculate the density of the liquid and of the metal slug.
92. Neutrinos are subatomic particles with a very low mass. Recent
work at CERN, Europe’s particle-physics lab near Geneva, Switzerland, suggests that neutrinos may have the ability to travel
faster than the speed of light. If the speed of light is 1.86 * 108
mi/hr, how many nanoseconds should it take for light to travel from
CERN to the Gran Sasso National Lab in Italy, a 730.0-km journey? If a neutrino can travel the same distance 60 nsec faster, how
many significant figures would you need to detect the difference
in speed?
Answers to Practice Exercises
2.1
(a) 1200 = 1.200 * 103
(left means positive exponent)
2.8
(a) 0.560 g; (b) 0.525 kg; (c) 1.75 * 105 mg
2.9
(a) 0.025 L; (b) 1550 mL; (c) 45 dL
(b) 6,600,000 = 6.6 * 106
(left means positive exponent)
2.10 0.14 lb
(c) 0.0468 = 4.68 * 10-2
(right means negative exponent)
2.12 30 m3 or 3 * 101 m3
2.11 (a) 1.69 * 104 m; (b) 3.5 * 10-2 m2
(d) 0.00003 = 3 * 10-5
(right means negative exponent)
2.13 165 lb
2.2
(a) the estimated digits are 4, 1, 0; (b) 3
2.15 1.89 L (the number of significant figures is arbitrary)
2.3
(a) 2; (b) 4; (c) 4; (d) 1; (e) 3; (f) 4; (g) 1; (h) 3
2.16
2.4
(a) 42.25 (Rule 2); (b) 88.0 (Rule 1); (c) 0.0897 (Rule 2);
(d) 0.090 (Rule 2); (e) 225 (Rule 1); (f) 14.2 (Rule 2)
2.17 37.0°C, 310. K
2.5
(a) 3350 in.2 = 3.4 * 103 in.2; (b) 50.7 mi>hr; (c) 0.79;
(d) 1.3; (e) 20.5; (f) 3.71
2.6
(a) 2; (b) 2; (c) 1; (d) 2; (e) 4; (f) 2; (g) 2; (h) 2
2.7
(a) 0.567 m; (b) 680; (c) 0.125
2.14 0.793 qt
-269°C, - 452°F
2.18 The density is 9.35 g>mL; therefore the ring is not pure silver.
The density of silver is 10.5 g>mL.
2.19 0.88 g>mL
chapter 3
• Elements and Compounds
This reclining Buddha in the Grand
Palace in Bangkok, Thailand, is
made of gold.
C h a p te r
3
Elements and
Compounds
I
n Chapter 1 we learned that matter can be divided into the
broad categories of pure substances and mixtures. We further learned that pure substances are either elements or
compounds. The chemical elements are very important to us
in our daily lives. In tiny amounts they play a large role in our
health and metabolism. Metallic elements are used for the skin
of ­airplanes, buildings, and sculpture. In this chapter we explore
the nature of the chemical elements and begin to learn how
chemists classify them.
Chapter Outline
3.1 Elements
3.2 Introduction to the Periodic Table
3.3 Compounds and Formulas
Bill Bachmann/PhotoEdit
44 3.1
• Elements 45
3.1 Elements
Define an element and write the chemical symbol for an element when given its
name.
Learning objective
All words in English are formed from an alphabet consisting of only 26 letters. All known substances on Earth—and most probably in the universe, too—are formed from a sort of “chemical
alphabet” consisting of over 100 known elements. An element is a fundamental or elementary
substance that cannot be broken down by chemical means to simpler substances. Elements are
the building blocks of all substances. The elements are numbered in order of increasing complexity beginning with hydrogen, number 1. Of the first 92 elements, 88 are known to occur in nature.
The other four—technetium (43), promethium (61), astatine (85), and francium (87)—either do
not occur in nature or have only transitory existences during radioactive decay. With the exception of number 94, plutonium, elements above number 92 are not known to occur naturally but
have been synthesized, usually in very small quantities, in laboratories. The discovery of trace
amounts of element 94 (plutonium) in nature has been reported. No elements other than those
on Earth have been detected on other bodies in the universe.
Most substances can be decomposed into two or more simpler substances. Water can be
decomposed into hydrogen and oxygen. Sugar can be decomposed into carbon, hydrogen, and
oxygen. Table salt is easily decomposed into sodium and chlorine. An element, however, cannot be decomposed into simpler substances by ordinary chemical changes.
If we could take a small piece of an element, say copper, and divide it and subdivide it into
smaller and smaller particles, we would finally come to a single unit of copper that we could no
longer divide and still have copper (see Figure 3.1). This smallest particle of an element that can
exist is called an atom, which is also the smallest unit of an element that can enter into a chemical reaction. Atoms are made up of still smaller subatomic particles. However, these subatomic
particles (described in Chapter 5) do not have the properties of elements.
key terms
element
atom
symbols
See the periodic table on the
inside front cover.
Figure 3.1
Photodisc/Getty Images, Inc.
The surface of a penny is made
up of tiny identical copper
atoms packed tightly together.
Natural States of the Elements
Nitrogen and oxygen gases are
composed of molecules (N2, )
and (O2, ).
David Muir/Getty Images, Inc.
Most substances around us are mixtures or compounds. Elements tend to be reactive, and they
combine with other elements to form compounds. It is rare to find elements in nature in pure
form. There are some exceptions, however. Gold, for example, can be found as nuggets. Silver
and platinum can also be found in nature in pure form. In fact, these metals are sometimes
called the noble metals since they have a low reactivity. The noble gases are also not reactive
and can be found in nature in uncombined form. Helium gas, for example, consists of tiny
helium atoms moving independently.
Gold is one of the few
elements found in nature in an
uncombined state.
46 chapter 3
• Elements and Compounds
Air can also be divided into its component gases. It is mainly composed of nitrogen and
oxygen gases. But when we “look inside” these gases, we find tiny molecules (N2 and O2)
instead of independent atoms like we see in the noble gases.
© Ashok Rodrigues/iStockphoto
Distribution of Elements
Helium causes these balloons to
float in air.
At the present time, 118 elements are known and only 88 of these occur naturally. At
normal room temperature only two of the elements, bromine and mercury, are liquids
(see Figure 3.2a). Eleven elements—hydrogen, nitrogen, oxygen, fluorine, chlorine, helium,
neon, argon, krypton, xenon, and radon—are gases (see Figure 3.2b). The elements are distributed unequally in nature.
Ten elements make up about 99% of the mass of the Earth’s crust, seawater, and atmosphere. The distribution of the elements is listed in order of their abundance in Table 3.1.
Oxygen, the most abundant of these elements, accounts for about 20% of the atmosphere
and is found in virtually all rocks, sand, and soil. In these places, oxygen is not present as O2
molecules but as part of compounds usually containing silicon and aluminum atoms. The mass
percents given in Table 3.1 include the Earth’s crust to a depth of 10 miles, the oceans, fresh
Table 3.1 Mass Percent of the Most Abundant Elements in the Earth’s Crust,
Oceans, and Atmosphere
Element
Oxygen
49.2
Silicon
25.7
Aluminum
7.5
Iron
4.7
Calcium
3.4
Sodium
2.6
Potassium
2.4
Magnesium
1.9
Hydrogen
0.9
Element
Titanium
Chlorine
Phosphorus
Manganese
Carbon
Sulfur
Barium
Nitrogen
Fluorine
All others
© Valerie Loiseleux/iStockphoto
Mg 1.9% H 0.9%
Others 1%
K 2.4%
Na 2.6%
Ca 3.4%
Fe 4.7%
Al 7.5%
O2
49.2%
Si
25.7%
Mass percent
Mark Evans/iStockphoto
(b) Neon creates the light in these signs.
(a) Mercury is sometimes used in
thermometers.
Figure 3.2
Examples of a liquid and
a gaseous element.
Mass percent
0.6
0.19
0.11
0.09
0.08
0.06
0.04
0.03
0.03
0.49
3.1
water, and the atmosphere. It does not include the mantle and core of the Earth, which consist
primarily of nickel and iron.
The list of elements found in living matter is very different from those of the Earth’s crust,
oceans, and atmosphere. The major elements in living organisms are shown in Table 3.2.
The major biologically important molecules are formed primarily from oxygen, carbon,
hydrogen, and nitrogen. Some elements found in the body are crucial for life, although they are
present in very tiny amounts (trace elements). Some of these trace elements include ­chromium,
copper, fluorine, iodine, and selenium.
Names of the Elements
The names of the elements come to us from various sources. Many are derived from early Greek,
Latin, or German words that describe some property of the element. For example, iodine is taken
from the Greek word iodes, meaning “violetlike,” and iodine is certainly violet in the vapor state.
The name of the metal bismuth originates from the German words weisse masse, which means
“white mass.” Miners called it wismat; it was later changed to bismat, and finally to bismuth.
Some elements are named for the location of their discovery—for example, germanium, discovered in 1886 by a German chemist. Others are named in commemoration of famous scientists,
such as einsteinium and curium, named for Albert Einstein and Marie Curie, respectively.
Table 3.2 Mass Percent
of the Elements in the
Human Body
Element
Mass percent
Oxygen
Carbon
Hydrogen
Nitrogen
Calcium
Phosphorus
Magnesium
Potassium
Sulfur
Sodium
C
18%
O2
65%
Symbols of the Elements
• Elements 47
65
18
10
3
1.4
1
0.5
0.34
0.26
0.14
H2
10%
N2 3%
Ca 1.4%
P 1%
Others 1%
We all recognize Mr., N.Y., and Ave. as abbreviations for mister, New York, and avenue, respectively. In a like manner, each element also has an abbreviation; these are called symbols
of the elements. Fourteen elements have a single letter as their symbol, and the rest have two
letters. A symbol stands for the element itself, for one atom of the element, and (as we shall
see later) for a particular quantity of the element.
Rules for Symbols of Elements
1. Symbols have either one or two letters.
2. If one letter is used, it is capitalized.
3. If two letters are used, only the first is capitalized.
Examples:
Iodine I
Barium Ba
The symbols and names of all the elements are given in the table on the inside front cover
of this book. Table 3.3 lists the more commonly used elements and their symbols. Examine
Symbols of the Most Common Elements
Element
Symbol
Aluminum
Antimony
Argon
Arsenic
Barium
Bismuth
Boron
Bromine
Cadmium
Calcium
Carbon
Chlorine
Chromium
Cobalt
Copper
Fluorine
Al
Sb
Ar
As
Ba
Bi
B
Br
Cd
Ca
C
Cl
Cr
Co
Cu
F
Element
Gold
Helium
Hydrogen
Iodine
Iron
Lead
Lithium
Magnesium
Manganese
Mercury
Neon
Nickel
Nitrogen
Oxygen
Palladium
Phosphorus
Symbol
Element
Symbol
Au
He
H
I
Fe
Pb
Li
Mg
Mn
Hg
Ne
Ni
N
O
Pd
P
Platinum
Plutonium
Potassium
Radium
Silicon
Silver
Sodium
Strontium
Sulfur
Tin
Titanium
Tungsten
Uranium
Xenon
Zinc
Pt
Pu
K
Ra
Si
Ag
Na
Sr
S
Sn
Ti
W
U
Xe
Zn
© Getty Images, Inc.
Table 3.3
Aluminum is used for recyclable
cans.
48 chapter 3
• Elements and Compounds
© Photononstop/SuperStock
Table 3.4
Symbols of the Elements Derived from Early Names*
Present name
Chunks of sulfur are transported in
baskets from a mine in East Java.
Antimony
Copper
Gold
Iron
Lead
Mercury
Potassium
Silver
Sodium
Tin
Tungsten
Symbol
Former name
Sb
Cu
Au
Fe
Pb
Hg
K
Ag
Na
Sn
W
Stibium
Cuprum
Aurum
Ferrum
Plumbum
Hydrargyrum
Kalium
Argentum
Natrium
Stannum
Wolfram
*These symbols are in use today even though they do not correspond to the current
name of the element.
Table 3.3 carefully and you will note that most of the symbols start with the same letter as
the name of the element that is represented. A number of symbols, however, appear to have
no connection with the names of the elements they represent (see Table 3.4). These symbols
have been carried over from earlier names (usually in Latin) of the elements and are so firmly
implanted in the literature that their use is continued today.
Special care must be taken in writing symbols. Capitalize only the first letter, and use a
lowercase second letter if needed. This is important. For example, consider Co, the symbol
for the element cobalt. If you write CO (capital C and capital O), you will have written the
two elements carbon and oxygen (the formula for carbon monoxide), not the single element
cobalt. Also, make sure that you write the letters distinctly; otherwise, Co (for cobalt) may be
misread as Ca (for calcium).
Knowledge of symbols is essential for writing chemical formulas and equations and will
be needed in the remainder of this book and in any future chemistry courses you may take. One
>Chemistry in action
Naming Elements
Have you ever wondered where the names of the elements
came from? Some of the elements were named for famous
scientists or where they were first discovered and others for
their characteristics. Some of the elemental symbols refer
to old and rarely used names for the elements.
Elements named for places
• Americium (Am)—first created in America.
• Berkelium (Bk)—first created in Berkeley, Califorina.
• Francium (Fr)—discovered at Curie Institute in France.
• Scandium (Sc)—discovered and mined in Scandinavia.
Elements named for planets and the sun
• Helium (He)—named for the sun (helios) because this
was the first place helium was detected.
• Uranium (U)—named for Uranus, which was discovered
just before uranium.
• Plutonium (Pu)—named for the former planet Pluto.
Elements named for mythological characters
• Thorium (Th)—named after Thor, the Scandinavian god
of war. Thorium is used as a fuel for nuclear weapons.
• Titanium (Ti)—named after Titans, supermen of Greek
mythology. Titanium is a super-element because it is
very resistant to acid.
anadium (V)—named after the Scandinavian goddess
• V
of beauty, Vandis, because vanadium compounds form
such beautiful colors.
• Tantalum (Ta) and Niobium (Nb)—named after the
­mythological Greek king, Tantalus, and his daughter,
Princess Niobe. Like the king and his daughter, tantalum and niobium are often found together.
Elements named for their properties
• C
hlorine (Cl)—from Greek chloros meaning “green.”
Chlorine is a greenish yellow gas.
• Iodine (l)—from Greek iodos meaning “violet.”
• A
rgon (Ar)—from Greek argos meaning “lazy.” Argon is
an extremely unreactive gas, so it was thought too lazy
to react.
obalt (Co)—from Greek kobold meaning “goblin”
• C
or “evil spirit.” Cobalt miners often died suddenly
and unexpectedly, which was probably due to ingesting arsenic found with the cobalt ore in the German
mines.
3.2
• Introduction to the Periodic Table 49
way to learn the symbols is to practice a few minutes a day by making flash cards of names
and symbols and then practicing daily. Initially, it is a good plan to learn the symbols of the
most common elements shown in Table 3.3.
Example 3.1
Write the names and symbols for the elements that have only one letter as their symbol.
(Use the periodic table on the inside front cover of your text.)
ENHANCED EXAMPLE
Solution
Boron, B; Carbon, C; Fluorine, F; Hydrogen, H; Iodine, I; Nitrogen, N; Oxygen, O;
­Phosphorus, P; Potassium, K; Sulfur, S; Uranium, U; Vanadium, V; Tungsten, W;
­Yttrium, Y.
Practice 3.1
Write the name, symbol, and vertical column location for the elements whose symbols
begin with C. [Use the periodic table on the inside front cover of your text. (1A, 2A, etc.)]
3.2 Introduction to the Periodic Table
Explain the arrangement of the elements on the periodic table and classify elements
as metal, nonmetal, or metalloid.
Learning objective
key terms
Almost all chemistry classrooms have a chart called the periodic table hanging on the wall. It
shows all the chemical elements and contains a great deal of useful information about them.
As we continue our study of chemistry, we will learn much more about the periodic table. For
now let’s begin with the basics.
A simple version of the periodic table is shown in Table 3.5. Notice that in each box there
is the symbol for the element and, above it, a number called the atomic number. For example
nitrogen is 7 and gold is 79 .
N
Table 3.5
groups
noble gases
alkali metals
alkaline earth metals
halogens
representative elements
transition elements
metals
nonmetals
metalloids
diatomic molecules
Au
The Periodic Table
1
H
Noble
gases
2
He
Metals
3
Li
4
Be
5
B
6
C
7
N
8
O
9
F
10
Ne
11
Na
12
Mg
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
55
Cs
56
Ba
57
La*
72
Hf
73
Ta
74
W
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
87
Fr
88
Ra
89
Ac†
104
Rf
105
Db
106
Sg
107
Bh
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
113
Uut
114
Fl
115
Uup
116
Lv
117
Uus
118
Uuo
*
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
65
Tb
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
†
90
Th
91
Pa
92
U
93
Np
94
Pu
95
Am
96
Cm
97
Bk
98
Cf
99
Es
100
Fm
101
Md
102
No
103
Lr
Metalloids
Nonmetals
chapter 3
• Elements and Compounds
Some groups of elements have
special names.
Noble Gases
2
He
10
Ne
18
Ar
36
Kr
2A
3A
Transition
metals
4A
5A
6A
7A
Halogens
H
Alkaline Earth Metals
Figure 3.3
8A
Alkali metals
1A
Noble gases
50 The elements are placed in the table in order of increasing atomic number in a particular
arrangement designed by Dimitri Mendeleev in 1869. His arrangement organizes the elements
with similar chemical properties in columns called families or groups. An example of this is
the column in the margin.
These elements are all gases and nonreactive. The group is called the noble gases. Other
groups with special names are the alkali metals (under 1A on the table), alkaline earth metals
(Group 2A), and halogens (Group 7A).
The tall columns of the periodic table (1A–7A and the noble gases) are known as the
­representative elements. Those elements in the center section of the periodic table are
called transition elements. Figure 3.3 shows these groups on a periodic table.
54
Xe
Courtesy IBM Research Division
86
Rn
These tiny chips containing silicon,
a metalloid, are being glued
together to form a microprocessor
many times more powerful than
current chips.
Metals, Nonmetals, and Metalloids
The elements can be classified as metals, nonmetals, and metalloids. Most of the elements
are metals. We are familiar with them because of their widespread use in tools, construction
materials, automobiles, and so on. But nonmetals are equally useful in our everyday life as
major components of clothing, food, fuel, glass, plastics, and wood. Metalloids are often used
in the electronics industry.
The metals are solids at room temperature (mercury is an exception). They have high
luster, are good conductors of heat and electricity, are malleable (can be rolled or hammered
into sheets), and are ductile (can be drawn into wires). Most metals have a high melting point
and a high density. Familiar metals are aluminum, chromium, copper, gold, iron, lead, magnesium, mercury, nickel, platinum, silver, tin, and zinc. Less familiar but still important metals
are calcium, cobalt, potassium, sodium, uranium, and titanium.
Metals have little tendency to combine with each other to form compounds. But many
metals readily combine with nonmetals such as chlorine, oxygen, and sulfur to form compounds such as metallic chlorides, oxides, and sulfides. In nature, minerals are composed of
the more reactive metals combined with other elements. A few of the less reactive metals such
as copper, gold, and silver are sometimes found in a native, or free, state. Metals are often
mixed with one another to form homogeneous mixtures of solids called alloys. Some examples
are brass, bronze, steel, and coinage metals.
Nonmetals, unlike metals, are not lustrous, have relatively low melting points and densities, and are generally poor conductors of heat and electricity. Carbon, phosphorus, sulfur,
selenium, and iodine are solids; bromine is a liquid; and the rest of the nonmetals are gases.
Common nonmetals found uncombined in nature are carbon (graphite and diamond), nitrogen, oxygen, sulfur, and the noble gases (helium, neon, argon, krypton, xenon, and radon).
Nonmetals combine with one another to form molecular compounds such as carbon dioxide (CO2), methane (CH4), butane (C4H10), and sulfur dioxide (SO2). Fluorine, the most
reactive nonmetal, combines readily with almost all other elements.
Several elements (boron, silicon, germanium, arsenic, antimony, tellurium, and polonium)
are classified as metalloids and have properties that are intermediate between those of metals and
those of nonmetals. The intermediate position of these elements is shown in Table 3.5. Certain
metalloids, such as boron, silicon, and germanium, are the raw materials for the ­semiconductor
devices that make the electronics industry possible.
3.2
• Introduction to the Periodic Table 51
Example 3.2
ENHANCED EXAMPLE
Which of the following elements are not metals?
Na, Mo, Cl, S, Mg, Pt, Kr, I, C, Cu
Solution
Nonmetals are generally found on the right-hand side of the periodic table: Cl, S, Kr, I, and C.
Practice 3.2
Write the chemical symbols for
(a) five elements that are metals.
(b) two elements that are liquids at normal room temperature.
(c)five elements that are gases (not noble gases) at normal room temperature.
Diatomic Elements
Diatomic molecules each contain exactly two atoms (alike or different). Seven elements in
their uncombined state are diatomic molecules. Their symbols, formulas, and brief descriptions are listed in Table 3.6. Whether found free in nature or prepared in the laboratory, the
molecules of these elements always contain two atoms. The formulas of the free elements are
therefore always written to show this molecular composition: H2 , N2 , O2 , F2 , Cl2 , Br2 , and I2 .
Table 3.6
Elements That Exist as Diatomic Molecules
Element
Symbol
Molecular formula
Hydrogen
Nitrogen
Oxygen
Fluorine
Chlorine
Bromine
Iodine
H
N
O
F
Cl
Br
I
H2
N2
O2
F2
Cl2
Br2
I2
Normal state
Container of hydrogen molecules.
Colorless gas
Colorless gas
Colorless gas
Pale yellow gas
Greenish-yellow gas
Reddish-brown liquid
Bluish-black solid
It is important to see that symbols can designate either an atom or a molecule of an element. Consider hydrogen and oxygen. Hydrogen gas is present in volcanic gases and can be
prepared by many chemical reactions. Regardless of their source, all samples of free hydrogen
gas consist of diatomic molecules.
Free hydrogen is designated by the formula H2 , which also expresses its composition. Oxygen makes up about 20% by volume of the air that we breathe. This free oxygen is constantly
being replenished by photosynthesis; it can also be prepared in the laboratory by several reactions. H2 (gray) and O2 (red) molecules.
The majority of free oxygen is diatomic and is designated by the
formula O2 . Now consider water, a compound designated by the formula H2O (sometimes HOH). Water contains neither free hydrogen
Elements
(H2) nor free oxygen (O2). The H2 part of the formula H2O simply
indicates that two atoms of hydrogen are combined with one atom of
oxygen to form water.
Atomic
Molecular
Symbols are used to designate elements, show the composition
of molecules of elements, and give the elemental composition of
compounds.
Figure 3.4 summarizes the classification of elements.
Figure 3.4
Elements fall into two
broad categories.
Example: Ar
Example: N2
52 chapter 3
• Elements and Compounds
>Chemistry in action
Atomic Oxygen, Friend or Foe?
When scientists at Nasa first began sending spacecraft
into orbit, they discovered that the spacecraft were being
corroded by the environment of the upper atmosphere.The
culprit was the atomic oxygen formed by the interaction of
ultraviolet radiation with oxygen gas, or O2. Researchers
using atomic oxygen to test the durability of satellite coatings found that atomic oxygen could remove carbon-based
materials from the surface of objects without damaging
them. Furthermore, NASA engineers discovered that
atomic oxygen would not react with and damage
glass composed of silicon dioxide because
this silicon had already reacted with oxygen
and would not react further. Armed with
this knowledge, they began to coat the
space station and shuttle parts with a
thin layer of silicon dioxide to protect
them from the corrosive effects of
atomic oxygen.
Back in the lab, NASA researchers
looked for new ways to utilize the
power of atomic oxygen. They discovered several useful applications.
Surgical Implants When implanting foreign objects in a patient, a surgeon
must be sure the parts are clean and sterile. It
Damaged
is easy to sterilize these parts and kill all bacteria, but occasionally there is organic debris left behind from the dead
bacteria. This organic material can cause postoperative
inflammation. Treatment of the sterile implants with atomic
oxygen dissolves away the organic matter, leading to more
successful and less painful recovery times.
Cleaning of Damaged Art Artwork that has been
damaged by soot can be cleaned using atomic oxygen. The
oxygen atoms react with the soot and cause it to vaporize
away. The chief conservator at the Cleveland Museum of Art
tried the atomic oxygen treatment on two paintings damaged in a church fire. Although the
paintings were not extremely valuable (and
so were good subjects for an experiment),
all other attempts to restore them had
failed. Atomic oxygen proved to work
wonders, and the soot and char were
removed to reveal the original image.
Because the treatment is a gas, the
underlying layers were not harmed.
The treatment doesn’t work on everything and won’t replace other techniques altogether, but the conservator
was impressed enough to continue to
work with NASA on the process.
Restored
NASA
P r a ct i c e 3 . 3
Identify the physical state of each of the following elements at room temperature (20°C):
H, Na, Ca, N, S, Fe, Cl, Br, Ne, Hg
Hint: You may need to use a resource (such as the Internet or a chemical handbook) to
assist you.
P r a ct i c e 3 . 4
Identify each of the following elements as a nonmetal, metal, or metalloid:
Na, F, Cr, Mo, Kr, Si, Cu, Sb, I, S
3.3 Compounds and formulas
Learning objective
key terms
compound
molecule
ion
cation
anion
chemical formula
subscripts
natural law
law of definite composition
law of multiple proportions
Distinguish between molecular and ionic compounds and write chemical formulas
for compounds.
A compound is a distinct substance that contains two or more elements chemically combined
in a definite proportion by mass. Compounds, unlike elements, can be decomposed chemically into simpler substances—that is, into simpler compounds and/or elements. Atoms of
the ­elements in a compound are combined in whole-number ratios, never as fractional parts.
Molecular and Ionic Compounds
Compounds fall into two general types, molecular and ionic. Figure 3.5 illustrates the
c­ lassification of compounds.
3.3
A molecule is the smallest uncharged individual unit of a compound formed by the union of two or more atoms. Water is a typical molecular compound. If we divide a drop of water into smaller
and smaller particles, we finally obtain a single molecule of water
consisting of two hydrogen atoms bonded to one oxygen atom, as
shown in Figure 3.6. This molecule is the ultimate particle of water;
it cannot be further subdivided without destroying the water molecule and forming hydrogen and oxygen.
An ion is a positively or negatively charged atom or group of
atoms. An ionic compound is held together by attractive forces that
exist between positively and negatively charged ions. A positively
charged ion is called a cation (pronounced cat-eye-on); a negatively
charged ion is called an anion (pronounced an-eye-on).
Sodium chloride is a typical ionic compound. The ultimate particles of sodium chloride are positively charged sodium ions and
negatively charged chloride ions, shown in Figure 3.6. Sodium chloride is held together in a crystalline structure by the attractive forces existing between these oppositely charged ions. Although ionic
compounds consist of large aggregates of cations and anions, their
formulas are normally represented by the simplest possible ratio of
the atoms in the compound. For example, in sodium chloride the
ratio is one sodium ion to one chloride ion, so the formula is NaCl.
O
H
Na+
H
Two hydrogen atoms combined
with an oxygen atom to form a
molecule of water (H2O).
• Compounds and Formulas 53
Compounds
Molecular
Ionic
Example: H2O
Example: NaCl
Figure 3.5
Compounds can be classified as molecular and ionic. Ionic
compounds are held together by attractive forces between
their positive and negative charges. Molecular compounds
are held together by covalent bonds.
Cl –
A positively charged sodium
ion and a negatively charged
chloride ion form the compound
sodium chloride (NaCl).
Figure 3.6
Representation of molecular and ionic (nonmolecular) compounds.
There are more than 50 million known registered compounds, with no end in sight as to
the number that will be prepared in the future. Each compound is unique and has characteristic
properties. Let’s consider two compounds, water and sodium chloride, in some detail. Water
is a colorless, odorless, tasteless liquid that can be changed to a solid (ice) at 0°C and to a gas
(steam) at 100°C. Composed of two atoms of hydrogen and one atom of oxygen per molecule,
water is 11.2% hydrogen and 88.8% oxygen by mass. Water reacts chemically with sodium
to produce hydrogen gas and sodium hydroxide, with lime to produce calcium hydroxide, and
with sulfur trioxide to produce sulfuric acid. When water is decomposed, it forms hydrogen
and oxygen molecules (see Figure 3.7). No other compound has all these exact physical and
chemical properties; they are characteristic of water alone.
Sodium chloride is a colorless crystalline substance with a ratio of one atom of sodium to
one atom of chlorine. Its composition by mass is 39.3% sodium and 60.7% chlorine. It does
not conduct electricity in its solid state; it dissolves in water to produce a solution that conducts electricity. When a current is passed through molten sodium chloride, solid sodium and
Figure 3.7
¡
Water molecules ¡ Oxygen molecule
+
+ Hydrogen molecules
A representation of the
decomposition of water into
oxygen and hydrogen molecules.
• Elements and Compounds
Andrew Lambert Photography/
Photo Researchers, Inc.
¡
¡
Sodium chloride
Charles D. Winters/Photo
Researchers, Inc.
chapter 3
Christoph Ermel/iStockphoto
54 +
Sodium metal
+
Chlorine gas
Figure 3.8
When sodium chloride is decomposed, it forms sodium metal and chlorine gas.
gaseous chlorine are produced (see Figure 3.8). These specific properties belong to sodium
chloride and to no other substance. Thus, a compound may be identified and distinguished
from all other compounds by its characteristic properties. We consider these chemical properties
further in Chapter 4.
Writing Formulas of Compounds
Chemical formulas are used as abbreviations for compounds. A chemical formula shows
the symbols and the ratio of the atoms of the elements in a compound. Sodium chloride
contains one atom of sodium per atom of chlorine; its formula is NaCl. The formula for
water is H2O; it shows that a molecule of water contains two atoms of hydrogen and one
atom of oxygen.
The formula of a compound tells us which elements it is composed of and how many
atoms of each element are present in a formula unit. For example, a unit of sulfuric acid is
composed of two atoms of hydrogen, one atom of sulfur, and four atoms of oxygen. We could
express this compound as HHSOOOO, but this is cumbersome, so we write H2SO4 instead.
The formula may be expressed verbally as “H-two-S-O-four.” Numbers that appear partially
below the line and to the right of a symbol of an element are called subscripts. Thus, the 2
Figure 3.9
Explanation of the formulas NaCl,
H2SO4, and Ca(NO3)2.
NaCl
Indicates the
element sodium
(one atom)
Indicates the
element chlorine
(one atom)
H 2SO4
Indicates the
element
hydrogen
Indicates the
element sulfur
(one atom)
Indicates four
atoms of
oxygen
Indicates the
element
oxygen
Indicates two
atoms of
hydrogen
Ca(NO3)2
Indicates the
element calcium
(one atom)
Indicates the nitrate
group composed of
one nitrogen atom and
three oxygen atoms
Indicates two
nitrate (NO–3 )
groups
3.3
and the 4 in H2SO4 are subscripts (see Figure 3.9). Characteristics of chemical formulas are
as follows:
1. The formula of a compound contains the symbols of all the elements in the compound.
2. When the formula contains one atom of an element, the symbol of that element represents
that one atom. The number 1 is not used as a subscript to indicate one atom of an element.
3. When the formula contains more than one atom of an element, the number of atoms is
indicated by a subscript written to the right of the symbol of that atom. For example, the
2 in H2O indicates two atoms of H in the formula.
4. When the formula contains more than one of a group of atoms that occurs as a unit,
parentheses are placed around the group, and the number of units of the group is indicated by a subscript placed to the right of the parentheses. Consider the nitrate group,
NO3- . The formula for sodium nitrate, NaNO3 , has only one nitrate group, so no parentheses are needed. Calcium nitrate, Ca(NO3)2 , has two nitrate groups, as indicated
by the use of parentheses and the subscript 2. Ca(NO3)2 has a total of nine atoms: one
Ca, two N, and six O atoms. The formula Ca(NO3)2 is read as “C-A [pause] N-O-three
taken twice.”
5. Formulas written as H2O, H2SO4 , Ca(NO3)2 , and C12H22O11 show only the number and
kind of each atom contained in the compound; they do not show the arrangement of the
atoms in the compound or how they are chemically bonded to one another.
Practice 3.5
How would you read these formulas aloud?
(a) KBr
(b) PbCl2
(c) CaCO3
(d) Mg(OH)2
Example 3.3
Write formulas for the following compounds; the atomic composition is given.
(a) hydrogen chloride: 1 atom hydrogen + 1 atom chlorine;
(b) methane: 1 atom carbon + 4 atoms hydrogen;
(c) glucose: 6 atoms carbon + 12 atoms hydrogen + 6 atoms oxygen.
SOLUTION
(a) First write the symbols of the atoms in the formula: H Cl.
Since the ratio of atoms is one to one, we bring the symbols together to give the formula
for hydrogen chloride as HCl.
(b) Write the symbols of the atoms: C H.
Now bring the symbols together and place a subscript 4 after the hydrogen atom. The
formula is CH4 .
(c) Write the symbols of the atoms: C H O.
Now write the formula, bringing together the symbols followed by the correct subscripts according to the data given (six C, twelve H, six O). The formula is C6H12O6 .
Practice 3.6
Write formulas for the following compounds from the compositions given.
(a) potassium bromide: 1 atom potassium + 1 atom bromine
(b) sodium chlorate: 1 atom sodium + 1 atom chlorine + 3 atoms oxygen
(c) sulfurous acid: 2 atoms hydrogen + 1 atom sulfur + 3 atoms oxygen
(d) aluminum carbonate: 2 aluminum atoms + 3 carbonate ions
Composition of Compounds
A large number of experiments extending over a long period have established the fact
that a particular compound always contains the same elements in the same proportions by
mass. For example, water always contains 11.2% hydrogen and 88.8% oxygen by mass
• Compounds and Formulas 55
56 chapter 3
• Elements and Compounds
O
H
H
H
H
O
O
O
O
H
H
O
H
H
H
O
O
H
H
(c)
(b)
(a)
H
H
H
O
O
O
Figure 3.10
(a) Dalton’s atoms were individual particles, the atoms of each element being alike in mass
and size but different in mass and size from other elements. (b) and (c) Dalton’s atoms
combine in specific ratios to form compounds.
(see Figure 3.10). The fact that water contains hydrogen and oxygen in this particular ratio does
not mean that hydrogen and oxygen cannot combine in some other ratio but rather that a compound with a different ratio would not be water. In fact, hydrogen peroxide is made up of two
atoms of hydrogen and two atoms of oxygen per molecule and contains 5.9% hydrogen and 94.1%
oxygen by mass; its properties are markedly different from those of water (see Figure 3.10).
Water
Atomic composition
11.2% H
88.8% O
2H + 1O
Hydrogen peroxide
5.9% H
94.1% O
2H + 2O
We often summarize our general observations regarding nature into a statement called a natural law. In the case of the composition of a compound, we use the law of definite composition, which states that a compound always contains two or more elements chemically
combined in a definite proportion by mass.
Let’s consider two elements, oxygen and hydrogen, that form more than one compound.
In water, 8.0 g of oxygen are present for each gram of hydrogen. In hydrogen peroxide, 16.0 g
of oxygen are present for each gram of hydrogen. The masses of oxygen are in the ratio of
small whole numbers, 16 : 8 or 2 : 1. Hydrogen peroxide has twice as much oxygen (by mass)
as does water. Using Dalton’s atomic model, we deduce that hydrogen peroxide has twice as
many oxygen atoms per hydrogen atom as water. In fact, we now write the formulas for water
as H2O and for hydrogen peroxide as H2O2. See Figure 3.10b and c.
The law of multiple proportions states that atoms of two or more elements may
combine in different ratios to produce more than one compound.
Some examples of the law of multiple proportions are given in Table 3.7. The reliability
of this law and the law of definite composition is the cornerstone of the science of chemistry.
In essence, these laws state that (1) the composition of a particular substance will always be
the same no matter what its origin or how it is formed, and (2) the composition of different
compounds formed from the same elements will always be unique.
Table 3.7 Selected Compounds Showing Elements That Combine to Give More
Than One Compound
Compound
Formula
Percent composition
Copper(I) chloride
Copper(II) chloride
CuCl
CuCl2
64.2% Cu, 35.8% Cl
47.3% Cu, 52.7% Cl
Methane
Octane
CH4
C8H18
74.9% C, 25.1% H
85.6% C, 14.4% H
Methyl alcohol
Ethyl alcohol
Glucose
CH4O
C2H6O
C6H12O6
37.5% C, 12.6% H, 49.9% O
52.1% C, 13.1% H, 34.7% O
40.0% C, 6.7% H, 53.3% O
Review 57
You need to recognize the difference between a law and a model (theory). A law is a summary of observed behavior. A model (theory) is an attempt to explain the observed behavior.
This means that laws remain constant—that is, they do not undergo modification—while theories (models) sometimes fail and are modified or discarded over time.
C h a p t e r
3 review
3.1 Elements
•
•
•
•
•
•
•
•
•
•
•
All matter consists of about 100 elements.
An element is a fundamental chemical substance.
The smallest particle of an element is an atom.
Elements cannot be broken down by chemical means to a simpler substance.
Most elements are found in nature combined with other elements.
Elements that are found in uncombined form in nature include gold, silver, copper, and platinum as
well as the noble gases (He, Ne, Ar, Kr, Xe, Rn).
Chemical elements are not distributed equally in nature.
Hydrogen is the most abundant element in the universe.
Oxygen is the most abundant element on the Earth and in the human body.
Names for the chemical elements come from a variety of sources, including Latin, location of
d­ iscovery, and famous scientists.
Rules for writing symbols for the elements are:
• One or two letters
• If one letter, use a capital
• If two letters, only the first is a capital
key terms
element
atom
symbols
3.2 Introduction to the Periodic Table
• T
he periodic table was designed by Dimitri Mendeleev and arranges the elements according to their
atomic numbers and in groups by their chemical properties.
• Elements can be classified as representative or as transition elements.
• Elements can also be classified by special groups with similar chemical properties. Such groups
include the noble gases, alkali metals, alkaline earth metals, and halogens.
• Elements can be classified as metals, nonmetals, or metalloids.
• Most elements are metals.
• Metals have the following properties:
• High luster
• Good conductors of heat and electricity
• Malleable
• Nonmetals have the following properties:
• Not lustrous
• Poor conductors of heat and electricity
• Diatomic molecules contain exactly two atoms (alike or different).
• Seven elements exist as diatomic molecules—H2, N2, O2, F2, Cl2, Br2, and I2.
Elements
Atomic
Molecular
Example: Ar
Example: N2
key terms
groups
noble gases
alkali metals
alkaline earth metals
halogens
representative elements
transition elements
metals
nonmetals
metalloids
diatomic molecules
58 chapter 3
• Elements and Compounds
3.3 Compounds and formulas
key terms
compound
molecule
ion
cation
anion
chemical formula
subscripts
natural law
law of definite composition
law of multiple proportions
• A
compound is a substance that contains two or more elements chemically combined in a definite
proportion by mass.
• There are two general types of compounds:
• Molecular—formed of individual molecules composed of atoms
• Ionic—formed from ions that are either positive or negative
• Cation—positively charged ion
• Anion—negatively charged ion
• A chemical formula shows the symbols and the ratios of atoms for the elements in a chemical compound.
• Characteristics of a chemical formula include:
• It contains symbols of all elements in the compound.
• The symbol represents one atom of the element.
• If more than one atom of an element is present, the number of atoms is indicated by a subscript.
• Parentheses are used to show multiple groups of atoms occurring as a unit in the compound.
• A formula does not show the arrangement of the atoms in the compound.
• The law of definite composition states that a compound always contains two or more elements
­combined in a definite proportion by mass.
• The law of multiple proportions states that atoms of two or more elements may combine in different
ratios to form more than one compound.
Compounds
Molecular
Ionic
Example: H2O
Example: NaCl
review Questions
1. Are there more atoms of silicon or hydrogen in the Earth’s crust,
seawater, and atmosphere? Use Table 3.1 and the fact that the mass
of a silicon atom is about 28 times that of a hydrogen atom.
2. Give the symbols for each of the following elements:
(a) manganese
(e) chlorine
(b) fluorine
(f) vanadium
(c) sodium
(g) zinc
(d) helium
(h) nitrogen
3. Give the names for each of the following elements:
(a) Fe
(e) Be
(b) Mg
(f) Co
(c) C
(g) Ar
(d) P
(h) Hg
4. What does the symbol of an element stand for?
5. List six elements and their symbols in which the first letter of the
symbol is different from that of the name. (Table 3.4)
6. Write the names and symbols for the 14 elements that have only
one letter as their symbol. (See periodic table on inside front cover.)
7. Of the six most abundant elements in the human body, how many
are metals? nonmetals? metalloids? (Table 3.2)
8. Write the names and formulas for the elements that exist as diatomic
molecules. (Table 3.6)
9. Interpret the difference in meaning for each pair:
(a) CO and Co
(c) S8 and 8 S
(b) H2 and 2 H
(d) CS and Cs
10. Distinguish between an element and a compound.
11. How many metals are there? nonmetals? metalloids? (Table 3.5)
12. Of the ten most abundant elements in the Earth’s crust, seawater,
and atmosphere, how many are metals? nonmetals? metalloids?
(Table 3.1)
13. Give the names of (a) the solid diatomic nonmetal and (b) the liquid
diatomic nonmetal. (Table 3.6)
14. Distinguish between a compound and a mixture.
15. What are the two general types of compounds? How do they differ
from each other?
16. What is the basis for distinguishing one compound from another?
17. What is the major difference between a cation and an anion?
Paired Exercises 59
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
P A I R E D E x er c i s e s
1. Which of the following are diatomic molecules?
(a) HCl
(e) SiO2
(b) O2
(f) H2O2
(c) N2O
(g) CH4
(d) P4
(h) CIF
2. Which of the following are diatomic molecules?
(a) O3
(e) S8
(b) H2O
(f) Cl2
(c) CO2
(g) CO
(d) HI
(h) NH3
3. What elements are present in each compound?
(a) potassium iodide KI
(b) sodium carbonate Na2CO3
(c) aluminum oxide Al2O3
(d) calcium bromide CaBr2
(e) acetic acid HC2H3O2
4. What elements are present in each compound?
(a) magnesium bromide MgBr2
(b) carbon tetrachloride CCl4
(c) nitric acid HNO3
(d) barium sulfate BaSO4
(e) aluminum phosphate AlPO4
5. Write the formula for each compound (the composition is given
after each name):
(a) zinc oxide
1 atom Zn, 1 atom O
(b) potassium chlorate
1 atom K, 1 atom Cl, 3 atoms O
(c) sodium hydroxide
1 atom Na, 1 atom O, 1 atom H
(d) ethyl alcohol
2 atoms C, 6 atoms H, 1 atom O
6. Write the formula for each compound (the composition is given
after each name):
(a) aluminum bromide
1 atom Al, 3 atoms Br
(b) calcium fluoride
1 atom Ca, 2 atoms F
(c) lead(II) chromate
1 atom Pb, 1 atom Cr, 4 atoms O
(d) benzene
6 atoms C, 6 atoms H
7. Many foods contain interesting compounds that give them their
flavors and smells. Some of these compounds are listed below with
the chemical composition given after each name. Write the formulas for these compounds.
(a) Allicin gives garlic its flavor (6 carbons, 10 hydrogens, 1 oxygen,
and 2 sulfurs).
(b) Capsaicin gives peppers their heat (18 carbons, 27 hydrogens,
1 nitrogen, and 3 oxygens).
(c) Limonene gives oranges their fragrance (10 carbons and 16
hydrogens).
8. Many plants contain interesting compounds that sometimes have
medicinal properties. Some of these compounds are listed below
with the chemical composition given after each name. Write the
formulas for these compounds.
(a) Aescin from horse chestnuts has anti-inflammatory properties
(55 carbons, 86 hydrogens, and 24 oxygens).
(b) Proanthocyanidins found in cranberries help to prevent urinary
tract infections. Proanthocyanidins are polymers composed of
epicatechin units (15 carbons, 14 hydrogens, and 6 oxygens).
(c) Betulinic acid found in the common birch tree is an antimalarial
drug (30 carbons, 48 hydrogens, and 3 oxygens).
9. Write the name and number of atoms of each element in each of the
following compounds:
(a) Fe2O3
(e) K2CO3
(b) Ca(NO3)2
(f) Cu3(PO4)2
(c) Co(C2H3O2)2
(g) C2H5OH
(d) CH3COCH3
(h) Na2Cr2O7
10. Write the name and number of atoms of each element in each of the
following compounds:
(a) HC2H3O2
(e) NiCO3
(b) (NH4)3PO4
(f) KMnO4
(c) Mg(HSO3)2
(g) CH3CH2CH2CH3
(d) ZnCl2
(h) PbCrO4
11. How many total atoms are represented in each formula?
(a) Co(ClO3)2
(c) CH3CH2COOH
(b) (NH4)2SO3
(d) C12H22O11
12. How many total atoms are represented in each formula?
(a) CH3CH2OH
(c) NH4C2H3O2
(b) KAl(SO4)2
(d) C6H4Cl2
13. How many hydrogen atoms are represented in each formula?
(a) Al(C2H3O2)3
(c) NH4OH
(b) CH3CH2CH2OH
(d) C6H5CH2CH3
14. How many oxygen atoms are represented in each formula?
(a) Fe(C2H3O2)3
(c) Ba(ClO3)2
(b) H3PO4
(d) Fe2(Cr2O7)3
15. Determine whether each of the following is a pure substance or a
mixture:
(a) hot tea
(d) cement
(b) beach sand
(e) zinc
(c) carbon dioxide
(f) vinegar
16. Determine whether each of the following is a pure substance or a
mixture:
(a) dirt
(d) dinitrogen monoxide
(b) salad dressing
(e) brass
(c) tungsten
(f) egg
17. For Question 15, state whether each pure substance is an element
or a compound.
18. For Question 16, state whether each pure substance is an element
or a compound.
60 chapter 3
• Elements and Compounds
19. Classify each of the following as an element, a compound, or a
mixture:
(a)
(c)
(b)
20. Classify each of the following as an element, a compound, or a
mixture:
(a)
(b)
(c)
21. Is there a pattern to the location of the gaseous elements on the
periodic table? If so, describe it.
22. Is there a pattern to the location of the liquid elements on the periodic
table? If so, describe it.
23. What percent of the first 36 elements on the periodic table are
metals?
24. What percent of the first 36 elements on the periodic table are
­solids at room temperature?
25. In Section 3.3, there is a statement about the composition of water.
It says that water (H2O) contains 8 grams of oxygen for every
1 gram of hydrogen. Show why this statement is true.
26. In Section 3.3, there is a statement about the composition of hydrogen peroxide. It says that hydrogen peroxide (H2O2) contains
16 grams of oxygen for every 1 gram of hydrogen. Show why this
statement is true.
Add i t i o n a l E x e rc i s e s
27. You accidentally poured salt into your large grind pepper shaker.
This is the only pepper that you have left and you need it to cook
a meal, but you don’t want the salt to be mixed in. How can you
successfully separate these two components?
28. On the periodic table at the front of this book, do you notice anything
about the atoms that make up the following ionic compounds:
NaCl, KI, and MgBr2? (Hint: Look at the position of the atoms in
the given compounds on the periodic table.)
29. How many total atoms are present in each of the following
compounds?
(a) CO
(d) KMnO4
(b) BF3
(e) Ca(NO3)2
(c) HNO3
(f) Fe3(PO4)2
30. The formula for vitamin B12 is C63H88CoN14O14P.
(a) How many atoms make up one molecule of vitamin B12 ?
(b) What percentage of the total atoms are carbon?
(c) What fraction of the total atoms are metallic?
31. It has been estimated that there is 4 * 10-4 mg of gold per liter of
seawater. At a price of $19.40>g, what would be the value of the
gold in 1 km3(1 * 1015 cm3) of the ocean?
32. Calcium dihydrogen phosphate is an important fertilizer. How many
atoms of hydrogen are there in ten formula units of Ca(H2PO4)2 ?
33. How many total atoms are there in one molecule of C145H293O168 ?
34. Name the following:
(a) three elements, all metals, beginning with the letter M
(b) four elements, all solid nonmetals
(c) five elements, all solids in the first five rows of the periodic
table, whose symbols start with letters different from the
element name
35. How would you separate a mixture of sugar and sand and isolate
each in its pure form?
36. How many total atoms are there in seven dozen formulas of nitric
acid, HNO3 ?
37. Make a graph using the following data. Plot the density of air in
grams per liter along the x-axis and temperature along the y-axis.
Temperature (°C)
0
10
20
40
80
Density (g/L)
1.29
1.25
1.20
1.14
1.07
(a) What is the relationship between density and temperature according to your graph?
(b) From your plot, find the density of air at these temperatures:
5°C
25°C
70°C
38. These formulas look similar but represent different things.
8S
S8
Compare and contrast them. How are they alike? How are they
different?
39. Write formulas for the following compounds that a colleague read
to you:
(a) NA9CL
(f) CA (pause) CN taken twice
(b) H29S9O4
(g) C69H129O6
(c) K29O
(h) C29 H5 (pause) OH
(d) Fe29S3
(i) CR (pause) NO3 taken three times
(e) K39P9O4
Answers to Practice Exercises 61
40. The abundance of iodine in seawater is 5.0 * 10-8 percent by
mass. How many kilograms of seawater must be treated to obtain
1.0 g of iodine?
41. When the tongue detects something sweet, it begins a chain of reactions to tell the body to expect calorie-containing foods. If we consume noncaloric foods containing artificial sweeteners, our bodies
expect food and will tell our brains to find more if the sweet food
does not contain anything to digest. Some common sweeteners are
listed below. Write the name and number of atoms of each element
in the sweeteners.
Compound
(a) Sucrose
(b) Saccharin
(c) Aspartame
(d) Acesulfame-K
(e) Sucralose
Common name
Table sugar
Sweet ’N Low
NutraSweet, Equal
Sunett, Sweet One
Splenda
42. Should the elemental symbol for cobalt be written as Co or CO?
What confusion could arise if the wrong notation were used?
43. Caffeine has the chemical structure illustrated in the image given
below. Carbon atoms are represented as grey spheres, nitrogen
atoms as blue spheres, oxygen atoms as red spheres, and hydrogen
atoms as white spheres. Write a chemical formula for caffeine.
Chemical formula
C12H22O11
C7H5O3NS
C14H18O5N2
C4H4O3NS-K
C12H19O8Cl3
C h a l l e n g e E x e rc i s e s
44. Write the chemical formulas of the neutral compounds that would
result if the following ions were combined. Use the Charges of Common Ions table on the inside back cover of the book to help you.
(a) ammonium ion and chloride ion
(b) hydrogen ion and hydrogen sulfate ion
(c) magnesium ion and iodide ion
(d) iron(II) ion and fluoride ion
(e) lead(II) ion and phosphate ion
(f) aluminum ion and oxide ion
45. Write formulas of all the compounds that will form between the
first five of the Group 1A and 2A metals and the oxide ion.
46. Unsaturated fats and oils are often used in cooking. For each pair
of fatty acids listed below, tell the number of carbon and hydrogen
atoms in each molecule.
(a) arachidic acid (saturated) CH3(CH2)18COOH and arachidonic
acid (unsaturated) CH3(CH2)4(CH “ CHCH2)4(CH2)2COOH
(b) stearic acid (saturated) CH3(CH2)16COOH and linoleic acid
(unsaturated) CH3(CH2)4(CH “ CHCH2)2(CH2)6COOH
(c) What is the ratio of hydrogen to carbon atoms for each pair?
(d) Propose a possible explanation for describing linoleic and stearic
acids as being saturated.
A n s w e r s t o P r a ct i c e E x e rc i s e s
3.1
Cesium, Cs, IA; Calcium, Ca, 2A; Chromium, Cr, 6B; Cobalt, Co,
8B; Copper, Cu, 1B; Cadmium, Cd, 2B; Carbon, C, 4A; Chlorine,
Cl, 7A; Cerium, Ce, 4B; Copernicium, Cn, 2B; Curium, Cm, 8B;
Californium, Cf, 1B.
3.5
(a) K9BR
(b) PB–CL2
(c) CA (pause) CO3
(d) MG (pause) OH taken twice
3.2
(a) any of the transition metals
(b) Br and Hg
(c) H, N, O, F, Cl
3.6
3.3
gases
liquids
solids
H, N, Cl, Ne
Br, Hg
Na, Ca, S, Fe
(a) KBr
(b) NaClO3
(c) H2SO3
(d) Al2(CO3)3
3.4
nonmetal
metal
metalloid
F, Kr, I, S
Na, Cr, Mo, Cu
Si, Sb
Chapter
4
Properties of matter
T
he world we live in is a kaleidoscope of sights, sounds,
smells, and tastes. Our senses help us to describe these
objects in our lives. For example, the smell of freshly
baked cinnamon rolls creates a mouthwatering desire to gobble
down a sample. Just as sights, sounds, smells, and tastes form
the properties of the objects around us, each substance in
chemistry has its own unique properties that allow us to identify
it and predict its interactions.
These interactions produce both physical and chemical
changes. When you eat an apple, the ultimate metabolic result
is carbon dioxide and water. These same products are achieved
by burning logs. Not only does a chemical change occur in these
cases, but an energy change occurs as well. Some reactions release energy (as does the apple or the log) whereas others require
energy, such as the production of steel or the melting of ice. Over
90% of our current energy comes from chemical reactions.
Chapter Outline
4.1 Properties of Substances
4.2 Physical and Chemical Changes
4.3 Learning to Solve Problems
4.4 Energy
4.5 Heat: Quantitative Measurement
4.6 Energy in the Real World
© Jenny Swanson/iStockphoto
A burning log undergoes chemical
change resulting in the release of
energy in the form of heat and
light. The physical properties of
the log change during the
chemical reaction.
4.1
• Properties of Substances 63
4.1 Properties of Substances
Compare the physical and chemical properties of a substance.
Learning objectiv e
How do we recognize substances? Each substance has a set of properties that is characteristic
of that substance and gives it a unique identity. Properties—the “personality traits” of
­substances—are classified as either physical or chemical. Physical properties are the inherent characteristics of a substance that can be determined without altering its composition; they
are associated with its physical existence. Common physical properties include color, taste, odor,
state of matter (solid, liquid, or gas), density, melting point, and boiling point (see Figure 4.1).
Chemical properties describe the ability of a substance to form new substances, either by
reaction with other substances or by decomposition.
Let’s consider a few of the physical and chemical properties of chlorine. Physically,
chlorine is a gas at room temperature about 2.4 times heavier than air. It is greenish
yellow in color and has a disagreeable odor. Chemically, chlorine will not burn but
will support the combustion of certain other substances. It can be used as a bleaching
agent, as a disinfectant for water, and in many chlorinated substances such as refrigerants and insecticides. When chlorine combines with the metal sodium, it forms a salt
called sodium chloride (see Figure 4.2). These properties, among many others, help us
characterize and identify chlorine.
key terms
Martin Dohrn/Photo Researchers, Inc.
properties
physical properties
chemical properties
Figure 4.1
Physical property. The boiling point of water is a
physical property. At its boiling point water changes
from a liquid to a gas, but the molecules remain water
molecules. They are still water but are farther apart.
+
¡
Chlorine gas
Sodium chloride
Christoph Ermel/iStockphoto
+
Charles D. Winters/Photo
Researchers, Inc.
Andrew Lambert Photography/
Photo Researchers, Inc.
Sodium metal
¡
Cl –
Na +
Na atom
Cl2 molecules
Na +
Figure 4.2
Chemical property. When sodium metal reacts with chlorine gas, a new substance called sodium chloride forms.
Cl –
64 chapter 4
• Properties of Matter
Table 4.1 Physical Properties of Selected Substances
Substance
Color
Odor
Physical state
Chlorine
Water
Sugar
Acetic acid
Nitrogen dioxide
Oxygen
Greenish yellow
Colorless
White
Colorless
Reddish brown
Colorless
Many chemists have reference
books such as the Handbook of
Chemistry and Physics to use
as a resource.
Sharp, suffocating
Odorless
Odorless
Like vinegar
Sharp, suffocating
Odorless
Gas (20°C)
Liquid
Solid
Liquid
Gas
Gas
Melting
point (°C)
-101.6
0.0
—
16.7
-11.2
-218.4
Boiling
point (°C)
-34.6
100.0
Decomposes 170–186
118.0
21.2
2183
Substances, then, are recognized and differentiated by their properties. Table 4.1 lists
six substances and several of their common physical properties. Information about physical
properties, such as that given in Table 4.1, is ­available in handbooks of chemistry and physics.
Scientists don’t pretend to know all the answers or to remember voluminous amounts of data,
but it is important for them to know where to look for data in the literature and on the Internet.
No two substances have identical physical and chemical properties.
>Chemistry in action
Making Money
Chemists are heavily involved in the manufacture of our
currency. In fact, in a very real way, the money industry
depends on chemistry and finding substances with the
correct properties. The most common paper currency in
the United States is the dollar bill. Chemistry is used to
form the ink and paper and in processes used to defeat
counterfeiters. The ink used on currency has to do a
variety of things. It must be just the right consistency to
fill the fine lines of the printing plate and release onto the
paper without smearing. The ink must dry almost immediately, since the sheets of currency fly out of the press and
into stacks 10,000 sheets tall. The pressure at the bottom
of the stack is large, and the ink must not stick to the back
of the sheet above it.
Once a bill is in circulation, Federal Reserve banks screen
it using light to measure wear. If the bill gets too dirty, it is
shredded and sent to a landfill. The average $1 bill stays
in circulation for only 18 months. No wonder Congress
decided in 1997 to revive the dollar coin by creating a
new dollar coin to succeed the Susan B. Anthony dollar.
Sacagawea dollar coin, which succeeded the Susan B.
Anthony dollar, is still not in wide circulation today.
Corbis Digital Stock
Ryan McVay/Getty
Images, Inc.
The security of our currency also depends on substances
in the ink, which is optically variable. This color-changing
ink shifts from green to black depending on how the bill
is tilted. The ink used for the numbers in the lower right
corner on the front of $10, $20, $50, and $100 bills shows
this color change.
Once the currency is printed, it has to undergo durability
tests. The bills are soaked in different solvents for 24 hours
to make sure that they can stand dry cleaning and chemical exposure to household items such as gasoline. Then
the bills must pass a washing-machine test to be sure that,
for example, your $20 bill is still intact in your pocket after
the washer is through with it. Paper currency is a blend of
75% cotton and 25% linen. Last, the bills must pass the
crumple test, in which they are rolled up, put in a metal
tube, crushed by a plunger, removed, and flattened. This
process is repeated up to 50 times.
Currency in stacks.
4.2
• Physical and Chemical Changes 65
Example 4.1
You have three flasks that contain chlorine gas, nitrogen dioxide gas, and oxygen gas.
Describe how you would use physical properties to identify the contents in each flask.
ENHANCED EXAMPLE
Solution
Use Table 4.1 to compare the compounds and note their colors. You will find that chlorine
is a greenish yellow color, nitrogen dioxide is reddish brown, and oxygen is colorless.
Now you have the information to identify the contents in each flask.
Practice 4.1
You are given three samples of silver compounds: AgCl, AgBr, and AgI. Arrange these
silver compounds in the order of their percent silver by mass (highest to lowest). Describe
how you arrived at your arrangement. Hint: Use the periodic table inside the front cover
of your text.
4.2 Physical and chemical Changes
Learning objective
Matter can undergo two types of changes, physical and chemical. Physical changes are
changes in physical properties (such as size, shape, and density) or changes in the state of
matter without an accompanying change in composition. The changing of ice into water and
water into steam are physical changes from one state of matter into another. No new substances
are formed in these physical changes.
When a clean platinum wire is heated in a burner flame, the appearance of the platinum
changes from silvery metallic to glowing red. This change is ­physical because the platinum
can be restored to its original metallic ­appearance by cooling and, more importantly, because
the composition of the platinum is not changed by heating and cooling.
In a chemical change, new substances are formed that have different properties and
composition from the original material. The new substances need not resemble the original
material in any way.
When a clean copper wire is heated in a burner flame, the appearance of the copper
changes from coppery metallic to glowing red. Unlike the platinum wire, the copper wire is
not restored to its original appearance by cooling but instead becomes a black material. This
black material is a new substance called copper(II) oxide. It was formed by a chemical change
when copper combined with oxygen in the air during the heating process. The unheated wire
key terms
© Josh McCulloch/All Canada Photos/SuperStock
Compare the physical and chemical changes in a substance.
Sawing wood produces
physical change.
physical change
chemical change
chemical equation
reactants
products
• Properties of Matter
After heating, the wire is
black copper(II) oxide
(79.9% copper, 20.1%
oxygen) (1.251 g).
Copper and oxygen from
the air combine chemically
when the wire is heated.
Tom Pantages
Before heating, the wire is
100% copper (1.000 g).
Tom Pantages
chapter 4
Tom Pantages
66 Figure 4.3
Chemical change: Forming of
copper(II) oxide from copper and
oxygen.
Cu atoms
Cu atoms
O2 molecules
–
O2
+
Cu2
Charles D. Winters/Photo Researchers, Inc.
was ­essentially 100% copper, but the copper(II) oxide is 79.9% copper and 20.1% oxygen.
One gram of copper will yield 1.251 g of copper(II) oxide (see Figure 4.3). The platinum is
changed only physically when heated, but the copper is changed both physically and chemically when heated.
Water can be decomposed chemically into hydrogen and oxygen. This is usually accomplished by passing electricity through the water in a process called electrolysis. Hydrogen collects at one electrode while oxygen collects at the other (see Figure 4.4). The composition and
the physical appearance of the ­hydrogen and the oxygen
are quite different from those of water. They are both colorless gases, but each behaves differently when a burning
wooden splint is placed into the sample: The hydrogen
explodes with a pop while the flame brightens considerably in the oxygen (oxygen supports and intensifies
O2 molecules
H2 molecules
H2O molecules
Figure 4.4
Electrolysis of water produces hydrogen
gas (on the right) and oxygen gas (on the
left). Note the ratio of the gases is 2 : 1.
4.2
• Physical and Chemical Changes 67
the combustion of the wood). From these observations, we conclude that a chemical change
has taken place.
Chemists have devised a shorthand method for expressing chemical changes in the form
of chemical equations. The two previous examples of chemical changes can be represented
by the following molecular representations, word and symbol equations. In the electrolysis
reaction, water decomposes into hydrogen and oxygen when electrolyzed. In the formation
reaction, copper plus oxygen when heated produce copper(II) oxide. The arrow means “produces,” and it points to the products. The Greek letter delta ()) represents heat. The starting
substances (water, copper, and oxygen) are called the reactants, and the substances produced
[hydrogen, oxygen, and copper(II) oxide] are called the products. We will learn more about
writing chemical equations in later chapters.
Formation of Copper(II) Oxide
Type of
equation
Reactants
Word
copper
+
Products
oxygen
Symbol
(formula)
2 Cu
+

copper(II) oxide
99:

+
Molecular
99:
O2
99:

2 CuO
Electrolysis of Water
Type of
equation
Reactants
Word
water
Products
99:
electrical
energy
hydrogen
99:
electrical
Molecular
2 H2O
99:
electrical
energy
oxygen
+
energy
Symbol
(formula)
+
2 H2
+
O2
Physical change usually accompanies a chemical change. Table 4.2 lists some common
physical and chemical changes; note that wherever a chemical change occurs, a physical change
also occurs. However, wherever a physical change is listed, only a physical change occurs.
Table 4.2
Physical or Chemical Changes of Some Common Processes
Process
taking place
Type of
change
Rusting of iron
Boiling of water
Burning of sulfur in air
Boiling of an egg
Combustion of gasoline
Digestion of food
Sawing of wood
Burning of wood
Heating of glass
Chemical
Physical
Chemical
Chemical
Chemical
Chemical
Physical
Chemical
Physical
Accompanying observations
Shiny, bright metal changes to reddish brown rust.
Liquid changes to vapor.
Yellow, solid sulfur changes to gaseous, choking sulfur dioxide.
Liquid white and yolk change to solids.
Liquid gasoline burns to gaseous carbon monoxide, carbon dioxide, and water.
Food changes to liquid nutrients and partially solid wastes.
Smaller pieces of wood and sawdust are made from a larger piece of wood.
Wood burns to ashes, gaseous carbon dioxide, and water.
Solid becomes pliable during heating, and the glass may change its shape.
68 chapter 4
• Properties of Matter
E x a mp l e 4 . 2
In a distillation, a chemist is boiling pure ethyl alcohol and observes that the boiling
point is 78.5°C. Using an apparatus to condense and recover the alcohol vapors back to
a liquid, she finds that the boiling point of the condensed liquid also is 78.5°C. Describe
whether this experiment is a physical or chemical change.
Solution
The process of changing a liquid to a vapor and back to a liquid is a physical change
­verified by the fact that both liquids have the same boiling points.
Practice 4.2
When ice melts, is this a physical or chemical change?
4.3 Learning to Solve Problems
Learning objective
List the basic steps in solving chemistry problems.
Now that we have learned some of the basics of chemistry including names and symbols for
the elements, dimensional analysis, and a good amount of chemical terminology, we are ready
to begin learning to solve chemistry problems.
One of the great joys of studying chemistry is learning to be a good problem solver. The
ability to solve complicated problems is a skill that will help you greatly throughout your life.
It is our goal to help you learn to solve problems by thinking through the problems on your
own. To this end the basic steps in solving problems are:
ReadRead the problem carefully. Determine what is known and what is to be solved
for and write them down. It is important to label all factors and measurements
with the proper units.
PlanDetermine which principles are involved and which unit relationships are
needed to solve the problem. You may refer to tables for needed data. Set
up the problem in a neat, organized, and logical fashion, making sure all unwanted units cancel. Use the examples as guides for setting up the problem.
CalculateProceed with the necessary mathematical operations. Make certain that your
answer contains the proper number of significant figures.
Check
Check the answer to see if it is reasonable.
At first we will help you through this process in each example. Then later in the text as you
learn more about solving problems and have lots of practice, we’ll help you less. By the end
of your course you’ll be solving problems on your own.
A few more words about problem solving. Don’t allow any formal method of problem
solving to limit your use of common sense and intuition. If a problem is clear to you and its
solution seems simpler by another method, by all means use it.
If you approach problem solving using this general method, you will be able to learn
to solve more and more difficult problems on your own. As you gain confidence you will
become an independent, creative problem solver and be able to use this skill ­wherever you
go in “real life.”
4.4 Energy
Learning objective
key terms
energy
potential energy
kinetic energy
law of conservation of energy
List the various forms of energy, explain the role of energy in chemical changes, and
state the law of conservation of energy.
From the early discovery that fire can warm us and cook our food to our ­discovery that nuclear
reactors can be used to produce vast amounts of controlled energy, our technical progress
has been directed by our ability to produce, harness, and utilize energy. Energy is the capacity of matter to do work. Energy exists in many forms; some of the more familiar forms are
mechanical, chemical, electrical, heat, nuclear, and radiant or light energy. Matter
can have both potential and kinetic energy.
Potential energy (PE) is stored energy, or energy that an object possesses
due to its relative position. For example, a ball located 20 ft above the ground
has more potential energy than when located 10 ft above the ground and will
bounce higher when allowed to fall. Water backed up behind a dam represents
potential energy that can be converted into useful work in the form of electrical
or mechanical energy. Gasoline is a source of chemical potential energy. When
gasoline burns (combines with oxygen), the heat released is associated with a
decrease in potential energy. The new substances formed by burning have less
chemical potential energy than the gasoline and oxygen.
Kinetic energy (KE) is energy that matter possesses due to its motion.
When the water behind the dam is released and allowed to flow, its potential
energy is changed into kinetic energy, which can be used to drive generators and
produce electricity. Moving bodies possess kinetic energy. We all know the results when two
moving vehicles collide: Their kinetic energy is expended in the crash that occurs. The pressure exerted by a confined gas is due to the kinetic energy of rapidly moving gas particles.
Energy can be converted from one form to another form. Some kinds of ­energy can be
converted to other forms easily and efficiently. For example, mechanical energy can be converted to electrical energy with an electric generator at better than 90% ­efficiency. On the other
hand, solar energy has thus far been directly converted to electrical energy at an efficiency
approaching 30%. In chemistry, energy is most frequently released as heat.
• Energy 69
Brian Yaruin/The Image Works
4.4
The mechanical energy of falling
water is converted to electrical
energy at the hydroelectric plant
at Niagara Falls. As the water falls,
potential energy is converted to
kinetic energy and turns a turbine
to produce electrical energy.
Energy in Chemical Changes
Figure 4.5
E x a mp l e 4 . 3
Methane (CH4) is the major compound in natural gas and is used in our homes for heat
and cooking. The combustion of methane in oxygen (air) produces considerable heat
energy. The chemical reaction is
CH4
methane
+
NASA
In all chemical changes, matter either absorbs or releases energy. Chemical changes can produce different forms of energy. For example, electrical energy to start
automobiles is produced by chemical changes in the lead storage battery. Light
energy is produced by the chemical change that occurs in a light stick. Heat and
light are ­released from the combustion of fuels. For example, the space shuttle uses
the ­reaction between hydrogen and oxygen-producing water vapor to help power it
(see Figure 4.5). All the energy needed for our life processes—breathing, muscle
contraction, blood circulation, and so on—is produced by chemical changes occurring within the cells of our bodies.
Conversely, energy is used to cause chemical changes. For example, a chemical change occurs in the electroplating of metals when electrical energy is passed
through a salt solution in which the metal is submerged. A chemical change occurs
when radiant energy from the sun is used by green plants in the process of photosynthesis. And as we saw, a chemical change occurs when electricity is used to decompose water into hydrogen and oxygen. Chemical changes are often used primarily to
produce energy rather than to produce new substances. The heat or thrust generated
by the combustion of fuels is more ­important than the new substances formed.
2 O2
oxygen
S
CO2
carbon dioxide
+
2 H2 O
water
+
802.5 kJ
heat
If it were possible to make methane from carbon dioxide and water, how much heat
energy would it take to make methane? Write the equation.
SOLUTION
From the equation it takes 802.5 kJ to make methane. The chemical equation for the reaction is
CO2 + 2 H2O + 802.5 kJ S CH4 + 2 O2
The external fuel tank (orange)
contained liquid hydrogen and
liquid oxygen which fueled the
shuttle’s main engine. The exhaust
gas from the shuttle was water.
70 chapter 4
• Properties of Matter
H2 + O2
Potential energy
Potential energy
H2 + O2
H2O
H2O
Time
Time
(a)
(b)
Figure 4.6
(a) In electrolysis of water, energy is absorbed by the system, so the products H2 and O2 have
a higher potential energy. (b) When hydrogen is used as a fuel, energy is released and the
product (H2O) has lower potential energy.
Conservation of Energy
An energy transformation occurs whenever a chemical change occurs (see Figure 4.6). If
energy is absorbed during the change, the products will have more chemical potential energy
than the reactants. Conversely, if energy is given off in a chemical change, the products will
have less chemical potential energy than the reactants. Water can be decomposed in an electrolytic cell by absorbing electrical energy. The products, hydrogen and oxygen, have a greater
chemical potential energy level than that of water (see Figure 4.6a). This potential energy is
released in the form of heat and light when the hydrogen and oxygen are burned to form water
again (see Figure 4.6b). Thus, energy can be changed from one form to another or from one
substance to another and, therefore, is not lost.
The energy changes occurring in many systems have been thoroughly studied. No ­system
has ever been found to acquire energy except at the expense of energy possessed by another system. This is the law of conservation of ­energy: Energy can be neither created nor
­destroyed, though it can be transformed from one form to ­another.
4.5 Heat: Quantitative Measurement
L earning obje ctive
Calculate the amount of heat lost or gained in a given system.
key terms
The SI-derived unit for energy is the joule (pronounced jool, rhyming with tool, and abbreviated J). Another unit for heat energy, which has been used for many years, is the calorie (abbreviated cal). The relationship between joules and calories is
joule
calorie
specific heat
1 kJ = 1000 J
1 kcal = 1000 cal = 1 Cal
4.184 J = 1 cal (exactly)
To give you some idea of the magnitude of these heat units, 4.184 joules, or 1 calorie,
is the quantity of heat energy required to change the temperature of 1 gram of water by 1°C,
usually measured from 14.5°C to 15.5°C.
Since joules and calories are rather small units, kilojoules (kJ) and kiloca­lories (kcal) are
used to express heat energy in many chemical processes. The kilocalorie is also known as the
nutritional Calorie (spelled with a capital C and abbreviated Cal). In this book, heat energy
will be expressed in joules.
The difference in the meanings of the terms heat and temperature can be seen by this
­example: Visualize two beakers, A and B. Beaker A contains 100 g of water at 20°C, and beaker B contains 200 g of water also at 20°C. The beakers are heated until the temperature of the
water in each reaches 30°C. The ­temperature of the water in the beakers was raised by exactly
the same amount, 10°C. But twice as much heat (8368 J) was required to raise the temperature
of the water in beaker B as was required in beaker A (4184 J).
4.5
Table 4.3
• Heat: Quantitative Measurement 71
Specific Heat of Selected Substances
Substance
Specific heat (J/g°C)
Specific heat (cal/g°C)
4.184
2.138
2.059
0.900
0.473
0.385
0.131
0.128
1.000
0.511
0.492
0.215
0.113
0.0921
0.0312
0.0305
Water
Ethyl alcohol
Ice
Aluminum
Iron
Copper
Gold
Lead
In the middle of the eighteenth century, Joseph Black (1728–1799), a Scottish chemist,
was ­experimenting with the heating of elements. He heated and cooled equal masses of iron
and lead through the same temperature range. Black noted that much more heat was needed for
the iron than for the lead. He had discovered a fundamental property of ­matter—namely, that
every substance has a characteristic heat capacity. Heat capacities may be compared in terms of
specific heats. The specific heat of a substance is the quantity of heat (lost or gained) required
to change the temperature of 1 g of that substance by 1°C. It follows then that the specific heat of
liquid water is 4.184 J>g°C (or 1.000 cal>g°C). The specific heat of water is high compared with
that of most substances. Aluminum and copper, for example, have specific heats of 0.900 J>g°C
and 0.385 J>g°C, ­respectively (see Table 4.3). The relation of mass, specific heat, temperature
change (t), and quantity of heat lost or gained by a system is expressed by this general equation:
a
mass of
specific heat
ba
b (Dt)  heat
substance of substance
Thus the amount of heat needed to raise the temperature of 200. g of water by 10.0°C can be
calculated as follows:
(200. g )a
The greek letter D is used to
mean “change in.”
Lowercase t is used for Celsius
and upper case T for Kelvin.
Mass is in grams, specific heat is
in cal/g°C or J/g°C and Dt is
in °C.
4.184 J
b (10.0 °C ) = 8.37 * 103 J
g° C
Example 4.4
Calculate the specific heat of a solid in J>g°C and cal>g°C if 1638 J raises the temperature of 125 g of the solid from 25.0°C to 52.6°C.
SOLUTION
Read •
Knowns: 125 g of the solid
t = 52.6 - 25.0 = 27.6°C
Heat = 1638 J
Solving for: specific heat of the solid
Plan •
Use the equation
(mass)(specific heat)(t) = heat
solving for specific heat
heat
specific heat =
g * t
Calculate • specific heat =
1638 J
= 0.475 J>g°C
125 g * 27.6°C
Convert joules to calories using 1.00 cal>4.184 J
Check •
specific heat = a
0.475 J
1.000 cal
ba
b = 0.114 cal>g°C
g°C
4.184 J
Note that the units in the answer agree with the units for specific heat.
ENHANCED EXAMPLE
72 chapter 4
• Properties of Matter
Example 4.5
A sample of a metal with a mass of 212 g is heated to 125.0°C and then dropped into
375 g water at 24.0°C. If the final temperature of the water is 34.2°C, what is the specific
heat of the metal? (Assume no heat losses to the surroundings.)
SOLUTION
Read •
Knowns:
mass (metal) = 212 g
mass (water) = 375 g
t (water) = 34.2 - 24.0 = 10.2°C
t (metal) = 125.0 - 34.2 = 90.8°C
Solving for: specific heat of the metal
Plan •
When the metal enters the water, it begins to cool, losing heat to the water.
At the same time, the temperature of the water rises. This continues until the
temperature of the metal and the water is equal (34.2°C), after which no net
flow of heat occurs.
Heat lost or gained by a system is given by
(mass)(specific heat)(t) = energy change (heat)
heat gained by the water  heat lost by the metal
heat gained by the water = (mass water)(specific heat)(t)
= (375 g )a
4.184 J
b (10.2°C) = 1.60 * 104 J
g °C
Calculate • heat gained by the water = heat lost by the metal = 1.60 * 104 J
(mass metal)(specific heat metal)(t) = 1.60 * 104 J
(212 g)(specific heat metal)(90.8°C) = 1.60 * 104 J
To determine the specific heat of the metal, we rearrange the equation
(mass)(specific heat)(t) = heat
solving for the specific heat of the metal
specific heat metal =
Check •
1.60 * 104 J
= 0.831 J>g°C
(212 g)(90.8°C)
Note the units in the answer agree with units for specific heat.
Practice 4.3
Calculate the quantity of energy needed to heat 8.0 g of water from 42.0°C to 45.0°C.
Practice 4.4
A 110.0-g sample of metal at 55.5°C raises the temperature of 150.0 g of water from
23.0°C to 25.5°C. Determine the specific heat of the metal in J>g°C.
4.6 Energy in the Real World
Learning objective
Define a hydrocarbon compound and explain its role in the world’s energy supply.
Coal, petroleum, natural gas, and woody plants provide us with a vast resource of energy, all
of which is derived from the sun. Plants use the process of photosynthesis to store the sun’s
energy, and we harvest that energy by burning the plants or using the decay products of the
plants. These decay products have been converted over millions of years to fossil fuels. As the
plants died and decayed, natural processes changed them into petroleum deposits that we now
use in the forms of gasoline and natural gas.
4.6
• Energy in the Real World 73
>Chemistry in action
Getting the most pops comes from the molecular changes
resulting from heating the kernel. Energy and chemistry
combine to produce a tasty snack!
The pericarp is the hard casing on the popcorn kernel. It
is made of long polymers of cellulose. When the kernel is
heated up, the long chains of cellulose line up and make
© EuToch/iStockphoto
Researchers in Indiana tested 14 types of popcorn to determine the most poppable. They found a range of 4–47%
of unpopped kernels and determined that the kernels that
held moisture the best produced the fewest unpopped
kernels. Further analysis led them to the conclusion that
the determining factor was the pericarp.
Popcorn
Although we do not really understand completely how petroleum deposits were
formed, they most likely came from the remains of marine organisms living over 500
million years ago. Petroleum is composed of hydrocarbon compounds (which contain
only carbon and hydrogen). Carbon is a unique element in that it can produce chains
of atoms in a molecule of different lengths. Table 4.4 gives the names and formulas of
some simple hydrocarbons. Natural gas is associated with petroleum deposits and consists mainly of methane with some ethane, propane, and butane mixed in. Large resources
of natural gas are now being made available in the United States through a process called
fracking, (hydraulic fracturing).
Coal was formed from the remains of plants that were buried and subjected to high
pressure and heat over many years. Chemical changes lowered the oxygen and hydrogen
content of these plant tissues and produced the black solid we recognize as coal. The
energy available from a mass of coal increases as the carbon content increases. Coal is an
important energy source in the United States.
© WillSelarep/iStockphoto
Alternative
energy sources
include wind,
solar, and
biofuels.
Table 4.4 Names and Formulas
for Simple Hydrocarbons
Names
Formula
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Nonane
Decane
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
C7H16
C8H18
C9H20
C10H22
© Elena Elisseeva/iStockphoto
What’s the secret to getting all those kernels in a bag
of microwave popcorn to pop? Food scientists have
looked at all sorts of things to improve the poppability of
popcorn. In every kernel of popcorn there is starch and a
little moisture. As the kernel heats up, the water vaporizes and the pressure builds up in the kernel. Finally, the
kernel bursts and the starch expands into the white foam
of popped corn.
a strong crystalline structure. This strong structure keeps
the water vapor inside the kernel, allowing the pressure to
rise high enough to pop the kernel.
© Derek Dammann/iStockphoto
Popping Popcorn
74 chapter 4
• Properties of Matter
We are currently facing an energy crisis that has caused us to search for new renewable
energy sources for our future. We will need to consider economic, supply, and climatic factors
to find fuels to serve us as we continue to grow our human population. There are a variety of
potential resources: solar, nuclear, biomass, wind, and synthetic fuels. We now are beginning
to focus on developing these resources and using them in the most economic way to lower our
energy dependence globally.
Example 4.6
Much of the energy used in the world is dependent on coal. Large deposits of coal are
found all over the world. Coal supplies about 20% of the energy used in the United States.
The combustion of coal, which is mainly carbon and reacts with oxygen in the air, produces carbon dioxide and much heat energy. Write the chemical equation for this reaction.
Solution
C + O2 S CO2 + heat (393 kJ)
Practice 4.5
The major compounds in gasoline are hydrocarbons. Without petroleum (from which
hydrocarbons are distilled to make gasoline), industry would come to a halt and millions
of automobiles would be idled. Devise a general formula (CxHy) that represents the
hydrocarbons in Table 4.4.
C h a p t e r
4 review
4.1 Properties of Substances
key terms
properties
physical properties
chemical properties
• E
ach substance has a set of properties that is ­charac­teristic of the substance and gives it a unique
identity.
• Physical properties are inherent in the substance and can be determined without altering the composition of the substance:
• Color
• Taste
• Odor
• State of matter
• Density
• Melting point
• Boiling point
• Chemical properties describe the ability of a ­substance to interact with other substances to form
­different ­substances.
4.2 Physical and Chemical Changes
key terms
physical change
chemical change
chemical equations
reactants
products
• A
physical change is a change in the physical properties or a change in the state of matter for a substance without altering the composition of the substance.
• In a chemical change different substances are formed that have different properties and composition
from the original material.
• Chemical changes can be represented by chemical equations:
• Word equation
• Molecular equation
• Symbol (formula) equation
4.3 Learning to solve problems
Read
Read the problem carefully, determining what is known and what is to be solved for .
PlanDetermine what is required to solve the problem and set up the problem in a neat, organized manner using the examples as guides.
CalculateComplete the necessary calculations making sure your answer contains the proper units
and significant figures.
Check
Check to see if your answer is reasonable.
Review Questions 75
4.4 Energy
•
•
•
•
•
•
•
•
•
•
•
Energy is the capacity to do work.
Potential energy is energy that results from position or is stored within a substance.
Kinetic energy is the energy matter possesses as a ­result of its motion.
Energy can be converted from one form to another.
Common forms of energy:
• Mechanical
• Chemical
• Electrical
• Heat
• Nuclear
• Light
In chemistry, energy is most frequently expressed as heat.
In all chemical changes matter either absorbs or ­releases energy.
Energy can be used to cause chemical change.
Chemical changes are often used to produce energy rather than new substances.
Energy can neither be created nor destroyed. It can be transformed from one form to another.
The energy released or absorbed in a chemical reaction can be summarized in graphical form.
H2O
energy
potential energy
kinetic energy
law of conservation of energy
H2 + O2
Potential energy
Potential energy
H2 + O2
key terms
H2O
Time
Time
(a)
(b)
4.5 Heat: Quantitative Measurement
• T
he SI unit for heat is the joule
4.184 J = 1 cal
• The calorie is defined as the amount of heat required to change the temperature of 1 gram of water 1°C.
• Every substance has a characteristic heat capacity:
• The specific heat of a substance is a measure of its heat capacity.
• The specific heat is the quantity of heat required to change the temperature of 1 gram of the
substance by 1°C.
• Heat lost or gained by a system can be calculated by
heat = (mass of substance)(specific heat of substance)(t).
key terms
joule
calorie
specific heat
review Questions
1. In what physical state does acetic acid exist at 393 K? (Table 4.1)
2. In what physical state does chlorine exist at - 65°C? (Table 4.1)
3. Calculate the boiling point of acetic acid in
(a) kelvins
(b) degrees Fahrenheit (Table 4.1)
4. Calculate the melting point of acetic acid in (Table 4.1)
(a) degrees Fahrenheit
(b) kelvins
5. What evidence of chemical change is visible when electricity is run
through water? (Figure 4.4)
6. What is the fundamental difference between a chemical change and
a physical change?
7. What is the critical first step in solving any chemistry problem?
8. Is the calculation step the last step in solving a chemistry problem?
Why or why not?
9. Distinguish between potential and kinetic energy.
10. How is chemistry calorie (cal) different from a food calorie (Cal)?
11. Which requires more energy to heat from 20ºC to 30ºC, iron or
gold? (See Table 4.3.)
12. Predict which molecule would produce more carbon dioxide when
burned, a molecule of ethane or a molecule of octane.
13. What is the common characteristic of all hydrocarbons?
14. Which is the most common element used to produce energy for
human use?
15. Why might the combustion of fossil fuels not be the best solution
to the planet’s energy needs? What are some other energy sources
that may become important in the future?
76 chapter 4
• Properties of Matter
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
P AIRE D E x e r c i s e s
1. Determine whether each of the following represents a physical
property or a chemical property:
(a) Vinegar has a pungent odor.
(b) Carbon cannot be decomposed.
(c) Sulfur is a bright yellow solid.
(d) Sodium chloride is a crystalline solid.
(e) Water does not burn.
(f) Mercury is a liquid at 25°C.
(g) Oxygen is not combustible.
(h)Aluminum combines with oxygen to form a protective oxide coating.
2. Determine whether each of the following represents a physical
property or a chemical property:
(a) Chlorine gas has a greenish-yellow tint.
(b) The density of water at 4°C is 1.000 g/mL.
(c) Hydrogen gas is very flammable.
(d) Aluminum is a solid at 25°C.
(e) Water is colorless and odorless.
(f) Lemon juice tastes sour.
(g) Gold does not tarnish.
(h) Copper cannot be decomposed.
3. Cite the evidence that indicates that only physical changes occur
when a platinum wire is heated in a ­Bunsen burner flame.
4. Cite the evidence that indicates that both physical and chemical
changes occur when a copper wire is heated in a Bunsen burner flame.
5. Identify the reactants and products when a copper wire is heated in
air in a Bunsen burner flame.
7. State whether each of the following represents a chemical change
or a physical change:
(a) A steak is cooked on a grill until well done.
(b) In the lab, students firepolish the end of a glass rod. The jagged
edge of the glass has become smooth.
(c) Chlorine bleach is used to remove a coffee stain on a white lab
coat.
(d) When two clear and colorless aqueous salt solutions are mixed
together, the solution turns cloudy and yellow.
(e) One gram of an orange crystalline solid is heated in a test tube,
producing a green powdery solid whose volume is 10 times the
volume of the original ­substance.
(f) In the lab, a student cuts a 20-cm strip of ­magnesium metal into
1-cm pieces.
6. Identify the reactants and products for the electrolysis of water.
9. Are the following examples of potential energy or ­kinetic energy?
(a) fan blades spinning (d) a person napping
(b) a bird flying (e) ocean waves rippling
(c) sodium hydroxide in a sealed jar
10. Are the following examples of potential energy or ­kinetic energy?
(a) a fish swimming
(d) water starting to boil
(b) a skier at the top of a hill (e) a leaf unfolding
(c) a bird’s egg in a nest
11. What happens to the kinetic energy of a speeding car when the car
is braked to a stop?
12. What energy transformation is responsible for the fiery reentry of
the space shuttle into Earth’s atmosphere?
13. Indicate with a plus sign ( + ) any of these processes that require
energy and a negative sign ( - ) any that release energy.
(a) arctic ice melting
(d) dry ice changing to vapor
(b) starting a car
(e) blowing up a balloon
(c) flash of lightning
15. How many joules of heat are required to heat 125 g aluminum from
19.0°C to 95.5°C? (Table 4.3)
14. Indicate with a plus sign ( +) any of these processes that require
energy and a negative sign ( -) any that release energy.
(a) riding a bike
(d) tires deflating
(b) fireworks bursting
(e) wood burning in a fireplace
(c) water evaporating
16. How many joules of heat are required to heat 65 g lead from 22.0°C
to 98.5°C? (Table 4.3)
17. How many joules of heat are required to heat 25.0 g of ethyl alcohol
from the prevailing room temperature, 22.5°C, to its boiling point,
78.5°C?
18. How many joules of heat are required to heat 35.0 g of isopropyl
alcohol from the prevailing room temperature, 21.2°C, to its boiling
point, 82.4°C? (The specific heat of isopropyl alcohol is 2.604 J/g°C.)
19. A 135-g sample of a metal requires 2.50 kJ to change its temperature from 19.5°C to 100.0°C. What is the specific heat of this
metal?
20. A 275-g sample of a metal requires 10.75 kJ to change its temperature from 21.2°C to its melting temperature, 327.5°C. What is the
specific heat of this metal?
21. A 155-g sample of copper was heated to 150.0°C, then placed into
250.0 g water at 19.8°C. (The specific heat of copper is 0.385 J>g°C.)
Calculate the final temperature of the mixture. (Assume no heat
loss to the surroundings.)
22. A 225-g sample of aluminum was heated to 125.5°C, then placed
into 500.0 g water at 22.5°C. (The specific heat of aluminum is
0.900 J>g°C). Calculate the final temperature of the mixture. (Assume
no heat loss to the surroundings.)
8. State whether each of the following represents a chemical change
or a physical change:
(a) A few grams of sucrose (table sugar) are placed in a small beaker of deionized water; the sugar crystals “disappear,” and the
liquid in the beaker remains clear and colorless.
(b) A copper statue, over time, turns green.
(c) When a teaspoon of baking soda (sodium ­bicarbonate) is placed
into a few ounces of vinegar (acetic acid), volumes of bubbles
(effervescence) are produced.
(d) When a few grams of a blue crystalline solid are placed into
a beaker of deionized water, the crystals “disappear” and the
liquid becomes clear and blue in color.
(e) In the lab, a student mixes 2 mL of sodium ­hydroxide with 2
mL of hydrochloric acid in a test tube. He notices that the test
tube has become very warm to the touch.
(f) A woman visits a hairdresser and has her hair ­colored a darker shade of brown. After several weeks the hair, even though
washed several times, has not changed back to the original color.
Additional Exercises 77
Add i t i o n a l E x e r c i s e s
23. Read the following passage and identify at least two physical and
two chemical properties of zeolites.
Zeolites are crystalline solids composed of silicon, aluminum,
and oxygen with very porous structures. They are very useful
in a variety of applications. Zeolites are formed naturally when
volcanic rocks and ash react with alkaline groundwater. They
may also be synthesized in the laboratory by allowing aluminosilicate crystals to form around small organic molecules.
Zeolites generally have very low density due to their porous
nature. They are often used as molecular sieves because they
have large pores that can trap molecules of certain sizes and
shapes while excluding others. Bandages used by the military
sometimes contain zeolites, which absorb water from the blood,
thus accelerating the natural clotting ­process. Zeolites have also
replaced phosphates in many detergents, which is an ecological
benefit, because zeolites do not ­encourage the growth of algae.
Interestingly, the name ­zeolite comes from the fact that zeolites
give off water in the form of steam when they are heated; thus,
they are called boiling (zeo) stones (lithos).
24. A 110.0-g sample of a gray-colored, unknown, pure metal was
heated to 92.0°C and put into a coffee-cup calorimeter ­containing
75.0 g of water at 21.0°C. When the heated metal was put into the
water, the temperature of the water rose to a final temperature of
24.2°C. The specific heat of water is 4.184 J/g°C. (a) What is the
specific heat of the metal. (b) Is it possible that the metal is either
iron or lead? Explain.
25. The specific heat of a human is approximately 3.47 J/g°C. Given
this information,
(a) If a 165-lb man eats a candy bar containing 262 Cal, how much
will his body temperature increase if all the calories from the
candy bar are converted into heat energy? Remember that a
food calorie (Cal) is equal to 1 kcal.
(b) If a 165-lb man eats a roll of candy containing 25.0 Cal, how
much will his body temperature increase if all the calories
from the candy bar are converted into heat energy?
26. A sample of water at 23.0°C required an input of 1.69 * 104 J of
heat to reach its boiling point, 100°C. What was the mass of the
water? (Table 4.3)
27. A sample of gold required 3.25 * 104 J of heat to melt it from
room temperature, 23.2°C, to its melting point, 1064.4°C. How
many pounds of gold were in the sample?
28. If 40.0 kJ of energy are absorbed by 500.0 g of water at 10.0°C,
what is the final temperature of the water?
29. A sample of iron was heated to 125.0°C, then placed into 375 g of
water at 19.8°C. The temperature of the water rose to 25.6°C. How
many grams of iron were in the sample? (Table 4.3)
30. A sample of copper was heated to 275.1°C and placed into 272 g of
water at 21.0°C. The temperature of the water rose to 29.7°C. How
many grams of copper were in the sample? (Table 4.3)
31. A 100.0-g sample of copper is heated from 10.0°C to 100.0°C.
(a) Determine the number of calories needed.
(The specific heat of copper is 0.0921 cal>g°C.)
(b) The same amount of heat is added to 100.0 g of Al at 10.0°C.
(The specific heat of Al is 0.215 cal>g°C.) Which metal gets
hotter, the copper or the aluminum?
32. A 500.0-g piece of iron is heated in a flame and dropped into 400.0 g
of water at 10.0°C. The temperature of the water rises to 90.0°C.
How hot was the iron when it was first removed from the flame?
(The specific heat of iron is 0.473 J>g°C.)
33. A 20.0-g piece of metal at 203°C is dropped into 100.0 g of water
at 25.0°C. The water temperature rose to 29.0°C. Calculate the
specific heat of the metal (J>g°C). Assume that all of the heat lost
by the metal is transferred to the water and no heat is lost to the
­surroundings.
34. Assuming no heat loss by the system, what will be the final temperature when 50.0 g of water at 10.0°C are mixed with 10.0 g of
water at 50.0°C?
35. Three 500.0-g pans of iron, aluminum, and copper are each used to
fry an egg. Which pan fries the egg (105°C) the quickest? Explain.
36. At 6:00 p.m., you put a 300.0-g copper pan containing 800.0 g
of water (all at room temperature, which is 25°C) on the stove.
The stove supplies 628 J>s. When will the water reach the boiling
point? (Assume no heat loss.)
37. Why does blowing gently across the surface of a cup of hot coffee
help to cool it? Why does inserting a spoon into the coffee do the
same thing?
38. If you are boiling some potatoes in a pot of water, will they cook
faster if the water is boiling vigorously than if the water is only
gently boiling? Explain your reasoning.
39. Homogenized whole milk contains 4% butterfat by ­volume. How
many milliliters of fat are there in a glass (250 mL) of milk? How
many grams of butterfat (d = 0.8 g>mL) are in this glass of milk?
40. Gloves are often worn to protect the hands from being burned when
they come in contact with very hot or very cold objects. Gloves are
often made of cotton or wool, but many of the newer heat-resistant
gloves are made of silicon rubber. The specific heats of these materials are listed below:
Material
wool felt
cotton
paper
rubber
silicon rubber
Specific heat (J/g°C)
1.38
1.33
1.33
3.65
1.46
(a) If a glove with a mass of 99.3 grams composed of cotton increases in temperature by 15.3°F, how much energy was absorbed by the glove?
(b) A glove with a mass of 86.2 grams increases in temperature by
25.9°F when it absorbs 1.71 kJ of energy. Calculate the specific
heat of the glove and predict its composition.
(c) If a glove with a mass of 50.0 grams needs to absorb 1.65 kJ of
energy, how much will the temperature of the glove increase for
each of the materials listed above?
(d) Which is the best material for a heat-resistant glove?
(e) If you were designing a heat-resistant glove, what kind of specific heat would you look for?
41. If solar panels are placed in the Mojave Desert in California that
generate 100 megawatts on a 1.3-square-mile site with an average
of 7.50 hours of productive daylight each day, how many tons of
coal are equivalent to the energy produced in one day by the solar
panels? (One ton of coal will produce 26.6 gigajoules of energy,
1 Watt = 1 J/s.)
78 chapter 4
• Properties of Matter
Challenge Exercises
42. Suppose a ball is sitting at the top of a hill. At this point, the ball
has potential energy. The ball rolls down the hill, so the potential
energy is converted into kinetic energy. When the ball reaches the
bottom of the hill, it goes halfway up the hill on the other side and
stops. If energy is supposed to be conserved, then why doesn’t the
ball go up the other hill to the same level as it started from?
43. From the following molecular picture, determine whether a physical or a chemical change has occurred. Justify your ­answer.
45. Describe the change(s) that you see in the following ­illustration.
Was this a physical or a chemical change?
¡
44. From the illustration below (a) describe the change that has
­occurred and (b) determine whether the change was a physical
change or a chemical change. Justify your answer.
Answers to Practice Exercises
4.1
Each compound has one atom of silver to one atom of the other
elements. Using the periodic table, we find the atomic masses
of these elements to be Ag (107.9), Cl (35.45), Br (79.90), and
I (126.8). Each compound has the same amount of silver, but the
percent of silver will be different. The highest percent of silver
will be in AgCl, the middle will be in AgBr, and the lowest will
be in AgI. So the order is AgCl, AgBr, AgI.
4.2
Physical change
4.3
1.0 * 102 J = 24 cal
4.4
0.477 J > g°C
4.5
CnH2n + 2 where n is the number of carbons in each formula
PUTTING It
Putting
IT Together:
TOGETHER:
Review for Chapters 1–4
Answers for Putting It Together Reviews are found in Appendix VII.
Multiple Choice
Multiple
Choose the Choice
correct answer to each of the following.
Choose
eachmeters?
of the following.
1. 1.00the
cmcorrect
is equalanswer
to howtomany
1. (a)
What
is meant by(b)
pressure
gas)?
2.54
100. (for a (c)
10.0
(d) 0.0100
2.1.00
How cm
does
airto
pressure
inside
the balloon shown in Figure 12.3
2.
is the
equal
how many
inches?
compare
air0.10
pressure outside
Explain.
(a)
0.394with the(b)
(c) 12the balloon?(d)
2.54
According
to Table
what two gases are the major constituents
3. 4.50
ft is how
many12.1,
centimeters?
of dry
air?
(a)
11.4
(b) 21.3
(c) 454
(d) 137
does the0.0048
pressure
represented
by 1 torr
compare
in magnitude
Hownumber
4. The
contains
how many
significant
figures?
to the
by 1 mm
(a)
1 pressure represented
(b) 2
(c) 3Hg? See Table
(d)12.2.
4
In which0.00382
container
in Figure 12.6 are the ­molecules of
5. Express
in illustrated
scientific notation.
-2­hydrogen.
3
gas 3.82
moving
faster?
Assume both
gases* to10be
(a)
(c) 3.82
* 10
-3
-3
3.8is *meant
10 by pressure (for
(d) a3.82
1. (b)
What
gas)?* 10
6.
is equivalent
to
2.42.0°C
How does
the air pressure
inside the balloon shown in Figure 12.3
(a)
273 Kwith the air pressure
(c)outside
108°Fthe balloon? Explain.
compare
5.55°F to Table 12.1, what
(d)two
53.3°F
3. (b)
According
gases are the major constituents
7. 267°F
is equivalent to
of dry air?
(a) 404 K
(c) 540 K
(b) 116°C
(d) 389 K
8. An object has a mass of 62 g and a volume of 4.6 mL. Its density is
(a) 0.074 mL>g
(c) 7.4 g>mL
(b) 285 g>mL
(d) 13 g>mL
9. The mass of a block is 9.43 g and its density is 2.35 g >mL. The
volume of the block is
(a) 4.01 mL
(c) 22.2 mL
(b) 0.249 mL
(d) 2.49 mL
10. The density of copper is 8.92 g/mL. The mass of a piece of copper
that has a volume of 9.5 mL is
(a) 2.6 g
(c) 0.94 g
(b) 85 g
(d) 1.1 g
11. An empty graduated cylinder has a mass of 54.772 g. When filled
with 50.0 mL of an unknown liquid, it has a mass of 101.074 g. The
density of the liquid is
(a) 0.926 g>mL
(c) 2.02 g>mL
(b) 1.08 g>mL
(d) 1.85 g>mL
12. The conversion factor to change grams to milligrams is
100 mg
1g
(a)
(c)
1g
1000 mg
(b)
1g
100 mg
(d)
1000 mg
1g
13. What Fahrenheit temperature is twice the Celsius temperature?
(a) 64°F
(c) 200°F
(b) 320°F
(d) 546°F
14. A gold alloy has a density of 12.41 g>mL and contains 75.0% gold
by mass. The volume of this alloy that can be made from 255 g of
pure gold is
(a) 4.22 * 103 mL
(c) 27.4 mL
(b) 2.37 * 103 mL
(d) 20.5 mL
15. A lead cylinder (V = r2 h) has radius 12.0 cm and length 44.0 cm
4. How
the of
pressure
represented
by 1oftorr
and adoes
density
11.4 g/mL.
The mass
thecompare
cylinder in
is magnitude
3 Table 12.2.
to
pressure
by
mm Hg?
(a)the
2.27
* 105represented
g
(c)1 1.78
* 10See
g
5
5. (b)
In which
Figure
12.6
1.89 *container
105 g illustrated
(d)in3.50
* 10
g are the ­molecules of
gas
moving
faster?
Assume
both
gases
to
be
­hydrogen.
16. The following units can all be used for density
except
3
3
1. (a)
Whatg>cm
is meant by(b)
pressure
gas)?g>L
kg>m (for a (c)
(d) kg>m2
2.H37.4
ow does
the2.2
aircm
pressure
17.
cm *
equalsinside the balloon shown in Figure 12.3
2
compare
balloon? Explain.
(a) 82.28with
cm2 the air pressure
(c)outside
82 cmthe
2
2
(b)
82.3
cm
(d)
82.2
cm
3. According to Table 12.1, what two gases are the major constituents
of dry
air?
18. The
following
elements are among the five most abundant by mass
thedoes
Earth’s
seawater,
and atmosphere
except in magnitude
4. in
How
the crust,
pressure
represented
by 1 torr compare
(a)
oxygen
(c)
silicon
to the pressure represented by 1 mm Hg? See Table 12.2.
hydrogen
(d)inaluminum
5. (b)
In which
container illustrated
Figure 12.6 are the ­molecules of
19. Which
of
the
following
is
a
compound?
gas moving faster? Assume both gases to be ­hydrogen.
(a) lead
(c) potassium
(b) wood
(d) water
20. Which of the following is a mixture?
(a) water
(c) wood
(b) chromium
(d) sulfur
21. How many atoms are represented in the formula Na2CrO4?
(a) 3
(b) (5)
(c) (7)
(d) 8
22. Which of the following is a characteristic of metals?
(a) ductile
(c) extremely strong
(b) easily shattered
(d) dull
23. Which of the following is a characteristic of nonmetals?
(a) always a gas
(b) poor conductor of electricity
(c) shiny
(d) combines only with metals
24. When a pure substance was analyzed, it was found to contain carbon and chlorine. This substance must be classified as
(a) an element
(b) a mixture
(c) a compound
(d) both a mixture and a compound
25. Chromium, fluorine, and magnesium have the symbols
(a) Ch, F, Ma
(c) Cr, F, Mg
(b) Cr, Fl, Mg
(d) Cr, F, Ma
26. Sodium, carbon, and sulfur have the symbols
(a) Na, C, S
(c) Na, Ca, Su
(b) So, C, Su
(d) So, Ca, Su
27. Coffee is an example of
(a) an element
(c) a homogeneous mixture
(b) a compound
(d) a heterogeneous mixture
28. The number of oxygen atoms in Al(C2H3O2)3 is
(a) 2
(b) 3
(c) 5
(d) 6
29. Which of the following is a mixture?
(a) water
(c) sugar solution
(b) iron(II) oxide
(d) iodine
79
80 Putting It Together
30. Which is the most compact state of matter?
(a) solid
(c) gas
(b) liquid
(d) amorphous
31. Which is not characteristic of a solution?
(a) a homogeneous mixture
(b) a heterogeneous mixture
(c) one that has two or more substances
(d) one that has a variable composition
32. A chemical formula is a combination of
(a) symbols
(c) elements
(b) atoms
(d) compounds
33. The number of nonmetal atoms in Al2(SO3)3 is
(a) 5
(b) 7
(c) 12
(d) 14
34. Which of the following is not a physical property?
(a) boiling point
(c) bleaching action
(b) physical state
(d) color
35. Which of the following is a physical change?
(a) A piece of sulfur is burned.
(b) A firecracker explodes.
(c) A rubber band is stretched.
(d) A nail rusts.
36. Which of the following is a chemical change?
(a) Water evaporates.
(b) Ice melts.
(c) Rocks are ground to sand.
(d) A penny tarnishes.
37. The changing of liquid water to ice is known as a
(a) chemical change
(b) heterogeneous change
(c) homogeneous change
(d) physical change
38. Which of the following does not represent a chemical change?
(a) heating of copper in air
(b) combustion of gasoline
(c) cooling of red-hot iron
(d) digestion of food
39. Heating 30. g of water from 20.°C to 50.°C requires
(a) 30. cal
(c) 3.8 * 103 J
(b) 50. cal
(d) 6.3 * 103 J
40. The specific heat of aluminum is 0.900 J>g°C. How many joules
of energy are required to raise the temperature of 20.0 g of Al from
10.0°C to 15.0°C?
(a) 79 J
(b) 90. J
(c) 100. J
(d) 112 J
41. A 100.-g iron ball (specific heat = 0.473 J>g°C) is heated to
125°C and is placed in a calorimeter holding 200. g of water at
25.0°C. What will be the highest temperature reached by the water?
(a) 43.7°C
(c) 65.3°C
(b) 30.4°C
(d) 35.4°C
42. Which has the highest specific heat?
(a) ice
(b) lead
(c) water
(d) aluminum
43. When 20.0 g of mercury are heated from 10.0°C to 20.0°C, 27.6 J
of energy are absorbed. What is the specific heat of mercury?
(a) 0.726 J>g°C
(c) 2.76 J>g°C
(b) 0.138 J>g°C
(d) no correct answer given
44. Changing hydrogen and oxygen into water is a
(a) physical change
(b) chemical change
(c) conservation reaction
(d) no correct answer given
Free Response Questions
Answer each of the following. Be sure to include your work and explanations in a clear, logical form.
1. You decide to go sailing in the tropics with some friends. Once
there, you listen to the marine forecast, which predicts in-shore
wave heights of 1.5 m, offshore wave heights of 4 m, and temperature of 27°C. Your friend the captain is unfamiliar with the metric
system, and he needs to know whether it is safe for your small boat
and if it will be warm. He asks you to convert the measurements to
feet and degrees Fahrenheit, respectively.
2. Jane is melting butter in a copper pot on the stove. If she knows
how much heat her stove releases per minute, what other measurements does she need to determine how much heat the butter
­absorbed? She assumes that the stove does not lose any heat to the
surroundings.
3. Julius decided to heat 75 g CaCO3 to determine how much carbon
dioxide is produced. (Note: When CaCO3 is heated, it produces
CaO and carbon dioxide.) He collected the carbon dioxide in a balloon. Julius found the mass of the CaO remaining was 42 g. If 44 g
of carbon dioxide take up 24 dm3 of space, how many liters of gas
were trapped in the balloon?
Use these pictures to answer Question 4.
(1)
(2)
(3)
(4)
4. (a) Which picture best describes a homogeneous mixture?
(b) How would you classify the contents of the other containers?
(c) Which picture contains a compound? Explain how you made
your choice.
Use these pictures to answer Question 5.
(1)
(2)
(3)
(4)
5. (a) Which picture best represents fluorine gas? Why?
(b) Which other elements could that picture also represent?
(c) Which of the pictures could represent SO3 gas?
6. Sue and Tim each left a one-quart bowl outside one night. Sue’s
bowl was full of water and covered, while Tim’s was empty and
open. The next day there was a huge snowstorm that filled Tim’s
bowl with snow. The temperature that night went down to 12°F.
(a) Which bowl would require less energy to bring its contents to
room temperature (25°C)? Why?
(b) What temperature change (°C) is required to warm the bowls to
25°C?
(c) How much heat (in kJ) is required to raise the temperature of
the contents of Sue’s bowl to 0°C (without converting the ice to
water)?
(d) Did the water in Sue’s bowl undergo chemical or physical
changes or both?
Putting It Together 81
7. One cup of Raisin Bran provides 60.% of the U.S. recommended
daily allowance (RDA) of iron.
(a) If the cereal provides 11 mg of iron, what is the U.S. RDA for
Fe?
(b) When the iron in the cereal is extracted, it is found to be the
pure element. What is the volume of iron in a cup of the cereal?
8. Absent-minded Alfred put down a bottle containing silver on the
table. When he went to retrieve it, he realized he had forgotten to
label the bottle. Unfortunately, there were two full bottles of the
same size side-by-side. Alfred realized he had placed a bottle of
mercury on the same table last week. State two ways Alfred can
determine which bottle contains silver without opening the bottles.
9. Suppose 25 g of solid sulfur and 35 g of oxygen gas are placed in
a sealed container.
(a) Does the container hold a mixture or a compound?
(b) After heating, the container was weighed. From a comparison
of the total mass before heating to the total mass after heating,
can you tell whether a reaction took place? Explain.
(c) After the container is heated, all the contents are gaseous. Has the
density of the container including its contents changed? Explain.
chap t e r
5
Early Atomic Theory
and Structure
P
ure substances are classified as elements or compounds,
but just what makes a substance possess its unique
properties? How small a piece of salt will still taste salty?
Carbon dioxide puts out fires, is used by plants to produce
oxygen, and forms dry ice when solidified. But how small a mass of
this material still behaves like carbon dioxide? Substances are in
their simplest identifiable form at the atomic, ionic, or molecular
level. Further division produces a loss of characteristic properties.
What particles lie within an atom or ion? How are these tiny
particles alike? How do they differ? How far can we continue to
divide them? Alchemists began the quest, early chemists laid the
foundation, and modern chemists continue to build and expand
on models of the atom.
Chapter Outline
5.1 Dalton’s Model of the Atom
5.2 Electric Charge
5.3 Subatomic Parts of the Atom
5.4 The Nuclear Atom
5.5 Isotopes of the Elements
5.6 Atomic Mass
SuperStock
Lightning occurs when electrons
move to neutralize a charge
difference between the clouds
and the Earth.
• Dalton’s Model of the Atom 83
5.1
5.1 Dalton’s Model of the Atom
Learning objective
Describe Dalton’s model of the atom and compare it to the earlier concepts of matter.
The structure of matter has long intrigued and engaged us. The earliest models of the atom were
developed by the ancient Greek philosophers. About 440 b.c. Empedocles stated that all matter
was composed of four “elements”—earth, air, water, and fire. Democritus (about 470–370 b.c.)
thought that all forms of matter were composed of tiny indivisible particles, which he called
atoms, derived from the Greek word atomos, meaning “indivisible.” He held that atoms were in
constant motion and that they combined with one another in various ways. This hypothesis was
not based on scientific observations. Shortly thereafter, Aristotle (384–322 b.c.) opposed the
theory of Democritus and instead endorsed and advanced the Empedoclean theory. So strong
was the influence of Aristotle that his theory dominated the thinking of scientists and philosophers until the beginning of the seventeenth century.
More than 2000 years after Democritus, the English schoolmaster John Dalton (1766–1844)
revived the concept of atoms and proposed an atomic model based on facts and experimental
evidence (Figure 5.1). His theory, described in a series of papers published from 1803 to
1810, rested on the idea of a different kind of atom for each element. The essence of Dalton’s
atomic model may be summed up as follows:
Key tErm
Dalton’s atomic model
1. Elements are composed of minute, indivisible particles called atoms.
2. Atoms of the same element are alike in mass and size.
3. Atoms of different elements have different masses and sizes.
4. Chemical compounds are formed by the union of two or more atoms of different
elements.
5. Atoms combine to form compounds in simple numerical ratios, such as one to one, one to
two, two to three, and so on.
6. Atoms of two elements may combine in different ratios to form more than one compound.
O
H
H
H
H
O
O
O
O
H
H
O
H
O
H
(a)
H
O
H
O
O
O
H
(b)
H
H
H
(c)
Figure 5.1
(a) Dalton’s atoms were individual particles, the atoms of each element being alike in mass
and size but different in mass and size from other elements; (b) and (c) Dalton’s atoms
combine in specific ratios to form compounds.
Dalton’s atomic model stands as a landmark in the development of chemistry. The major
premises of his model are still valid, but some of his statements must be modified or qualified
because later investigations have shown that (1) atoms are composed of subatomic particles;
(2) not all the atoms of a specific element have the same mass; and (3) atoms, under special
circumstances, can be decomposed.
In chemistry we use models (theories) such as Dalton’s atomic model to explain the behavior of atoms, molecules, and compounds. Models are modified to explain new information.
We frequently learn the most about a system when our models (theories) fail. That is the time
when we must rethink our explanation and determine whether we need to modify our model
or propose a new or different model to explain the behavior.
84 chapter 5
• Early Atomic Theory and Structure
Example 5.1
ENHANCED EXAMPLE
Which statement of Dalton’s model of the atom does this series of compounds represent:
NO2, N2O, N2O3, N2O5?
SOLUTION
Atoms combine to form compounds in simple numerical ratios. Atoms of two elements
combine in different ratios to form more than one compound.
Practice 5.1
List another series of compounds that illustrates the statements in Example 5.1.
5.2Electric Charge
Lear ning obje ctive
Recognize the force between particles and distinguish between a cation and an anion.
You’ve probably received a shock after walking across a carpeted area on a dry day. You may
have also experienced the static electricity associated with combing your hair and have had
your clothing cling to you. These phenomena result from an accumulation of electric charge.
This charge may be transferred from one object to another. The properties of electric charge
are as follows:
1. Charge may be of two types, positive and negative.
2. Unlike charges attract (positive attracts negative), and like charges repel (negative repels
negative and positive repels positive).
3. Charge may be transferred from one object to another by contact or induction.
4. The less the distance between two charges, the greater the force of attraction between
unlike charges (or repulsion between identical charges). The force of attraction (F) can be
expressed using the following equation:
F =
kq1q2
r2
where q1 and q2 are the charges, r is the distance between the charges, and k is a constant.
Discovery of Ions
Richard Megna/Fundamental Photographs
When ions are present in a
solution of salt water and an
electric current is passed through
the solution, the light bulb glows.
English scientist Michael Faraday (1791–1867) made the discovery that certain substances
when dissolved in water conduct an electric current. He also noticed that certain compounds
decompose into their elements when an electric current is passed through the compound.
Atoms of some elements are attracted to the positive electrode, while atoms of other elements
are attracted to the negative electrode. Faraday concluded that these atoms are electrically
charged. He called them ions after the Greek word meaning “wanderer.”
Any moving charge is an electric current. The electrical charge must travel through a
substance known as a conducting medium. The most familiar conducting media are metals
formed into wires.
The Swedish scientist Svante Arrhenius (1859–1927) extended Faraday’s work. Arrhenius reasoned that an ion is an atom (or a group of atoms) carrying a positive or negative
charge. When a compound such as sodium chloride (NaCl) is melted, it conducts electricity.
Water is unnecessary. Arrhenius’s explanation of this conductivity was that upon melting,
the sodium chloride dissociates, or breaks up, into charged ions, Na + and Cl - . The Na +
ions move toward the negative electrode (cathode), whereas the Cl - ions migrate toward
the positive electrode (anode). Thus positive ions are called cations, and negative ions are
called anions.
From Faraday’s and Arrhenius’s work with ions, Irish physicist G. J. Stoney (1826–1911)
realized there must be some fundamental unit of electricity associated with atoms.
5.3
• Subatomic Parts of the Atom 85
He named this unit the electron in 1891. Unfortunately, he had no means of supporting his idea
with experimental proof. Evidence remained elusive until 1897, when English physicist J. J.
Thomson (1856–1940) was able to show experimentally the existence of the electron.
Example 5.2
ENHANCED EXAMPLE
What is a cation? What is an anion?
SOLUTION
A cation is a positively charged ion. An anion is a negatively charged ion.
Practice 5.2
Which of the following ions are cations and which are anions?
Na + , Cl - , NH4 + , Ca2 + , Br - , Al3 + , NO3-, SO42 - , PO43 - , K + , Ag + , Fe2 + , Fe3 + , S2 -
5.3 Subatomic Parts of the Atom
Describe the three basic subatomic particles and how they changed Dalton’s model
of the atom.
Learning objecti ve
The concept of the atom—a particle so small that until recently it could not be seen even
with the most powerful microscope—and the subsequent determination of its structure stand
among the greatest creative intellectual human achievements.
Any visible quantity of an element contains a vast number of identical atoms. But when we
refer to an atom of an element, we isolate a single atom from the multitude in order to present
the element in its simplest form. Figure 5.2 shows individual atoms as we can see them today.
What is this tiny particle we call the atom? The diameter of a single atom ranges from
0.1 to 0.5 nanometer (1 nm = 1 * 10-9 m). Hydrogen, the smallest atom, has a diameter of
about 0.1 nm. To arrive at some idea of how small an atom is, consider this dot (•), which has
a diameter of about 1 mm, or 1 * 106 nm. It would take 10 million hydrogen atoms to form a
line of atoms across this dot. As inconceivably small as atoms are, they contain even smaller
particles, the subatomic particles, including electrons, protons, and neutrons.
The development of atomic theory was helped in large part by the invention of new instruments. For example, the Crookes tube, developed by Sir William Crookes (1832–1919)
in 1875, opened the door to the subatomic structure of the atom (Figure 5.3). The emissions
generated in a Crookes tube are called cathode rays. J. J. Thomson demonstrated in 1897 that
cathode rays (1) travel in straight lines, (2) are negative in charge, (3) are deflected by electric
and magnetic fields, (4) produce sharp shadows, and (5) are capable of moving a small paddle
wheel. This was the experimental discovery of the fundamental unit of charge—the electron.
Key tErms
subatomic particles
electron
proton
Thomson model of the atom
neutron
Figure 5.2
Courtesy of IBM Corporation
A scanning tunneling microscope
shows an array of copper atoms.
86 chapter 5
• Early Atomic Theory and Structure
Figure 5.3
Power source
(electrical)
Cathode ray tube. A stream
of electrons passes between
electrodes. The fast-moving
particles excite the gas inside the
tube, creating a greenish glow
between the electrodes.
(–)
(+)
Metal Plate
Metal Plate
Stream of electrons
(negative particles)
Figure 5.4
Thomson model of the atom.
In this early model of the atom,
negative particles (electrons)
were thought to be embedded
in a positively charged sphere.
It is sometimes called the plum
pudding model.
The electron (e - ) is a particle with a negative electrical charge and a mass of
9.110 * 10-28 g. This mass is 1/1837 the mass of a hydrogen atom. Although the actual charge
of an electron is known, its value is too cumbersome for practical use and has therefore been
assigned a relative electrical charge of -1. The size of an electron has not been determined
exactly, but its diameter is believed to be less than 10 - 12 cm.
Protons were first observed by German physicist Eugen Goldstein (1850–1930) in 1886.
However, it was Thomson who discovered the nature of the proton. He showed that the proton
is a particle, and he calculated its mass to be about 1837 times that of an electron. The proton
(p) is a particle with actual mass of 1.673 * 10-24 g. Its relative charge (+1) is equal in magnitude, but opposite in sign, to the charge on the electron. The mass of a proton is only very
slightly less than that of a hydrogen atom.
Thomson had shown that atoms contain both negatively and positively charged
particles. Clearly, the Dalton model of the atom was no longer acceptable. Atoms are not indivisible but are instead composed of smaller parts. Thomson proposed a new model of the atom.
In the Thomson model of the atom, the electrons are negatively charged particles
embedded in the positively charged atomic sphere (see Figure 5.4). A neutral atom could
become an ion by gaining or losing electrons.
Positive ions were explained by assuming that a neutral atom loses electrons. An atom
with a net charge of +1 (for example, Na + or Li + ) has lost one electron. An atom with a net
charge of +3 (for example, Al3 + ) has lost three electrons (Figure 5.5a).
Negative ions were explained by assuming that additional electrons can be added to atoms.
A net charge of -1 (for example, Cl - or F - ) is produced by the addition of one electron.
A net charge of -2 (for example, O2 - or S2 - ) requires the addition of two electrons (Figure 5.5b).
The third major subatomic particle was discovered in 1932 by James Chadwick
(1891–1974). This particle, the neutron (n), has neither a positive nor a negative charge and
e–
Na
e–
Cl
Na+
Cl–
Anion
Cation
e–
e–
e–
S
Al3+
Al
Cation
e–
(a)
S 2–
e–
Anion
(b)
Figure 5.5
(a) When one or more electrons are lost from a neutral atom, a cation is formed. (b) When
one or more electrons are added to a neutral atom, an anion is formed.
5.4
Table 5.1
• The Nuclear Atom 87
Electrical Charge and Relative Mass of Electrons, Protons, and Neutrons
Particle
Symbol
ep
n
Electron
Proton
Neutron
Relative
electrical charge
Actual mass
(g)
-1
+1
0
9.110 * 10 - 28
1.673 * 10 - 24
1.675 * 10 - 24
has an actual mass (1.675 * 10 - 24 g) that is only very slightly greater than that of a proton.
The properties of these three subatomic particles are summarized in Table 5.1.
Nearly all the ordinary chemical properties of matter can be explained in terms of atoms
consisting of electrons, protons, and neutrons. The discussion of atomic structure that follows
is based on the assumption that atoms contain only these principal subatomic particles. Many
other subatomic particles, such as mesons, positrons, neutrinos, and antiprotons, have been
discovered, but it is not yet clear whether all these particles are actually present in the atom
or whether they are produced by reactions occurring within the nucleus. The fields of atomic
and high-energy physics have produced a long list of subatomic particles. Descriptions of the
properties of many of these particles are to be found in physics textbooks.
Example 5.3
The mass of a helium atom is 6.65 * 10 - 24 g. How many atoms are in a 4.0-g sample
of helium?
ENHANCED EXAMPLE
SOLUTION
(4.0 g)a
1 atom He
b = 6.0 * 1023 atoms He
6.65 * 10 - 24g
Practice 5.3
The mass of an atom of hydrogen is 1.673 * 10 - 24 g. How many atoms are in a 10.0-g
sample of hydrogen?
5.4 The Nuclear Atom
Explain how the nuclear model of the atom differs from Dalton’s and Thomson’s models.
Learning objective
The discovery that positively charged particles are present in atoms came soon after the discovery of radioactivity by Henri Becquerel (1852–1908) in 1896. Radioactive elements spontaneously emit alpha particles, beta particles, and gamma rays from their nuclei (see Chapter 18).
By 1907 Ernest Rutherford (1871–1937) had established that the positively charged
alpha particles emitted by certain radioactive elements are ions of the element helium.
Rutherford used these alpha particles to establish the nuclear nature of atoms. In experiments performed in 1911, he directed a stream of positively charged helium ions (alpha
particles) at a very thin sheet of gold foil (about 1000 atoms thick). See Figure 5.6a.
He observed that most of the alpha particles passed through the foil with little or no deflection;
but a few of the particles were deflected at large angles, and occasionally one even bounced
back from the foil (Figure 5.6b). It was known that like charges repel each other and that an
electron with a mass of 1/1837 that of a proton could not possibly have an appreciable effect on
the path of an alpha particle, which is about 7350 times more massive than an electron. Rutherford therefore reasoned that each gold atom must contain a positively charged mass occupying a
relatively tiny volume and that, when an alpha particle approaches close enough to this positive
mass, it is deflected. Rutherford spoke of this positively charged mass as the nucleus of the atom.
Because alpha particles have relatively high masses, the extent of the deflections (some actually
key terms
nucleus
atomic number
88 chapter 5
• Early Atomic Theory and Structure
Source of
alpha particles
Deflected
particles
Scattered
alpha
particles
Thin gold
foil
Beam of
alpha
particles
Circular
fluorescent
screen
Most particles
are undeflected
(a)
(b)
Figure 5.6
(a) Rutherford’s experiment on alpha-particle scattering, where positive alpha particles (a),
emanating from a radioactive source, were directed at a thin gold foil. (b) Deflection (red)
and scattering (blue) of the positive alpha particles by the positive nuclei of the gold atoms.
bounced back) indicated to Rutherford that the nucleus is very heavy and dense. (The density of
the nucleus of a hydrogen atom is about 1012 g/cm3—about 1 trillion times the density of water.)
Because most of the alpha particles passed through the thousand or so gold atoms without any
apparent deflection, he further concluded that most of an atom consists of empty space.
When we speak of the mass of an atom, we are referring primarily to the mass of the
nucleus. The nucleus contains all the protons and neutrons, which represent more than 99.9%
of the total mass of any atom. By way of illustration, the largest number of electrons known
to exist in an atom is 118. The mass of even 118 electrons is only about 1/17 of the mass of a
single proton or neutron. The mass of an atom therefore is primarily determined by the combined masses of its protons and neutrons.
General Arrangement of Subatomic Particles
The alpha-particle scattering experiments of Rutherford established that the atom contains a
dense, positively charged nucleus. The later work of Chadwick demonstrated that the atom contains neutrons, which are particles with mass, but no charge. Rutherford also noted that light,
negatively charged electrons are present and offset the positive charges in the nucleus. Based on
this experimental evidence, a model of the atom and the location of its subatomic particles was
devised in which each atom consists of a nucleus surrounded by electrons (see Figure 5.7). The
nucleus contains protons and neutrons but does not contain electrons. In a neutral atom the positive charge of the nucleus (due to protons) is exactly offset by the negative electrons. Because the
charge of an electron is equal to, but of opposite sign than, the charge of a proton, a neutral atom
must contain exactly the same number of electrons as protons. However, this model of atomic
structure provides no information on the arrangement of electrons within the atom.
A neutral atom contains the same number of protons and electrons.
Figure 5.7
In the nuclear model of the atom,
protons and neutrons are located
in the nucleus. The electrons are
found in the remainder of the
atom (which is mostly empty space
because electrons are very tiny).
10–13 cm
Electron region
Nucleus
Neutron
Proton
10–8 cm
5.5
• Isotopes of the Elements 89
Example 5.4
What are the major subatomic particles and where are they located in an atom?
ENHANCED EXAMPLE
SOLUTION
Protons, neutrons, and electrons are the three major subatomic particles in an atom. The
protons and neutrons are located in the nucleus and the electrons are located outside the
nucleus and around the nucleus.
Atomic Numbers of the Elements
The atomic number (Z) of an element is the number of protons in the nucleus of an atom of that
element. The atomic number determines the identity of an atom. For example, every atom with an
atomic number of 1 is a hydrogen atom; it contains one proton in its nucleus. Every atom with an
atomic number of 6 is a carbon atom; it contains 6 protons in its nucleus. Every atom with an atomic
number of 92 is a uranium atom; it contains 92 protons in its nucleus. The atomic number tells us
not only the number of positive charges in the nucleus but also the number of electrons in the neutral
atom, since a neutral atom contains the same number of electrons and protons.
atomic number = number of protons in the nucleus
You don’t need to memorize the atomic numbers of the elements because a periodic table
is usually provided in texts, in laboratories, and on examinations. The atomic numbers of all
elements are shown in the periodic table on the inside front cover of this book and are also
listed in the table of atomic masses on the inside front endpapers.
Practice 5.4
Look in the periodic table and determine how many protons and electrons are in a
Cu atom and an Xe atom.
5.5 Isotopes of the Elements
Define the terms atomic number, mass number, and isotope.
Learning objecti ve
Shortly after Rutherford’s conception of the nuclear atom, experiments were performed to
determine the masses of individual atoms. These experiments showed that the masses of nearly
all atoms were greater than could be accounted for by simply adding up the masses of all the
protons and electrons that were known to be present in an atom. This fact led to the concept
of the neutron, a particle with no charge but with a mass about the same as that of a proton.
Because this particle has no charge, it was very difficult to detect, and the existence of the
neutron was not proven experimentally until 1932. All atomic nuclei except that of the simplest
hydrogen atom contain neutrons.
All atoms of a given element have the same number of protons. Experimental evidence
has shown that, in most cases, all atoms of a given element do not have identical masses. This
is because atoms of the same element may have different numbers of neutrons in their nuclei.
Atoms of an element having the same atomic number but different atomic masses are
called isotopes of that element. Atoms of the various isotopes of an element therefore have
the same number of protons and electrons but different numbers of neutrons.
Three isotopes of hydrogen (atomic number 1) are known. Each has one proton in the nucleus and one electron. The first isotope (protium), without a neutron, has a mass number of 1;
the second isotope (deuterium), with one neutron in the nucleus, has a mass number of 2; the
third isotope (tritium), with two neutrons, has a mass number of 3 (see Figure 5.8).
The three isotopes of hydrogen may be represented by the symbols 11H, 21H and 31H,
indicating an atomic number of 1 and mass numbers of 1, 2, and 3, respectively. This method
Key tErms
isotope
mass number
© 2010 Paul Silverman/Fundamental Photographs
An uncut diamond from a
Colorado mine. Diamonds are
composed of carbon.
90 chapter 5
• Early Atomic Theory and Structure
Figure 5.8
The isotopes of hydrogen. The
number of protons (purple) and
neutrons (blue) are shown within
the nucleus. The electron (e−)
exists outside the nucleus.
3
1H
2
1H
1
H
1
or T
Tritium
or D
Deuterium
Protium
of representing atoms is called isotopic notation. The subscript (Z) is the atomic number; the
superscript (A) is the mass number, which is the sum of the number of protons and the number of neutrons in the nucleus. The hydrogen isotopes may also be referred to as hydrogen-1,
hydrogen-2, and hydrogen-3.
The mass number of an element
is the sum of the protons and
neutrons in the nucleus.
Mass number
(sum of protons and
neutrons in the nucleus)
Atomic number
(number of protons
in the nucleus)
A
ZE
Symbol of element
Most of the elements occur in nature as mixtures of isotopes. However, not all
isotopes are stable; some are radioactive and are continuously decomposing to form other
elements. For example, of the seven known isotopes of carbon, only two, carbon-12 and
carbon-13, are stable. Of the seven known isotopes of oxygen, only three— 168O, 178O,
and 188O—are stable. Of the fifteen known isotopes of arsenic, 75
33As is the only one that
is stable.
The relationship between mass number and atomic number is such that if we subtract the
atomic number from the mass number of a given isotope, we obtain the number of neutrons
in the nucleus of an atom of that isotope. Table 5.2 shows this method of determining the
number of neutrons. For example, the fluorine atom (199 F), atomic number 9, having a mass of
19 contains 10 neutrons:
mass number
19
Use four significant figures
for atomic masses in this text.
-
atomic number
9
=
=
number of neutrons
10
The atomic masses given in the table on the front endpapers of this book are values accepted by international agreement. You need not memorize atomic masses. In the calculations
in this book, the use of atomic masses rounded to four significant figures will give results of
sufficient accuracy. (See periodic table.)
Table 5.2 Determination of the Number of Neutrons in an Atom by Subtracting
Atomic Number from Mass Number
Hydrogen
(11H)
Mass number
Atomic number
Number of neutrons
1
( -)1
0
Oxygen
(168O)
16
( -)8
8
Sulfur
(32
16S)
32
( -)16
16
Fluorine
(199F)
19
( -)9
10
Iron
(56
26Fe)
56
( - )26
30
5.5
• Isotopes of the Elements 91
Example 5.5
How many protons, neutrons, and electrons are found in an atom of 146C?
ENHANCED EXAMPLE
SOLUTION
The element is carbon, atomic number 6. The number of protons or electrons equals the
atomic number and is 6. The number of neutrons is determined by subtracting the atomic
number from the mass number: 14 - 6 = 8.
Practice 5.5
How many protons, neutrons, and electrons are in each of these isotopes?
(c) 235
(a) 168O
(b) 80
(d) 64
35Br
92 U
29Cu
Practice 5.6
What are the atomic number and the mass number of the elements that contain
(a) 9 electrons
(b) 24 protons and 28 neutrons
(c) 197
79 X
What are the names of these elements?
>Chemistry in action
Isotope Detectives
Scientists are learning to use isotopes to determine the origin of drugs and gems. It turns out that isotope ratios similar
to those used in carbon dating can also identify the source
of cocaine or the birthplace of emeralds.
It turns out that a similar isotopic analysis of oxygen has led
researchers in France to be able to track the birthplace of
emeralds. Very high quality emeralds have few inclusions (microscopic cavities). Gemologists use these inclusions and the
material trapped in them to identify the source of the gem.
High-quality gems can now also be identified by using an
oxygen isotope ratio. These tests use an ion microscope that
blasts a few atoms from the gems’ surface (with virtually undetectable damage). The tiny sample is analyzed for its oxygen
isotope ratio and then compared to a database from emerald
mines around the world. Using the information, gemologists
can determine the mine from which the emerald came. Since
emeralds from Colombian mines are valued much more highly
than those from other countries, this technique can be used to
help collectors know just what they are paying for, as well as
to identify the history of treasured emeralds.
© PjrStudio/Alamy
Researchers with the Drug Enforcement Agency (DEA)
have created a database of the origin of coca leaves that
pinpoints the origin of the leaves with a 90% accuracy.
Cocaine keeps a chemical signature of the environment
where it grew. Isotopes of carbon and nitrogen are found in
a particular ratio based on climatic conditions in the growing region. These ratios correctly identified the source of
90% of the samples tested, according to James Ehleringer
of the University of Utah, Salt Lake City. This new method
can trace drugs a step further back than current techniques,
which mainly look at chemicals introduced by processing
practices in different locations. This could aid in tracking the
original exporters and stopping production at the source.
Raw and cut emeralds
92 chapter 5
• Early Atomic Theory and Structure
5.6 Atomic Mass
L earning obje ctive
Explain the relationship between the atomic mass of an element and the masses of
its isotopes.
Key terms
The mass of a single atom is far too small to measure on a balance, but fairly precise
determinations of the masses of individual atoms can be made with an instrument called
a mass spectrometer. The mass of a single hydrogen atom is 1.673 * 10 - 24 g. However,
it is neither convenient nor practical to compare the actual masses of atoms expressed in
grams; therefore, a table of relative atomic masses using atomic mass units was devised.
(The term atomic weight is sometimes used instead of atomic mass.) The carbon isotope
having six protons and six neutrons and designated carbon-12, or 126C, was chosen as
the standard for atomic masses. This reference isotope was assigned a value of exactly
12 atomic mass units (amu). Thus, 1 atomic mass unit (amu) is defined as equal to
exactly 1/12 of the mass of a carbon-12 atom. The actual mass of a carbon-12 atom is
1.9927 * 10 - 23 g and that of one atomic mass unit is 1.6606 * 10 - 24 g. In the table of
atomic masses, all elements then have values that are relative to the mass assigned to the
reference isotope, carbon-12.
A table of atomic masses is given on the inside front cover of this book. Hydrogen atoms,
with a mass of about 1/12 that of a carbon atom, have an average atomic mass of 1.00794 amu
on this relative scale. Magnesium atoms, which are about twice as heavy as carbon, have an
average mass of 24.305 amu. The average atomic mass of oxygen is 15.9994 amu.
Since most elements occur as mixtures of isotopes with different masses, the atomic mass
determined for an element represents the average relative mass of all the naturally occurring
isotopes of that element. The atomic masses of the individual isotopes are approximately
whole numbers, because the relative masses of the protons and neutrons are approximately
1.0 amu each. Yet we find that the atomic masses given for many of the elements deviate
considerably from whole numbers.
For example, the atomic mass of rubidium is 85.4678 amu, that of copper is 63.546 amu,
and that of magnesium is 24.305 amu. The deviation of an atomic mass from a whole number
is due mainly to the unequal occurrence of the various isotopes of an element.
65
The two principal isotopes of copper are 63
29Cu and 29Cu. Copper used in everyday objects
and the Liberty Bell contains a mixture of these two isotopes. It is apparent that copper-63
atoms are the more abundant isotope, since the atomic mass of copper, 63.546 amu, is closer
to 63 than to 65 amu (see Figure 5.9). The actual values of the copper isotopes observed by
mass spectra determination are shown in the following table:
atomic mass unit (amu)
atomic mass
1 amu = 1.6606 * 10 - 24 g
Tetra Images /Getty Images, Inc.
The copper used in casting the
Liberty Bell contains a mixture of
the isotopes of copper.
Isotope
63
29Cu
65
29Cu
The average atomic mass is a
weighted average of the masses
of all the isotopes present in the
sample.
Isotopic mass
(amu)
Abundance
(%)
62.9298
69.09
64.9278
30.91
Average atomic mass
(amu)
63.55
The average atomic mass can be calculated by multiplying the atomic mass of each isotope by the fraction of each isotope present and adding the results. The calculation for
copper is
(62.9298 amu)(0.6909) = 43.48 amu
(64.9278 amu)(0.3091) = 20.07 amu
63.55 amu
The atomic mass of an element is the average relative mass of the isotopes of that element
compared to the atomic mass of carbon-12 (exactly 12.0000 . . . amu).
Review 93
Figure 5.9
69.09%
A typical reading from a mass
spectrometer. The two principal
isotopes of copper are shown with
the abundance (%) given.
Abundance
30.91%
62
63
64
Mass
65
66
Example 5.6
35
Chlorine is found in nature as two isotopes, 37
17Cl (24.47%) and 17Cl (75.53%). The
atomic masses are 36.96590 and 34.96885 amu, respectively. Determine the average
atomic mass of chlorine.
SOLUTION
Multiply each mass by its percentage and add the results to find the average:
(0.2447)(36.96590 amu) + (0.7553)(34.96885 amu)
= 35.4575 amu
= 35.46 amu (four significant figures)
Practice 5.7
Silver occurs as two isotopes with atomic masses 106.9041 and 108.9047 amu, respectively.
The first isotope represents 51.82% and the second 48.18%. Determine the average atomic
mass of silver.
C h a p t e r
5 review
5.1 Dalton’s Model of the Atom
• Greek model of matter:
• Four elements—earth, air, water, fire
• Democritus—atoms (indivisible particles) make up matter
• Aristotle—opposed atomic ideas
• Summary of Dalton’s model of the atom:
• Elements are composed of atoms.
• Atoms of the same element are alike (mass and size).
• Atoms of different elements are different in mass and size.
• Compounds form by the union of two or more atoms of different elements.
• Atoms form compounds in simple numerical ratios.
• Atoms of two elements may combine in different ratios to form different compounds.
5.2 Electric Charge
• The properties of electric charge:
• Charges are one of two types—positive or negative.
• Unlike charges attract and like charges repel.
Key Term
Dalton’s atomic model
94 chapter 5
• Early Atomic Theory and Structure
• Charge is transferred from one object to another by contact or by induction.
• The force of attraction between charges is expressed by
F =
kq1q2
r2
• Michael Faraday discovered electrically charged ions.
• Svante Arrhenius explained that conductivity results from the dissociation of compounds into ions:
• Cation—positive charge—attracted to negative electrode (cathode)
• Anion—negative charge—attracted to positive electrode (anode)
5.3 Subatomic Parts of the Atom
Key Terms
subatomic particles
electron
proton
Thomson model of the atom
neutron
• Atoms contain smaller subatomic particles:
• Electron—negative charge, 1/1837 mass of proton
• Proton—positive charge, 1.673 * 10 - 24 g
• Neutron—no charge, 1.675 * 10 - 24 g
• Thomson model of the atom:
• Negative electrons are embedded in a positive atomic sphere.
• Number of protons = number of electrons in a neutral atom.
• Ions are formed by losing or gaining electrons.
5.4 The Nuclear Atom
Key Terms
nucleus
atomic number
• Rutherford gold foil experiment modified the Thomson model to a nuclear model of the atom:
• Atoms are composed of a nucleus containing protons and/or neutrons surrounded by electrons,
which occupy mostly empty space.
• Neutral atoms contain equal numbers of protons and electrons.
5.5 Isotopes of the Elements
Key Terms
isotope
mass number
• The mass number of an element is the sum of the protons and neutrons in the nucleus.
Mass number
(sum of protons and
neutrons in the nucleus)
Atomic number
(number of protons
in the nucleus)
A
ZE
Symbol of element
• The number of neutrons in an atom is determined by
mass number 2 atomic number = number of neutrons
A
2
Z
= neutrons
5.6 Atomic Mass
Key TermS
• The average atomic mass is a weighted average of the masses of all the isotopes present in the sample.
atomic mass unit (amu)
atomic mass
Review Questions
1. List the main features of Dalton’s model of the atom.
2. What elements of Democritus’ theory did Dalton incorporate into his
model of the atom?
3. How does Dalton’s theory explain why chemical formulas are always written with whole-number values?
4. Distinguish between an atom and an ion.
5. Does the force of attraction increase, decrease, or stay the same as
two oppositely charged particles approach each other?
6. How do cations and anions differ?
7. A neutron is approximately how many times heavier than an electron?
8. From the chemist’s point of view, what are the essential differences
among a proton, a neutron, and an electron?
9. What are the atomic numbers of (a) copper, (b) nitrogen, (c) phosphorus, (d) radium, and (e) zinc?
10. What letters are used to designate atomic number and mass number in
isotopic notation of atoms?
11. In what ways are isotopes alike? In what ways are they different?
12. Explain why the mass number of an element is always a whole number.
Paired Exercises 95
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.
com). All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1.Identify the ions in a mixture containing Cu2 + , N3 - , O2 - , Ca2 + , Al3 + ,
Te3 - , and H + that would be attracted to the positive end of an electric field.
2.Identify the ions in a mixture containing Mg2 + , Br - , Cr3 + , Ba2 + ,
P3-, Ca2 + , Se2 - , and Y3 + that would be attracted to the anode of a
battery.
3.Explain why, in Rutherford’s experiments, some alpha particles were
scattered at large angles by the gold foil or even bounced back.
4.What experimental evidence led Rutherford to conclude the
following?
(a) The nucleus of the atom contains most of the atomic mass.
(b) The nucleus of the atom is positively charged.
(c) The atom consists of mostly empty space.
5.Describe the general arrangement of subatomic particles in the atom.
6.What part of the atom contains practically all its mass?
7.What contribution did these scientists make to atomic models of the
atom?
(a) Dalton (b) Thomson (c) Rutherford
8. Consider the following models of the atom: (a) Dalton, (b)
Thomson, (c) Rutherford. How does the location of the electrons in
an atom vary? How does the location of the atom’s positive matter
compare?
9. Explain why the atomic masses of elements are not whole numbers.
10. Is the isotopic mass of a given isotope ever an exact whole number?
Is it always? (Consider the masses of 126C and 63
29Cu.)
11. What special names are given to the isotopes of hydrogen?
12. List the similarities and differences in the three isotopes of hydrogen.
13.What is the nuclear composition of the five naturally occurring isotopes of germanium having mass numbers 70, 72, 73, 74, and 76?
14.What is the nuclear composition of the five naturally occurring
isotopes of zinc having mass numbers 64, 66, 67, 68, and 70?
15. Write isotopic notation symbols for each of the following:
(a) Z = 29,
A = 65
(b) Z = 20,
A = 45
(c) Z = 36,
A = 84
16. Write isotopic notation symbols for each of the following:
(a) Z = 47,
Z = 109
(b) Z = 8,
A = 18
(c) Z = 26,
A = 57
17. Give the number of protons, neutrons, and electrons in the following isotopes.
(a) 63Cu
(b) 32S
(c) manganese-55 (55Mn)
(d) potassium-39 (39K)
18. Tell the number of protons, neutrons, and electrons in the following
isotopes.
(a) 54Fe
(b) 23Na
(c) bromine-79 (79Br)
(d) phosphorus-31 (31P)
19. An unknown element contains 33 protons, 36 electrons, and has a
mass number 76. Answer the following questions:
(a)What is the atomic number of this element?
(b)What is the name and symbol of this element?
(c) How many neutrons does it contain?
(d)What is the charge on this ion? Is it a cation or an anion?
(e)Write the symbolic notation for this element.
20. An unknown element contains 56 protons, 54 electrons, and has a
mass number 135. Answer the following questions:
(a)What is the atomic number of this element?
(b)What is the name and symbol of this element?
(c) How many neutrons does it contain?
(d) What is the charge on this ion? Is it a cation or an anion?
(e)Write the symbolic notation for this element.
21. Naturally occurring zirconium exists as five stable isotopes: 90Zr
with a mass of 89.905 amu (51.45%); 91Zr with a mass of 90.906 amu
(11.22%); 92Zr with a mass of 91.905 amu (17.15%); 94Zr with a
mass of 93.906 amu (17.38%); and 96Zr with a mass of 95.908 amu
(2.80%). Calculate the average mass of zirconium.
22. Naturally occurring titanium exists as five stable isotopes. Four of
the isotopes are 46Ti with a mass of 45.953 amu (8.0%); 47Ti with
a mass of 46.952 amu (7.3%); 48Ti with a mass of 47.948 amu
(73.8%); and 49Ti with a mass of 48.948 amu (5.5%). The average
mass of an atom of titanium is 47.9 amu. Determine the mass of the
fifth isotope of titanium.
23. A particular element exists in two stable isotopic forms. One isotope has a mass of 62.9296 amu (69.17% abundance). The other
isotope has a mass of 64.9278 amu. Calculate the average mass of
the element and determine its identity.
24. A particular element exists in two stable isotopic forms. One
isotope has a mass of 34.9689 amu (75.77% abundance). The
other isotope has a mass of 36.9659 amu. Calculate the average
mass of the element and determine its identity.
25. An average dimension for the radius of an atom is 1.0 * 10 - 8 cm,
and the average radius of the nucleus is 1.0 * 10 - 13 cm. Determine the ratio of atomic volume to nuclear volume. Assume that
the atom is spherical [V = (4/3)pr3 for a sphere].
26.An aluminum atom has an average diameter of about 3.0 * 10 - 8 cm.
The nucleus has a diameter of about 2.0 * 10 - 13 cm. Calculate the
ratio of the atom’s diameter to its nucleus.
96 chapter 5
• Early Atomic Theory and Structure
Add i t i o n a l E x e r c i s e s
27. What experimental evidence supports these statements?
33. The actual mass of one atom of an element is 3.27 * 10 - 19 mg.
(a) The nucleus of an atom is small.
(a) Calculate the atomic mass of the element.
(b) The atom consists of both positive and negative charges.
(b) Identify the element, giving the symbol and the name. (There
(c) The nucleus of the atom is positive.
is only one stable isotope of this element.)
28. Given the following information, determine which, if any, elements
34. The mass of an atom of silver is 1.79 * 10 - 22 g. How many atoms
are isotopes.
would be in a 0.52-lb sample of silver?
(a) mass number = 32; number of neutrons = 17
35. Using the periodic table inside the front cover of the book, de(b) mass number = 32; number of protons = 16
termine which of the first 20 elements have isotopes that you
(c) number of protons = 15; number of neutrons = 16
would expect to have the same number of protons, neutrons, and
29. The diameter of a silicon atom is 2.34 * 10 - 8 cm. If the silicon
electrons.
atoms were placed next to each other in a straight line across an
36. Solar winds that send streams of charged particles toward the Earth
8.5-in. width of paper, how many silicon atoms would be there?
cause the aurora borealis or the northern lights. As these charged
30. How is it possible for there to be more than one kind of atom of the
particles hit the upper atmosphere of the Earth or the ionosphere (so
same element?
named because it is composed of a plasma or gaseous soup of ions),
31. Place each of the following elements in order of increasing
they collide with ions located in this region. These high-energy ions
number of neutrons (least to most); 156Dy, 160Gd, 162Er, 165Ho.
then lose energy in the form of light, which is visible in the northDoes this order differ from the elements’ positions on the peernmost and southernmost regions of the planet as ribbons of light
riodic table?
in the night sky. Some of the ions responsible for the colors seen are
32. An unknown element Q has two known isotopes: 60Q and 63Q. If
C + , O + , and O2 + . Determine the number of protons and electrons
the average atomic mass is 61.5 amu, what are the relative percentin each of these ions.
ages of the isotopes?
37. Often the minerals we need in our diet are provided to us in vitamin tablets as cations. Following is a list of ingredients for a vitamin and mineral supplement. The table below highlights some of these minerals and identifies the ion provided for each mineral. Determine the number of
protons and electrons in each of these ions.
Mineral supplement
Mineral use
Ion provided
Number of protons
Number of electrons
2+
Calcium carbonate
Bones and Teeth
Ca
Iron(II) sulfate
Hemoglobin
Fe2 +
Chromium(III) nitrate
Insulin
Cr3 +
Magnesium sulfate
Bones
Mg2 +
Zinc sulfate
Cellular metabolism
Zn2 +
Potassium iodide
Thyroid function
I-
38.Complete the following table with the appropriate data for each isotope given (all are neutral atoms):
Element
Symbol
Atomic number
Mass number
Number of protons
Number of neutrons
Number of electrons
36
Cl
Gold
197
56
79
58
20
18
Number of neutrons
Number of electrons
143
92
28
39. Complete the following table with the appropriate data for each isotope given (all are neutral atoms):
Element
Symbol
Atomic number
Mass number
Number of protons
134
Xe
Silver
107
9
92
41
19
Answers to Practice Exercises 97
40. Draw diagrams similar to those shown in Figure 5.5 for the following ions:
(a) O2 (b) Na+ (c) P3 - (d) Ca2+
41. Draw pictures similar to those in Figure 5.8 for the following
isotopes:
(a) 3He and 4He
(c) 10B and 11B
6
7
(b) Li and Li
(d) 12C and 13C
42. What percent of the total mass of one atom of each of the following
elements comes from electrons?
(a) aluminum (mass of one atom = 4.480 * 10 - 23 g)
(b) phosphorus (mass of one atom = 5.143 * 10 - 23 g)
(c) krypton (mass of one atom = 1.392 * 10 - 22 g)
(d) platinum (mass of one atom = 3.240 * 10 - 22 g)
43. What percent of the total mass of one atom of each of the following
elements comes from protons?
(a) selenium (mass of one atom = 1.311 * 10 - 22 g)
(b) xenon (mass of one atom = 2.180 * 10 - 22 g)
(c) chlorine (mass of one atom = 5.887 * 10 - 23 g)
(d) barium (mass of one atom = 2.280 * 10 - 22 g)
44. Figure 5.7 is a representation of the nuclear model of the atom. The
area surrounding the nucleus is labeled as the electron region. What
is the electron region?
Challenge Exercise
45. You have discovered a new element and are trying to determine
where on the periodic table it would fit. You decide to do a mass
spectrometer analysis of the sample and discover that it contains three isotopes with masses of 270.51 amu, 271.23 amu, and
269.14 amu and relative abundances of 34.07%, 55.12%, and
10.81%, respectively. Sketch the mass spectrometer reading,
determine the average atomic mass of the element, estimate its
atomic number, and determine its approximate location on the periodic table.
Answers to Practice Exercises
5.1
CH4, C2H2, C2H4, C2H6
5.2
+
cations: Na , NH4 , Ca Al , K , Ag , Fe , Fe
anions: Cl - , Br - , NO3- , S2 - , SO42 - , PO43 -
5.3
5.98 * 1024 atoms
5.4
Cu, 29 protons and 29 electrons; Xe, 36 protons and 36 electrons
5.5
protons
(a) 8
(b) 35
(c) 92
(d) 29
+
neutrons
8
45
143
35
2+,
3+
+
electrons
8
35
92
29
+
2+
5.6
(a) atomic number
mass number
name
(b) atomic number
mass number
name
(c) atomic number
mass number
name
5.7
107.9 amu
3+
9
19
fluorine
24
52
chromium
79
179
gold
Chapter
6
Nomenclature of
Inorganic Compounds
A
s children, we begin to communicate with other
people in our lives by learning the names of objects
around us. As we continue to develop, we learn to
speak and use language to complete a wide variety of tasks. As
we enter school, we begin to learn of other languages—the languages of mathematics, of other cultures, of computers. In each
case, we begin by learning the names of the building blocks
and then proceed to more abstract concepts. Chemistry has a
language all its own—a whole new way of describing the objects
so familiar to us in our daily lives. Once we learn the language,
we can begin to understand the macroscopic and microscopic
world of chemistry.
Chapter Outline
6.1 Common and Systematic Names
6.2 Elements and Ions
6.3 Writing Formulas from Names
of Ionic Compounds
6.4 Naming Binary Compounds
6.5 Naming Compounds Containing
Polyatomic Ions
6.6 Acids
© Exactostock/SuperStock
This seashell is formed from the
chemical calcium carbonate,
commonly called limestone. It is
the same chemical used in many
calcium supplements for our diets.
6.1
• Common and Systematic Names 99
6.1 Common and Systematic Names
Distinguish between the common and systematic names of chemical substances.
Learning objectiv e
Chemical nomenclature is the system of names that chemists use to identify compounds. When
a new substance is formulated, it must be named in order to distinguish it from all other substances (see Figure 6.1). In this chapter, we will restrict our discussion to the nomenclature of
inorganic compounds—compounds that do not generally contain carbon.
Number of Elements Present
1
2
3 or more
Element
Binary
compounds
Polyatomic
compounds
See Sections
3.1–3.4
See Section
6.4
See Section
6.5
Figure 6.1
Where to find rules for naming
inorganic substances in this book.
Common names are arbitrary names that are not based on the chemical composition of
compounds. Before chemistry was systematized, a substance was given a name that generally
associated it with one of its outstanding physical or chemical properties. For example, quicksilver
is a common name for mercury, and nitrous oxide (N2O), used as an anesthetic in dentistry,
has been called laughing gas because it induces laughter when inhaled. Water and ammonia
are also common names because neither provides any information about the chemical composition of the compounds. If every substance were assigned a common name, the amount of
memorization required to learn over 50 million names would be astronomical.
Common names have distinct limitations, but they remain in frequent use. Common names
continue to be used because the systematic name is too long or too technical for everyday use.
For example, calcium oxide (CaO) is called lime by plasterers; photographers refer to hypo
rather than sodium thiosulfate (Na2S2O3); and nutritionists use the name vitamin D3, instead
of 9,10-secocholesta-5,7,10(19)-trien-3-ß-ol (C27H44O). Table 6.1 lists the common names,
formulas, and systematic names of some familiar substances.
Table 6.1
Water
Ammonia
Water (H2O) and ammonia (NH3)
are almost always referred to by
their common names.
Common Names, Formulas, and Chemical Names of Familiar Substances
Common names Formula
Chemical names
Acetylene
C2H2
ethyne
Lime
CaO
calcium oxide
Slaked lime
Ca(OH)2
calcium hydroxide
Water
H2O
water
Galena
PbS
lead(II) sulfide
Alumina
Al2O3
aluminum oxide
Baking soda
NaHCO3
sodium hydrogen carbonate
Cane or beet sugar C12H22O11
sucrose
Borax
Na2B4O7 # 10 H2O sodium tetraborate
decahydrate
Brimstone
S
sulfur
Calcite, marble,
CaCO3
calcium carbonate
limestone
Cream of tartar
KHC4H4O6
potassium hydrogen tartrate
Epsom salts
MgSO4 # 7 H2O
magnesium sulfate
heptahydrate
Common names
Formula
Chemical names
Gypsum
Grain alcohol
Hypo
Laughing gas
Lye, caustic soda
Milk of magnesia
Muriatic acid
Plaster of paris
Potash
Pyrite (fool’s gold)
Quicksilver
Saltpeter (chile)
Table salt
Vinegar
Washing soda
Wood alcohol
CaSO4 # 2 H2O
C2H5OH
Na2S2O3
N2O
NaOH
Mg(OH)2
HCl
CaSO4 # 1/2 H2O
K2CO3
FeS2
Hg
NaNO3
NaCl
HC2H3O2
Na2CO3 # 10 H2O
CH3OH
calcium sulfate dihydrate
ethanol, ethyl alcohol
sodium thiosulfate
dinitrogen monoxide
sodium hydroxide
magnesium hydroxide
hydrochloric acid
calcium sulfate hemihydrate
potassium carbonate
iron disulfide
mercury
sodium nitrate
sodium chloride
acetic acid
sodium carbonate decahydrate
methanol, methyl alcohol
100 chapter 6
• Nomenclature of Inorganic Compounds
Chemists prefer systematic names that precisely identify the chemical composition of
chemical compounds. The system for inorganic nomenclature was devised by the International
Union of Pure and Applied Chemistry (IUPAC), which was founded in 1921. The IUPAC
meets regularly and constantly reviews and updates the system.
6.2Elements and Ions
Learning objective
Discuss the formation, charge, and naming of simple ions.
In Chapter 3, we studied the names and symbols for the elements as well as their location on
the periodic table. In Chapter 5, we investigated the composition of the atom and learned that
all atoms are composed of protons, electrons, and neutrons; that a particular element is defined
by the number of protons it contains; and that atoms are uncharged because they contain equal
numbers of protons and electrons.
The formula for most elements is simply the symbol of the element. In chemical reactions or mixtures, an element behaves as though it were a collection of individual particles.
A small number of elements have formulas that are not single atoms at normal temperatures.
Seven of the elements are diatomic molecules—that is, two atoms bonded together to form a
molecule. These diatomic elements are hydrogen, H2 , oxygen, O2 , nitrogen, N2 , fluorine, F2 ,
chlorine, Cl2 , bromine, Br2 , and iodine, I2 . Three other elements that are commonly polyatomic are S8 , sulfur, Se8, selenium and P4 , phosphorus.
Elements occurring as polyatomic molecules
Hydrogen
H2
Oxygen
O2
Nitrogen
N2
Chlorine
Fluorine
Bromine
Iodine
Cl2
F2
Br2
I2
Sulfur
Phosphorus
Selenium
S8
P4
Se8
We have learned that a charged particle, known as an ion, can be produced by adding
or removing one or more electrons from a neutral atom. For example, potassium atoms contain 19 protons and 19 electrons. To make a potassium ion, we remove one electron, leaving
19 protons and only 18 electrons. This gives an ion with a positive one (11) charge:
19p
19e–
19p
18e–
K+ ion
K atom
Written in the form of an equation, K S K+ + e-. A positive ion is called a cation. Any
neutral atom that loses an electron will form a cation. Sometimes an atom may lose one electron, as in the potassium example. Other atoms may lose more than one electron:
Mg S Mg2+ + 2e-
or
Al S Al3+ + 3e-
Cations are named the same as their parent atoms, as shown here:
Atom
Ion
K
Mg
Al
K+
Mg2+
Al3+
potassium
magnesium
aluminum
potassium ion
magnesium ion
aluminum ion
6.2
• Elements and Ions 101
>Chemistry in action
What’s in a Name?
Other names of elements recognize the contributions
of great scientists and researchers such as the American
Glenn Seaborg, the first living scientist to have an element named after him. The British recognized Ernest
Rutherford, who discovered the atom’s nucleus. The
Germans recognized Lise Meitner who co-discovered
atomic fusion. Both the Germans and the Russians won
recognition for Niels Bohr whose model of the atom
led to modern ideas about atomic structure. Wilhelm
Roentgen was honored for discovering X-rays and winning the first Nobel Prize in physics. Most recently, the
astronomer Copernicus was honored for changing our
model of the solar system to a heliocentric model and
to highlight the link between astronomy and nuclear
science. The Russians honored Georgy Flerov, who
developed the Soviet nuclear program, with the name
for element 114.
When a scientist discovered a new element in the early
days of chemistry, he or she had the honor of naming it.
Now researchers must submit their choices for a name to
an international committee called the International Union of
Pure and Applied Chemistry (IUPAC) before the name can
be placed on the periodic table. Since 1997 the IUPAC has
decided on names for 11 new elements from 104 through
118. The new elements are called rutherfordium (Rf), dubnium (Db), seaborgium (Sg), bohrium (Bh), hessium (Hs),
meitnerium (Mt), darmstadtium (Ds), roentgenium (Rg),
copernicium (Cn), flerovium (Fv), and livermorium (Lv).
The new names are compromises among the choices presented by different research teams. Some names given to
the elements recognize places where much of the
research to find these elements is done. The Russians
gained recognition for work done at a laboratory in
Dubna, the Germans for their laboratories in Darmstadt
and Hesse, and the Americans for the Lawrence Livermore
National Laboratory (LLNL).
Elements 113, 115, 117, and 118 do not yet have
official names and are currently shown on the periodic
table using three letter symbols representing their
numbers.
Science Source/PhotoResearchers, Inc.
Copernicus
107
Bh
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
Glenn Seaborg
113
114
Fl
RIA Novosti/Photo Researchers, Inc.
106
Sg
© Hulton/Getty Images, Inc.
Ernest Rutherford
105
Db
Science Source/Photo Researchers, Inc.
© Corbis
104
Rf
Lise Meitner
© Corbis
The Image Works
Niels Bohr
Wilhelm Roentgen
Flerov
102 chapter 6
• Nomenclature of Inorganic Compounds
Ions can also be formed by adding electrons to a neutral atom. For example, the chlorine atom contains 17 protons and 17 electrons. The equal number of positive charges and
negative charges results in a net charge of zero for the atom. If one electron is added to the
chlorine atom, it now contains 17 protons and 18 electrons, resulting in a net charge of negative one (21) on the ion:
17p
17p
18e–
17e–
Cl− ion
Cl atom
In a chemical equation, this process is summarized as Cl + e- S Cl-. A negative ion is called
an anion. Any neutral atom that gains an electron will form an anion. Atoms may gain more
than one electron to form anions with different charges:
O + 2e- S O2N + 3e- S N3Anions are named differently from cations. To name an anion consisting of only one element, use the stem of the parent element name and change the ending to -ide. For example,
the Cl- ion is named by using the stem chlor- from chlorine and adding -ide to form chloride
ion. Here are some examples:
Symbol
F
Br
Cl
I
O
N
Name of atom
fluorine
bromine
chlorine
iodine
oxygen
nitrogen
Ion
-
F
BrClIO2N3-
Name of ion
fluoride ion
bromide ion
chloride ion
iodide ion
oxide ion
nitride ion
Example 6.1
ENHANCED EXAMPLE
What is the general basis for naming a negative ion in a binary compound?
SOLUTION
Establish an identifying stem from the name of the negative element; then add the suffix
-ide to that stem. For example, the stem for chlorine is chlor-. Add the suffix -ide to get
the name of the ion, chloride. The stem for oxygen is ox-. Add the suffix -ide to get the
name of the ion, oxide.
Practice 6.1
Derive the names of these ions when used as the negative ion in a binary compound: P, N, B, C.
Ions are formed by adding or removing electrons from an atom. Atoms do not form ions on
their own. Most often ions are formed when metals combine with nonmetals.
The charge on an ion can often be predicted from the position of the element on the
periodic table. Figure 6.2 shows the charges of selected ions from several groups on the
periodic table. Notice that all the metals and hydrogen in the far left column (Group 1A)
are (11), all those in the next column (Group 2A) are (21), and the metals in the next
tall column (Group 3A) form (31) ions. The elements in the lower center part of the table
are called transition metals. These elements tend to form cations with various positive
charges. There is no easy way to predict the charges on these cations. All metals lose
electrons to form positive ions.
6.3
• Writing Formulas from Names of Ionic Compounds 103
1A
H+
2A
Li
+
3A
4A
2+
Be
Na+
Mg2+
K+
Ca2+
Rb+
Sr2+
Cs+
Ba2+
Al3+
Cr2+
Cr3+
Fe2+
Fe3+
Transition
metals
5A
6A
7A
3–
N
2–
O
F–
P3–
S2–
Cl–
Cu+
Cu2+
Zn2+
Br–
Ag+
Cd2+
I–
Figure 6.2
Charges of selected ions in the periodic table.
In contrast, the nonmetals form anions by gaining electrons. On the right side of the periodic
table in Figure 6.2, you can see that the atoms in Group 7A form (12) ions. The nonmetals in
Group 6A form (22) ions. It’s important to learn the charges on the ions shown in Figure 6.2 and
their relationship to the group number at the top of the column. For nontransition metals the charge
is equal to the group number. For nonmetals, the charge is equal to the group number minus 8. We
will learn more about why these ions carry their particular charges later in the course.
6.3 Writing Formulas from Names
of Ionic Compounds
Write the chemical formula for an ionic compound from the name of the
compound.
Learning objective
Name of
compound
Ions
Na+
Least common
multiple
Sum of charges
on ions
Formula
Sodium bromide
Na+, Br-
1
( +1) + ( -1) = 0
NaBr
Potassium sulfide
K+, S2-
2
2( +1) + ( -2) = 0
K2S
Zinc sulfate
Zn2+, SO24
2
( +2) + ( -2) = 0
ZnSO4
Ammonium phosphate
NH4+, PO34
3
3( +1) + ( -3) = 0
(NH4)3PO4
6
2( +3) + 3( - 2) = 0
Al2(CrO4)3
Aluminum chromate
3+
Al ,
CrO24
Ken Karp
In Chapters 3 and 5, we learned that compounds can be composed of ions. These substances
are called ionic compounds and will conduct electricity when dissolved in water. An excellent example of an ionic compound is ordinary table salt composed of crystals of sodium
chloride. When dissolved in water, sodium chloride conducts electricity very well, as shown
in Figure 6.3.
A chemical compound must have a net charge of zero. If it contains ions, the charges on
the ions must add up to zero in the formula for the compound. This is relatively easy in the
case of sodium chloride. The sodium ion (11) and the chloride ion (12) add to zero, resulting
in the formula NaCl. Now consider an ionic compound containing calcium (Ca2+) and fluoride (F-) ions. How can we write a formula with a net charge of zero? To do this we need one
Ca2+ and two F- ions. The correct formula is CaF2. The subscript 2 indicates that two fluoride
ions are needed for each calcium ion. Aluminum oxide is a bit more complicated because
it consists of Al3+ and O2- ions. Since 6 is the least common multiple of 3 and 2, we have
2(3+) + 3(2-) = 0, or a formula containing 2 Al3+ ions and 3 O2- ions for Al2O3. Here are
a few more examples of formula writing for ionic compounds:
Cl –
H2O
Figure 6.3
A solution of salt water contains
Na+ and Cl− ions in addition to
water molecules. The ions cause
the solution to conduct electricity,
lighting the bulb.
104 chapter 6
• Nomenclature of Inorganic Compounds
Rules for Writing Formulas for Ionic Compounds
1. Write the formula for the metal ion followed by the formula for the nonmetal ion.
2. Combine the smallest numbers of each ion needed to give the charge sum
equal to zero.
3. Write the formula for the compound as the symbol for the metal and nonmetal,
each followed by a subscript of the number determined in 2.
Example 6.2
ENHANCED EXAMPLE
Write formulas for (a) calcium chloride, (b) magnesium oxide, and (c) barium phosphide.
Solution
(a) Use the following steps for calcium chloride.
1. From the name, we know that calcium chloride is composed of calcium and chloride
ions. First write down the formulas of these ions:
Ca2+
Cl-
2. To write the formula of the compound, combine the smallest numbers of Ca2+ and Clions to give the charge sum equal to zero. In this case the lowest common multiple
of the charges is 2:
(Ca2+) + 2(Cl-) = 0
(2+) + 2(1-) = 0
3. Therefore, the formula is CaCl2.
(b) Use the same procedure for magnesium oxide:
1. From the name, we know that magnesium oxide is composed of magnesium and
oxide ions. First write down the formulas of these ions:
Mg2+
O2-
2. To write the formula of the compound, combine the smallest numbers of Mg2+ and
O2- ions to give the charge sum equal to zero:
(Mg2+) + (O2-) = 0
(2+) + (2-) = 0
3. The formula is MgO.
(c) Use the same procedure for barium phosphide:
1. From the name, we know that barium phosphide is composed of barium and phosphide ions. First write down the formulas of these ions:
Ba2+
P3-
2. To write the formula of this compound, combine the smallest numbers of Ba2+ and
P3- ions to give the charge sum equal to zero. In this case the lowest common multiple of the charges is 6:
3(Ba2+) + 2(P3-) = 0
3(2+) + 2(3-) = 0
3. The formula is Ba3P2.
Pr a c t i c e 6 . 2
Write formulas for compounds containing the following ions:
(a) K+ and F(d) Na+ and S22+
(b) Ca and Br
(e) Ba2+ and O22+
3(c) Mg and N
(f) As5+ and CO23
6.4
• Naming Binary Compounds 105
6.4 Naming Binary Compounds
Name binary ionic and nonionic compounds.
Learni ng objecti ve
Binary compounds contain only two different elements. Many binary compounds are
formed when a metal combines with a nonmetal to form a binary ionic compound. The metal
loses one or more electrons to become a cation while the nonmetal gains one or more electrons
to become an anion. The cation is written first in the formula, followed by the anion.
key terms
Binary Ionic Compounds Containing a Metal Forming
Only One Type of Cation
The chemical name is composed of the name of the metal followed by the name of the nonmetal, which has been modified to an identifying stem plus the suffix -ide.
For example, sodium chloride, NaCl, is composed of one atom each of sodium and chlorine. The name of the metal, sodium, is written first and is not modified. The second part of the
name is derived from the nonmetal, chlorine, by using the stem chlor- and adding the ending
-ide; it is named chloride. The compound name is sodium chloride.
NaCl
Elements:
Sodium (metal)
Chlorine (nonmetal)
name modified to the stem chlor- + -ide
Name of compound:
Sodium chloride
Stems of the more common negative-ion-forming elements are shown in Table 6.2.
Table 6.2
Examples of Elements Forming Anions
Symbol
Element
Stem
Anion name
Br
Cl
F
H
I
N
O
P
S
bromine
chlorine
fluorine
hydrogen
iodine
nitrogen
oxygen
phosphorus
sulfur
brom
chlor
fluor
hydr
iod
nitr
ox
phosph
sulf
bromide
chloride
fluoride
hydride
iodide
nitride
oxide
phosphide
sulfide
Rules for Naming Binary Ionic Compounds of Metal
Forming One Type of Cation
1. Write the name of the cation.
2. Write the stem for the anion and add the suffix -ide.
Compounds may contain more than one atom of the same element, but as long as they
contain only two different elements and only one compound of these two elements exists, the
name follows the rules for binary compounds:
Examples:
CaBr2
Mg3N2
Li2O
calcium bromide
magnesium nitride
lithium oxide
binary compounds
Stock System
106 chapter 6
• Nomenclature of Inorganic Compounds
Table 6.3 lists some compounds with names ending in -ide.
Table 6.3
Formula
AlCl3
Al2O3
CaC2
HCl
HI
Examples of Compounds with Names Ending in -ide
Name
Formula
aluminum chloride
aluminum oxide
calcium carbide
hydrogen chloride
hydrogen iodide
Name
BaS
LiI
MgBr2
NaH
Na2O
barium sulfide
lithium iodide
magnesium bromide
sodium hydride
sodium oxide
Example 6.3
ENHANCED EXAMPLE
F–
Ca2+
F–
Name the compound CaF2.
Solution
From the formula it is a two-element compound and follows the rules for binary
compounds.
1. The compound is composed of a metal, Ca, and a nonmetal, F. Elements in the 2A column of the periodic table form only one type of cation. Thus, we name the positive part
of the compound calcium.
2. Modify the name of the second element to the stem fluor- and add the binary ending -ide
to form the name of the negative part, fluoride.
Therefore, the name of the compound is calcium fluoride.
Practice 6.3
Write formulas for these compounds:
(a) strontium chloride
(b) potassium iodide
(c) aluminum nitride
(d) calcium sulfide
(e) sodium oxide
Binary Ionic Compounds Containing a Metal
That Can Form Two or More Types of Cations
The metals in the center of the periodic table (including the transition metals) often form more
than one type of cation. For example, iron can form Fe2+ and Fe3+ ions, and copper can form
Cu+ and Cu2+ ions. This can be confusing when you are naming compounds. For example,
copper chloride could be CuCl2 or CuCl. To resolve this difficulty the IUPAC devised a system,
known as the Stock System, to name these compounds. This system is currently recognized
as the official system to name these compounds, although another older system is sometimes
used. In the Stock System, when a compound contains a metal that can form more than one
type of cation, the charge on the cation of the metal is designated by a Roman numeral placed
in parentheses immediately following the name of the metal. The negative element is treated
in the usual manner for binary compounds:
The Roman numeral is part of the
cation name.
Cation charge
+1
+2
+3
+4
+5
Roman numeral
(I)
(II)
(III)
(IV)
(V)
Examples:
FeCl2
FeCl3
CuCl
CuCl2
iron(II) chloride
iron(III) chloride
copper(I) chloride
copper(II) chloride
Fe2+
Fe3+
Cu+
Cu2+
6.4
• Naming Binary Compounds 107
The fact that FeCl2 has two chloride ions, each with a 21 charge, establishes that the charge
on Fe is 12. To distinguish between the two iron chlorides, FeCl2 is named iron(II) chloride
and FeCl3 is named iron(III) chloride:
iron(II) chloride
FeCl2
�2
iron(II)
charge:
name:
�1
chloride
iron(III) chloride
FeCl3
�3
iron(III)
�1
chloride
When a metal forms only one possible cation, we need not distinguish one cation from another, so Roman numerals are not needed. Thus we do not say calcium(II) chloride for CaCl2,
but rather calcium chloride, since the charge of calcium is understood to be 12.
Rules for Naming Binary Ionic Compounds of Metal Forming
Two or More Types of Cations (Stock System)
1. Write the name of the cation.
2. Write the charge on the cation as a Roman numeral in parentheses.
3. Write the stem of the anion and add the suffix -ide.
In classical nomenclature, when the metallic ion has only two cation types, the name of
the metal (usually the Latin name) is modified with the suffixes -ous and -ic to distinguish
between the two. The lower-charge cation is given the -ous ending, and the higher one, the -ic
ending.
Examples:
FeCl2
FeCl3
CuCl
CuCl2
Fe2 +
Fe3 +
Cu +
Cu2 +
ferrous chloride
ferric chloride
cuprous chloride
cupric chloride
(lower-charge cation)
(higher-charge cation)
(lower-charge cation)
(higher-charge cation)
Table 6.4 lists some common metals that have more than one type of cation.
Table 6.4 Names and Charges of Some Common Metal Ions That Have More Than
One Type of Cation
Formula
Cu1+
Cu2+
Hg1+ (Hg2)2+
Hg2+
Fe2+
Fe3+
Sn2+
Stock System
name
copper(I)
copper(II)
mercury(I)
mercury(II)
iron(II)
iron(III)
tin(II)
Classical
name
Formula
cuprous
cupric
mercurous
mercuric
ferrous
ferric
stannous
Sn4+
Pb2+
Pb4+
As3+
As5+
Ti3+
Ti4+
Stock System
name
Classical
name
tin(IV)
lead(II)
lead(IV)
arsenic(III)
arsenic(V)
titanium(III)
titanium(IV)
stannic
plumbous
plumbic
arsenous
arsenic
titanous
titanic
Notice that the ous–ic naming system does not give the charge of the cation of an element
but merely indicates that at least two types of cations exist. The Stock System avoids any possible uncertainty by clearly stating the charge on the cation.
In this book we will use mainly
the Stock System.
108 chapter 6
• Nomenclature of Inorganic Compounds
Example 6.4
Name the compound FeS.
S2–
Fe2+
Solution
This compound follows the rules for a binary compound.
1, 2. It is a compound of Fe, a metal, and S, a nonmetal, and Fe is a transition metal that
has more than one type of cation. In sulfides, the charge on the S is 22. Therefore, the
charge on Fe must be 12, and the name of the positive part of the compound is iron(II).
3. We have already determined that the name of the negative part of the compound will
be sulfide.
The name of FeS is iron(II) sulfide.
P r a ct i c e 6 . 4
Write the name for each of the following compounds using the Stock System:
(a) PbI2
(b) SnF4
(c) Fe2O3
(d) CuO
P r a ct i c e 6 . 5
Write formulas for the following compounds:
(c) tin(II) fluoride
(a) tin(IV) chromate
(b) chromium(III) bromide
(d) copper(I) oxide
Prefix
Number
mono 1
di 2
tri 3
tetra 4
penta 5
hexa 6
hepta 7
octa 8
nona 9
deca
10
Binary Compounds Containing Two Nonmetals
Compounds between nonmetals are molecular, not ionic. Therefore, a different system for
naming them is used. In a compound formed between two nonmetals, the element that occurs
first in this series is written and named first:
Si, B, P, H, C, S, I, Br, N, Cl, O, F
The name of the second element retains the -ide ending as though it were an anion. A Latin or
Greek prefix (mono-, di-, tri-, and so on) is attached to the name of each element to indicate
the number of atoms of that element in the molecule. The prefix mono- is rarely used for naming the first element. Some common prefixes and their numerical equivalences are as shown
in the margin table.
Rules for Naming Binary Compounds Containing Two Nonmetals
1. Write the name for the first element using a prefix if there is more than one
atom of this element.
2. Write the stem of the second element and add the suffix -ide. Use a prefix to
indicate the number of atoms for the second element.
Here are some examples of compounds that illustrate this system:
N2O3
CO
CO2
PCl3
SO2
P2O5
CCl4
di nitrogen
tri oxide
Indicates two
nitrogen atoms
Indicates three
oxygen atoms
carbon monoxide
carbon dioxide
phosphorus trichloride
sulfur dioxide
diphosphorus pentoxide
carbon tetrachloride
N2O
N2O4
NO
N2O3
S2Cl2
S2F10
dinitrogen monoxide
dinitrogen tetroxide
nitrogen monoxide
dinitrogen trioxide
disulfur dichloride
disulfur decafluoride
6.5
• Naming Compounds Containing Polyatomic Ions 109
These examples illustrate that we sometimes drop the final o (mono) or a (penta) of the prefix
when the second element is oxygen. This avoids creating a name that is awkward to pronounce.
For example, CO is carbon monoxide instead of carbon monooxide.
Example 6.5
Name the compound PCl5.
Solution
1. Phosphorus and chlorine are nonmetals, so the rules for naming binary compounds
containing two nonmetals apply. Phosphorus is named first. Therefore, the compound
is a chloride.
2. No prefix is needed for phosphorus because each molecule has only one atom of phosphorus. The prefix penta- is used with chloride to indicate the five chlorine atoms.
(PCl3 is also a known compound.)
The name for PCl5 is phosphorus pentachloride.
Phosphorus pentachloride
ENHANCED EXAMPLE
Practice 6.6
Name these compounds:
(a) Cl2O
(b) SO2 (c) CBr4
(d) N2O5
(e) NH3
(d) SnF2
(e) CuCl2
(f) ICl3
Practice 6.7
Name these compounds:
(a) KBr (b) Ca3N2
(c) SO3
(f) N2O4
6.5 Naming Compounds Containing
Polyatomic Ions
Recognize names, formulas, and charges of polyatomic ions, name compounds containing polyatomic ions, and write formulas from names of compounds containing
polyatomic ions.
Learning objective
A polyatomic ion is an ion that contains two or more elements. Compounds containing polyatomic ions are composed of three or more elements and usually consist of one or more
cations combined with a negative polyatomic ion. In general, naming compounds containing
polyatomic ions is similar to naming binary compounds. The cation is named first, followed
by the name for the negative polyatomic ion.
To name these compounds, you must learn to recognize the common polyatomic ions
(Table 6.5) and know their charges. Consider the formula KMnO4. You must be able to recognize that it consists of two parts KMnO4. These parts are composed of a K+ ion and a MnO4-
key term
Table 6.5
polyatomic ion
Names, Formulas, and Charges of Some Common Polyatomic Ions
Name
Formula
Acetate
Ammonium
Arsenate
Hydrogen carbonate
Hydrogen sulfate
Bromate
Carbonate
Chlorate
Chromate
C2H3O2NH4+
AsO34
HCO3HSO4BrO3CO23
ClO3CrO24
Charge
-1
+1
-3
-1
-1
-1
-2
-1
-2
Name
Cyanide
Dichromate
Hydroxide
Nitrate
Nitrite
Permanganate
Phosphate
Sulfate
Sulfite
Formula
-
CN
Cr2O27
OHNO3NO2MnO4PO34
SO24
SO23
Charge
-1
-2
-1
-1
-1
-1
-3
-2
-2
More ions are listed on the back
endpapers.
110 chapter 6
• Nomenclature of Inorganic Compounds
Ken Karp
ion. The correct name for this compound is potassium permanganate. Many polyatomic ions
that contain oxygen are called oxy-anions and generally have the suffix -ate or -ite. Unfortunately, the suffix doesn’t indicate the number of oxygen atoms present. The -ate form contains
2more oxygen atoms than the -ite form. Examples include sulfate (SO24 ), sulfite (SO3 ), nitrate
(NO3 ), and nitrite (NO2 ).
Rules for Naming Compounds Containing Polyatomic Ions
1. Write the name of the cation.
2. Write the name of the anion.
Potassium permanganate
crystals are dark purple.
Only four of the common negatively charged polyatomic ions do not use the ate–ite system. These exceptions are hydroxide (OH−) hydrogen sulfide (HS−) peroxide (O22 ), and cyanide (CN−). Care must be taken with these, as their endings can easily be confused with the
-ide ending for binary compounds (Section 6.4).
There are three common positively charged polyatomic ions as well—the ammonium, the
+
+
mercury(I) (Hg2+
2 ), and the hydronium (H3O ) ions. The ammonium ion (NH4 ) is frequently
+
found in polyatomic compounds, whereas the hydronium ion (H3O ) is usually associated with
aqueous solutions of acids (Chapter 15).
EXAMPLE 6.6
ENHANCED EXAMPLE
What is the minimum number of different elements in a compound that contains a polyatomic ion?
SOLUTION
Most polyatomic ions are anions and contain two different elements. Adding a cation to a
polyatomic ion to form a compound means the minimum is three different elements in a
polyatomic compound.
Practice 6.8
Name these compounds:
(a) NaNO3
(b) Ca3(PO4)2
Table 6.6 Names of Selected Compounds
That Contain More Than One Kind
of Positive Ion
Formula
Name of compound
KHSO4
Ca(HSO3)2
NH4HS
MgNH4PO4
NaH2PO4
Na2HPO4
KHC2O4
KAl(SO4)2
Al(HCO3)3
potassium hydrogen sulfate
calcium hydrogen sulfite
ammonium hydrogen sulfide
magnesium ammonium phosphate
sodium dihydrogen phosphate
sodium hydrogen phosphate
potassium hydrogen oxalate
potassium aluminum sulfate
aluminum hydrogen carbonate
(c) KOH
(d) Li2CO3
(e) NaClO3
Inorganic compounds are also formed from more than three elements
(see Table 6.6). In these cases, one or more of the ions is often a polyatomic
ion. Once you have learned to recognize the polyatomic ions, naming these
compounds follows the patterns we have already learned. First identify the
ions. Name the cations in the order given and follow them with the names of
the anions. Study the following examples:
Compound
NaHCO3
NaHS
MgNH4PO4
NaKSO4
Ions
Na ; HCO3+
-
Name
+
Na ; HS
Mg2+; NH4+; PO34
Na+; K+; SO24
sodium hydrogen carbonate
sodium hydrogen sulfide
magnesium ammonium phosphate
sodium potassium sulfate
Practice 6.9
Name these compounds:
(a) KHCO3 (b) NaHC2O4
(c) BaNH4PO4
(d) NaAl(SO4)2
6.6
• Acids 111
6.6Acids
Use the rules to name an acid from its formula and to write the formula for an acid
from its name.
Learning objectiv e
While we will learn much more about acids later (see Chapter 15), it is helpful to be able to
recognize and name common acids both in the laboratory and in class. The simplest way to
recognize many acids is to know that acid formulas often begin with hydrogen.
Binary Acids
Certain binary hydrogen compounds, when dissolved in water, form solutions that have
acid properties. Because of this property, these compounds are given acid names in addition
to their regular -ide names. For example, HCl is a gas and is called hydrogen chloride, but
its water solution is known as hydrochloric acid. Binary acids are composed of hydrogen
and one other nonmetallic element. However, not all binary hydrogen compounds are acids.
To express the formula of a binary acid, it’s customary to write the symbol of hydrogen
first, followed by the symbol of the second element (e.g., HCl, HBr, or H 2S). When we
see formulas such as CH4 or NH3 , we understand that these compounds are not normally
considered to be acids.
To name a binary acid, place the prefix hydro- in front of, and the suffix -ic after, the stem
of the nonmetal name. Then add the word acid:
Examples:
HCl
Hydro-chlor-ic acid
H2S
Hydro-sulfur-ic acid
(hydrochloric acid)
(hydrosulfuric acid)
Rules for Naming Binary acids
1. Write the prefix hydro- followed by the stem of the second element and add
the suffix -ic.
2. Add the word acid.
Example 6.7
Name the acid HI.
SOLUTION
The compound follows the Rules for Naming Binary Acids.
1. Write the prefix hydro- and the stem of the second element with the suffix -ic: hydroiodic.
2. Add the word acid: hydroiodic acid.
P r a ct i c e 6 . 1 0
Name these binary acids:
(a) HCl (b) HBr
Acids are hydrogen-containing substances that liberate hydrogen ions when dissolved in
water. The same formula is often used to express binary hydrogen compounds, such as HCl,
regardless of whether or not they are dissolved in water. Table 6.7 shows several examples
of binary acids.
Table 6.7 Names
and Formulas of
Selected Binary Acids
Formula Acid name
HF
HCl
HBr
HI
H2S
H2Se
Hydrofluoric acid
Hydrochloric acid
Hydrobromic acid
Hydroiodic acid
Hydrosulfuric acid
Hydroselenic acid
112 chapter 6
• Nomenclature of Inorganic Compounds
Figure 6.4 summarizes how to name binary compounds.
Binary compounds
Usually end in -ide
Two nonmetals
Metal/nonmetal
Use prefixes to
indicate number
of atoms for
each element
Metal with
one type
of cation
Metal with
varying types
of cations
1. Name metal
2. Stem name of
nonmetal, -ide
Determine the
charge of
the cation
1. Choose
appropriate
-ous or -ic
ending on metal
2. Stem name of
nonmetal, -ide
Figure 6.4
Flow diagram for naming binary
compounds.
1. Use Roman
numeral after
cation name
2. Stem name of
nonmetal, -ide
Naming Oxy-Acids
Inorganic compounds containing hydrogen, oxygen, and one other element are called oxyacids. The first step in naming these acids is to determine that the compound in question
is really an oxy-acid. The keys to identification are (1) hydrogen is the first element in the
compound’s formula and (2) the second part of the formula consists of a polyatomic ion
containing oxygen.
Hydrogen in an oxy-acid is not specifically designated in the acid name. The presence
of hydrogen in the compound is indicated by the use of the word acid in the name of the substance. To determine the particular type of acid, the polyatomic ion following hydrogen must
be examined. The name of the polyatomic ion is modified in the following manner: (1) -ate
changes to an -ic ending; (2) -ite changes to an -ous ending. (See Table 6.8.) The compound
with the -ic ending contains more oxygen than the one with the -ous ending.
Table 6.8
Comparison of Acid and Anion Names for Selected Oxy-Acids
Acid
Anion
H2SO4
Sulfuric acid
H2SO3
Sulfurous acid
HNO3
Nitric acid
HNO2
Nitrous acid
SO24
Sulfate ion
SO23
Sulfite ion
NO3Nitrate ion
NO2Nitrite ion
Acid
Anion
Acid
Anion
H2CO3
Carbonic acid
H3BO3
Boric acid
H3PO4
Phosphoric acid
H3PO3
Phosphorous acid
CO23
HIO3
Iodic acid
HC2H3O2
Acetic acid
H2C2O4
Oxalic acid
HBrO3
Bromic acid
IO3Iodate ion
C2H3O2Acetate ion
C2O24
Oxalate ion
BrO3Bromate ion
Carbonate ion
BO33
Borate ion
PO34
Phosphate ion
PO33
Phosphite ion
6.6
• Acids 113
Example 6.8
Name the acids H2SO4 and H2SO3.
SOLUTION
The names are written by modifying the name of the polyatomic ion in the acid.
For H2SO4 the polyatomic ion is the SO24 - sulfate ion. The -ate ending changes to an -ic
ending in the acid. The stem of the sulfate ion is sulfur so the name for the compound is
sulfuric and the word acid is added to show the presence of hydrogen in the compound:
sulfuric acid.
For H2SO3 the polyatomic ion is the SO23 - sulfite ion. The -ite ending changes to an
-ous ending in the acid, sulfurous, and the word acid is added to show the presence of
hydrogen in the compound: sulfurous acid.
P r a ct i c e 6 . 1 1
Name these polyatomic acids:
(a) H2CO3
(b) HC2H3O2 (c) HNO3
(d) HNO2
Some elements form more than two different polyatomic ions containing oxygen. To name
these ions, prefixes are used in addition to a suffix. To indicate more oxygen than in the -ate
form we add the prefix per-, which is a short form of hyper- meaning “more.” The prefix
hypo-, meaning “less” (oxygen in this case), is used for the ion containing less oxygen than
the -ite form. An example of this system is shown for the oxy-acids containing chlorine and
oxygen in Table 6.9. The prefixes are also used with other similar ions, such as iodate (IO3- ),
bromate (BrO3- ).
We have now looked at ways of naming a variety of inorganic compounds—binary compounds consisting of a metal and a nonmetal and of two nonmetals, acids, and polyatomic
compounds. (See Figure 6.5.) These compounds are just a small part of the classified chemical
compounds. Most of the remaining classes are in the broad field of organic chemistry under such
categories as hydrocarbons, alcohols, ethers, aldehydes, ketones, phenols, and carboxylic acids.
Table 6.9
Oxy-Acids of Chlorine
Acid formula
Anion formula
ClO ClO2ClO3ClO4-
HClO
HClO2
HClO3
HClO4
Anion name Acid name
hypochlorite
chlorite
chlorate
perchlorate
hypochlorous acid
chlorous acid
chloric acid
perchloric acid
Figure 6.5
Acid
(begins with H)
Flow diagram for naming acids.
Anion contains oxygen
Anion does not contains oxygen
Check ending on anion
1. Use prefix hydro- and suffix -ic
2. Add word acid
-ite
Change anion
name to end
in -ous acid
-ate
Change anion
name to end
in -ic acid
114 chapter 6
• Nomenclature of Inorganic Compounds
C h a p t e r
6 review
6.1 Common and Systematic Names
• Common names are arbitrary and do not describe the chemical composition of compounds.
• Examples—ammonia, water
• A standard system of nomenclature was devised and is maintained by the IUPAC.
6.2 Elements and Ions
•
•
•
•
Diatomic elements—H2, O2, N2, F2, Cl2, Br2, I2.
Polyatomic elements—P4, S8.
Ions form through the gain or loss of electrons from neutral atoms.
The charge on ions can often be predicted from the periodic table.
6.3 Writing Formulas from Names of Ionic Compounds
• Compounds must have a net charge of zero.
• To write a formula from a name:
• Identify and write the symbols for the elements in the compound.
• Combine the smallest number of ions required to produce a net charge of zero.
6.4 Naming Binary Compounds
Key Terms
• The rules for naming binary compounds can be summarized in the following chart.
binary compounds
Stock System
Binary compounds
Usually end in -ide
Two nonmetals
Use prefixes to
indicate number
of atoms for
each element
Metal/nonmetal
Metal with
one type
of cation
Metal with
varying types
of cations
1. Name metal
2. Stem name of
nonmetal, -ide
Determine the
charge of
the cation
1. Choose
appropriate
-ous or -ic
ending on metal
2. Stem name of
nonmetal, -ide
1. Use Roman
numeral after
cation name
2. Stem name of
nonmetal, -ide
Review Questions 115
6.5 Naming Compounds Containing Polyatomic Ions
Key Term
• The rules for naming these compounds:
1. Write the name of the cation.
2. Write the name of the anion.
polyatomic ion
6.6 Acids
• The rules for naming acids can be summarized in the chart that follows:
Acid
(begins with H)
Anion contains oxygen
Anion does not contains oxygen
Check ending on anion
1. Use prefix hydro- and suffix -ic
2. Add word acid
-ite
Change anion
name to end
in -ous acid
-ate
Change anion
name to end
in -ic acid
review Questions
1. List two compounds almost always referred to by their common
names. What would be their systematic names?
2. What must happen in order to convert an atom into an anion? Into
a cation?
3. Does the fact that two elements combine in a one-to-one
atomic ratio mean that the charges on their ions are both 1?
Explain.
4. When naming a binary ionic compound, how do you determine
which element to name first?
5. Use the common ion table on the back endpapers of your text to
determine the formulas for compounds composed of the following
ions:
(a) potassium and sulfide
(b) cobalt(II) and bromate
(c) ammonium and nitrate
(d) hydrogen and phosphate
(e) iron(III) oxide
(f) magnesium and hydroxide
6. Explain why P2O5 is named dinitrogen pentoxide, using prefixes,
but Al2O3 is named aluminum oxide, without prefixes.
7. Explain why FeCl2 is named iron(II) chloride, using a Roman
numeral, and BaCl2 is named barium chloride, using no Roman
numeral. Why does neither use prefixes?
8. The names of the nitrogen oxides are nitrogen dioxide (NO2) and
dinitrogen monoxide (N2O). Why is the mono- prefix used only on
the second molecule?
9. What is the advantage of using the Stock System for naming metals
with variable charge instead of using the common name?
10. Write formulas for the compounds formed when a nickel(II) ion is
combined with
(a) sulfate
(f) acetate
(b) phosphide
(g) dichromate
(c) chromate
(h) bromide
(d) hydroxide
(i) nitrate
(e) iodite
(j) hypochlorite
(Use the common ion table on the back endpapers of your text.)
11. Write the names and formulas for the four oxy-acids containing
(a) bromine, (b) iodine.
12. How do you differentiate between different anions formed from a
polyprotic acid?
116 chapter 6
• Nomenclature of Inorganic Compounds
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VII.
P AIRED E x e r c i s e s
1. Write the formula of the compound that will be formed between
these elements:
(a) Ba and S
(d) Mg and N
(b) Cs and P
(e) Ca and I
(c) Li and Br
(f) H and Cl
2. Write the formula of the compound that will be formed between
these elements:
(a) Al and S
(d) Sr and O
(b) H and F
(e) Cs and P
(c) K and N
(f) Al and Cl
3. Write formulas for the following cations:
(a) potassium
(h) calcium
(b) ammonium
(i) lead(II)
(c) copper(I)
(j) zinc
(d) titanium(IV)
(k) silver
(e) nickel(III)
(l) hydrogen
(f) cesium
(m) tin(II)
(g) mercury(II)
(n) iron(III)
4. Write formulas for the following anions:
(a) fluoride
(h) oxide
(b) acetate
(i) dichromate
(c) iodide
(j) hydrogen carbonate
(d) carbonate
(k) phosphate
(e) sulfide
(l) sulfate
(f) nitrate
(m) nitride
(g) phosphide
(n) chloride
5. Write the systematic names for the following:
(a) baking soda (NaHCO3)
(d) vinegar (HC2H3O2)
(b) quicksilver (Hg)
(e) Epsom salts (MgSO4 # 7 H2O)
(c) lime (CaO)
(f) lye (NaOH)
6. Write the systematic names for the following:
(a) hypo (Na2S2O3)
(d) table salt (NaCl)
(b) laughing gas (N2O)
(e) milk of magnesia (Mg(OH)2)
(c) alumina (Al2O3)
(f) galena (PbS)
7. Complete the table, filling in each box with the proper formula.
8. Complete the table, filling in each box with the proper formula.
Anions
Anions
Br
O
2-
NO 3-
PO34
SO24
CO23
KBr
2+
Mg Al3+ Zn2+ Zn3(PO4)2
+
H Cations
Cations
K
+
-
OH
NH4+ 2+
-
AsO34
C2H3O2-
CrO24
(NH4)3AsO4
Ca Fe3+ Fe2(SO4)3
Ag+ Cu2+ 9. Write the names of each of the compounds formed in Question 7.
10. Write the names of each of the compounds formed in Question 8.
11. Write formulas for each of the following binary compounds, all of
which are composed of nonmetals:
(a) diphosphorus pentoxide
(e) carbon tetrachloride
(b) carbon dioxide
(f) dichlorine heptoxide
(c) tribromine octoxide
(g) boron trifluoride
(d) sulfur hexachloride
(h) tetranitrogen hexasulfide
12. Write formulas for each of the following binary compounds, all of
which are composed of a metal and nonmetal:
(a) potassium nitride
(e) calcium nitride
(b) barium oxide
(f) cesium bromide
(c) iron(II) oxide
(g) manganese(III) iodide
(d) strontium phosphide
(h) sodium selenide
13. Name each of the following binary compounds, all of which are
composed of a metal and a nonmetal:
(a) BaO
(e) Al2O3
(b) K2S
(f) CaBr2
(c) CaCl2
(g) SrI2
(d) Cs2S
(h) Mg3N2
14. Name each of the following binary compounds, all of which are
composed of nonmetals:
(a) PBr5
(e) SiCl4
(b) I4O9
(f) ClO2
(c) N2S5
(g) P4S7
(d) S2F10
(h) IF6
15. Name these compounds by the Stock System (IUPAC):
(a) CuCl2
(d) FeCl3
(b) FeCl2
(e) SnF2
(c) Fe(NO3)2
(f) VPO4
16. Write formulas for these compounds:
(a) tin(IV) bromide
(d) mercury(II) nitrite
(b) copper(I) sulfate
(e) cobalt(III) carbonate
(c) nickel(II) borate
(f) iron(II) acetate
17. Write formulas for these acids:
(a) hydrochloric acid
(d) carbonic acid
(b) chloric acid
(e) sulfurous acid
(c) nitric acid
(f) phosphoric acid
18. Write formulas for these acids:
(a) acetic acid
(d) boric acid
(b) hydrofluoric acid
(e) nitrous acid
(c) hydrosulfuric acid
(f) hypochlorous acid
Additional Exercises 117
19. Name these acids:
(a) HNO2
(b) H2SO4
(c) H2C2O4
(d) HBr
(e) H3PO3
(f) HC2H3O2
20. Name these acids:
(a) H3PO4
(b) H2CO3
(c) HIO3
(g) HF
(h) HBrO3
(i) HIO4
(d) HCl
(e) HClO
(f) HNO3
(g) HI
(h) HClO4
(i) H2SO3
21. Write formulas for these compounds:
(a) silver sulfite
(b) cobalt(II) bromide
(c) tin(II) hydroxide
(d) aluminum sulfate
(e) lead(II) chloride
(f) ammonium carbonate
(g) chromium(III) oxide
(h) copper(II) chloride
(i) potassium permanganate
(j) arsenic(V) sulfite
(k) sodium peroxide
(l) iron(II) sulfate
(m) potassium dichromate
(n) bismuth(III) chromate
22. Write formulas for these compounds:
(a) sodium chromate
(b) magnesium hydride
(c) nickel(II) acetate
(d) calcium chlorate
(e) magnesium bromate
(f) potassium dihydrogen phosphate
(g) manganese(II) hydroxide
(h) cobalt(II) hydrogen carbonate
(i) sodium hypochlorite
(j) barium perchlorate
(k) chromium(III) sulfite
(l) antimony(III) sulfate
(m) sodium oxalate
(n) potassium thiocyanate
23. Write the name of each compound:
(a) ZnSO4
(f) CoF2
(b) Hg2S
(g) Cr(ClO3)3
(c) CuCO3
(h) Ag3PO4
(d) Cd(NO3)2
(i) MnS
(e) Al(C2H3O2)3
(j) BaCrO4
24. Write the name of each compound:
(a) Ca(HSO4)2
(f) BiAsO4
(b) As2(SO3)3
(g) (NH4)2CO3
(c) Sn(NO2)2
(h) (NH4)2HPO4
(d) CuI
(i) NaClO
(e) KHCO3
(j) KMnO4
25. Write the chemical formulas for these substances:
(a) slaked lime
(e) cane sugar
(b) fool’s gold
(f) borax
(c) washing soda
(g) wood alcohol
(d) calcite
(h) acetylene
26. Write the chemical formulas for these substances:
(a) grain alcohol
(e) muriatic acid
(b) cream of tartar
(f) plaster of paris
(c) gypsum
(g) lye
(d) brimstone
(h) laughing gas
ADDI T IO N AL EXER C ISES
27. Write equations similar to those found in Section 6.2 for the formation of
(a) potassium ion
(d) iron(II) ion
(b) iodide ion
(e) calcium ion
(c) bromide ion
(f) oxide ion
28. Name each of the following polyatomic ions:
(a)
2–
=S
(b)
=P
(d)
=C
=O
(e)
=Br
=Cl
(b)
–
=O
=H
=O
(c)
(a)
–
=O
3–
31. Write the formula and name for each of the following compounds:
=B
=F
2–
–
=N
=O
(f)
=C
=O
29. Write formulas for all possible compounds formed between the
calcium ion and the anions shown in Question 28.
30. Write formulas for all possible compounds formed between the
potassium ion and the anions shown in Question 28.
(c)
=P
=Cl
118 chapter 6
• Nomenclature of Inorganic Compounds
32. Write the formula and name for the following compound:
K+
Cl –
33. State how each of the following is used in naming inorganic compounds: ide, ous, ic, hypo, per, ite, ate, Roman numerals.
34. Several excerpts from newspaper articles follow. Tell the chemical formula of the chemical spilled and indicate whether this spill
should raise concern in the local area.
(a) A mistake at the Columbia Water and Light Department’s West
Ash Street pumping station sent an estimated 7200 gallons of
calcium-carbonate-laden water into Harmony Creek early yesterday, forcing crews to spend hours cleaning up the mess.
(b) A tanker truck carrying 6000 gallons of acetic acid overturned
on the I-5 freeway in San Diego, California.
(c) A university in New Hampshire was discovered dumping dihydrogen oxide into the local sewer system.
35. How many of each type of subatomic particle (protons and electrons) is in
(a) an atom of tin?
(b) a Sn2+ ion?
(c) a Sn4+ ion?
36. The compound X2 Y3 is a stable solid. What ionic charge do you
expect for X and Y? Explain.
37. The ferricyanide ion has the formula Fe(CN)36 . Write the formula
for the compounds that ferricyanide would form with the cations of
elements 3, 13, and 30.
38. Compare and contrast the formulas of
(a) nitride with nitrite
(b) nitrite with nitrate
(c) nitrous acid with nitric acid
CHALLENGE Exercise
39. After studying chemistry, you should be able to recognize more of
the substances listed on consumer products. A list of ingredients for
dog food follows:
Acid, Vitamin A Acetate, Calcium Pantothenate, Biotin, Thiamine
Mononitrate (source of Vitamin B1), Vitamin B12 Supplement,
Niacin, Riboflavin Supplement (source of Vitamin B2), Inositol,
Pyridoxine Hydrochloride (source of Vitamin B6), Vitamin D3
Supplement, Folic Acid), Beta-Carotene, L-Carnitine, Marigold,
Citric Acid, Rosemary Extract.
Chicken By-Product Meal (Natural source of Chondroitin
Sulfate and Glucosamine), Corn Meal, Ground Whole Grain
Sorghum, Ground Whole Grain Barley, Fish Meal (source of fish
oil), Chicken, Chicken Fat (preserved with mixed Tocopherols,
a source of Vitamin E), Dried Beet Pulp, Chicken Flavor, Dried
Egg Product, Potassium Chloride, Brewers Dried Yeast, Salt, Sodium Hexametaphosphate, Fructooligosaccharides, Fish Oil (preserved with mixed Tocopherols, a source of Vitamin E), Calcium
Carbonate, Flax Meal, Choline Chloride, Minerals (Ferrous Sulfate, Zinc Oxide, Manganese Sulfate, Copper Sulfate, Manganous
Oxide, Potassium Iodide, Cobalt Carbonate), Vitamin E Supplement, Dried Chicken Cartilage (Natural source of Chondroitin
Sulfate and Glucosamine), DL-Methionine, Vitamins (Ascorbic
Many of the substances in this bag of dog food are ionic compounds that you should be able to recognize. The manufacturers of
this dog food did not completely identify some of the compounds.
Write the chemical formula of the following compounds found in
this ingredient list.
(a) potassium chloride
(f) copper sulfate
(b) calcium carbonate
(g) manganous oxide
(c) ferrous sulfate
(h) potassium iodide
(d) zinc oxide
(i) cobalt carbonate
(e) manganese sulfate
(j) sodium chloride
A n s w e r s t o P r ac t i ce E x e r c i s e s
6.1
phosphorus stem is phosph- 1 -ide 5 phosphide; nitrogen stem
is nitr- 1 -ide 5 nitride; boron stem is bor- 1 -ide 5 boride;
carbon stem is carb- 1 -ide 5 carbide
6.7
( a) potassium bromide; (b) calcium nitride;
(c) sulfur trioxide; (d) tin(II) fluoride; (e) copper(II) chloride;
(f) dinitrogen tetroxide
6.2
(a) KF; (b) CaBr2; (c) Mg3N2; (d) Na2S; (e) BaO; (f) As2(CO3)5
6.8
6.3
(a) SrCl2; (b) KI; (c) AlN; (d) CaS; (e) Na2O
( a) sodium nitrate; (b) calcium phosphate; (c) potassium
­hydroxide; (d) lithium carbonate; (e) sodium chlorate
6.4
(a) lead(II) iodide; (b) tin(IV) fluoride; (c) iron(III) oxide;
(d) copper(II) oxide
6.9
( a) potassium hydrogen carbonate; (b) sodium hydrogen oxalate; (c) barium ammonium phosphate; (d) sodium aluminum
sulfate
6.5
(a) Sn(CrO4)2; (b) CrBr3; (c) SnF2; (d) Cu2O
6.6
(a) dichlorine monoxide; (b) sulfur dioxide; (c) carbon tetrabromide; (d) dinitrogen pentoxide; (e) ammonia; (f) iodine
trichloride
6.10(a) hydrochloric acid; (b) hydrobromic acid
6.11(a) carbonic acid; (b) acetic acid; (c) nitric acid;
(d) nitrous acid
PUTTING IT TOGETHER:
Review for Chapters 5–6
Multiple Choice
Choose the correct answer to each of the following.
1. The concept of positive charge and a small, “heavy” nucleus surrounded by electrons was the contribution of
(a) Dalton
(c) Thomson
(b) Rutherford
(d) Chadwick
2. The neutron was discovered in 1932 by
(a) Dalton
(c) Thomson
(b) Rutherford
(d) Chadwick
3. An atom of atomic number 53 and mass number 127 contains how
many neutrons?
(a) 53
(c) 127
(b) 74
(d) 180
40
18Ar?
4. How many electrons are in an atom of
(a) 20
(c) 40
(b) 22
(d) no correct answer given
5. The number of neutrons in an atom of 139
56Ba is
(a) 56
(c) 139
(b) 83
(d) no correct answer given
6. The name of the isotope containing one proton and two neutrons is
(a) protium
(c) deuterium
(b) tritium
(d) helium
7. Each atom of a specific element has the same
(a) number of protons
(c) number of neutrons
(b) atomic mass
(d) no correct answer given
8. Which pair of symbols represents isotopes?
23
29
(c) 63
(a) 23
11Na and 12Na
29Cu and 64Cu
7
6
12
(b) 3Li and 3Li
(d) 24Mg and 12
26Mg
9. Two naturally occurring isotopes of an element have masses
and abundance as follows: 54.00 amu (20.00%) and 56.00 amu
(80.00%). What is the relative atomic mass of the element?
(a) 54.20
(c) 54.80
(b) 54.40
(d) 55.60
10. Substance X has 13 protons, 14 neutrons, and 10 electrons. Determine its identity.
(c) 27Al3+
(a) 27Mg
27
(b) Ne
(d) 27Al
11. The mass of a chlorine atom is 5.90 * 10-23g. How many atoms
are in a 42.0-g sample of chlorine?
(a) 2.48 * 10-21
(c) 1.40 * 10-24
23
(b) 7.12 * 10
(d) no correct answer given
12. The number of neutrons in an atom of 108
47Ag is
(a) 47
(c) 155
(b) 108
(d) no correct answer given
13. The number of electrons in an atom of 27
13Al is
(a) 13
(c) 27
(b) 14
(d) 40
14. The number of protons in an atom of 65
30Zn is
(a) 65
(c) 30
(b) 35
(d) 95
15. The number of electrons in the nucleus of an atom of 24
12Mg is
(a) 12
(c) 36
(b) 24
(d) no correct answer given
Names and Formulas
In which of the following is the formula correct for the name given?
1. copper(II) sulfate, CuSO4
2. ammonium hydroxide, NH4OH
3. mercury(I) carbonate, HgCO3
4. phosphorus triiodide, PI3
5. calcium acetate, Ca(C2H3O2)2
6. hypochlorous acid, HClO
7. dichlorine heptoxide, Cl2O7
8. magnesium iodide, MgI
9. sulfurous acid, H2SO3
10. potassium manganate, KMnO4
11. lead(II) chromate, PbCrO4
12. ammonium hydrogen carbonate, NH4HCO3
13. iron(II) phosphate, FePO4
14. calcium hydrogen sulfate, CaHSO4
15. mercury(II) sulfate, HgSO4
16. dinitrogen pentoxide, N2O5
17. sodium hypochlorite, NaClO
18. sodium dichromate, Na2Cr2O7
19. cadmium cyanide, Cd(CN)2
20. bismuth(III) oxide, Bi3O2
21. carbonic acid, H2CO3
22. silver oxide, Ag2O
23. ferric iodide, FeI2
24. tin(II) fluoride, TiF2
25. carbon monoxide, CO
26. phosphoric acid, H3PO3
119
120 Putting It Together
27. sodium bromate, Na2BrO3
28. hydrosulfuric acid, H2S
29. potassium hydroxide, POH
30. sodium carbonate, Na2CO3
31. zinc sulfate, ZnSO3
32. sulfur trioxide, SO3
33. tin(IV) nitrate, Sn(NO3)4
34. ferrous sulfate, FeSO4
35. chloric acid, HCl
36. aluminum sulfide, Al2S3
37. cobalt(II) chloride, CoCl2
38. acetic acid, HC2H3O2
39. zinc oxide, ZnO2
40. stannous fluoride, SnF2
Free Response Questions
Answer each of the following. Be sure to include your work and
explanations in a clear, logical form.
1. (a) What is an ion?
(b) The average mass of a calcium atom is 40.08 amu. Why do we
also use 40.08 amu as the average mass of a calcium ion (Ca2+)?
2. Congratulations! You discover a new element you name wyzzlebium
(Wz). The average atomic mass of Wz was found to be 303.001 amu,
and its atomic number is 120.
(a) If the masses of the two isotopes of wyzzlebium are 300.9326
amu and 303.9303 amu, what is the relative abundance of each
isotope?
(b) What are the isotopic notations of the two isotopes? (e.g., A
Z Wz)
(c) How many neutrons are in one atom of the more abundant
isotope?
3. How many protons are in one molecule of dichlorine heptoxide? Is
it possible to determine precisely how many electrons and neutrons
are in a molecule of dichlorine heptoxide? Why or why not?
4. An unidentified metal forms an ionic compound with phosphate.
The metal forms a 2 + cation. If the minimum ratio of protons in the
metal to the phosphorus is 6:5, what metal is it? (Hint: First write
the formula for the ionic compound formed with phosphate anion.)
5. For each of the following compounds, indicate what is wrong with
the name and why. If possible, fix the name.
(a) iron hydroxide
(b) dipotassium dichromium heptoxide
(c) sulfur oxide
6. Sulfur dioxide is a gas formed as a by-product of burning coal.
Sulfur trioxide is a significant contributor to acid rain. Does the
existence of these two substances violate the law of multiple proportions? Explain.
7. (a) Which subatomic particles are not in the nucleus?
(b) What happens to the size of an atom when it becomes an anion?
(c) What do an ion of Ca and an atom of Ar have in common?
8. An unidentified atom is found to have an atomic mass 7.18 times
that of the carbon-12 isotope.
(a) What is the mass of the unidentified atom?
(b) What are the possible identities of this atom?
(c) Why are you unable to positively identify the element based on
the atomic mass and the periodic table?
(d) If the element formed a compound M2O, where M is the unidentified atom, identify M by writing the isotopic notation for
the atom.
9. Scientists such as Dalton, Thomson, and Rutherford proposed
important models, which were ultimately challenged by later
technology. What do we know to be false in Dalton’s atomic
model? What was missing in Thomson’s model of the atom?
What was Rutherford’s experiment that led to the current model
of the atom?
• Calculating Empirical Formulas 121
© Hemis/Alamy
7.4
These black pearls are made
of layers of calcium ­carbonate.
They can be measured by
­counting or weighing.
ch a p te r
7
Quantitative Composition
of Compounds
C
ereals, cleaning products, and pain remedies all list their
ingredients on the package label. The ingredients are
­listed in order from most to least, but the amounts are
rarely given. However, it is precisely these amounts that give products
their desired properties and distinguish them from the competition.
Understandably, manufacturers carefully regulate the amounts of ingredients to maintain quality and hopefully their customers’ loyalty. In
the medicines we purchase, these quantities are especially important
for safety reasons—for ­example, they determine whether a medicine
is given to ­children or is safe only for adults.
The composition of compounds is an important concept in
chemistry. ­Determining numerical relationships among the elements
in compounds and measuring exact quantities of particles are fundamental tasks that chemists routinely perform in their daily work.
Chapter Outline
7.1 The Mole
7.2 Molar Mass of Compounds
7.3 Percent Composition
of ­Compounds
7.4 Calculating Empirical Formulas
7.5 Calculating the Molecular Formula
from the Empirical Formula
122 chapter 7
• Quantitative Composition of Compounds
7.1 The Mole
L earning obje ctive
Apply the concepts of the mole, molar mass, and Avogadro’s number to solve
­chemistry problems.
Key terms
The atom is an incredibly tiny object. Its mass is far too small to measure on an ordinary balance. In Chapter 5 (Section 5.6), we learned to compare atoms using a table of atomic mass
units. These units are valuable when we compare the masses of individual atoms (mentally),
but they have no practical use in the laboratory. The mass in grams for an “average” carbon
atom (atomic mass 12.01 amu) would be 2.00 * 10-23 g, which is much too tiny for the best
laboratory balance.
So how can we confidently measure masses for these very tiny atoms? We increase the
number of atoms in a sample until we have an amount large enough to measure on a laboratory
balance. The problem then is how to count our sample of atoms.
Consider for a moment the produce in a supermarket. Frequently, apples and oranges
are sorted by size and then sold by weight, not by the piece of fruit. The grocer is counting
by weighing. To do this, he needs to know the mass of an “average” apple (235 g) and the
mass of an “average” orange (186 g). Now suppose he has an order from the local college for
275 apples and 350 oranges. It would take a long time to count and package this order. The
grocer can quickly count fruit by weighing. For example,
Avogadro’s number
mole
molar mass
PhotoDisc, Inc./Getty Images
(275 apples) a
Oranges can be “counted” by
weighing them in the store.
(350 oranges) a
235 g
b = 6.46 * 104 g = 64.6 kg
apple
186 g
b = 6.51 * 104 g = 65.1 kg
orange
He can now weigh 64.6 kg of apples and 65.1 kg of oranges and pack them without actually counting them. Manufacturers and suppliers often count by weighing. Other examples of
counting by weighing include nuts, bolts, and candy.
Chemists also count atoms by weighing. We know the “average” masses of atoms, so
we can count atoms by defining a unit to represent a larger number of atoms. Chemists have
chosen the mole (mol) as the unit for counting atoms. The mole is a unit for counting just as a
dozen or a ream or a gross is used to count:
1 dozen = 12 objects
1 ream = 500 objects
1 gross = 144 objects
1 mole = 6.022 * 1023 objects
Eggs are measured by the dozen.
Paper is measured by the ream (500 sheets).
Units of measurement need to be appropriate for the object being measured.
Tom Pantages
Royalty-Free/Corbis Images
Copyright John Wiley & Sons, Inc.
Note that we use a unit only when it is appropriate. A dozen eggs is practical in our kitchen,
a gross might be practical for a restaurant, but a ream of eggs would not be very practical.
Chemists can’t use dozens, grosses, or reams because atoms are so tiny that a dozen, gross, or
ream of atoms still couldn’t be measured in the ­laboratory.
Pencils are measured by the gross (144).
7.1
• The Mole 123
The number represented by 1 mol, 6.022 * 1023, is called Avogadro’s number, in honor
of Amadeo Avogadro (1776–1856), who investigated several quantitative aspects in chemistry.
It’s difficult to imagine how large Avogadro’s number really is, but this example may help: If
10,000 people started to count Avogadro’s number, and each counted at the rate of 100 numbers
per minute each minute of the day, it would take them over 1 trillion (1012) years to count the
total number. So even the tiniest amount of matter contains extremely large numbers of atoms.
Average atomic mass
12.01 amu
Mass of sample
12.01 g
Number of atoms in sample 6.022 * 1023 atoms
Pb (lead)
Richard Megna/Fundamental Photographs
Al (aluminum)
Richard Megna/Fundamental Photographs
Element
C (carbon)
Richard Megna/Fundamental Photographs
Symbol name
26.98 amu
26.98 g
6.022 * 1023 atoms
207.2 amu
207.2 g
6.022 * 1023 atoms
Avogadro’s number has been experimentally determined by several methods. How does it
relate to atomic mass units? Remember that the atomic mass for an element is the average relative mass of all the isotopes for the element. The atomic mass (expressed in grams) of 1 mole of
any element contains the same number of particles (Avogadro’s number) as there are in exactly
12 g of 12C. Thus, 1 mole of anything is the amount of the substance that contains the same
number of items as there are atoms in exactly 12 g of 12C.
Remember that 12C is the
reference isotope for atomic
masses.
1 mole = 6.022 * 1023 items
From the definition of mole, we can see that the atomic mass in grams of any element
contains 1 mol of atoms. The term mole is so commonplace in chemistry that chemists use it as
freely as the words atom or molecule. A mole of atoms, molecules, ions, or electrons represents
Avogadro’s number of these particles. If we can speak of a mole of atoms, we can also speak
of a mole of molecules, a mole of electrons, or a mole of ions, understanding that in each case
we mean 6.022 * 1023 particles:
1 mol of atoms = 6.022 * 1023 atoms
1 mol of molecules = 6.022 * 1023 molecules
1 mol of ions = 6.022 * 1023 ions
Practice 7.1
How many atoms are in 1.000 mole of the following?
(a) Fe
(b) H2
(c) H2SO4
The atomic mass of an element in grams contains Avogadro’s number of atoms and is
defined as the molar mass of that element. To determine molar mass, we change the units of
the atomic mass (found in the periodic table) from atomic mass units to grams. For example,
sulfur has an atomic mass of 32.07 amu, so 1 mol of sulfur has a molar mass of 32.07 g and
contains 6.022 * 1023 atoms of sulfur. Here are some other examples:
Element
H
Mg
Na
Atomic mass
Molar mass
Number of atoms
1.008 amu
24.31 amu
22.99 amu
1.008 g
24.31 g
22.99 g
6.022 * 1023
6.022 * 1023
6.022 * 1023
The abbreviation for mole is mol,
both singular and plural.
124 chapter 7
• Quantitative Composition of Compounds
In this text molar masses of
­elements are given to four
­significant figures.
To summarize:
1. The atomic mass expressed in grams is the molar mass of an element. It is different for
each element. In this text, molar masses are expressed to four significant figures.
1 molar mass 5 atomic mass of an element in grams
2. One mole of any element contains Avogadro’s number of atoms.
1 mol of atoms = 6.022 * 1023 atoms
We can use these relationships to make conversions between number of atoms, mass, and
moles, as shown in the following examples.
32.07 g
6.022 * 1023 atoms
126.9 g
6.022 * 1023 atoms
55.85 g
6.022 * 1023 atoms
Richard Megna/Fundamental Photographs
Mercury
Richard Megna/Fundamental Photographs
Iron
Richard Megna/Fundamental Photographs
Iodine
Richard Megna/Fundamental Photographs
Sulfur
200.6 g
6.022 * 1023 atoms
Example 7.1
ENHANCED EXAMPLE
How many moles of iron does 25.0 g of iron (Fe) represent?
SOLUTION
Plan •
Solution map: g Fe S moles Fe
1 mole Fe = 55.85 g (from periodic table)
The conversion factor needed is
55.85 g Fe
1 mole Fe
or
55.85 g Fe
1 mole Fe
1 mole Fe
b = 0.448 mol Fe
55.85 g Fe
Since 25.0 g has three significant figures, the number of significant figures
allowed in the answer is three. The answer is reasonable since 0.448 mol is
about ½ mole and 25.0 g is about ½ of 55.85 g (one mole Fe).
Calculate • (25.0 g Fe)a
Check •
Example 7.2
How many magnesium atoms are contained in 5.00 g Mg?
SOLUTION
Plan •
Solution map: g Mg S atoms Mg
1 mol Mg 5 24.31 g (from periodic table)
The conversion factor needed is
6.022 * 10 23 atoms Mg
24.31 g Mg
24.31 g Mg
6.022 * 1023 atoms Mg
6.022 * 1023 atoms Mg
b = 1.24 * 1023 atoms Mg
24.31 g Mg
Since 5.00 g has three significant figures, the number of significant ­figures
allowed in the answer is three.
Calculate • (5.00 g Mg)a
Check •
or
7.1
Example 7.3
What is the mass, in grams, of one atom of carbon (C)?
SOLUTION
Plan •
Solution map: atoms C S grams C
1 mol C 5 12.01 g (from periodic table)
The conversion factor needed is
6.022 * 1023 atoms C
12.01 g C
Calculate • (1 atom C)a
Check •
12.01 g C
or
6.022 * 1023 atoms C
12.01 g C
6.022 * 1023 atoms C
b = 1.994 * 10-23 g C
Since 1 atom is an exact number, the number of significant figures ­allowed
in the answer is four based on the number of significant figures in the molar
mass of C.
Example 7.4
What is the mass of 3.01 * 1023 atoms of sodium (Na)?
SOLUTION
Plan •
Solution map: atoms Na S grams Na
1 mol Na = 22.99 g (from periodic table)
The conversion factor needed is
6.022 * 1023 atoms Na
or
22.99 g Na
Calculate • (3.01 * 1023 atoms Na)a
Check •
22.99 g Na
6.022 * 1023 atoms Na
22.99 g Na
6.022 * 1023 atoms Na
b = 11.5 g Na
ince 3.01 * 1023 has three significant figures, the answer (mass) should
S
have three as well.
Example 7.5
What is the mass of 0.252 mol of copper (Cu)?
SOLUTION
Plan •
Solution map: mol Cu S grams Cu
1 mol Cu 5 63.55 g (from periodic table)
The conversion factor needed is
1 mol Cu
63.55 g Cu
Calculate • (0.252 mol Cu)¢
Check •
or
63.55 g Cu
1 mol Cu
63.55 g Cu
≤ = 16.0 g Cu
1 mol Cu
The number of significant figures in both the question and answer is
three.
• The Mole 125
126 chapter 7
• Quantitative Composition of Compounds
Example 7.6
How many oxygen atoms are present in 1.00 mol of oxygen molecules?
SOLUTION
Plan •
Oxygen is a diatomic molecule with the formula O2. Therefore, a molecule
2 atoms O
of ­oxygen contains 2 oxygen atoms:
.
1 molecule O2
Solution map: moles O2 S molecules O2 S atoms O
The conversion factors needed are:
6.022 * 1023 molecules O2
1 mole O2
Calculate • (1.00 mol O2)a
and
2 atoms O
1 molecule O2
6.022 * 1023 molecules O2
2 atoms O
ba
b
1 mole O2
1 molecule O2
= 1.20 * 1024 atoms O
Pr a c t i ce 7 . 2
What is the mass of 2.50 mol of helium (He)?
Pr a c t i ce 7 . 3
How many atoms are present in 0.025 mol of iron?
7.2 Molar Mass of Compounds
Learning objective
Calculate the molar mass of a compound.
A formula unit is indicated by the
formula, for example, Mg, MgS,
H2O, NaCI.
One mole of a compound contains Avogadro’s number of formula units of that compound. The
terms molecular weight, molecular mass, formula weight, and ­formula mass have been used in
the past to refer to the mass of 1 mol of a compound. However, the term molar mass is more
inclusive because it can be used for all types of ­compounds.
If the formula of a compound is known, its molar mass can be determined by adding the
molar masses of all the atoms in the formula. If more than one atom of any element is present,
its mass must be added as many times as it ­appears in the compound.
Example 7.7
The formula for water is H2O. What is its molar mass?
SOLUTION
Read •
Known:
Plan •
The formula contains 1 atom O and 2 atoms H. We need to use the atomic
masses from the inside front cover of the text.
Calculate •
H2 O
O = 16.00 g = 16.00 g
2 H = 2(1.008 g) = 2.016 g
18.02 g = molar mass of H2O
7.2
• Molar Mass of Compounds 127
Example 7.8
Calculate the molar mass of calcium hydroxide, Ca(OH)2.
SOLUTION
Read •
Known:
Plan •
he formula contains 1 atom Ca and 2 atoms each of O and H. We need to use
T
the atomic masses from the inside front cover of the text.
Calculate •
Ca(OH)2
1 Ca = 1(40.08 g) = 40.08 g
2 O = 2(16.00 g) = 32.00 g
2 H = 2(1.008 g) = 2.016 g
74.10 g = molar mass of Ca(OH)2
In this text we round all molar
masses to four significant figures,
although you may need to use a
different number of significant
figures for other work (in the lab).
Practice 7.4
Calculate the molar mass of KNO3.
The mass of 1 mol of a compound contains Avogadro’s number of formula units.
Consider the compound hydrogen chloride (HCl). One atom of H combines with one atom
of Cl to form HCl. When 1 mol of H (1.008 g of H or 6.022 * 1023 H atoms) combines
with 1 mol of Cl (35.45 g of Cl or 6.022 * 1023 Cl atoms), 1 mol of HCl (36.46 g of HCl
or 6.022 * 1023 HCl molecules) is produced. These relationships are summarized in the
following table:
H
6.022 * 1023 H atoms
1 mol H atoms
1.008 g H
1 molar mass H atoms
Cl
6.022 * 1023 Cl atoms
1 mol Cl atoms
34.45 g Cl
1 molar mass Cl atoms
HCl
6.022 * 1023 HCl molecules
1 mol HCl molecules
36.46 g HCl
1 molar mass HCl molecules
In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), we must take special
care to distinguish between a mole of atoms and a mole of molecules. For example, consider
1 mol of oxygen molecules, which has a mass of 32.00 g. This quantity is equal to 2 mol of
Tom Pantages
A mole of table salt (in front of a salt shaker)
and a mole of water (in the plastic container)
have different sizes but both contain
Avogadro’s number of formula units.
128 chapter 7
• Quantitative Composition of Compounds
oxygen atoms. Remember that 1 mol represents Avogadro’s number of the particular chemical
entity that is under consideration:
= 6.022 * 1023 molecules
= 6.022 * 1023 formula units
= 6.022 * 1023 molecules
1 mol HNO3 = 63.02 g HNO3 = 6.022 * 1023 molecules
1 mol K2SO4 = 174.3 g K2SO4 = 6.022 * 1023 formula units
1 mol = 6.022 * 1023 formula unit or molecules
= 1 molar mass of a compound
1 mol H2O = 18.02 g H2O
1 mol NaCl = 58.44 g NaCl
1 mol H2 = 2.016 g H2
Formula units are often used in
place of molecules for substances
that contain ions.
Example 7.9
ENHANCED EXAMPLE
What is the mass of 1 mol of sulfuric acid (H2SO4)?
SOLUTION
Read •
Known:
H2so4
Plan •We need to use the atomic masses from the inside front cover of the text.
Calculate •
2 H = 2(1.008 g) = 2.016 g
1 S = 32.07 g
= 32.07 g
4 O = 4(16.00 g) = 64.00 g
98.09 g = mass of 1 mol of H2SO4
Example 7.10
How many moles of sodium hydroxide (NaOH) are there in 1.00 kg of sodium
­hydroxide?
SOLUTION
Read •
Known:
1.00 kg NaOH
Plan •We need to use the atomic masses from the inside front cover of the text to
find the molar mass of NaOH.
Na
O
H
22.99 g + 16.00 g + 1.008 g = 40.00 g NaOH = Molar mass NaOH
Solution map: kg NaOH S g NaOH S mol NaOH
Create the appropriate
­conversion factor by placing the
unit desired in the numerator
and the unit to be eliminated
in the denominator.
We need two conversion factors:
Calculate • (1.00 kg NaOH)a
1000 g
1 kg
and
1000 g NaOH
1 mol NaOH
ba
b
1 kg NaOH
40.00 g NaOH
= 25.00 mol NaOH
Therefore, 1.00 kg NaOH = 25.00 mol NaOH.
Example 7.11
What is the mass of 5.00 mol water?
SOLUTION
Read •
Known:
1 mol NaOH
.
40.00 g NaOH
5.00 mol H2O
7.3
• Percent Composition of Compounds 129
Plan •We need to use the atomic masses from the inside front cover of the text to
find the molar mass of H2O.
Solution map: mol H2O S g H2O
18.02 g H2O
(conversion factor)
1 mol H2O
Calculate • (5.00 mol H2O)a
18.02 g H2O
b = 90.1 g H2O
1 mol H2O
Therefore, 5.00 mol H2O = 90.1 g H2O.
Example 7.12
How many molecules of hydrogen chloride (HCl) are there in 25.0 g of hydrogen chloride?
SOLUTION
Read •
Known:
25.0 g HCl
Plan •We need to use the atomic masses from the inside front cover of the text to
find the molar mass of HCl.
H
Cl
1.008 g + 34.45 g = 36.46 g = molar mass HCl
Solution map: g HCl S mol HCl S molecules HCl
1 mol HCl
6.022 * 1023 molecules HCl
and
(conversion factors)
36.46 g HCl
1 mol HCl
Calculate • (25.0 g HCl)a
1 mol HCl
6.022 * 1023 molecules HCl
ba
b
36.46 g HCl
1 mol HCl
= 4.13 * 1023 molecules HCl
Practice 7.5
What is the mass of 0.150 mol of Na2SO4?
Practice 7.6
How many moles and molecules are there in 500.0 g of HC2H3O2?
7.3Percent Composition of Compounds
Calculate the percent composition of a compound from its chemical formula and from
experimental data.
Percent means parts per 100 parts. Just as each piece of pie is a percent of the whole
pie, each element in a compound is a percent of the whole compound. The percent
composition of a compound is the mass percent of each element in the compound.
The molar mass represents the total mass, or 100%, of the compound. Thus, the percent
composition of water, H 2O, is 11.19% H and 88.79% O by mass. According to the law
of definite composition, the percent composition must be the same no matter what size
sample is taken.
The percent composition of a compound can be determined (1) from knowing its formula or
(2) from experimental data.
Learning objectiv e
Key term
percent composition of a compound
H
11.19%
O
88.79%
130 chapter 7
• Quantitative Composition of Compounds
Percent Composition from Formula
If the formula of a compound is known, a two-step process is needed to determine the percent composition:
Problem-Solving Strategy: For Calculating Percent Composition
from Formula
1. Calculate the molar mass (Section 7.2).
2. Divide the total mass of each element in the formula by the molar mass and multiply by 100.
This gives the percent ­composition:
total mass of the element
* 100 = percent of the element
molar mass
Example 7.13
ENHANCED EXAMPLE
Calculate the percent composition of sodium chloride (NaCl).
SOLUTION
Read •
Known:
Plan •
se the Problem-Solving Strategy for Calculating Percent Composition from
U
Formula.
NaCl
Calculate • 1. We need to use the atomic masses from the inside front cover of the text to
find the molar mass of NaCl.
Na
Na
39.34%
Na:
Cl
60.66%
Cl:
NOTE
Cl
22.99 g + 34.45 g = 58.44 g = molar mass NaCl
2. Calculate the percent composition for each element.
a
a
22.99 g Na
b (100) =
58.44 g
35.45 g Cl
b (100) =
58.44 g
39.34% Na
60.66% Cl
100.00% total
In any two-component system, if the percent of one component is known, the
other is automatically defined by difference; that is, if Na is 39.43%, then Cl
is 100% - 39.34% = 60.66%. However, the calculation of the percent should
be carried out for each component, since this provides a check against possible error. The percent composition should add to 100% { 0.2%.
Example 7.14
Calculate the percent composition of potassium chloride (KCl).
SOLUTION
Read •
Known:
Plan •
Use the Problem-Solving Strategy for Calculating Percent Composition from
Formula.
KCl
Calculate • 1.We need to use the atomic masses from the inside front cover of the text to
find the molar mass of KCl.
K
Cl
39.10 g + 35.45 g = 74.55 g = molar mass KCl
7.3
• Percent Composition of Compounds 131
2. Calculate the percent composition for each element.
39.10 g K
b (100) =
K: a
74.55 g
35.45 g Cl
Cl: a
b (100) =
74.55 g
Cl
47.55%
52.45% K
K
52.45%
47.55% Cl
100.00% total
Comparing the results calculated for NaCl and for KCl, we see that NaCl contains a higher percentage of Cl by mass, although each compound has a one-to-one atom ratio of Cl to Na and Cl to K.
The reason for this mass percent difference is that Na and K do not have the same atomic masses.
It is important to realize that when we compare 1 mol of NaCl with 1 mol of KCl, each
quantity contains the same number of Cl atoms—namely, 1 mol of Cl atoms. However, if we
compare equal masses of NaCl and KCl, there will be more Cl atoms in the mass of NaCl,
since NaCl has a higher mass percent of Cl.
1 mol NaCl
contains
100.00 g NaCl
contains
1 mol Na
1 mol Cl
1 mol KCl
contains
39.34 g Na
1 mol K
60.66 g Cl
1 mol Cl
60.66% Cl
100.00 g KCl
contains
52.45 g K
47.55 g Cl
47.55% Cl
Example 7.15
Calculate the percent composition of potassium sulfate (K2SO4).
SOLUTION
Read •
Plan •
Known: K2SO4
Use the Problem-Solving Strategy for Calculating Percent Composition from
Formula.
Calculate • 1. We need to use the atomic masses from the inside front cover of the text to
find the molar mass of K2SO4.
K
S
O
2(39.10) g + 32.07 g + 4(16.00) = 174.3 g = molar mass K2SO4
2. Calculate the percent composition for each element.
K:
S:
O:
a
78.20 g K
b (100) = 44.87% K
174.3 g
a
64.00 g O
b (100) = 36.72% O
174.3 g
99.99% total
a
32.07 g S
b (100) = 18.40% S
174.3 g
Practice 7.7
Calculate the percent composition of Ca(NO3)2.
Practice 7.8
Calculate the percent composition of K2CrO4.
Problem-Solving Strategy: For Percent Composition
from Experimental Data
1. Calculate the mass of the compound formed.
2. Divide the mass of each element by the total mass of the compound and ­multiply by 100.
S
18.40%
O
36.72%
K
44.87%
132 chapter 7
• Quantitative Composition of Compounds
Example 7.16
Zinc oxide is a compound with many uses from preventing sunburn to a pigment in white
paint. When heated in air, 1.63 g of zinc (Zn) combines with 0.40 g of oxygen (O2) to
form zinc oxide. Calculate the percent composition of the compound formed.
SOLUTION
Read •
Knowns: 1.63 g Zn
0.40 g O2
Plan •
Use the Problem-Solving Strategy for Calculating Percent Composition from
Experimental Data.
Calculate • 1. We need to calculate the mass of the product formed.
O
20%
1.63 g Zn + 0.40 g O2 = 2.03 g product
Zn
80.3%
2. Calculate the percent for each element.
1.63 g Zn
b (100) = 80.3% Zn
Zn: a
2.03 g
O: a
0.40 g O
20. % O
b (100) =
2.03 g
100.3% total
Practice 7.9
Aluminum chloride is formed by reacting 13.43 g aluminum with 53.18 g chlorine. What
is the percent composition of the compound?
> Chemistry in action
Vanishing Coins?
Modern technology is changing our coins in some pretty
interesting ways. The first U.S. coins were produced
during the late 1700s from “coin silver,” an alloy of 90%
silver and 10% copper. Coins were made following a
very simple rule: The mass of the coin reflects its relative value. Therefore, a half dollar weighed half as much
as a dollar coin, and a quarter weighed 1>4 as much as a
dollar coin.
In the twentieth century our society began using machines
to collect coins. We use vending machines, parking meters, toll baskets, laundromats, slot machines, and video
games to make our lives more convenient and fun. This
change produced a huge demand for coins. At the same
time, the price of silver rose rapidly.
The solution? Make coins from a different alloy that would
still work in all of our machines. This was a tricky process
since the machines require a specific mass and electric
“resistivity.” In the mid-1960s new coins that sandwiched
a layer of copper between layers of an alloy of copper
and nickel began to replace quarters and dimes. And a
new Susan B. Anthony dollar appeared briefly in 1979.
Finally, in 1997 Congress decided to revive the dollar coin
once more.
Congress specified that the new dollar coin had to look
and feel different to consumers, be golden in color, and
have a distinctive edge. At the same time, it needed to
fool vending machines so that both the Susan B. Anthony
dollar and the new coin would work. Designers decided
on the portrait of Sacagawea, a Shoshone woman and her
child. She was a guide on the Lewis and Clark expedition.
On the back is an eagle. The chemists struggled to find a
golden alloy for the coin. It is slightly larger than a quarter
and weighs 8.1 grams. Vending machines determine a
coin’s value by its size, weight, and electromagnetic signature. This electromagnetic signature is hard to duplicate.
All the golden coins tested had three times the conductivity of the silver Anthony coin. Metallurgists finally tried
adding manganese and zinc to the copper core and found
a “golden” alloy that fooled the vending machines! Sacagawea dollars are made of an alloy that is 77% copper,
12% zinc, 7% manganese, and 4% nickel. And on top of it
all, the new coins only cost 12 cents each to make, leaving
an 88 cent profit for the mint on each coin!
The future of coins is becoming less certain, however, as our
machines are converted to electronic devices that use magnetic strips and a swipe of a card in place of “coins” and scan
our cars for toll pass information. Some day soon, coins may
vanish altogether from our pockets into collectors’ albums.
7.4
• Calculating Empirical Formulas 133
7.4 Calculating Empirical Formulas
Determine the empirical formula for a compound from its percent composition.
Learning objectiv e
The empirical formula, or simplest formula, gives the smallest whole-number ratio of atoms
present in a compound. This formula gives the relative number of atoms of each element in
the compound.
We can establish empirical formulas because (1) individual atoms in a compound are
combined in whole-number ratios and (2) each element has a specific atomic mass.
To calculate an empirical formula, we need to know (1) the elements that are combined,
(2) their atomic masses, and (3) the ratio by mass or percentage in which they are combined. If
elements A and B form a compound, we may represent the empirical formula as Ax By ,where
x and y are small whole numbers that represent the atoms of A and B. To write the empirical
formula, we must determine x and y:
Key term
empirical formula
Problem-Solving Strategy: For Calculating an Empirical Formula
1. Assume a definite starting quantity (usually 100.0 g) of the compound, if not given, and
express the mass of each element in grams.
2. Convert the grams of each element into moles using each element’s molar mass. This
conversion gives the number of moles of atoms of each element in the quantity assumed in
Step 1. At this point, these numbers will usually not be whole numbers.
3. Divide each value obtained in Step 2 by the smallest of these values. If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. If the
numbers obtained are not whole numbers, go on to Step 4.
4. Multiply the values obtained in Step 3 by the smallest number that will convert them to
whole numbers. Use these whole numbers as the subscripts in the empirical formula. For
example, if the ratio of A to B is 1.0 : 1.5, multiply both numbers by 2 to obtain a ratio of
2 : 3. The empirical formula then is A2 B3 .
In many of these calculations, results will vary somewhat from an exact whole number;
this can be due to experimental errors in obtaining the data or from rounding off numbers.
Calculations that vary by no more than {0.1 from a whole number usually are rounded off to
the nearest whole number. Deviations greater than about 0.1 unit usually mean that the calculated ratios need to be multiplied by a factor to make them all whole numbers. For example, an
atom ratio of 1 : 1.33 should be multiplied by 3 to make the ratio 3 : 4. Let’s do a few examples
to see how it works.
Example 7.17
Calculate the empirical formula of a compound containing 11.19% hydrogen (H) and
88.79% oxygen (O).
Some common fractions and
their decimal equivalents are
1
/4 =
/3 =
2
/3 5
1
/2 =
3
/4 =
1
0.25
0.333 . . .
0.666 . . .
0.5
0.75
Multiply the decimal equivalent by the number in the
­denominator of the fraction to
get a whole number: 4(0.75) = 3.
SOLUTION
Read •
Knowns: 11.19% H
88.79% O
Plan •
Use the Problem-Solving Strategy for Calculating an Empirical Formula.
Calculate • 1.Assume 100.0 g of material. We know that the percent of each ­element
equals the grams of each element:
11.19 g H
88.79 g O
ENHANCED EXAMPLE
134 chapter 7
• Quantitative Composition of Compounds
2. Convert grams of each element to moles:
H: (11.19 g)a
1 mol H atoms
b = 11.10 mol H atoms
1.008 g H
1 mol O atoms
b = 5.549 mol O atoms
16.00 g O
The formula could be expressed as H11.10O5.549. However, it’s customary
to use the smallest whole-number ratio of atoms.
O: (88.79 g)a
3. Change the number of moles to whole numbers by dividing by the smallest
number.
H =
11.10 mol
= 2.000
5.549 mol
5.549 mol
= 1.000
5.549 mol
The simplest ratio of H to O is 2:1.
Empirical formula = H2O
O =
Example 7.18
The analysis of a salt shows that it contains 56.58% potassium (K), 8.68% carbon (C),
and 34.73% oxygen (O). Calculate the empirical formula for this substance.
SOLUTION
Read •
Knowns: 56.58% K
8.68% C
34.73% O
Plan •
Use the Problem-Solving Strategy for Calculating an Empirical Formula.
Calculate • 1. Assume 100.0 g of material. We know that the percent of each element
equals the grams of each element:
56.58 g K
8.68 g C
34.73 g O
2. Convert the grams of each element to moles:
1 mol K atoms
b = 1.447 mol K atoms
39.10 g K
1 mol C atoms
C: (8.68 g C)a
b
= 0.723 mol C atoms
12.01 g C
1 mol O atoms
O: (34.73 g O)a
b = 2.171 mol O atoms
16.00 g O
3. Change the number of moles to whole numbers by dividing by the smallest
number.
K: (56.58 g K)a
1.447 mol
= 2.00
0.723 mol
0.723 mol
C =
= 1.00
0.723 mol
2.171 mol
O =
= 3.00
0.723 mol
The simplest ratio of K:C:O is 2:1:3.
Empirical formula = K2CO3
K =
7.5
• Calculating the Molecular Formula from the Empirical Formula 135
Example 7.19
A sulfide of iron was formed by combining 2.233 g of iron (Fe) with 1.926 g of ­sulfur
(S). What is the empirical formula for the compound?
SOLUTION
Read •
Knowns: 2.233 g Fe
1.926 g S
Plan •
Use the Problem-Solving Strategy for Calculating an Empirical Formula.
Calculate • 1. The mass of each element is known, so we use it directly.
2. Convert grams of each element to moles.
Fe: (2.233 g Fe)a
S: (1.926 g S)a
1 mol Fe atoms
b = 0.03998 mol Fe atoms
55.85 g Fe
1 mol S atoms
b
32.07 g S
= 0.06006 mol S atoms
3. Change the number of moles to whole numbers by dividing by the smallest number.
Fe =
0.03998 mol
= 1.000
0.03998 mol
S =
0.06006 mol
= 1.502
0.03998 mol
4. We still have not reached a ratio that gives whole numbers in the formula, so
we multiply by a number that will give us whole numbers.
Fe: (1.000)2 = 2.000
S: (1.502)2 = 3.004
Empirical formula = Fe2S3
Practice 7.10
Calculate the empirical formula of a compound containing 52.14% C, 13.12% H, and
34.73% O.
Practice 7.11
Calculate the empirical formula of a compound that contains 43.7% phosphorus and
56.3% O by mass.
7.5 Calculating the Molecular
Formula from the Empirical Formula
Compare an empirical formula to a molecular formula and calculate a molecular
­formula from the empirical formula of the compound and its molar mass.
Learning objectiv e
The molecular formula is the true formula, representing the total number of atoms of
each element present in one molecule of a compound. It is entirely possible that two or
more substances will have the same percent composition yet be distinctly different compounds. For example, acetylene (C2H2) is a common gas used in welding; benzene (C6H6)
is an important solvent obtained from coal tar and is used in the synthesis of styrene and
nylon. Both acetylene and benzene contain 92.3% C and 7.7% H. The smallest ratio of C
and H corresponding to these percentages is CH (1:1) Therefore, the empirical formula
for both acetylene and benzene is CH, even though the molecular formulas are C2H2 and
Key term
molecular formula
136 chapter 7
• Quantitative Composition of Compounds
C6H6 , respectively. Often the molecular formula is the same as the empirical formula. If
the molecular formula is not the same, it will be an integral (whole number) multiple of
the empirical formula. For example,
CH = empirical formula
(CH)2 = C2H2 = acetylene (molecular formula)
(CH)6 = C6H6 = benzene
(molecular formula)
Table 7.1 compares the formulas of these substances. Table 7.2 shows empirical and
molecular formula relationships of other compounds.
Table 7.1 Molecular Formulas of Two Compounds Having an Empirical Formula
with a 1:1 Ratio of Carbon and Hydrogen Atoms
Composition
Formula
%C
%H
Molar mass
CH (empirical)
C2H2 (acetylene)
C6H6 (benzene)
92.3
92.3
92.3
7.7
7.7
7.7
13.02 (empirical)
26.04 (2 * 13.02)
78.12 (6 * 13.02)
Table 7.2
Some Empirical and Molecular Formulas
Substance
Acetylene
Benzene
Ethylene
Formaldehyde
Acetic acid
Glucose
Hydrogen chloride
Carbon dioxide
Empirical
formula
CH
CH
CH2
CH2O
CH2O
CH2O
HCl
CO2
Molecular
formula
Substance
C2H2
C6H6
C2H4
CH2O
C2H4O2
C6H12O6
HCl
CO2
Diborane
Hydrazine
Hydrogen
Chlorine
Bromine
Oxygen
Nitrogen
Iodine
Empirical
formula
Molecular
formula
BH3
NH2
H
Cl
Br
O
N
I
B2H6
N2H4
H2
Cl2
Br2
O2
N2
I2
The molecular formula can be calculated from the empirical formula if the molar mass is
known. The molecular formula will be equal either to the empirical formula or some multiple
of it. For example, if the empirical formula of a compound of hydrogen and fluorine is HF, the
molecular formula can be expressed as (HF)n , where n = 1, 2, 3, 4, . . . . This n means that
the molecular formula could be HF, H2F2 , H3F3 , H4F4 , and so on. To determine the molecular
formula, we must evaluate n:
molar mass
= number of empirical formula units
n=
mass of empirical formula
What we actually calculate is the number of units of the empirical formula contained in the
molecular formula.
Example 7.20
ENHANCED EXAMPLE
A compound of nitrogen and oxygen with a molar mass of 92.00 g was found to have an
empirical formula of NO2. What is its molecular ­formula?
SOLUTION
Read •
Knowns: NO2 = empirical formula
Molar mass = 92.00
7.5
Plan •
• Calculating the Molecular Formula from the Empirical Formula 137
Let n be the number of NO2 units in a molecule; then the molecular formula
is (NO2)n. Find the molar mass of an NO2 unit and determine the numbers of
units needed for the molar mass of the compound.
Calculate • The molar mass of NO2 is 14.01 g + 2(16.00 g) = 46.01 g.
92.00 g
= 2 (empirical formula units)
46.01 g
The molecular formula is (NO2)2 or N2O4.
n=
N2O4
Example 7.21
Propylene is a compound frequently polymerized to make polypropylene, which is used
for rope, laundry bags, blankets, carpets, and textile fibers. Propylene has a molar mass
of 42.08 g and contains 14.3% H and 85.7% C. What is its molecular formula?
SOLUTION
Read •
Knowns: Molar mass = 42.08 g
14.3% H
85.7% C
First find the empirical formula using the Problem-Solving Strategy for
­Calculating an Empirical Formula. Then use n to determine the molecular
formula from the empirical formula.
Tony Freeman/PhotoEdit
Plan •
Calculate • 1. Assume 100.0 g of material. We know that the percent of each element
equals the grams of each element:
14.3 g H
85.7 g C
2. Convert the grams of each element to moles:
1 mol H atoms
b = 14.2 mol H atoms
H: (14.3 g H)a
1.008 g H
1 mol C atoms
b = 7.14 mol C atoms
12.01 g C
3. Change the number of moles to whole numbers by dividing by the smallest
number.
14.2 mol
H=
= 1.99
7.14 mol
When propylene is polymerized
into polypropylene, it can be used
for rope and other recreational
products.
C: (85.7 g C)a
C=
7.14 mol
= 1.00
7.14 mol
Empirical formula = CH2
We determine the molecular formula from the empirical formula and the
molar mass:
Molecular formula = (CH2)n
CH2 molar mass = 12.01 g + 2(1.008 g) = 14.03 g
42.08 g
= 3 (empirical formula units)
14.03 g
The molecular formula is (CH2)3 = C3H6.
n=
Practice 7.12
Calculate the empirical and molecular formulas of a compound that contains 80.0% C,
20.0% H, and has a molar mass of 30.00 g.
C3H6
138 chapter 7
• Quantitative Composition of Compounds
C h a p te r
7 review
7.1 The Mole
Key Terms
Avogadro’s number
mole
molar mass
• We count atoms by weighing them since they are so tiny.
• 1 mole = 6.022 * 1023 items.
• Avogadro’s number is 6.022 * 1023.
7.2 Molar Mass of Compounds
•
•
•
•
One mole of a compound contains Avogadro’s number of formula units of that compound.
The mass (grams) of one mole of a compound is the molar mass.
Molar mass is determined by adding the molar masses of all the atoms in a formula.
Molar masses are given to four significant figures in this text.
7.3 Percent Composition of Compounds
Key Term
percent composition
of a compound
• To determine the percent composition from a formula:
• Calculate the molar mass.
• For each element in the formula
total mass of the element
* 100 = percent of the element
molar mass of the compound
• To determine percent composition from experimental data:
• Calculate the mass of the compound formed.
• For each element in the formula
mass of the element
* 100 = percent of the element
mass of the compound formed
7.4 Calculating Empirical Formulas
Key Term
empirical formula
• T
he empirical formula is the simplest formula giving the smallest whole-number ratio
of atoms ­present in a compound.
• To determine the empirical formula for a compound you need to know:
• The elements that are combined
• Their atomic masses
• The ratio of masses or percentage in which they are combined
• Empirical formulas are represented in the form Ax By . To determine the empirical formula
of this ­compound:
• Assume a starting quantity (100.0 g is a good choice).
• Convert mass (g) to moles for each element.
• Divide each element’s moles by the smallest number of moles.
• If the ratios are whole numbers, use them as subscripts and write the empirical formula.
• If the ratios are not whole numbers, multiply them all by the smallest number, which will
­convert them to whole numbers.
• Use the whole numbers to write the empirical ­formula.
7.5 Calculating the Molecular Formula
from the Empirical Formula
Key Term
molecular formula
• T
he molecular formula is the true formula representing the total number of atoms of each element
present in one molecule of the compound.
• Two or more substances may have the same empirical formulas but different molecular formulas.
• The molecular formula is calculated from the empirical formula when the molar mass is known:
• n =
molar mass
= number of empirical formula units
mass of empirical formula
• The molecular formula is (Ax By)n .
Paired Exercises 139
Review Questions
1.What is a mole?
2. Which would have a higher mass: a mole of K atoms or a mole of
Au atoms?
3. Which would contain more atoms: a mole of K atoms or a mole of
Au atoms?
4. Which would contain more electrons: a mole of K atoms or a mole
of Au atoms?
5. What is the numerical value of Avogadro’s number?
6. What is the relationship between Avogadro’s number and the mole?
7. Which has the greater number of oxygen atoms, 1 mole of oxygen
gas (O2) or 1 mole of ozone gas (O3)?
8. What is molar mass?
9. If the atomic mass scale had been defined differently, with an atom
of 126C defined as a mass of 50 amu, would this have any effect on
the value of Avogadro’s number? Explain.
10. Complete these statements, supplying the proper ­quantity.
(a) A mole of O atoms contains _____ atoms.
(b) A mole of O2 molecules contains _____ molecules.
(c) A mole of O2 molecules contains _____ atoms.
(d) A mole of O atoms has a mass of _____ grams.
(e) A mole of O2 molecules has a mass of _____ grams.
11. How many molecules are present in 1 molar mass of sulfuric acid
(H2SO4)? How many atoms are present?
12. What are the two types of information that can be used to calculate
the percent composition of a compound?
13. A compound is composed of only nitrogen and oxygen. If the compound contains 43.8% nitrogen, what is the percent oxygen in the
compound?
14. Write the steps used to calculate percent composition of an element
in a compound.
15. In calculating the empirical formula of a ­compound from its percent composition, why do we choose to start with 100.0 g of the
­compound?
16. Which of the following compounds have the same empirical­
formula? What is the empirical formula? CH4, C2H4, C4H6, C4H10,
C8H12, C8H18
17. What is the difference between an empirical formula and a molecular formula?
18. What data beyond the percent composition are generally used to
determine the molecular formula of a compound?
molar mass
19. What information does the ratio
provide?
mass of emprical formula
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1. Determine the molar masses of these ­compounds:
(a) KBr
(f) Fe3O4
(b) Na2SO4
(g) C12H22O11
(c) Pb(NO3)2
(h) Al2(SO4)3
(d) C2H5OH
(i) (NH4)2HPO4
(e) HC2H3O2
2.Determine the molar masses of these ­compounds:
(a) NaOH
(f) C6H5COOH
(b) Ag2CO3
(g) C6H12O6
(c) Cr2O3
(h) K4Fe(CN)6
(d) (NH4)2CO3
(i) BaCl2 # 2H2O
(e) Mg(HCO3)2
3. How many moles of atoms are contained in the following?
(a) 22.5 g Zn
(d) 382 g Co
(b) 0.688 g Mg
(e) 0.055 g Sn
(c) 4.5 * 1022 atoms Cu (f) 8.5 * 1024 molecules N2
4.How many moles of atoms are contained in the following?
(a) 25.0 g NaOH
(d) 14.8 g CH3OH
(b) 44.0 g Br2
(e) 2.88 g Na2SO4
(c) 0.684 g MgCl2
(f) 4.20 lb ZnI2
5. Calculate the number of grams in each of the following:
(a) 0.550 mol Au
(c) 12.5 mol Cl2
(b) 15.8 mol H2O
(d) 3.15 mol NH4NO3
6.Calculate the number of grams in each of the following:
(a) 4.25 * 10-4 mol H2SO4
(c) 0.00255 mol Ti
(b) 4.5 * 1022 molecules CCl4 (d) 1.5 * 1016 atoms S
7. How many molecules are contained in each of the following?
(a) 2.5 mol S8
(c) 17.5 g C2H5OH
(b) 7.35 mol NH3
(d) 225 g Cl2
8. How many molecules are contained in each of the following?
(a) 9.6 mol C2H4
(c) 23.2 g CH3OH
(b) 2.76 mol N2O
(d) 32.7 g CCl4
9.How many atoms are contained in each of the following?
(a) 25 molecules P2O5
(d) 1.25 g Na
(b) 3.62 mol O2
(e) 2.7 g CO2
(c) 12.2 mol CS2
(f) 0.25 g CH4
10. How many atoms are contained in each of the following?
(a) 2 molecules CH3COOH
(d) 92.5 g Au
(b) 0.75 mol C2H6
(e) 75 g PCl3
(c) 25 mol H2O
(f) 15 g C6H12O6
11.Calculate the mass in grams of each of the following:
(a) 1 atom He
(c) 4 molecules N2O5
(b) 15 atoms C
(d) 11 molecules C6H5NH2
12.Calculate the mass in grams of each of the following:
(a) 1 atom Xe
(c) 9 molecules CH3COOH
(b) 22 atoms Cl
(d) 15 molecules C4H4O2(NH2)2
13.Make the following conversions:
(a) 25 kg CO2 to mol CO2
(c) 6 mol O2 to atoms oxygen
(b) 5 atoms Pb to mol Pb
(d) 25 molecules P4 to g P4
14. Make the following conversions:
(a) 275 atoms W to mol W
(c) 12 molecules SO2 to g SO2
(b) 95 atoms H2O to kg H2O
(d) 25 mol Cl2 to atoms chlorine
15.1 molecule of tetraphosphorus decoxide contains
(a) how many moles?
(d) how many oxygen atoms?
(b) how many grams?
(e) how many total atoms?
(c) how many phosphorus atoms?
16.125 grams of disulfur decafluoride contain
(a) how many moles?
(d) how many atoms of sulfur?
(b) how many molecules? (e) how many atoms of fluorine?
(c) how many total atoms?
140 chapter 7
• Quantitative Composition of Compounds
17.How many atoms of hydrogen are contained in each of the ­following?
(a) 25 molecules C6H5CH3 (c) 36 g CH3CH2OH
(b) 3.5 mol H2CO3
18.How many atoms of hydrogen are contained in each of the ­following?
(a) 23 molecules CH3CH2COOH (c) 57 g C6H5ONH2
(b) 7.4 mol H3PO4
19.Calculate the number of
(a) grams of silver in 25.0 g AgBr
(b) grams of nitrogen in 6.34 mol (NH4)3PO4
(c)grams of oxygen in 8.45 * 1022 molecules SO3
20. Calculate the number of
(a) grams of chlorine in 5.0 g PbCl2
(b) grams of hydrogen in 4.50 mol H2SO4
(c) grams of hydrogen in 5.45 * 1022 molecules NH3
21.Calculate the percent composition by mass of these ­compounds:
(a) NaBr
(d) SiCl4
(b) KHCO3
(e) Al2(SO4)3
(c) FeCl3
(f) AgNO3
22.Calculate the percent composition by mass of these
­compounds:
(a) ZnCl2
(d) (NH4)2SO4
(b) NH4C2H3O2
(e) Fe(NO3)3
(c) MgP2O7
(f) ICl3
23.Calculate the percent of iron in the following compounds:
(a) FeO
(c) Fe3O4
(b) Fe2O3
(d) K4Fe(CN)6
24.Which of the following chlorides has the highest and which has the
lowest percentage of chlorine, by mass, in its formula?
(a) KCl
(c) SiCl4
(b) BaCl2
(d) LiCl
25. A 73.16-g sample of an interesting barium silicide reported to have
superconducting properties was found to contain 33.62 g barium
and the remainder silicon. Calculate the percent composition of the
compound.
26.One of the compounds that give garlic its characteristic odor is
ajoene. A 7.52-g sample of this compound contains 3.09 g sulfur,
0.453 g hydrogen, 0.513 g oxygen, and the remainder is carbon.
Calculate the percent composition of ajoene.
27. E
xamine the following formulas. Which compound has the
(a)higher percent by mass of hydrogen: H2O or H2O2 ?
(b)lower percent by mass of nitrogen: NO or N2O3 ?
(c)higher percent by mass of oxygen: NO2 or N2O4 ?
28.Examine the following formulas. Which compound has the
(a)lower percent by mass of chlorine: NaClO3 or KClO3 ?
(b)higher percent by mass of sulfur: KHSO4 or K2SO4 ?
(c)lower percent by mass of chromium: Na2CrO4 or Na2Cr2O7 ?
Check your answers by calculation if you wish.
Check your answers by calculation if you wish.
29. Calculate the empirical formula of each compound from the ­percent
compositions given:
(a) 63.6% N, 36.4% O
(d) 43.4% Na, 11.3% C, 45.3% O
(b) 46.7% N, 53.3% O
(e) 18.8% Na, 29.0% Cl, 52.3% O
(c) 25.9% N, 74.1% O
(f) 72.02% Mn, 27.98% O
30.Calculate the empirical formula of each compound from the percent compositions given:
(a) 64.1% Cu, 35.9% Cl (d) 55.3% K, 14.6% P, 30.1% O
(b) 47.2% Cu, 52.8% Cl (e) 38.9% Ba, 29.4% Cr, 31.7% O
(c) 51.9% Cr, 48.1% S
(f) 3.99% P, 82.3% Br, 13.7% Cl
31.Determine the empirical formula for each of the following, using
the given masses:
(a)a compound containing 26.08 g zinc, 4.79 g carbon, and 19.14 g
oxygen
(b)a 150.0-g sample of a compound containing 57.66 g ­carbon,
7.26 g hydrogen, and the rest, chlorine
(c)a 75.0-g sample of an oxide of vanadium, containing 42.0 g V
and the rest, oxygen
(d)a compound containing 67.35 g nickel, 43.46 g oxygen, and
23.69 g phosphorus
32.Determine the empirical formula for each of the following, using
the given masses:
(a)a compound containing 55.08 g carbon, 3.85 g hydrogen, and
61.07 g bromine
(b)a 65.2-g sample of a compound containing 36.8 g silver, 12.1 g
chlorine, and the rest, oxygen
(c) a 25.25-g sample of a sulfide of vanadium, containing 12.99 g V,
and the rest, sulfur
(d) a compound containing 38.0 g zinc and 12.0 g phosphorus
33. A 15.267-g sample of a sulfide of gold was found to contain 12.272 g
gold, and the rest, sulfur. Calculate the empirical formula of this
sulfide of gold.
34.A 10.724-g sample of an oxide of titanium was found to ­contain
7.143 g titanium, and the rest, oxygen. Calculate the empirical formula of this oxide of titanium.
35. Lenthionine, a naturally occurring sulfur compound, gives shitake
mushrooms their interesting flavor. This compound is composed of
carbon, sulfur, and hydrogen. A 5.000-g sample of lenthionine contains 0.6375 g carbon and 0.1070 g hydrogen, and the remainder
is sulfur. The molar mass of lenthionine is 188.4 g/mol. Determine
the empirical formula for lenthionine.
36. A 5.276-g sample of an unknown compound was found to contain
3.989 g of mercury and the rest, chlorine. Calculate the empirical
formula of this chloride of mercury.
37. Traumatic acid is a plant hormone that induces injured cells to
begin to divide and repair the trauma they have experienced. Traumatic acid contains 63.13% carbon, 8.830% hydrogen, and 28.03%
oxygen. Its molar mass is 228 g/mol. Determine the empirical and
molecular formulas of traumatic acid.
38. Dixanthogen, an herbicide, is composed of 29.73% carbon, 4.16%
hydrogen, 13.20% oxygen, and 52.91% sulfur. It has a molar
mass of 242.4 g/mol. Determine the empirical and the molecular
­formulas.
39.Ethanedioic acid, a compound that is present in many vegetables, has
a molar mass of 90.04 g/mol and a composition of 26.7% C, 2.2% H,
and 71.1% O. What is the molecular formula for this substance?
40. Butyric acid is a compound that is present in butter. It has a molar
mass of 88.11 g/mol and is composed of 54.5% C, 9.2% H, and
36.3% O. What is the molecular formula for this ­substance?
Additional Exercises 141
41. Calculate the percent composition and determine the molecular formula and the empirical formula for the ­nitrogen-oxygen compound
that results when 12.04 g of nitrogen are reacted with enough oxygen to produce 39.54 g of product. The molar mass of the product
is 92.02 g.
42. Calculate the percent composition and determine the molecular formula and the empirical formula for the carbon-hydrogen-oxygen
compound that results when 30.21 g of carbon, 40.24 g of oxygen,
and 5.08 g of hydrogen are reacted to produce a product with a
molar mass of 180.18 g.
43. The compound XYZ3 has a molar mass of 100.09 g and a percent
composition (by mass) of 40.04% X, 12.00% Y, and 47.96% Z.
What is the formula of the compound?
44. The compound X2(YZ3)3 has a molar mass of 282.23 g and a percent composition (by mass) of 19.12% X, 29.86% Y, and 51.02% Z.
What is the formula of the compound?
Additional Exercises
45. White phosphorus is one of several forms of phosphorus and exists
as a waxy solid consisting of P4 molecules. How many atoms are
present in 0.350 mol of P4?
46. How many grams of sodium contain the same number of atoms as
10.0 g of potassium?
47.One molecule of an unknown compound is found to have a mass of
3.27 * 10-22 g. What is the molar mass of this compound?
48. How many molecules of sugar are in a 5-lb bag of sugar? The formula for table sugar or sucrose is C12H22O11.
49.If a stack of 500 sheets of paper is 4.60 cm high, what will be the
height, in meters, of a stack of Avogadro’s number of sheets of
paper?
50.There are about 7.0 billion (7.0 * 109) people on Earth. If exactly
1 mol of dollars were distributed equally among these people, how
many dollars would each person ­receive?
51. If 20. drops of water equal 1.0 mL (1.0 cm3),
(a) how many drops of water are there in a cubic mile of water?
(b) what would be the volume in cubic miles of a mole of drops of
water?
52.Silver has a density of 10.5 g>cm3. If 1.00 mol of silver were
shaped into a cube,
(a) what would be the volume of the cube?
(b) what would be the length of one side of the cube?
53.Given 1.00-g samples of each of the compounds CO2 , O2 , H2O,
and CH3OH,
(a) which sample will contain the largest number of ­molecules?
(b)which sample will contain the largest number of atoms?
Show proof for your answers.
54.How many grams of Fe2S3 will contain a total number of atoms
equal to Avogadro’s number?
55.How many grams of calcium must be combined with 1 g phosphorus to form the compound Ca3P2 ?
56.An iron ore contains 5% FeSO4 by mass. How many grams of iron
could be obtained from 1.0 ton of this ore?
57. How many grams of lithium will combine with 20.0 g of sulfur to
form the compound Li2S?
58.Calculate the percentage of
(a) mercury in HgCO3
(b) oxygen in Ca(ClO3)2
(c) nitrogen in C10H14N2 (nicotine)
(d) Mg in C55H72MgN4O5 (chlorophyll)
59. Zinc and sulfur react to form zinc sulfide, ZnS. If we mix 19.5 g of
zinc and 9.40 g of sulfur, have we added sufficient sulfur to fully
react all the zinc? Show evidence for your answer.
60.Vomitoxin is produced by some fungi that grow on wheat and
barley. It derives its name from the fact that it causes pigs that
eat contaminated wheat to vomit. The chemical formula for
vomitoxin is C15H20O6. What is the percent composition of this
­compound?
61.Diphenhydramine hydrochloride, a drug used commonly as an antihistamine, has the formula C17H21NO # HCl. What is the percent
composition of each element in this compound?
62.What is the percent composition of each element in sucrose,
C12H22O11?
63.Aspirin is well known as a pain reliever (analgesic) and as a fever
reducer (antipyretic). It has a molar mass of 180.2 g>mol and a
composition of 60.0% C, 4.48% H, and 35.5% O. Calculate the
molecular formula of aspirin.
64. How many grams of oxygen are contained in 8.50 g of Al2(SO4)3 ?
65. Gallium arsenide is one of the newer materials used to make semiconductor chips for use in supercomputers. Its composition is
48.2% Ga and 51.8% As. What is the empirical ­formula?
66. Calcium tartrate is used as a preservative for certain foods and as
an antacid. It contains 25.5% C, 2.1% H, 21.3% Ca, and 51.0% O.
What is the empirical formula for calcium tartrate?
67.The compositions of four different compounds of carbon and chlorine follow. Determine the empirical formula and the molecular
formula for each compound.
Percent C
Percent Cl
Molar mass (g)
(a) 7.79
92.21
153.8
(b) 10.13
89.87
236.7
(c) 25.26
74.74
284.8
(d) 11.25
88.75
319.6
68.How many years is a mole of seconds?
69.A normal penny has a mass of about 2.5 g. If we assume the penny
to be pure copper (which means the penny is very old since newer
pennies are a mixture of copper and zinc), how many atoms of copper does it contain?
70.What would be the mass (in grams) of one thousand trillion molecules of glycerin (C3H8O3)?
71.If we assume that there are 7.0 billion people on the Earth, how
many moles of people is this?
72.An experimental catalyst used to make polymers has the following
composition: Co, 23.3%; Mo, 25.3%; and Cl, 51.4%. What is the
empirical formula for this compound?
73.If a student weighs 18 g of aluminum and needs twice as many
atoms of magnesium as she has of aluminum, how many grams of
Mg does she need?
74. If 10.0 g of an unknown compound composed of carbon, hydrogen,
and nitrogen contains 17.7% N and 3.8 * 1023 atoms of hydrogen,
what is its empirical formula?
75.A substance whose formula is A2O (A is a mystery element) is
60.0% A and 40.0% O. Identify the element A.
142 chapter 7
• Quantitative Composition of Compounds
76.For the following compounds whose molecular formulas are given,
indicate the empirical formula:
(a) C6H12O6
glucose
(b) C8H18
octane
(c) C3H6O3
lactic acid
(d) C25H52
paraffin
(e) C12H4Cl4O2
dioxin (a powerful poison)
77.Copper is often added to paint for ocean-going ships and boats to
discourage the growth of marine organisms on their hulls. Unfortunately the copper dissolves into the water and is toxic to the marine
life in harbors where these boats are docked. When copper(II) ions
reach a concentration of 9.0 microgram/L, the growth of some marine life slows. How many copper ions are found in a liter of water
containing 9.0 micrograms of Cu2 + /L?
78.Molecules of oxygen gas have been very difficult to find in space.
Recently the European Space Agency’s Herschel space observatory
has found molecular oxygen in the Orion star-forming complex.
It is present at very low concentrations, 1 molecule of oxygen per
every million (1.0 * 106 ) molecules of hydrogen, but it does exist.
How many molecules of oxygen would exist in a 3.0-g sample of
hydrogen gas?
79. Scientists from the European Space Agency have been studying geysers located around the southern pole of one of Saturn’s
moons. These geysers send 250.0 kg of water into the atmosphere every second. Scientists trying to discover what happens to this water have found that it forms a very thin vapor
ring around Saturn. How many moles of water are shot into the
­atmosphere in a Saturn day? (One Saturn day is equal to 10 hours
and 45 minutes.)
80. Researchers from the University of Cape Town in South Africa
have discovered that penguins are able to detect a chemical released
by plankton as they are being consumed by schools of fish. By following this scent they are able to find these schools of fish and feast
on them. The chemical they smell is composed of 38.65% carbon,
9.74% hydrogen, and 51.61% sulfur. Determine the empirical formula of this compound.
Challenge Exercises
81. The compound A(BC)3 has a molar mass of 78.01 g and a percent
composition of 34.59% A, 61.53% B, and 3.88% C. Determine the
identity of the elements A, B, and C. What is the percent composition by mass of the compound A2 B3 ?
82.A 2.500-g sample of an unknown compound containing only C, H,
and O was burned in O2. The products were 4.776 g of CO2 and
2.934 g of H2O.
(a) What is the percent composition of the original compound?
(b) What is the empirical formula of the compound?
83.Researchers at Anna Gudmundsdottir’s laboratory at the University of
Cincinnati have been studying extremely reactive chemicals known
as radicals. One of the interesting phenomena they have discovered
is that these radicals can be chemically attached to fragrance molecules, effectively tethering them to a solution. When light strikes
these tethered molecules, the fragrance is released. This property
would allow us to produce perfumes, cleansers, and other consumer
products that release fragrance only when exposed to light. If limonene, C10H16, the molecule that gives fruits their citrus scent, were
able to be tethered to one of these radicals and every photon of
light would release one molecule of limonene, calculate the time in
seconds required to release 1.00 picogram of limonene if ambient
light releases 2.64 * 1018 photons/sec.
Answers to Practice Exercises
7.1
(a) 6.022 * 1023 atoms Fe
7.6
8.326 mol and 5.014 * 1024 molecules HC2H3O2
(b) 1.204 * 1024 atoms H
7.7
24.42% Ca; 17.07% N; 58.50% O
(c) 4.215 * 10 atoms
7.8
40.27% K; 26.78% Cr; 32.96% O
10.0 g helium
7.9
20.16% Al; 79.84% Cl
24
7.2
22
7.3
1.5 * 10 atoms
7.10
C2H6O
7.4
101.1 g KNO3
7.11
P2O5
7.5
21.3 g Na2SO4
7.12 The empirical formula is CH3; the molecular formula is C2H6 .
• Properties of Gases 143
Flames and sparks result when
aluminum foil is dropped into
liquid bromine.
chapter
8
Chemical Equations
I
n the world today, we continually strive to express information
in a concise, useful manner. From early childhood, we are
taught to translate our ideas and desires into sentences. In
mathematics, we learn to describe numerical ­relationships and
situations through mathematical expressions and equations.
Historians describe thousands of years of history in 500-page text-
books. ­Filmmakers translate entire events, such as the Olympics,
into a few hours of entertainment.
Chemists use chemical equations to describe reactions they
observe in the laboratory or in nature. Chemical equations provide us with the means to (1) summarize the reaction, (2) display
the substances that are reacting, (3) show the products, and
(4) indicate the amounts of all component substances in a reaction.
Chapter Outline
8.1 The Chemical Equation
8.2 Writing and Balancing Chemical
Equations
8.3 Types of Chemical Equations
8.4 Heat in Chemical Reactions
8.5 Global Warming:
The Greenhouse Effect
Charles D. Winters/Photo Researchers
5.1
144 chapter 8
• Chemical Equations
8.1The Chemical Equation
Learning objective
Describe the information present in a chemical equation.
Key terms
Chemical reactions always involve change. Atoms, molecules, or ions rearrange to form different substances, sometimes in a spectacular manner. For example, the thermite reaction is a
reaction between aluminum metal and iron(III) oxide, which produces molten iron and aluminum oxide. The substances entering the reaction are called the reactants, and the substances
formed are called the products. In our example,
reactants
products
chemical equation
law of conservation of mass
© Phil Degginger/Alamy
reactantsaluminum
iron(III) oxide
productsiron
aluminum oxide
During reactions, chemical bonds are broken and new bonds are formed. The reactants and
products may be present as solids, liquids, gases, or in solution.
In a chemical reaction atoms are neither created nor destroyed. All atoms present in the
reactants must also be present in the ­products.
A chemical equation is a shorthand expression for a chemical change or reaction. A chemical
equation uses the chemical symbols and formulas of the reactants and products and other symbolic
terms to represent a chemical ­reaction. The equations are written according to this general format:
1. Reactants are separated from products by an arrow (¡) that indicates the direction of
the reaction. The reactants are placed to the left and the products to the right of the arrow.
A plus sign (+) is placed between ­reactants and between products when needed.
Al + Fe2O3 ¡ Fe + Al2O3
reactants
Molten iron falls from the reaction
container during the thermite
reaction.
Table 8.1 Symbols Commonly
Used in Chemical Equations
Symbol
Meaning
+ Plus or added to
(placed between
substances)
¡ Yields; produces
(points to products)
(s)Solid state (written
after a substance)
(l)Liquid state (written
after a substance)
(g)Gaseous state (written
after a substance)
(aq)Aqueous solution
(substance dissolved
in water)
Heat is added (when
written above or
below arrow)
products
2. Coefficients (whole numbers) are placed in front of substances to ­balance the equation and
to indicate the number of units (atoms, ­molecules, moles, ions) of each substance reacting or being produced. When no number is shown, it is understood that one unit of the
­substance is indicated.
2 Al + Fe2O3 ¡ 2 Fe + Al2O3
3. Conditions required to carry out the reaction may, if desired, be placed above or below

the arrow or equality sign. For example, a delta sign placed over the arrow (S) indicates
that heat is supplied to the reaction.

2 Al + Fe2O3 ¡ 2 Fe + Al2O3
4. The physical state of a substance is indicated by the following symbols: (s) for solid state;
(l) for liquid state; (g) for gaseous state; and (aq) for substances in aqueous solution. States
are not always given in chemical equations.

2 Al(s) + Fe2O3(s) ¡ Fe(l) + Al2O3(s)
Symbols commonly used in chemical equations are given in Table 8.1.
Example 8.1
Classify each formula in the chemical reactions below as a reactant or a product.
(a) MnO2 + 4 HCl ¡ MnCl2 + Cl2 + 2 H2O
(b) Na2SO3 + H2SO4 ¡ Na2SO4 + SO2 + H2O
SOLUTION
(a)Reactants: MnO2 and HCl
Products: MnCl2, Cl2, and H2O
(b)Reactants: Na2SO3, and H2SO4
Products: Na2SO4, SO2, and H2O
Practice 8.1
How is a substance in a chemical equation identified as a solid, a gas, or in a water solution?
8.2
• Writing and Balancing Chemical Equations 145
Conservation of Mass
Yuri Gripas/Reuters/Landov LLC
The law of conservation of mass states that no change is observed in the total mass of
the substances involved in a chemical change. This law, tested by extensive laboratory
experimentation, is the basis for the quantitative mass relationships among reactants and
products.
The decomposition of water into hydrogen and oxygen illustrates this law. Thus, 100.0 g
of water decompose into 11.2 g of hydrogen and 88.8 g of oxygen:
water
100.0 g
100.0 g
reactant
¡
hydrogen
11.2 g
¡
+
oxygen
88.8 g
100.0 g
products
In a chemical reaction
mass of reactants 5 mass of products
Combustion of gasoline is a
chemical change.
8.2 Writing and Balancing
Chemical Equations
Write and balance chemical equations.
L earning objectiv e
To represent the quantitative relationships of a reaction, the chemical ­equation must
be balanced. A balanced equation contains the same number of each kind of atom on
each side of the equation. The balanced equation therefore obeys the law of conservation
of mass.
Every chemistry student must learn to balance equations. Many equations are balanced by
trial and error, but care and attention to detail are still required. The way to balance an equation
is to adjust the number of atoms of each ­element so that they are the same on each side of the
equation, but a correct formula is never changed in order to balance an equation. The general
procedure for balancing equations is as follows:
Key term
balanced equation
Correct formulas are not changed
to balance an equation.
Problem-Solving Strategy: For Writing and Balancing a Chemical Equation
1. Identify the reaction. Write a description or word equation for the reaction. For example,
let’s consider mercury(II) oxide ­decomposing into mercury and oxygen.

mercury(II) oxide ¡
mercury + oxygen
2. Write the unbalanced (skeleton) equation. Make sure that the formula for each substance
is correct and that reactants are written to the left and products to the right of the arrow. For
our example,
Study this procedure carefully
and refer to it when you work
examples.

HgO ¡
Hg + O2
The correct formulas must be known or determined from the ­periodic table, lists of ions, or
experimental data.
3. Balance the equation. Use the following process as necessary:
(a)Count and compare the number of atoms of each element on each side of the equation
and determine those that must be balanced:
Hg is balanced
(1 on each side)
O needs to be balanced
(1 on reactant side, 2 on product side)
(b) B
alance each element, one at a time, by placing whole numbers (coefficients) in front
of the formulas containing the unbalanced element. It is usually best to balance metals
first, then nonmetals, then hydrogen and oxygen. Select the smallest coefficients that
will give the same number of atoms of the element on each side. A coefficient placed in
front of a formula multiplies every atom in the formula by that number (e.g., 2 H2SO4
Leave elements that are in two or
more formulas (on the same side
of the equation) unbalanced until
just before balancing hydrogen
and oxygen.
146 chapter 8
• Chemical Equations
means two units of sulfuric acid and also means four H atoms, two S atoms, and eight
O atoms). Place a 2 in front of HgO to balance O:

2 HgO ¡ Hg + O2
(c)Check all other elements after each individual element is ­balanced to see whether, in
balancing one element, other ­elements have become unbalanced. Make adjustments as
needed. Now Hg is not balanced. To adjust this, we write a 2 in front of Hg:

2 HgO ¡ 2 Hg + O2
(balanced)
(d)Do a final check, making sure that each element and /or polyatomic ion is balanced and
that the smallest possible set of whole-number coefficients has been used:

(correct form)

(incorrect form)
2 HgO ¡ 2 Hg + O2
4 HgO ¡ 4 Hg + 2 O2
Not all chemical equations can be balanced by the simple method of ­inspection just
described. The following examples show stepwise sequences leading to balanced equations.
Study each one carefully.
Example 8.2
ENHANCED EXAMPLE
Write the balanced equation for the reaction that takes place when magnesium metal is
burned in air to produce magnesium oxide.
SOLUTION
Use the Problem-Solving Strategy for Writing and Balancing a ­Chemical Equation.
Elements shown in color are
­balanced.
R 5 reactant
P 5 product
(a)
(b)
(c)
R
1 Mg
2O
P
1 Mg
1O
R
1 Mg
2O
P
2 Mg
2O
R
2 Mg
2O
P
2 Mg
2O
1. Word equation:
magnesium + oxygen ¡ magnesium oxide
reactants (R)
product (P)
2. Skeleton equation:
Mg + O2 ¡ MgO
(unbalanced)
3. Balance:
(a) Mg is balanced.
Oxygen is not balanced. Two O atoms appear on the left side and one on the right side.
(b) Place the coefficient 2 in front of MgO to balance oxygen:
Mg + O2 ¡ 2 MgO
(unbalanced)
(c) Now Mg is not balanced. One Mg atom appears on the left side and two on the right
side. Place a 2 in front of Mg:
2 Mg + O2 ¡ 2 MgO
(balanced)
(d) Check: Each side has two Mg and two O atoms.
Example 8.3
When methane, CH4, undergoes complete combustion, it reacts with oxygen to produce
carbon dioxide and water. Write the balanced ­equation for this ­reaction.
(a)
R
1 C
4 H
2O
P
1 C
2 H
3O
SOLUTION
Use the Problem-Solving Strategy for Writing and Balancing a ­Chemical Equation.
1. Word equation:
methane + oxygen ¡ carbon dioxide + water
2. Skeleton equation:
CH4 + O2 ¡ CO2 + H2O (unbalanced)
3. Balance:
(a) Carbon is balanced.
Hydrogen and oxygen are not balanced.
8.2
• Writing and Balancing Chemical Equations 147
(b) Balance H atoms by placing a 2 in front of H2O:
CH4 + O2 ¡ CO2 + 2 H2O (unbalanced)
Each side of the equation has four H atoms; oxygen is still not balanced. Place a
2 in front of O2 to balance the oxygen atoms:
CH4 + 2 O2 ¡ CO2 + 2 H2O (balanced)
(c) The other atoms remain balanced.
(d) Check: The equation is correctly balanced; it has one C, four O, and four H atoms
on each side.
(b)
(c)
R
1 C
4 H
2O
P
1 C
4 H
4O
R
1 C
4 H
4O
P
1 C
4 H
4O
Example 8.4
Oxygen and potassium chloride are formed by heating potassium ­chlorate. Write a
balanced equation for this reaction.
SOLUTION
Use the Problem-Solving Strategy for Writing and Balancing a ­Chemical Equation.
1. Word equation:

potassium chlorate ¡ potassium chloride + oxygen
2. Skeleton equation:

KClO3 ¡ KCl + O2 (unbalanced)
3. Balance:
(a) Potassium and chlorine are balanced.
Oxygen is unbalanced (three O atoms on the left and two on the right side).
(b) How many oxygen atoms are needed? The subscripts of oxygen (3 and 2) in KClO3
and O2 have a least common multiple of 6. Therefore, coefficients for KClO3 and
O2 are needed to get six O atoms on each side. Place a 2 in front of KClO3 and a 3
in front of O2 to balance oxygen:

2 KClO3 ¡ KCl + 3 O2 (unbalanced)
(c) Now K and Cl are not balanced. Place a 2 in front of KCl, which balances both K
and Cl at the same time:

2 KClO3 ¡ 2 KCl + 3 O2 (balanced)
(d) Check: Each side now contains two K, two Cl, and six O atoms.
(a)
(b)
(c)
R
1K
1 Cl
3O
P
1K
1 Cl
2O
R
2K
2 Cl
6O
P
1K
1 Cl
6O
R
2K
2 Cl
6O
P
2K
2 Cl
6O
R
1 Ag
2 H
1 S
1 NO3
P
2 Ag
1 H
1 S
1 NO3
R
2 Ag
2 H
1 S
2 NO3
P
2 Ag
1 H
1 S
1 NO3
R
2 Ag
2 H
1 S
2 NO3
P
2 Ag
2 H
1 S
2 NO3
Example 8.5
Silver nitrate reacts with hydrogen sulfide to produce silver sulfide and nitric acid. Write
a balanced equation for this reaction.
SOLUTION
Use the Problem-Solving Strategy for Writing and Balancing a ­Chemical Equation.
1. Word equation:
silver nitrate + hydrogen sulfide ¡ silver sulfide + nitric acid
2. Skeleton equation:
AgNO3 + H2S ¡ Ag2S + HNO3
(unbalanced)
3. Balance:
(a) Ag and H are unbalanced.
(b) Place a 2 in front of AgNO3 to balance Ag:
2 AgNO3 + H2S ¡ Ag2S + HNO3 (unbalanced)
(c) H and NO3- are still unbalanced. Balance by placing a 2 in front of HNO3:
2 AgNO3 + H2S ¡ Ag2S + 2 HNO3
(balanced)
In this example, N and O atoms are balanced by balancing the NO3- ion as a unit.
(d) The other atoms remain balanced.
(e) Check: Each side has two Ag, two H, and one S atom. Also, each side has two NO3- ions.
(a)
(b)
(c)
148 chapter 8
• Chemical Equations
Example 8.6
When aluminum hydroxide is mixed with sulfuric acid, the products are aluminum sulfate
and water. Write a balanced equation for this ­reaction.
SOLUTION
Use the Problem-Solving Strategy for Writing and Balancing a ­Chemical Equation.
1. Word equation:
aluminum hydroxide + sulfuric acid ¡ aluminum sulfate + water
2. Skeleton equation:
Al(OH)3 + H2SO4 ¡ Al2(SO4)3 + H2O
(a)
(b)
(c)
R
1 AI
1 SO4
3 O
5H
P
2 AI
3 SO4
1 O
2H
R
2 AI
3 SO4
6 O
12 H
P
2 AI
3 SO4
1 O
2H
R
2 AI
3 SO4
6 O
12 H
P
2 AI
3 SO4
6 O
12 H
(unbalanced)
3. Balance:
(a) All elements are unbalanced.
(b) Balance Al by placing a 2 in front of Al(OH)3. Treat the unbalanced SO24 ion as a
unit and balance by placing a 3 in front of H2SO4:
2 Al(OH)3 + 3 H2SO4 ¡ Al2(SO4)3 + H2O
(unbalanced)
Balance the unbalanced H and O by placing a 6 in front of H2O:
2 Al(OH)3 + 3 H2SO4 ¡ Al2(SO4)3 + 6 H2O (balanced)
(c) The other atoms remain balanced.
(d) Check: Each side has 2 Al, 12 H, 3 S, and 18 O atoms.
Example 8.7
When the fuel in a butane gas stove undergoes complete combustion, it reacts with
oxygen to form carbon dioxide and water. Write the balanced equation for this ­reaction.
SOLUTION
Use the Problem-Solving Strategy for Writing and Balancing a ­Chemical Equation.
1. Word equation:
Butane
Butane
C4H10
butane + oxygen ¡ carbon dioxide + water
2. Skeleton equation:
C4H10 + O2 ¡ CO2 + H2O
(a)
(b)
(c)
R
4 C
10 H
2O
P
1 C
2 H
3O
R
4 C
10 H
2O
P
4 C
10 H
13 O
R
8 C
20 H
26 O
P
8 C
20 H
26 O
(unbalanced)
3. Balance:
(a) All elements are unbalanced.
(b) Balance C by placing a 4 in front of CO2:
C4H10 + O2 ¡ 4 CO2 + H2O
(unbalanced)
Balance H by placing a 5 in front of H2O:
C4H10 + O2 ¡ 4 CO2 + 5 H2O
(unbalanced)
Oxygen remains unbalanced. The oxygen atoms on the right side are fixed because
4 CO2 and 5 H2O are derived from the single C4H10 molecule on the left. When
we try to balance the O atoms, we find that there is no whole number that can be
placed in front of O2 to bring about a balance, so we double the coefficients of each
substance, and then balance the oxygen:
2 C4H10 + 13 O2 ¡ 8 CO2 + 10 H2O (balanced)
(c) The other atoms remain balanced.
(d) Check: Each side now has 8 C, 20 H, and 26 O atoms.
Practice 8.2
Write a balanced formula equation for:
aluminum + oxygen ¡ aluminum oxide
8.2
• Writing and Balancing Chemical Equations 149
Practice 8.3
Write a balanced formula equation for:
magnesium hydroxide + phosphoric acid ¡ magnesium phosphate + water
Information in a Chemical Equation
Depending on the particular context in which it is used, a formula can have different meanings.
A formula can refer to an individual chemical entity (atom, ion, molecule, or formula unit) or
to a mole of that chemical entity. For example, the formula H2O can mean any of the following:
•
•
•
•
•
2 H atoms and 1 O atom
1 molecule of water
1 mol of water
6.022 * 1023 molecules of water
18.02 g of water
Formulas used in equations can represent units of individual chemical entities or moles,
the latter being more commonly used. For example, in the reaction of hydrogen and
oxygen to form water,
+
2 H2
2 molecules hydrogen
2 mol hydrogen
¡
O2
1 molecule oxygen
1 mol oxygen
2 H2 O
2 molecules water
2 mol water
As indicated earlier, a chemical equation is a shorthand description of a chemical reaction.
Interpretation of a balanced equation gives us the following information:
1.
2.
3.
4.
5.
What the reactants are and what the products are
The formulas of the reactants and products
The number of molecules or formula units of reactants and products in the ­reaction
The number of atoms of each element involved in the reaction
The number of moles of each substance
Consider the reaction that occurs when propane gas (C3H8) is burned in air; the products
are carbon dioxide (CO2) and water (H2O). The balanced ­equation and its interpretation are
as follows:
Propane
C3H8(g)
Oxygen
�
Carbon
dioxide
Water
�
5 O2(g)
3 CO2(g)
4 H2O(g)
1 molecule
5 molecules
3 molecules
4 molecules
3 atoms C
8 atoms H
10 atoms O
3 atoms C
6 atoms O
8 atoms H
4 atoms O
1 mol
5 mol
3 mol
4 mol
44.09 g
5(32.00 g)
= (160.0 g)
3(44.01 g)
= (132.0 g)
4(18.02 g)
= (72.08 g)
We generally use moles in
equations because molecules
are so small.
150 chapter 8
• Chemical Equations
Practice 8.4
Consider the reaction that occurs when hydrogen gas reacts with chlorine gas to produce
gaseous hydrogen chloride:
(a) Write a word equation for this reaction.
(b) Write a balanced formula equation including the state for each ­substance.
(c) Label each reactant and product to show the relative amounts of each ­substance.
(1) number of molecules (3) number of moles
(2) number of atoms
(4) mass
(d)What mass of HCl would be produced if you reacted 2 mol hydrogen gas with 2 mol
chlorine gas?
The quantities involved in chemical reactions are important when working in industry or
the laboratory. We will study the relationship among quantities of reactants and products in
the next chapter.
8.3Types of Chemical Equations
Learning objective
Give examples of a combination reaction, decomposition reaction, single-­
displacement reaction, and double-displacement reaction.
Key tERMs
Chemical equations represent chemical changes or reactions. Reactions are classified into
types to assist in writing equations and in predicting other reactions. Many chemical reactions fit one or another of the four principal reaction types that we discuss in the following
paragraphs. Reactions are also classified as oxidation–reduction. Special methods are used to
balance complex ­oxidation–reduction equations. (See Chapter 17.)
combination reaction
decomposition reaction
single-displacement reaction
double-displacement reaction
Combination Reaction
In a combination reaction, two reactants combine to give one product. The general form of
the equation is
A + B ¡ AB
Charles D. Winters/Photo Researchers, Inc.
in which A and B are either elements or compounds and AB is a compound. The formula of
the compound in many cases can be determined from knowledge of the ionic charges of the
reactants in their combined states. Some ­reactions that fall into this category are given here:
1. metal + oxygen ¡ metal oxide:

2 Mg(s) + O2(g) ¡ 2 MgO(s)

4 Al(s) + 3 O2(g) ¡ 2 Al2O3(s)
2. nonmetal + oxygen ¡ nonmetal oxide:

A strip of Mg burning in air
produces MgO (left of the flame).
S(s) + O2(g) ¡ SO2(g)

N2(g) + O2(g) ¡ 2 NO(g)
3. metal + nonmetal ¡ salt:
2 Na(s) + Cl2(g) ¡ 2 NaCl(s)
2 Al(s) + 3 Br2(l) ¡ 2 AlBr3(s)
4. metal oxide + water ¡ metal hydroxide:
Na2O(s) + H2O(l) ¡ 2 NaOH(aq)
CaO(s) + H2O(l) ¡ Ca(OH)2(aq)
5. nonmetal oxide + water ¡ oxy-acid:
SO3(g) + H2O(l) ¡ H2SO4(aq)
N2O5(s) + H2O(l) ¡ 2 HNO3(aq)
8.3
• Types of Chemical Equations 151
>Chemistry in Action
Carbon monoxide (CO) is one possible product when
carbon-containing fuels such as coal, natural gas,
propane, heating oil, and gasoline are burned. For
­example, when methane, the main component of
­natural gas, burns, the reaction is
CH4(g) + 2 O2(g) ¡ 2 H2O(g) + CO2(g)
However, if not enough oxygen is available,
carbon monoxide will be produced instead of
carbon dioxide:
2 CH4(g) + 3 O2(g) ¡ 4 H2O(g) + 2 CO(g)
Carbon monoxide is colorless and has no flavor or
smell. It often can poison several people at once,
since it can build up undetected in enclosed spaces.
The symptoms of carbon monoxide poisoning are so
easily missed that people often don’t realize that they
have been poisoned. Common symptoms are head­­
ache, ringing in ears, nausea, dizziness, weakness,
confusion, and drowsiness. A low ­concentration of
CO in the air may go completely unnoticed while
the toxicity level increases in the blood.
Carbon monoxide in the blood is poisonous because
of its ability to bond to a hemoglobin molecule. Both
CO and O2 bond to the same place on the hemoglobin
molecule. The CO bonds more tightly than O2. We use
hemoglobin to transport oxygen to the various tissues
from the lungs. Unfortunately, CO also bonds to hemoglobin, forming a molecule called carboxyhemoglobin.
In fact, hemoglobin prefers CO to O2. Once a CO is
bound to a hemoglobin molecule, it cannot carry an
oxygen molecule. So, until the CO is released, that
molecule of hemoglobin is lost as an O2 carrier. If you
breathe normal air (about 20% oxygen) containing
0.1% CO, in just one hour the CO bonds to 50% of the
hemoglobin molecules. To keep CO from being a silent
killer, there are three possible actions:
If a person already has CO poisoning, the treatment must
focus on getting CO released from the hemoglobin. This
can be done by giving the victim oxygen or by placing the
victim in a hyperbaric chamber (at about 3 atm pressure).
These treatments reduce the time ­required for carboxyhemoglobin to release CO from between 4 to 6 hours to
30 to 90 min. This treatment also supplies more oxygen to
the ­system to help keep the victim’s brain functioning.
A CO detector can sound an alarm before the CO level
becomes toxic in a home. Current CO detectors sound an
alarm when levels increase rapidly or when lower concentrations are present over a long time. Inside a CO detector
is a beam of infrared light that shines on a chromophore
(a substance that turns darker with increasing CO levels).
If the infrared light passing through the chromophore is
too low, the alarm sounds. Typically, these alarms ring
when CO levels reach 70 ppm, or greater than 30 ppm
for 30 days.
The final option is to convert CO to CO2 before toxicity
occurs. David Schre­yer from NASA’s Langley Research Center has devised a catalyst (tin hydroxide with small amounts
of Pd) that speeds up the conversion of CO to CO2 at room
temperature. He and his colleagues have adapted their
catalyst to home ventilation systems to eliminate even low
levels of CO before it has a chance to build up. Using all of
these approaches can help to eliminate 500–10,000 cases
of CO poisoning each year.
Mark Harmel/Photo Researchers
CO Poisoning—A Silent Killer
1. Immediate treatment of victims to restore the
hemoglobin
2. Detection of CO levels before ­poison­ing occurs
3. Conversion of CO to CO2 to ­elimi­nate the threat
People in a hyperbaric chamber.
Decomposition Reaction
In a decomposition reaction, a single substance is decomposed, or broken down, to give two
or more different substances. This reaction may be considered the reverse of combination. The
starting material must be a compound, and the products may be elements or compounds.
The general form of the equation is
AB ¡ A + B
Predicting the products of a decomposition reaction can be difficult and ­requires an understanding of each individual reaction. Heating oxygen-­containing compounds often results in
decomposition. The following reactions fall into this category.
152 chapter 8
• Chemical Equations
Richard Megna/Fundamental Photographs
1. Some metal oxides decompose to yield the free metal plus oxygen; others give another
oxide, and some are very stable, resisting ­decomposition by heating:

2 HgO(s) ¡
2 Hg(l) + O2(g)

2 PbO2(s) ¡ 2 PbO(s) + O2(g)
2. Carbonates and hydrogen carbonates decompose to yield CO2 when ­heated:

CaCO3(s) ¡
CaO(s) + CO2(g)

Na2CO3(s) + H2O(g) + CO2(g)
2 NaHCO3(s) ¡
3. Miscellaneous reactions in this category:

2 KClO3(s) ¡
2 KCl(s) + 3 O2(g)

2 NaNO3(s) ¡ 2 NaNO2(s) + O2(g)

2 H2O2(l) ¡
2 H2O(l) + O2(g)
Single-Displacement Reaction
Hydrogen peroxide decomposes
to steam (H2O(g)) and oxygen.
In a single-displacement reaction, one element reacts with a compound to replace one of
the elements of that compound, yielding a different element and a different compound. The
general forms of the equation follow.
If A is a metal, A will replace B to form AC, provided that A is a more ­reactive metal than B.
A + BC ¡ B + AC
If A is a halogen, it will replace C to form BA, provided that A is a more reactive halogen than C.
Charles D. Winters/Photo Researchers
A + BC ¡ C + BA
When pieces of zinc metal are
placed in hydrochloric acid,
hydrogen bubbles form
immediately.
Table 8.2 Activity Series
increasing activity
Metals
K
Ca
Na
Mg
Al
Zn
Fe
Ni
Sn
Pb
H
Cu
Ag
Hg
Au
Halogens
F2
Cl2
Br2
I2
A brief activity series of selected metals (and hydrogen) and halogens is shown in Table 8.2.
This series is listed in descending order of chemical activity, with the most active metals and
halogens at the top. Many chemical reactions can be predicted from an activity series because
the atoms of any element in the series will replace the atoms of those elements below it. For
example, zinc metal will replace hydrogen from a hydrochloric acid solution. But copper metal,
which is below hydrogen on the list and thus less reactive than hydrogen, will not replace hydrogen from a hydrochloric acid solution. Here are some reactions that fall into this category:
1. metal + acid ¡ hydrogen + salt
Zn(s) + 2 HCl(aq) ¡ H2(g) + ZnCl2(aq)
2 Al(s) + 3 H2SO4(aq) ¡ 3 H2(g) + Al2(SO4)3(aq)
2. metal + water ¡ hydrogen + metal hydroxide or metal oxide
2 Na(s) + 2 H2O ¡ H2(g) + 2 NaOH(aq)
Ca(s) + 2 H2O ¡ H2(g) + Ca(OH)2(aq)
3 Fe(s) + 4 H2O(g) ¡ 4 H2(g) + Fe3O4(s)
steam
3. metal + salt ¡ metal + salt
Fe(s) + CuSO4(aq) ¡ Cu(s) + FeSO4(aq)
Cu(s) + 2 AgNO3(aq) ¡ 2 Ag(s) + Cu(NO3)2(aq)
4. halogen + halide salt ¡ halogen + halide salt
Cl2(g) + 2 NaBr(aq) ¡ Br2(l) + 2 NaCl(aq)
Cl2(g) + 2 KI(aq) ¡ I2(s) + 2 KCl(aq)
A common chemical reaction is the displacement of hydrogen from water or acids
(shown in 1 and 2 above). This reaction is a good illustration of the relative reactivity of
metals and the use of the activity series. For example,
•
•
•
•
K, Ca, and Na displace hydrogen from cold water, steam (H2O), and acids.
Mg, Al, Zn, and Fe displace hydrogen from steam and acids.
Ni, Sn, and Pb displace hydrogen only from acids.
Cu, Ag, Hg, and Au do not displace hydrogen.
8.3
• Types of Chemical Equations 153
Example 8.8
Will a reaction occur between (a) nickel metal and hydrochloric acid and (b) tin metal
and a solution of aluminum chloride? Write balanced equations for the ­reactions.
SOLUTION
(a)Nickel is more reactive than hydrogen, so it will displace hydrogen from ­hydrochloric
acid. The products are hydrogen gas and a salt of Ni2+ and Cl- ions:
Ni(s) + 2 HCl(aq) ¡ H2(g) + NiCl2(aq)
(b) According to the activity series, tin is less reactive than aluminum, so no reaction will occur:
Sn(s) + AlCl3(aq) ¡ no reaction
Practice 8.5
Write balanced equations for these reactions:
(a) iron metal and a solution of magnesium chloride
(b) zinc metal and a solution of lead(II) nitrate
Double-Displacement Reaction
In a double-displacement reaction, two compounds exchange partners with each other to
produce two different compounds. The general form of the equation is
AB + CD ¡ AD + CB
This reaction can be thought of as an exchange of positive and negative groups, in which
A combines with D and C combines with B. In writing formulas for the products, we must
account for the charges of the combining groups.
It’s also possible to write an equation in the form of a double-displacement reaction when
a reaction has not occurred. For example, when solutions of sodium chloride and potassium
nitrate are mixed, the following equation can be written:
NaCl(aq) + KNO3(aq) ¡ NaNO3(aq) + KCl(aq)
When the procedure is carried out, no physical changes are observed, indicating that no chemical reaction has taken place.
A double-displacement reaction is accompanied by evidence of reaction such as:
1. The evolution of heat
2. The formation of an insoluble precipitate
3. The production of gas bubbles
Let’s look at some of these reactions more closely:
Neutralization of an acid and a base The production of a molecule of water from
an H + and an OH- ion is accompanied by a release of heat, which can be detected by
touching the reaction container or by using a thermometer. For neutralization reactions,
H+ + OH- ¡ H2O:
acid + base ¡ salt + water + heat
HCl(aq) + NaOH(aq) ¡ NaCl(aq) + H2O(l) + heat
H2SO4(aq) + Ba(OH)2(aq) ¡ BaSO4(s) + 2 H2O(l) + heat
Metal oxide + acid Heat is released by the production of a molecule of water.
metal oxide + acid ¡ salt + water + heat
CuO(s) + 2 HNO3(aq) ¡ Cu(NO3)2(aq) + H2O(l) + heat
CaO(s) + 2 HCl(aq) ¡ CaCl2(aq) + H2O(l) + heat
ENHANCED EXAMPLE
chapter 8
• Chemical Equations
Richard Megna/Fundamental Photographs
154 Formation of an insoluble precipitate The solubilities of the products can be determined
by consulting the solubility table in Appendix V to see whether one or both of the products
are insoluble in water. An insoluble product (precipitate) is indicated by placing (s) after its
formula in the equation.
BaCl2(aq) + 2 AgNO3(aq) ¡ 2 AgCl(s) + Ba(NO3)2(aq)
Pb(NO3)2(aq) + 2 KI(aq) ¡ PbI2(s) + 2 KNO3(aq)
Formation of a gas A gas such as HCl or H2S may be produced directly, as in these two
examples:
H2SO4(l) + NaCl(s) ¡ NaHSO4(s) + HCl(g)
2 HCl(aq) + ZnS(s) ¡ ZnCl2(aq) + H2S(g)
A double-displacement reaction
A gas can also be produced indirectly. Some unstable compounds formed in a double-displacement
results from pouring a clear,
reaction, such as H2CO3, H2SO3, and NH4OH, will decompose to form water and a gas:
colorless solution of Pb(NO3)2
into a clear, colorless solution 2 HCl(aq) + Na2CO3(aq) ¡ 2 NaCl(aq) + H2CO3(aq) ¡ 2 NaCl(aq) + H2O(l) + CO2(g)
of KI, forming a yellow
2 HNO3(aq) + K2SO3(aq) ¡ 2 KNO3(aq) + H2SO3(aq) ¡ 2 KNO3(aq) + H2O(l) + SO2(g)
precipitate of Pbl2.
Pb(NO3)2 1 2KI ¡
NH4Cl(aq) + NaOH(aq) ¡ NaCl(aq) + NH4OH(aq) ¡ NaCl(aq) + H2O(l) + NH3(g)
PbI2 1 2KNO3.
Example 8.9
Write the equation for the reaction between aqueous solutions of ­hydrobromic acid and
potassium hydroxide.
SOLUTION
First we write the formulas for the reactants. They are HBr and KOH. Then we classify
the type of reaction that would occur between them. Because the ­reactants are compounds,
one an acid and the other a base, the reaction will be of the neutralization type:
acid + base ¡ salt + water
Now rewrite the equation using the formulas for the known substances:
HBr(aq) + KOH(aq) ¡ salt + H2O
In this reaction, which is a double-displacement type, the H+ from the acid combines with
the OH− from the base to form water. The ionic compound must be composed of the other
two ions, K+ and Br-. We determine the formula of the ionic compound to be KBr from
the fact that K is a +1 cation and Br is a −1 anion. The final balanced equation is
Richard Megna/Fundamental Photographs
HBr(aq) + KOH(aq) ¡ KBr(aq) + H2O(l)
When barium chloride is poured
into a solution of sodium sulfate, a
white precipitate of barium sulfate
forms.
Example 8.10
Complete and balance the equation for the reaction between aqueous solutions of barium
chloride and sodium sulfate.
SOLUTION
First determine the formulas for the reactants. They are BaCl2 and Na2SO4. Then classify these substances as acids, bases, or ionic compounds. Both substances are ionic
compounds. Since both substances are compounds, the reaction will be of the doubledisplacement type. Start writing the equation with the reactants:
BaCl2(aq) + Na2SO4(aq) ¡
If the reaction is double-displacement, Ba2 + will be written combined with SO24 - , and
Na + with Cl - as the products. The balanced equation is
BaCl2(aq) + Na2SO4(aq) ¡ BaSO4(s) + 2 NaCl(aq)
8.3
• Types of Chemical Equations 155
The final step is to determine the nature of the products, which controls whether or not
the reaction will take place. If both products are soluble, we have a mixture of all the ions
in solution. But if an insoluble precipitate is formed, the reaction will definitely occur.
We know from experience that NaCl is fairly soluble in water, but what about BaSO4?
Consulting the solubility table in Appendix V, we see that BaSO4 is insoluble in water, so
it will be a precipitate in the reaction. Thus, the reaction will occur, forming a precipitate.
The equation is
BaCl2(aq) + Na2SO4(aq) ¡ BaSO4(s) + 2 NaCl(aq)
Practice 8.6
Complete and balance the equations for these reactions in water:
(a) potassium phosphate + barium chloride
(b) hydrochloric acid + nickel carbonate
(c) ammonium chloride + sodium nitrate
Some of the reactions you attempt may fail because the substances are not reactive or
because the proper conditions for reaction are not present. For example, mercury(II) oxide
does not decompose until it is heated; magnesium does not burn in air or oxygen until
the temperature reaches a certain point. When silver is placed in a ­solution of copper(II)
sulfate, no reaction occurs. When copper wire is placed in a solution of silver nitrate,
a single-displacement reaction takes place because copper is a more reactive metal than
silver. (See Figure 8.1.)
Michael Watson
Michael Watson
Michael Watson
Cu(s) + 2 AgNO3(aq) ¡ 2Ag(s) + Cu(NO3)2 (aq)
Ag+
NO3–
Cu2+
A solution of silver nitrate
contains Ag+ ions and
NO3– ions.
A copper wire is placed in a
solution of silver nitrate.
After 24 hours, crystals of silver
are seen hanging on the copper
wire and the solution has turned
blue, indicating copper ions are
present there.
Figure 8.1
Reaction of a copper wire and silver nitrate solution.
The silver metal clings to
the copper wire, and the
blue solution contains Cu2+
and NO3– ions.
156 chapter 8
• Chemical Equations
The successful prediction of the products of a reaction is not always easy. The ability to
predict products correctly comes with knowledge and experience. Although you may not be able
to predict many reactions at this point, as you continue to experiment you will find that reactions
can be categorized and that prediction of the products becomes easier, if not always certain.
8.4 Heat in Chemical Reactions
Lear ning objective
Explain the following terms and how they relate to a chemical reaction: exothermic
reaction, endothermic reaction, heat of reaction, and activation energy.
Key Terms
Energy changes always accompany chemical reactions. One reason reactions occur is that
the products attain a lower, more stable energy state than the reactants. When the reaction
leads to a more stable state, energy is released to the surroundings as heat and/or work. When
a solution of a base is neutralized by the addition of an acid, the liberation of heat energy is
signaled by an immediate rise in the temperature of the solution. When an automobile engine
burns gasoline, heat is certainly liberated; at the same time, part of the liberated energy does
the work of moving the automobile.
Reactions are either exothermic or endothermic. Exothermic reactions liberate heat;
endothermic reactions absorb heat. In an exothermic reaction, heat is a product and may be
written on the right side of the equation for the reaction. In an endothermic reaction, heat can
be regarded as a reactant and is written on the left side of the equation. Here are two examples:
exothermic reactions
endothermic reactions
heat of reaction
hydrocarbons
activation energy
H2(g) + Cl2(g) ¡ 2 HCl(g) + 185 kJ
N2(g) + O2(g) + 181 kJ ¡ 2 NO(g)
(exothermic)
(endothermic)
The quantity of heat produced by a reaction is known as the heat of reaction. The units
used can be kilojoules or kilocalories. Consider the reaction represented by this equation:
C(s) + O2(g) ¡ CO2(g) + 393 kJ
=O
=C
=H
Glucose
Glucose
C6H12CO
6 12O6
6H
When the heat released is expressed as part of the equation, the substances are expressed in
units of moles. Thus, when 1 mol (12.01 g) of C combines with 1 mol (32.00 g) of O2, 1 mol
(44.01 g) of CO2 is formed and 393 kJ of heat are released. In this reaction, as in many others,
the heat energy is more useful than the chemical products.
Aside from relatively small amounts of energy from nuclear processes, the sun is the
major provider of energy for life on Earth. The sun maintains the temperature necessary for life
and also supplies light energy for the endothermic photosynthetic reactions of green plants. In
photosynthesis, carbon dioxide and water are converted to free oxygen and glucose:
6 CO2 + 6 H2O + 2519 kJ ¡ C6H12O6 + 6 O2
glucose
This cornfield is a good example of the
endothermic reactions happening through
photosynthesis in plants.
Paul Damien/National Geographic Society
Nearly all of the chemical energy used by living organisms is obtained from glucose or
compounds derived from glucose.
8.4
• Heat in Chemical Reactions 157
Richard Megna/Fundamental Photographs
The major source of energy for modern technology is fossil fuel—coal, petroleum, and
natural gas. The energy is obtained from the combustion (burning) of these fuels, which are
converted to carbon dioxide and water. Fossil fuels are mixtures of hydrocarbons, compounds
containing only hydrogen and carbon.
Natural gas is primarily methane, CH4. Petroleum is a mixture of hydrocarbons (compounds of carbon and hydrogen). Liquefied petroleum gas (LPG) is a mixture of propane
(C3H8) and butane (C4H10).
Here are some examples:
CH4(g) + 2 O2(g) ¡ CO2(g) + 2 H2O(g) + 890 kJ
C3H8(g) + 5 O2(g) ¡ 3 CO2(g) + 4 H2O(g) + 2200 kJ
Phil Degginger/Alamy
Potential energy
The combustion of these fuels releases a tremendous amount of energy, but reactions
won’t occur to a significant extent at ordinary temperatures. A spark or a flame must be present before methane will ignite. The amount of energy that must be supplied to start a chemical
reaction is called the activation energy. In an exothermic reaction, once this activation energy
is provided, enough energy is then generated to keep the reaction going.
Be careful not to confuse an exothermic reaction that requires heat (activation energy) to get it started with an endothermic process that requires energy to keep it going.
The combustion of magnesium, for example, is highly exothermic, yet magnesium must
be heated to a fairly high temperature in air before combustion begins. Once started,
however, the combustion reaction goes very vigorously until either the magnesium or
the available supply of oxygen is exhausted. The electrolytic decomposition of water to
hydrogen and oxygen is highly endothermic. If the electric current is shut off when this
process is going on, the reaction stops instantly. The relative energy levels of reactants
and products in exothermic and endothermic processes are presented graphically in
Figures 8.2 and 8.3.
Examples of endothermic and exothermic processes can be easily demonstrated.
In Figure 8.2, the products are at a higher potential energy than the reactants. Energy has
therefore been absorbed, and the reaction is endothermic. An endothermic reaction takes
place when you apply a cold pack to an injury. When a cold pack is activated, ammonium
chloride (NH4Cl) dissolves in water. For example, temperature changes from 24.5ºC to
18.1ºC result when 10 g of NH4Cl are added to 100 mL of water. Energy, in the form of
heat, is taken from the immediate surroundings (water), causing the salt solution to become cooler.
Combustion of Mg inside a block
of dry ice (CO2) makes a glowing
lantern.
Activation energy
needed to start
the reaction
Products
Heat (energy) of reaction–
net energy absorbed
Reactants
Time (reaction progress)
Endothermic reaction
Figure 8.2
An endothermic process occurs when a cold pack containing an ampule of solid NH4CI
is broken, releasing the crystals into water in the pack. The dissolving of NH4CI in water
is endothermic, producing a salt solution that is cooler than the surroundings. This
process is represented graphically to show the energy changes between the reactants
and the products.
• Chemical Equations
Potential energy
chapter 8
Activation energy needed
to start the reaction
Reactants
Heat (energy) of reaction–
net energy released
Products
Tom Pantages
Time (reaction progress)
Exothermic reaction
Figure 8.3
Exothermic reaction between KCIO3 and sugar. A sample of KCIO3 and sugar is well mixed
and placed on a fireproof pad. Several drops of concentrated H2SO4 are used to ignite
the mixture.
Example 8.11
Label the graph in Figure 8.2 showing the specific reactants and products as well as the
heat of reaction for the reaction.
SOLUTION
For Figure 8.2 the chemical equation is NH4Cl(s) + H2O(l) ¡ NH4 + (aq) + Cl-(aq) +
H2O(l). The reactants are NH4Cl(s) + H2O(l) and the products are NH4+(aq) + Cl−(aq) +
H2O(l). The graph is labeled and shown here:
Potential energy
158 Activation energy needed
to start the reaction
NH4+(aq) + Cl–(aq) + H2O(l)
Heat (energy) of reaction–
net energy released
NH4Cl(s) + H2O(l)
Time (reaction progress)
Practice 8.7
For each of the following reactions, draw a reaction graph (like those in Figure 8.2 or 8.3)
and label the parts of the graph with the information from the chemical equation.
(a) 2 Na + Cl2 ¡ 2 NaCl + 822 kJ
(b) 2 BrF3 + 300.8 kJ/mol ¡ Br2 + 3 F2
In Figure 8.3, the products are at a lower potential energy than the reactants. Energy
(heat) is given off, producing an exothermic reaction. Here, potassium chlorate (KClO3) and
sugar are mixed and placed into a pile. A drop of concentrated sulfuric acid is added, creating
a spectacular exothermic reaction.
8.5
• Global Warming: The Greenhouse Effect 159
Discuss the possible causes and results of the greenhouse effect.
Learning objectiv e
8.5 global Warming:
The Greenhouse effect
Fossil fuels, derived from coal and petroleum, provide the energy we use to power our industries, heat and light our homes and workplaces, and run our cars. As these fuels are burned,
they produce carbon dioxide and water, releasing over 50 billion tons of carbon dioxide into
our atmosphere each year.
The concentration of CO2 has been monitored by scientists since 1958. Each week a scientist walks outdoors at a particular location and, carefully holding his breath, opens a basketballsized glass sphere and collects 5 liters of the atmosphere. At these same remote sites all over
the globe (and at dozens of others), instruments sniff the air adding more measurements of the
atmosphere to the data. These data are summarized in Figure 8.4. Several trends have been
discovered in these data.
1.In the Northern Hemisphere the concentration of CO2 rises and falls 7 ppm over a year,
peaking in May and dropping as plants use the CO2 to produce growth till October, when
a second peak occurs as a result of the fallen leaves decaying.
2.The 7-ppm annual variation is superimposed on an average increase in the concentration of
CO2. Our current average is more then 380 ppm, or an increase of more than 65 ppm over
the last 50 years. It continues to rise about 2 ppm annually.
Carbon dioxide is a minor component in our atmosphere and is not usually considered
to be a pollutant. The concern expressed by scientists arises from the dramatic increase
occurring in the Earth’s atmosphere. Without the influence of humans in the environment,
the exchange of carbon dioxide between plants and animals would be relatively balanced.
Our continued use of fossil fuels led to an increase of 7.4% in carbon dioxide between
1900 and 1970 and an additional 3.5% increase during the 1980s. Continued increases were
observed in the 1990s.
Besides our growing consumption of fossil fuels, other factors contribute to increased
carbon dioxide levels in the atmosphere: Rain forests are being destroyed by cutting and
burning to make room for increased population and agricultural needs. Carbon dioxide is
added to the atmosphere during the burning, and the loss of trees diminishes the uptake of
carbon dioxide by plants.
About half of all the carbon dioxide released into our atmosphere each year remains there,
thus increasing its concentration. The other half is absorbed by plants during photosynthesis
or is dissolved in the ocean to form hydrogen carbonates and carbonates.
400.00
Figure 8.4
CO2 concentration
390.00
Concentration of CO2 in the atmosphere.
Source: Data from the National Oceanic
and Atmospheric Administration,
http://www.esrl.noaa.gov/gmd/ccgg/trends/
#mol-full, February 28, 2012.
380.00
370.00
360.00
350.00
340.00
330.00
320.00
2011
2007
2003
1999
1995
1991
1987
1983
1979
1975
1971
1967
1963
300.00
1959
310.00
160 chapter 8
• Chemical Equations
>Chemistry in action
In order to slow down global warming, we need to stop
the increase of carbon dioxide (CO2) in the air. We can, of
course, continue to try to limit our dependence on fossil
fuels. But scientists are also working to capture carbon dioxide emissions. Capturing CO2 presents two challenges:
how to remove the CO2 from the air and what to do with
it once it has been collected.
One way of capturing and storing carbon dioxide is to
let nature work. Plants capture 505 million tons of carbon
dioxide in North America through photosynthesis by storing it as carbon in woody plants and trees or as organic
material in the soil and wetlands or sediments in rivers
and lakes. The difficulty with this method is that as the
decay process continues in an oxygen-rich environment,
the carbon is returned to the atmosphere as CO2. Ning
Zing, from the University of Maryland, suggests that
we could prevent the release of this CO2 from forests
by burying wood in an oxygen-poor environment. The
carbon would then remain in the ground and not in the
atmosphere. He notes that it is possible to accomplish this
process at a cost of $14 per ton of wood, but that would
mean collecting wood and burying it in trenches, which is
not an environment-friendly method of capture and storage. Other scientists have suggested seeding the oceans
to increase the algae and phytoplankton. This would also
pull CO2 from the atmosphere, but many scientists think
that this method would not have much effect on climate
since the oceans could only capture a small fraction of the
CO2 emitted by humans.
Coal and petroleum each account for about 40% of global
CO2 emissions. Coal is the resource that is the greatest
threat to our climate since it produces more CO2 per unit
of energy than any other fossil fuel. Each year coal-fired
power plants produce 8 billion tons of CO2 (containing
2.2 billion tons of C). Future coal plants (planned in the
next 25 years) will generate 660 billion tons of CO2. This
is 25% more than all the CO2 that humans have produced
by burning coal since 1751. How can we reduce these
numbers? The current goal is to capture 90% of the CO2
emissions from a power plant but only increase the cost
of electricity by 20%. Scientists are developing materials
that act as CO2 sponges. These materials, called zeolites,
are very porous and have large surface areas where CO2
molecules can attach themselves.
The question here is that once the CO2 is captured, how
can it be stored? Many scientists propose to lock it underground or in deep ocean. Since CO2 is a liquid at high pressure (as in ocean depths below 500 meters) and does not
mix very well with water, it could be stored as large pool
on the ocean floor. This storage method raises a number
of environmental concerns. In some places there are saline
aquifers underground. The water from these aquifers is not
drinkable, and so some of them might be a suitable storage
place for CO2. Another possibility is using volcanic rock
as a storage site. Tests indicate that liquid CO2 will react
with basalt to produce minerals such as calcium carbonate. These deposits contain many holes and cracks where
the CO2 might be injected. It is thought that in the United
States these saline aquifers and geologic formations could
store 150 years’ worth of power plant ­emissions.
We are now looking seriously at ways to slow down the rise
in atmospheric CO2 and to decrease our carbon footprints
without losing our fossil fuels as one of our energy sources.
Steven Raniszewski /Design Pics/Newscom
Decreasing Carbon Footprints
Zeolite (a basaltic volcanic rock) could be used as a storage
place for carbon dioxide from power plants.
Methane is another important greenhouse gas. Its concentration in the atmosphere has also
increased significantly since the 1850s, as shown in Figure 8.5. Methane is produced by cows,
termites, agriculture, and anaerobic bacteria. Coal mining and oil wells also release methane
into the atmosphere. The greenhouse effect of methane is 20 times more than that of CO2, but
there is less methane in the atmosphere.
Carbon dioxide and other greenhouse gases, such as methane and water, act to warm our
atmosphere by trapping heat near the surface of the Earth. Solar radiation strikes the Earth and
warms the surface. The warmed surface then reradiates this energy as heat. The greenhouse
gases absorb some of this heat energy from the surface, which then warms our atmosphere.
(See Figure 8.6.) A similar principle is illustrated in a greenhouse, where sunlight comes
through the glass yet heat cannot escape. The air in the greenhouse warms, producing a climate
considerably different than that in nature.
Review 161
Methane concentration (ppbv)
2000
Figure 8.5
1750
Atmospheric
measurements
1500
Ice core data
Concentration of methane in
the atmosphere.
1250
1000
Infrared radiation
from the Earth
750
500
1600
1650
1700
1750
1800
Year
1850
1900
1950
CO2 and H2O
molecules
2000
The long-term effects of global warming are still a matter of debate. One consideration is whether the polar ice caps will melt; this
would cause a rise in sea level and lead to major flooding on the coasts
of our continents. Further effects could include shifts in rainfall patterns, producing droughts and extreme seasonal change in such major
agricultural regions as California. To reverse these trends will require
major efforts in the following areas:
• The development of new energy sources to cut our dependence on
fossil fuels
• An end to deforestation worldwide
• Intense efforts to improve conservation
On an individual basis each of us play a significant role. Recycling,
switching to more fuel-efficient cars, and using energy-efficient appliances, heaters, and air conditioners all would result in decreased energy
consumption.
C h a p t e r
Earth’s
atmosphere
Figure 8.6
Visible light from the sun (red arrows) penetrates the
atmosphere and strikes the Earth. Part of this light is
changed to infrared radiation. Molecules of CO2 and
H2O in the atmosphere absorb infrared radiation, acting
like the glass of a greenhouse by trapping the energy.
8 review
8.1 The Chemical Equation
• A chemical equation is shorthand for expressing a chemical change or reaction.
• In a chemical reaction atoms are neither created nor ­destroyed.
• All atoms in the reactants must be present in the ­products.
8.2 Writing and Balancing Chemical Equations
• To balance a chemical equation:
• Identify the reaction.
• Write the unbalanced (skeleton) equation.
• Balance the equation:
• Count the number of atoms of each element on each side and determine which need to be
­balanced.
• Balance each element (one at a time) by placing whole numbers (coefficients) in front of the
formulas containing the unbalanced element:
• Begin with metals, then nonmetals, and then H and O.
• Check the other elements to see if they have ­become unbalanced in the process of balancing
the chosen element. If so, rebalance as needed.
• Do a final check to make sure all elements are ­balanced.
• The following information can be found in a chemical equation:
• Identity of reactants and products
• Formulas for reactants and products
• Number of formula units for reactants and products
• Number of atoms of each element in the reaction
• Number of moles of each substance
Key Terms
reactants
products
chemical equation
law of conservation of mass
Key Term
balanced equation
162 chapter 8
• Chemical Equations
8.3 Types of Chemical Equations
Key Terms
combination reaction
decomposition reaction
single-displacement reaction
double-displacement reaction
• Combination reactions
• Decomposition reactions
• Single-displacement reactions:
• In which A is a metal
• In which A is a halogen
• Double-displacement reactions
• Evidence for a chemical reaction:
• Evolution of heat
• Formation of an insoluble precipitate
• Production of a gas
A + B → AB
AB → A + B
A + BC → B + AC
A + BC → C + BA
AB + CD → AD + CB
8.4 Heat in Chemical Reactions
Potential energy
exothermic reactions
endothermic reactions
heat of reaction
hydrocarbons
activation energy
• Exothermic reactions release heat.
Activation energy needed
to start the reaction
Reactants
Heat (energy) of reaction–
net energy released
Products
Time (reaction progress)
• Endothermic reactions absorb heat.
Potential energy
Key Terms
Activation energy
needed to start
the reaction
Products
Heat (energy) of reaction–
net energy absorbed
Reactants
Time (reaction progress)
• The amount of heat released or absorbed in a chemical reaction is called the heat of reaction:
• It can be written as a reactant or product in the chemical equation.
• Units are joules (J) or kilojoules (kJ).
• The major source of energy for modern technology is hydrocarbon combustion.
• To initiate a chemical reaction, activation energy is ­required:
• In exothermic reactions, this energy is returned and more is released, which allows the reaction
to ­continue on its own.
• In endothermic reactions, energy must be added to start and continue to be added to sustain the
­reaction.
8.5 Global Warming: The Greenhouse Effect
• Carbon dioxide levels are now increasing on our ­planet every year.
• Carbon dioxide and other greenhouse gases warm the atmosphere by trapping heat near the surface
of the Earth.
Paired Exercises 163
R e v i e w Q u e st i o n s
1. What is represented by the numbers (coefficients) that are placed in
front of the formulas in a balanced equation?
2. What is meant by the physical state of a substance? What ­symbols are
used to represent these physical states, and what does each symbol mean?
3. What is the purpose of balancing chemical equations?
4. In a balanced chemical equation:
(a) Are atoms conserved?
(b) Are molecules conserved?
(c) Are moles conserved?
Explain your answers briefly.
5. In Section 8.2, there are small charts in color along the margins.
What information is represented in those charts?
6. Why is it incorrect to balance chemical reactions by changing the
subscripts on the reactants and products?
7. What is a combustion reaction? How can you identify whether a
reaction is a combustion reaction?
8. What information does the activity series shown in Table 8.2
give?
9. What major types of chemical reactions are studied in this
chapter?
10. What three observations indicate a chemical reaction has taken
place?
11. Explain how endothermic reactions differ from exothermic
­reactions.
12. Why does an exothermic reaction require heat to occur?
13. Name several greenhouse gases.
14. What factors are responsible for the annual rise and fall in carbon
dioxide levels?
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1. Classify the following as an endothermic or exothermic ­reaction:
(a) freezing water
(b) the reaction inside an ice pack
(c) burning wood
(d) combustion of Mg in dry ice
(e) melting ice
2. Classify the following as an endothermic or exothermic ­reaction:
(a) making popcorn in a microwave oven
(b) a burning match
(c) boiling water
(d) burning rocket fuel
(e) the reaction inside a heat pack
3.Balance each of the following equations. Classify each ­reaction
as combination, decomposition, ­single-displacement, or doubledisplacement.
4.Balance each of the following equations. Classify each reaction
as combination, decomposition, ­single-displacement, or doubledisplacement.
(a) H2 + O2 ¡ H2O
(a) H2 + Br2 ¡ HBr
(b) N2H4(l) ¡ NH3(g) + N2(g)
(b) BaO2(s) + H2SO4(aq) ¡ BaSO4(s) + H2O2(aq)
(c) H2SO4 + NaOH ¡ H2O + Na2SO4

(d) Al2(CO3)3 ¡
Al2O3 + CO2
(e) NH4I + Cl2 ¡ NH4Cl + I2

(c) Ba(ClO3)2 ¡ BaCl2 + O2
(d) CrCl3 + AgNO3 ¡ Cr(NO3)3 + AgCl
(e) H2O2 ¡ H2O + O2
5. What reactant(s) is (are) required to form an oxide ­product?
6. What reactant(s) is (are) required to form a salt?
7. Balance the following equations:
8. Balance the following equations:
(a) MnO2 + CO ¡ Mn2O3 + CO2
(a) SO2 + O2 ¡ SO3
(b) Cu2O(s) + C(s) ¡ Cu(s) + CO(g)
(b) Li2O(s) + H2O(l) ¡ LiOH(aq)
(c) C3H5(NO3)3 ¡ CO2 + H2O + N2 + O2
(c) Na + H2O ¡ NaOH + H2
(d) FeS + O2 ¡ Fe2O3 + SO2
(d) AgNO3 + Ni ¡ Ni(NO3)2 + Ag
(e) Cu(NO3)2 ¡ CuO + NO2 + O2
(e) Bi2S3 + HCl ¡ BiCl3 + H2S
(f) NO2 + H2O ¡ HNO3 + NO

PbO + O2
(f) PbO2 ¡
(g) Hg2(C2H3O2)2(aq) + KCl(aq) ¡ Hg2Cl2(s) +
KC2H3O2(aq)
(g) Fe(s) + S(l) ¡ Fe2S3(s)
(h) HCN + O2 ¡ N2 + CO2 + H2O
(i) B5H9 + O2 ¡ B2O3 + H2O
(h) KI + Br2 ¡ KBr + I2
(i) K3PO4 + BaCl2 ¡ KCl + Ba3(PO4)2
164 chapter 8
• Chemical Equations
9.Change these word equations into formula equations and balance
them. Be sure to use the proper symbols to indicate the state of each
substance, as given.
(a)Magnesium metal is placed into hydrobromic acid solution,
forming hydrogen gas and aqueous magnesium bromide.
(b)When heated, solid calcium chlorate decomposes into calcium
chloride solid, releasing oxygen gas.
(c)Lithium metal reacts with oxygen gas to form solid ­lithium
oxide.
(d)Solutions of barium bromate and sodium phosphate combine to
form solid barium phosphate and aqueous sodium bromate.
(e)Solutions of acetic acid and sodium carbonate are mixed together, forming a solution of sodium acetate, along with carbon
dioxide gas and liquid water.
(f)Solutions of silver nitrate and aluminum iodide are mixed together, forming solid silver iodide and aqueous aluminum nitrate.
10.Change these word equations into formula equations and balance
them. Be sure to use the proper symbols to indicate the state of each
substance, as given.
(a)Upon heating, solid magnesium carbonate decomposes into
solid magnesium oxide and carbon dioxide gas.
(b)Solid calcium hydroxide reacts with aqueous chloric acid to
form a solution of calcium chlorate along with liquid water.
(c)Solutions of iron(III) sulfate and sodium hydroxide are mixed
together, forming solid iron(III) hydroxide and a solution of
sodium sulfate.
(d)Zinc metal is placed into a solution of acetic acid, ­producing
hydrogen gas and aqueous zinc acetate.
(e)Gaseous sulfur trioxide reacts with liquid water to form a solution of sulfuric acid.
(f)Solutions of sodium carbonate and cobalt(II) chloride react to
form solid cobalt(II) carbonate and a solution of sodium ­chloride.
11.For each of the following reactions, predict the products, converting
each to a balanced formula ­equation:
(a)Aqueous solutions of sulfuric acid and sodium ­hydroxide are
mixed together. (Heat is released during the reaction.)
(b)Aqueous solutions of lead(II) nitrate and potassium ­bromide
are mixed together. (The solution turns cloudy white during the
reaction.)
(c)Aqueous solutions of ammonium chloride and silver ­nitrate are
mixed together. (The solution turns cloudy white during the
reaction.)
(d)Solid calcium carbonate is mixed with acetic acid. (Bubbles of
gas are formed during the ­reaction.)
13.Use the activity series to predict which of the following reactions
will occur. Complete and balance the equations. If no reactions occurs, write “no reaction” as the product.
(a) Ca(s) + H2O(l) S
(c) Cu(s) 1 HCl(aq) S
(b) Br2(l) 1 KI(aq) S
(d) Al(s) 1 H2SO4(aq) S
12.For each of the following reactions, predict the products, converting
each to a balanced formula equation:
(a)Aqueous solutions of copper(II) sulfate and potassium hydroxide are mixed together. (The solution turns cloudy and light
blue during the reaction.)
(b)Aqueous solutions of phosphoric acid and sodium hydroxide
are mixed together. (Heat is produced during the reaction.)
(c)Solid sodium bicarbonate is mixed with phosphoric acid.
(Bubbles of gas are formed during the ­reaction.)
(d)Aqueous solutions of aluminum chloride and lead(II) ­nitrate
are mixed together. (The solution turns cloudy white during the
reaction.)
15.Complete and balance the equations for each of the following reactions. All yield products.
(a) Sr(s) + H2O(l) S
(c) Mg(s) + ZnBr2(aq) S
(b) BaCl2(aq) + AgNO3(aq) S (d) K(s) + Cl2(g) S
16.Complete and balance the equations for each of the following reactions. All yield products.
(a) Li2O(s) + H2O(l) S
(c) Zn(s) + CuSO4(aq) S
(b) Na2SO4(aq) + Pb(NO3)2(aq) S (d) Al(s) + O2(g) S
17.Complete and balance the equations for these reactions. All reactions yield products.
(a) Ba + O2 ¡
(d) MgO + HCl ¡

(b) NaHCO3 ¡ Na2CO3 + (e) H3PO4 + KOH ¡
(c) Ni + CuSO4 ¡
19.Interpret these chemical reactions in terms of the number of moles
of each reactant and product:
(a) MgBr2 + 2 AgNO3 ¡ Mg(NO3)2 + 2 AgBr
(b) N2 + 3 H2 ¡ 2 NH3
(c) 2 C3H7OH + 9 O2 ¡ 6 CO2 + 8 H2O
18.Complete and balance the equations for these reactions. All reactions yield products.
(a) C + O2 ¡
(d) SbCl3 + (NH4)2S ¡

(b) Al(ClO3)3 ¡ O2 + (e) NaNO3 ¡ NaNO2 +
(c) CuBr2 + Cl2 ¡
20.Interpret these equations in terms of the relative number of moles
of each substance involved and indicate whether the ­reaction is
exothermic or endothermic:
(a) 2 Na + Cl2 ¡ 2 NaCl + 822 kJ
(b) PCl5 + 92.9 kJ ¡ PCl3 + Cl2
(c) S(s) + 2 CO(g) ¡ SO2(g) + 2 C(s) + 76 kJ
21.Write balanced equations for each of these reactions, ­including the
heat term:
(a)Solid mercury(II) oxide decomposes into liquid mercury and
oxygen gas upon the absorption of 90.8 kJ for each mole of
mercury(II) oxide decomposed.
(b)Hydrogen gas reacts with oxygen gas to form liquid water.
The reaction produces 285.8 kJ of heat for each mole of water
formed
23.Determine what reactants would form the given products. Give a balanced equation for each and classify the reaction as a combination,
decomposition, single-­displacement, or ­double-displacement reaction.
(a) AgCl(s) + O2(g)
(c) ZnCl2(s)
(b) H2(g) + FeSO4(aq)
(d) KBr(aq) + H2O(l)
22.Write balanced equations for each of these reactions, ­including the
heat term:
(a)Calcium metal reacts with water to produce calcium ­hydroxide
and hydrogen gas, releasing 635.1 kJ of heat for every mole of
calcium that reacts.
(b)Bromine trifluoride decomposes into bromine and fluorine
upon the absorption of 300.8 kJ of heat for each mole of bromine trifluoride decomposed.
24.Determine what reactants would form the given products. Give a balanced equation for each and classify the reaction as a combination,
decomposition, single-displacement, or ­double-displacement reaction.
(a) Pb(s) + Ni(NO3)2(aq) (c) Hg(l) + O2(g)
(b) Mg(OH)2(s)
(d) PbCO3(s) + NH4Cl(aq)
14.Use the activity series to predict which of the following reactions
will occur. Complete and balance the equations. If no reaction occurs, write “no reaction” as the product.
(a) Cu(s) + NiCl2(aq) S (c) I2(s) + CaCl2(aq) S
(b) Rb(s) + H2O(l) S
(d) Mg(s) + Al(NO3)3(aq) S
Additional Exercises 165
Add i t i o n a l E x e r c i s e s
25. An aqueous mixture of the nitrate salts of sodium, calcium, and silver is
mixed with ammonium salts of chloride, sulfate, and carbonate. Write
balanced equations for any combination that will form a precipitate.
26. Balance this equation, using the smallest possible whole numbers.
Then determine how many atoms of oxygen appear on each side of
the equation:
P4O10 + HClO4 ¡ Cl2O7 + H3PO4
27. Suppose that in a balanced equation the term 5 Ni3(PO4)2 appears.
(a) How many atoms of nickel are represented?
(b) How many atoms of phosphorus are represented?
(c) How many atoms of oxygen are represented?
(d) How many atoms of all kinds are represented?
28. In a lab experiment, a student places a few stones, composed mainly
of calcium carbonate, into a beaker and then adds some acetic
acid. The student observes that effervescence ­occurs immediately.
Explain this observation by writing a ­balanced formula equation.
29. Make a drawing to show the combustion reaction of one ­molecule
of methane, CH4.
30. A variety of iron-containing ores can be converted into iron and
steel in a blast furnace. The reactions that take place are outlined
below. Write balanced chemical equations for each reaction.
(a) Pure carbon is reacted with oxygen gas to form carbon dioxide.
(b) Carbon dioxide is reacted with more carbon to form monoxide.
(c)Carbon monoxide reacts with the iron ore to form elemental
iron and carbon dioxide. There are two main types of ore that
may be used: hematite (Fe2O3) and magnetite (Fe3O4).
31. The box at the left represents the reactant atoms in blue (B) and
molecules in green (G2). After reacting, these reactants from the
compound pictured in the box at the right. Write a balanced chemical reaction for this process.
32. Students in a lab exercise were asked to choose among a sample of
different metals to react with a solution of nickel(II) chloride. Which
of the following metals should they choose? Write balanced equations for all that will react and explain why each will or will not react.
(a) copper
(d) lead
(b) zinc
(e) calcium
(c) aluminum
33. A student does an experiment to determine where titanium metal
should be placed on the activity series chart. He places newly cleaned
pieces of titanium into solutions of nickel(II) ­nitrate, lead(II) nitrate,
and magnesium ­nitrate. He finds that the titanium reacts with the
­nickel(II) nitrate and lead(II) nitrate solutions, but not with the magnesium nitrate solution. From this information, place titanium in the
activity series in a position ­relative to these ions.
34. Predict the products of each of the following combination reactions in
words. Then change each equation into a balanced formula equation.
(a) Cesium reacts with oxygen.
(b) Aluminum metal reacts with sulfur.
(c) Sulfur trioxide reacts with water.
(d) Sodium oxide reacts with water.
35. Automobile designers are currently working on cars that will run
on hydrogen. Hydrogen-fueled cars have the advantage that they
produce only water as a waste product. In these cars, hydrogen gas
reacts with oxygen gas to form gaseous water.
(a) Write a balanced chemical equation for this reaction.
(b)There are many ways to generate hydrogen to fuel these cars.
The most commonly used method of making hydrogen gas is
to react methane gas (CH4) with gaseous water at 1000ºC to
form gaseous carbon monoxide and hydrogen gas. The carbon
monoxide gas can then be reacted with more gaseous water to
produce more hydrogen gas and carbon dioxide gas. Write balanced chemical equations for these two reactions.
(c)One of the products of burning fossil fuels to provide energy
is carbon dioxide. This carbon dioxide may be increasing the
temperature of the planet. Methane can also be used to provide
energy by reacting it with oxygen gas to form carbon dioxide and water. Write the balanced chemical equation for this
­reaction.
(d)Compare the number of molecules of carbon dioxide produced
out of one molecule of methane by burning it directly to produce energy and by converting it to hydrogen gas and using this
hydrogen gas to produce energy. Is there an advantage to using
hydrogen gas as the fuel source? Explain.
36. Many paint pigments are composed of ionic compounds. For each
of the pigments listed below, write a complete balanced equation
to synthesize the pigment using the reagents listed in parentheses.
Write the correct IUPAC name of each pigment.
(a) chrome yellow, PbCrO4 (Pb(NO3)2(aq) + K2CrO4(aq))
(b) cadmium yellow, CdS (CdCl2(aq) + Li2S(aq))
(c) white lead, PbCO3 (Pb(C2H3O2)2(aq) + H2CO3(aq))
(d) vermillion, HgS (Hg(NO3)2(aq) + Na2S(aq))
(e) titanium black, Ti2O3 (Ti(s) + O2(g))
(f) iron oxide red, Fe2O3 (Fe(s) + O2(g))
37. Predict the products of each of the following decomposition reactions in words. Then change each equation into a balanced formula
equation.
(a) Zinc oxide decomposes upon heating strongly.
(b) Tin(IV) oxide decomposes into two products when ­heated.
(c) Sodium carbonate decomposes when heated.
(d) Magnesium chlorate decomposes when heated.
38. Predict the products of each of the following single-displacement
reactions in words. Then change each ­equation into a balanced formula equation. All reactions occur in aqueous ­solution.
(a)Magnesium metal reacts with hydrochloric acid.
(b)Sodium bromide reacts with chlorine.
(c)Zinc metal reacts with iron(III) nitrate.
(d)Aluminum reacts with copper(II) sulfate.
39. Predict the products of each of the following double-displacement
reactions in words. Then change each ­equation into a balanced formula equation. All reactants are in aqueous solution.
(a) Ammonium phosphate reacts with barium nitrate.
(b) Sodium sulfide reacts with lead(II) acetate.
(c) Copper(II) sulfate reacts with calcium chlorate.
(d) Barium hydroxide reacts with oxalic acid.
(e) Phosphoric acid reacts with potassium hydroxide.
(f) Sulfuric acid reacts with sodium carbonate.
40. Predict which of the following double-displacement reactions will
occur. For those that do, write a balanced formula equation. For any
166 chapter 8
• Chemical Equations
reaction that does not occur, write “no reaction” as the product. All
reactants are in aqueous solution.
(a) potassium sulfate and barium acetate
(b) sulfuric acid and lithium hydroxide
(c) ammonium phosphate and sodium bromide
(d) calcium iodide and silver nitrate
(e) nitric acid and strontium hydroxide
(f) cesium nitrate and calcium hydroxide
41.Write balanced equations for each of the following combustion
­reactions:
+
+
¡
42. Write balanced equations for the complete combustion of the following hydrocarbons:
(a) CH4
(b) C3H6
(c) C6H5CH3
43. List the factors that contribute to an increase in carbon dioxide in
our atmosphere.
44. List three gases considered to be greenhouse gases. ­Explain why
they are given this name.
45. How can the effects of global warming be reduced?
46. What happens to carbon dioxide released into our ­atmosphere?
47. In the Northern Hemisphere, the concentration of CO2 peaks twice
a year. When do these peaks occur, and what causes them?
H
C
+
¡
O
Challenge Exercise
48. You are given a solution of the following cations in water: Ag+,
Co2 + , Ba2 + , Zn2 + , and Sn2 + . You want to separate the ions out one
at a time by precipitation using the ­following reagents: NaF, NaI,
Na2SO4, and NaCl.
(a) In what order would you add the reagents to ensure that only
one cation is precipitating out at a time? Use the solubility
table in Appendix V to help you. (Note: A precipitate that is
slightly soluble in water is still considered to precipitate out of
the ­solution.)
(b) Why are the anionic reagents listed above all ­sodium salts?
A n s w e r s to P r a c t i c e E x e r c i s e s
8.1
To show that a substance is a solid, write (s) after the formula
of the substance. When the substance is a gas, write (g) after the
formula, and when the substance is in an aqueous solution write
(aq) after the formula.
8.2
4 Al + 3 O2 ¡ 2 Al2O3
8.3
3 Mg(OH)2 + 2 H3PO4 ¡ Mg3(PO4)2 + 6 H2O
8.4
(a) hydrogen gas + chlorine gas ¡ hydrogen chloride gas
(b) H2(g) + Cl2(g) ¡ 2 HCl(g)
(c)
8.6
¡
2 HCl(g)
1 molecule
1 molecule
2 molecules
2 atoms H
2 atoms Cl
2 atoms H
2 atoms Cl
1 mol
2.016 g
1 mol
70.90 g
2 mol
2 (36.46 g)
(72.92 g)
2 Na + Cl2
822 kJ
2 NaCl
Time (reaction progress)
(b)
(d) 2 mol H2 + 2 mol Cl2 ¡ 4 mol HCl(145.8 g)
(a) Fe + MgCl2 ¡ no reaction
(b) Zn(s) + Pb(NO3)2(aq) ¡ Pb(s) + Zn(NO3)2(aq)
(a) 2 K3PO4(aq) + 3 BaCl2(aq) ¡
Ba3(PO4)2(s) + 6 KCl(aq)
(b) 2 HCl(aq) + NiCO3(aq) ¡
NiCl2(aq) + H2O(l) + CO2(g)
(c) NH4Cl(aq) + NaNO3(aq) ¡ no reaction
Potential energy
8.5
� Cl2(g)
(a)
Activation energy
Potential energy
H2(g)
8.7
Activation energy
Br2 + 3F2
300.8 kJ/mole
2BrF3
Time (reaction progress)
• Properties of Gases 167
Accurate calculations of dose
and accurate measurement of
chemicals are required in order
for a pharmacist to dispense
the correct dosage of medicine
to her patients.
C h apt e r
9
Calculations from
Chemical Equations
T
he old adage “waste not, want not” is appropriate in our
daily life and in the laboratory. Determining correct
amounts comes into play in almost all professions.
Chapter Outline
9.1 Introduction to Stoichiometry
9.2 Mole–Mole Calculations
9.3 Mole–Mass Calculations
A seamstress determines the amount of material, lining, and trim
9.4 Mass–Mass Calculations
necessary to produce a gown for her client by relying on a pattern
9.5 Limiting Reactant and Yield
Calculations
or her own experience to guide the selection. A carpet layer determines the correct amount of carpet and padding necessary to
recarpet a customer’s house by calculating the floor area. The IRS
determines the correct deduction for federal income taxes from
your paycheck based on your expected annual income.
Corbis RF/Age Fotostock America, Inc.
12.1
168 chapter 9
• Calculations from Chemical Equations
9.1 Introduction to Stoichiometry
L earning obje ctive
Define stoichiometry and describe the strategy required to solve problems based on
chemical equations.
key terms
We often need to calculate the amount of a substance that is either produced from, or needed to
react with, a given quantity of another substance. The area of chemistry that deals with quantitative
relationships among reactants and products is known as stoichiometry (stoy-key-ah-meh-tree).
Solving problems in stoichiometry requires the use of moles in the form of mole ratios.
The chemist also finds it necessary to calculate amounts of products or reactants by using
a balanced chemical equation. With these calculations, the chemist can control the amount of
product by scaling the reaction up or down to fit the needs of the laboratory and can thereby
minimize waste or excess materials formed during the reaction.
stoichiometry
molar mass
mole ratio
A Short Review
Molar mass The sum of the atomic masses of all the atoms in an element or compound is
called molar mass. The term molar mass also applies to the mass of a mole of any formula
unit—atoms, molecules, or ions; it is the atomic mass of an atom or the sum of the atomic
masses in a molecule or an ion (in grams).
Relationship between molecule and mole A molecule is the smallest unit of a molecular substance (e.g., Br2), and a mole is Avogadro’s number (6.022 * 1023) of molecules
of that substance. A mole of bromine (Br2) has the same number of molecules as a mole of
carbon dioxide, a mole of water, or a mole of any other molecular substance. When we relate
molecules to molar mass, 1 molar mass is equivalent to 1 mol, or 6.022 * 1023 molecules.
The term mole also refers to any chemical species. It represents a quantity (6.022 * 1023
particles) and may be applied to atoms, ions, electrons, and formula units of nonmolecular
substances. In other words,
Richard Megna/Fundamental Photographs
6.022
6.022
1 mole = d
6.022
6.022
*
*
*
*
1023 molecules
1023 formula units
1023 atoms
1023 ions
Other useful mole relationships are
molar mass =
grams of a substance
number of moles of the substance
molar mass =
grams of a monatomic element
number of moles of the element
number of moles =
A mole of water, salt, and any
gas all have the same number
of particles (6.022 3 1023).
number of molecules
6.022 * 1023 molecules>mole
Balanced equations When using chemical equations for calculations of mole–mass–volume
relationships between reactants and products, the equations must be balanced. Remember: The
number in front of a formula in a balanced chemical equation represents the number of moles
of that substance in the chemical reaction.
A mole ratio is a ratio between the number of moles of any two species involved in a
chemical reaction. For example, in the reaction
2 H 2 + O2 ¡ 2 H 2 O
2 mol 1 mol
2 mol
six mole ratios can be written:
2 mol H2
1 mol O2
2 mol H2
2 mol H2O
1 mol O2
2 mol H2O
1 mol O2
2 mol H2
2 mol H2O
2 mol H2
2 mol H2O
1 mol O2
9.1
• Introduction to Stoichiometry 169
We use the mole ratio to convert the number of moles of one substance to the corresponding
number of moles of another substance in a chemical reaction. For example, if we want to calculate the number of moles of H2O that can be obtained from 4.0 mol of O2 , we use the mole
ratio 2 mol H2O>1 mol O2 :
(4.0 mol O2 )¢
2 mol H2O
≤ = 8.0 mol H2O
1 mol O2
Example 9.1
Use the following equation CO2 + 4 H2 S CH4 + 2 H2O to write the mole ratio needed
to calculate
(a) the number of moles of water that can be produced from 3 moles of CO2.
(b) the number of moles of hydrogen required to make 3 moles of water.
SOLUTION
In order to write the mole ratio we need to determine the relationship between the desired
substance and the starting substance.
(a) In this case the desired substance is water. The moles of water belong in the numerator
of the mole ratio. The starting substance is CO2. The moles of CO2 then belong in the
denominator of the mole ratio. In the mole ratio the coefficients of the balanced equation
2 mol H2O
are used. Therefore, the mole ratio is
.
1 mol CO2
(b) In this case the desired substance is hydrogen. The moles of hydrogen belong in the
numerator of the mole ratio. The starting substance is water. The moles of water then
belong in the denominator of the mole ratio. In the mole ratio the coefficients of the
4 mol H2
balanced equation are used. Therefore, the mole ratio is
.
2 mol H2O
Practice 9.1
Use the equation 2 KClO3 S 2 KCl + 3 O2 to write the mole ratio needed to calculate
(a) moles of KCl produced from 3.5 moles of KClO3.
(b) moles of KCl produced when 4.5 moles of O2 are produced.
Problem-Solving Strategy: For Stoichiometry Problems
1. Convert the quantity of starting substance to moles (if it is not given in moles).
2. Convert the moles of starting substance to moles of desired substance.
3. Convert the moles of desired substance to the units specified in the problem.
Like balancing chemical equations, making stoichiometric calculations requires practice. Several worked examples follow. Study this material and practice on the problems at the end of
this chapter.
Problem-Solving Strategy: For Stoichiometry Problems
Use a balanced equation.
1. Determine the number of moles of starting substance. Identify the starting substance
from the data given in the problem statement. Convert the quantity of the starting substance
to moles, if it is not already done:
moles = (grams)a
1 mole
b
molar mass
Write and balance the equation
before you begin the problem.
170 chapter 9
• Calculations from Chemical Equations
As in all problems with units, the
desired quantity is in the numerator, and the quantity to be eliminated is in the denominator.
2. Determine the mole ratio of the desired substance to the starting ­substance. The number of moles of each substance in the balanced equation is indicated by the coefficient in
front of each substance. Use these coefficients to set up the mole ratio:
moles of desired substance in the equation
mole ratio =
moles of starting substance in the equation
Multiply the number of moles of starting substance (from Step 1) by the mole ratio to
obtain the number of moles of desired ­substance:
From Step 1
Units of moles of starting
substance cancel in the
numerator and denominator.
moles of desired
�
substance
�
moles of starting
substance
�
�
moles of desired substance
in the balanced equation
moles of starting substance
in the balanced equation
�
3. Calculate the desired substance in the units specified in the problem. If the answer is
to be in moles, the calculation is complete. If units other than moles are wanted, multiply the
moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the
units required. For example, if grams of the desired substance are wanted:
From Step 2
grams = (moles) a
molar mass
b
1 mol
If moles ¡ atoms, use
6.022 * 1023 atoms
1 mol
If moles ¡ molecules, use
6.022 * 1023 molecules
1 mol
The steps for converting the mass of a starting substance A to either the mass, atoms, or
molecules of desired substance B are summarized in Figure 9.1.
Figure 9.1
Steps for converting starting
substance A to mass, atoms, or
molecules of desired substance B.
Grams of A
Grams of B
Step 1
Moles of A
Step 2
Mole ratio
Moles of B
Atoms or
molecules of A
Step 3
Atoms or
molecules of B
9.2 Mole–Mole Calculations
Lear ning objec ti ve
Solve problems in which the reactants and products are both in moles.
Let’s solve stoichiometric problems for mole–mole calculations. The quantity of starting
­substance is given in moles, and the quantity of desired substance is ­requested in moles.
Example 9.2
ENHANCED EXAMPLE
How many moles of carbon dioxide will be produced by the complete reaction of 2.0 mol
of glucose (C6H12O6) according to the following equation?
C6H12O6 + 6 O2 S 6 CO2 + 6 H2O
1 mol
6 mol
6 mol
6 mol
SOLUTION
Read •
Knowns:
2.0 mol of glucose
C6H12O6 + 6 O2 S 6 CO2 + 6 H2O
Solving for: moles of carbon dioxide
9.2
Plan •
The balanced equation states that 6 mol CO2 will be produced from 1 mol
C6H12O6. Even though we can readily see that 12 mol CO2 will be formed,
let’s use the Problem-Solving Strategy for Stoichiometry Problems.
• Mole–Mole Calculations 171
The mole ratio is exact and does
not affect the number of significant figures in the answer.
Solution map: moles of C6H12O6 S moles of CO2
The mole ratio from the balanced equation is
Calculate • (2.0 mol C6H12O6)a
Check •
6 mol CO2
.
1 mol C6H12O6
6 mol CO2
b = 12 mol CO2
1 mol C6H12O6
Note that the moles C6H12O6 cancel and that the answer agrees with our
r­ easoned answer.
Example 9.3
How many moles of ammonia can be produced from 8.00 mol of hydrogen reacting with
nitrogen? The balanced equation is
3 H2 + N2 S 2 NH3
SOLUTION
Read •
Knowns:
8.00 mol of hydrogen
3 H2 + N2 S 2 NH3
Solving for: moles of ammonia
Plan •
The balanced equation states that we get 2 mol NH3 for every 3 mol H2 that
react. Let’s use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: moles of H2 S moles of NH3
The mole ratio from the balanced equation is
Calculate • (8.00 mol H2)a
Check •
2 mol NH3
.
3 mol H2
2 mol NH3
b = 5.33 mol NH3
3 mol H2
Note that the moles H2 cancel and that the answer is less than the starting
8.00 mol, which makes sense since our mole ratio is 2:3.
Example 9.4
Given the balanced equation
K2Cr2O7 + 6 KI + 7 H2SO4 S Cr2(SO4)3 + 4 K2SO4 + 3 I2 + 7 H2O
1 mol
6 mol
3 mol
Calculate (a) the number of moles of potassium dichromate (K2Cr2O7) that will react with
2.0 mol of potassium iodide (KI) and (b) the number of moles of iodine (I2) that will be
produced from 2.0 mol of potassium iodide.
Solution
Read •
Knowns:
K2Cr2O7 + 6 KI + 7 H2SO4 S Cr2(SO4)3 + 4 K2SO4 + 3 I2 + 7 H2O
(a) 2.0 mol KI
(b) 2.0 mol KI
Ammonia
NH3
172 chapter 9
• Calculations from Chemical Equations
(a) Solving for: moles of K2Cr2O7
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: moles of KI S moles of K2Cr2O7
The mole ratio from the balanced equation is
Calculate • (2.0 mol KI)a
Check •
1 mol K2Cr2O7
.
6 mol KI
1 mol K2Cr2O7
b = 0.33 mol K2Cr2O7
6 mol KI
Note that the moles KI cancel and that the answer is less than the starting
2.0 mol, which makes sense since our mole ratio is 1:6.
(b) Solving for: moles of I2
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: mole of KI S moles of l2
The mole ratio from the balanced equation is
Calculate • (2.0 mol KI)a
Check •
3 mol I2
.
6 mol KI
3 mol I2
b = 1.0 mol I2
6 mol KI
Note that the moles KI cancel and that the answer is half the starting 2.0 mol,
which makes sense since our mole ratio is 1:2.
Example 9.5
How many molecules of water can be produced by reacting 0.010 mol of oxygen with
hydrogen? The balanced equation is 2 H2 + O2 S 2 H2O.
Solution
Read •
Knowns:
0.010 mol O2
2 H 2 + O2 S 2 H 2 O
Solving for: molecules H2O
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: moles of O2 S moles of H2O S molecules of H2O
The mole ratio from the balanced equation is
Calculate • (0.010 mol O2)a
2 mol H2O
.
1 mol O2
2 mol H2O
b = 0.020 mol H2O
1 mol O2
Conversion factor for moles H2O S molecules H2O
6.022 * 1023 molecules
1 mol
(0.020 mol H2O)a
Note
6.022 * 1023 molecules
b = 1.2 * 1022 molecules H2O
1 mol
Note that 0.020 mol is still quite a large number of water molecules.
9.3
• Mole–Mass Calculations 173
Practice 9.2
How many moles of aluminum oxide will be produced from 0.50 mol of ­oxygen?
4 Al + 3 O2 ¡ 2 Al2O3
Practice 9.3
How many moles of aluminum hydroxide are required to produce 22.0 mol of water?
2 Al(OH)3 + 3 H2SO4 ¡ Al2(SO4)3 + 6 H2O
9.3 Mole–Mass Calculations
Solve problems in which mass is given and the answer is to be determined in moles
or the moles are given and mass is to be determined.
Learning objective
The object of this type of problem is to calculate the mass of one substance that reacts with or
is produced from a given number of moles of another substance in a chemical reaction. If the
mass of the starting substance is given, we need to convert it to moles. We use the mole ratio to
convert moles of starting substance to moles of desired substance. We can then change moles
of desired substance to mass.
Example 9.6
What mass of hydrogen can be produced by reacting 6.0 mol of aluminum with ­hydrochloric
acid? The balanced equation is 2 Al(s) + 6 HCl(aq) S 2 AlCl3(aq) + H2(g).
Solution
Read •
Knowns:
6.0 mol Al
2 Al(s) + 6 HCl(aq) S 2 AlCl3(aq) + 3 H2(g)
Solving for: mass H2
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: moles of Al S moles of H2 S grams H2
The mole ratio from the balanced equation is
Calculate • (6.0 mol Al) a
3 mol H2
b = 9.0 mol H2
2 mol Al
Conversion factor for moles H2 S grams H2
(9.0 mol H2)a
Note
3 mol H2
.
2 mol Al
2.016 g H2
1 mol H2
2.016 g H2
b = 18 g H2
1 mol H2
You can keep the answer for the first calculation (9.0 mol H2) on your calculator and continue the calculation to the answer of 18 g H2 in one continuous
calculation as shown below:
moles of Al S moles of H2 S grams H2
(6.0 mol Al) a
3 mol H2 2.016 g H2
ba
b = 18 g H2
2 mol Al
1 mol H2
ENHANCED EXAMPLE
174 chapter 9
• Calculations from Chemical Equations
Example 9.7
How many moles of water can be produced by burning 325 g of octane (C8H18)? The balanced equation is 2 C8H18(l) + 25 O2(g) S 16 CO2(g) + 18 H2O(g).
Solution
Read •
Knowns:
325 g C8H18
2 C8H18(l) + 25 O2(g) S 16 CO2(g) + 18 H2O(g)
Solving for: mol H2O
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: grams of C8H18 S moles of C8H18 S moles of H2O
Conversion factor for grams C8H18 S moles C8H18
Octane
C8H18
The mole ratio from the balanced equation is
Calculate • (325 g C8H18)a
18 mol H2O
.
2 mol C8H18
1 mol C8H18
b = 2.85 mol C8H18
114.2 g C8H18
(2.85 mol C8H18)a
Check •
1 mol C8H18
114.2 g C8H18
18 mol H2O
b = 25.7 mol H2O
2 mol C8H18
On your calculator in continuous calculation you might get 25.6 mol H2O for
your answer. The difference in the last digit is due to rounding off at different
points in the calculation. You should check with your ­instructor to find out the
appropriate rules for your course. In our ­examples we round at the end of each
step to show the proper number of significant figures for each calculation.
Practice 9.4
How many moles of potassium chloride and oxygen can be produced from 100.0 g of
potassium chlorate? The balanced equation is:
2 KClO3 ¡ 2 KCl + 3 O2
Practice 9.5
How many grams of silver nitrate are required to produce 0.25 mol of silver sulfide? The
balanced equation is:
2 AgNO3 + H2S ¡ Ag2S + 2 HNO3
9.4Mass–Mass Calculations
L earning obje ctive
Solve problems in which mass is given and the answer is to be determined as mass.
Solving mass–mass stoichiometry problems requires all the steps of the mole-ratio method.
The mass of starting substance is converted to moles. The mole ratio is then used to determine
moles of desired substance, which, in turn, is converted to mass of desired substance.
Example 9.8
What mass of carbon dioxide is produced by the complete combustion of 100. g of the
hydrocarbon pentane, C5H12? The balanced equation is
C5H12(l) + 8 O2(g) S 5 CO2(g) + 6 H2O(g).
Solution
Read •
Pentane
C5H12
Knowns:100. g C5H12
C5H12(l) + 8 O2(g) S 5 CO2(g) + 6 H2O(g)
Solving for: mass CO2
9.4
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: grams of C5H12 S moles of C5H12 S moles CO2 S g CO2
Conversion factor for grams C5H12 S moles C5H12
The mole ratio from the balanced equation is
1 mol C5H12
72.15 g C5H12
5 mol CO2
.
1 mol C5H12
44.01 g CO2
1 mol CO2
Conversion factor for moles CO2 S g CO2
Calculate • grams of C5H12 S moles of C5H12
(100. g C5H12) a
1 mol C5H12
b = 1.39 mol C5H12
72.15 g C5H12
moles of C5H12 S moles CO2
(1.39 mol C5H12) a
5 mol CO2
b = 6.95 mol CO2
1 mol C5H12
moles of CO2 S g CO2
(6.95 mol CO2) a
44.01 g CO2
b = 306 g CO2
1 mol CO2
Example 9.9
How many grams of nitric acid (HNO3) are required to produce 8.75 g of dinitrogen
­monoxide (N2O) according to the following equation?
4 Zn(s) + 10 HNO3(aq) S 4 Zn(NO3)2(aq) + N2O(g) + 5 H2O(l)
Solution
Read •
Knowns: 8.75 g N2O
4 Zn(s) + 10 HNO3(aq) S 4 Zn(NO3)2(aq) + N2O(g) + 5 H2O(l)
Solving for: g HNO3
Plan •
Use the Problem-Solving Strategy for Stoichiometry Problems.
Solution map: grams of N2O S moles N2O S moles HNO3 S g HNO3
Conversion factor for grams N2O S moles N2O
The mole ratio from the balanced equation is
10 mol HNO3
.
1 mol N2O
Conversion factor for moles HNO3 S g HNO3
Calculate • grams of N2O S moles of N2O
1 mol N2O
(8.75 g N2O)a
b = 0.199 mol N2O
44.02 g N2O
moles N2O S moles HNO3
(0.199 mol N2O)a
10 mol HNO3
b = 1.99 mol HNO3
1 mol N2O
moles HNO3 S g HNO3
(1.99 mol HNO3)a
1 mol N2O
44.02 g N2O
63.02 g HNO3
b = 125 g HNO3
1 mol HNO3
63.02 g HNO3
1 mol HNO3
• Mass–Mass Calculations 175
176 chapter 9
• Calculations from Chemical Equations
Practice 9.6
How many grams of chromium(III) chloride are required to produce 75.0 g of silver
­chloride? The balanced equation is:
CrCl3 + 3 AgNO3 ¡ Cr(NO3)3 + 3 AgCl
Practice 9.7
What mass of water is produced by the complete combustion of 225.0 g of ­butane (C4H10)?
The balanced equation is:
2 C4H10 + 13 O2 ¡ 8 CO2 + 10 H2O
9.5Limiting Reactant
and Yield ­Calculations
L earning obje ctive
Solve problems involving limiting reactants and yield.
Key terms
In many chemical processes, the quantities of the reactants used are such that one reactant is
in excess. The amount of the product(s) formed in such a case depends on the reactant that is
not in excess. This reactant is called the ­limiting reactant—it limits the amount of product
that can be formed.
Consider the case illustrated in Figure 9.2. How many bicycles can be ­assembled from
the parts shown? The limiting part in this case is the number of pedal assemblies; only three
bicycles can be built because there are only three pedal assemblies. The wheels and frames
are parts in excess.
Let’s consider a chemical example at the molecular level in which seven molecules of H2
are combined with four molecules of Cl2 (Figure 9.3 before reaction). How many molecules
of HCl can be produced according to this reaction?
limiting reactant
theoretical yield
actual yield
percent yield
H2 + Cl2 ¡ 2 HCl
Figure 9.2
8 wheels
4 frames
The number of bicycles that
can be built from these parts
is determined by the “limiting
reactant” (the pedal assemblies).
3 pedal
assemblies
3 bicycles
9.5
• Limiting Reactant and Yield Calculations 177
>Chemistry in action
The microchip has revolutionized the field of electronics. Engineers at Bell Laboratories, Massachusetts
Institute of Technology, the University of California, and
Stanford University are racing to produce parts for tiny
­machi­nes and robots. New techniques now produce
gears smaller than a grain of sand and motors lighter
than a speck of dust.
To produce ever smaller computers, calculators, and
even microbots (micro­sized robots), precise quantities of
chemicals in exact proportions are required. The secret to
producing minute circuits is to print the entire ­circuit or
blueprint at one time. Computers are used to draw a chip.
This image is then transferred onto a pattern, or mask,
with details finer than a human hair. In a process similar to
photography, light is then shined through the mask onto
a silicon-coated surface. The areas created on the ­silicon
exhibit high or low resistance to chemical etching. Chemicals then etch away the silicon.
Micromachinery is produced in the same way. First a thin
layer of silicon dioxide is applied (sacrificial material); then
a layer of polysilicon is carefully ­applied (structural material). Next, a mask is applied, and the whole structure is
covered with plasma (excited gas). The plasma acts as a
tiny sandblaster, ­removing everything the mask doesn’t
protect. This process is repeated as the entire machine
is constructed. When the entire assembly is complete,
the whole machine is placed in hydrofluoric acid, which
dissolves all the sacrificial material and permits the various
parts of the machine to move.
To turn these micromachines into true microbots, ­current
research is ­focusing on locomotion and sensing imaging systems. Possible uses for these microbots include
“smart” pills, which could contain sensors or drug
reservoirs (currently used in birth control). Tiny pumps,
once inside the body, will dispense the proper amount of
medication at precisely the correct site. These microbots
are currently in production for the treatment of diabetes
(to release ­insulin).
© Michael Knight/iStockphoto
A Shrinking Technology
These nanorobots attach themselves to red blood cells.
If the molecules of H2 and Cl2 are taken apart and recombined as HCl ­(Figure 9.3 after
r­ eaction), we see that eight molecules of HCl can be formed before we run out of Cl2 . Therefore,
the Cl2 is the limiting reactant and H2 is in excess—three molecules of H2 remain unreacted.
Before reaction
H
Figure 9.3
Cl
Reaction of H2 with Cl2. Eight
molecules of HCl are formed, and
three molecules of H2 remain. The
limiting reactant is Cl2.
After reaction
When problem statements give the amounts of two reactants, one of them is usually a
limiting reactant. We can identify the limiting reactant using the following method:
Problem-Solving Strategy: For Limiting Reactant Problems
1. Calculate the amount of product (moles or grams, as needed) formed from each reactant.
2. Determine which reactant is limiting. (The reactant that gives the least amount of product
is the limiting reactant; the other reactant is in ­excess.)
3. Once we know the limiting reactant, the amount of product formed can be ­determined. It is
the amount determined by the limiting reactant.
4. If we need to know how much of the other reactant remains, we calculate the amount of the
other reactant required to react with the limiting reactant, then subtract this amount from
the starting quantity of the reactant. This gives the amount of that substance that remains
­unreacted.
178 chapter 9
• Calculations from Chemical Equations
Example 9.10
ENHANCED EXAMPLE
How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2?
Which compound is the limiting reactant? The equation is H2(g) + Cl2(g) S 2 HCl(g).
Solution
Read •
Knowns:
4.0 mol H2
3.5 mol Cl2
H2(g) + Cl2(g) S 2 HCl(g)
Solving for: mol HCl
Plan •
Use the Problem-Solving Strategy for Limiting Reactant Problems.
Calculate the mol HCl formed from each reactant.
The mole ratios from the balanced equation are
2 mol HCl
1 mol H2
2 mol HCl
1 mol Cl2
Calculate • moles H2 S moles HCl
(4.0 mol H2)a
2 mol HCl
b = 8.0 mol HCl
1 mol H2
moles of Cl2 S moles HCl
2 mol HCl
b = 7.0 mol HCl
1 mol Cl2
Determine the limiting reactant. The limiting reactant is Cl2 because it
­produces less HCl than H2. The H2 is in excess. The yield of HCl is 7.0 mol.
(3.5 mol Cl2)a
Check •
In this reaction we can decide on the limiting reactant by inspection. From the
equation you can see that 1 mol H2 reacts with 1 mol Cl2. Therefore, when we
react 4.0 mol H2 with 3.5 mol Cl2, Cl2 is the reactant that limits the amount
of HCl produced, since it is present in smaller amount.
Example 9.11
How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe with 10.0 g H2O?
Which substance is the limiting reactant? Which substance is in excess?
3 Fe(s) + 4 H2O(g) 9: Fe3O4(s) + 4 H2(g)

Solution
Read •
Knowns:
16.5 g Fe
10.0 g H2O

3 Fe(s) + 4 H2O(g) 9: Fe3O4(s) + 4 H2(g)
Solving for: mol Fe3O4
Plan •
Use the Problem-Solving Strategy for Limiting Reactant Problems.
Calculate the mol Fe3O4 formed from each reactant:
g reactant S mol reactant S mol Fe3O4
Calculate • For Fe
(16.8 g Fe)a
1 mol Fe3O4
1 mol Fe
ba
b = 0.100 mol Fe3O4
55.85 g Fe
3 mol Fe
9.5
• Limiting Reactant and Yield Calculations 179
For H2O
(10.0 g H2O)a
1 mol Fe3O4
1 mol H2O
ba
b = 0.139 mol Fe3O4
18.02 g H2O
4 mol H2O
Determine the limiting reactant. The limiting reactant is Fe because it
­produces less Fe3O4 than H2O. The H2O is in excess. The yield of Fe3O4 is
0.100 mol.
Example 9.12
How many grams of silver bromide (AgBr) can be formed when solutions containing
50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of excess
reactant remain unreacted?
Solution
Read •
Knowns:
50.0 g MgBr2
100.0 g AgNO3
MgBr2(aq) + 2 AgNO3(aq) S 2 AgBr(s) + Mg(NO3)2(aq)
Solving for: g AgBr
Plan •
Use the Problem-Solving Strategy for Limiting Reactant Problems.
Calculate the g AgBr formed from each reactant:
g reactant S mol reactant S mol AgBr S g AgBr
Calculate • For MgBr2
(50.0 g MgBr2)a
1 mol MgBr2
2 mol AgBr
187.8 g AgBr
ba
ba
b
184.1 g MgBr2 1 mol MgBr2
1 mol AgBr
= 102 g AgBr
For AgNO3
(100.0 g AgNO3)a
1 mol AgNO3
2 mol AgBr
187.8 g AgBr
ba
ba
b
169.9 g AgNO3 2 mol AgNO3
1 mol AgBr
= 110.5 g AgBr
Determine the limiting reactant. The limiting reactant is MgBr2 because it
produces less AgBr than AgNO3. The AgNO3 is in excess. The yield of AgBr
is 102 g.
Calculate the mass of unreacted AgNO3. We must first determine the g AgNO3
that react with 50.0 g MgBr2.
g MgBr2 S mol MgBr2 S mol AgNO3 S g AgNO3
2 mol AgNO3
169.9 g AgNO3
1 mol MgBr2
ba
ba
b
184.1 g MgBr2
1 mol MgBr2
1 mol AgNO3
= 92.3 g AgNO3
(50.0 g MgBr2)a
Unreacted AgNO3 = 100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3
The final mixture will contain:
102 g AgBr(s) 7.7 g AgNO3
Practice 9.8
How many grams of hydrogen chloride can be produced from 0.490 g of hydrogen and
50.0 g of chlorine? The balanced equation is:
H2(g) + Cl2(g) ¡ 2 HCl(g)
180 chapter 9
• Calculations from Chemical Equations
Practice 9.9
How many grams of barium sulfate will be formed from 200.0 g of barium ­nitrate and
100.0 g of sodium sulfate? The balanced equation is:
Ba(NO3)2(aq) + Na2SO4(aq) ¡ BaSO4(s) + 2 NaNO3(aq)
The quantities of the products we have been calculating from chemical equations ­represent
the maximum yield (100%) of product according to the reaction represented by the equation.
Many reactions, especially those involving organic substances, fail to give a 100% yield of
product. The main reasons for this ­failure are the side reactions that give products other than
the main product and the fact that many reactions are reversible. In addition, some product
may be lost in handling and transferring from one vessel to another. The theore­tical yield
of a reaction is the calculated amount of product that can be obtained from a given amount of
reactant, according to the chemical equation. The ­actual yield is the amount of product that
we finally obtain.
The percent yield is the ratio of the actual yield to the theoretical yield multiplied
by 100. Both the theoretical and the actual yields must have the same units to obtain a
percent:
actual yield
* 100 = percent yield
theoretical yield
For example, if the theoretical yield calculated for a reaction is 14.8 g, and the amount of
product obtained is 9.25 g, the percent yield is
Round off as appropriate for your
particular course.
percent yield = a
9.25 g
b (100) = 62.5% yield
14.8 g
Example 9.13
ENHANCED EXAMPLE
Carbon tetrachloride (CCl4) was prepared by reacting 100. g of carbon disulfide and
100. g of chlorine. Calculate the percent yield if 65.0 g of CCl4 were obtained from the
­reaction CS2 + 3 Cl2 S CCl4 + S2Cl2.
Solution
Read •
Knowns:
100. g CS2
100. g Cl2
65.0 g CCl4 obtained
CS2 + 3 Cl2 S CCl4 + S2Cl2
Solving for: % yield of CCl4
Plan •
Use the equation for percent yield
actual yield
* 100 = percent yield
theoretical yield
Determine the theoretical yield. Calculate the g CCl4 formed from each
­reactant.
g reactant S mol reactant S mol CCl4 S g CCl4
Calculate • For CS2
153.8 g CCl4
1 mol CS2
1 mol CCl4
ba
ba
b
76.15 g CS2
1 mol CS2
1 mol CCl4
= 202 g CCl4
(100. g CS2)a
9.5
For Cl2
• Limiting Reactant and Yield Calculations 181
153.8 g CCl4
1 mol Cl2
1 mol CCl4
ba
ba
b
70.90 g Cl2
3 mol Cl2
1 mol CCl4
= 72.3 g CCl4
(100. g Cl2)a
Determine the limiting reactant. The limiting reactant is Cl 2 because it
­produces less CCl4 than CS2. The CS2 is in excess. The theoretical yield of
CCl4 is 72.3 g.
Calculate the percent yield
65.0 g
actual yield
* 100 = a
b (100) = 89.9% yield
theoretical yield
72.3 g
Example 9.14
Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an ­adequate
amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide were
obtained from the reaction:
MgBr2 + 2 AgNO3 S Mg(NO3)2 + 2 AgBr
Solution
Read •
Knowns:
200.0 g MgBr2
375.0 g AgBr
MgBr2 + 2 AgNO3 S Mg(NO3)2 + 2 AgBr
Solving for: % yield AgBr
Plan •
Use the equation for percent yield
actual yield
* 100 = percent yield
theoretical yield
Determine the theoretical yield. Calculate the grams of AgBr formed:
g MgBr2 S mol MgBr2 S mol AgBr S g AgBr
Calculate • (200.0 g MgBr2)a
1 mol MgBr2
2 mol AgBr
187.8 g AgBr
ba
ba
b
184.1 g MgBr2 1 mol MgBr2
1 mol AgBr
= 408.0 g AgBr
The theoretical yield is 408.0 g AgBr.
Calculate the percent yield.
actual yield
375.0 g
* 100 = a
b (100) = 91.91% yield
theoretical yield
408.0 g
Practice 9.10
Aluminum oxide was prepared by heating 225 g of chromium(II) oxide with 125 g of
aluminum. Calculate the percent yield if 100.0 g of aluminum oxide were obtained. The
balanced equation is:
2 Al + 3 CrO ¡ Al2O3 + 3 Cr
182 chapter 9
• Calculations from Chemical Equations
C h a p t e r
9 review
9.1 Introduction to Stoichiometry
Key Terms
stoichiometry
molar mass
mole ratio
• Solving stoichiometry problems requires the use of moles and mole ratios.
• To solve a stoichiometry problem:
• Convert the quantity of starting substance to moles.
• Convert the moles of starting substance to moles of desired substance.
• Convert the moles of desired substance to the ­appropriate unit.
Grams of A
Grams of B
Step 1
Moles of A
Step 2
Mole ratio
Moles of B
Step 3
Atoms or
molecules of A
Atoms or
molecules of B
9.2 Mole–Mole Calculations
• If the starting substance is given in moles and the unit specified for the answer is moles:
• Convert the moles of starting substance to moles of desired substance.
9.3 Mole–Mass Calculations
• If the starting substance is given in moles and the unit specified for the answer is grams:
• Convert the moles of starting substance to moles of desired substance.
• Convert the moles of desired substance to grams.
9.4 Mass–Mass Calculations
• If the starting substance is given in grams and the unit specified for the answer is grams:
• Convert the quantity of starting substance to moles.
• Convert the moles of starting substance to moles of desired substance.
• Convert the moles of desired substance to grams.
9.5 Limiting Reactant and Yield Calculations
Key Terms
limiting reactant
theoretical yield
actual yield
percent yield
• To identify the limiting reactant in a reaction:
• Calculate the amount of product formed from each reactant.
• Determine the limiting reactant by selecting the one that gives the least amount of product.
• To determine the actual amount of product formed in a limiting reactant situation:
• Use the result calculated that is the least amount of product.
• To determine the amount of the other reactants ­required to react with the limiting reactant:
• Calculate the amount of the other reactant needed to react with the limiting reactant.
• Subtract this amount from the original amount of the other reactant to find the amount of excess
r­ eactant (unreacted).
• The theoretical yield of a chemical reaction is the calculated amount of product from a given amount
of ­reactant (the limiting reactant).
• The actual yield is the amount of product actually ­obtained experimentally.
• The percent yield is
actual yield
* 100 = percent yield
theoretical yield
Paired Exercises 183
Review Questions
1. W
hat is a mole ratio?
2. What piece of information is needed to convert grams of a compound to moles of the same compound?
3. Phosphine (PH3) can be prepared by the reaction of calcium phosphide, Ca3P2 :
Ca3P2 + 6 H2O ¡ 3 Ca(OH)2 + 2 PH3
Based on this equation, which of the following statements are correct? Show evidence to support your ­answer.
(a)One mole of Ca3P2 produces 2 mol of PH3 .
(b)One gram of Ca3P2 produces 2 g of PH3 .
(c)Three moles of Ca(OH)2 are produced for each 2 mol of PH3
produced.
(d)The mole ratio between phosphine and ­calcium phosphide is
2 mol PH3
1 mol Ca3P2
(e)When 2.0 mol of Ca3P2 and 3.0 mol of H2O react, 4.0 mol of
PH3 can be formed.
(f)When 2.0 mol of Ca3P2 and 15.0 mol of H2O react, 6.0 mol of
Ca(OH)2 can be formed.
(g)When 200. g of Ca3P2 and 100. g of H2O react, Ca3P2 is the
limiting reactant.
(h)When 200. g of Ca3P2 and 100. g of H2O react, the theoretical
yield of PH3 is 57.4 g.
4.The equation representing the reaction used for the commercial
preparation of hydrogen cyanide is
2 CH4 + 3 O2 + 2 NH3 ¡ 2 HCN + 6 H2O
Based on this equation, which of the following statements are
­correct? Rewrite incorrect statements to make them correct.
(a)Three moles of O2 are required for 2 mol of NH3 .
(b)Twelve moles of HCN are produced for every 16 mol of O2 that
react.
(c)The mole ratio between H2O and CH4 is
6 mol H2O
2 mol CH4
(d)When 12 mol of HCN are produced, 4 mol of H2O will be formed.
(e)When 10 mol of CH4 , 10 mol of O2 , and 10 mol of NH3 are
mixed and reacted, O2 is the limiting ­reactant.
(f)When 3 mol each of CH4 , O2 , and NH3 are mixed and reacted,
3 mol of HCN will be produced.
5.What information beyond the mole ratio is needed to convert from
moles of reactant to grams of product?
6.Draw a flowchart showing the conversions necessary to determine
grams of ammonia formed from 10 g of hydrogen gas.
3 H2 + 2 N2 ¡ 2 NH3
7.What is the difference between the theoretical and the ­actual yield
of a chemical reaction?
8. How is the percent yield of a reaction calculated?
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1. Calculate the number of moles in these quantities:
(a) 25.0 g KNO3
(b) 56 millimol NaOH
(c) 5.4 * 102 g (NH4)2C2O4
(d)16.8 mL H2SO4 solution
(d = 1.727 g>mL, 80.0% H2SO4 by mass)
3. Calculate the number of grams in these quantities:
(a) 2.55 mol Fe(OH)3
(b) 125 kg CaCO3
(c) 10.5 mol NH3
(d) 72 millimol HCl
(e) 500.0 mL of liquid Br2 (d = 3.119 g>mL)
2.Calculate the number of moles in these quantities:
(a) 2.10 kg NaHCO3
(b) 525 mg ZnCl2
(c) 9.8 * 1024 molecules CO2
(d) 250 mL ethyl alcohol, C2H5OH (d = 0.789 g>mL)
5. Which contains the larger number of molecules, 10.0 g H2O or
10.0 g H2O2 ? Show evidence for your answer.
6. Which contains the larger numbers of molecules, 25.0 g HCl or
85.0 g C6H12O6 ? Show evidence for your answer.
7. Balance the equation for the synthesis of sucrose
8. Balance the equation for the combustion of butanol
CO2 + H2O ¡ C12H22O11 + O2
and set up the mole ratio of
(a) CO2 to H2O
(b) H2O to C12H22O11
(c) O2 to CO2
4.Calculate the number of grams in these quantities:
(a) 0.00844 mol NiSO4
(b) 0.0600 mol HC2H3O2
(c) 0.725 mol Bi2S3
(d) 4.50 * 1021 molecules glucose, C6H12O6
(e)75 mL K2CrO4 solution
(d = 1.175 g>mL, 20.0% K2CrO4 by mass)
C4H9OH + O2 ¡ CO2 + H2O
and set up the mole ratio of
(d) C12H22O11 to CO2
(e) H2O to O2
(f) O2 to C12H22O11
(a) O2 to C4H9OH
(b) H2O to O2
(c) CO2 to H2O
(d) C4H9OH to CO2
(e) H2O to C4H9OH
(f) CO2 to O2
184 chapter 9
• Calculations from Chemical Equations
10. Given the unbalanced equation
9. Given the unbalanced equation
CO2 + H2 ¡ CH4 + H2O
H2SO4 + NaOH ¡ Na2SO4 + H2O
(a)How many moles of water can be produced from 25 moles of
carbon dioxide?
(b)How many moles of CH4 will be produced along with 12 moles
of water?
11. Given the equation
MnO2(s) + HCl(aq) ¡ Cl2(g) + MnCl2(aq) + H2O(l)
(unbalanced)
(a) How many moles of HCl will react with 1.05 mol of MnO2?
(b)How many moles of MnCl2 will be produced when 1.25 mol of
H2O are formed?
(c)How many grams of Cl2 will be produced when 3.28 mol of
MnO2 react with excess HCl?
(d)How many moles of HCl are required to produce 15.0 kg of MnCl2?
(a)How many moles of NaOH will react with 17 mol of
H2SO4?
(b)How many moles of Na2SO4 will be produced when 21 mol
of NaOH react?
12.Given the equation
13.Carbonates react with acids to form a salt, water, and ­carbon dioxide gas. When 50.0 g of calcium carbonate are reacted with sufficient hydrochloric acid, how many grams of calcium chloride will
be produced? (Write a balanced equation first.)
15. I n a blast furnace, iron(III) oxide reacts with coke (carbon) to produce molten iron and carbon monoxide:
Fe2O3 + 3 C 9: 2 Fe + 3 CO
Al4C3 + 12 H2O ¡ 4 Al(OH)3 + 3 CH4
(a)How many moles of water are needed to react with 100. g of Al4C3 ?
(b)How many moles of Al(OH)3 will be produced when 0.600 mol
of CH4 is formed?
(c)How many moles of CH4 will be formed by the reaction of
275 grams of Al4C3 with excess water?
(d)How many grams of water are required to produce 4.22 moles
of Al(OH)3?
14.Certain metals can displace hydrogen from acids to produce hydrogen gas and a salt. When 25.2 g of aluminum metal are placed into
hydrobromic acid, how many grams of aluminum bromide will be
produced? (Write the balanced equation first.)
16.How many grams of steam and iron must react to produce 375 g of
magnetic iron oxide, Fe3O4 ?
3 Fe(s) + 4 H2O(g) 9: Fe3O4(s) + 4 H2(g)


How many kilograms of iron would be formed from 125 kg of Fe2O3 ?
17.Polychlorinated biphenyls (PCBs), which were formerly used in
the manufacture of electrical transformers, are environmental and
health hazards. They break down very slowly in the environment.
The decomposition of PCBs can be represented by the equation
2 C12H4Cl6 + 23 O2 + 2 H2O ¡ 24 CO2 + 12 HCl
(a)How many moles of water are needed to react 10.0 mol O2?
(b)How many grams of HCl are produced when 15.2 mol H2O react?
(c)How many moles of CO2 are produced when 76.5 g HCl are
produced?
(d)How many grams of C12H4Cl6 are reacted when 100.25 g CO2
are produced?
(e)How many grams of HCl are produced when 2.5 kg C12H4Cl6 react?
19. Draw the molecules of product(s) formed for each of the following
reactions; then determine which substance is the limiting reactant:
(a)
18.Mercury can be isolated from its ore by the reaction
4 HgS + 4 CaO ¡ 4 Hg + 3 CaS + CaSO4
(a) How many moles of CaS are produced from 2.5 mol CaO?
(b)How many grams of Hg are produced along with 9.75 mol of
CaSO4?
(c)How many moles of CaO are required to react with 97.25 g of
HgS?
(d) How many grams of Hg are produced from 87.6 g of HgS?
(e)How many grams of CaS are produced along with 9.25 kg
of Hg?
20.Draw the molecules of product(s) formed for the following reactions; then determine which substance is the limiting ­reactant:
(a)
¡
¡
Li
H2
O2
I2
(b)
(b)
¡
¡
H2
Ag
Br2
Cl2
21.Draw pictures similar to those in Question 19 and ­determine which
is the limiting reactant in the following reactions:
(a)Eight atoms of potassium react with five molecules of chlorine
to produce potassium chloride.
(b)Ten atoms of aluminum react with three molecules of oxygen
to produce aluminum oxide.
22.Draw pictures similar to those in Question 20 and ­determine which
is the limiting reactant in the following reactions:
(a)Eight molecules of nitrogen react with six ­molecules of oxygen
to produce nitrogen dioxide.
(b)Fifteen atoms of iron react with twelve molecules of water to
produce magnetic iron oxide (Fe3O4) and hydrogen.
Additional Exercises 185
23.In the following equations, determine which reactant is the limiting reactant and which reactant is in excess. The amounts used are
given below each reactant. Show evidence for your answers.
(a) KOH + HNO3 ¡ KNO3 + H2O
16.0 g
12.0 g
(b) 2 NaOH + H2SO4 ¡ Na2SO4 + H2O
10.0 g
10.0 g
25. Aluminum sulfate has many uses in industry. It can be prepared by
reacting aluminum hydroxide and sulfuric acid:
2 Al(OH)3 + 3 H2SO4 ¡ Al2(SO4)3 + 6 H2O
24.In the following equations, determine which reactant is the ­limiting
reactant and which reactant is in excess. The amounts used are
given below each reactant. Show evidence for your answers.
(a) 2 Bi(NO3)3 + 3 H2S ¡ Bi2S3 + 6 HNO3
50.0 g
6.00 g
(b) 3 Fe + 4 H2O ¡ Fe3O4 + 4 H2
40.0 g 16.0 g
26. Phosphorous trichloride, a precursor for the manufacture of certain
pesticides and herbicides, can be produced by the reaction of phosphorous and chlorine:
P4 + 6 Cl2 ¡ 4 PCl3
(a)When 35.0 g of Al(OH)3 and 35.0 g of H2SO4 are reacted, how
many moles of Al2(SO4)3 can be produced? Which substance
is the limiting reactant?
(b)When 45.0 g of H2SO4 and 25.0 g of Al(OH)3 are reacted, how
many grams of Al2(SO4)3 are produced? Which reactant is in
excess?
(c)In a reaction vessel, 2.5 mol of Al(OH)3 and 5.5 mol of H2SO4
react. Determine which substances will be present in the container when the reaction goes to ­completion and how many
moles of each.
(a)When 20.5 g of P4 and 20.5 g of Cl2 are reacted, how many
moles of PCl3 can be produced?
(b)When 55 g of Cl2 are reacted with 25 g of P4, how many grams
of PCl3 can be produced? Which reactant is in ­excess?
(c)In a reaction vessel, 15 mol of P4 and 35 mol of Cl2 react.
­Determine which substances will be present in the container
after the reaction goes to completion and how many moles of
each substance is present.
27.When a certain nonmetal whose formula is X8 burns in air, XO3
forms. Write a balanced equation for this reaction. If 120.0 g of
oxygen gas are consumed completely, along with 80.0 g of X8 ,
identify the element X.
28.When a particular metal X reacts with HCl, the resulting products
are XCl2 and H2 . Write and balance the equation. When 78.5 g of
the metal react completely, 2.42 g of hydrogen gas result. Identify
the element X.
29.Elemental silicon can be produced by the reduction of silicon dioxide, or sand, with carbon:
30. Chromium metal can be produced by the reduction of Cr2O3 with
elemental silicon:
SiO2 + 2 C ¡ Si + 2 CO
When 35.0 kg of SiO2 reacts with 25.3 kg of carbon, and 14.4 kg
of silicon is recovered, what is the percent yield for the reaction?
31. In a laboratory experiment, 27.5 g of Cu were reacted with 125 g
of HNO3:
2 Cr2O3 + 3 Si ¡ 4 Cr + 3 SiO2
If 350.0 grams of Cr2O3 are reacted with 235.0 grams of elemental
silicon and 213.2 grams of chromium metal are recovered, what is
the percent yield?
32.In aqueous solution, metal oxides can react with acids to form a salt
and water:
3 Cu + 8 HNO3 ¡ 3 Cu(NO3)2 + 4 H2O + 2 NO
Fe2O3(s) + 6 HCI(aq) ¡ 2 FeCl3(aq) + 3 H2O(l)
(a) How many grams of Cu(NO3)2 can be theoretically produced?
(b)How many grams of the excess reactant remain?
(c)If the percent yield of Cu(NO3)2 is 87.3%, what is the actual
yield in grams?
(a)How many moles of each product will be formed when 35 g of
Fe2O3 react with 35 g of HCl?
(b)From part (a), how many grams of the excess reactant remain?
(c) If the percent yield of FeCl3 is 92.5%, what is the actual yield
in grams?
Additional Exercises
33.A tool set contains 6 wrenches, 4 screwdrivers, and 2 pliers. The
manufacturer has 1000 pliers, 2000 screwdrivers, and 3000 wrenches
in stock. Can an order for 600 tool sets be filled? Explain briefly.
34.Write a balanced equation for the fermentation of glucose,
C6H12O6 , into ethyl alcohol, C2H5OH and CO2. Determine how
many liters of ethyl alcohol (d = 0.789 g/mL) can be produced from
575 lb of glucose.
35.For mass–mass calculation, why is it necessary to convert the grams
of starting material to moles of starting material, then determine the
moles of product from the moles of starting material, and convert
the moles of product to grams of product? Why can’t you calculate
the grams of product directly from the grams of starting material?
36.Oxygen masks for producing O2 in emergency situations contain
potassium superoxide (KO2). It reacts according to this equation:
4 KO2 + 2 H2O + 4 CO2 ¡ 4 KHCO3 + 3 O2
(a)If a person wearing such a mask exhales 0.85 g of CO2 every
minute, how many moles of KO2 are consumed in 10.0 minutes?
(b)How many grams of oxygen are produced in 1.0 hour?
37. In the presence of concentrated H2SO4, sucrose, C12H22O11, undergoes dehydration, forming carbon and water:
C12H22O11 99: 12 C + 11 H2O
H2SO4
(a)How many grams of carbon and how many grams of water can
be produced from 2.0 lb of sucrose?
(b)If 25.2 g sucrose are dehydrated, how many grams of liquid
water with a density of 0.994 g/mL can be collected?
38.Alfred Nobel extended Sobrero’s work on nitroglycerin and discovered that when mixed with diatomaceous earth it would form a
stable mixture. This mixture is now known as dynamite. The reaction of nitroglycerin is
4 C3H5(NO3)3(l) ¡ 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)
(a)Calculate the number of grams of each product that are produced by the decomposition of 45.0 g of nitroglycerin.
(b)Calculate the total number of moles of gas produced.
186 chapter 9
• Calculations from Chemical Equations
39. The methyl alcohol (CH3OH) used in alcohol burners combines
with oxygen gas to form carbon dioxide and water. How many
grams of oxygen are required to burn 60.0 mL of methyl alcohol
(d 5 0.787 g/mL)?
40. Hydrazine (N2H4) and hydrogen peroxide (H2O2) have been used as
rocket propellants. They react according to the following equation:
7 H2O2 + N2H4 ¡ 2 HNO3 + 8 H2O
(a)When 75 kg of hydrazine react, how many grams of nitric acid
can be formed?
(b)When 250 L of hydrogen peroxide (d 5 1.41 g/mL) react, how
many grams of water can be formed?
(c)How many grams of hydrazine will be required to react with
725 g hydrogen peroxide?
(d)How many grams of water can be produced when 750 g hydrazine combine with 125 g hydrogen peroxide?
(e)How many grams of the excess reactant in part (d) are left
­unreacted?
41. Chlorine gas can be prepared according to the reaction
16 HCl + 2 KMnO4 ¡ 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O
(a)How many moles of MnCl2 can be produced when 25 g KMnO4
are mixed with 85 g HCl?
(b)How many grams of water will be produced when 75 g KCl are
produced?
(c)What is the percent yield of Cl2 if 150 g HCl are reacted, producing 75 g of Cl2?
(d)When 25 g HCl react with 25 g KMnO4, how many grams of
Cl2 can be produced?
(e) How many grams of the excess reactant in part (d) are left
unreacted?
42.Silver tarnishes in the presence of hydrogen sulfide (which smells
like rotten eggs) and oxygen because of the reaction
4 Ag + 2 H2S + O2 ¡ 2 Ag2S + 2 H2O
(a)How many grams of silver sulfide can be formed from a mixture of 1.1 g Ag, 0.14 g H2S, and 0.080 g O2?
(b)How many more grams of H2S would be needed to completely
react all of the Ag?
43.Humans long ago discovered that they could make soap by cooking
fat with ashes from a fire. This happens because the ash contains
potassium hydroxide that reacts with the triglycerides in the fat to
form soap. The chemical reaction for this process is
C3H5(C15H31CO2)3 + 3 KOH ¡ C3H5(OH)3 + 3 C15H31CO2K
(triglyceride)
(soap)
If 500.0 g of this triglyceride are mixed with 500.0 g of potassium
hydroxide and 361.7 g of soap are produced, what is the percent
yield for the reaction?
44.Group 2A metal oxides react with water to produce hydroxides. When
35.55 g of CaO are placed into 125 mL water (d 5 1.000 g/mL), how
many grams of Ca(OH)2 can be produced?
45.When a solution of lead(II) nitrate is mixed with a solution of sodium chromate, a yellow precipitate forms.
(a)Write the balanced equation for the reaction, including the
states of all the substances.
(b)When 26.41 g of lead(II) nitrate are mixed with 18.33 g of sodium chromate, what is the percent yield of the solid if 21.23 g
are recovered?
46. Ethyl alcohol burns, producing carbon dioxide and water:
C2H5OH + 3 O2 ¡ 2 CO2 + 3 H2O
(a)If 2.5 mol of C2H5OH react with 7.5 mol of O2 in a closed
container, what substances and how many moles of each will
be present in the container when the reaction has gone to
completion?
(b)When 225 g of C2H5OH are burned, how many grams of CO2
and how many grams of H2O will be produced?
47.After 180.0 g of zinc were dropped into a beaker of hydrochloric acid
and the reaction ceased, 35 g of unreacted zinc remained in the beaker:
Zn + HCl ¡ ZnCl2 + H2
(a)How many grams of hydrogen gas were produced?
(b)How many grams of HCl were reacted?
(c)How many more grams of HCl would be required to completely
react with the original sample of zinc?
48.Use this equation to answer (a) and (b):
Fe(s) + CuSO4(aq) ¡ Cu(s) + FeSO4(aq)
(a)When 2.0 mol of Fe and 3.0 mol of CuSO4 are reacted, what
substances will be present when the reaction is over? How
many moles of each substance are present?
(b)When 20.0 g of Fe and 40.0 g of CuSO4 are reacted, what substances will be present when the reaction is over? How many
grams of each substance are present?
49.Methyl alcohol (CH3OH) is made by reacting carbon monoxide
and hydrogen in the presence of certain metal oxide catalysts. How
much alcohol can be obtained by reacting 40.0 g of CO and 10.0 g
of H2? How many grams of excess reactant remain ­unreacted?
CO(g) + 2 H2(g) ¡ CH3OH(l)
50. Ethyl alcohol (C2H5OH), also called grain alcohol, can be made by
the fermentation of glucose:
C6H12O6 ¡ 2 C2H5OH + 2 CO2
glucose
ethyl alcohol
If an 84.6% yield of ethyl alcohol is obtained,
(a)what mass of ethyl alcohol will be produced from 750 g of
glucose?
(b)what mass of glucose should be used to produce 475 g of
C2H5OH?
51. Both CaCl2 and MgCl2 react with AgNO3 to precipitate AgCl.
When solutions containing equal masses of CaCl2 and MgCl2 are
reacted with AgNO3 , which salt solution will produce the larger
amount of AgCl? Show proof.
52.An astronaut excretes about 2500 g of water a day. If lithium oxide
(Li2O) is used in the spaceship to absorb this water, how many
kilograms of Li2O must be carried for a 30-day space trip for three
astronauts?
Li2O + H2O ¡ 2 LiOH
53. Much commercial hydrochloric acid is prepared by the reaction of
concentrated sulfuric acid with sodium chloride:
H2SO4 + 2 NaCl ¡ Na2SO4 + 2 HCl
How many kilograms of concentrated H2SO4, 96% H2SO4 by mass,
are required to produce 20.0 L of concentrated hydrochloric acid
(d = 1.20 g/mL, 42.0% HCl by mass)?
54.Silicon carbide or carborundum is a widely used ceramic material in car brakes and bullet-proof vests because of its hardness.
Carborundum is found naturally in the mineral moissanite, but the
demand for it exceeds the supply. For this reason it is generally
synthesized in the laboratory. The reaction is
SiO2 + 3 C ¡ SiC + 2 CO
How many kg of silicon carbide can be formed by the reaction of
250.0 kg of sand (SiO2) with excess carbon?
Answers to Practice Exercises 187
55. Three chemical reactions that lead to the formation of ­sulfuric
acid are
S + O2 ¡ SO2
56. A 10.00-g mixture of NaHCO3 and Na2CO3 was heated and yielded 0.0357 mol H2O and 0.1091 mol CO2. Calculate the percent
composition of the mixture:
NaHCO3 ¡ Na2O + CO2 + H2O
2 SO2 + O2 ¡ 2 SO3
Na2CO3 ¡ Na2O + CO2
SO3 + H2O ¡ H2SO4
Starting with 100.0 g of sulfur, how many grams of sulfuric acid
will be formed, assuming a 10% loss in each step? What is the
percent yield of H2SO4?
57. Draw a picture showing the reaction of 2 molecules CH4 with 3 molecules of O2 to form water and carbon dioxide gas. Also show any
unreacted CH4 and O2. Does one of the reactants limit the amount of
products formed? If so, which one?
Challenge Exercises
58. When 12.82 g of a mixture of KClO3 and NaCl are heated strongly,
the KClO3 reacts according to this equation:
begin your drive west, but upon arriving in Kansas you discover to
your dismay that your funds have dwindled. You meet up with the
manager of a feed mill who has just lost his formulations chemist
and needs help determining how many bags of feed supplement
he can produce with the oats, molasses, alfalfa, and apples that
are arriving on Monday. Luckily you just finished your chemistry
class and remember learning about limiting reactants. If you can
use these ideas to solve the following problem your money worries may be over! Here is the problem: A new customer just ordered a new formulation of supplement containing 25.0 lb of oats,
17.5 lb of molasses, 30.0 lb of alfalfa, and 28.5 lb of apples per
bag. If four train cars of materials are due in the plant on Monday
containing the components listed below, how much money will you
make for the remainder of your trip if you earn $2.25 for every bag
of supplement produced?
2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g)
The NaCl does not undergo any reaction. After the heating, the
mass of the residue (KCl and NaCl) is 9.45 g. Assuming that all the
loss of mass represents loss of oxygen gas, calculate the percent of
KClO3 in the original mixture.
59.Gastric juice contains about 3.0 g HCl per liter. If a person produces
about 2.5 L of gastric juice per day, how many antacid tablets, each
containing 400 mg of Al(OH)3, are needed to neutralize all the HCl
produced in one day?
Al(OH)3(s) + 3 HCl(aq) ¡ AlCl3(aq) + 3 H2O(l)
60. Phosphoric acid, H3PO4, can be synthesized from phosphorus,
­oxygen, and water according to these two reactions:
4 P + 5 O2 ¡ P4O10
Raw materials
1 boxcar of oats containing 4.20 tons of oats
1 tanker of molasses containing 2490 gallons of molasses (molasses has a density of 1.46 g/mL)
250.0 crates of apples, each containing 12.5 kg of apples
350.0 bales of alfalfa, each weighing 62.8 kg
P4O10 + 6 H2O ¡ 4 H3PO4
Starting with 20.0 g P, 30.0 g O2, and 15.0 g H2O, what is the mass
of phosphoric acid that can be formed?
61.You have just finished your first year of college in South Carolina
and decide to take a cross-country trip to visit California. You
Answers to Practice Exercises
9.1
9.2
2 mol KCl
2 mol KClO3
0.33 mol Al2O3
(a)
(b)
2 mol KCl
3 mol O2
9.3
7.33 mol Al(OH)3
9.4
0.8157 mol KCl, 1.223 mol O2
9.5
85 g AgNO3
9.6
27.6 g CrCl3
9.7
348.6 g H2O
9.8
17.7 g HCl
9.9
164.3 g BaSO4
9.10 89.1% yield
PUTTING IT TOGETHER:
Review for Chapters 7–9
Multiple Choice
Choose the correct answer to each of the following.
1. 4.0 g of oxygen contain
(a) 1.5 * 1023 atoms of oxygen
(b) 4.0 molar masses of oxygen
(c) 0.50 mol of oxygen
(d) 6.022 * 1023 atoms of oxygen
2. One mole of hydrogen atoms contains
(a) 2.016 g of hydrogen
(b) 6.022 * 1023 atoms of hydrogen
(c) 1 atom of hydrogen
(d) 12 g of carbon-12
3. The mass of one atom of magnesium is
(a) 24.31 g
(c) 12.00 g
(b) 54.94 g
(d) 4.037 * 10-23 g
4. Avogadro’s number of magnesium atoms
(a) has a mass of 1.0 g
(b) has the same mass as Avogadro’s number of sulfur atoms
(c) has a mass of 12.0 g
(d) is 1 mol of magnesium atoms
5. Which of the following contains the largest number of moles?
(a) 1.0 g Li
(c) 1.0 g Al
(b) 1.0 g Na
(d) 1.0 g Ag
6. The number of moles in 112 g of acetylsalicylic acid (aspirin),
C9H8O4 , is
(a) 1.61
(c) 112
(b) 0.622
(d) 0.161
7. How many moles of aluminum hydroxide are in one antacid tablet
containing 400 mg of Al(OH)3 ?
(a) 5.13 * 10-3
(c) 5.13
(b) 0.400
(d) 9.09 * 10-3
8. How many grams of Au2S can be obtained from 1.17 mol of Au?
(a) 182 g
(c) 364 g
(b) 249 g
(d) 499 g
9. The molar mass of Ba(NO3)2 is
(a) 199.3
(c) 247.3
(b) 261.3
(d) 167.3
10. A 16-g sample of oxygen
(a) is 1 mol of O2
(b) contains 6.022 * 1023 molecules of O2
(c) is 0.50 molecule of O2
(d) is 0.50 molar mass of O2
11. What is the percent composition for a compound formed from 8.15 g
of zinc and 2.00 g of oxygen?
(a) 80.3% Zn, 19.7% O
(c) 70.3% Zn, 29.7% O
(b) 80.3% O, 19.7% Zn
(d) 65.3% Zn, 34.7% O
188
12. Which of these compounds contains the largest percentage of
oxygen?
(a) SO2
(c) N2O3
(b) SO3
(d) N2O5
13. 2.00 mol of CO2
(a) have a mass of 56.0 g
(b) contain 1.20 * 1024 molecules
(c) have a mass of 44.0 g
(d) contain 6.00 molar masses of CO2
14. In Ag2CO3 , the percent by mass of
(a) carbon is 43.5%
(c) oxygen is 17.4%
(b) silver is 64.2%
(d) oxygen is 21.9%
15. The empirical formula of the compound containing 31.0% Ti and
69.0% Cl is
(a) TiCl
(c) TiCl3
(b) TiCl2
(d) TiCl4
16. A compound contains 54.3% C, 5.6% H, and 40.1% Cl. The empirical formula is
(a) CH3Cl
(c) C2H4Cl2
(b) C2H5Cl
(d) C4H5Cl
17. A compound contains 40.0% C, 6.7% H, and 53.3% O. The molar
mass is 60.0 g>mol. The molecular formula is
(a) C2H3O2
(c) C2HO
(b) C3H8O
(d) C2H4O2
18. How many chlorine atoms are in 4.0 mol of PCl3 ?
(a) 3
(c) 12
(b) 7.2 * 1024
(d) 2.4 * 1024
19. What is the mass of 4.53 mol of Na2SO4 ?
(a) 142.1 g
(c) 31.4 g
(b) 644 g
(d) 3.19 * 10-2 g
20. The percent composition of Mg3N2 is
(a) 72.2% Mg, 27.8% N (c) 83.9% Mg, 16.1% N
(b) 63.4% Mg, 36.6% N (d) no correct answer given
21. How many grams of oxygen are contained in 0.500 mol of
Na2SO4 ?
(a) 16.0 g
(c) 64.0 g
(b) 32.0 g
(d) no correct answer given
22. The empirical formula of a compound is CH. If the molar mass of
this compound is 78.11, then the molecular formula is
(a) C2H2
(c) C6H6
(b) C5H18
(d) no correct answer given
23. The reaction
BaCl2 + (NH4)2CO3 ¡ BaCO3 + 2 NH4Cl
is an example of
(a) combination
(b) decomposition
(c) single displacement
(d) double displacement
Putting It Together 189
34. When the equation
24. When the equation
Al + O2 ¡ Al2O3
is properly balanced, which of the following terms ­appears?
(a) 2 Al
(b) 2 Al2O3
(c) 3 Al
(d) 2 O2
25. Which equation is incorrectly balanced?

(a) 2 KNO3 ¡ 2 KNO2 + O2
(b) H2O2 ¡ H2O + O2
(c) 2 Na2O2 + 2 H2O ¡ 4 NaOH + O2
(d) 2 H2O 9999: 2 H2 + O2
electrical energy
26. The reaction
H2SO4
2 Al + 3 Br2 ¡ 2 AlBr3
is an example of
(a) combination
(b) single displacement
Fe2(SO4)3 + Ba(OH)2 ¡
is completed and balanced, one term in the balanced equation is
(c) 2 Fe2(SO4)3
(a) Ba2(SO4)3
(b) 2 Fe(OH)2
(d) 2 Fe(OH)3
35. For the reaction
2 H2 + O2 ¡ 2 H2O + 572.4 kJ
which of the following is not true?
(a) The reaction is exothermic.
(b) 572.4 kJ of heat are liberated for each mole of water formed.
(c) 2 mol of hydrogen react with 1 mol of oxygen.
(d) 572.4 kJ of heat are liberated for each 2 mol of hydrogen
reacted.
(c) decomposition
(d) double displacement
36. How many moles are 20.0 g of Na2CO3 ?
(a) 1.89 mol
(c) 212 mol
(d) 0.189 mol
(b) 2.12 * 103 mol

37. What is the mass in grams of 0.30 mol of BaSO4 ?
(a) 7.0 * 103 g
(c) 70. g
(b) 0.13 g
(d) 700.2 g
27. When the equation
PbO2 ¡ PbO + O2
is balanced, one term in the balanced equation is
(a) PbO2
(c) 3 PbO
(b) 3 O2
(d) O2
28. When the equation
Cr2S3 + HCl ¡ CrCl3 + H2S
is balanced, one term in the balanced equation is
(a) 3 HCl
(c) 3 H2S
(b) CrCl3
(d) 2 Cr2S3
29. When the equation
F2 + H2O ¡ HF + O2
is balanced, one term in the balanced equation is
(a) 2 HF
(c) 4 HF
(b) 3 O2
(d) 4 H2O
30. When the equation
NH4OH + H2SO4 ¡
is completed and balanced, one term in the balanced ­equation is
(a) NH4SO4
(c) H2OH
(b) 2 H2O
(d) 2 (NH4)2SO4
31. When the equation
H2 + V2O5 ¡ V +
is completed and balanced, one term in the balanced ­equation is
(a) 2 V2O5
(c) 2 V
(b) 3 H2O
(d) 8 H2
32. When the equation
Al(OH)3 + H2SO4 ¡ Al2(SO4)3 + H2O
is balanced, the sum of the coefficients will be
(a) 9
(c) 12
(b) 11
(d) 15
33. When the equation
H3PO4 + Ca(OH)2 ¡ H2O + Ca3(PO4)2
is balanced, the proper sequence of coefficients is
(a) 3, 2, 1, 6
(c) 2, 3, 1, 6
(b) 2, 3, 6, 1
(d) 2, 3, 3, 1
38. How many molecules are in 5.8 g of acetone, C3H6O ?
(a) 0.10 molecule
(b) 6.0 * 1022 molecules
(c) 3.5 * 1024 molecules
(d) 6.0 * 1023 molecules
Problems 39–45 refer to the reaction
2 C2H4 + 6 O2 ¡ 4 CO2 + 4 H2O
39. If 6.0 mol of CO2 are produced, how many moles of O2 were
reacted?
(a) 4.0 mol
(c) 9.0 mol
(b) 7.5 mol
(d) 15.0 mol
40. How many moles of O2 are required for the complete ­combustion of
45 g of C2H4 ?
(a) 1.3 * 102 mol
(c) 112.5 mol
(b) 0.64 mol
(d) 4.8 mol
41. If 18.0 g of CO2 are produced, how many grams of H2O are produced?
(a) 7.37 g
(c) 9.00 g
(b) 3.68 g
(d) 14.7 g
42. How many moles of CO2 can be produced by the reaction of 5.0 mol
of C2H4 and 12.0 mol of O2 ?
(a) 4.0 mol
(c) 8.0 mol
(b) 5.0 mol
(d) 10. mol
43. How many moles of CO2 can be produced by the reaction of
0.480 mol of C2H4 and 1.08 mol of O2 ?
(a) 0.240 mol
(c) 0.720 mol
(b) 0.960 mol
(d) 0.864 mol
44. How many grams of CO2 can be produced from 2.0 g of C2H4 and
5.0 g of O2 ?
(a) 5.5 g
(c) 7.6 g
(b) 4.6 g
(d) 6.3 g
45. If 14.0 g of C2H4 are reacted and the actual yield of H2O is 7.84 g,
the percent yield in the reaction is
(a) 0.56%
(c) 87.1%
(b) 43.6%
(d) 56.0%
190 Putting It Together
Problems 46–48 refer to the equation
H3PO4 + MgCO3 ¡ Mg3(PO4)2 + CO2 + H2O
46. The sequence of coefficients for the balanced equation is
(a) 2, 3, 1, 3, 3
(c) 2, 2, 1, 2, 3
(b) 3, 1, 3, 2, 3
(d) 2, 3, 1, 3, 2
47. If 20.0 g of carbon dioxide are produced, the number of moles of
magnesium carbonate used is
(a) 0.228 mol
(c) 0.910 mol
(b) 1.37 mol
(d) 0.454 mol
48. If 50.0 g of magnesium carbonate react completely with H3PO4 , the
number of grams of carbon dioxide produced is
(a) 52.2 g
(c) 13.1 g
(b) 26.1 g
(d) 50.0 g
49. When 10.0 g of MgCl2 and 10.0 g of Na2CO3 are reacted in
MgCl2 + Na2CO3 ¡ MgCO3 + 2 NaCl
the limiting reactant is
(a) MgCl2
(b) Na2CO3
(c) MgCO3
(d) NaCl
50. When 50.0 g of copper are reacted with silver nitrate ­solution
Cu + 2 AgNO3 ¡ Cu(NO3)2 + 2 Ag
148 g of silver are obtained. What is the percent yield of silver
obtained?
(a) 87.1%
(c) 55.2%
(b) 84.9%
(d) no correct answer given
Free Response Questions
Answer each of the following. Be sure to include your work and
explanations in a clear, logical form.
1. Compound X requires 104 g of O2 to produce 2 moles of CO2 and
2.5 moles of H2O.
(a) What is the empirical formula for X?
(b) What additional information would you need to ­determine the
molecular formula for X?
2. Consider the reaction of sulfur dioxide with oxygen to form sulfur
trioxide taking place in a closed container.
(b) Is Ca3(PO4)2 an element, a mixture, or a compound?
(c) Milk is a good source of calcium in the diet. If 120 mL of skim
milk provide 13% of the U.S. RDA for Ca, how many cups of
milk should you drink per day if that is your only source of
calcium?
5. Compound A decomposes at room temperature in an exothermic
reaction, while compound B requires heating before it will decompose in an endothermic reaction.
(a) Draw reaction profiles (potential energy vs. reaction progress)
for both reactions.
(b) The decomposition of 0.500 mol of NaHCO3 to form sodium
carbonate, water, and carbon dioxide requires 85.5 kJ of heat.
How many grams of water could be ­collected, and how many
kJ of heat are absorbed when 24.0 g of CO2 are ­produced?
(c) Could NaHCO3 be compound A or compound B? ­Explain your
reasoning.
6. Aqueous ammonium hydroxide reacts with aqueous cobalt(II)
sulfate to produce aqueous ammonium sulfate and solid cobalt(II)
hydroxide. When 38.0 g of one of the reactants is fully reacted
with enough of the other ­reactant, 8.09 g of ammonium sulfate is
obtained, which corresponded to a 25.0% yield.
(a) What type of reaction took place?
(b) Write the balanced chemical equation.
(c) What is the theoretical yield of ammonium sulfate?
(d) Which reactant was the limiting reagent?
7. C6H12O6 ¡ 2 C2H5OH + 2 CO2(g)
(a) If 25.0 g of C6H12O6 were used, and only 11.2 g of C2H5OH
were produced, how much reactant was left over and what volume of gas was produced? (Note: One mole of gas occupies
24.0 L of space at room temperature.)
(b) What was the yield of the reaction?
(c) What type of reaction took place?
8. When solutions containing 25 g each of lead(II) nitrate and potassium iodide were mixed, a yellow precipitate ­resulted.
(a) What type of reaction occurred?
(b) What are the name and formula for the solid product?
(c)If after filtration and drying, the solid product weighed 7.66 g,
what was the percent yield for the reaction?
9.Consider the following unbalanced reaction:
O2
SO3
(a) Draw what you would expect to see in the box at the completion
of the reaction.
(b) If you begin with 25 g of SO2 and 5 g of oxygen gas, which is
the limiting reagent?
(c) Is the following statement true or false? “When SO2 is converted into SO3 , the percent composition of S in the compounds
changes from 33% to 25%.” ­Explain.
3. The percent composition of compound Z is 63.16% C and 8.77%
H. When compound Z burns in air, the only products are carbon
dioxide and water. The molar mass for Z is 114.
(a) What is the molecular formula for compound Z?
(b) What is the balanced reaction for Z burning in air?
4. 16% of the U.S. RDA for Ca is 162 mg.
(a) What mass of calcium phosphate, Ca3(PO4)2 , provides 162 mg
of calcium?
XNO3 + CaCl2 ¡ XCl + Ca(NO3)2
(a) If 30.8 g of CaCl2 produced 79.6 g of XCl, what is X?
(b)Would X be able to displace hydrogen from an acid?
10. Consider the following reaction: H2O2 ¡ H2O + O2
(a)If at the end of the reaction there are eight water molecules and
eight oxygen molecules, what was in the flask at the start of the
reaction?
(b)Does the following reaction profile indicate the reaction is exothermic or endothermic?
Energy
SO2
Reaction progress
(c) What type of reaction is given above?
(d) What is the empirical formula for hydrogen peroxide?
C ha p t e r
10
Modern Atomic Theory
and the Periodic Table
to see? Think back to that birthday present you could
H
10.1 Electromagnetic Radiation
look at but not yet open. Judging from the wrapping
10.2 The Bohr Atom
lifting the package all gave indirect clues to its contents. After
10.4Atomic Structures of the
First 18 Elements
ow do we go about studying an object that is too small
and size of the box was not very useful, but shaking, turning, and
all your experiments were done, you could make a fairly good
guess about the contents. But was your guess correct? The only
way to know for sure would be to open the package.
Chemists have the same dilemma when they study the
atom. Atoms are so very small that it isn’t possible to use the
normal senses to describe them. We are essentially working
in the dark with this package we call the atom. However, our
improvements in instruments (X-ray machines and scanning
tunneling microscopes) and measuring devices (spectrophotometers and magnetic resonance imaging, MRI) as well as in
our mathematical skills are bringing us closer to revealing the
secrets of the atom.
Chapter Outline
10.3 Energy Levels of Electrons
10.5Electron Structures and the
Periodic Table
© Hannu Viitanen/iStockphoto
The amazing colors in these
fireworks explosions are the result
of electrons transferring between
energy levels in atoms.
192 chapter 10
• Modern Atomic Theory and the Periodic Table
10.1 Electromagnetic Radiation
Learning objectiv e
List the three basic characteristics of electromagnetic radiation.
Key Terms
In the last 200 years, vast amounts of data have been accumulated to support the atomic
theory. When atoms were originally suggested by the early Greeks, no physical evidence
existed to support their ideas. Early chemists did a variety of experiments, which culminated in Dalton’s model of the atom. Because of the limitations of Dalton’s model,
modifications were proposed first by Thomson and then by Rutherford, which eventually
led to our modern concept of the nuclear atom. These early models of the atom work
reasonably well—in fact, we continue to use them to visualize a variety of chemical concepts. There remain questions that these models cannot answer, including an explanation
of how atomic structure relates to the periodic table. In this chapter, we will present our
modern model of the atom; we will see how it varies from and improves upon the earlier
atomic models.
wavelength
frequency
speed
photons
Electromagnetic Radiation
Tony Freeman/PhotoEdit
Scientists have studied energy and light for centuries, and several models have been proposed
to explain how energy is transferred from place to place. One way energy travels through space
is by electromagnetic radiation. Examples of electromagnetic radiation include light from the
sun, X-rays in your dentist’s office, microwaves from your microwave oven, radio and television waves, and radiant heat from your fireplace. While these examples seem quite different,
they are all similar in some important ways. Each shows wavelike behavior, and all travel at
the same speed in a vacuum (3.00 * 108 m/s).
The study of wave behavior is a topic for another course, but we need some basic
terminology to understand atoms. Waves have three basic characteristics: wavelength, frequency, and speed. Wavelength (lambda, l) is the distance between consecutive peaks (or
troughs) in a wave, as shown in Figure 10.1. Frequency (nu, n) tells how many waves pass a
particular point per second. Speed (n) tells how fast a wave moves through space.
λ
Surfers judge the wavelength,
frequency, and speed of waves
to get the best ride.
λ
Figure 10.1
The wavelength of this wave is shown by l . It can be measured
from peak to peak or trough to trough.
Light is one form of electromagnetic radiation and is usually classified by its wavelength, as shown in Figure 10.2. Visible light, as you can see, is only a tiny part of the
electromagnetic spectrum. Some examples of electromagnetic radiation involved in
energy transfer outside the visible region are hot coals in your backyard grill, which transfer infrared radiation to cook your food, and microwaves, which transfer energy to water molecules
in the food, causing them to move more quickly and thus raise the temperature of the food.
Gamma
rays
10 –10
X-rays
10 –8
4 x 10 –7
7 x 10–7
Ultraviolet
Visible
10 –12
10 – 4
10 –2
Infrared
Microwave
Figure 10.2
The electromagnetic spectrum.
Wavelength in meters
1
10 2
10 4
Radio waves
10.2
• The Bohr Atom 193
>Chemistry in action
You Light Up My Life
Courtesy of NJ Marshall,
University of Queensland
Birds in the parrot family have an unusual way to attract
their mates—their feathers glow in the dark! This phenomenon is called fluorescence. It results from the absorption
of ultraviolet (UV) light, which is then reemitted at longer
wavelengths that both birds and people can see. In everyday life this happens in a fluorescent bulb or in the many
glow-in-the-dark products such as light sticks.
Perhaps we may find that the glow of candlelight really
does add to the radiance of romance. Ultraviolet light
certainly does for parrots!
Courtesy of NJ Marshall,
University of Queensland
Kathleen Arnold from the University of Glasgow, Scotland,
discovered that the feathers of parrots that fluoresced were
only those used in display or those shown off during courtship. She decided to experiment using budgerigars with
their natural colors. The researchers offered birds a choice
of two companion birds, which were smeared with petroleum jelly. One of the potential companions also had a UV
blocker in the petroleum jelly. The birds clearly preferred
companions without the UV blocker, leading the researchers to conclude that the parrots prefer to court radiant
partners. The researchers also tested same-sex companions
and discovered they did not prefer radiant companions.
Budgerigars under normal light and under UV light showing
the glow used to attract mates.
We have evidence for the wavelike nature of light. We also know that a beam of
light behaves like a stream of tiny packets of energy called photons. So what is light
exactly? Is it a particle? Is it a wave? Scientists have agreed to explain the properties of
electromagnetic radiation by using both wave and particle properties. Neither explanation
is ideal, but currently these are our best models.
Example 10.1
What is a wavelength?
Solution
A wavelength is the distance between consecutive peaks or troughs of a wave.
The symbol used for wavelength is lambda (l).
Practice 10.1
Draw a wave function and indicate the distance of two wavelengths (2 lambda).
10.2 The Bohr Atom
Explain the relationship between the line spectrum and the quantized energy levels
of an electron in an atom.
As scientists struggled to understand the properties of electromagnetic radiation, evidence began to accumulate that atoms could radiate light. At high temperatures, or when
subjected to high voltages, elements in the gaseous state give off colored light. Brightly
colored neon signs illustrate this property of matter very well. When the light emitted by
a gas is passed through a prism or diffraction grating, a set of brightly colored lines called
Learning objectiv e
Key Terms
line spectrum
quanta
ground state
orbital
194 chapter 10
• Modern Atomic Theory and the Periodic Table
410
434
486
656
(in nanometers)
Figure 10.3
Line spectrum of hydrogen. Each line corresponds to the wavelength of the energy
emitted when the electron of a hydrogen atom, which has absorbed energy, falls back
to a lower principal energy level.
Possible
electron
orbits
Nucleus
Figure 10.4
The Bohr model of the hydrogen
atom described the electron
revolving in certain allowed
circular orbits around the nucleus.
Excited
state 2
a line spectrum results (Figure 10.3). These colored lines indicate that the light is being
emitted only at certain wavelengths, or frequencies, that correspond to specific colors.
Each element possesses a unique set of these spectral lines that is different from the sets
of all the other elements.
In 1912–1913, while studying the line spectrum of hydrogen, Niels Bohr (1885–1962), a
Danish physicist, made a significant contribution to the rapidly growing knowledge of atomic
structure. His research led him to believe that electrons exist in specific regions at various
distances from the nucleus. He also visualized the electrons as revolving in orbits around the
nucleus, like planets rotating around the sun, as shown in Figure 10.4.
Bohr’s first paper in this field dealt with the hydrogen atom, which he described as a
single electron revolving in an orbit about a relatively heavy nucleus. He applied the concept of energy quanta, proposed in 1900 by the German physicist Max Planck (1858–1947),
to the observed line spectrum of hydrogen. Planck stated that energy is never emitted in
a continuous stream but only in small, discrete packets called quanta (Latin, quantus,
“how much”). From this, Bohr theorized that electrons have several possible energies corresponding to several possible orbits at different distances from the nucleus. Therefore, an
electron has to be in one specific energy level; it cannot exist between energy levels. In
other words, the energy of the electron is said to be quantized. Bohr also stated that when
a hydrogen atom absorbed one or more quanta of energy, its electron would “jump” to a
higher energy level.
Bohr was able to account for spectral lines of hydrogen this way. A number of energy levels are available, the lowest of which is called the ground state. When an electron falls from
a high energy level to a lower one (say, from the fourth to the second), a quantum of energy is
emitted as light at a specific frequency, or wavelength (Figure 10.5). This light corresponds
to one of the lines visible in the hydrogen spectrum (Figure 10.3). Several lines are visible
in this spectrum, each one corresponding to a specific electron energy-level shift within the
hydrogen atom.
Example 10.2
Excited
state 1
What does the line spectrum of a hydrogen atom illustrate?
Solution
Each line shows the frequency of light energy emitted when an electron that has
­absorbed energy falls from a higher energy level to a lower energy level.
Practice 10.2
Ground
state
What is a photon?
Figure 10.5
When an excited electron returns
to the ground state, energy
emitted as a photon is released.
The color (wavelength) of the light
is determined by the difference
in energy between the two states
(excited and ground).
The chemical properties of an element and its position in the periodic table depend on
electron behavior within the atoms. In turn, much of our knowledge of the behavior of electrons
within atoms is based on spectroscopy. Niels Bohr contributed a great deal to our knowledge
of atomic structure by (1) suggesting quantized energy levels for electrons and (2) showing
that spectral lines result from the radiation of small increments of energy (Planck’s quanta)
when electrons shift from one energy level to another. Bohr’s calculations succeeded very well
10.3
• Energy Levels of Electrons 195
in correlating the experimentally observed spectral lines with electron energy levels for the
hydrogen atom. However, Bohr’s methods of calculation did not succeed for heavier atoms.
More theoretical work on atomic structure was needed.
In 1924, the French physicist Louis de Broglie (1892–1957) suggested a surprising hypothesis: All objects have wave properties. Louis de Broglie used sophisticated
mathematics to show that the wave properties for an object of ordinary size, such as
a baseball, are too small to be observed. But for smaller objects, such as an electron,
the wave properties become significant. Other scientists confirmed de Broglie’s hypothesis, showing that electrons do exhibit wave properties. In 1926, Erwin Schrödinger
(1887–1961), an Austrian physicist, created a mathematical model that described electrons
as waves. Using Schrödinger’s wave mechanics, we can determine the probability of finding an electron in a certain region around the nucleus of the atom.
This treatment of the atom led to a new branch of physics called wave mechanics or
quantum mechanics, which forms the basis for our modern understanding of atomic structure.
Although the wave-mechanical description of the atom is mathematical, it can be translated,
at least in part, into a visual model. It is important to recognize that we cannot locate an electron precisely within an atom; however, it is clear that electrons are not revolving around the
nucleus in orbits as Bohr postulated. The electrons are instead found in orbitals. An orbital
is pictured in Figure 10.6 as a region in space around the nucleus where there is a high probability of finding a given electron.
Figure 10.6
An orbital for a hydrogen atom.
The intensity of the dots shows
that the electron spends more
time closer to the nucleus.
10.3 Energy Levels of Electrons
Describe the principal energy levels, sublevels, and orbitals of an atom.
Learning objective
One of the ideas Bohr contributed to the modern concept of the atom was that the energy
of the electron is quantized—that is, the electron is restricted to only certain allowed energies. The wave-mechanical model of the atom also predicts discrete principal energy levels
within the atom. These energy levels are designated by the letter n, where n is a positive
integer (Figure 10.7). The lowest principal energy level corresponds to n = 1, the next to
n = 2, and so on. As n increases, the energy of the electron increases, and the electron is found
on average farther from the nucleus.
Each principal energy level is divided into sublevels, which are illustrated in Figure 10.8.
The first principal energy level has one sublevel. The second principal energy level has two
sublevels, the third energy level has three sublevels, and so on. Each of these sublevels contains
spaces for electrons called orbitals.
In each sublevel the electrons are found within specified orbitals (s, p, d, f ). Let’s consider each principal energy level in turn. The first principal energy level (n = 1) has one
Key Terms
n=4
n=3
n=2
Number of sublevels
Energy
4f
4d
3d
4p
3p
2p
4s
3s
2s
1s
n=1
Figure 10.7
Figure 10.8
The first four principal energy levels in the
hydrogen atom. Each level is assigned a
principal quantum number n.
The types of orbitals on each
of the first four principal
energy levels.
principal energy levels
sublevels
spin
Pauli exclusion principle
196 chapter 10
• Modern Atomic Theory and the Periodic Table
Figure 10.9
z
z
z
Perspective representation of the
px, py, and pz atomic orbitals.
x
x
y
y
px
Notice that there is a correspondence between the energy level
and number of sublevels.
py
x
y
pz
sublevel or type of orbital. It is spherical in shape and is designated as 1s. It is important
to understand what the spherical shape of the 1s orbital means. The electron does not move
around on the surface of the sphere, but rather the surface encloses a space where there is
a 90% probability where the electron may be found. It might help to consider these orbital
shapes in the same way we consider the atmosphere. There is no distinct dividing line
between the atmosphere and “space.” The boundary is quite fuzzy. The same is true for
atomic orbitals. Each has a region of highest density roughly corresponding to its shape.
The probability of finding the electron outside this region drops rapidly but never quite
reaches zero. Scientists often speak of orbitals as electron “clouds” to emphasize the fuzzy
nature of their boundaries.
How many electrons can fit into a 1s orbital? To answer this question, we need to consider one more property of electrons. This property is called spin. Each electron appears
to be spinning on an axis, like a globe. It can only spin in two directions. We represent this
spin with an arrow: c or T . In order to occupy the same orbital, electrons must have opposite spins. That is, two electrons with the same spin cannot occupy the same orbital. This
gives us the answer to our question: An atomic orbital can hold a maximum of two electrons, which must have opposite spins. This rule is called the Pauli exclusion principle.
The first principal energy level contains one type of orbital (1s) that holds a maximum of
two electrons.
What happens with the second principal energy level (n = 2)? Here we find two sublevels, 2s and 2p. Like 1s in the first principal energy level, the 2s orbital is spherical in
shape but is larger in size and higher in energy. It also holds a maximum of two electrons.
The second type of orbital is designated by 2p. The 2p sublevel consists of three orbitals:
2px, 2py, and 2pz. The shape of p orbitals is quite different from the s orbitals, as shown in
Figure 10.9.
Each p orbital has two “lobes.” Remember, the space enclosed by these surfaces represents
the regions of probability for finding the electrons 90% of the time. There are three separate
p orbitals, each oriented in a different direction, and each p orbital can hold a maximum of
two electrons. Thus the total number of electrons that can reside in all three p orbitals is six.
To summarize our model, the first principal energy level of an atom has a 1s orbital. The second principal energy level has a 2s and three 2p orbitals labeled 2px, 2py, and 2pz, as shown
in Figure 10.10.
The third principal energy level has three sublevels labeled 3s, 3p, and 3d. The 3s orbital
is spherical and larger than the 1s and 2s orbitals. The 3px, 3py, and 3pz orbitals are shaped
Figure 10.10
z
z
Orbitals on the second principal
energy level are one 2s and three
2p orbitals.
x
x
y
y
s orbital
p orbitals
10.3
• Energy Levels of Electrons 197
>Chemistry in action
Atomic Clocks
level to keep the cesium atoms moving from one level to
the next. One second is equal to 9,192,631,770 of these
vibrations. The clock is set to this frequency and can keep
accurate time for over a million years.
Imagine a clock that keeps time to within 1 second over
a million years. The National Institute of Standards and
Technology in Boulder, Colorado, has an atomic clock that
does just that—a little better than your average alarm,
grandfather, or cuckoo clock! This atomic clock serves as
the international standard for time and frequency. How
does it work?
Clayton Hansen/iStockphoto
Within the case are several layers of magnetic shielding.
In the heart of the clock is a small oven that heats cesium
metal to release cesium atoms, which are collected into
a narrow beam (1 mm wide). The beam of atoms passes
down a long evacuated tube while being excited by a laser
until all the cesium atoms are in the same electron state.
The atoms then enter another chamber filled with reflecting microwaves. The frequency of the microwaves
(9,192,631,770 cycles per second) is exactly the same
frequency required to excite a cesium atom from its
ground state to the next higher energy level. These
excited cesium atoms then release electromagnetic radiation in a process known as fluorescence. Electronic circuits
maintain the microwave frequency at precisely the right
This clock automatically updates itself by comparing time
with an atomic clock by radio signal.
like those of the second level, only larger. The five 3d orbitals have the shapes shown in
Figure 10.11. You don’t need to memorize these shapes, but notice that they look different
from the s or p orbitals.
Each time a new principal energy level is added, we also add a new sublevel. This makes
sense because each energy level corresponds to a larger average distance from the nucleus,
which provides more room on each level for new sublevels containing more orbitals.
The pattern continues with the fourth principal energy level. It has 4s, 4p, 4d, and 4f
orbitals. There are one 4s, three 4p, five 4d, and seven 4f orbitals. The shapes of the s, p, and
d orbitals are the same as those for lower levels, only larger. We will not consider the shapes of
the f orbitals. Remember that for all s, p, d, and f orbitals, the maximum number of electrons
per orbital is two. We summarize each principal energy level:
n
n
n
n
=
=
=
=
1
2
3
4
1s
2s
3s
4s
2p 2p 2p
3p 3p 3p
4p 4p 4p
z
3d 3d 3d 3d 3d
4d 4d 4d 4d 4d
4f 4f 4f 4f 4f 4f 4f
z
z
dxz orbital
x
x
dxy orbital
dyz orbital
z
y
y
y
x
z
y
x
dz2 orbital
Figure 10.11
The five d orbitals are found in the third principal energy level along with one 3s orbital
and three 3p orbitals.
x
dx2 –y2 orbital
198 chapter 10
• Modern Atomic Theory and the Periodic Table
Example 10.3
Energy levels of the atoms are designated as principal energy levels and energy sublevels.
(a) What labels are used for these energy levels?
(b) How many of each of these labels are there?
Solution
(a) The letter n is used to designate the principal energy levels. The letters s, p, d, and f
are used to designate the energy sublevels.
(b) According to the periodic table, there are seven (7) principal energy levels and 2s, 6p,
10d, and 14f energy sublevels.
Practice 10.3
(a) How many electrons can occupy a single subenergy level?
(b) What is the maximum number of electrons that can occupy the third principal energy
level? List them.
Figure 10.12
The modern concept of a hydrogen
atom consists of a proton and an
electron in an s orbital. The shaded
area represents a region where the
electron may be found with
90% probability.
The hydrogen atom consists of a nucleus (containing one proton) and one electron occupying a region outside of the nucleus. In its ground state, the electron occupies a 1s orbital,
but by absorbing energy the electron can become excited and move to a higher energy level.
The hydrogen atom can be represented as shown in Figure 10.12. The diameter of the nucleus
is about 10-13 cm, and the diameter of the electron orbital is about 10-8 cm. The diameter of the
electron cloud of a hydrogen atom is about 100,000 times greater than the diameter of the nucleus.
10.4 Atomic Structures
of the First 18 Elements
L earning obje ctive
Use the guidelines to write electron configurations.
Key Terms
We have seen that hydrogen has one electron that can occupy a variety of orbitals in different
principal energy levels. Now let’s consider the structure of atoms with more than one electron.
Because all atoms contain orbitals similar to those found in hydrogen, we can describe the
structures of atoms beyond hydrogen by systematically placing electrons in these hydrogenlike orbitals. We use the following guidelines:
electron configuration
orbital diagram
valence electrons
1. No more than two electrons can occupy one orbital.
2.Electrons occupy the lowest energy orbitals available. They enter a higher energy orbital
only when the lower orbitals are filled. For the atoms beyond hydrogen, orbital energies
vary as s 6 p 6 d 6 f for a given value of n.
3.Each orbital in a sublevel is occupied by a single electron before a second electron enters.
For example, all three p orbitals must contain one electron before a second electron enters
a p orbital.
We can use several methods to represent the atomic structures of atoms, depending on what
we are trying to illustrate. When we want to show both the nuclear makeup and the electron
structure of each principal energy level (without orbital detail), we can use a diagram such as
Figure 10.13.
Figure 10.13
Atomic structure diagrams of fluorine, sodium, and
magnesium atoms. The number of protons and
neutrons is shown in the nucleus. The number of
electrons is shown in each principal energy level
outside the nucleus.
9p
10n
n= 1
2e–
Fluorine atom
2
7e–
11p
12n
n= 1
2e–
Sodium atom
2
3
8e– 1e–
12p
12n
n= 1
2e–
2
3
8e– 2e–
Magnesium atom
10.4
• Atomic Structures of the First 18 Elements 199
Often we are interested in showing the arrangement of the electrons in an atom in their
orbitals. There are two ways to do this. The first method is called the electron ­configuration.
In this method, we list each type of orbital, showing the number of electrons in it as an exponent. An electron configuration is read as follows:
Number of electrons
in sublevel orbitals
6
2p
Principal
energy level
Type of orbital
We can also represent this configuration with an orbital diagram in which boxes represent
the orbitals (containing small arrows indicating the electrons). When the orbital contains one
electron, an arrow, pointing upward ( c ), is placed in the box. A second arrow, pointing downward ( T ), indicates the second electron in that orbital.
Let’s consider each of the first 18 elements on the periodic table in turn. The order of filling for the orbitals in these elements is 1s, 2s, 2p, 3s, 3p, and 4s. Hydrogen, the first element,
has only one electron. The electron will be in the 1s orbital because this is the most favorable
position (where it will have the greatest attraction for the nucleus). Both representations are
shown here:
c 
1s1
Orbital
diagram
Electron
configuration
H
Helium, with two electrons, can be shown as
He
cT
1s2
Orbital
diagram
Electron
configuration
The first energy level, which can hold a maximum of two electrons, is now full. An atom
with three electrons will have its third electron in the second principal energy level. Thus, in
lithium (atomic number 3), the first two electrons are in the 1s orbital, and the third electron
is in the 2s orbital of the second energy level.
Lithium has the following structure:
Li
cT c 
1s
1s22s1
2s
All four electrons of beryllium are s electrons:
Be
cT cT
1s
1s22s2
2s
The next six elements illustrate the filling of the p orbitals. Boron has the first p electron.
Because p orbitals all have the same energy, it doesn’t matter which of these orbitals fills first:
B
cT
cT
1s
2s
c 


1s22s22p1
2p
Carbon is the sixth element. It has two electrons in the 1s orbital, two electrons in the
2s orbital, and two electrons to place in the 2p orbitals. Because it is more difficult for the
p electrons to pair up than to occupy a second p orbital, the second p electron is located in a
different p orbital. We could show this by writing 2p1x 2p1y , but we usually write it as 2p2; it is
understood that the electrons are in different p orbitals. The spins on these electrons are alike,
for reasons we will not explain here.
C
cT
cT
1s
2s
cc
2p

1s22s22p2
200 chapter 10
• Modern Atomic Theory and the Periodic Table
Nitrogen has seven electrons. They occupy the 1s, 2s, and 2p orbitals. The third p electron
in nitrogen is still unpaired and is found in the 2pz orbital:
N
cT
cT
c c c 
1s
2s
2p
1s22s22p3
Oxygen is the eighth element. It has two electrons in both the 1s and 2s orbitals and four
electrons in the 2p orbitals. One of the 2p orbitals is now occupied by a second electron, which
has a spin opposite the electron already in that orbital:
O
cT
cT
cTc c 
1s
2s
2p
1s22s22p4
The next two elements are fluorine with nine electrons and neon with ten electrons:
F
cT
cT
cTcTc 
1s
2s
2p
Ne
1s22s22p5
cT
cT
cTcTcT
1s
2s
2p
1s22s22p6
With neon, the first and second energy levels are filled as shown in Table 10.1. The second
energy level can hold a maximum of eight electrons, 2s22p6.
Sodium, element 11, has two electrons in the first energy level and eight electrons in
the second energy level, with the remaining electron occupying the 3s orbital in the third
energy level:
Na
Table 10.1
cT
cT
cTcTcT
c 
1s
2s
2p
3s
1s22s22p63s1
Orbital Filling for the First Ten Elements*
Atomic
number
Element
Orbitals
1s
Electron
configuration
2s
2p
1
H
c 
1s1
2
He
cT
1s2
3
Li
cT
c 
1s22s1
4
Be
cT
cT
1s22s2
5
B
cT
cT
c 


1s22s22p1
6
C
cT
cT
c c 

1s22s22p2
7
N
cT
cT
c c c 
1s22s22p3
8
O
cT
cT
cTc c 
1s22s22p4
9
F
cT
cT
cTcTc 
1s22s22p5
Ne
cT
cT
cTcTcT
1s22s22p6
10
*
Boxes represent the orbitals grouped by sublevel. Electrons are shown by arrows.
10.5
Table 10.2
• Electron Structures and the Periodic Table 201
Orbital Diagrams and Electron Configurations for Elements 11–18
Atomic number
Element
Orbitals
1s
2s
2p
Electron configuration
3s
3p
11
Na
 c T   c T   c T  c T  c T   c  1s22s22p63s1
12
Mg
 c T   c T   c T  c T  c T   c T  1s22s22p63s2
13
Al
cT cT cTcTcT cT c 


1s22s22p63s23p1
14
Si
cT cT cTcTcT cT c c 

1s22s22p63s23p2
15
P
cT cT cTcTcT cT c c c 
1s22s22p63s23p3
16
S
cT cT cTcTcT cT cTc c 
1s22s22p63s23p4
17
Cl
cT cT cTcTcT cT cTcTc 
1s22s22p63s23p5
18
Ar
cT cT cTcTcT cT cTcTcT
1s22s22p63s23p6
Magnesium (12), aluminum (13), silicon (14), phosphorus (15), sulfur (16), chlorine (17), and
argon (18) follow in order. Table 10.2 summarizes the filling of the orbitals for elements 11–18.
The electrons in the outermost (highest) energy level of an atom are called the
valence electrons . For example, oxygen, which has the electron configuration of
1s22s22p4, has electrons in the first and second energy levels. Therefore, the second
principal energy level is the valence level for oxygen. The 2s and 2p electrons are the
valence electrons. Valence electrons are involved in bonding atoms together to form compounds and are of particular interest to chemists, as we will see in Chapter 11.
The number of the column in
1A–7A of the periodic table gives
the number of valence electrons
of the element in those columns.
Example 10.4
What is the electron configuration for the valence electrons in magnesium?
ENHANCED EXAMPLE
Solution
Magnesium has the electron configuration of 1s22s22p63s2. The electrons are in the
first, second, and third energy levels. This means the valence electrons are in the third
energy level. The 3s electrons are the valence electrons for magnesium. The electron
configuration for the valence electrons in 3s2.
Practice 10.4
Give the electron configuration for the valence electrons in these elements.
(a) B (b) N (c) Na (d) Cl
10.5 Electron Structures
and the Periodic Table
Describe how the electron configurations of the atoms relate to their position on the
periodic table and write electron configurations for elements based on their position
on the periodic table.
We have seen how the electrons are assigned for the atoms of elements 1–18. How do the
electron structures of these atoms relate to their position on the periodic table? To answer this
question, we need to look at the periodic table more closely.
Learning objective
Key Terms
period
groups or families (of elements)
representative elements
transition elements
202 chapter 10
• Modern Atomic Theory and the Periodic Table
>Chemistry in action
Collecting the Elements
now has an assistant and utilizes Mathematica, a computer
program from Wolfram Research, to help him. He has also
created a beautiful periodic table poster from the element
images.
Theodore Gray loves collecting samples of elements.
He had an idea several years ago to design a conference
table for his company, Wolfram Research in Champaign,
Illinois. He built the table in the shape of the periodic
table and made little boxes below each element to
house a sample of the element. As he was collecting the
element samples, he began to realize that he needed to
keep track of where each sample had come from before
he forgot. So he began to keep notes on each element
sample, where he obtained it, what it was, and why it
was there. Gray says, “Once I had collected three or four
dozen such descriptions, I thought, “‘Well I had better
put this on a website, because that’s what you do when
you have anything these days—you put it on your own
website.” His element collection is ever expanding and
contains more than 1400 samples.
© Mike Walker Photography
Take a couple of minutes and spend some time with Gray
on his websites enjoying the photos and stories of the
elements. He has some truly amazing examples of the
elements.
Theodore Gray’s websites, http://www.theodoregray.
com/PeriodicTable and periodictable.com, both include
elemental data and many stories that are both entertaining and very distracting. Gray also has created video for
each element from all angles in stunning high-resolution
photography, formed into video loops. He used to take all
the photos himself, but the project has grown so big he
Theodore Gray and his periodic table conference table
containing samples of each element.
The periodic table represents the efforts of chemists to organize the elements logically.
Chemists of the early nineteenth century had sufficient knowledge of the properties of
elements to recognize similarities among groups of elements. In 1869, Dimitri Mendeleev
(1834–1907) of Russia and Lothar Meyer (1830–1895) of Germany independently published
periodic arrangements of the elements based on increasing atomic masses. Mendeleev’s
arrangement is the precursor to the modern periodic table, and his name is associated with it.
The modern periodic table is shown on the inside front cover of this book.
Each horizontal row in the periodic table is called a period, as shown in Figure 10.14.
There are seven periods of elements. The number of each period corresponds to the outermost
Noble
gases
8A
Period
Group number
1A
1
1
H
2A
9
F
Atomic number
Symbol
2
3
Li
4
Be
3
11
Na
12
Mg
3B
4B
5B
6B
7B
4
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
5
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
6
55
Cs
56 57–71 72
Ba La–Lu Hf
73
Ta
74
W
75
Re
7
87
Fr
88 89–103 104
Ra Ac–Lr Rf
105
Db
106
Sg
107
Bh
3A
4A
5A
6A
7A
2
He
5
B
6
C
7
N
8
O
9
F
10
Ne
1B
2B
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
113
Uut
114
Fl
115
Uup
116
Lv
117
Uus
118
Uuo
8B
Figure 10.14
The periodic table of the elements.
10.5
• Electron Structures and the Periodic Table 203
1A
Noble
gases
1
H
1s1
2
He
1s 2
3
Li
2s1
2A
4
Be
2s2
3A
5
B
2s22p 1
4A
6
C
2s22p 2
5A
7
N
2s22p 3
6A
8
O
2s22p 4
7A
9
F
2s22p 5
10
Ne
2s22p 6
11
Na
3s1
12
Mg
3s2
13
Al
3s23p 1
14
Si
3s23p 2
15
P
3s23p 3
16
S
3s23p 4
17
Cl
3s23p 5
18
Ar
3s23p 6
energy level that contains electrons for elements in that period. Those in Period 1 contain
electrons only in energy level 1, while those in Period 2 contain electrons in levels 1 and 2. In
Period 3, electrons are found in levels 1, 2, 3, and so on.
Elements that behave in a similar manner are found in groups or families. These form
the vertical columns on the periodic table. Two systems exist for numbering the groups. In
one system, the columns are numbered from left to right using the numbers 1–18. However,
we use a system that numbers the columns with numbers and the letters A and B, as shown in
Figure 10.14. The A groups are known as the representative elements. The B groups are
called the transition elements. In this book we will focus on the representative elements.
The groups (columns) of the periodic table often have family names. For example, the group
on the far right side of the periodic table (He, Ne, Ar, Kr, Xe, and Rn) is called the noble
gases. Group 1A is called the alkali metals, Group 2A the alkaline earth metals, and Group
7A the halogens.
How is the structure of the periodic table related to the atomic structures of the elements? We’ve just seen that the periods of the periodic table are associated with the energy
level of the outermost electrons of the atoms in that period. Look at the valence electron
configurations of the elements we have just examined (Figure 10.15). Do you see a pattern?
The valence electron configuration for the elements in each column is the same. The chemical behavior and properties of elements in a particular family must therefore be associated
with the electron configuration of the elements. The number for the principal energy level is
different. This is expected since each new period is associated with a different energy level
for the valence electrons.
The electron configurations for elements beyond these first 18 become long and tedious to
write. We often abbreviate the electron configuration using the following notation:
Na
[Ne]3s1
Look carefully at Figure 10.15 and you will see that the p orbitals are full at the noble gases.
By placing the symbol for the noble gas in square brackets, we can abbreviate the complete
electron configuration and focus our attention on the valence electrons (the electrons we will
be interested in when we discuss bonding in Chapter 11). To write the abbreviated electron
configuration for any element, go back to the previous noble gas and place its symbol in square
brackets. Then list the valence electrons. Here are some examples:
B
1s22s22p1 [He] 2s22p1
Cl
1s22s22p63s23p5 [Ne] 3s23p5
Na
1s22s22p63s1
[Ne] 3s1
The sequence for filling the orbitals is exactly as we would expect up through the
3p orbitals. The third energy level might be expected to fill with 3d electrons before electrons enter the 4s orbital, but this is not the case. The behavior and properties of the next two
elements, potassium (19) and calcium (20), are very similar to the elements in Groups 1A
and 2A, respectively. They clearly belong in these groups. The other elements in Group 1A
and Group 2A have electron configurations that indicate valence electrons in the s orbitals. For example, since the electron configuration is connected to the element’s properties,
Figure 10.15
Valence electron configurations
for the first 18 elements.
Both numbering systems are
shown in the periodic table on
the inside front cover.
chapter 10
• Modern Atomic Theory and the Periodic Table
we should place the last electrons for potassium and calcium in the 4s orbital. Their electron
configurations are
K
1s22s22p63s23p64s1 or
[Ar] 4s1
Ca
1s22s22p63s23p64s2 or
[Ar] 4s2
Practice 10.5
Write the abbreviated electron configuration for the following elements:
(a) Br (b) Sr (c) Ba (d) Te
Elements 21–30 belong to the elements known as transition elements. Electrons are
placed in the 3d orbitals for each of these elements. When the 3d orbitals are full, the
electrons fill the 4p orbitals to complete the fourth period. Let’s consider the overall
relationship between orbital filling and the periodic table. Figure 10.16 illustrates the
type of orbital filling and its location on the periodic table. The tall columns on the table
(labeled 1A–7A, and noble gases) are often called the representative elements. Valence
electrons in these elements occupy s and p orbitals. The period number corresponds to
the energy level of the valence electrons. The elements in the center of the periodic table
(shown in ) are the transition elements where the d orbitals are being filled. Notice that
the number for the d orbitals is one less than the period number. The two rows shown
at the bottom of the table in Figure 10.16 are called the inner transition elements or the
lanthanide and actinide series. The last electrons in these elements are placed in the
f orbitals. The number for the f orbitals is always two less than that of the s and p orbitals. A periodic table is almost always available to you, so if you understand the relationship between the orbitals and the periodic table, you can write the electron configuration
for any element. There are several minor variations to these rules, but we won’t concern
ourselves with them in this course.
Noble
gases
8A
1A
Period
204 1
1s
2
2s
3
3s
3B
4
4s
3d
4p
5
5s
4d
5p
6
6s
5d
6p
7
7s
6d
7p
2A
3A
5B
6B
7B
8B
5A
2p
Transition elements
4B
4A
1B
2B
3p
Inner transition elements
s block
4f
p block
5f
d block
f block
Figure 10.16
Arrangement of elements according to the electron sublevel being filled.
6A
7A
1s
10.5
• Electron Structures and the Periodic Table 205
Example 10.5
ENHANCED EXAMPLE
Use the periodic table to write the electron configuration for phosphorus and tin.
Solution
Phosphorus is element 15 and is located in Period 3, Group 5A. The electron configuration must have a full first and second energy level:
1s22s22p63s23p3 or
P
[Ne]3s23p3
Sn
P
You can determine the electron configuration by looking across the period and counting
the element blocks.
Tin is element 50 in Period 5, Group 4A, two places after the transition metals. It
must have two electrons in the 5p series. Its electron configuration is
Sn 1s22s22p63s23p64s23d104p65s24d105p2
or [Kr]5s24d105p2
Notice that the d series of electrons always has a principal energy level number one less
than its period.
Practice 10.6
Use the periodic table to write the complete electron configuration for (a) O, (b) Ca,
and (c) Ti.
The early chemists classified the elements based only on their observed properties, but
modern atomic theory gives us an explanation for why the properties of elements vary periodically. For example, as we “build” atoms by filling orbitals with electrons, the same orbitals occur on each energy level. This means that the same electron configuration reappears
regularly for each level. Groups of elements show similar chemical properties because of the
similarity of these outermost electron configurations.
In Figure 10.17, only the electron configuration of the outermost electrons is given. This
periodic table illustrates these important points:
1. The number of the period corresponds with the highest energy level occupied by
electrons in that period.
Noble
gases
8A
Group number
1A
1
1
2
1s
2A
3A
4A
5A
6A
7A
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
2s1
2s2
2s22p1
2s22p2
2s22p3
2s22p4
2s22p5
2s22p6
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
3s1
3s2
3B
4B
5B
6B
7B
3s23p6
19
20
21
22
23
24
25
K
Ca
Sc
1
2
1
2
Period
3
4
5
6
7
He
H
8B
2
Ti
1
2
V
2
2
Cr
3
1
26
Mn
5
2
27
Fe
5
28
Co
2
6
2
Ni
7
2
1B
2B
3s23p1
3s23p2
3s23p3
3s23p4
3s23p5
29
30
31
32
33
34
35
Cu
8
1s2
1
10
Zn
2
Ga
10
2
Ge
1
2
As
2
2
Se
3
2
36
Br
4
2
Kr
5
4s24p6
4s
4s
4s 3d
4s 3d
4s 3d
4s 3d
4s 3d
4s 3d
4s 3d
4s 3d
4s 3d
4s 3d
4s 4p
4s 4p
4s 4p
4s 4p
4s 4p
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
5s1
5s2
5s24d1
5s24d2
5s14d4
5s14d5
5s14d6
5s14d7
5s14d8
5s04d10
5s14d10
5s24d10
5s25p1
5s25p2
5s25p3
5s25p4
5s25p5
5s25p6
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
6s1
6s2
6s25d1
6s25d2
6s25d3
6s25d4
6s25d5
6s25d6
6s25d7
6s15d9
6s15d10
6s25d10
6s26p1
6s26p2
6s26p3
6s26p4
6s26p5
6s26p6
87
88
89
104
105
106
107
108
109
110
111
112
113
114
115
116
117
Fr
Ra
Ac
1
2
7s
7s
2
7s 6d
Rf
1
2
Db
2
7s 6d
2
Sg
3
7s 6d
Figure 10.17
Outermost electron configurations.
2
Bh
4
7s 6d
2
Hs
5
7s 6d
2
Mt
7s 6d
6
2
Ds
7
7s 6d
1
7s 6d
Rg
9
1
Cn
7s 6d
10
2
10
7s 5d
Uut
2
1
7s 7p
Fl
2
Uup
2
7s 7p
2
3
7s 7p
Lv
2
118
Uus
7s 7p
4
2
5
7s 7p
Uuo
7s27p6
206 chapter 10
• Modern Atomic Theory and the Periodic Table
2. The group numbers for the representative elements are equal to the total number of outermost
electrons in the atoms of the group. For example, elements in Group 7A always have the
electron configuration ns2np5. The d and f electrons are always in a lower energy level than the
highest principal energy level and so are not considered as outermost (valence) electrons.
3. The elements of a family have the same outermost electron configuration except that the
electrons are in different principal energy levels.
4. The elements within each of the s, p, d, f blocks are filling the s, p, d, f orbitals, as shown
in Figure 10.16.
5. Within the transition elements, some discrepancies in the order of filling occur. (Explanation of these discrepancies and similar ones in the inner transition elements is beyond the
scope of this book.)
Example 10.6
Write the complete electron configuration of a zinc atom and a rubidium atom.
Rb
Zn
Solution
The atomic number of zinc is 30; it therefore has 30 protons and 30 electrons in a neutral atom. Using Figure 10.14, we see that the electron configuration of a zinc atom is
1s22s22p63s23p64s23d10. Check by adding the superscripts, which should equal 30.
The atomic number of rubidium is 37; therefore, it has 37 protons and 37 electrons
in a neutral atom. With a little practice using a periodic table, you can write the
electron configuration directly. The electron configuration of a rubidium atom is
1s22s22p63s23p64s23d104p65s1. Check by adding the superscripts, which should equal 37.
Practice 10.7
Write the complete electron configuration for a gallium atom and a lead atom.
C h apt e r
10 review
10.1 Electromagnetic Radiation
Key Terms
wavelength
frequency
speed
photons
• Atomic theory has changed over the past 200 years:
• Early Greek ideas
• Dalton’s model
• Thomson’s model
• Rutherford’s model
• Modern atomic theory
• Basic wave characteristics include
• Wavelength (l)
• Frequency (n)
• Speed (y)
• Light is a form of electromagnetic radiation.
• Light can also be considered to be composed of energy packets called photons, which behave like
particles.
10.2 The Bohr Atom
Key Terms
line spectrum
quanta
ground state
orbital
• The study of atomic spectra led Bohr to propose that
• Electrons are found in quantized energy levels in an atom.
• Spectral lines result from the radiation of quanta of energy when the electron moves from a
higher level to a lower level.
• The chemical properties of an element and its position on the periodic table depend on electrons.
• Louis de Broglie suggested that all objects have wave properties.
• Schrödinger created a mathematical model to describe electrons as waves.
• Electrons are located in orbitals, or regions of probability, around the nucleus of an atom.
Review Questions 207
10.3 Energy Levels of Electrons
• Modern atomic theory predicts that
• Electrons are found in discrete principal energy levels (n 5 1, 2, 3 . . . ):
• Energy levels contain sublevels.
Number of sublevels
4f
4d
3d
4p
3p
2p
Key Terms
principal energy levels
sublevels
spin
Pauli exclusion principle
4s
3s
2s
1s
• Two electrons fit into each orbital but must have opposite spin to do so.
10.4 Atomic Structures of the First 18 Elements
• Guidelines for writing electron configurations:
• Not more than two electrons per orbital
• Electrons fill lowest energy levels first:
• s 6 p 6 d 6 f for a given value of n
• Orbitals on a given sublevel are each filled with a single electron before pairing of electrons
begins to occur
• For the representative elements, only electrons in the outermost energy level (valence electrons) are
involved in bonding.
Key Terms
electron configuration
orbital diagram
valence electrons
10.5 Electron Structures and the Periodic Table
• E
lements in horizontal rows on the periodic table contain elements whose valence electrons
(s and p) are generally on the same energy level as the number of the row.
• Elements that are chemically similar are arranged in columns (groups) on the periodic table.
• The valence electron configurations of elements in a group or family are the same, but they are
located in different principal energy levels.
Key Terms
period
groups or families (of elements)
representative elements
transition elements
review Questions
1. Define the terms wavelength and frequency. What symbols do
chemists generally use to represent these quantities?
2. What is the wavelength range for visible light? Which has a longer
wavelength, red light or blue light?
3. What is the name given to a packet of energy?
4. What is an orbital?
5. What is meant when we say the electron structure of an atom is in
its ground state?
6. What is the major difference between an orbital and a Bohr orbit?
7. Explain how and why Bohr’s model of the atom was modified to
include the cloud model of the atom.
8. How do 1s and 2s orbitals differ? How are they alike?
9. What letters are used to designate the types of orbitals?
10. List the following orbitals in order of increasing energy: 2s, 2p, 4s,
1s, 3d, 3p, 4p, 3s.
11. How many s electrons, p electrons, and d electrons are possible in
any energy level?
12. Sketch the s, px, py, and pz orbitals.
13. Under what conditions can a second electron enter an orbital
already containing one electron?
14. What is a valence shell?
15. What are valence electrons, and why are they important?
16. A lanthanide element has the designation 4f 3 in its electron structure. What is the significance of the 4, the f, and the 3?
17. Of the elements Ir, Pb, Xe, Zr, and Ag, which ones are not contained in the groups of representative elements?
18. From the standpoint of electron structure, what do the elements in
the p block have in common?
19. Write symbols for the elements with atomic numbers 6, 7, 8, 15, and 33.
Which of those elements have something in common? Explain.
20. Write the symbols of the first three elements that have six electrons
in their outermost energy level.
21. What is the greatest number of elements to be found in any period?
Which periods have this number?
22. From the standpoint of energy level, how does the placement of
the last electron in the Group A elements differ from that of the
Group B elements?
23. Find the places in the periodic table where elements are not in
proper sequence according to atomic mass. (See inside of front
cover for periodic table.)
24. What are the names of the two scientists who independently published results that have led to the establishment of the periodic table?
25. Which scientist is credited with being the author of the modern
periodic table?
208 chapter 10
• Modern Atomic Theory and the Periodic Table
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
P AIRE D E x e r c i s e s
1. How many total and how many valence electrons are in each of the
following neutral atoms?
(a) lithium
(c) calcium
(b) magnesium
(d) fluorine
2. How many total and how many valence electrons are in each of the
following neutral atoms?
(a) sodium
(c) phosphorus
(b) arsenic
(d) aluminum
3. Write the complete electron configuration for each of the following
elements:
(a) scandium
(c) bromine
(b) rubidium
(d) sulfur
4. Write the complete electron configuration for each of the following
elements:
(a) manganese
(c) gallium
(b) krypton
(d) boron
5. Explain how the spectral lines of hydrogen occur.
6. Explain how Bohr used the data from the hydrogen spectrum to
support his model of the atom.
7. How many orbitals exist in the fourth principal energy level?
What are they, and in what periods can they be found?
8. How many total electrons can be present in the third principal
energy level? In what orbitals are they found?
10. Write orbital diagrams for these elements:
(a) Si (b) S (c) Ar (d) V (e) P
9. Write orbital diagrams for these elements:
(a) N (b) Cl (c) Zn (d) Zr (e) I
11. For each of the orbital diagrams given, write out the corresponding electron configurations.
(a) O
cT cT cTc c 
(b) Ca
cT cT cTcTcT cT
cTcTcT
(c) Ar
cT cT cTcTcT cT
cTcTcT
(d) Br
cT cT cTcTcT cT
cTcTcT
cT cTcTcTcTcT
(e) Fe
cT cT cTcTcT cT
cTcTcT
cT cTc c c c 
cT
cTcTc 
12. For each of the orbital diagrams given, write out the corresponding electron configurations.
(a) Li
cT c 
(b) P
cT cT cTcTcT cT
c c c 
(c) Zn
cT cT cTcTcT cT
cTcTcT
cT cTcTcTcTcT
(d) Na
cT cT cTcTcT c 
(e) K
cT cT cTcTcT cT
cTcTcT
c 
13. Identify the correct orbital diagrams below. For the incorrect diagrams redraw the correct orbital diagram.
1s
2s
2p
3s
3p
4s
3d
4p
(a)  c T 
cT cTcTc 
c 
(b)  c T 
cT cTcTcT
cT c c c 
(c)  c T 
cT cTcTcT
cT cTcTcT
cT
cTcTcTcTcT c 
(d)  c T 
cT cTcTcT
cT cTcTcT
cT
cTcTc 




14. Identify the correct orbital diagrams below. For the incorrect diagrams redraw the correct orbital diagram.
1s
2s
2p
4s
4p
cT cTcTcT
cT
cTcTcT
cT cTcTcT
cT cTcTcT
cc
cT cTcTcT
cT cTc c 
c
c
3s
(a)  c T 
cT
c
(b)  c T 
cT cTcTcT
(c)  c T 
(d)  c T 
3p
3d

c c 



Paired Exercises 209
15. Identify the element represented by each of the orbital diagrams in
Problem 13.
16. Identify the element represented by each of the orbital diagrams in
Problem 14.
17. Which elements have these electron configurations? Give both the
symbol and the name:
(a) 1s22s22p5
(c) [Ne]3s23p4
2 2 6 1
(b) 1s 2s 2p 3s
(d) [Ar]4s23d8
18. Which of the elements have these electron configurations? Give
both the symbol and the name:
(a) 1s22s22p1
(c) [Xe]6s24f 145d106p2
2 2 6 2 2
(b) 1s 2s 2p 3s 3p
(d) [Kr]5s24d105p4
19. Identify the element and draw orbital diagrams for elements with
these atomic numbers:
(a) 22 (b) 18 (c) 33 (d) 35 (e) 25
20. Identify the element and draw orbital diagrams for elements with
these atomic numbers:
(a) 15 (b) 30 (c) 20 (d) 34 (e) 19
21. For each of the electron configurations given, write the corresponding orbital diagrams.
(a) F
1s22s22p5
(b) S
1s22s22p63s23p4
(c) Co
1s22s22p63s23p64s23d7
(d) Kr
1s22s22p63s23p64s23d104p6
(e) Ru
1s22s22p63s23p64s23d104p65s24d 6
22. For each of the electron configuration given, write the corresponding orbital diagrams.
(a) Cl
1s22s22p63s23p5
(b) Mg
1s22s22p63s2
(c) Ni
1s22s22p63s23p64s23d 8
(d) Cu
1s22s22p63s23p64s23d 9
(e) Ba
1s22s22p63s23p64s23d104p65s24d105p66s2
23. Identify these elements from their atomic structure diagrams:
24. Diagram the atomic structures (as you see in Exercise 23) for these
elements:
(a) 27
13Al
16p
(a) 16n 2e-
8e-
6e-
(b) 28p 2e32n
8e-
16e-
2e-
(b)
48
22Ti
25. Why is the eleventh electron of the sodium atom located in the third
energy level rather than in the second energy level?
26. Why is the last electron in potassium located in the fourth energy
level rather than in the third energy level?
27. What electron structure do the noble gases have in common?
28. What is the unique about the noble gases from an electron point of view?
29. How are elements in a period related to one another?
30. How are elements in a group related to one another?
31. How many valence electrons do each of the following elements
have?
(a) C (b) S
(c) K
(d) I
(e) B
32. How many valence electrons do each of the following elements
have?
(a) N
(b) P (c) O
(d) Ba (e) Al
33. What do the electron structures of the alkali metals have in
­common?
34. Why would you expect elements zinc, cadmium, and mercury to be
in the same chemical family?
35. Pick the electron structures that represent elements in the same
chemical family:
(a) 1s22s1
(e) 1s22s22p63s23p6
(b) 1s22s22p4
(f) 1s22s22p63s23p64s2
2 2 2
(c) 1s 2s 2p
(g) 1s22s22p63s23p64s1
2 2 6 2 4
(d) 1s 2s 2p 3s 3p
(h) 1s22s22p63s23p64s23d1
36. Pick the electron structures that represent elements in the same
chemical family:
(a) [He]2s22p6
(e) [Ar]4s23d10
(b) [Ne]3s1
(f) [Ar]4s23d104p6
2
(c) [Ne]3s
(g) [Ar]4s23d 5
2 3
(d) [Ne]3s 3p
(h) [Kr]5s24d10
37. In the periodic table, calcium, element 20, is surrounded by elements 12, 19, 21, and 38. Which of these have physical and chemical properties most resembling calcium?
38. In the periodic table, phosphorus, element 15, is surrounded by elements 14, 7, 16, and 33. Which of these have physical and chemical
properties most resembling phosphorus?
39. Classify the following elements as metals, nonmetals, or metalloids
(review Chapter 3 if you need help):
(a) potassium
(c) sulfur
(b) plutonium
(d) antimony
40. Classify the following elements as metals, nonmetals, or metalloids
(review Chapter 3 if you need help):
(a) iodine
(c) molybdenum
(b) tungsten
(d) germanium
41. In which period and group does an electron first appear in an f orbital?
42. In which period and group does an electron first appear in a d ­orbital?
43. How many electrons occur in the valence level of Group 7A and
7B elements? Why are they different?
44. How many electrons occur in the valence level of Group 3A and
3B elements? Why are they different?
45. Determine the identity of the element that contains exactly
(a) three 4p electrons in the ground state
(b) seven 3d electrons in the ground state
(c) one 2s electrons in the ground state
(d) five 3p electrons in the ground state
46. Determine the identity of the element that contains exactly
(a) two 6p electrons in the ground state
(b) five 4f electrons in the ground state
(c) one 4p electrons in the ground state
(d) seven 5d electrons in the ground state
210 chapter 10
• Modern Atomic Theory and the Periodic Table
A D D ITIONAL EXERCISES
47. Using only the periodic table, explain how the valence energy level
and the number of valence electrons could be determined.
48. Using only a periodic table, write the full electron configuration for
each of the following elements. Circle the valence electrons.
(a) Mg (b) P (c) K (d) F (e) Se (f) N
49. Which of the following would have the same number of valence
electrons?
(a) Na+ (b) O (c) Li (d) F- (e) Ne
50. Name the group in which each of the following elements appear:
(a) 1s22s22p5
(b) 1s22s22p63s23p64s2
(c) 1s22s22p63s1
(d) 1s22s22p63s23p64s23d104p6
(e) 1s2
(f) 1s22s22p63s23p64s23d104p65s1
O. Louis Mazzatenta/NG Image Collection
51. Chromium is a lustrous silver-colored metal that has been used to
prevent corrosion for centuries. Bronze swords and other weapons
discovered in burial pits from the Qin Dynasty were coated with
chromium and had not corroded at all since their entombing. Today
many items are coated with a layer of chromium as a decorative and
protective covering.
(a) Sometimes the actual electron configurations of the elements
differ from those predicted by the periodic table. The experimentally determined electron configuration for chromium
is 1s22s22p63s23p64s13d5. Is this the electronic configuration you would predict based on the periodic table? If not,
what is the configuration predicted by the periodic table for
chromium?
(b) Chromium has a density of 7.19 g/cm3. How many atoms of
chromium are contained in a 5.00-cm3 sample of chromium?
(c) If the radius of a chromium atom is 1.40 * 10 - 8 cm, what is
4
the volume ¢V = pr 3 ≤ of a single chromium atom?
3
(d) How many chromium atoms occupy a volume of 5.00 cm3?
52. Anyone watching fireworks enjoys the beautiful colors produced.
The colors are produced by adding different elements to the mix of
fuels in the fireworks. Red is produced by adding calcium, yellow
by adding sodium, and blue by adding copper. Explain why adding
different elements might give fireworks their characteristic colors.
53. There has long been speculation that Wolfgang Amadeus Mozart may have been poisoned by a rival. There is some evidence
that he may have been poisoned by an antimony compound. This
compound would have been prescribed by his doctor, however,
­suggesting that there was no foul play. What is the electronic configuration of antimony.
54. A dog is suspected of attacking a runner in the park. This runner was wearing a pearlescent lip gloss containing bismuth
hypochlorite, a compound often used in cosmetics to give this
pearlescent effect. Upon analyzing a residue on the dog’s fur,
this compound is detected giving additional support to the victim’s claim. Write the complete and shorthand electronic configurations for bismuth.
55. Write the electron configuration of each of the following neutral
atoms. (You may need to refer to Figure 3.2.)
(a) the four most abundant elements in the Earth’s crust, seawater,
and air
(b) the five most abundant elements in the human body
56. Give the maximum number of electrons that can reside in the
­following:
(a) a p orbital
(b) a d sublevel
(c) the third principal energy level
(d) an s orbital
(e) an f sublevel
57. Give the names of each of the following elements based on the
information given:
(a) the second element in Period 3
(b) [Ne]3s23p3
(c)  c T 
1s
cT
cTcTcT
cT
cTcTcT
2s
2p
3s
3p
58. Why does the emission spectrum for nitrogen reveal many more
spectral lines than that for hydrogen?
59. List the first element on the periodic table that satisfies each of
these conditions:
(a) a completed set of p orbitals
(b) two 4p electrons
(c) seven valence electrons
(d) three unpaired electrons
60. In Group 7A, the first two elements are gases, the next one is a
liquid, and the fourth one is a solid. What do they have in common
that causes them to be in the same group?
61. In which groups are the transition elements located?
62. How do the electron structures of the transition elements differ
from those of the representative elements?
63. The atomic numbers of the noble gases are 2, 10, 18, 36, 54 and 86.
Without looking at a periodic table, determine the atomic numbers
for the elements that have five electrons in their valence shell.
64. What is the family name for
(a) Group 1A?
(c) Group 7A?
(b) Group 2A?
65. What sublevel is being filled in
(a) Period 3, Group 3A to 7A?
(b) Period 5, transition elements?
(c) the lanthanide series?
Answers to Practice Exercises 211
66. Classify each of the following as a noble gas, a representative element, or a transition metal. Also indicate whether the element is a
metal, nonmetal, or metalloid.
(a) Na
(d) Ra
(b) N
(e) As
(c) Mo
(f) Ne
67. If element 36 is a noble gas, in which groups would you expect
elements 35 and 37 to occur?
69. What is the relationship between two elements if
(a) one of them has 10 electrons, 10 protons, and 10 neutrons and
the other has 10 electrons, 10 protons, and 12 neutrons?
(b) one of them has 23 electrons, 23 protons, and 27 neutrons and
the other has 24 electrons, 24 protons, and 26 neutrons?
70. Is there any pattern for the location of gases on the periodic table?
for the location of liquids? for the location of solids?
68. Write a paragraph describing the general features of the periodic table.
CHALLEN G E E x e r c i s e s
71. A valence electron in an atom of sulfur is excited by heating a
sample. The electron jumps from the s orbital to the p orbital. What
is the electron configuration of the excited sulfur atom, and what
would the orbital diagram look like?
73. Use the periodic table to explain why metals tend to lose electrons
and nonmetals tend to gain electrons.
74. Show how the periodic table helps determine the expected electron
configuration for any element.
72. Element 87 is in Group 1A, Period 7. In how many principal energy
levels are electrons located? Describe its outermost energy level.
Answers to Practice Exercises
10.1
2λ
10.4 (a) 2s22p1 (c) 3s1
(b) 2s22p3 (d) 3s23p5
10.5 (a) [Ar]4s24p5
(b) [Kr]5s2
10.6 (a) O
(b) Ca
(c) Ti
10.2 A
photon is a little packet of electromagnetic radiation energy
and has the property of both a particle and a wave. It travels at
the speed of light, 3.00 × 108 meters per second.
10.3 ( a) Two electrons. Wolfgang Pauli’s exclusion principle.
(b) 1s22s22p63s23p6 for a total of 18 electrons.
(c) [Xe]6s2
(d) [Kr]5s25p4
1s22s22p4
1s22s22p63s23p64s2
1s22s22p63s23p64s23d2
10.7 Ga, 1s22s22p63s23p64s23d104p1
Pb, 1s22s22p63s23p64s23d104p65s24d105p66s24f 145d106p2
C ha p t e r
Pasieka/Photo Researchers, Inc.
The atoms in tartaric acid bond
together in a very specific
orientation to form the shape
of the molecule. The molecules
collect together to form a crystal,
photographed here in a polarized
light micrograph. Tartaric acid is
used in baking powder and as a
food additive.
11
Chemical Bonds:
The Formation of
­Compounds from Atoms
F
or centuries we’ve been aware that certain metals cling to a
magnet. We’ve seen balloons sticking to walls. Why? Highspeed levitation trains are heralded to be the wave of the
future. How do they function? In each case, forces of attraction
and repulsion are at work.
Human interactions also suggest that “opposites attract” and
“likes repel.” Attractions draw us into friendships and significant relationships, whereas repulsive forces may produce debate
and antagonism. We form and break apart interpersonal bonds
throughout our lives.
In chemistry, we also see this phenomenon. Substances form
chemical bonds as a result of electrical ­attractions. These bonds
provide the tremendous diversity of compounds found in nature.
This chapter is one of the most significant and useful chapters in
the book—chemical bonding between atoms. This is what chemistry is really all about. Study it carefully.
Chapter Outline
11.1 Periodic Trends in Atomic ­Properties
11.2 Lewis Structures of Atoms
11.3 The Ionic Bond: Transfer of ­Electrons
from One Atom to Another
11.4 Predicting Formulas of Ionic
­Compounds
11.5 The Covalent Bond: Sharing
­Electrons
11.6 Electronegativity
11.7 Lewis Structures of Compounds
11.8 Complex Lewis Structures
11.9 Compounds Containing
­Polyatomic Ions
11.10 Molecular Shape
11.1
• Periodic Trends in Atomic ­Properties 213
11.1 Periodic Trends in Atomic
­Properties
Discuss the atomic trends for metals and nonmetals, atomic radius, and ionization
energy.
Learning objectiv e
Although atomic theory and electron configuration help us understand the arrangement and
behavior of the elements, it’s important to remember that the design of the periodic table is
based on observing properties of the elements. Before we use the concept of atomic structure to explain how and why atoms combine to form compounds, we need to understand the
characteristic properties of the elements and the trends that occur in these properties on the
periodic table. These trends allow us to use the periodic table to accurately predict properties
and reactions of a wide variety of substances.
Key term
ionization energy
Metals and Nonmetals
In Section 3.5, we classified elements as metals, nonmetals, or metalloids. The heavy stair-step
line beginning at boron and running diagonally down the periodic table separates the elements
into metals and nonmetals. Metals are usually lustrous, malleable, and good conductors of
heat and electricity. Nonmetals are just the opposite—nonlustrous, brittle, and poor conductors. Metalloids are found bordering the heavy diagonal line and may have properties of both
metals and nonmetals.
Most elements are classified as metals (see Figure 11.1). Metals are found on the left side
of the stair-step line, while the nonmetals are located toward the upper right of the table. Note
that hydrogen does not fit into the division of metals and nonmetals. It displays nonmetallic
properties under normal conditions, even though it has only one outermost electron like the
alkali metals. Hydrogen is considered to be a unique element.
It is the chemical properties of metals and nonmetals that interest us most. Metals tend to
lose electrons and form positive ions, while nonmetals tend to gain electrons and form negative ions. When a metal reacts with a nonmetal, electrons are often transferred from the metal
to the nonmetal.
1
H
2
He
Metals
3
Li
4
Be
5
B
6
C
7
N
8
O
9
F
10
Ne
11
Na
12
Mg
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
55
Cs
56
Ba
57
La*
72
Hf
73
Ta
74
W
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
87
Fr
88
Ra
89
Ac†
104
Rf
105
Db
106
Sg
107
Bh
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
113
Uut
114
Fl
115
Uup
116
Lv
117
Uus
118
Uuo
*
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
65
Tb
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
†
90
Th
91
Pa
92
U
93
Np
94
Pu
95
Am
96
Cm
97
Bk
98
Cf
99
Es
100
Fm
101
Md
102
No
103
Lr
Metalloids
Nonmetals
Figure 11.1
The elements are classified as metals, nonmetals, and metalloids.
214 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
1A
2A
3A
4A
5A
6A
7A
H
Li
Noble
gases
He
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Ga
Ge
As
Se
Br
Kr
Rb
Sr
In
Sn
Sb
Te
I
Xe
Cs
Ba
Tl
Pb
Bi
Po
At
Rn
Figure 11.2
Relative atomic radii for the
representative elements. Atomic
radius decreases across a period
and increases down a group in the
periodic table.
Atomic Radius
The relative radii of the representative elements are shown in Figure 11.2. Notice that the
radii of the atoms tend to increase down each group and that they tend to decrease from left
to right across a period.
The increase in radius down a group can be understood if we consider the electron
­structure of the atoms. For each step down a group, an additional energy level is added to the
atom. The average distance from the nucleus to the outside edge of the atom must increase
as each new energy level is added. The atoms get bigger as electrons are placed in these new
higher-energy levels.
Understanding the decrease in atomic radius across a period requires more thought, however. As we move from left to right across a period, electrons within the same block are being
added to the same principal energy level. Within a given energy level, we expect the orbitals
to have about the same size. We would then expect the atoms to be about the same size across
the period. But each time an electron is added, a proton is added to the nucleus as well. The
increase in positive charge (in the nucleus) pulls the electrons closer to the nucleus, which
results in a gradual decrease in atomic radius across a period.
Ionization Energy
The ionization energy of an atom is the energy required to remove an electron from the atom.
For example,
Na + ionization energy ¡ Na+ + eThe first ionization energy is the amount of energy required to remove the first electron from
an atom, the second is the amount required to remove the second electron from that atom, and
so on.
Table 11.1 gives the ionization energies for the removal of one to five electrons from
several elements. The table shows that even higher amounts of energy are needed to remove the
second, third, fourth, and fifth electrons. This makes sense because removing electrons leaves
fewer electrons attracted to the same positive charge in the nucleus. The data in Table 11.1
also show that an extra-large ionization energy (blue) is needed when an electron is ­removed
11.1
Table 11.1
• Periodic Trends in Atomic ­Properties 215
Ionization Energies for Selected Elements*
Required amounts of energy (kJ>mol)
Element
H
He
Li
Be
B
C
Ne
Na
1st
e
2nd
e
3rd
e
4th
e
5th
e
1,314
2,372
520
900
800
1,088
2,080
496
5,247
7,297
1,757
2,430
2,352
3,962
4,565
11,810
14,845
3,659
4,619
6,276
6,912
21,000
25,020
6,222
9,376
9,540
32,810
37,800
12,190
13,355
*Values are expressed in kilojoules per mole, showing energies required to remove 1 to 5 electrons per atom.
Blue type indicates the energy needed to remove an electron from a noble gas electron structure.
from a noble gas-like structure, clearly showing the stability of the electron structure of the
noble gases.
First ionization energies have been experimentally determined for most elements.
­Figure 11.3 plots these energies for representative elements in the first four periods. Note
these important points:
1. Ionization energy in Group A elements decreases from top to bottom in a group.
For example, in Group 1A the ionization energy changes from 520 kJ>mol for Li to
419 kJ>mol for K.
2. Ionization energy gradually increases from left to right across a period. Noble gases have
a relatively high value, confirming the nonreactive nature of these elements.
Metals don’t behave in exactly the same manner. Some metals give up electrons much
more easily than others. In the alkali metal family, cesium gives up its 6s electron much more
easily than the metal lithium gives up its 2s electron. This makes sense when we consider that
the size of the atoms increases down the group. The distance between the nucleus and the outer
electrons increases and the ionization energy decreases. The most chemically active metals are
located at the lower left of the periodic table.
2400
He
Figure 11.3
Ne
Periodic relationship of the
first ionization energy for
representative elements in the
first four periods.
Ionization energy (kJ/mol)
2000
F
Ar
1600
N
H
1200
C
O
P
S
As
Se
0
Si
Al
Ge
Ca
Na
400
B
Mg
Li
5A
6A
Kr
Br
Be
800
Cl
Ga
K
1A
2A
3A
4A
7A
Noble
gases
216 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Nonmetals have relatively large ionization energies compared to metals. Nonmetals tend
to gain electrons and form anions. Since the nonmetals are located at the right side of the periodic table, it is not surprising that ionization energies tend to increase from left to right across
a period. The most active nonmetals are found in the upper right corner of the periodic table
(excluding the noble gases).
Example 11.1
Why is there such a large increase in the ionization energy to remove the fifth electron
from a carbon atom? (See Table 11.1.)
Solution
Two main factors apply here. (1) The remaining carbon atom has six protons attracting
only two electrons and (2) these two electrons constitute the stable noble gas helium.
Both factors make it much more difficult to withdraw the fifth electron.
Practice 11.1
(a)What are the trends in the first ionization energy for the groups of elements and for
the periods of elements in the periodic table?
(b)Consider your answer to part (a) and state why this is the case for the groups of
elements.
11.2 Lewis Structures of Atoms
L earning obje ctive
Draw the Lewis structure for a given atom.
Key tErm
Metals tend to form cations (positively charged ions) and nonmetals form anions (negatively charged ions) in order to attain a stable valence electron structure. For many elements this stable valence level contains eight electrons (two s and six p), identical to
the valence electron configuration of the noble gases. Atoms undergo rearrangements of
electron structure to lower their chemical potential energy (or to become more stable).
These rearrangements are accomplished by losing, gaining, or sharing electrons with other
atoms. For example, a hydrogen atom could accept a second electron and attain an electron
structure the same as the noble gas helium. A fluorine atom could gain an electron and
attain an electron structure like neon. A sodium atom could lose one electron to attain an
electron structure like neon.
The valence electrons in the outermost energy level of an atom are responsible for the
electron activity that occurs to form chemical bonds. The Lewis structure of an atom is
a representation that shows the valence electrons for that atom. American chemist Gilbert
N. Lewis (1875–1946) proposed using the symbol for the element and dots for electrons.
The number of dots placed around the symbol equals the number of s and p electrons in the
­outermost energy level of the atom. Paired dots represent paired electrons; unpaired dots
represent unpaired ­electrons. For example, HD is the Lewis symbol for a hydrogen atom, 1s1;
D is the Lewis symbol for a boron atom, with valence electrons 2s22p1. In the case of boron,
CB
the symbol B represents the boron nucleus and the 1s2 electrons; the dots represent only the
2s22p1 electrons.
Lewis structure
Paired electrons
B
Unpaired electron
Symbol of the element
The Lewis method is used not only because of its simplicity of expression but also because much of the chemistry of the atom is directly associated with the electrons in the outermost energy level. Figure 11.4 shows Lewis structures for the elements hydrogen through
calcium.
11.3
1A
2A
3A
4A
5A
• The Ionic Bond: Transfer of ­Electrons from One Atom to Another 217
6A
Noble
gases
7A
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Figure 11.4
Lewis structures of the first
20 elements. Dots represent
electrons in the outermost s and p
energy levels only.
Example 11.2
Write the Lewis structure for a phosphorus atom.
ENHANCED EXAMPLE
Solution
First establish the electron structure for a phosphorus atom, which is 1s22s22p63s23p3.
Note that there are five electrons in the outermost energy level; they are 3s23p3. Write the
symbol for phosphorus and place the five electrons as dots around it.
A quick way to determine
the correct number of dots
­(electrons) for a Lewis structure is
to use the group number. For the
A groups on the periodic table,
the group number is the same as
the number of electrons in the
Lewis structure.
P
The 3s2 electrons are paired and are represented by the paired dots. The 3p3 electrons,
which are unpaired, are represented by the single dots.
Practice 11.2
Write the Lewis structure for the following elements:
(a) N
(b) Al
(c) Sr
(d) Br
11.3 The Ionic Bond: Transfer of
­Electrons from One Atom to Another
Discuss the formation of an ionic bond and the chemical change that results from
the bond.
Learning objective
The chemistry of many elements, especially the representative ones, is to attain an outer electron structure like that of the chemically stable noble gases. With the exception of helium,
this stable structure consists of eight electrons in the outermost energy level (see Table 11.2).
Let’s look at the electron structures of sodium and chlorine to see how each element can
attain a structure of 8 electrons in its outermost energy level. A sodium atom has 11 ­electrons:
Key term
Table 11.2
ionic bond
Arrangement of Electrons in the Noble Gases*
Electron structure
Noble gas
Helium
Neon
Argon
Krypton
Xenon
Radon
Symbol
n  1
2
3
He
Ne
Ar
Kr
Xe
Rn
1s 2
1s 2
1s 2
1s 2
1s 2
1s 2
2s 2 2p6
2s 2 2p 6
2s 2 2p 6
2s 2 2p 6
2s 2 2p 6
3s 2 3p6
3s 2 3p 6 3d 10
3s 2 3p 6 3d 10
3s 2 3p 6 3d 10
*Each gas except helium has eight electrons in its outermost energy level.
4
4s 2 4p6
4s 2 4p 6 4d 10
4s 2 4p 6 4d 10 4f 14
5
5s 2 5p6
5s 2 5p 6 5d 10
6
6s 2 6p6
218 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
2 in the first energy level, 8 in the second energy level, and 1 in the third energy level. A chlorine atom has 17 electrons: 2 in the first energy level, 8 in the second energy level, and 7 in
the third energy level. If a sodium atom transfers or loses its 3s electron, its third energy level
becomes vacant, and it becomes a sodium ion with an electron configuration identical to that
of the noble gas neon. This process requires energy:
11+
11+
1le–
10e–
+ l e–
Na+ ion
(1s22s22p6)
Na atom
(1s22s22p63s1)
An atom that has lost or gained electrons will have a positive or negative charge, depending on which particles (protons or electrons) are in excess. Remember that a charged particle
or group of particles is called an ion.
By losing a negatively charged electron, the sodium atom becomes a positively charged particle known as a sodium ion. The charge, +1, results because the nucleus still contains 11 positively
charged protons, and the electron orbitals contain only 10 negatively charged electrons. The charge
is indicated by a plus sign (+) and is written as a superscript after the symbol of the element (Na+).
A chlorine atom with seven electrons in the third energy level needs one electron to pair
up with its one unpaired 3p electron to attain the stable outer electron structure of argon. By
gaining one electron, the chlorine atom becomes a chloride ion (Cl-), a negatively charged
particle containing 17 protons and 18 electrons. This process releases energy:
17e–
17 +
18e–
17 +
+ 1e–
Cl atom
Cl– ion
(1s22s22p63s23p5)
(1s22s22p63s23p6)
Consider sodium and chlorine atoms reacting with each other. The 3s electron from the
sodium atom transfers to the half-filled 3p orbital in the chlorine atom to form a positive sodium ion and a negative chloride ion. The compound sodium chloride results because the Na+
and Cl- ions are strongly attracted to each other by their opposite electrostatic charges. The
force holding the oppositely charged ions together is called an ionic bond:
le–
17+
11 +
17 +
1 l e–
Na atom
17e–
11+
18 e–
10 e–
Cl atom
+
Cl–
Na
sodium chloride
The Lewis representation of sodium chloride formation is
Na � Cl ¡ [Na]� Cl
�
The chemical reaction between sodium and chlorine is a very vigorous one, producing
considerable heat in addition to the salt formed. When energy is released in a chemical reaction, the products are more stable than the reactants. Note that in NaCl both atoms attain a
noble gas electron structure.
Na +
Cl –
• The Ionic Bond: Transfer of ­Electrons from One Atom to Another 219
Na +
© Elena Moiseeva/iStockphoto
11.3
Cl –
Figure 11.5
Sodium chloride crystal. Diagram represents a small fragment of sodium chloride, which
forms cubic crystals. Each sodium ion is surrounded by six chloride ions, and each chloride
ion is surrounded by six sodium ions. The tiny NaCl crystals show the cubic crystal structure
of salt.
Sodium chloride is made up of cubic crystals in which each sodium ion is surrounded by
six chloride ions and each chloride ion by six sodium ions, except at the crystal surface. A visible crystal is a regularly arranged aggregate of millions of these ions, but the ratio of sodium
to chloride ions is 1:1, hence the formula NaCl. The cubic crystalline lattice arrangement of
sodium chloride is shown in Figure 11.5.
Figure 11.6 contrasts the relative sizes of sodium and chlorine atoms with those of their
ions. The sodium ion is smaller than the atom due primarily to two factors: (1) The sodium
atom has lost its outermost electron, thereby reducing its size; and (2) the 10 remaining electrons are now attracted by 11 protons and are thus drawn closer to the nucleus. Conversely,
the chloride ion is larger than the atom because (1) it has 18 electrons but only 17 protons and
(2) the nuclear attraction on each electron is thereby decreased, allowing the chlorine atom to
expand as it forms an ion.
We’ve seen that when sodium reacts with chlorine, each atom becomes an ion. Sodium
chloride, like all ionic substances, is held together by the attraction existing between positive
and negative charges. An ionic bond is the attraction between oppositely charged ions.
Ionic bonds are formed whenever one or more electrons are transferred from one atom to
another. Metals, which have relatively little attraction for their valence electrons, tend to form
ionic bonds when they combine with nonmetals.
It’s important to recognize that substances with ionic bonds do not exist as molecules.
In sodium chloride, for example, the bond does not exist solely between a single sodium ion
and a single chloride ion. Each sodium ion in the crystal attracts six near-neighbor negative
chloride ions; in turn, each negative chloride ion attracts six near-neighbor positive sodium
ions (see Figure 11.5).
A metal will usually have one, two, or three electrons in its outer energy level. In reacting, metal atoms characteristically lose these electrons, attain the electron structure of a noble
gas, and become positive ions. A nonmetal, on the other hand, is only a few electrons short
of having a noble gas electron structure in its outer energy level and thus has a tendency to
gain electrons. In reacting with metals, nonmetal atoms characteristically gain one to four
electrons; attain the electron structure of a noble gas; and become negative ions. The ions
0.186 nm
0.099 nm
0.095 nm
0.181 nm
These tiny NaCl crystals show
the cubic structure illustrated in
Figure 11.5.
Remember: A cation is always
smaller than its parent atom,
whereas an anion is always larger
than its parent atom.
Figure 11.6
Relative radii of sodium and
chlorine atoms and their ions.
Na atom
Na + ion
Cl atom
Cl – ion
chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Table 11.3
Change in Atomic Radii (nm) of Selected Metals and Nonmetals*
Atomic
radius
Li
Na
K
Mg
Al
Ionic
radius
0.152
0.186
0.227
0.160
0.143
Li+
Na+
K+
Mg 2+
Al3+
Atomic
radius
0.060
0.095
0.133
0.065
0.050
F
Cl
Br
O
S
Ionic
radius
FClBrO2S2-
0.071
0.099
0.114
0.074
0.103
0.136
0.181
0.195
0.140
0.184
*The metals shown lose electrons to become positive ions. The nonmetals gain electrons to become negative ions.
formed by loss of electrons are much smaller than the corresponding metal atoms; the ions
formed by gaining electrons are larger than the corresponding nonmetal atoms. The dimensions of the atomic and ionic radii of several metals and nonmetals are given in Table 11.3.
Practice 11.3
What noble gas structure is formed when an atom of each of these metals loses all its
valence electrons? Write the formula for the metal ion formed.
(a) K
(b) Mg
(c) Al
(d) Ba
Study the following examples. Note the loss and gain of electrons between atoms; also
note that the ions in each compound have a noble gas electron structure.
Example 11.3
Explain how magnesium and chlorine combine to form magnesium chloride, MgCl2.
SOLUTION
A magnesium atom of electron structure 1s22s22p63s2 must lose two electrons or gain six
to reach a stable electron structure. If magnesium reacts with chlorine and each chlorine
atom can accept only one electron, two chlorine atoms will be needed for the two electrons
from each magnesium atom. The compound formed will contain one magnesium ion and
two chloride ions. The magnesium atom, having lost two electrons, becomes a magnesium
ion with a +2 charge. Each chloride ion will have a -1 charge. The transfer of electrons
from a magnesium atom to two chlorine atoms is shown in the following illustration:
1e–
17+
17+
12+
12+
17e–
18e–
17+
17+
10e–
12e–
17e–
18e–
1e–
1 Mg atom
. Cl..
+ . Cl..
2 Cl atoms
Mg
2+
..Cl..
..Cl..
.. .. .. ..
+
.. ..
Mg ..
.. ..
220 1–
1–
or MgCl 2
Magnesium chloride
• The Ionic Bond: Transfer of ­Electrons from One Atom to Another 221
11.3
Example 11.4
Explain the formation of sodium fluoride (NaF) from its elements.
SOLUTION
le–
9+
11+
9e–
10 e–
. F ..
.. ..
Na .
9+
�
Na atom
10e–
+
Na
F atom
.. F ..
.. ..
11e–
11+
–
or NaF
Sodium fluoride
The fluorine atom, with seven electrons in its outer energy level, behaves similarly to the
chlorine atom.
Example 11.5
Explain the formation of aluminum fluoride (AlF3) from its elements.
SOLUTION
Al
1 Al atom
�
F
F
F
¡ [Al]3� F
�
�
or
AlF3
�
F
F
3 F atoms
Aluminium fluoride
Each fluorine atom can accept only one electron. Therefore, three fluorine atoms are
needed to combine with the three valence electrons of one aluminum atom. The aluminum atom has lost three electrons to become an aluminum ion (Al3+) with a +3 charge.
Example 11.6
Explain the formation of magnesium oxide (MgO) from its elements.
SOLUTION
1e–
12+
8+
12e–
8e–
12+
8+
10e–
10 e–
1e–
Mg atom
. O. ..
O atom
Mg
2+
..O..
.. ..
�
..
Mg ..
2–
or MgO
Magnesium oxide
222 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
The magnesium atom, with two electrons in the outer energy level, exactly fills the need
of one oxygen atom for two electrons. The resulting compound has a ratio of one magnesium atom to one oxygen atom. The oxygen (oxide) ion has a -2 charge, having gained
two electrons. In combining with oxygen, magnesium behaves the same way as when it
combines with chlorine—it loses two electrons.
Example 11.7
Explain the formation of sodium sulfide (Na2S) from its elements.
SOLUTION
Na
�
Na
2 Na atoms
S
¡
1 S atom
[Na] �
[Na] �
S
2�
or
Na 2S
Sodium sulfide
Two sodium atoms supply the electrons that one sulfur atom needs to make eight in its
outer energy level.
Example 11.8
Explain the formation of aluminum oxide (Al2O3) from its elements.
SOLUTION
Al
O
�
O ¡
Al
O
2 Al atoms
3 O atoms
[Al]3�
[Al]
3�
O
O
O
2�
2�
or Al2O3
2�
Aluminium oxide
One oxygen atom, needing two electrons, cannot accommodate the three electrons from
one aluminum atom. One aluminum atom falls one electron short of the four electrons
needed by two oxygen atoms. A ratio of two atoms of aluminum to three atoms of oxygen, involving the transfer of six electrons (two to each oxygen atom), gives each atom a
stable electron configuration.
Note that in each of these examples, outer energy levels containing eight electrons were
formed in all the negative ions. This formation resulted from the pairing of all the s and p
electrons in these outer energy levels.
11.4Predicting Formulas of Ionic
­Compounds
Learning objective
Predict the formulas of ionic compounds from their position on the periodic table.
In previous examples, we learned that when a metal and a nonmetal react to form an ionic
compound, the metal loses one or more electrons to the nonmetal. In Chapter 6, where we
learned to name compounds and write formulas, we saw that Group 1A metals always form
+1 cations, whereas Group 2A elements form +2 cations. Group 7A elements form -1 anions
and Group 6A elements form -2 anions.
11.4
• Predicting Formulas of Ionic ­Compounds 223
It stands to reason, then, that this pattern is directly related to the stability of the noble
gas configuration. Metals lose electrons to attain the electron configuration of a noble gas (the
previous one on the periodic table). A nonmetal forms an ion by gaining enough electrons to
achieve the electron configuration of the noble gas following it on the periodic table. These
observations lead us to an important chemical principle:
In almost all stable chemical compounds of representative elements, each atom attains a
noble gas electron configuration. This concept forms the basis for our understanding of
chemical bonding.
We can apply this principle in predicting the formulas of ionic compounds. To predict
the formula of an ionic compound, we must recognize that chemical compounds are always
electrically neutral. In addition, the metal will lose electrons to achieve noble gas configuration and the nonmetal will gain electrons to achieve noble gas configuration. Consider the
compound formed between barium and sulfur. Barium has two valence electrons, whereas
sulfur has six valence electrons:
Ba [Xe]6s2 S [Ne]3s23p4
If barium loses two electrons, it will achieve the configuration of xenon. By gaining two
electrons, sulfur achieves the configuration of argon. Consequently, a pair of electrons is transferred between atoms. Now we have Ba2+ and S2-. Since compounds are electrically neutral,
there must be a ratio of one Ba to one S, giving the formula BaS.
The same principle works for many other cases. Since the key lies in the electron configuration, the periodic table can be used to extend the prediction even further. Because of similar
electron structures, the elements in a family generally form compounds with the same atomic
ratios. In general, if we know the atomic ratio of a particular compound—say, NaCl—we can
predict the atomic ratios and formulas of the other alkali metal chlorides. These formulas are
LiCl, KCl, RbCl, CsCl, and FrCl (see Table 11.4).
Similarly, if we know that the formula of the oxide of hydrogen is H2O, we can predict
that the formula of the sulfide will be H2S, because sulfur has the same valence electron
structure as oxygen. Recognize, however, that these are only predictions; it doesn’t necessarily follow that every element in a group will behave like the others or even that a
predicted compound will actually exist. For example, knowing the formulas for potassium
chlorate, bromate, and iodate to be KClO3, KBrO3, and KIO3, we can correctly predict
the corresponding sodium compounds to have the formulas NaClO3, NaBrO3, and NaIO3.
Fluorine belongs to the same family of elements (Group 7A) as chlorine, bromine, and
iodine, so it would appear that fluorine should combine with potassium and sodium to give
fluorates with the formulas KFO3 and NaFO3. However, potassium and sodium fluorates
are not known to exist.
In the discussion in this section, we refer only to representative metals (Groups 1A, 2A,
and 3A). The transition metals (Group B) show more complicated behavior (they form multiple ions), and their formulas are not as easily predicted.
Table 11.4
Formulas of Compounds Formed by Alkali Metals
Lewis
structure
Oxides
Chlorides
Bromides
Sulfates
LiD
NaD
KD
RbD
CsD
Li2O
Na2O
K2O
Rb2O
Cs2O
LiCl
NaCl
KCl
RbCl
CsCl
LiBr
NaBr
KBr
RbBr
CsBr
Li2SO4
Na2SO4
K2SO4
Rb2SO4
Cs2SO4
224 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Example 11.9
ENHANCED EXAMPLE
The formula for calcium sulfide is CaS and that for lithium phosphide is Li3P. ­Predict
formulas for (a) magnesium sulfide, (b) potassium phosphide, and (c) magnesium
­selenide.
SOLUTION
(a) Look up calcium and magnesium in the periodic table; they are both in Group 2A. The
formula for calcium sulfide is CaS, so it’s reasonable to predict that the formula for
magnesium sulfide is MgS.
(b) Find lithium and potassium in the periodic table; they are in Group 1A. Since the formula for lithium phosphide is Li3P, it’s reasonable to predict that K3P is the formula
for potassium phosphide.
(c) Find selenium in the periodic table; it is in Group 6A just below sulfur. Therefore, it’s
reasonable to assume that selenium forms selenide in the same way that sulfur forms
sulfide. Since MgS was the predicted formula for magnesium sulfide in part (a), we can
reasonably assume that the formula for magnesium selenide is MgSe.
Practice 11.4
The formula for sodium oxide is Na2O. Predict the formula for
(a) sodium sulfide
(b) rubidium oxide
Practice 11.5
The formula for barium phosphide is Ba3P2. Predict the formula for
(a) magnesium nitride
(b) barium arsenide
11.5The Covalent Bond:
Sharing ­Electrons
Lea rning obje ctive
Draw the electron structure of a covalent bond.
Key tERMs
Some atoms do not transfer electrons from one atom to another to form ions. Instead they form
a chemical bond by sharing pairs of electrons between them. A covalent bond consists of
a pair of electrons shared between two atoms. This bonding concept was introduced in 1916
by G. N. Lewis. In the millions of known compounds, the covalent bond is the predominant
chemical bond.
True molecules exist in substances in which the atoms are covalently bonded. It is proper
to refer to molecules of such substances as hydrogen, chlorine, hydrogen chloride, carbon dioxide, water, or sugar (Figure 11.7). These substances contain only covalent bonds and exist
as aggregates of molecules. We don’t use the term molecule when
talking about ionically bonded compounds, such as sodium chloride, because such substances exist as large aggregates of positive
and negative ions, not as molecules (Figure 11.7).
A study of the hydrogen molecule gives us an insight into the
nature of the covalent bond and its formation. The formation of a
hydrogen molecule (H2) involves the overlapping and pairing of 1s
electron orbitals from two hydrogen atoms, shown in Figure 11.8.
Each atom contributes one electron of the pair that is shared jointly
by two hydrogen nuclei. The orbital of the electrons now includes
both hydrogen nuclei, but probability factors show that the most
covalent bond
polar covalent bond
Figure 11.7
Na +
O
C
Dry ice
Cl –
Sodium chloride
Solid carbon dioxide (dry ice) is composed of individual covalently
bonded molecules of CO2 closely packed together. Table salt is a
large aggregate of Na + and Cl- ions instead of molecules.
11.5
1s orbitals
H.
• The Covalent Bond: Sharing ­Electrons 225
Figure 11.8
Overlapping 1s orbitals
H .. H
H.
Hydrogen atoms
The formation of a hydrogen
molecule from two hydrogen
atoms. The two 1s orbitals
overlap, forming the H2 molecule.
In this molecule the two electrons
are shared between the atoms,
forming a covalent bond.
Shared pair
of electrons
Hydrogen molecule
likely place to find the electrons (the point of highest electron density) is between the two
nuclei. The two nuclei are shielded from each other by the pair of electrons, allowing the two
nuclei to be drawn very close to each other.
The formula for chlorine gas is Cl2. When the two atoms of chlorine combine to form
this molecule, the electrons must interact in a manner similar to that shown in the hydrogen
example. Each chlorine atom would be more stable with eight electrons in its outer energy
level. But chlorine atoms are identical, and neither is able to pull an electron away from the
other. What happens is this: The unpaired 3p electron orbital of one chlorine atom overlaps the
unpaired 3p electron orbital of the other atom, resulting in a pair of electrons that is mutually
shared between the two atoms. Each atom furnishes one of the pair of shared electrons. Thus,
each atom attains a stable structure of eight electrons by sharing an electron pair with the other
atom. The pairing of the p electrons and the formation of a chlorine molecule are illustrated in
Figure 11.9. Neither chlorine atom has a positive or negative charge because both contain the
same number of protons and have equal attraction for the pair of electrons being shared. Other
examples of molecules in which electrons are equally shared between two atoms are hydrogen
(H2), oxygen (O2), nitrogen (N2), fluorine (F2), bromine (Br2), and iodine (I2). Note that more
than one pair of electrons may be shared between atoms:
H H
F F
Br Br
I I
O O
N
hydrogen
fluorine
bromine
iodine
oxygen
nitrogen
Tom Pantages
N
The Lewis structure given for oxygen does not adequately account for all the properties of
the oxygen molecule. Other theories explaining the bonding in oxygen molecules have been
advanced, but they are complex and beyond the scope of this book.
In writing structures, we commonly replace the pair of dots used to represent a shared pair
of electrons with a dash (—). One dash represents a single bond; two dashes, a double bond;
and three dashes, a triple bond. The six structures just shown may be written thus:
H
H
F
F
Br
Br
I
I
O
O
N
N
The ionic bond and the covalent bond represent two extremes. In ionic bonding the atoms
are so different that electrons are transferred between them, forming a charged pair of ions.
In covalent bonding, two identical atoms share electrons equally. The bond is the mutual attraction of the two nuclei for the shared electrons. Between these extremes lie many cases in
which the atoms are not different enough for a transfer of electrons but are different enough
that the electron pair cannot be shared equally. This unequal sharing of electrons results in the
formation of a polar covalent bond.
Molecular models for F2 (green,
single bond), O2 (black, double
bond), and N2 (blue, triple bond).
Remember: A dash represents a
shared pair of electrons.
Figure 11.9
Overlap of p orbitals
+
. Cl..
.. ..
.. ..
.. Cl.
Unshared p orbitals
Paired p orbital
Chlorine molecule
.. Cl ..
Cl..
.. ..
Chlorine atoms
.. ..
p orbitals
Shared pair of p electrons
Pairing of p electrons in the
formation of a chlorine molecule.
226 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Example 11.10
What is the basic concept of a covalent bond?
SOLUTION
The basic concept of a covalent bond is the sharing of one or more pairs of electrons
between two atoms to form a covalent bond between the two atoms.
Practice 11.6
The atoms in the two molecules HBr and C2H6 are bonded by covalent bonds. Draw
Lewis structures showing these covalent bonds.
11.6Electronegativity
Learning objective
Explain how electronegativities of component atoms in a molecule determine the
polarity of the molecule.
Key tERMs
When two different kinds of atoms share a pair of electrons, a bond forms in which electrons
are shared unequally. One atom assumes a partial positive charge and the other a partial negative charge with respect to each other. This difference in charge occurs because the two atoms
exert unequal attraction for the pair of shared electrons. The attractive force that an atom of an
element has for shared electrons in a molecule or polyatomic ion is known as its electronegativity. Elements differ in their electronegativities. For example, both hydrogen and chlorine
need one electron to form stable electron configurations. They share a pair of electrons in
hydrogen chloride (HCl). Chlorine is more electronegative and therefore has a greater attraction for the shared electrons than does hydrogen. As a result, the pair of electrons is displaced
toward the chlorine atom, giving it a partial negative charge and leaving the hydrogen atom
with a partial positive charge. Note that the electron is not transferred entirely to the chlorine
atom (as in the case of sodium chloride) and that no ions are formed. The entire molecule, HCl,
is electrically neutral. A partial charge is usually indicated by the Greek letter delta, d. Thus, a
partial positive charge is represented by d+ and a partial negative charge by d-.
electronegativity
nonpolar covalent bond
dipole
+δ
−δ
H :Cl
��
��
H Cl
Hydrogen chloride
The pair of shared electrons in HCl is closer to the
more electronegative chlorine atom than to the
hydrogen atom, giving chlorine a partial negative
charge with respect to the hydrogen atom.
A scale of relative electronegativities, in which the most electronegative element, fluorine,
is assigned a value of 4.0, was developed by the Nobel Laureate (1954 and 1962) Linus Pauling
(1901–1994). Table 11.5 shows that the relative electronegativity of the nonmetals is high and
that of the metals is low. These electronegativities indicate that atoms of metals have a greater
tendency to lose electrons than do atoms of nonmetals and that nonmetals have a greater
tendency to gain electrons than do metals. The higher the electronegativity value, the greater
the attraction for electrons. Note that electronegativity generally increases from left to right
across a period and decreases down a group for the representative elements. The highest electronegativity is 4.0 for fluorine, and the lowest is 0.7 for francium and cesium. It’s important
to remember that the higher the electronegativity, the more strongly an atom attracts electrons.
Example 11.11
What are the electronegativity trends for the groups of representative elements and the
periods of representative elements in the periodic table?
SOLUTION
In general, electronegativity decreases from top to bottom for a group of representative
elements and increases from left to right across a period of representative elements.
• Electronegativity 227
11.6
Table 11.5
Three-Dimensional Representation of Electronegativity
Atomic number
Symbol
Electronegativity
9
1
F
H
4.0
2.1
11
12
13
Na
Mg
Al
0.9
1.2
20
K
Ca
0.8
1.0
1.5
21
22
Sc
Ti
Mn
1.6
1.5
41
42
43
Mo
Tc
1.6
1.8
1.9
73
75
72
74
Hf
Ta
Re
La–Lu
W
1.5
1.9
1.3
1.7
1.1–1.2
1.3
1.4
40
Rb
Sr
Y
Zr
0.8
1.0
1.2
1.4
56
0.7
0.9
25
Cr
38
Ba
24
V
37
55
23
1.6
39
Cs
2.5
2.0
1.5
19
C
B
Be
Li
1.0
6
5
4
3
57–71
87
88
89–103
Fr
Ra
Ac–Lr
0.7
0.9
1.1–1.7
Nb
28
29
Co
Ni
Cu
1.8
1.8
1.9
26
27
Fe
1.8
44
45
46
47
Ru
Rh
Pd
Ag
2.2
2.2
2.2
1.9
76
77
78
79
Os
Ir
Pt
Au
2.2
2.2
2.2
2.4
14
Si
1.8
P
2.1
33
F
O
N
4.0
3.5
3.0
15
9
8
7
16
S
2.5
34
17
Cl
3.0
35
Br
30
31
32
Zn
Ga
Ge
As
1.6
1.6
1.8
2.0
48
49
50
51
Cd
In
Sn
Sb
Te
1.7
1.7
1.8
1.9
2.1
80
81
82
83
84
Hg
Tl
Pb
Bi
Po
At
1.9
1.8
1.8
1.9
2.0
2.2
Se
2.4
52
2.8
53
I
2.5
85
Practice 11.7
Explain the term electronegativity and how it relates to the elements in the periodic table.
The polarity of a bond is determined by the difference in electronegativity values of the
atoms forming the bond (see Figure 11.10). If the electronegativities are the same, the bond
is nonpolar covalent and the electrons are shared equally. If the atoms have greatly different
electronegativities, the bond is very polar. At the extreme, one or more electrons are actually
transferred and an ionic bond results.
A dipole is a molecule that is electrically asymmetrical, causing it to be oppositely
charged at two points. A dipole is often written as ± – . A hydrogen chloride molecule is
Cl. The arrow
polar and behaves as a small dipole. The HCl dipole may be written as H
points toward the negative end of the dipole. Molecules of H2O, HBr, and ICl are polar:
O
H
Cl
H
Br
I
Cl
H
H
How do we know whether a bond between two atoms is ionic or covalent? The difference in electronegativity between the two atoms determines the character of the bond formed
between them. As the difference in electronegativity increases, the polarity of the bond (or
percent ionic character) increases.
In general, if the electronegativity difference between two bonded atoms is greater than
1.7–1.9, the bond will be more ionic than covalent.
If the electronegativity difference is greater than 2.0, the bond is strongly ionic. If the electronegativity difference is less than 1.5, the bond is strongly covalent.
Figure 11.10
–
+
Cl2
H2
Nonpolar
molecules
HCl
NaCl
Polar covalent
molecule
Ionic
compound
Nonpolar, polar covalent, and
ionic compounds.
228 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
>Chemistry in Action
H
O
C“C
CH3 ¬ (CH2)7
(CH2)7 ¬ C
≈
Trans fats contain a particular kind of unsaturated fatty
acid. When there is a double bond between carbon
atoms, the molecule bends in one of two ways: the cis
or the trans direction. In cis configuration, the carbon
chain on both sides of the double bond bends in to the
same side of the double bond (see structure). In the trans
configuration, the chain on either side of the double bond
bends toward opposite sides of the double bond (see
structure). Most trans fats come from processing oils for
prepared foods and from solid fats such as margarine.
“
H
≈ ≈
Different categories of fat can be identified from the pattern of bonds and hydrogen atoms in the molecule. Fatty
acids are one component of fats and contain long chains of
carbon atoms with hydrogen atoms bonded to some or all
of the carbon atoms. Unsaturated fats (such as corn or soybean oil) contain double bonds between some of the carbon atoms in the chains. A carbon that is doubly bonded
to another carbon usually also bonds to a hydrogen atom.
If the fatty acid has only one double bond, it is monounsaturated (such as olive oil). If the fatty acid contains more
than one double bond, it is polyunsaturated. All fat with no
double bonds is saturated—all the carbon atoms have the
maximum possible hydrogen atoms bonded to them.
≈ ≈
Trans fats are virtually everywhere in the American diet.
These fats remain when vegetable oils are converted into
solid substances that are used in many processed foods.
In fact, beginning January 1, 2006, manufacturers were
required to label products to show their trans fat content.
So just what is a trans fat?
When an oil is converted into a solid fat, some of the
double bonds are converted to single bonds by adding hydrogen (hydrogenation). This process is easier at
cis double bonds and, therefore, the remaining double
bonds are mainly in the trans configuration. These trans
fatty acids tend to stack together, making a solid easier
than the cis forms. Studies have linked diets high in trans
fats to poor health, high cholesterol, heart disease, and
diabetes. Food producers are working on ways to lower
the trans fat content of foods. Gary List at USDA in Peoria,
Illinois, has used high-pressure hydrogen gas on soybean
oil at 140°–170°C to hydrogenate the oil, producing a soft
margarine containing 5%–6% trans fat instead of the ,40%
from the standard hydrogenation techniques. This could
lead to a product that qualifies for a label of 0 g trans fat.
Look for lots of new products and labels in your grocery
stores as manufacturers trans-form products into ones with
less trans fat.
OH
cis fatty acid
“
O
≈ ≈
≈
(CH2)7 ¬ C
OH
C“C
H
CH3 ¬ (CH2)7
H
≈ ≈
Trans-forming Fats
trans fatty acid
Practice 11.8
Which of these compounds would you predict to be ionic and which would be covalent?
(a) SrCl2
(d) RbBr
(b) PCl3
(e) LiCl
(c) NH3
(f) CS2
Care must be taken to distinguish between polar bonds and polar molecules. A covalent
bond between different kinds of atoms is always polar. But a molecule containing different
kinds of atoms may or may not be polar, depending on its shape or geometry. Molecules of HF,
HCl, HBr, HI, and ICl are all polar because each contains a single polar bond. However, CO2,
CH4, and CCl4 are nonpolar molecules despite the fact that all three contain polar bonds. The
carbon dioxide molecule O “ C “ O is nonpolar because the carbon–oxygen dipoles cancel
each other by acting in opposite directions.
O
C
O
dipoles in equal and opposite directions
Spacefilling molecular model CCl4
(top) and methane (Ch4).
Carbon tetrachloride (CCl4) is nonpolar because the four C i Cl polar bonds are identical, and
since these bonds emanate from the center to the corners of a tetrahedron in the molecule, their
polarities cancel one another. Methane has the same molecular structure and is also nonpolar.
We will discuss the shapes of molecules later in this chapter.
11.7
• Lewis Structures of Compounds 229
Bond type
Covalent
Polar covalent
Ionic
Figure 11.11
0
Intermediate
3.3
Electronegativity difference
Relating bond type to
electronegativity difference
between atoms.
We have said that water is a polar molecule. If the atoms in water were linear like those in
carbon dioxide, the two O i H dipoles would cancel each other, and the molecule would be
nonpolar. However, water is definitely polar and has a nonlinear (bent) structure with an angle
of 105° between the two O i H bonds.
The relationships among types of bonds are summarized in Figure 11.11. It is important
to realize that bonding is a continuum; that is, the difference between ionic and covalent is a
gradual change.
11.7 Lewis Structures of Compounds
Draw the Lewis structure of a covalent compound.
Learning objectiv e
As we have seen, Lewis structures are a convenient way of showing the covalent bonds in
many molecules or ions of the representative elements. In writing Lewis structures, the most
important consideration for forming a stable compound is that the atoms attain a noble gas
configuration.
The most difficult part of writing Lewis structures is determining the arrangement of
the atoms in a molecule or an ion. In simple molecules with more than two atoms, one atom
will be the central atom surrounded by the other atoms. Thus, Cl2O has two possible arrangements, Cl i Cl i O or Cl i O i Cl. Usually, but not always, the single atom in the formula
(except H) will be the central atom.
Although Lewis structures for many molecules and ions can be written by inspection, the
following procedure is helpful for learning to write them:
Problem-Solving Strategy: Writing a Lewis Structure
1. Obtain the total number of valence electrons to be used in the structure by adding the
number of valence electrons in all the atoms in the molecule or ion. If you are writing the
structure of an ion, add one electron for each negative charge or subtract one electron for
each positive charge on the ion.
2. Write the skeletal arrangement of the atoms and connect them with a single covalent bond
(two dots or one dash). Hydrogen, which contains only one bonding electron, can form
only one covalent bond. Oxygen atoms are not normally bonded to each other, except in
compounds known to be peroxides. Oxygen atoms normally have a maximum of two covalent bonds (two single bonds or one double bond).
3. Subtract two electrons for each single bond you used in Step 2 from the total number of
electrons calculated in Step 1. This gives you the net number of electrons available for
completing the structure.
4. Distribute pairs of electrons (pairs of dots) around each atom (except hydrogen) to give
each atom a noble gas structure.
5. If there are not enough electrons to give these atoms eight electrons, change single bonds
between atoms to double or triple bonds by shifting unbonded pairs of electrons as needed.
Check to see that each atom has a noble gas electron structure (two electrons for hydrogen
and eight for the others). A double bond counts as four electrons for each atom to which it
is bonded.
Remember: The number of
valence electrons of Group A
­elements is the same as their
group number in the periodic
table.
230 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Example 11.12
ENHANCED EXAMPLE
How many valence electrons are in each of these atoms: Cl, H, C, O, N, S, P, I?
SOLUTION
You can look at the periodic table to determine the electron structure, or, if the element
is in Group A of the periodic table, the number of valence electrons is equal to the group
number:
Atom
Group
Valence electrons
Cl
H
C
O
N
S
P
I
7A
1A
4A
6A
5A
6A
5A
7A
7
1
4
6
5
6
5
7
Example 11.13
Use the Writing a Lewis Structure Problem-Solving Strategy for water (H2O).
SOLUTION
1. The total number of valence electrons is eight, two from the two hydrogen atoms and
six from the oxygen atom.
2. The two hydrogen atoms are connected to the oxygen atom. Write the skeletal structure:
or
HO
H
HOH
Place two dots between the hydrogen and oxygen atoms to form the covalent bonds:
or
H O
H
H O H
3. Subtract the four electrons used in Step 2 from eight to obtain four electrons yet to be
used.
4. Distribute the four electrons in pairs around the oxygen atom. Hydrogen atoms cannot
accommodate any more electrons:
H
O
or
H
O
H
H
These arrangements are Lewis structures because each atom has a noble gas electron
structure. Note that the shape of the molecule is not shown by the Lewis structure.
Example 11.14
Use the Writing a Lewis Structure Problem-Solving Strategy for a molecule of methane
(CH4).
SOLUTION
1. The total number of valence electrons is eight, one from each hydrogen atom and four
from the carbon atom.
2. The skeletal structure contains four H atoms around a central C atom. Place two electrons between the C and each H.
H
H C H
H
H
H C H
H
11.7
• Lewis Structures of Compounds 231
3. Subtract the eight electrons used in Step 2 from eight (obtained in Step 1) to obtain zero
electrons yet to be placed. Therefore, the Lewis structure must be as written in Step 2:
H
H
H C H
H C H
or
H
H
Example 11.15
Use the Writing a Lewis Structure Problem-Solving Strategy for a molecule of carbon
tetrachloride (CCl4).
SOLUTION
1. The total number of valence electrons to be used is 32, 4 from the carbon atom and 7
from each of the four chlorine atoms.
2. The skeletal structure contains the four Cl atoms around a central C atom. Place 2 electrons between the C and each Cl:
Cl
Cl C Cl
Cl
Cl
Cl C Cl
Cl
3. Subtract the 8 electrons used in Step 2 from 32 (obtained in Step 1) to obtain 24 electrons yet to be placed.
4. Distribute the 24 electrons (12 pairs) around the Cl atoms so that each Cl atom has 8
electrons around it:
Cl
Cl C Cl
Cl
Cl
or
Cl
C
Cl
Cl
This arrangement is the Lewis structure; CCl4 contains four covalent bonds.
Example 11.16
Use the Writing a Lewis Structure Problem-Solving Strategy for CO2.
SOLUTION
1. The total number of valence electrons is 16, 4 from the C atom and 6 from each O atom.
2. The two O atoms are bonded to a central C atom. Write the skeletal structure and place
2 electrons between the C and each O atom.
O •• C •• O
3. Subtract the 4 electrons used in Step 2 from 16 (found in Step 1) to obtain 12 electrons
yet to be placed.
4. Distribute the 12 electrons (six pairs) around the C and O atoms. Several possibilities
exist:
O CO
O CO
O CO
I
II
III
5. Not all the atoms have 8 electrons around them (noble gas structure). Remove one pair
of unbonded electrons from each O atom in structure I and place one pair between each
O and the C atom, forming two double bonds:
O C O
or
O
C
O
Each atom now has 8 electrons around it. The carbon is sharing four pairs of electrons,
and each oxygen is sharing two pairs. These bonds are known as double bonds because
each involves sharing two pairs of electrons.
232 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
>Chemistry in Action
What do color-changing pens, bullet-resistant vests, and
calculators have in common? The chemicals that make
each of them work are liquid crystals. These chemicals
find numerous applications; you are probably most familiar with liquid crystal displays (LCDs) and color-changing
products, but these chemicals are also used to make
superstrong synthetic fibers.
Molecules in a normal crystal remain in an orderly arrangement, but in a liquid crystal the molecules can flow
and maintain an orderly arrangement at the same time.
Liquid crystal molecules are linear and polar. Since the
atoms tend to lie in a relatively straight line, the molecules
are generally much longer than they are wide. These polar
molecules are attracted to each other and are able to line
up in an orderly fashion without solidifying.
Liquid crystals with twisted arrangements of molecules
give us novelty color-changing products. In these liquid
crystals the molecules lie side by side in a nearly flat layer.
The next layer is similar but at an angle to the one below.
The closely packed flat layers have a special effect on
light. As the light strikes the surface, some of it is
­reflected from the top layer and some from lower layers.
When the same wavelength is reflected from many layers,
we see a color. (This is similar to the rainbow of colors
formed by oil in a puddle on the street or the film of a
soap bubble.) As the temperature is increased, the molecules move faster, causing a change in the angle and the
space between the layers. This results in a color change
in the reflected light. Different compounds change color
within different temperature ranges, allowing a variety of
practical and amusing applications.
Liquid crystal (nematic) molecules that lie parallel to one another are used to manufacture very strong synthetic fibers.
Perhaps the best example of these liquid crystals is Kevlar,
a synthetic fiber used in bullet-resistant vests, canoes, and
parts of the space shuttle. Kevlar is a synthetic polymer, like
nylon or polyester, that gains strength by passing through a
liquid crystal state during its manufacture.
In a typical polymer, the long molecular chains are jumbled
together, somewhat like spaghetti. The strength of the material is limited by the disorderly arrangement. The trick is
to get the molecules to line up parallel to each other. Once
the giant molecules have been synthesized, they are dissolved in sulfuric acid. At the proper concentration the molecules align, and the solution is forced through tiny holes
in a nozzle and further aligned. The sulfuric acid is removed
in a water bath, thereby forming solid fibers in near-perfect
alignment. One strand of Kevlar is stronger than an equalsized strand of steel. It has a much lower density as well,
making it a material of choice in bullet-resistant vests.
© Haku Nellies/iStockphoto
Strong Enough to Stop a Bullet?
Kevlar is used to make protective vests for police.
Practice 11.9
Write the Lewis structures for the following:
(a) PBr3
(b) CHCl3
(c) HF
(d) H2CO
(e) N2
Although many compounds attain a noble gas structure in covalent bonding, there are
numerous exceptions. Sometimes it’s impossible to write a structure in which each atom has
8 electrons around it. For example, in BF3 the boron atom has only 6 electrons around it, and
in SF6 the sulfur atom has 12 electrons around it.
Although there are exceptions, many molecules can be described using Lewis structures
where each atom has a noble gas electron configuration. This is a useful model for understanding chemistry.
11.8Complex Lewis Structures
L earning obje ctive
Draw the resonance structures for a polyatomic ion.
Key tERM
Most Lewis structures give bonding pictures that are consistent with experimental information on bond strength and length. There are some molecules and polyatomic ions for which no
single Lewis structure consistent with all characteristics and bonding information can be written. For example, consider the nitrate ion, NO3-. To write a Lewis structure for this polyatomic
ion, we use the following steps.
resonance structure
11.8
1. The total number of valence electrons is 24, 5 from the nitrogen atom, 6 from each oxy-
gen atom, and 1 from the -1 charge.
2. The three O atoms are bonded to a central N atom. Write the skeletal structure and place
two electrons between each pair of atoms. Since we have an extra electron in this ion,
resulting in a -1 charge, we enclose the group of atoms in square brackets and add
a - charge as shown.
O
O N O
�
3. Subtract the 6 electrons used in Step 2 from 24 (found in Step 1) to obtain 18 elec-
trons yet to be placed.
4. Distribute the 18 electrons around the N and O atoms:
O
O N O
electron deficient
5. One pair of electrons is still needed to give all the N and O atoms a noble gas structure.
Move the unbonded pair of electrons from the N atom and place it between the N and the
electron-deficient O atom, making a double bond.
�
O
O
N
or
O
�
O
O
N
or
O
�
O
O
N
O
Are these all valid Lewis structures? Yes, so there really are three possible Lewis structures
for NO3-.
A molecule or ion that has multiple correct Lewis structures shows resonance. Each of
these Lewis structures is called a resonance structure. In this book, however, we will not be
concerned with how to choose the correct resonance structure for a molecule or ion. Therefore,
any of the possible resonance structures may be used to represent the ion or molecule.
Example 11.17
Write the Lewis structure for a carbonate ion (CO23 ).
SOLUTION
1. These four atoms have 22 valence electrons plus 2 electrons from the -2 charge, which
makes 24 electrons to be placed.
2. In the carbonate ion, the carbon is the central atom surrounded by the three oxygen
atoms. Write the skeletal structure and place 2 electrons (or a single line) between each
C and O:
O
C
O
O
3. Subtract the 6 electrons used in Step 2 from 24 (from Step 1) to give 18 electrons yet to
be placed.
4. Distribute the 18 electrons around the three oxygen atoms and indicate that the carbonate ion has a -2 charge:
2
O
C
O
O
The difficulty with this structure is that the carbon atom has only six electrons around it
instead of a noble gas octet.
• Complex Lewis Structures 233
234 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
5. Move one of the nonbonding pairs of electrons from one of the oxygens and place them
between the carbon and the oxygen. Three Lewis structures are possible:
2
O
or
C
O
2
O
or
C
O
O
O
2
O
C
O
O
Practice 11.10
Write the Lewis structure for each of the following:
(a) NH3
(b) H3O+
(c) NH4+
(d) HCO3-
11.9Compounds Containing
­Polyatomic Ions
Learning obje ctive
Describe a compound that contains both ionic and covalent bonds.
A polyatomic ion is a stable group of atoms that has either a positive or a negative charge and
behaves as a single unit in many chemical reactions. Sodium carbonate, Na2CO3, contains two
sodium ions and a carbonate ion. The carbonate ion (CO23 ) is a polyatomic ion composed of
one carbon atom and three oxygen atoms and has a charge of -2. One carbon and three oxygen
atoms have a total of 22 electrons in their outer energy levels. The carbonate ion contains 24
outer electrons and therefore has a charge of -2. In this case, the 2 additional electrons come
from the two sodium atoms, which are now sodium ions:
2�
O
[Na] �
[Na] �
C
O
2�
O
C
O
O
Sodium carbonate
O
Carbonate ion
Sodium carbonate has both ionic and covalent bonds. Ionic bonds exist between each of
the sodium ions and the carbonate ion. Covalent bonds are pres­ent between the carbon and
oxygen atoms within the carbonate ion. One important difference between the ionic and covalent bonds in this compound can be demonstrated by dissolving sodium carbonate in water.
It dissolves in water, forming three charged particles—two sodium ions and one carbonate
ion—per formula unit of Na2CO3:
Na2CO3(s)
sodium carbonate
water
¡
2 Na+ (aq)
sodium ions
+
CO23 (aq)
carbonate ion
The CO23 ion remains as a unit, held together by covalent bonds; but where the bonds are
ionic, dissociation of the ions takes place. Do not think, however, that polyatomic ions are
so stable that they cannot be altered. Chemical reactions by which polyatomic ions can be
changed to other substances do exist.
Practice 11.11
Write Lewis structures for the calcium compounds of the following ­anions: nitrate and
sulfate.
11.10
• Molecular Shape 235
>Chemistry in Action
Chemistry or Art?
Scientists at the University of Edinburgh have synthesized
an interesting knotted molecule called a pentafoil. Its
structure is very symmetric and looks a bit like a fun sculpture. But what makes this sculpture special is that it is
composed of atoms! This molecule is composed of 160 atoms that twist to form a pentagonal or five-pointed
star. The inspiration for this molecule comes from
the ­variety of biological polymers, which
also incorporate twists and knots into their
structures.
form new materials with interesting and useful properties.
Natural rubber is composed of polymers or long-chain molecules with knots between the chains. Perhaps a molecule
like this could be part of synthesizing a more flexible and
stronger rubberlike material.
Until chemists figure out a way to utilize these molecules to
design more useful materials, we can simply enjoy their
form as chemical art.
Why would a chemist be interested in synthesizing a knotted structure such as this? Of course,
the first answer is to prove that it can be done.
But beyond that, such molecules could potentially
Reprinted by permission from Macmillan Publishers
Ltd: JF Ayme et al, Nature Chemistry
11.10 Molecular Shape
Determine the shape of a compound by using VSEPR method.
Learning objec ti ve
So far in our discussion of bonding we have used Lewis structures to represent valence electrons in molecules and ions, but they don’t indicate anything regarding the molecular or
geometric shape of a molecule. The three-dimensional arrangement of the atoms within a
molecule is a significant feature in understanding molecular interactions. Let’s consider several
examples illustrated in Figure 11.12.
Water is known to have the geometric shape known as “bent” or “V-shaped.” Carbon dioxide exhibits a linear shape. BF3 forms a third molecular shape called trigonal planar since
all the atoms lie in one plane in a triangular arrangement. One of the more common molecular
shapes is the tetrahedron, illustrated by the molecule methane (CH4).
How do we predict the geometric shape of a molecule? We will now study a model developed to assist in making predictions from the Lewis structure.
Key terms
linear structure
trigonal planar structure
tetrahedral structure
bent structure
The Valence Shell Electron Pair Repulsion (VSEPR) Model
The chemical properties of a substance are closely related to the structure of its molecules.
A change in a single site on a large biomolecule can make a difference in whether or not a
particular reaction occurs.
Figure 11.12
Geometric shapes of common
molecules. Each molecule is
shown as a ball-and-stick model
(showing the bonds) and as a
spacefilling model (showing the
shape).
Water
H 2O
(V-shaped)
Carbon dioxide
CO2
(linear shape)
Boron trifluoride
BF3
(trigonal planar)
Methane
CH4
(tetrahedral)
236 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Nonbonding pairs of electrons are
not shown here, so you can focus
your attention on the shapes, not
electron arrangement.
Instrumental analysis can be used to determine exact spatial arrangements of atoms. Quite
often, though, we only need to be able to predict the approximate structure of a molecule. A
relatively simple model has been developed to allow us to make predictions of shape from
Lewis structures.
The VSEPR model is based on the idea that electron pairs will repel each other electrically
and will seek to minimize this repulsion. To accomplish this minimization, the electron pairs
will be arranged around a central atom as far apart as possible. Consider BeCl2, a molecule
with only two pairs of electrons surrounding the central atom. These electrons are arranged
180° apart for maximum separation:
180
Cl
Be
Cl
This molecular structure can now be labeled as a linear structure. When only two pairs of
electrons surround a central atom, they should be placed 180° apart to give a linear structure.
What occurs when there are only three pairs of electrons around the central atom?
­Consider the BF3 molecule. The greatest separation of electron pairs occurs when the angles
between atoms are 120°:
120
F
B
120
F
120
F
This arrangement of atoms is flat (planar) and, as noted earlier, is called trigonal planar
structure. When three pairs of electrons surround an atom, they should be placed 120° apart
to show the trigonal planar structure.
Now consider the most common situation (CH4), with four pairs of electrons on the central carbon atom. In this case the central atom exhibits a noble gas electron structure. What
arrangement best minimizes the electron pair repulsions? At first, it seems that an obvious
choice is a 90° angle with all the atoms in a single plane:
H
90
H
90
C
90
H
90
H
However, we must consider that molecules are three-dimensional. This concept results in a
structure in which the electron pairs are actually 109.5° apart:
H
109.5
H
C
H
Figure 11.13
H
In this diagram the wedged line seems to protrude from the page, whereas the dashed line
recedes. Two representations of this arrangement, known as tetrahedral structure, are illustrated in Figure 11.13. When four pairs of electrons surround a central atom, they should
be placed 109.5° apart to give them a tetrahedral structure.
Ball-and-stick models
of methane and carbon
tetrachloride. Methane
and carbon tetrachloride
H
are nonpolar molecules
because their polar
bonds cancel each
other in the tetrahedral
C
H
arrangement of their
atoms. The carbon
atoms are located
H
in the centers of the
tetrahedrons.
Methane, CH 4
Cl
Cl
H
CH 4
C
Cl
Cl
Carbon tetrachloride, CCl 4
CCl 4
11.10
• Molecular Shape 237
Figure 11.14
N
(a) The tetrahedral arrangement
of electron pairs around the
N atom in the nh3 molecule.
(b) Three pairs are shared and one
is unshared. (c) The nh3 molecule
is pyramidal.
N
H
H
H
(a)
(b)
(c)
The VSEPR model is based on the premise that we are counting electron pairs. It’s quite
possible that one or more of these electron pairs may be nonbonding (lone) pairs. What happens to the molecular structure in these cases?
Consider the ammonia molecule. First we draw the Lewis structure to determine the number of electron pairs around the central atom:
H N H
H
Since there are four pairs of electrons, the arrangement of electrons around the central atom
will be tetrahedral (Figure 11.14a). However, only three of the pairs are bonded to another
atom, so the molecule itself is pyramidal. It is important to understand that the placement of the
electron pairs determines the shape but the name for the molecule is determined by the position
of the atoms themselves. Therefore, ammonia is pyramidal. See Figure 11.14c.
Now consider the effect of two unbonded pairs of electrons in the water molecule. The
Lewis structure for water is
H
O
H
The four electron pairs indicate that a tetrahedral electron arrangement is necessary (see
Figure 11.15a). The molecule is not called tetrahedral because two of the electron pairs are
unbonded pairs. The water molecule displays a bent structure.
Let’s summarize the VSEPR model.
Problem-Solving Strategy: Determining Molecular Shape Using VSEPR
1. Draw the Lewis structure for the molecule.
2. Count the electron pairs around the central atom and arrange them to minimize repulsions
(as far apart as possible). This determines the electron pair arrangement.
The arrangement of electron
pairs around an atom determines
its shape, but we name the shape
of molecules by the position of
the atoms.
3. Determine the positions of the atoms.
4. Name the molecular structure from the position of the atoms.
It is important to recognize that the placement of the electron pairs determines the structure
but the name of the molecular structure is determined by the position of the atoms. Table 11.6
shows the results of this process. Note that when the number of electron pairs is the same as the
number of atoms, the electron pair arrangement and the molecular structure are the same. But
when the number of atoms and the number of electron pairs are not the same, the molecular
structure is different from the electron pair arrangement. This is illustrated when the number
of electron pairs is four (a tetrahedral arrangement) in Table 11.6.
Figure 11.15
O
(a)
H
O
H
(b)
(c)
(a) The tetrahedral arrangement
of the four electron pairs around
oxygen in the h2O molecule.
(b) Two of the pairs are shared
and two are unshared. (c) The
h2O molecule is bent.
238 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
Table 11.6
Arrangement of Electron Pairs and Molecular Structure
Number of
Electron electron
pair
Ball-and-stick
Molecular
pairs
arrangement
model
Bonds structure
Molecular
structure
model
180°
2
Linear
3
Trigonal planar
4
Tetrahedral
4
Tetrahedral
4
Tetrahedral
120°
109.5°
109.5°
109.5°
2
Linear
3
Trigonal planar
4
Tetrahedral
3
Pyramidal
2
Bent
Example 11.18
ENHANCED EXAMPLE
Predict the molecular shape for these molecules: H2S, CCl4, AlCl3.
SOLUTION
1. Draw the Lewis structure.
2. Count the electron pairs around the central atom and determine the electron arrangement
that will minimize repulsions.
3. Determine the positions of the atoms and name the shape of the molecule.
Molecule
Lewis structure
Number of
electron pairs
Electron pair
arrangement
Molecular
shape
H2S
H S H
4
tetrahedral
bent
CCl4
Cl
Cl C Cl
Cl
4
tetrahedral
tetrahedral
AlCl3
Cl
Cl Al Cl
3
trigonal planar
trigonal planar
Practice 11.12
Predict the shape for CF4 , NF3 , and BeI2.
Review 239
C ha p t e r
11 review
11.1 Periodic Trends in Atomic Properties
• Metals and nonmetals
• Atomic radius:
• Increases down a group
• Decreases across a row
• Ionization energy:
• Energy required to remove an electron from an atom
• Decreases down a group
• Increases across a row
Key Term
ionization energy
11.2 Lewis Structures of Atoms
• A Lewis structure is a representation of the atom where the symbol represents the element and dots
around the symbol represent the valence electrons.
• To determine a Lewis structure for representative elements, use the group number as the number of
electrons to place around the symbol for the element.
Key Term
Lewis structure
11.3 The Ionic Bond: Transfer of Electrons
from One Atom to Another
• The goal of bonding is to achieve stability:
• For representative elements, this stability can be achieved by attaining a valence electron
­structure of a noble gas.
• In an ionic bond stability is attained by transferring an electron from one atom to another:
• The atom that loses an electron becomes a cation:
• Positive ions are smaller than their parent atoms.
• Metals tend to form cations.
• The atom gaining an electron becomes an anion:
• Negative ions are larger than their parent atoms.
• Nonmetals tend to form anions.
• Ionic compounds do not exist as molecules:
• Ions are attracted by multiple ions of the opposite charge to form a crystalline structure.
Key Term
ionic bond
11.4 Predicting Formulas of Ionic Compounds
• Chemical compounds are always electrically neutral.
• Metals lose electrons and nonmetals gain electrons to form compounds.
• Stability is achieved (for representative elements) by attaining a noble gas electron configuration.
11.5 The Covalent Bond: Sharing Electrons
• Covalent bonds are formed when two atoms share a pair of electrons between them:
• This is the predominant type of bonding in compounds.
• True molecules exist in covalent compounds.
• Overlap of orbitals forms a covalent bond.
• Unequal sharing of electrons results in a polar covalent bond.
Key Terms
covalent bond
polar covalent bond
11.6 Electronegativity
• Electronegativity is the attractive force an atom has for shared electrons in a molecule or polyatomic ion.
• Electrons spend more time closer to the more electronegative atom in a bond forming a polar bond.
• The polarity of a bond is determined by the electronegativity difference between the atoms involved
in the bond:
• The greater the difference, the more polar the bond is.
• At the extremes:
• Large differences result in ionic bonds.
• Tiny differences (or no difference) result(s) in a nonpolar covalent bond.
Key Terms
electronegativity
nonpolar covalent bond
dipole
240 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
• A molecule that is electrically asymmetrical has a dipole, resulting in charged areas within the
molecule.
+δ
��
−δ
��
H Cl
H :Cl
hydrogen chloride
• If the electronegativity difference between two bonded atoms is greater than 1.7–1.9, the bond
will be more ionic than covalent.
• Polar bonds do not always result in polar molecules.
11.7 Lewis Structures of Compounds
Problem-Solving Strategy: Writing a Lewis Structure
1. Obtain the total number of valence electrons to be used in the structure by adding the number of valence electrons in all the atoms in the molecule or ion. If you are writing the structure of an ion, add
one electron for each negative charge or subtract one electron for each positive charge on the ion.
2. Write the skeletal arrangement of the atoms and connect them with a single covalent bond (two
dots or one dash). Hydrogen, which contains only one bonding electron, can form only one covalent bond. Oxygen atoms are not normally bonded to each other, except in compounds known
to be peroxides. Oxygen atoms normally have a maximum of two covalent bonds (two single
bonds or one double bond).
3. Subtract two electrons for each single bond you used in Step 2 from the total number of electrons calculated in Step 1. This gives you the net number of electrons available for completing
the structure.
4. Distribute pairs of electrons (pairs of dots) around each atom (except hydrogen) to give each
atom a noble gas structure.
5. If there are not enough electrons to give these atoms eight electrons, change single bonds between atoms to double or triple bonds by shifting unbonded pairs of electrons as needed. Check
to see that each atom has a noble gas electron structure (two electrons for hydrogen and eight for
the others). A double bond counts as four electrons for each atom to which it is bonded.
11.8 Complex Lewis Structures
Key Term
resonance structure
• When a single unique Lewis structure cannot be drawn for a molecule, resonance structures (multiple Lewis structures) are used to represent the molecule.
11.9 Compounds Containing Polyatomic Ions
• Polyatomic ions behave like a single unit in many chemical reactions.
• The bonds within a polyatomic ion are covalent.
11.10 Molecular Shape
Key Terms
linear structure
trigonal planar structure
tetrahedral structure
bent structure
• Lewis structures do not indicate the shape of a molecule.
Problem-Solving Strategy: Determining Molecular Shape Using VSEPR
1. Draw the Lewis structure for the molecule.
2. Count the electron pairs around the central atom and arrange them to minimize repulsions (as far
apart as possible). This determines the electron pair arrangement.
3. Determine the positions of the atoms.
4. Name the molecular structure from the position of the atoms.
Review Questions
1. Rank these elements according to the radii of their atoms, from
smallest to largest: Na, Mg, Cl, K, and Rb. (Figure 11.2)
2. Explain why much more ionization energy is required to remove
the first electron from neon than from sodium. (Table 11.1)
3. Explain the large increase in ionization energy needed to remove
the third electron from beryllium compared with that needed for the
second electron. (Table 11.1)
Paired Exercises 241
4. Does the first ionization energy increase or decrease from top to
bottom in the periodic table for the alkali metal family? Explain.
(Figure 11.3)
5. Does the first ionization energy increase or decrease from top to
bottom in the periodic table for the noble gas family? Explain.
­(Figure 11.3)
6. Explain the reason that an atom of helium has a much higher first
ionization energy than does an atom of hydrogen. (Table 11.1)
7. Why is there such a large increase in the energy required to remove
a second electron from an atom of lithium? ( Table 11.1)
8. Which element has the larger radius?
(a) Li or Be
(d) F or Cl
(b) K or Rb
(e) O or Se
(c) Al or P
(f) As or Kr
9. In Groups 1A–7A, which element in each group has the smallest
atomic radius? (Figure 11.2)
10. Why does the atomic size increase in going down any family of the
periodic table?
11. Explain why potassium usually forms a K+ ion but not a K2+ ion.
12. Why does an aluminum ion have a + 3 charge?
13. Why are only valence electrons represented in a Lewis structure?
14. All the atoms within each Group A family of elements can be represented by the same Lewis structure. Complete the following table,
expressing the Lewis structure for each group. (Use E to represent
the elements.) (Figure 11.4)
Group
1A
2A
3A
4A
5A
6A
7A
ED
15. Draw the Lewis structure for Cs, Ba, Tl, Pb, Po, At, and Rn. How
do these structures correlate with the group in which each element
occurs?
16. What are valence electrons?
17. Why do metals tend to lose electrons and nonmetals tend to gain
electrons when forming ionic bonds?
18. State whether the elements in each group gain or lose electrons in
order to achieve a noble gas configuration. Explain.
(a) Group 1A
(c) Group 6A
(b) Group 2A
(d) Group 7A
19. What is the overall charge on an ionic compound?
20. Which family of elements tends to form ionic compounds with a
1 : 1 ratio of cation to anion with sulfur?
21. If the formula for calcium phosphate is Ca3(PO4)2, predict the formulas of magnesium phosphate, beryllium phosphate, strontium
phosphate, and barium phosphate.
22. Explain why magnesium tends to lose two electrons when forming
an ionic compound.
23. Why is the term molecule used to describe covalent compounds but
not ionic compounds?
24. How many shared electrons make up a single covalent bond? What
is the maximum number of covalent bonds that can be formed between any two atoms?
25. Explain in simple terms how a chemist describes the formation of
a covalent bond.
26. What does the dash between atoms represent in a Lewis electron
dot structure?
27. Are all molecules that contain polar bonds polar molecules?
­Explain.
28. In a polar covalent bond, how do you determine which atom has a
partial negative charge (d -) and which has a partial positive charge
(d +)?
29. In which general areas of the periodic table are the elements with
(a) the highest and (b) the lowest electronegativities located?
30. What is the purpose of a Lewis structure?
31. In a Lewis structure, what do the dots represent and what do the
lines represent?
32. Can there be more than one correct Lewis structure for a compound? Explain.
33. If a molecule has more than one correct Lewis structure, what is the
term used to describe these structures?
34. When drawing a Lewis structure for an ion, how is the charge
represented?
35. Write two examples of molecules that have both ionic and covalent
bonds.
36. What is the difference between electron pair arrangement and molecular shape?
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1. An atom of potassium and a K + ion are drawn below. Identify each
and explain your choice.
2. An atom of chlorine and a Cl - ion are drawn below. Identify each
and explain your choice.
1
1
2
2
3. Which one in each pair has the larger radius? Explain.
(a) a calcium atom or a calcium ion
(b) a chlorine atom or a chloride ion
(c) a magnesium ion or an aluminum ion
(d) a sodium atom or a silicon atom
(e) a potassium ion or a bromide ion
4. Which one in each pair has the larger radius? Explain.
(a) Fe2+ or Fe3+
(b) a potassium atom or a potassium ion
(c) a sodium ion or a chloride ion
(d) a strontium atom or an iodine atom
(e) a rubidium ion or a strontium ion
242 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
5. Using the table of electronegativity values (Table 11.5), indicate
which element is more positive and which is more negative in these
compounds:
(a) H2O
(d) PbS
(b) RbCl
(e) PF3
(c) NH3
(f) CH4
6. Using the table of electronegativity values (Table 11.5), indicate
which element is more positive and which is more negative in these
­compounds:
(a) HCl
(d) IBr
(b) SO2
(e) CsI
(c) CCl4
(f) OF2
7. Classify the bond between these pairs of elements as principally
ionic or principally covalent (use Table 11.5):
(a) sulfur and oxygen
(b) barium and nitrogen
(c) potassium and bromine
(d) carbon and chlorine
8. Classify the bond between these pairs of elements as principally
ionic or principally covalent (use Table 11.5):
(a) sodium and oxygen
(b) nitrogen and hydrogen
(c) oxygen and hydrogen
(d) phosphorus and chlorine
9. Write an equation representing each of the following:
(a) the change of a magnesium atom to a magnesium ion
(b) the change of a bromine atom to a bromide ion
10. Write an equation representing each of the following:
(a) the change of a potassium atom to a potassium ion
(b) the change of a sulfur atom to a sulfide ion
11. Use Lewis structures to show the electron transfer that enables
these ionic compounds to form:
(a) Li2O
(b) K3N
12. Use Lewis structures to show the electron transfer that enables
these ionic compounds to form:
(a) K2S
(b) Ca3N2
13. State the number of valence electrons in an atom of each of the
following elements:
(a) selenium
(d) magnesium
(b) phosphorus
(e) helium
(c) bromine
(f) arsenic
14. State the number of valence electrons in an atom of each of the
following elements:
(a) lead
(d) cesium
(b) lithium
(e) gallium
(c) oxygen
(f) argon
15. How many electrons must be gained or lost for the following to
achieve a noble gas electron configuration?
(a) a potassium atom
(c) a bromine atom
(b) an aluminum ion
(d) a selenium atom
16. How many electrons must be gained or lost for the following to
achieve a noble gas electron structure?
(a) a sulfur atom
(c) a nitrogen atom
(b) a calcium atom
(d) an iodide ion
17. Determine whether the following atoms will form an ionic
­compound or a molecular compound, and give the formula of the
compound.
(a) sodium and chlorine
(b) carbon and 4 hydrogen
(c) magnesium and bromine
(d) 2 bromine
(e) carbon and 2 oxygen
18. Determine whether each of the following atoms will form a nonpolar covalent compound or a polar covalent compound, and give the
formula of the compound.
(a) 2 oxygen
(b) hydrogen and bromine
(c) oxygen and 2 hydrogen
(d) 2 iodine
19. Let E be any representative element. Following the pattern in the
table, write formulas for the hydrogen and oxygen compounds of
the following:
(a) Na
(c) Al
(b) Ca
(d) Sn
20. Let E be any representative element. Following the pattern in the
table, write formulas for the hydrogen and oxygen compounds of
the following:
(a) Sb
(c) Cl
(b) Se
(d) C
Group
Group
1A
2A
3A
4A
5A
6A
7A
1A
2A
3A
4A
5A
6A
7A
EH
EH2
EH3
EH4
EH3
H2E
HE
EH
EH2
EH3
EH4
EH3
H2E
HE
E2O
EO
E2O3
EO2
E2O5
EO3
E2O7
E2O
EO
E2O3
EO2
E2O5
EO3
E2O7
21. The formula for sodium sulfate is Na2SO4. Write the names and
formulas for the other alkali metal sulfates.
22. The formula for calcium bromide is CaBr2. Write the names and
formulas for the other alkaline earth metal bromides.
23. Write Lewis structures for the following:
(a) Na
(b) Br-
24. Write Lewis structures for the following:
(a) Ga
(b) Ga3+
(c) O2-
(c) Ca2+
25. Classify the bonding in each compound as ionic or covalent:
(a) H2O
(c) MgO
(b) NaCl
(d) Br2
26. Classify the bonding in each compound as ionic or covalent:
(a) HCl
(c) NH3
(b) BaCl2
(d) SO2
27. Predict the type of bond that would be formed between the following pairs of atoms:
(a) P and I
(b) N and S
(c) Br and I
28. Predict the type of bond that would be formed between the
­following pairs of atoms:
(a) H and Si
(b) O and F
(c) Si and Br
Additional Exercises 243
29. Draw Lewis structures for the following:
(a) H2
(b) N2
(c) Cl2
30. Draw Lewis structures for the following:
(a) O2
(b) Br2
(c) I2
31. Draw Lewis structures for the following:
(a) NCl3
(c) C2H6
(b) H2CO3
(d) NaNO3
32. Draw Lewis structures for the following:
(a) H2S
(c) NH3
(b) CS2
(d) NH4Cl
33. Draw Lewis structures for the following:
(a) Ba2+
(d) CN(b) Al3+
(e) HCO32(c) SO3
34. Draw Lewis structures for the following:
(a) I(d) ClO3(b) S2(e) NO3(c) CO23
35. Classify each of the following molecules as polar or nonpolar:
(a) CH3Cl
(c) OF2
(b) Cl2
(d) PBr3
36. Classify each of the following molecules as polar or nonpolar:
(a) H2
(c) CH3OH
(b) NI3
(d) CS2
37. Give the number and arrangement of the electron pairs around the
central atom:
(a) C in CCl4
(c) Al in AlH3
(b) S in H2S
38. Give the number and arrangement of the electron pairs around the
central atom:
(a) Ga in GaCl3
(c) Cl in ClO3(b) N in NF3
39. Use VSEPR theory to predict the structure of these polyatomic
ions:
(a) sulfate ion
(c) periodate ion
(b) chlorate ion
40. Use VSEPR theory to predict the structure of these polyatomic
ions:
(a) ammonium ion
(c) phosphate ion
(b) sulfite ion
41. Use VSEPR theory to predict the shape of these molecules:
(a) SiH4
(b) PH3
(c) SeF2
42. Use VSEPR theory to predict the shape of these molecules:
(a) SiF4
(b) OF2
(c) Cl2O
43. Element X reacts with sodium to form the compound Na2X and is
in the second period on the periodic table. Identify this element.
44. Element Y reacts with oxygen to form the compound Y2O and has
the lowest ionization energy of any Period 4 element on the periodic table. Identify this element.
Additional Exercises
45. Atoms of sodium, cesium, and argon are drawn below. Identify
each and explain your choice.
1
2
3
2+
2+
46. Spheres representing Br , Se , Rb , Sr , and Kr are drawn
below. Identify each and explain your choice.
1
2
3
4
5
47. Write Lewis structures for hydrazine (N2H4) and hydrazoic acid
(HN3).
48. Draw Lewis structures and give the shape of each of the following
compounds:
(a) NO2(c) SOCl2
(b) SO2(d) Cl2O
4
49. Draw Lewis structures for each of the following compounds:
(a) ethane (C2H6)
(b) ethylene (C2H4)
(c) acetylene (C2H2)
50. Using the periodic table, identify each element given the
­description:
(a) the most electronegative element of the alkaline earth metals
(b) the noble gas with the highest first ionization energy
(c) the element in Period 4 with the lowest first ionization energy
(d) the element with the highest electronegativity
(e) the alkali metal with the largest radius
(f) the element in Period 2 with the smallest radius
51. Choose the element that fits each description:
(a) the higher electronegativity
Cl or Br
(b) the higher first ionization energy O or S
(c) the lower electronegativity
Ca or C
52. Why do you think there are no electronegativity values given for
the noble gases in Table 11.5?
53. When one electron is removed from an atom of Li, it has two
left. Helium atoms also have two electrons. Why is more energy
­required to remove the second electron from Li than to remove the
first from He?
54. Group 1B elements (see the periodic table on the inside cover of
your book) have one electron in their outer energy level, as do
Group 1A elements. Would you expect them to form compounds
such as CuCl, AgCl, and AuCl? Explain.
55. The formula for lead(II) bromide is PbBr2 . Predict formulas for
tin(II) and germanium(II) bromides.
56. Why is it not proper to speak of sodium chloride molecules?
57. What is a covalent bond? How does it differ from an ionic bond?
58. Briefly comment on the structure NaCOC
A Na for the compound
Na2O.
A
59. What are the four most electronegative elements?
60. Rank these elements from highest electronegativity to lowest:
Mg, S, F, H, O, Cs.
61. Is it possible for a molecule to be nonpolar even though it contains
polar covalent bonds? Explain.
62. Why is CO2 a nonpolar molecule, whereas CO is a polar molecule?
244 chapter 11
• Chemical Bonds: The Formation of ­Compounds from Atoms
63. Rank the following bonds in order of increasing polarity:
(a) NaO, CO, NO
(b) CO, SiO, GeO
(c) BO, BS, BSe
64. Estimate the bond angle between atoms in these molecules:
(a) H2S
(b) NH3
(c) NH4+
(d) SiCl4
65. Consider the two molecules BF3 and NF3. Compare and contrast
them in terms of the following:
(a) valence-level orbitals on the central atom that are used for
bonding
(b) shape of the molecule
(c) number of lone electron pairs on the central atom
(d) type and number of bonds found in the molecule
66. With respect to electronegativity, why is fluorine such an important
atom? What combination of atoms on the periodic table results in
the most ionic bond?
67. Why does the Lewis structure of each element in a given group of
representative elements on the periodic table have the same number
of dots?
68. A sample of an air pollutant composed of sulfur and oxygen was found
to contain 1.40 g sulfur and 2.10 g oxygen. What is the empirical formula for this compound? Draw a Lewis structure to represent it.
69. A dry-cleaning fluid composed of carbon and chlorine was found to
have the composition 14.5% carbon and 85.5% chlorine. Its known
molar mass is 166 g/mol. Draw a Lewis structure to represent the
compound.
70. Identify each of the following drawings as representing ionic compounds, covalent compounds, or compounds composed of both
ionic and covalent portions.
(a) (b) (c)
(d) 71. Identify each of the following compounds as ionic, covalent, or
both. Give the correct IUPAC name for each compound.
(a) Na3P
(c) SO2
(e) Cu(NO3)2
(b) NH4I
(d) H2S
(f) MgO
Challenge Exercises
72. Draw the following Lewis structures, showing any full charges
where they exist on ions:
(a) H2SO4
(d) HCN
(b) NaNO3
(e) Al2(SO4)3
(c) K2CO3
(f) CH3COOH
73. The first ionization energy for lithium is 520 kJ/mol. How much energy would be required to change 25 g of lithium atoms to lithium
ions? (Refer to Table 11.1.)
74. What is the total amount of energy required to remove the first two
electrons from 15 moles of sodium atoms? (Refer to Table 11.1.)
75. Sulfur trioxide gas reacts with water to form sulfuric acid. Draw the
Lewis structures of each substance to represent the reaction.
76. Compounds found in garlic have been shown to have significant
biological effects including immune system enhancement, anticancer activity, and cardiovascular activity. Garlic clearly seems to
have more value than as just a great tasting addition to spaghetti!
Some of the compounds in garlic that are currently being studied
are shown in the following figures.
O
H
HH
H
H
C
C O H
C
C
H
C
S
C
H
H
O
N
H
H
Alliin
H
H
H
C
H
C
C
H
O
H
C
S
S
C
H
H
C
H
H
Allicin (gives garlic its characteristic aroma)
Answers to Practice Exercises 245
Redraw these molecules adding in dots to represent any lone pairs
present on atoms. Draw a circle around atoms with tetrahedral
geometry, draw a square around atoms with trigonal planar geometry, draw a pentagon around atoms with a trigonal pyramidal
geometry, and draw a hexagon around atoms with a bent geometry. Draw a triangle around any atoms that have an unexpected
number of bonds.
77. Two of the most complex organic compounds found outside of
our solar system have been detected in a massive star-forming region near the heart of the Milky Way known as Sagittarius B2.
The skeleton structures of these two compounds, ethyl formate (the
c­ ompound that gives raspberries their flavor) and n-propyl cyanide,
are drawn below. Draw the complete Lewis electron dot structures
for these two compounds, identify the electron pair and molecular
geometry around each atom, and predict the bond angles.
H
H
H
C
C
H
O
O
C
H
H
H
Ethyl formate
H
H
H
C
C
C
H
H
H
C
H
n-propyl cyanide
Answers to Practice Exercises
11.1 (a)Ionization energy decreases from top to bottom for a group
of elements and increases from left to right across a period
of elements.
(b)For a group of elements the outer shell of electrons is increasingly further away from the nucleus as you go from
top to bottom of the group. Thus, they are not held as
strongly by the positive nucleus and are easier to ionize.
the greater the attraction for a pair of shared electrons. Consequently, when two different atoms with different electronegativities share a pair of electrons, one atom becomes partially
negative and the other atom becomes partially positive due to
the attractive forces for the pair of electrons. The bond formed
is said to be polar covalent.
11.8 ionic: (a), (d), (e)
11.2 (a) N
covalent: (b), (c), (f)
(b) Al
11.9 (a)
(c) Sr
(d) Br
11.3 (a)
(b)
(c)
(d)
K+
Mg2+
Al3+
Ba2+
Ar;
Ne;
Ne;
Xe;
or
Br
H
H
H
C
C
O
(d) H O C O
�
H
11.7 Electronegativity of an atom gives information about the
strength that the atom has for attracting a pair of shared electrons in a covalent bond. The higher the electronegativity value,
2�
O
H
Ca
H
�
�
11.11
�
O
2�
O
S
O
H H
H C C H
H H
H
(c) H N H
H
(b) H O H
H
H
(e) N N
O
(d) H C H
11.10 (a) H N H
H
11.5 (a) Mg3N2
(b) Ba3As2
or
(c) H F
H
(b) Cl C Cl
Cl
11.4 (a) Na2S
(b) Rb2O
11.6 H Br
Br
P Br
Br
Ca
N
O
O
�
O
2�
O
11.12 CF4, tetrahedral; NF3, pyramidal; BeI2, linear
N
O
O
PUTTING IT TOGETHER:
Review for Chapters 10–11
Answers for Putting It Together Reviews are found in Appendix VII.
Multiple Choice
Choose the correct answer to each of the following.
1. The concept of electrons existing in specific orbits around the nucleus was the contribution of
(a) Thomson
(c) Bohr
(b) Rutherford
(d) Schrödinger
2. The correct electron structure for a fluorine atom (F) is
(a) 1s22s22p5
(c) 1s22s22p43s1
(b) 1s22s22p23s23p1
(d) 1s22s22p3
3. The correct electron structure for 48Cd is
(a) 1s22s22 p63s23p64s23d10
(b) 1s22s22p63s23p64s23d104p65s24d10
(c) 1s22s22p63s23p64s23d104p64d4
(d) 1s22s22p63s23p64s24p64d105s25d10
5. Which of the following is the correct atomic structure for 48
22Ti?
22+
26n
22e–
22e–
22+
48n
(b)
26+
22n
(c)
22+
26n
(d)
22e–
7. The number of electrons in the third principal energy level in an
atom having the electron structure 1s22s22p63s23p2 is
(a) 2
(b) 4
(c) 6
(d) 8
8. The total number of orbitals that contain at least one electron in an
atom having the structure 1s22s22p63s23p2 is
(a) 5
(c) 14
(b) 8
(d) no correct answer given
9. Which of these elements has two s and six p electrons in its outer
energy level?
(a) He
(c) Ar
(b) O
(d) no correct answer given
(d) Ar
11. Which element has the largest number of unpaired electrons?
(a) F
(b) S
(c) Cu
(d) N
246
14. In moving down an A group on the periodic table, the number of
electrons in the outermost energy level
(a) increases regularly
(b) remains constant
(c) decreases regularly
(d) changes in an unpredictable manner
16. Elements of the noble gas family
(a) form no compounds at all
(b) have no valence electrons
(c) have an outer electron structure of ns2np6 (helium excepted),
where n is the period number
(d) no correct answer given
17. The lanthanide and actinide series of elements are
(a) representative elements
(b) transition elements
(c) filling in d-level electrons
(d) no correct answer given
18. The element having the structure 1s22s22p63s23p2 is in Group
(a) 2A
(b) 2B
(c) 4A
(d) 4B
19. In Group 5A, the element having the smallest atomic radius is
(a) Bi
(b) P
(c) As
(d) N
48e–
6. The number of orbitals in a d sublevel is
(a) 3
(c) 7
(b) 5
(d) no correct answer given
10. Which element is not a noble gas?
(b) Xe
(c) He
(a) Ra
13. Groups 3A–7A plus the noble gases form the area of the periodic
table where the electron sublevels being filled are
(a) p sublevels
(c) d sublevels
(b) s and p sublevels
(d) f sublevels
15. Which of the following is an incorrect formula?
(a) NaCl
(b) K2O
(c) AlO
(d) BaO
4. The correct electron structure of 23V is
(a) [Ar]4s23d3
(c) [Ar]4s24d3
2 3
(b) [Ar]4s 4p
(d) [Kr]4s23d3
(a)
12. How many unpaired electrons are in the electron structure of 24Cr,
[Ar]4s13d5?
(a) 2
(b) 4
(c) 5
(d) 6
20. In Group 4A, the most metallic element is
(a) C
(b) Si
(c) Ge
(d) Sn
21. Which group in the periodic table contains the least reactive
­elements?
(a) 1A
(b) 2A
(c) 3A
(d) noble gases
22. Which group in the periodic table contains the alkali metals?
(a) 1A
(b) 2A
(c) 3A
(d) 4A
23. An atom of fluorine is smaller than an atom of oxygen. One possible explanation is that, compared to oxygen, fluorine has
(a) a larger mass number
(b) a smaller atomic number
(c) a greater nuclear charge
(d) more unpaired electrons
24. If the size of the fluorine atom is compared to the size of the fluoride ion,
(a) they would both be the same size
(b) the atom is larger than the ion
(c) the ion is larger than the atom
(d) the size difference depends on the reaction
Putting It Together 247
25. Sodium is a very active metal because
(a) it has a low ionization energy
(b) it has only one outermost electron
(c) it has a relatively small atomic mass
(d) all of the above
26. Which of the following formulas is not correct?
(a) Na+
(b) S(c) Al3+
(d) F27. Which of the following molecules does not have a polar covalent
bond?
(a) CH4
(b) H2O
(c) CH3OH
(d) Cl2
28. Which of the following molecules is a dipole?
(a) HBr
(b) CH4
(c) H2
(d) CO2
29. Which of the following has bonding that is ionic?
(a) H2
(b) MgF2
(c) H2O
(d) CH4
30. Which of the following is a correct Lewis structure?
(a) O C O
(c) Cl Cl
Cl
(b) Cl C Cl
(d) N N
Cl
31. Which of the following is an incorrect Lewis structure?
H
(c) H C H
(a) H N H
H
(d) N N
(b) O H
H
32. The correct Lewis structure for SO2 is
(a) O S O
(c) O S O
(b) O S O
(d) O S O
33. Carbon dioxide (CO2) is a nonpolar molecule because
(a) oxygen is more electronegative than carbon
(b) the two oxygen atoms are bonded to the carbon atom
(c) the molecule has a linear structure with the carbon atom in the
middle
(d) the carbon–oxygen bonds are polar covalent
34. When a magnesium atom participates in a chemical reaction, it is
most likely to
(a) lose 1 electron
(c) lose 2 electrons
(b) gain 1 electron
(d) gain 2 electrons
35. If X represents an element of Group 3A, what is the general formula
for its oxide?
(a) X3O4
(b) X3O2
(c) XO
(d) X2O3
36. Which of the following has the same electron structure as an argon
atom?
(a) Ca2+
(b) Cl0
(c) Na+
(d) K0
37. As the difference in electronegativity between two elements decreases, the tendency for the elements to form a covalent bond
(a) increases
(b) decreases
(c) remains the same
(d) sometimes increases and sometimes decreases
38. Which compound forms a tetrahedral molecule?
(a) NaCl
(b) CO2
(c) CH4
(d) MgCl2
39. Which compound has a bent (V-shaped) molecular structure?
(a) NaCl
(b) CO2
(c) CH4
(d) H2O
40. Which compound has double bonds within its molecular structure?
(a) NaCl
(b) CO2
(c) CH4
(d) H2O
41. The total number of valence electrons in a nitrate ion, NO3- , is
(a) 12
(b) 18
(c) 23
(d) 24
42. The number of electrons in a triple bond is
(a) 3
(b) 4
(c) 6
(d) 8
43. The number of unbonded pairs of electrons in H2O is
(a) 0
(b) 1
(c) 2
(d) 4
44. Which of the following does not have a noble gas electron structure?
(a) Na
(b) Sc3+
(c) Ar
(d) O2-
Free Response Questions
Answer each of the following. Be sure to include your work and explanations in a clear, logical form.
1. An alkaline earth metal, M, combines with a halide, X. Will the
resulting compound be ionic or covalent? Why? What is the Lewis
structure for the compound?
2. “All electrons in atoms with even atomic numbers are paired.” Is this
statement true or false? Explain your answer using an example.
3. Discuss whether the following statement is true or false: “All nonmetals have two valence electrons in an s sublevel with the exception of the noble gases, which have at least one unpaired electron in
a p sublevel.”
4. The first ionization energy (IE) of potassium is lower than the first
IE for calcium, but the second IE of calcium is lower than the second
IE of potassium. Use an electron configuration or size argument to
explain this trend in ionization energies.
5. Chlorine has a very large first ionization energy, yet it forms a chloride ion relatively easily. Explain.
6. Three particles have the same electron configuration. One is a cation
of an alkali metal, one is an anion of the halide in the third period,
and the third particle is an atom of a noble gas. What are the identities of the three particles (including charges)? Which particle should
have the smallest atomic/ionic radius, which should have the largest,
and why?
7. Why is the Lewis structure of AlF3 not written as
F
F
Al
F
What is the correct Lewis structure, and which electrons are shown
in a Lewis structure?
8. Why does carbon have a maximum of four covalent bonds?
9. Both NCl3 and BF3 have a central atom bonded to three other atoms,
yet one is pyramidal and the other is trigonal planar. ­Explain.
10. Draw the Lewis structure of the atom whose electron configuration
is 1s22s22p63s23p64s23d104p5. Would you expect this atom to form
an ionic, nonpolar covalent, or polar covalent bond with sulfur?
C h a p te r
12
The Gaseous State
of Matter
O
ur atmosphere is composed of a mixture of gases,
including nitrogen, oxygen, argon, carbon dioxide,
ozone, and trace amounts of others. These gases
are essential to life, yet they can also create hazards for us. For
example, carbon dioxide is valuable when it is taken in by plants
and converted to carbohydrates, but it is also associated with
Chapter Outline
12.1 Properties of Gases
12.2 Boyle’s Law
12.3 Charles’ Law
12.4 Avogadro’s Law
12.5 Combined Gas Laws
12.6 Ideal Gas Law
12.7 Dalton’s Law of Partial Pressures
the potentially hazardous greenhouse effect. Ozone surrounds
12.8 Density of Gases
the Earth at high altitudes and protects us from harmful ultravio-
12.9 Gas Stoichiometry
let rays, but it also destroys rubber and plastics. We require air
to live, yet scuba divers must be concerned about oxygen poisoning and the “bends.”
In chemistry, the study of the behavior of gases allows us to
understand our atmosphere and the effects that gases have on
our lives.
Kees Van DenBerg/Photo Researchers, Inc.
The air in a hot air balloon
expands when it is heated. Some
of the air escapes from the top
of the balloon, lowering the air
density inside the balloon making
the balloon buoyant.
12.1
• Properties of Gases 249
12.1 Properties of Gases
Gases are the least dense and most mobile of the three states of matter (see Figure 12.1). A
solid has a rigid structure, and its particles remain in essentially fixed positions. When a solid
absorbs sufficient heat, it melts and changes into a liquid. Melting occurs because the molecules
(or ions) have absorbed enough energy to break out of the rigid structure of the solid. The molecules or ions in the liquid are more energetic than they were in the solid, as indicated by their
increased mobility. Molecules in the liquid state cling to one another. When the liquid absorbs
additional heat, the more energetic molecules break away from the liquid surface and go into
the gaseous state—the most mobile state of matter. Gas molecules move at very high velocities
and have high kinetic energy. The average velocity of hydrogen molecules at 0°C is over 1600
meters (1 mile) per second. Mixtures of gases are uniformly distributed within the container in
which they are confined.
The same quantity of a substance occupies a much greater volume as a gas than it does
as a liquid or a solid (see Figure 12.2). For example, 1 mol of water (18.02 g) has a volume
of 18 mL at 4°C. This same amount of liquid water would occupy about 22,400 mL in the
gaseous state—more than a 1200-fold increase in volume. We may assume from this difference
in volume that (1) gas molecules are relatively far apart, (2) gases can be greatly compressed,
and (3) the volume occupied by a gas is mostly empty space.
key terms
© CanBalcioglu/
iStockphoto
Gas (Steam)
(c)
Figure 12.1
The three states of matter. (a) Solid—water molecules are held together rigidly and
are very close to each other. (b) Liquid—water molecules are close together but are
free to move around and slide over each other. (c) Gas—water molecules are far
apart and move freely and randomly.
Tom Pantages
Liquid (Water)
(b)
pressure
atmospheric pressure
barometer
1 atmosphere
Figure 12.2
A mole of water occupies 18 mL
as a liquid but would fill this box
(22.4 L) as a gas at the same
­temperature.
© Jon Helgason/iStockphoto
Solid (Ice)
(a)
© Carlos Alvarez/
iStockphoto
Learning objectiv e
© Diane Diederich/
iStockphoto
Explain atmospheric pressure and how it is measured. Be able to convert among the
various units of pressure.
Measuring the Pressure of a Gas
Pressure is defined as force per unit area. When a rubber balloon is inflated with air,
it stretches and maintains its larger size because the pressure on the inside is greater
than that on the outside. Pressure results from the collisions of gas molecules with the
walls of the balloon (see Figure 12.3). When the gas is released, the force or pressure
of the air escaping from the small neck propels the balloon in a rapid, irregular flight. If
the balloon is inflated until it bursts, the gas escaping all at once causes an explosive noise.
The effects of pressure are also observed in the mixture of gases in the atmosphere,
which is composed of about 78% nitrogen, 21% oxygen, 1% argon, and other minor constituents by volume (see Table 12.1). The outer boundary of the atmosphere is not known
precisely, but more than 99% of the atmosphere is below an altitude of 32 km (20 miles). Thus,
the concentration of gas molecules in the atmosphere decreases with altitude, and at about
Figure 12.3
The pressure resulting from the
collisions of gas molecules with
the walls of the balloon keeps the
balloon inflated.
250 chapter 12
• The Gaseous State of Matter
Table 12.1
N2
78.1%
O2
20.9%
Others 1%
Table 12.2 Pressure Units
Equivalent to 1 Atmosphere
1 atm
760 torr
760 mm Hg
76 cm Hg
101.325 kPa
1013 mbar
29.9 in. Hg
14.7 lb>in.2
Average Composition of Dry Air
Gas
Percent by volume
N2
O2
Ar
CO2
Ne
78.08
20.95
0.93
0.033
0.0018
Gas
Percent by volume
He
CH4
Kr
Xe, H2 , and N2O
0.0005
0.0002
0.0001
Trace
6.4 km (4 miles) the amount of oxygen is insufficient to sustain human life. The gases in the
atmosphere exert a pressure known as atmospheric pressure. The pressure exerted by a gas
depends on the number of molecules of gas present, the temperature, and the volume in which
the gas is confined. Gravitational forces hold the atmosphere relatively close to Earth and
prevent air molecules from flying off into space. Thus, the atmospheric pressure at any point
is due to the mass of the atmosphere pressing downward at that point.
The pressure of the gases in the atmosphere can be measured with a barometer. A mercury barometer may be prepared by completely filling a long tube with pure, dry mercury and
inverting the open end into an open dish of mercury. If the tube is longer than 760 mm, the
mercury level will drop to a point at which the column of mercury in the tube is just supported
by the pressure of the atmosphere. If the tube is properly prepared, a vacuum will exist above
the mercury column. The weight of mercury, per unit area, is equal to the pressure of the
atmosphere. The column of mercury is supported by the pressure of the atmosphere, and the
height of the column is a measure of this pressure (see Figure 12.4). The mercury barometer
was invented in 1643 by the Italian physicist E. Torricelli (1608–1647), for whom the unit
of pressure torr was named.
Air pressure is measured and expressed in many units. The standard atmospheric pressure,
or simply 1 atmosphere (atm), is the pressure exerted by a column of mercury 760 mm high
at a temperature of 0°C. The normal pressure of the atmosphere at sea level is 1 atm, or 760
torr, or 760 mm Hg. The SI unit for pressure is the pascal (Pa), where 1 atm = 101,325 Pa, or
101.3 kPa. Other units for expressing pressure are inches of mercury, centimeters of mercury,
the millibar (mbar), and pounds per square inch (lb>in.2 or psi). The values of these units
equivalent to 1 atm are summarized in Table 12.2.
Atmospheric pressure varies with altitude. The average pressure at Denver, Colorado,
1.61 km (1 mile) above sea level, is 630 torr (0.83 atm). Atmospheric pressure is 380 torr
(0.5 atm) at about 5.5 km (3.4 miles) altitude.
Pressure is often measured by reading the height of a mercury column in millimeters on
a barometer. Thus pressure may be recorded as mm Hg (torr). In problems dealing with gases,
Figure 12.4
Preparation of a mercury
barometer. The full tube of
mercury at the left is inverted
and placed in a dish of mercury.
Vacuum
Hg
Hg
760 mm (height of Hg column
supported by atmospheric
pressure at sea level)
Atmospheric
pressure
Hg
12.1
• Properties of Gases 251
it is necessary to make conversions among the various pressure units. Since atm, torr, and mm
Hg are common pressure units, we give examples involving all three of these units:
1 atm = 760 torr = 760 mm Hg
example 12.1
The average atmospheric pressure at Walnut, California, is 740. mm Hg. Calculate this
pressure in (a) torr and (b) atmospheres.
ENHANCED EXAMPLE
Solution
Let’s use conversion factors that relate one unit of pressure to another.
(a) To convert mm Hg to torr, use the conversion factor
760 torr>760 mm Hg (1 torr>1 mm Hg):
(740. mm Hg)a
1 torr
b = 740. torr
1 mm Hg
(b) To convert mm Hg to atm, use the conversion factor
1 atm>760. mm Hg:
(740. mm Hg)a
1 atm
b = 0.974 atm
760. mm Hg
Practice 12.1
A barometer reads 1.12 atm. Calculate the corresponding pressure in (a) torr, (b) mm Hg,
and (c) kilopascals.
Pressure Dependence on the Number of Molecules
and the Temperature
Pressure is produced by gas molecules colliding with the walls of a container. At a specific
temperature and volume, the number of collisions depends on the number of gas molecules
present. The number of collisions can be increased by increasing the number of gas molecules
present. If we double the number of molecules, the frequency of collisions and the pressure
should double. We find, for an ideal gas, that this doubling is actually what happens. When the
temperature and mass are kept constant, the pressure is directly proportional to the number of
moles or molecules of gas present. Figure 12.5 illustrates this concept.
A good example of this molecule–pressure relationship may be observed in an ordinary
cylinder of compressed gas equipped with a pressure gauge. When the valve is opened, gas
escapes from the cylinder. The volume of the cylinder is constant, and the decrease in quantity
(moles) of gas is registered by a drop in pressure indicated on the gauge.
Figure 12.5
The pressure exerted by a gas
is directly proportional to the
number of molecules present.
In each case shown, the volume
is 22.4 L and the temperature
is 0°C.
2 mol H 2
P = 2 atm
1 mol H 2
P = 1 atm
0.5 mol H 2
P = 0.5 atm
252 chapter 12
• The Gaseous State of Matter
>Chemistry in action
What the Nose Knows
your shoes. If researchers at the University of Illinois
are successful, you may no longer have to remove your
shoes at the airport. A handheld sensor is able to quickly
and ­accurately detect the presence of a commonly used
explosive in shoes. Now as passengers walk through security, chemical sensors will look for triacetone triperoxide
or TATP vapors emanating from their shoes. This technology
holds the promise that other harmful compounds may
also be detected in this way.
The sense of smell is one of the most intriguing senses
of animals. Through the sense of smell dogs can detect
the presence of many different drugs, explosives, and
other substances. In some cases they may even be able to
detect certain human diseases. The air around us is filled
with traces of the chemicals composing our world. Much
work has gone into finding ways to identify the chemicals
present in the air. Developing methods to identify chemicals at low concentrations will have tremendous benefits.
The doctor nose you’re sick! Many human
diseases produce volatile compounds that are exhaled
with every breath you take. Physicians can often diagnose
diabetes simply by sniffing the breath of their patients for
the odor of acetone. While most diseases do not produce
easily detectable odors, physicians can take advantage
of technology to detect these compounds at the low
concentrations in which they are produced. For example,
lung cancer patients emit specific metabolic by-products
in their breath that can be detected by dogs. Researchers
are actively looking for techniques to detect these same
compounds so that physicians can have a quick, easy
screening tool for lung cancer.
Better Coffee? Kenneth Suslick at the University
of Illinois, Urbana-Champaign, and his colleagues have
developed an electronic nose. Molecules placed in polymer films change color when exposed to the aromatic
compounds found in coffee. By comparing the relative
amounts of each of these compounds to the amounts
found in an exceptional cup of coffee, coffee roasters are
able to easily identify problems with their coffee beans
and thereby ensure a consistently good cup of coffee.
No more stinky feet? If you have been flying
recently, you have probably been required to remove
3.06 atm
2.24 atm
100°C
0°C
Figure 12.6
The pressure of a gas in a fixed
volume increases with increasing
temperature. The increased
pressure is due to more frequent
and more energetic collisions of
the gas molecules with the walls
of the container at the higher
temperature.
Volume = 1 liter
0.1 mole gas
P = 2.24 atm
Volume = 1 liter
0.1 mole gas
P = 3.06 atm
The pressure of a gas in a fixed volume also varies with temperature. When the temperature is increased, the kinetic energy of the molecules increases, causing more frequent and
more energetic collisions of the molecules with the walls of the container. This increase in
collision frequency and energy results in a pressure increase (see Figure 12.6).
12.2 Boyle’s Law
Learning objective
Use Boyle’s law to calculate changes in pressure or volume of a sample of gas at a
constant temperature.
key term
Through a series of experiments, Robert Boyle (1627–1691) determined the relationship between the pressure (P) and volume (V) of a particular quantity of a gas. This relationship of P
and V is known as Boyle’s law:
Boyle’s law
12.2
8
At constant temperature (T ), the volume (V ) of a fixed mass of a gas is inversely proportional
to the pressure (P), which may be expressed as
1
P
6
or
P1 V1 = P2 V2
This law says that the volume varies () inversely with the pressure at constant mass and
temperature. When the pressure on a gas is increased, its volume will decrease, and vice versa.
The inverse relationship of pressure and volume is graphed in Figure 12.7.
When Boyle doubled the pressure on a specific quantity of a gas, keeping the temperature
constant, the volume was reduced to one-half the original volume; when he tripled the pressure on the system, the new volume was one-third the original volume; and so on. His work
showed that the product of volume and pressure is constant if the temperature is not changed:
PV = constant
PV = k (mass and temperature are constant)
or
7
Let’s demonstrate this law using a cylinder with a movable piston so that the volume of
gas inside the cylinder may be varied by changing the external pressure (see Figure 12.8).
Assume that the temperature and the number of gas molecules do not change. We start with
a volume of 1000 mL and a pressure of 1 atm. When we change the pressure to 2 atm, the
gas molecules are crowded closer together, and the volume is reduced to 500 mL. When we
increase the pressure to 4 atm, the volume becomes 250 mL.
Note that the product of the pressure times the volume is the same number in each case,
verifying Boyle’s law. We may then say that
Pressure (atm)
V 
• Boyle’s Law 253
5
4
3
2
1
0
0
1
2
3 4 5 6
Volume (liters)
Figure 12.7
Graph of pressure versus
volume showing the inverse PV
relationship of an ideal gas.
P1V1 = P2V2
where P1V1 is the pressure–volume product at one set of conditions, and P2V2 is the product at
another set of conditions. In each case, the new volume may be calculated by multiplying the
starting volume by a ratio of the two pressures involved.
V2 = V1 a
P1
b
P2
Of course, the ratio of pressures used must reflect the direction in which the volume should
change. When the pressure is changed from 1 atm to 2 atm, the ratio to be used is 1atm>2atm.
Now we can verify the volumes shown in Figure 12.8:
1. Starting volume, 1000 mL; pressure change, 1 atm h 2 atm
P1
P2
(1000 mL) a
P = 1 atm
1 atm
b = 500 mL
2 atm
P = 2 atm
P = 4 atm
Figure 12.8
The effect of pressure on
the volume of a gas.
V = 1000 mL
PV = 1 atm x 1000 mL
=
=
V = 500 mL
2 atm x 500 mL
=
=
V = 250 mL
4 atm x 250 mL
7
8
254 chapter 12
• The Gaseous State of Matter
2. Starting volume, 1000 mL; pressure change, 1 atm h 4 atm
P1
P2
(1000 mL) a
1 atm
b = 250 mL
4 atm
3. Starting volume, 500 mL; pressure change, 2 atm h 4 atm
P1
P2
(500 mL) a
2 atm
b = 250 mL
4 atm
In summary, a change in the volume of a gas due to a change in pressure can be calculated
by multiplying the original volume by a ratio of the two pressures. If the pressure is increased,
the ratio should have the smaller pressure in the numerator and the larger pressure in the denominator. If the pressure is decreased, the larger pressure should be in the numerator and the
smaller pressure in the denominator:
P1
new volume = original volume * ratio of pressures = V1 a b
P2
We use Boyle’s law in the following examples. If no mention is made of temperature, assume
that it remains constant.
ENHANCED EXAMPLE
example 12.2
What volume will 2.50 L of a gas occupy if the pressure is changed from
760. mm Hg to 630. mm Hg?
Solution
Read •
Knowns: V1 = 2.50 L
P1 = 760. mm Hg P2 = 630. mm Hg
Solving for: new volume (V2)
Plan •
Since the problem is concerned only with pressure and volume, Boyle’s law
can be used: P1V1 = P2V2
Solution map: pressure decreases S volume increases
P1V1 = P2V2 solving for V2 gives V2 =
Calculate • V2 =
P1V1
P2
(2.50 L)(760. mm Hg)
= 3.20 L
630. mm Hg
Another way to think about this is to use a ratio of pressures that will result
in an increase in volume.
V2 = (2.50 L) a
760. mm Hg
b = 3.20 L
630. mm Hg
Note this produces the same equation as we used above to solve the problem.
Check •
The final volume is larger than the original as predicted by Boyle’s law and
the solution map.
12.2
example 12.3
A given mass of hydrogen occupies 40.0 L at 700. torr. What volume will it occupy at
5.00 atm pressure?
Solution
Read •
Knowns: P1 = 700. torr = 700. torr a
P2 = 5.00 atm
1 atm
b = 0.921 atm
760 torr
V1 = 40.0 L
Solving for: new volume (V2)
Plan •
Since the problem is concerned only with pressure and volume, Boyle’s law
can be used: P1V1 = P2V2
Solution map: pressure increases S volume decreases
P1V1 = P2V2 solving for V2 gives V2 =
Calculate • V2 =
P1V1
P2
(0.921 atm) (40.0 L)
= 7.37 L
5.00 atm
We can also use a ratio of pressures to solve this problem.
Check •
V2 = (40.0 L) a
0.921 atm
b = 7.37 L
5.00 atm
The final volume is smaller than the original as predicted by Boyle’s law and
the solution map.
example 12.4
A gas occupies a volume of 200. mL at 400. torr pressure. To what pressure must the gas
be subjected in order to change the volume to 75.0 mL?
Solution
Read •
Knowns:
P1 = 400. torr
V1 = 200. mL
V2 = 75.0 mL
Solving for: new pressure (P2)
Plan •
Since the problem is concerned only with pressure and volume, Boyle’s law
can be used: P1V1 = P2V2
Solution map: volume decreases S pressure increases
Since P1V1 = P2V2 solving for P2 gives P2 =
Calculate • P2 =
P1V1
V2
(400. torr) (200. mL)
= 1.07 * 103 torr
75.0 mL
We can also think about this by using a ratio of volumes that will result in
an increase in pressure.
V2 = (400. torr) a
200. mL
b = 1.07 * 103 torr
75.0 mL
This result is the same as the one used above to solve the problem.
Check •
The final pressure is larger than the original as predicted by Boyle’s law and
the solution map.
• Boyle’s Law 255
256 chapter 12
• The Gaseous State of Matter
Practice 12.2
A gas occupies a volume of 3.86 L at 0.750 atm. At what pressure will the volume be
4.86 L?
12.3 Charles’ Law
Learning objective
Use Charles’ law to calculate changes in temperature or volume of a sample of gas
at constant pressure.
key terms
The effect of temperature on the volume of a gas was observed in about 1787 by the French
physicist J. A. C. Charles (1746–1823). Charles found that various gases expanded by the same
fractional amount when they underwent the same change in temperature. Later it was found
that if a given volume of any gas initially at 0°C was cooled by 1°C, the volume decreased
20
1
2
by 273
; if cooled by 2°C, it decreased by 273
; if cooled by 20°C, by 273
; and so on. Since each
1
degree of cooling reduced the volume by 273 , it was apparent that any quantity of any gas
would have zero volume if it could be cooled to -273C. Of course, no real gas can be cooled
to -273C for the simple reason that it would liquefy before that temperature is reached. However, -273C (more precisely, -273.15C) is referred to as absolute zero; this temperature
is the zero point on the Kelvin (absolute) temperature scale—the temperature at which the
volume of an ideal, or perfect, gas would become zero.
The volume–temperature relationship for methane is shown graphically in Figure 12.9.
Experimental data show the graph to be a straight line that, when extrapolated, crosses the
temperature axis at -273.15C, or absolute zero. This is characteristic for all gases.
absolute zero
Charles’ law
5
Figure 12.9
4
Volume (L)
Volume–temperature
relationship of methane
(CH4). Extrapolated portion
of the graph is shown by
the broken line.
CH4
3
2
–273°C
1
0
–300
–200
–100
0
100
200
Temperature (°C)
300
400
In modern form, Charles’ law is as follows:
At constant pressure the volume of a fixed mass of any gas is directly proportional to the
absolute temperature, which may be expressed as
V  T or A capital T is usually used for
absolute temperature (K) and a
small t for °C.
V1
V2
=
T1
T2
Mathematically, this states that the volume of a gas varies directly with the absolute temperature when the pressure remains constant. In equation form, Charles’ law may be written as
V = kT or V
= k (at constant pressure)
T
where k is a constant for a fixed mass of the gas. If the absolute temperature of a gas is doubled,
the volume will double.
To illustrate, let’s return to the gas cylinder with the movable or free-floating piston
(see Figure 12.10). Assume that the cylinder labeled (a) contains a quantity of gas and the
pressure on it is 1 atm. When the gas is heated, the molecules move faster, and their kinetic
12.3
T1
P = 1 atm
• Charles’ Law 257
T2
P = 1 atm
Increase in volume
due to increased
temperature
Free-floating
piston
Figure 12.10
The effect of temperature on the
volume of a gas. The gas in cylinder
(a) is heated from T1 to T2. With the
external pressure constant at
1 atm, the free-floating piston rises,
resulting in an increased volume,
shown in cylinder (b).
(a)
(b)
energy increases. This action should increase the number of collisions per unit of time and
therefore increase the pressure. However, the increased internal pressure will cause the piston
to rise to a level at which the internal and external pressures again equal 1 atm, as we see in
cylinder (b). The net result is an increase in volume due to an increase in temperature.
Another equation relating the volume of a gas at two different temperatures is
V1
V2
=
(constant P)
T1
T2
Charles D. Winters/Photo
Researchers, Inc.
where V1 and T1 are one set of conditions and V2 and T2 are another set of conditions.
A simple experiment showing the variation of the volume of a gas with temperature is illustrated in Figure 12.11. A balloon is placed in a beaker, and liquid N2 is poured over it. The
volume is reduced, as shown by the collapse of the balloon; when the balloon is removed from the
liquid N2, the gas expands as it warms back to room temperature and the balloon increases in size.
(a)
(b)
(c)
Figure 12.11
The air-filled balloons (a) are being placed into the beaker containing liquid nitrogen. Notice
that the blue and red balloons on the left in (a) have been placed in the beaker in (b) and are
much smaller. In (c) all of the balloons now fit inside the beaker.
example 12.5
Three liters of hydrogen at -20.°C are allowed to warm to a room temperature of 27°C.
What is the volume at room temperature if the pressure remains constant?
Solution
Read •
Knowns: V1 = 3.00 L T1 = -20.°C + 273 = 253 K
T2 = 27°C + 273 = 300. K
Solving for: new volume (V2)
ENHANCED EXAMPLE
Remember: Temperature must
be changed to Kelvin in gas law
problems. Note that we use 273
to convert instead of 273.15 since
our original measurements are to
the nearest degree.
258 chapter 12
• The Gaseous State of Matter
Plan •
Since the problem is concerned only with temperature and volume, Charles’
V1
V2
law can be used:
=
T1
T2
Solution map: temperature increases S volume increases
Since
Calculate • V2 =
V1
V2
= ,
T1
T2
V2 =
V1T2
T1
(3.00 L) (300 K)
= 3.56 L
253 K
Note the solution map also produces the same result by using a ratio of
temperatures that will result in an increase in volume.
300. K
b = 3.56 L
V2 = (3.00 L) a
253 K
Check •
The final volume is larger than the original as predicted by Charles’ law and
the solution map.
example 12.6
If 20.0 L of oxygen are cooled from 100.°C to 0.°C, what is the new volume?
Solution
Read •
Since no mention is made of pressure, we assume the pressure does not
change. Remember that temperature must be in kelvins.
Knowns:
V1 = 20.0 L
T1 = 100.°C = 373 K
T2 = 0.°C = 273 K
Solving for: new volume (V2)
Plan •
Since the problem is concerned only with temperature and volume, Charles’
V1
V2
law can be used:
=
T1
T2
Solution map: temperature decreases S volume decreases
Since
Calculate • V2 =
V1
V2
=
,
T1
T2
V2 =
V1T2
T1
(20.0 L) (273 K)
= 14.6 L
373 K
Or we can use the solution map to multiply the original volume by a ratio of
temperatures that will result in a decrease in volume.
Check •
V2 = (20.0 L) a
273 K
b = 14.6 L
373 K
The final volume is smaller than the original as predicted by Charles’ law and
the solution map.
Practice 12.3
A 4.50-L container of nitrogen gas at 28.0°C is heated to 56.0°C. Assuming that the
­volume of the container can vary, what is the new volume of the gas?
12.4
• Avogadro’s Law 259
12.4 Avogadro’s Law
Solve problems using the relationship among moles, mass, and volume of gases.
Learning objectiv e
Early in the nineteenth century, a French chemist, J.L. Gay-Lussac (1778–1850) studied the
volume relationships of reacting gases. His results, published in 1809, were summarized in a
statement known as Gay-Lussac’s law of combining volumes:
key terms
Gay-Lussac’s law of combining
volumes
Avogadro’s law
When measured at the same temperature and pressure, the ratios of the volumes of reacting
gases are small whole numbers.
Thus, H2 and O2 combine to form water vapor in a volume ratio of 2:1 (Figure 12.12); H2 and
Cl2 react to form HCl in a volume ratio of 1:1 and H2 , and N2 react to form NH3 in a volume
ratio of 3:1.
+
2 H 2 (g)
H2
H2
+
2 volumes
O 2 (g)
2 H 2 O(g)
O2
H 2O
1 volume
H 2O
2 volumes
Figure 12.12
Gay-Lussac’s law of combining volumes of
gases applied to the reaction of hydrogen
and oxygen. When measured at the same
temperature and pressure, hydrogen and
oxygen react in a volume ratio of 2:1.
Two years later, in 1811, Amedeo Avogadro (1776–1856) used the law of combining
volumes of gases to make a simple but significant and far-reaching generalization concerning
gases. Avogadro’s law states:
Equal volumes of different gases at the same temperature and pressure contain the same
number of molecules.
This law was a real breakthrough in understanding the nature of gases.
1. It offered a rational explanation of Gay-Lussac’s law of combining volumes of gases and
indicated the diatomic nature of such elemental gases as hydrogen, chlorine, and oxygen.
2. It provided a method for determining the molar masses of gases and for comparing the
densities of gases of known molar mass (see Section 12.8).
3. It afforded a firm foundation for the development of the kinetic-molecular theory.
By Avogadro’s law, equal volumes of hydrogen and chlorine at the same temperature and
pressure contain the same number of molecules. On a volume basis, hydrogen and chlorine
react thus (see Figure 12.13):
hydrogen + chlorine ¡ hydrogen chloride
1 volume
1 volume
2 volumes
Therefore, hydrogen molecules react with chlorine molecules in a 1:1 ratio. Since two volumes
of hydrogen chloride are produced, one molecule of hydrogen and one molecule of chlorine
must produce two molecules of hydrogen chloride. Therefore, each hydrogen molecule and
Figure 12.13
Avogadro’s law proved the
­concept of diatomic molecules
for hydrogen and chlorine.
+
H2
+
¡
Cl2
¡
2 HCl
260 chapter 12
• The Gaseous State of Matter
each chlorine molecule must consist of two atoms. The coefficients of the balanced equation for
the reaction give the correct ratios for volumes, molecules, and moles of reactants and products:
H2 1 volume
1 molecule
1 mol
+ Cl2 ¡ 2HCl
1 volume
1 molecule
1 mol
2 volumes
2 molecules
2 mol
By like reasoning, oxygen molecules also must contain at least two atoms because
one volume of oxygen reacts with two volumes of hydrogen to produce two volumes of
water vapor.
The volume of a gas depends on the temperature, the pressure, and the number of gas
molecules. Different gases at the same temperature have the same average kinetic energy.
Hence, if two different gases are at the same temperature, occupy equal volumes, and exhibit
equal pressures, each gas must contain the same number of molecules. This statement is true
because systems with identical PVT properties can be produced only by equal numbers of
molecules having the same average kinetic energy.
12.5 Combined Gas Laws
Learning obje ctiv e
Use the Combined gas law to calculate changes in pressure temperature, or volume
of a sample of gas.
key terms
To compare volumes of gases, common reference points of temperature and pressure were
selected and called standard conditions or standard temperature and pressure (abbreviated STP). Standard temperature is 273.15 K (0°C), and standard pressure is 1 atm (760 torr,
760 mm Hg, or 101.325 kPa). For purposes of comparison, volumes of gases are often changed
to STP conditions:
standard conditions
standard temperature and pressure
(STP)
molar volume
In this text we’ll use 273 K for
temperature conversions and
calculations. Check with your
instructor for rules in your class.
Remember that you determine
the correct ratios by thinking
about the final result. For example an increase in pressure should
­decrease the volume, and so on.
standard temperature = 273.15 K or 0.00°C
standard pressure = 1 atm or 760 torr, 760 mm Hg, or 101.325 kPa
When temperature and pressure change at the same time, the new volume may be calculated by multiplying the initial volume by the correct ratios of both pressure and temperature
as follows:
final volume = (initial volume) a
ratio of
ratio of
ba
b
pressures temperatures
The P, V, and T relationships for a given mass of any gas, in fact, may be expressed as a
single equation, PV/T = k. For problem solving, this equation is usually written
P1V1
P2V2
=
T1
T2
where P1 , V1 , and T1 are the initial conditions and P2 , V2 , and T2 are the final conditions.
This equation can be solved for any one of the six variables and is useful in dealing with
the pressure–volume–temperature relationships of gases. Note what happens to the combined
gas law when one of the variables is constant:
• T constant S P1V1 = P2V2
• P constant S
V1
V2
=
T1
T2
Boyle’s law
Charles’ law
12.5
• Combined Gas Laws 261
example 12.7
Given 20.0 L of ammonia gas at 5oC and 730. torr, calculate the volume at 50.oC
and 800. torr.
Solution
Read •
Remember that temperature must be in kelvins.
Knowns:
P1 = 730. torr
P2 = 800. torr
o
T2 = 50.oC = 323 K
T1 = 5 C = 278 K
V1 = 20.0 L
Solving for: new volume (V2)
Plan •
Since P, V, and T are all changing, we must use the combined gas law:
P1V1
P2V2
=
T1
T2
V2 =
Calculate • V2 =
Check •
V1P1T2
P2T1
(20.0 L)(730. torr)(323 K)
= 21.2 L
(800. torr)(278 K)
The number of significant figures should be three, and it is, as indicated by
the significant figures, in the knowns.
example 12.8
To what temperature (oC) must 10.0 L of nitrogen at 25oC and 700. torr be heated in
order to have a volume of 15.0 L and a pressure of 760. torr?
Solution
Read •
Remember that temperature must be in kelvins.
Knowns:
P1 = 700. torr
V1 = 10.0 L
T1 = 25oC = 298 K
P2 = 760. torr
V2 = 15.0 L
Solving for: new temperature (T2)
Plan •
Since P, V, and T are all changing, we use the combined gas law:
P1V1
P2V2
=
T1
T2
T2 =
Calculate • T2 =
T1P2V2
P1V1
(298 K)(760. torr)(15.0 L)
= 485 K
(700. torr)(10.0 L)
Since the problem asks for °C, we must convert our answer:
485 K – 273 = 212oC
ENHANCED EXAMPLE
262 chapter 12
• The Gaseous State of Matter
example 12.9
The volume of a gas-filled balloon is 50.0 L at 20.oC and 742 torr. What volume will it
occupy at standard temperature and pressure (STP)?
Solution
Read •
Remember that temperature must be in kelvins.
Knowns:
P1 = 742 torr
T1 = 20.°C = 293 K
V1 = 50.0 L
P2 = 760. torr (standard pressure)
T2 = 273 K (standard temperature)
Solving for: new volume (V2)
Plan •
Since P, V, and T are all involved, we use the combined gas law:
P1V1
P2V2
=
T1
T2
V2 =
Calculate • V2 =
P1V1T2
P2T1
(742 torr)(50.0 L)(273 K)
= 45.5 L
(760. torr)(293 K)
Practice 12.4
15.00 L of gas at 45.0°C and 800. torr are heated to 400.°C, and the pressure is changed
to 300. torr. What is the new volume?
Practice 12.5
To what temperature must 5.00 L of oxygen at 50.°C and 600. torr be heated in order to
have a volume of 10.0 L and a pressure of 800. torr?
Mole–Mass–Volume Relationships of Gases
As with many constants, the molar
volume is known more exactly to
be 22.414 L. We use 22.4 L in our
calculations, since the extra
­figures don’t often affect the
result, given the other measurements in the calculation.
Because a mole contains 6.022 * 1023 molecules (Avogadro’s number), a mole of any gas
will have the same volume as a mole of any other gas at the same temperature and pressure. It
has been experimentally determined that the volume occupied by a mole of any gas is 22.4 L
at STP. This volume, 22.4 L, is known as the molar volume of a gas. The molar volume is a
cube about 28.2 cm (11.1 in.) on a side. The molar masses of several gases, each occupying
22.4 L at STP, are shown in Table 12.3 and Figure 12.14:
One mole of a gas occupies 22.4 L at STP.
The molar volume is useful for determining the molar mass of a gas or of substances that
can be easily vaporized. If the mass and the volume of a gas at STP are known, we can calculate
its molar mass. For example, 1.00 L of pure oxygen at STP has a mass of 1.429 g. The molar
mass of oxygen may be calculated by multiplying the mass of 1.00 L by 22.4 L/mol:
Standard conditions apply only
to pressure, temperature, and
volume. Mass is not affected.
a
1.429 g
22.4 L
ba
b = 32.0 g/mol (molar mass)
1.00 L
1 mol
If the mass and volume are at other than standard conditions, we change the volume to STP
and then calculate the molar mass.
12.5
The molar volume, 22.4 L/mol, is used as a conversion factor to convert grams per liter to
grams per mole (molar mass) and also to convert liters to moles. The two conversion factors are
22.4 L
1 mol
and
1 mol
22.4 L
These conversions must be done at STP except under certain special circumstances.
example 12.10
If 2.00 L of a gas measured at STP has a mass of 3.23 g, what is the molar mass of
the gas?
Solution
Read •
Knowns: V = 2.00 L
T = 273 K
P = 1.00 atm
m = 3.23 g
Plan •
• Combined Gas Laws 263
Table 12.3 Mass of a Molar
Volume (22.4 L) of Various Gases
Gas
Mass of 22.4 L
Xe
O2
CO2
N2
NH3
Ar
H 2S
H2
SO2
HCl
CH4
Cl2
131.3 g
32.00 g
44.01 g
28.02 g
17.03 g
39.95 g
34.09 g
2.016 g
64.07 g
36.46 g
16.04 g
71.90 g
The unit for molar mass is g/mol.
Solution map: g/L S g/mol
1 mol = 22.4 L, so the conversion factor is 22.4 L/1 mol.
Calculate • a
3.23 g
22.4 L
ba
b = 36.2 g/mol (molar mass)
2.00 L
1 mol
example 12.11
Measured at 40.°C and 630. torr, the mass of 691 mL of diethyl ether is 1.65 g. Calculate
the molar mass of diethyl ether.
Solution
Read •
To find the molar mass we must first convert to STP.
Knowns:
P1 = 630. torr
P2 = 760 torr
T1 = 40.°C = 313 K
T2 = 273 K
V1 = 691 mL
Plan •
Figure 12.14
One mole of a gas occupies 22.4 L
at STP.
Since P, V, and T are all changing, we must use the combined gas law:
Diethyl ether
P1V1
P2V2
=
T1
T2
V2 =
Calculate • V2 =
V1P1T2
P2T1
(691 mL) (630. torr) (273 K)
= 500. mL = 0.500 L (at STP)
(760. torr) (313 K)
a
1.65 g
22.4 L
ba
b = 73.9 g/mol
0.500 L
1 mol
Practice 12.6
A gas with a mass of 86 g occupies 5.00 L at 25°C and 3.00 atm pressure. What is the
molar mass of the gas?
ENHANCED EXAMPLE
264 chapter 12
• The Gaseous State of Matter
12.6 Ideal Gas Law
Learning objective
Use the ideal gas law to solve problems involving pressure, volume, temperature,
and number of moles.
key terms
Now that we have considered all of the relationships used in gas calculations, we can simplify
our work by combining the relationships into a single equation describing a gas.
The four variables that characterize a gas are:
ideal gas law
kinectic-molecular theory (KMT)
ideal gas
•
•
•
•
volume (V )
pressure (P)
absolute temperature (T )
number of molecules of moles (n)
Combining these variables, we obtain
V 
nT
P
or
V =
nRT
P
where R is a proportionality constant known as the ideal gas constant. The equation is commonly written as
PV = nRT
and is known as the ideal gas law. This law summarizes in a single expression what we have
considered in our earlier discussions. The value and units of R depend on the units of P, V,
and T. We can calculate one value of R by taking 1 mol of a gas at STP conditions. Solve the
equation for R:
R =
(1 atm)(22.4 L)
PV
L # atm
=
= 0.0821
nT
(1 mol)(273 K)
mol # K
The units of R in this case are liter # atmospheres (L # atm) per mole kelvin (mol # K). When
the value of R = 0.0821 L # atm/mol # K, P is in atmospheres, n is in moles, V is in liters, and
T is in kelvins.
The ideal gas equation can be used to calculate any one of the four variables when the
other three are known.
example 12.12
ENHANCED EXAMPLE
What pressure will be exerted by 0.400 mol of a gas in a 5.00-L container at 17°C?
Solution
Read •
Knowns: V = 5.00 L
T = 17°C = 290. K
n = 0.400 mol
Plan •
We need to use the ideal gas law:
PV = nRT
P =
nRT
V
L # atm
b (290. K)
mol # K
= 1.90 atm
5.00 L
(0.400 mol)a 0.0821
Calculate • P =
12.6
• Ideal Gas Law 265
example 12.13
How many moles of oxygen gas are in a 50.0-L tank at 22°C if the pressure gauge reads
2000. lb/in.2?
Solution
Read •
Knowns: P = a
2000. lb
1 atm
b
= 136.1 atm
2
£
lb ≥
in.
14.7 2
in.
1 atm = 14.7 lb/in.2 from
Table 12.2.
V = 50.0 L
T = 22°C = 295 K
Plan •
We need to use the ideal gas law:
PV = nRT
n =
PV
RT
(136.1 atm)(50.0 L)
= 281 mol O2
L # atm
(0.0821
)(295 K)
mol # K
Calculate • n =
Practice 12.7
A 23.8-L cylinder contains oxygen gas at 20.0°C and 732 torr. How many moles of
oxygen are in the cylinder?
The molar mass of a gaseous substance can also be determined using the ideal gas law.
Since molar mass = g/mol it follows that mol = g/molar mass. Using M for molar mass and
g for grams, we can substitute g/M for n (moles) in the ideal gas law to get
PV =
g
RT
M
or
M =
gRT
(modified ideal gas law)
PV
which allows us to calculate the molar mass, M, for any substance in the gaseous state.
This form of the ideal gas law is
most useful in problems containing
mass instead of moles.
Calculate the molar mass of butane gas if 3.69 g occupy 1.53 L at 20.ºC and 1.00 atm.
Solution
Read •
Knowns:
Plan •
P
V
T
m
=
=
=
=
1.00 atm
1.53 L
20.ºC = 293 K
3.69 g (mass)
S
ince we know m and not the number of moles (n) of the gas, we need to use
the ideal gas law, modifying it since n = g/M, where M is the molar mass
of the gas:
PV =
M =
g
RT
M
gRT
PV
© Emil Schreiner/iStockphoto
example 12.14
266 chapter 12
• The Gaseous State of Matter
L # atm
b (293 K)
mol # K
= 58.0 g/mol
(1.00 atm)(1.53 L)
(3.69 g)a 0.0821
Calculate • M =
Check •
I f we look up the formula for butane, we find it is C4H10 with a molar mass
of 58.0 g/mol, so our calculation is correct.
Practice 12.8
Butane
A 0.286-g sample of a certain gas occupies 50.0 mL at standard temperature and 76.0 cm Hg.
Determine the molar mass of the gas.
The Kinetic-Molecular Theory
The data accumulated from the gas laws allowed scientists to formulate a general theory to explain the behavior and properties of gases. This theory is called the kinetic-molecular theory
(KMT). The KMT has since been extended to cover, in part, the behavior of liquids and solids.
The KMT is based on the motion of particles, particularly gas molecules. A gas that
behaves exactly as outlined by the theory is known as an ideal gas. No ideal gases exist, but
under certain conditions of temperature and pressure, real gases approach ideal behavior or at
least show only small deviations from it.
The principal assumptions of the kinetic-molecular theory are as follows:
1. Gases consist of tiny particles.
2. The distance between particles is large compared with the size of the particles themselves. The volume occupied by a gas consists mostly of empty space.
3. Gas particles have no attraction for one another.
4. Gas particles move in straight lines in all directions, colliding frequently with one another
and with the walls of the container.
5. No energy is lost by the collision of a gas particle with another gas particle or with the
walls of the container. All collisions are perfectly elastic.
6. The average kinetic energy for particles is the same for all gases at the same temperature,
and its value is directly proportional to the Kelvin temperature.
The kinetic energy (KE) of a particle is expressed by the equation
KE =
1 2
mv
2
where m is the mass and v is the velocity of the particle.
All gases have the same kinetic energy at the same temperature. Therefore, from the kinetic
energy equation we can see that, if we compare the velocities of the molecules of two gases, the
lighter molecules will have a greater velocity than the heavier ones. For example, calculations
show that the velocity of a hydrogen molecule is four times the velocity of an oxygen molecule.
Real Gases
All the gas laws are based on the behavior of an ideal gas—that is, a gas with a behavior that is
described exactly by the gas laws for all possible values of P, V, and T. Most real gases actually
do behave very nearly as predicted by the gas laws over a fairly wide range of temperatures
and pressures. However, when conditions are such that the gas molecules are crowded closely
together (high pressure and/or low temperature), they show marked deviations from ideal behavior. Deviations occur because molecules have finite volumes and also have intermolecular
attractions, which result in less compressibility at high pressures and greater compressibility at
low temperatures than predicted by the gas laws. Many gases become liquids at high pressure
and low temperature.
12.7
­
• Dalton’s Law of Partial Pressures 267
>Chemistry in action
Air Quality
Chemical reactions occur among the gases that are emitted into our atmosphere. In recent years, there has been
growing concern over the effects these reactions have on
our environment and our lives.
The outer portion (stratosphere) of the atmosphere plays
a significant role in determining the conditions for life at
the surface of the Earth. This stratosphere protects the
surface from the intense radiation and particles bombarding our planet. Some of the high-energy radiation from
the sun acts upon oxygen molecules, O2, in the stratosphere, converting them into ozone, O3.
O2 99: O + O
sunlight
oxygen atoms
O2 + O ¡ O3
ozone
In addition to ozone, the air in urban areas contains
nitrogen oxides, which are components of smog. The
term smog refers to air pollution in urban environments.
Often the chemical reactions occur as part of a photochemical process. Nitrogen monoxide (NO) is oxidized
in the air or in automobile engines to produce nitrogen
dioxide (NO2). In addition to nitrogen oxides, combustion of fossil fuels releases CO2, CO, and sulfur oxides.
Incomplete combustion releases unburned and partially
burned ­hydrocarbons. Society is continually attempting to
discover, understand, and control emissions that contribute to this sort of atmospheric chemistry.
EPA/Paul Hilton/NewsCom
Ultraviolet radiation from the sun is highly damaging to
living tissues of plants and animals. The ozone layer, however, shields the Earth by absorbing ultraviolet radiation
and thus prevents most of this lethal radiation from reaching the Earth’s surface.
In the lower atmosphere ozone is a harmful pollutant.
Ozone is formed in the atmosphere during electrical
storms and by the photochemical action of ultraviolet
radiation on a mixture of nitrogen dioxide and oxygen.
Areas with high air pollution are subject to high atmospheric ozone concentrations. Ozone is not a desirable
low-altitude constituent of the atmosphere because it is
known to cause extensive plant damage, cracking of rubber, and the formation of eye-irritating substances.
Hong Kong skyline comparing poor and good air quality.
12.7 Dalton’s Law of Partial Pressures
Use Dalton’s law of partial pressures to calculate the total pressure from a mixture of
gases or the pressure of a single gas in a mixture of gases.
Learning objectiv e
If gases behave according to the kinetic-molecular theory, there should be no difference in
the pressure–volume–temperature relationships whether the gas molecules are all the same or
different. This similarity in the behavior of gases is the basis for an understanding of Dalton’s
law of partial pressures:
key terms
The total pressure of a mixture of gases is the sum of the partial pressures exerted by each
of the gases in the mixture.
Dalton’s law of partial pressures
partial pressure
268 chapter 12
• The Gaseous State of Matter
0.5 mol H 2
P = 0.5 atm
0.5 mol O2
P = 0.5 atm
0.5 mol H 2 + 0.5 mol O2
P = 1 atm
Figure 12.15
Each gas in a mixture of gases (H2 and O2) has a partial pressure which is the
same as though the gas were alone in the container.
Each gas in the mixture exerts a pressure that is independent of the other gases pres­ent. These
pressures are called partial pressures (see Figure 12.15). Thus, if we have a mixture of
hydrogen and oxygen gases (H2 and O2), exerting partial pressures of 0.5 atm and 0.5 atm,
respectively, the total pressure will be 1.0 atm:
Ptotal = PH2 + PO2
Ptotal = 0.5 atm + 0.5 atm = 1.0 atm
We can see an application of Dalton’s law in the collection of insoluble gases over
water. When prepared in the laboratory, oxygen is commonly collected by the downward ­displacement of water. Thus the oxygen is not pure but is mixed with water vapor
(see Figure 12.16). When the water levels are adjusted to the same height inside and outside the bottle, the pressure of the oxygen plus water vapor inside the bottle is equal to the
­atmospheric pressure:
Patm = PO2 + PH2O
>Chemistry in action
Getting High to Lose Weight?
Researchers in Germany have reported an interesting
relationship between atmospheric pressure and weight
loss. They invited 20 obese men with an average mass of
105 kg to spend a week at a research station 2650 meters
above sea level. The men were instructed to maintain
the same activity level as they normally would and wore
pedometers to ensure that they were walking the same
number of steps each day. At the end of the week the
average weight loss was 1.5 kg, a half kilogram more than
would have been expected based on their lowered appetites at this altitude.
The research group hypothesized that some of the
extra weight loss could have been due to the low partial
­pressure of oxygen gas in the atmosphere. This would
have required the subjects’ hearts to work harder in order
to draw in enough oxygen to survive. This increase in heart
rate resulted in a higher energy consumption and net
weight loss. Comparing this result to data from similar experiments with athletes training at high altitude, a similar
weight loss was observed. In these experiments, however,
the relative amounts of fat and muscle tissue metabolized
to produce energy were compared. The percentage of
fat burned decreased and the amount of muscle burned
increased at high altitude. The result suggests that in an
oxygen-poor environment the body chooses to metabolize
the energy source requiring the least oxygen.
So, is moving to a high altitude the answer to your weight
loss struggles? Probably not, considering that you most
likely want to lose fat mass rather than muscle mass.
12.7
• Dalton’s Law of Partial Pressures 269
Oxygen plus
water vapor
Oxygen from
generator
Figure 12.16
Oxygen collected over water.
To determine the amount of O2 or any other gas collected over water, we subtract the pressure
of the water vapor (vapor pressure) from the total pressure of the gases:
PO2 = Patm - PH2O
The vapor pressure of water at various temperatures is tabulated in ­Appendix IV.
Example 12.15
A 500.-mL sample of oxygen was collected over water at 23ºC and 760. torr. What
volume will the dry O2 occupy at 23ºC and 760. torr? The vapor pressure of water at 23ºC
is 21.2 torr.
Solution
Read •
Knowns:
Voxygen and water vapor = 500. mL
Poxygen and water vapor = 760 torr
Pwater vapor = 21.2 torr
Temperature constant at 23°C
Solving for: Vdry oxygen (at 760 torr)
Plan •
F
irst, we must find the pressure of the dry oxygen, and then determine the volume of the dry oxygen using Boyle’s law (since the temperature is constant).
P1V1 = P2V2
Pdry oxygen = Poxygen and water vapor - Pwater vapor
= 760. torr – 21.2 torr = 739 torr
P1 = 739 torr P2 = 760. torr
V1 = 500. mL
Solution map: pressure increases S volume decreases
Since P1V1 = P2V2,
Calculate • V2 =
Check •
V2 =
P1V1
P2
(739 torr)(500. mL)
= 486 mL dry O2
760. torr
The final volume is smaller than the original as predicted by Dalton’s law.
Practice 12.9
Hydrogen gas was collected by downward displacement of water. A volume of 600.0 mL
of gas was collected at 25.0°C and 740.0 torr. What volume will the dry hydrogen occupy
at STP?
270 chapter 12
• The Gaseous State of Matter
12.8 Density of Gases
Learning obj ective
Calculate the density of a gas.
The density, d, of a gas is its mass per unit volume, which is generally expressed in grams per
liter as follows:
g
mass
d =
=
volume
L
Table 12.4 Density of
Common Gases at STP
Gas
Molar
mass
(g/mol)
Density
(g/L at STP)
H2
CH4
NH3
C2H2
HCN
CO
N2
air
O2
H2S
HCl
F2
CO2
C3H8
O3
SO2
Cl2
2.016
16.04
17.03
26.04
27.03
28.01
28.02
(28.9)
32.00
34.09
36.46
38.00
44.01
44.09
48.00
64.07
70.90
0.0900
0.716
0.760
1.16
1.21
1.25
1.25
(1.29)
1.43
1.52
1.63
1.70
1.96
1.97
2.14
2.86
3.17
Because the volume of a gas depends on temperature and pressure, both should be given when
stating the density of a gas. The volume of a solid or liquid is hardly affected by changes in pressure and is changed only slightly when the temperature is varied. Increasing the temperature from
0°C to 50°C will reduce the density of a gas by about 18% if the gas is allowed to expand, whereas
a 50°C rise in the temperature of water (0°C S 50°C) will change its density by less than 0.2%.
The density of a gas at any temperature and pressure can be determined by calculating the
mass of gas present in 1 L. At STP, in particular, the density can be calculated by multiplying
the molar mass of the gas by 1 mol>22.4 L:
dSTP = (molar mass) a
molar mass = (dSTP) a
Table 12.4 lists the densities of some common gases.
1 mol
b
22.4 L
22.4 L
b
1 mol
Example 12.16
Calculate the density of Cl2 at STP.
Solution
Read •
To find the density, we need the molar mass of Cl2. Using the periodic table,
we find it to be 70.90 g/mol.
Plan •
The unit for density for gases is g/L.
Solution map: g/mol S g/L
1 mol = 22.4 L, so the conversion factor is 1 mol /22.4 L.
Calculate • a
Check •
70.90 g
1 mol
ba
b = 3.165 g/L
1 mol
22.4 L
Compare the answer 3.165 g/L to Cl2 density in Table 12.4.
Practice 12.10
The molor mass of a gas is 20. g/mol. Calculate the density of the gas at STP.
Practice 12.11
List these gases in order of increasing density (same T and P): H2 , CO, C3H8 , NO2 , O2 ,
N2O, POF3 , and B2H6 .
12.9 Gas Stoichiometry
Learning objective
Solve stoichiometric problems involving gases.
Mole–Volume and Mass–Volume Calculations
Stoichiometric problems involving gas volumes can be solved by the general mole-ratio
method outlined in Chapter 9 (summarized in Figure 12.17). The factors 1 mol>22.4 L and
12.9
Volume A
• Gas Stoichiometry 271
Volume B
Grams A
Moles A
Moles B
Atoms or
molecules A
Grams B
Atoms or
molecules B
Figure 12.17
Summary of the primary
conversions involved in
stoichiometry. The conversion
for volumes of gases is included.
22.4 L>1 mol are used for converting volume to moles and moles to volume, respectively.
These conversion factors are used under the assumption that the gases are at STP and that they
behave as ideal gases. In actual practice, gases are measured at other than STP conditions, and
the volumes are converted to STP for stoichiometric calculations. The following are examples
of typical problems involving gases in chemical reactions.
Example 12.17
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium ­chlorate?
ENHANCED EXAMPLE
Solution
Read •
Knowns: P = 1.00 atm
T = 273 K
n = 0.500 mol KClO3
Plan •
e can use the balanced equation to solve this problem since moles of a gas
W
are directly convertible to L at STP.
2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g)
Solution map: mol KClO3 S mol O2 S L O2
We use our Problem-Solving Strategy for Stoichiometry Problems to find the
moles of oxygen produced:
(0.500 mol KClO3) a
3 mol O2
b = 0.750 mol O2
2 mol KClO3
Calculate • Now we can convert moles of O2 to liters of O2.
(0.750 mol O2)a
22.4 L
b = 16.8 L O2
1 mol
example 12.18
How many grams of aluminum must react with sulfuric acid to produce 1.25 L of hy­
drogen gas at STP?
Solution
Read •
Knowns: V = 1.25 L H2
T = 273 K
P = 1.00 atm
Plan •
We can use the balanced equation to solve this problem since moles of a gas
are directly convertible to liters at STP.
2 Al(s) + 3 H2SO4(aq) ¡ Al2(SO4)3(aq) + 3 H2(g)
Solution map: L H2 S mol H2 S mol Al S g Al
See Section 9.1 if you have
forgotten the Problem-Solving
Strategy.
272 chapter 12
• The Gaseous State of Matter
We first use the conversion factor of 1 mol/22.4 L to convert the liters H2 to
moles. Then we use our Problem-Solving Strategy for Stoichiometry Problems to find the moles of aluminum produced:
(1.25 L H2)a
1 mol
2 mol Al
ba
b = 0.0372 mol Al
22.4 L 3 mol H2
Calculate • Now we can convert moles of Al to grams of Al.
(0.0372 mol Al)a
26.98 g Al
b = 1.00 g Al
mol Al
example 12.19
What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting
50.0 g of aluminum with hydrochloric acid?
Solution
Read •
Knowns:
Plan •
We need the balanced equation to determine the number of moles of Al and
then the moles of H2.
T = 30.°C = 303 K
P = 700. torr = 0.921 atm
m = 50.0 g
2 Al(s) + 6 HCl(aq) S 2 AlCl3(aq) + 3 H2(g)
Solution map: g Al S mol Al S mol H2
We first use the molar mass for Al to convert the g Al to moles. Then we use the
mole ratio from the balanced equation to find the moles of hydrogen produced:
(50.0 g Al)a
3 mol H2
1 mol Al
ba
b = 2.78 mol H2
26.98 g Al
2 mol Al
Now we can use the ideal gas law to complete the calculation:
PV = nRT
L # atm
b (303 K)
mol # K
= 75.1 L H2
(0.921 atm)
(2.78 mol H2)a 0.0821
Calculate • V =
Practice 12.12
If 10.0 g of sodium peroxide (Na2O2) react with water to produce sodium hydroxide and
oxygen, how many liters of oxygen will be produced at 20.°C and 750. torr?
2 Na2O2(s) + 2 H2O(l) ¡ 4 NaOH(aq) + O2(g)
Volume–Volume Calculations
When all substances in a reaction are in the gaseous state, simplifications in the calculation
can be made. These are based on Avogadro’s law, which states that gases under identical
conditions of temperature and pressure contain the same number of molecules and occupy
the same volume. Under the standard conditions of ­temperature and pressure, the volumes of
gases reacting are proportional to the numbers of moles of the gases in the balanced equation.
Consider the reaction:
H2(g) + Cl2(g) ¡ 2 HCl(g)
1 mol
1 volume
Y volume
1 mol
1 volume
Y volume
2 mol
2 volumes
2 Y volumes
12.9
This statement is true because these volumes are equivalent to the number of reacting
moles in the equation. Therefore, Y volume of H2 will combine with Y volume of Cl2 to give
2 Y volumes of HCl. For example, 100 L of H2 react with 100 L of Cl2 to give 200 L of HCl.
For reacting gases at constant temperature and pressure, volume–volume relationships are
the same as mole–mole relationships.
example 12.20
What volume of oxygen will react with 150. L of hydrogen to form water vapor? What
volume of water vapor will be formed?
Solution
Assume that both reactants and products are measured at standard conditions. Calculate
by using reacting volumes:
2 H2(g) + O2(g) ¡ 2 H2O(g)
2 mol
2 volumes
150. L
1 mol
1 volume
75 L
2 mol
2 volumes
150. L
or every two volumes of H2 that react, one volume of O2 reacts and two volumes of
F
H2O(g) are produced:
(150. L H2)¢
(150. L H2)¢
1 volume O2
≤ = 75.0 L O2
2 volumes H2
2 volumes H2O
≤ = 150. L H2O
2 volumes H2
example 12.21
The equation for the preparation of ammonia is
3 H2(g) + N2(g) 99: 2 NH3(g)
400°C
Assuming that the reaction goes to completion, determine the following:
(a) What volume of H2 will react with 50.0 L of N2 ?
(b) What volume of NH3 will be formed from 50.0 L of N2 ?
(c) What volume of N2 will react with 100. mL of H2 ?
(d) What volume of NH3 will be produced from 100. mL of H2 ?
(e)If 600. mL of H2 and 400. mL of N2 are sealed in a flask and allowed to
react, ­what amounts of H2 , N2 , and NH3 are in the flask at the end of the reaction?
SOLUTION
The answers to parts (a)–(d) are shown in the boxes and can be determined from the
­equation by inspection, using the principle of reacting volumes:
3 H2(g)
3 volume
+
N2(g)
¡
2 NH3(g)
1 volume
2 volumes
(a)  150. L 
(b) 50.0 L
50.0 L
(c) 100. mL
 33.3 mL 
 100. L 
(d) 100. mL
 66.7 mL 
• Gas Stoichiometry 273
Remember: For gases at the
same T and P, equal volumes contain equal numbers of particles.
274 chapter 12
• The Gaseous State of Matter
(e) Volume ratio from the equation =
Volume ratio used =
3 volumes H2
1 volume N2
600. mL H2
3 volumes H2
=
400. mL N2
2 volumes N2
Comparing these two ratios, we see that an excess of N2 is present in the gas mixture.
Therefore, the reactant limiting the amount of NH3 that can be formed is H2 :
3 H2(g) + N2(g) ¡ 2 NH3(g)
600. mL
200. mL
400. mL
To have a 3:1 ratio of volumes reacting, 600. mL of H2 will react with 200. mL of N2 to
produce 400. mL of NH3 , leaving 200. mL of N2 unreacted. At the end of the reaction,
the flask will contain 400. mL of NH3 and 200. mL of N2 .
Practice 12.13
What volume of oxygen will react with 15.0 L of propane (C3H8) to form carbon dioxide
and water? What volume of carbon dioxide will be formed? What volume of water vapor
will be formed?
C3H8(g) + 5 O2 ¡ 3 CO2(g) + 4 H2O(g)
C h a p te r
12 review
12.1 Properties Of gases
key terms
pressure
atmospheric pressure
barometer
1 atmosphere
• Gases:
• Particles are relatively far apart.
• Particles are very mobile.
• Gases take the shape and volume of the container.
• Gases are easily compressible.
• Pressure is force per unit area.
• Pressure of the atmosphere is measured by using a barometer:
• Units of pressure include:
• Atmosphere (atm) = 760 mm Hg.
• Pascal, 1 atm = 101,325 Pa = 101.3 kPa.
• Torr, 1 atm = 760 torr.
• Pressure is directly related to the number of molecules in the sample.
• Pressure is directly related to the Kelvin temperature of the sample.
12.2 Boyle’s Law
key term
• At constant temperature, the volume of a gas is inversely proportional to the pressure of the gas:
Boyle’s law
V 
8
7
Pressure (atm)
6
5
4
3
2
1
0
0
1
2
3 4 5 6
Volume (liters)
7
8
1
P
or
P1V1 = P2V2
Review 275
12.3 Charles’ Law
• At constant pressure, the volume of a gas is directly proportional to the absolute temperature of the gas.
V  T or V1
V2
=
T1
T2
key terms
absolute zero
Charles’ law
5
Volume (L)
4
CH4
3
2
–273°C
1
0
–300
–200
–100
0
100
200
Temperature (°C)
300
400
12.4 Avogadro’s Law
• Gay-Lussac’s law of combining volumes states that when measured at constant T and P, the ratios of
the volumes of reacting gases are small whole numbers.
• Avogadro’s law states that equal volumes of different gases at the same T and P contain the same
number of particles.
key terms
Gay-Lussac’s law of
combining volumes
Avogadro’s law
12.5 Combined Gas Laws
• The P V T relationship for gases can be expressed in a single equation known as the combined gas
law:
P1V1
P2V2
=
T1
T2
key terms
standard conditions
standard temperature
and pressure (STP)
molar volume
• One mole of any gas occupies 22.4 L at STP.
12.6 Ideal Gas Law
• The ideal gas law combines all the variables involving gases into a single expression:
PV = nRT
• R is the ideal gas constant and can be expressed as
R = 0.0821
L # atm
mol # K
• Kinetic-molecular theory assumptions:
• Gases are tiny particles with no attraction for each other.
• The distance between particles is great compared to the size of the particles.
• Gas particles move in straight lines.
• No energy is lost in particle collisions (perfectly elastic collisions).
• The average kinetic energy for particles is the same for all gases at the same temperature and
pressure.
• A gas that follows the KMT is an ideal gas.
1
• The kinetic energy of a particle is expressed as KE = mv 2.
2
• Real gases show deviation from the ideal gas law:
• Deviations occur at:
• High pressure
• Low temperature
• These deviations occur because:
• Molecules have finite volumes.
• Molecules have intermolecular attractions.
key terms
ideal gas law
kinetic-molecular theory (KMT)
ideal gas
276 chapter 12
• The Gaseous State of Matter
12.7 Dalton’s Law of Partial Pressures
key terms
• The total pressure of a mixture of gases is the sum of the partial pressures of the component gases in
the mixture.
• When a gas is collected over water, the pressure of the collected gas is the difference between the
atmospheric pressure and the vapor pressure of water at that temperature.
Dalton’s law of partial
pressures
partial pressure
12.8 Density of Gases
• The density of a gas is usually expressed in units of g/L.
• Temperature and pressure are given for a density of a gas since the volume of the gas depends on
these conditions.
12.9 Gas Stoichiometry
• Stoichiometry problems involving gases are solved the same way as other stoichiometry problems.
See graphic below.
Volume A
Grams A
Atoms or
molecules A
Volume B
Moles A
Moles B
Grams B
Atoms or
molecules B
• For reacting gases (contant T and P) volume–volume relationships are the same as mole–mole
relationships.
review Questions
1. What is meant by pressure (for a gas)?
2.How does the air pressure inside the balloon shown in Figure 12.3
compare with the air pressure outside the balloon? Explain.
3. According to Table 12.1, what two gases are the major constituents
of dry air?
4. How does the pressure represented by 1 torr compare in magnitude
to the pressure represented by 1 mm Hg? See Table 12.2.
5. In which container illustrated in Figure 12.6 are the ­molecules of
gas moving faster? Assume both gases to be ­hydrogen.
6. In Figure 12.7, what gas pressure corresponds to a volume of 4 L?
7. How do the data illustrated in Figure 12.7 verify Boyle’s law?
8. What effect would you observe in Figure 12.10 if T2 were lower
than T1 ?
9. Explain how the reaction

N2(g) + O2(g) ¡ 2 NO(g)
proves that nitrogen and oxygen are diatomic molecules.
10. What is the reason for comparing gases to STP?
11. What are the four parameters used to describe the behavior of a gas?
12. What are the characteristics of an ideal gas?
13. How is Boyle’s law related to the ideal gas law?
14. How is Charles’ law related to the ideal gas law?
15. What are the basic assumptions of the kinetic-molecular theory?
16. Arrange the following gases, all at standard temperature, in order of
increasing relative molecular velocities: H2 , CH4 , Rn, N2 , F2 , and
He. What is your basis for determining the order?
17. List, in descending order, the average kinetic energies of the molecules in Question 16.
18. Under what condition of temperature, high or low, is a gas least
likely to exhibit ideal behavior? Explain.
19. Under what condition of pressure, high or low, is a gas least likely
to exhibit ideal behavior? Explain.
20. How does the kinetic-molecular theory account for the behavior of
gases as described by
(a) Boyle’s law?
(b) Charles’ law?
(c) Dalton’s law of partial pressures?
21. Is the conversion of oxygen to ozone an exothermic or endothermic
reaction? How do you know?
22. Write formulas for an oxygen atom, an oxygen molecule, and an
ozone molecule. How many electrons are in an oxygen molecule?
23. In the diagram shown in Figure 12.16, is the pressure of the oxygen
plus water vapor inside the bottle equal to, greater than, or less than
the atmospheric pressure outside the bottle? Explain.
24. List five gases in Table 12.4 that are more dense than air. Explain
the basis for your selection.
Paired Exercises 277
25. Compare, at the same temperature and pressure, equal volumes of
H2 and O2 as to the following:
(a)number of molecules
(b)mass
(c)number of moles
(d)average kinetic energy of the molecules
(e)density
26. When constant pressure is maintained, what effect does heating a
mole of N2 gas have on
(a)its density?
(b)its mass?
(c)the average kinetic energy of its molecules?
(d)the average velocity of its molecules?
(e)the number of N2 molecules in the sample?
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com)
All exercises with blue numbers have answers in Appendix VI.
P A I R E D E x er c i s e s
1. Fill in the table below with the missing information:
torr
(a) (b)
in. Hg
2. Fill in the table below with the missing information:
kilopascals (kPa)
30.2
mm Hg
(a)
752
99.3
atmospheres (atm)
789
(b) (c) lb/in.2
32
(c) 1.4
3. Perform the following pressure conversions:
(a) Convert 953 torr to kPa.
(b) Convert 2.98 kPa to atm.
(c) Convert 2.77 atm to mm Hg.
(d) Convert 372 torr to atm.
(e) Convert 2.81 atm to cm Hg.
4. Perform the following pressure in conversions:
(a) Convert 649 torr to kPa.
(b) Convert 5.07 kPa to atm.
(c) Convert 3.64 atm to mm Hg.
(d) Convert 803 torr to atm.
(e) Convert 1.08 atm to cm Hg.
5. This scuba diver watch shows the air pressure in the diver’s scuba
tank as 1920 lb/in.2. Convert this pressure to
(a) atm (b) torr (c) kPa
6. A pressure sensor shows tire pressure as 31 lb/in.2. Convert this
pressure to
(a) atm (b) torr (c) kPa
Courtesy Aeris
© Thomas Acop/iStockphoto
7. A sample of a gas occupies a volume of 725 mL at 825 torr. At
constant temperature, what will be the new pressure (torr) when the
volume changes to the following:
(a) 283 mL
(b) 2.87 L
8. A sample of a gas occupies a volume of 486 mL at 508 torr. At
constant temperature, what will be the new pressure (torr) when the
volume changes to the following:
(a) 185 mL
(b) 6.17 L
9. A sample of methane gas, CH4, occupies a volume (L) of 58.2 L
at a pressure of 7.25 atm. What volume will the gas occupy if the
pressure is lowered to 2.03 atm?
10. A sample of nitrous oxide gas, N2O, occupies a volume of 832 L
at a pressure of 0.204 atm. What volume (L) will the gas occupy if
the pressure is increased to 8.02 atm?
11. A sample of CO2 gas occupies a volume of 125 mL at 21°C. If pressure remains constant, what will be the new volume if temperature
changes to:
(a) - 5°C (b) 95°F (c) 1095 K
12. A sample of CH4 gas occupies a volume of 575 mL at - 25°C. If
pressure remains constant, what will be the new volume if temperature changes to:
(a) 298 K (b) 32°F (c) 45°C
278
chapter 12
• The Gaseous State of Matter
13. A sample of a gas occupies a volume of 1025 mL at 75°C and
0.75 atm. What will be the new volume if temperature decreases to
35°C and pressure increases to 1.25 atm?
14. A sample of a gas occupies a volume of 25.6 L at 19°C and 678 torr.
What will be the new volume if temperature increases to 35°C and
pressure decreases to 595 torr?
15. An expandable balloon contains 1400. L of He at 0.950 atm pressure and 18°C. At an altitude of 22 miles (temperature 2.0°C and
pressure 4.0 torr), what will be the volume of the balloon?
16. A gas occupies 22.4 L at 2.50 atm and 27°C. What will be its volume at 1.50 atm and - 5.00°C?
17. A 775-mL sample of NO2 gas is at STP. If the volume changes to
615 mL and the temperature changes to 25°C, what will be the new
pressure?
18. A 2.5-L sample of SO3 is at 19°C and 1.5 atm. What will be the new
temperature in °C if the volume changes to 1.5 L and the pressure
to 765 torr?
19. A sample of O2 gas was collected over water at 23°C and 772 torr.
What is the partial pressure of the O2?
(Refer to Appendix IV for the vapor pressure of water.)
20. A sample of CH4 gas was collected over water at 29°C and 749 mm
Hg. What is the partial pressure of the CH4? (Refer to Appendix IV
for the vapor pressure of water.)
21. A mixture contains H2 at 600. torr pressure, N2 at 200. torr pressure, and O2 at 300. torr pressure. What is the total pressure of the
gases in the system?
22. A mixture contains H2 at 325 torr pressure, N2 at 475 torr pressure,
and O2 at 650. torr pressure. What is the total pressure of the gases
in the system?
23. A sample of methane gas, CH4 was collected over water at 25.0°C
and 720. torr. The volume of the wet gas is 2.50 L. What will be
the volume of the dry methane at standard pressure?
24. A sample of propane gas, C3H8 was collected over water at 22.5°C
and 745 torr. The volume of the wet gas is 1.25 L. What will be the
volume of the dry propane at standard pressure?
25. Calculate the volume of 6.26 mol of nitrogen gas, N2, at STP.
26. Calculate the volume of 5.89 mol of carbon dioxide, CO2, at STP.
27. What volume will each of the following occupy at STP?
(a) 6.02 * 1023 molecules of CO2
(c) 12.5 g oxygen
(b) 2.5 mol CH4
28. What volume will each of the following occupy at STP?
(a) 1.80 * 1024 molecules of SO3
(c) 25.2 g chlorine
(b) 7.5 mol C2H6
29. How many grams of NH3 are present in 725 mL of the gas at STP?
30. How many grams of C3H6 are present in 945 mL of the gas at STP?
31. How many liters of CO2 gas at STP will contain 1025 molecules?
32. How many molecules of CO2 gas at STP are present in 10.5 L?
33. What volume would result if a balloon were filled with 10.0 grams
of chlorine gas at STP?
34. How many grams of methane gas were used to fill a balloon to a
volume of 3.0 L at STP?
35. At 22°C and 729 torr pressure, what will be the volume of 75 mol
of NH3 gas?
36. At 39°C and 1.5 atm, what will be the volume of 105 mol of
CH4?
37. How many moles of O2 are contained in 5.25 L at 26°C and 1.2 atm?
38. How many moles of CO2 are contained in 9.55 L at 45°C and
752 torr?
39. At what Kelvin temperature will 25.2 mol of Xe occupy a volume
of 645 L at a pressure of 732 torr?
40. At what Kelvin temperature will 37.5 mol of Ar occupy a volume
of 725 L at a pressure of 675 torr?
41. Calculate the density of each of the following gases at STP:
(a) He (b) HF (c) C3H6 (d) CCl2F2
42. Calculate the density of each of the following gases at STP:
(a) Rn (b) NO2 (c) SO3 (d) C2H4
43. Calculate the density of each of the following gases:
(a) NH3 at 25°C and 1.2 atm
(b) Ar at 75°C and 745 torr
44. Calculate the density of each of the following gases:
(a) C2H4 at 32°C and 0.75 atm
(b) He at 57°C and 791 torr
45. In the lab, students decomposed a sample of calcium carbonate
by heating it over a Bunsen burner and collected carbon dioxide
according to the following equation.
46. In the lab, students generated and collected hydrogen gas according­
to the following equation:
CaCO3(s) ¡ CaO(s) + CO2(g)
(a) How many mL of carbon dioxide gas were generated by the
decomposition of 6.24 g of calcium carbonate at STP?
(b) If 52.6 L of carbon dioxide at STP were needed, how many
moles of calcium carbonate would be required?
47. Consider the following equation:
Mg(s) + 2HCl(aq) ¡ MgCl2(aq) + H2(g)
(a) How many mL of hydrogen gas at STP were generated from
42.9 g of magnesium metal?
(b) If 825 mL of hydrogen gas at STP were needed, how many
moles of HCI would be required?
48. Consider the following equation:
4 NH3(g) + 5 O2(g) ¡ 4 NO(g) + 6 H2O(g)
C3H8(g) + 5 O2(g) ¡ 3 CO2(g) + 4 H2O(g)
(a) How many liters of oxygen are required to react with
2.5 L NH3? Both gases are at STP.
(b) How many grams of water vapor can be produced from 25 L
NH3 if both gases are at STP?
(c) How many liters of NO can be produced when 25 L O2 are
reacted with 25 L NH3? All gases are at the same ­temperature
and pressure.
(a) How many liters of oxygen are required to react with 7.2 L C3H8?
Both gases are at STP.
(b) How many grams of CO2 will be produced from 35 L C3H8 if
both gases are at STP?
(c) How many liters of water vapor can be produced when 15 L
C3H8 are reacted with 15 L O2? All gases are at the same temperature and pressure.
Additional Exercises 279
49. Oxygen gas can be generated by the decomposition of ­potassium
chlorate according to the following equation:
50. When glucose is burned in a closed container, carbon dioxide gas
and water are produced according to the ­following equation:
2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g)
C6H12O6(s) + 6 O2(g) ¡ 6 CO2(g) + 6 H2O(l)
ow many liters of oxygen at STP will be produced when 0.525 kg
H
of KCl is also produced?
How many liters of CO2 at STP can be produced when 1.50 kg of
glucose are burned?
ADDITIONAL EXERCISES
51. How is it possible that 1 mole of liquid water occupies a volume of
18 mL, but 1 mole of gaseous water occupies a volume of 22.4 L?
(See Figure 12.2.)
52. Explain why it is necessary to add air to a car’s tires during the
winter.
53. Look at the three pictures below. If each of these pictures represents
a gas at 85°C, which of the gases is exerting the greatest pressure?
(d) What will happen to the pressure if 1 mole of N2 gas is added
to the container while keeping the temperature and size of the
container the same?
56. Sketch a graph to show each of the following relationships:
(a) P vs. V at constant temperature and number of moles
(b) T vs. V at constant pressure and number of moles
(c) n vs. V at constant temperature and pressure
57. For gases the relationship among the factors P, V, T, and n can be
used to solve many different problems. Using the factors and the
graphs below, complete the table.
(a)
(b)
(c)
54. The balloon below is filled with helium at a temperature of 37°C.
If the balloon is placed into a freezer with a temperature of - 20°C,
which diagram best represents the molecules inside the balloon?
(a)
(c)
(b)
(d)
Law
Graph showing the
Factors that Factors that
relationship of
are constant are variable
variable factors
Boyle’s law
Charles’ law
Avogadro’s law
(a)
(b)
(c)
58. Why is it dangerous to incinerate an aerosol can?
55. You have a 10-L container filled with 0.5 mol of O2 gas at a temperature of 30.°C with a pressure of 945 torr.
(a) What will happen to the pressure if the container size is doubled
while keeping the temperature and number of moles constant?
(b) What will happen to the pressure when the temperature is doubled while keeping the size of the container and the number of
moles constant?
(c) What will happen to the pressure when the amount of O2 gas
is cut in half while keeping the size of the container and the
temperature constant?
59. What volume does 1 mol of an ideal gas occupy at standard ­conditions?
60. Which of these occupies the greatest volume?
(a) 0.2 mol of chlorine gas at 48°C and 80 cm Hg
(b) 4.2 g of ammonia at 0.65 atm and - 11°C
(c) 21 g of sulfur trioxide at 55°C and 110 kPa
61. Which of these contains the largest number of molecules?
(a) 1.00 L of CH4 at STP
(b) 3.29 L of N2 at 952 torr and 235°F
(c) 5.05 L of Cl2 at 0.624 atm and 0°C
280 chapter 12
• The Gaseous State of Matter
71. A gas sample at 22°C and 740 torr pressure is heated until its volume is doubled. What pressure would restore the sample to its
­original volume?
62. Which of these has the greatest density?
(a) SF6 at STP
(b) C2H6 at room conditions
(c) He at - 80°C and 2.15 atm
63. For each of the following pairs, predict which sample would have
the greater density.
72. A gas occupies 250. mL at 700. torr and 22°C. When the pressure is
changed to 500. torr, what temperature (°C) is needed to maintain
the same volume?
73. The tires on an automobile were filled with air to 30. psi at 71.0°F.
When driving at high speeds, the tires become hot. If the tires have
a bursting pressure of 44 psi, at what temperature (°F) will the tires
“blow out”?
(a)
or
74. What pressure will 800. mL of a gas at STP exert when its volume
is 250. mL at 30°C?
2
1
(b)
76. How many gas molecules are present in 600. mL of N2O at 40°C
and 400. torr pressure? How many atoms are present? What would
be the volume of the sample at STP?
or
1
75. How many L of SO2 gas at STP will contain 9.14 g of sulfur dioxide gas?
2
64. A chemist carried out a chemical reaction that produced a gas. It
was found that the gas contained 80.0% carbon and 20.0% hydrogen. It was also noticed that 1500 mL of the gas at STP had a mass
of 2.01 g.
(a) What is the empirical formula of the compound?
(b) What is the molecular formula of the compound?
(c) What Lewis structure fits this compound?
65. Three gases were added to the same 2.0-L container. The total pressure of the gases was 790 torr at room temperature (25.0°C). If the
mixture contained 0.65 g of oxygen gas, 0.58 g of carbon dioxide,
and an unknown amount of nitrogen gas, determine the following:
(a) the total number of moles of gas in the container
(b) the number of grams of nitrogen in the container
(c) the partial pressure of each gas in the mixture
66. When carbon monoxide and oxygen gas react, carbon dioxide
results. If 500. mL of O2 at 1.8 atm and 15°C are mixed with
500. mL of CO at 800 mm Hg and 60°C, how many milliliters of
CO2 at STP could possibly result?
67. One of the methods for estimating the temperature at the center of
the sun is based on the ideal gas law. If the center is assumed to be a
mixture of gases whose average molar mass is 2.0 g/mol, and if the
density and pressure are 1.4 g/cm3 and 1.3 * 109 atm, ­respectively,
calculate the temperature.
68. A soccer ball of constant volume 2.24 L is pumped up with air to
a gauge pressure of 13 lb/in.2 at 20.0°C. The molar mass of air is
about 29 g/mol.
(a) How many moles of air are in the ball?
(b) What mass of air is in the ball?
(c) During the game, the temperature rises to 30.0°C. What mass of
air must be allowed to escape to bring the gauge pressure back
to its original value?
77. An automobile tire has a bursting pressure of 60 lb/in.2. The normal
pressure inside the tire is 32 lb/in.2. When traveling at high speeds,
a tire can get quite hot. Assuming that the temperature of the tire is
25°C before running it, determine whether the tire will burst when
the inside temperature gets to 212°F. Show your ­calculations.
78. If you prepared a barometer using water instead of mercury, how
high would the column of water be at one atmosphere pressure?
(Neglect the vapor pressure of water.)
79. How many moles of oxygen are in a 55-L cylinder at 27°C and
2.20 * 103 lb>in.2?
80. How many moles of Cl2 are in one cubic meter (1.00 m3) of Cl2
gas at STP?
81. At STP, 560. mL of a gas have a mass of 1.08 g. What is the molar
mass of the gas?
82. A gas has a density at STP of 1.78 g/L. What is its molar mass?
83. Using the ideal gas law, PV = nRT, calculate the following:
(a) the volume of 0.510 mol of H2 at 47°C and 1.6 atm pressure
(b) the number of grams in 16.0 L of CH4 at 27°C and 600. torr
pressure
(c) the density of CO2 at 4.00 atm pressure and 20.0°C
(d) the molar mass of a gas having a density of 2.58 g/L at 27°C
and 1.00 atm pressure.
84. Acetylene (C2H2) and hydrogen fluoride (HF) react to give
difluoroethane:­
C2H2(g) + 2 HF(g) ¡ C2H4F2(g)
hen 1.0 mol of C2H2 and 5.0 mol of HF are reacted in a 10.0-L
W
flask, what will be the pressure in the flask at 0°C when the reaction is complete?
85. What volume of hydrogen at STP can be produced by reacting
8.30 mol of Al with sulfuric acid? The equation is
2 Al(s) + 3 H2SO4(aq) ¡ Al2(SO4)3(aq) + 3 H2(g)
69. A balloon will burst at a volume of 2.00 L. If it is partially filled at
20.0°C and 65 cm Hg to occupy 1.75 L, at what temperature will
it burst if the pressure is exactly 1 atm at the time that it bursts?
86. A gas has a percent composition by mass of 85.7% carbon and
14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is
the molecular formula of the gas?
70. Given a sample of a gas at 27°C, at what temperature would the volume of the gas sample be doubled, the pressure ­remaining ­constant?
87. Assume that the reaction
2 CO(g) + O2(g) ¡ 2 CO2(g)
Answers to Practice Exercises 281
goes to completion. When 10. mol of CO and 8.0 mol of O2 react
in a closed 10.-L vessel,
(a) how many moles of CO, O2 , and CO2 are present at the end of
the reaction?
(b) what will be the total pressure in the flask at 0°C?
88. If 250 mL of O2 , measured at STP, are obtained by the decomposition of the KClO3 in a 1.20-g mixture of KCl and KClO3 ,
2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g)
what is the percent by mass of KClO3 in the mixture?
CHALLENGE Exercises
89. Air has a density of 1.29 g/L at STP. Calculate the density of air
on Pikes Peak, where the pressure is 450 torr and the temperature
is 17°C.
90. Consider the arrangement of gases shown below. If the valve between the gases is opened and the temperature is held constant,
determine the following:
(a) the pressure of each gas
(b) the total pressure in the system
CO2
V = 3.0 L
P = 150. torr
H2
V = 1.0 L
P = 50. torr
93. A baker is making strawberry cupcakes and wants to ensure they
are very light. To do this, he must add enough baking soda to increase the volume of the cupcakes by 55.0%. All of the carbon
dioxide produced by the decomposition of the baking soda is incorporated into the cupcakes. The baking soda or sodium bicarbonate
will react with citric acid in the batter according to the reaction
3 NaHCO3 + H3C6H5O7 ¡ Na3C6H5O7 + 3 H2O + 3 CO2
which occurs with a 63.7% yield. How many grams of baking
soda must be added to 1.32 L of cupcake batter at a baking temperature of 325°F and a pressure of 738 torr to achieve the correct
­consistency?
91. A steel cylinder contains 50.0 L of oxygen gas under a pressure of
40.0 atm and at a temperature of 25°C. What was the pressure in
the cylinder during a storeroom fire that caused the temperature to
rise 152°C? (Be careful!)
92. A balloon has a mass of 0.5 g when completely deflated. When it is
filled with an unknown gas, the mass increases to 1.7 g. You notice
on the canister of the unknown gas that it occupies a volume of
0.4478 L at a temperature of 50°C. You note the temperature in the
room is 25°C. Identify the gas.
A n s w e r s t o P r ac t i ce E x e r c i s e s
12.1 (a) 851 torr, (b) 851 mm Hg, (c) 113 kPa
12.8 128 g/mol
12.2 0.596 atm
12.9 518 mL
12.3 4.92 L
12.10 0.89 g/L
12.4 84.7 L
12.11 H2 < B2H6 < CO < O2 < N2O < C3H8 < NO2 < POF3
12.5 861 K (588°C)
12.12 1.55 L O2
12.6 1.4 * 102 g/mol
12.13 75.0 L O2 , 45.0 L CO2 , 60.0 L H2O
12.7 0.953 mol
C h a pte r
13
Liquids
P
lanet Earth, that magnificient blue sphere we enjoy viewing
from space, is spectacular. Over 75% of Earth is covered
with water. We are born from it, drink it, bathe in it, cook
with it, enjoy its beauty in waterfalls and rainbows, and stand in
awe of the majesty of icebergs. Water supports and ­enhances life.
In chemistry, water provides the medium for numerous reactions. The shape of the water molecule is the basis for hydrogen bonds. These bonds determine the unique properties and
­reactions of water. The tiny water molecule holds the answers to
many of the mysteries of chemical reactions.
Chapter Outline
13.1 States of Matter: A Review
13.2 Properties of Liquids
13.3 Boiling Point and Melting Point
13.4 Changes of State
13.5 Intermolecular Forces
13.6 Hydrates
13.7 Water, a Unique Liquid
Ingram Publishing/SuperStock
Liquid water provides the base
for the recreation of windsurfing
and also for our bodies. Water is
a unique liquid and is the most
­common liquid on our blue planet.
13.2
• Properties of Liquids 283
13.1 States of matter: a review
In the last chapter, we found that gases contain particles that are far apart, in rapid random
motion, and essentially independent of each other. The kinetic-molecular theory, along with
the ideal gas law, summarizes the behavior of most gases at relatively high temperatures and
low pressures.
Solids are obviously very different from gases. Solids contain particles that are very close
together; solids have a high density, compress negligibly, and maintain their shape regardless
of container. These characteristics indicate large attractive forces between particles. The model
for solids is very different from the one for gases.
Liquids, on the other hand, lie somewhere between the extremes of gases and solids. Liquids contain particles that are close together; liquids are essentially incompressible and have
definite volume. These properties are very similar to those of solids. But liquids also take the
shape of their containers; this is closer to the model of a gas.
Although liquids and solids show similar properties, they differ tremendously from gases
(see Figure 13.1). No simple mathematical relationship, like the ideal gas law, works well for
liquids or solids. Instead, these models are directly related to the forces of attraction between
molecules. With these general statements in mind, let’s consider some specific properties of
liquids.
Figure 13.1
Solid
Liquid
Gas
The three states of matter.
Solid—water molecules are held
together rigidly and are very
close to each other. Liquid—water
molecules are close together
but are free to move around and
slide over each other. Gas—water
molecules are far apart and move
freely and randomly.
13.2 Properties of liquids
Surface Tension
Have you ever observed water and mercury in the form of small drops? These liquids form
drops because liquids have surface tension. A droplet of liquid that is not falling or under the
influence of gravity (as on the space shuttle) will form a sphere. Spheres minimize the ratio of
surface area to volume. The molecules within the liquid are attracted to the surrounding liquid
molecules, but at the liquid’s surface, the attraction is nearly all inward. This pulls the surface
into a spherical shape. The resistance of a liquid to an increase in its surface area is called
the surface tension of the liquid. Substances with large attractive forces between molecules
have high surface tensions. The effect of surface tension in water is illustrated by floating a
needle on the surface of still water. Other examples include a water strider walking across a
calm pond (see Figure 13.2) and water beading on a freshly waxed car. Surface tension is
temperature dependent, decreasing with increasing temperature.
Liquids also exhibit a phenomenon called capillary action, the spontaneous rising of a
liquid in a narrow tube. This action results from the cohesive forces within the liquid and the
adhesive forces between the liquid and the walls of the container. If the forces between the
liquid and the container are greater than those within the liquid itself, the liquid will climb
the walls of the container. For example, consider the California sequoia, a tree that reaches
over 200 feet in height. Although water rises only 33 feet in a glass tube (under atmospheric
pressure), capillary action causes water to rise from the sequoia’s roots to all its parts.
Learning objective
Key tErms
surface tension
capillary action
meniscus
evaporation
vaporization
sublimation
condensation
vapor pressure
volatile
Charles D. Winters/Photo Researchers
Explain why liquids tend to form drops and explain the process of evaporation and
its ­relationship to vapor pressure.
Mercury droplets.
chapter 13
• Liquids
Figure 13.2
Yoav levy/Phototake
A water strider skims the surface
of the water as a result of surface
tension. At the molecular level,
the surface tension results
from the net attraction of the
water molecules toward the liquid
below. In the interior of the water,
the forces are balanced in all
directions.
Figure 13.3
The meniscus of mercury (left) and
water (right). The meniscus is the
characteristic curve of the surface
of a liquid in a narrow tube.
Nature’s Images/Photo Researchers
284 The meniscus in liquids is further evidence of cohesive and adhesive forces. When a liquid
is placed in a glass cylinder, the surface of the liquid shows a curve called the meniscus (see
Figure 13.3). The concave shape of water’s meniscus shows that the adhesive forces between
the glass and water are stronger than the cohesive forces within the water. In a nonpolar
­substance such as mercury, the meniscus is convex, indicating that the cohesive forces within
mercury are greater than the adhesive forces between the glass wall and the mercury.
Evaporation
When beakers of water, ethyl ether, and ethyl alcohol (all liquids at room temperature) are
­allowed to stand uncovered, their volumes gradually decrease. The process by which this
change takes place is called evaporation.
Attractive forces exist between molecules in the liquid state. Not all of these molecules,
however, have the same kinetic energy. Molecules that have greater-than-average kinetic
­energy can overcome the attractive forces and break away from the surface of the liquid to
become a gas (see Figure 13.4). Evaporation, or vaporization, is the escape of molecules
from the liquid state to the gas or vapor state.
In evaporation, molecules of greater-than-average kinetic energy escape from a liquid,
leaving it cooler than it was before they escaped. For this reason, evaporation of perspiration is
one way the human body cools itself and keeps its temperature constant. When volatile liquids
such as ethyl chloride (C2H5Cl) are sprayed on the skin, they evaporate rapidly, cooling the
area by removing heat. The numbing effect of the low temperature produced by evaporation
of ethyl chloride allows it to be used as a local anesthetic for minor surgery.
Solids such as iodine, camphor, naphthalene (moth balls), and, to a small extent, even ice
will go directly from the solid to the gaseous state, bypassing the liquid state. This change is
a form of evaporation and is called sublimation:
evaporation
liquid ::::" vapor
sublimation
solid ::::" vapor
Figure 13.4
The high-energy molecules escape
from the surface of a liquid in a
process known as evaporation.
Example 13.1
Which will evaporate more quickly: 60 mL of water in a beaker with a diameter of 5 cm
or 60 mL of water in a shallow dish with a diameter of 15 cm? Why?
Solution
The water in the shallow dish will evaporate more quickly since there is more surface
area in the dish than in the beaker.
13.2
Evaporation
Condensation
begins
(a)
Rate of evaporation =
rate of condensation
(b)
Figure 13.5
(a) Molecules in an open flask
evaporate from the liquid and
disperse into the atmosphere.
(b) When the flask is stoppered,
the number of gaseous molecules
increases inside the flask. Some
gaseous molecules collide with
the surface of the liquid and
stick—condensation occurs.
(c) When the rate of evaporation
equals the rate of condensation,
an equilibrium between the liquid
and vapor is established.
(c)
Vapor Pressure
When a liquid vaporizes in a closed system like that shown in Figure 13.5b, some of the molecules in the vapor or gaseous state strike the surface and return to the liquid state by the process
of condensation. The rate of condensation increases until it’s equal to the rate of vaporization.
At this point, the space above the liquid is said to be saturated with vapor, and an equilibrium,
or steady state, exists between the liquid and the vapor. The equilibrium equation is
99991 vapor
liquid 29999
condensation
evaporation
This equilibrium is dynamic; both processes—vaporization and condensation—are taking
place, even though we cannot see or measure a change. The number of molecules leaving
the liquid in a given time interval is equal to the number of molecules returning to the liquid.
At equilibrium, the molecules in the vapor exert a pressure like any other gas. The pressure exerted by a vapor in equilibrium with its liquid is known as the vapor pressure of the
liquid. The vapor pressure may be thought of as a measure of the “escaping” tendency of molecules to go from the liquid to the vapor state. The vapor pressure of a liquid is independent
of the amount of liquid and vapor present, but it increases as the temperature rises. We can
measure the vapor pressure of a liquid by using a simple barometer as shown in Figure 13.6.
H2O vapor
Vacuum
Ethyl
alcohol
vapor
Diethyl
ether
vapor
Original
Hg level
18 mm
44 mm
Patm =
760 mm Hg
442 mm
742 mm
716 mm
318 mm
(a)
(b)
(c)
• Properties of Liquids 285
(d)
Figure 13.6
(a)A barometer measures atmospheric pressure. (Values given for 20°C)
To measure the vapor pressure of a liquid, a tiny amount is injected into the barometer.
(b) The water vapor presses the Hg down 18 mm (vapor pressure = 18 mm).
(c) Ethyl alcohol presses the Hg level down 44 mm (vapor pressure = 44 mm).
(d) Ethyl ether has a vapor pressure of 442 mm Hg.
286 chapter 13
• Liquids
Mercury is so dense that if we inject a sample of liquid at the bottom of the Hg tube, the liquid
rises to the top of the Hg column. It is then trapped in a closed space where it produces a vapor
that pushes down the Hg column. When equilibrium between the liquid and vapor is reached,
we can measure the vapor pressure by the change in height of the Hg column.
When equal volumes of water, ethyl ether, and ethyl alcohol are placed in beakers and
allowed to evaporate at the same temperature, we observe that the ether evaporates faster than
the alcohol, which evaporates faster than the water. This order of evaporation is consistent with
the fact that ether has a higher vapor pressure at any particular temperature than ethyl alcohol
or water. One reason for this higher vapor pressure is that the attraction is less between ether
molecules than between alcohol or water molecules.
Substances that evaporate readily are said to be volatile. A volatile liquid has a relatively
high vapor pressure at room temperature. Ethyl ether is a very volatile liquid; water is not
too volatile; and mercury, which has a vapor pressure of 0.0012 torr at 20°C, is essentially a
nonvolatile liquid. Most substances that are normally in a solid state are nonvolatile (solids
that sublime are exceptions).
Practice 13.1
How is the vapor pressure of a substance defined? How does it depend on temperature?
Practice 13.2
A liquid in a closed container has a constant vapor pressure. What is the relationship
between the rate of evaporation of the liquid and the rate of condensation of the vapor in
the container?
13.3 Boiling Point and melting point
Learning objective
Define boiling point and melting point and determine the boiling point of a liquid
from a graph of temperature versus vapor pressure.
Key terms
The boiling temperature of a liquid is related to its vapor pressure. We’ve seen that vapor
pressure increases as temperature increases. When the internal or vapor pressure of a liquid
becomes equal to the external pressure, the liquid boils. (By external pressure we mean the
pressure of the atmosphere above the liquid.) The boiling temperature of a pure liquid remains
constant as long as the external pressure does not vary.
The boiling point (bp) of water is 100°C at 1 atm pressure. Figure 13.7 shows that the
vapor pressure of water at 100°C is 760 torr. The significant fact here is that the boiling point
is the temperature at which the vapor pressure of the water or other liquid is equal to standard,
or atmospheric, pressure at sea level. Boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure above the liquid.
We can readily see that a liquid has an infinite number of boiling points. When we give
the boiling point of a liquid, we should also state the pressure. When we express the boiling
point without stating the pressure, we mean it to be the normal boiling point at standard
pressure (760 torr). Using Figure 13.7 again, we see that the normal boiling point of ethyl
ether is between 30°C and 40°C, and for ethyl alcohol it is between 70°C and 80°C because
for each compound 760 torr lies within these stated temperature ranges. At the normal boiling
point, 1 g of a liquid changing to a vapor (gas) absorbs an amount of energy equal to its heat
of vaporization (see Table 13.1).
The boiling point at various pressures can be evaluated by plotting the vapor pressures of
various liquids on the graph in Figure 13.7, where temperature is plotted horizontally along the
x-axis and vapor pressure is plotted vertically along the y-axis. The resulting curve is known
as vapor pressure curve. Any point on this curve represents a vapor–liquid equilibrium at a
particular temperature and pressure. We can find the boiling point at any pressure by tracing a
horizontal line from the designated pressure to a point on the vapor pressure curve. From this
point, we draw a vertical line to obtain the boiling point on the temperature axis. Three such
points are shown in Figure 13.7; they represent the normal boiling points of the three compounds at 760 torr. By reversing this process, you can ascertain at what pressure a substance
boiling point
normal boiling point
vapor pressure curve
melting point (or freezing point)
13.3
• Boiling Point and Melting Point 287
1300
1200
1100
1000
Ethyl
ether
Vapor pressure (torr)
900
Ethyl
alcohol
Water
800
700
600
500
400
300
200
Figure 13.7
100
0
–10
0
10
20
30
40
50
60
Temperature (°C)
70
80
90
110
100
Vapor pressure–temperature
curves for ethyl ether, ethyl
alcohol, and water.
will boil at a specific temperature. The boiling point is one of the most commonly used physical properties for characterizing and identifying substances.
Example 13.2
ENHANCED EXAMPLE
Determine the boiling point of ethyl alcohol at a vapor pressure of 500 torr.
Solution
Use Figure 13.7. Trace a horizontal line from 500 torr to the green ethyl alcohol curve.
Then trace a vertical line down from the curve to the x-axis. The boiling point is 69°C.
Practice 13.3
Use the graph in Figure 13.7 to determine the boiling points of ethyl ether, ethyl alcohol,
and water at 600 torr.
Practice 13.4
The average atmospheric pressure in Denver is 0.83 atm. What is the boiling point of
water in Denver?
As heat is removed from a liquid, the liquid becomes colder and colder, until a temperature
is reached at which it begins to solidify. A liquid changing into a solid is said to be freezing, or
solidifying. When a solid is heated continuously, a temperature is reached at which the solid
begins to liquefy. A solid that is changing into a liquid is said to be melting. The temperature
Table 13.1 Physical Properties of Ethyl Chloride, Ethyl Ether, Ethyl Alcohol,
and Water
Substance
Ethyl chloride
Ethyl ether
Ethyl alcohol
Water
Boiling
point
(°C)
Melting
point
(°C)
12.3
34.6
78.4
100.0
-139
-116
-112
0
Heat of
vaporization
J/g (cal/g)
Heat of
fusion
J/g (cal/g)
387 (92.5)
351 (83.9)
855 (204.3)
2259 (540)
—
—
104 (24.9)
335 (80)
288 chapter 13
• Liquids
>Chemistry in Action
The chemical reaction between Mentos and diet soda
has been seen everywhere from chemistry classrooms to
David Letterman to YouTube. The Mentos candy fizzes,
the diet soda sprays out in a tall fountain, and everyone
laughs. But just what is happening during the reaction?
Lots of informal explanations have been given including one in Chemical and Engineering News and also one
on “MythBusters” on the Discovery Channel. Yet none
of these have been systematic experiments to discover
the details of the reaction. Now Tonya Coffey and her
students at Appalachian State University have explained
the details of the reaction’s parameters.
Coffey and her students discovered that the type of
Mentos did not matter and neither did the type of diet
soda (caffeinated or clear). They also found that the reaction was not a simple acid–base reaction; the acidity of
the soda was the same before and after the reaction, and
Mentos candy is not basic. The key component of the reaction is the ability of some of the reactants to reduce the
surface tension of the soda, allowing the carbon dioxide
to escape more quickly, producing the fountain. It turns
out that the gum Arabic in the Mentos’s outer coating
is an excellent surfactant (a substance that reduces the
surface tension of a liquid). In addition, two chemicals
in the diet soda (aspartame and potassium benzoate)
also reduce surface tension and assist in the formation of
bubbles. Coffey and her students also determined (using
a scanning electron microscope) that the surface of the
Mentos candy is very rough, providing many “nucleation
sites” for carbon dioxide bubbles. The Appalachian State
University students tested other materials (sand, dishwashing detergent, salt) to determine whether they might
produce fountains and found that they did but not as
­spectacular as the ones Mentos produced.
Several other factors contribute to the exciting nature of
the reaction. Mentos candy is very dense, so it sinks to the
bottom of the soda rapidly. The students also found that
temperature matters: Warm soda leads to a higher fountain.
Charles D. Winters/Photo Researchers, Inc.
Chemical Eye Candy
The Mentos in Diet Coke reactions
at which the solid phase of a substance is in equilibrium with its liquid phase is known as the
melting point or freezing point of that substance. The equilibrium equation is
9991 liquid
solid 2999
freezing
melting
When a solid is slowly and carefully heated so that a solid–liquid equilibrium is achieved and
then maintained, the temperature will remain constant as long as both phases are present. The
energy is used solely to change the solid to the liquid. The melting point is another physical
property that is commonly used for characterizing substances.
The most common example of a solid–liquid equilibrium is ice and water. In a well-stirred
system of ice and water, the temperature remains at 0°C as long as both phases are present.
The melting point changes only slightly with pressure unless the pressure change is very large.
13.4Changes of State
Learning objective
Calculate the amount of energy involved in a change of state.
Key terms
The majority of solids undergo two changes of state upon heating. A solid changes to a liquid
at its melting point, and a liquid changes to a gas at its boiling point. This warming process
can be represented by a graph called a heating curve (Figure 13.8). This figure shows ice
being heated at a constant rate. As energy flows into the ice, the vibrations within the crystal
increase and the temperature rises (A ¡ B). Eventually, the molecules begin to break free
heat of fusion
heat of vaporization
13.4
120
• Changes of State 289
F
Vapor (steam)
D
100
E
Liquid to vapor
(water to steam)
Temperature (°C )
80
60
Liquid (water)
40
Solid to liquid
(ice to water)
20
0
B
A
C
Solid (ice)
– 20
Heat added
Figure 13.8
Heating curve for a pure
substance—the absorption of
heat by a substance from the solid
state to the vapor state. Using
water as an example, the AB
interval represents the ice phase;
BC interval, the melting of ice to
water; CD interval, the elevation
of the temperature of water from
0°C to 100°C; DE interval, the
boiling of water to steam; and EF
interval, the heating of steam.
from the crystal and melting occurs (B ¡ C). During the melting process, all energy goes
into breaking down the crystal structure; the temperature remains constant.
The energy required to change exactly one gram of a solid at its melting point into a
liquid is called the heat of fusion. When the solid has completely melted, the temperature
once again rises (C ¡ D); the energy input is increasing the molecular motion within the
water. At 100°C, the water reaches its boiling point; the temperature remains constant while
the added energy is used to vaporize the water to steam (D ¡ E). The heat of vaporization is the energy required to change exactly one gram of liquid to vapor at its normal boiling
point. The attractive forces between the liquid molecules are overcome during vaporization.
Beyond this temperature, all the water exists as steam and is being heated further (E ¡ F).
Example 13.3
How many joules of energy are needed to change 10.0 g of ice at 0.00°C to water at
20.0°C?
Solution
Read •
Knowns:
m = 10.0 g ice
tinitial = 0.00°C
tfinal = 20.0°C
Solving for: energy
Plan •
The process involves two steps:
1. melting the ice (absorbing 335 J/g) and
2. warming the water from 0.00°C - 20.0°C, which requires 4.184 J/g°C.
Setup •
1. Energy needed to melt the ice
335 J
b = 3350 J = 3.35 * 103 J
1g
2. Energy needed to warm the water (mass)(specific heat)(t) = energy
(10.0 g) a
(10.0 g)a
4.184 J
b (20.0 °C ) = 837 J
1 g°C
Calculate • The total amount of energy required for the process is the sum of the two
steps:
3350 J + 837 J = 4.19 * 103 J
ENHANCED EXAMPLE
290 chapter 13
• Liquids
Example 13.4
How many kilojoules of energy are needed to change 20.0 g of water at 20.°C to steam at
100.°C?
Solution
Read •
Knowns:
m = 20.0 g water
tinitial = 20.°C
tfinal = 100.°C
Solving for: energy
Plan •
The process involves two steps:
1. warming the water from 20°C to 100°C, which requires 4.184 J/g°C, and
2. vaporizing the water to steam (absorbing 2.26 kJ/g).
Setup •
1. Energy needed to warm the water
(mass)(specific heat) (t) = energy
(20.0 g)a
4.184 J
b (80. °C ) = 6.7 * 103 J = 6.7 kJ
1 g°C
2. Energy needed to vaporize the water to steam
2.26 kJ
(20.0 g)a
b = 45.2 kJ
1g
Calculate • The total amount of energy required for the process is the sum of the two
steps:
6.7 kJ + 45.2 kJ = 51.9 kJ
Practice 13.5
How many kilojoules of energy are required to change 50.0 g of ethyl alcohol from
60.0°C to vapor at 78.4°C? The specific heat of ethyl alcohol is 2.138 J/g°C.
13.5 Intermolecular Forces
Learning objecti ve
Describe the three types of intermolecular forces and explain their significance in
liquids.
Key terms
Why do molecules of a liquid stay together instead of floating away from each other like
molecules of a gas? Intermolecular forces are the forces of attraction between molecules.
These forces hold molecules close together and allow for the formation of liquids and solids.
These intermolecular forces can also help us to predict some of the properties of liquids. For
example, if the molecules of a liquid are strongly held together, it will be difficult to pull the
molecules apart, resulting in a low vapor pressure and a high boiling point. Molecules with
strong intermolecular forces are also likely to have strong cohesive forces resulting in a high
surface tension as well.
Molecules are held together by bonds, sometimes called intramolecular forces, that
occur inside the molecules. These covalent bonding forces arise from sharing electrons between atoms. Intermolecular forces occur between molecules. How do these forces happen?
There are several different kinds of intermolecular forces. Some of the most common are
dipole-dipole attractions, hydrogen bonds, and London dispersion forces. Let’s look at each
of these types of intermolecular forces in more detail.
intermolecular forces
intramolecular forces
dipole–dipole attractions
hydrogen bond
London dispersion forces
Dipole–Dipole Attractions
As we learned in Chapter 11, some covalent bonds do not share electrons equally because of
differences in the electronegativity of the bonded atoms. When this happens, the polar bonds
13.5
• Intermolecular Forces 291
that form may create polar molecules or molecules with a positive end and a negative end.
Some examples of polar molecules, such as HF, CO, and ICl, are easy to identify because they
have only two atoms. Others are more difficult because of their complex molecular geometry.
Example 13.5
Determine which of the following molecules are polar: HCl, CO2, and NCl3.
Solution
The Lewis structures and three-dimensional drawings for these three molecules are
shown here. Arrows represent the dipole moments of the individual bonds.
O==C==O
..
The carbon dioxide molecule
is nonpolar because the ­carbon–
oxygen dipoles cancel each other
by acting in equal and opposite
directions.
The nitrogen–chlorine bonds are
all polar and all three dipoles
point toward the nitrogen atom
resulting in a net dipole up. This
is a polar molecule.
.. ..
.. Cl—H
Chlorine is more
electronegative than
hydrogen so both the
H i Cl bond and the
molecule are polar.
.. ..
.. ..
—
.. ..
..
.. ..
..Cl—N—Cl..
.. Cl..
Practice 13.6
Determine which of the following molecules are polar: SO2, Br2, and CH3F.
–
+
–
–
+
–
+
–
+
–
+
+
+
–
–
–
+
+
As we saw in Chapter 11 water is a polar molecule—it has a dipole moment.
When molecules with dipole moments are put together, they orient themselves to take
advantage of their charge distribution. The positive ends of the molecules are attracted
to the negative ends of other molecules. Because of this attraction, the molecules tend
to stick together creating dipole–dipole attraction.
The Hydrogen Bond
Table 13.2 compares the physical properties of H2O, H2S, H2Se, and H2Te. From this comparison, it is apparent that four physical properties of water—melting point, boiling point,
heat of fusion, and heat of vaporization—are extremely high and do not fit the trend relative to
the molar masses of the four compounds. For example, if the properties of water followed the
progression shown by the other three compounds, we would expect the melting point of water
to be below -85°C and the boiling point to be below -60°C.
Why does water exhibit these anomalies? Because liquid water molecules are held together more strongly than other molecules in the same family. The intermolecular force acting
Table 13.2
Physical Properties of Water and Other Hydrogen Compounds of Group 6A Elements
Formula
Color
Molar
mass
(g/mol)
Melting
point
(°C)
Boiling point,
1 atm
(°C)
Heat of
fusion
J/g (cal/g)
Heat of
vaporization
J/g (cal/g)
H2O
H2S
H2Se
H2Te
18.02
34.09
80.98
129.6
0.00
- 85.5
- 65.7
- 49
100.0
- 60.3
- 41.3
-2
335 (80.1)
69.9 (16.7)
31 (7.4)
—
2 .26 * 103 (540)
548 (131)
238 (57.0)
179 (42.8)
Colorless
Colorless
Colorless
Colorless
292 chapter 13
• Liquids
Oxygen
Hydrogen
Hydrogen bonds
Figure 13.9
Hydrogen bonding. Water in the
liquid and solid states exists as
aggregates in which the water
molecules are linked together by
hydrogen bonds.
between water molecules is called a hydrogen bond, which acts like a very weak bond between two polar molecules. A hydrogen bond is formed between polar molecules that contain
hydrogen covalently bonded to a small, highly electronegative atom such as fluorine, oxygen,
or nitrogen (F i H, O i H, N i H). A hydrogen bond is actually the dipole–dipole attraction
between polar molecules containing these three types of polar bonds.
Compounds that have significant hydrogen-bonding ability are those that contain H
­covalently bonded to F, O, or N.
Because a hydrogen atom has only one electron, it forms only one covalent bond. When
it is bonded to a strong electronegative atom such as oxygen, a hydrogen atom will also be
attracted to an oxygen atom of another molecule, forming a dipole–dipole attraction (H bond)
between the two molecules. Water has two types of bonds: covalent bonds that exist between
hydrogen and oxygen atoms within a molecule and hydrogen bonds that exist between hydrogen and oxygen atoms in different water molecules.
Hydrogen bonds are intermolecular bonds; that is, they are formed between atoms
in different molecules. They are somewhat ionic in character because they are formed by
electrostatic attraction. Hydrogen bonds are much weaker than the ionic or covalent bonds
that unite atoms to form compounds. Despite their weakness, they are of great chemical
importance.
The oxygen atom in water can form two hydrogen bonds—one through each of the
unbonded pairs of electrons. Figure 13.9 shows two water molecules linked by a hydrogen bond and eight water molecules linked by hydrogen bonds. A dash (9) is used for the
covalent bond and a dotted line (DDDD) for the hydrogen bond. In water each molecule is
linked to others through hydrogen bonds to form a three-dimensional aggregate of water
molecules. This intermolecular hydrogen bonding effectively gives water the properties
of a much larger, heavier molecule, explaining in part its relatively high melting point,
boiling point, heat of fusion, and heat of vaporization. As water is heated and energy
absorbed, hydrogen bonds are continually being broken until at 100°C, with the absorption of an additional 2.26 kJ/g, water separates into individual molecules, going into the
gaseous state. Sulfur, selenium, and tellurium are not sufficiently electronegative for their
hydrogen compounds to behave like water. The lack of hydrogen bonding is one reason
H2S is a gas, not a liquid, at room temperature.
Fluorine, the most electronegative element, forms the strongest hydrogen bonds. This
bonding is strong enough to link hydrogen fluoride molecules together as dimers, H2F2 , or as
larger (HF)n molecular units. The dimer structure may be represented in this way:
F
H
H
F
H bond
13.5
• Intermolecular Forces 293
>Chemistry in Action
How Sweet It Is!
Did you think artificial sweeteners were a product of the
post–World War II chemical industry? Not so—many of
them have been around a long time, and several of the
important ones were discovered quite by accident. In 1878,
Ira Remsen was working late in his laboratory and realized
he was about to miss a dinner with friends. In his haste to
leave the lab, he forgot to wash his hands. Later at dinner
he broke a piece of bread and tasted it only to discover that
it was very sweet. The sweet taste had to be the chemical
he had been working with in the lab. Back at the lab, he
isolated saccharin—the first of the artificial sweeteners.
contain i NH and i OH groups (with hydrogen available
to bond) as well as C “ O groups (providing oxygen for
hydrogen bonding). “Sweet molecules” also contain
H-bonding groups including i OH, i NH2, and O or N.
These molecules not only must have the proper atoms to
form hydrogen bonds, but they must also contain a hydrophobic region (repels H2O). A new model for binding to a
sweetness receptor has been developed at Senomyx in La
Jolla, California. The model shows four binding sites that
can act independently. Small molecules bind to a pocket on
a subunit as shown in the model. Large molecules (such as
proteins) bind to a different site above one of the pockets.
In 1965, James Schlatter was researching antiulcer drugs
for the pharmaceutical firm G. D. Searle. In the course
of his work, he accidentally ingested a small amount of
a preparation and found to his surprise that it had an
extremely sweet taste. He had discovered aspartame, a
molecule consisting of two amino acids joined together.
Since only very small quantities of aspartame are necessary to produce sweetness, it proved to be an excellent
low-calorie artificial sweetener. More than 50 different
molecules have a sweet taste, and it is difficult to find a
single binding site that could interact with all of them.
Our taste receptors are composed of proteins that can
form hydrogen bonds with other molecules. The proteins
J. Mol. Graphics 1996, 14, 51
In 1937, Michael Sveda was smoking a cigarette in his
laboratory (a very dangerous practice to say the least!).
He touched the cigarette to his lips and was surprised by
the exceedingly sweet taste. The chemical on his hands
turned out to be cyclamate, which soon became a staple
of the artificial sweetener industry.
Sweet bondage. Model shows how the sweetener
aspartame binds to a site on the sweetness receptor’s
T1R3 subunit. Red and blue are hydrogen-bond donor and
acceptor residues, respectively; aspartame is in gold, except
for its carboxylate (red) and ammonium (blue) groups. Model
prepared with MOLMOL. Adapted from J. Med. Chem.
Hydrogen bonding can occur between two different atoms that are capable of forming
H bonds. Thus, we may have an ODDDDH i N or O i HDDDDN linkage in which the hydrogen
atom forming the H bond is between an oxygen and a nitrogen atom. This form of H bond
­exists in certain types of protein molecules and many biologically active substances.
Example 13.6
Would you expect hydrogen bonding to occur between molecules of these substances?
H
H
H H
(a) H
C
C
H
H
O
(b) H
H
C
O
H
C
H
H
dimethyl ether
ethyl alcohol
SOLUTION
(a) Hydrogen bonding should occur in ethyl alcohol because one hydrogen atom is bonded
to an oxygen atom:
H H
H H H
H
C
C
H
H
O
O
C
C
H bond
H
H
H
H
ENHANCED EXAMPLE
294 chapter 13
• Liquids
(b) There is no hydrogen bonding in dimethyl ether because all the hydrogen atoms are
bonded only to carbon atoms.
Both ethyl alcohol and dimethyl ether have the same molar mass (46.07 g/mol).
Although both compounds have the same molecular formula, C2H6O , ethyl alcohol has a
much higher boiling point (78.4°C) than dimethyl ether (-23.7°C) because of hydrogen
bonding between the alcohol molecules.
Practice 13.7
Would you expect hydrogen bonding to occur between molecules of these substances?
H H H
H
H H H
(a) H
C
C
H
H
N
(b) H
H
C
H
N
C
(c)
H
H
H
H
H H
C
C
C
N
H
H H
C
H
H
H
London Dispersion Forces
Even molecules without dipole moments must exert forces on each other. We know this ­because
all substances, even noble gases, exist in the liquid and solid states (at very low temperatures).
The forces that exist between nonpolar molecules and also between noble gas molecules are
called London dispersion forces. Let’s consider a couple of noble gas atoms. We would
­assume the electrons are evenly distributed around the nucleus.
No polarity
Instantaneous dipole
δ–
δ+
+
+
But the electrons are in motion around the nucleus, and so this even distribution is not true all
of the time. Atoms can develop an instantaneous dipolar arrangement of charge. This instantaneous dipole can induce a similar dipole in a nearby atom.
Instantaneous dipole on A
induces a dipole on B.
Instantaneous dipole
δ–
δ+
δ–
δ+
δ–
δ+
+
+
+
+
Atom A
Atom B
Atom A
Atom B
The interaction is weak and short-lived but can be important in large atoms and molecules.
London dispersion forces become more significant as the atoms or molecules get larger. Larger
size means more electrons are available to form dipoles.
The strength of this attraction depends on the molecular mass of the substance. Nonpolar
molecules with very high molar masses such as paraffin wax, C25H52, have strong enough
intermolecular forces to form soft solids.
Example 13.7
Predict which of the following compounds will have the strongest London dispersion
forces: C4H10, C15H32, and C8H18.
Solution
London dispersion forces depend on molar mass. The molar masses of these compounds
are 58.12 amu, 114.2 amu, and 212.4 amu, so C15H32 will have the strongest London
dispersion forces.
13.6
• Hydrates 295
Practice 13.8
Predict which of the following compounds will have the strongest London dispersion
forces: F2, Br2, and I2.
13.6Hydrates
Explain what hydrates are, write formulas for hydrates, and calculate the percent
water in a hydrate.
Learning objectiv e
When certain solutions containing ionic compounds are allowed to evaporate, some water
molecules remain as part of the crystalline compound that is left after evaporation is complete. Solids that contain water molecules as part of their crystalline structure are known as
hydrates. Water in a hydrate is known as water of hydration or water of crystallization.
Formulas for hydrates are expressed by first writing the usual anhydrous (without water)
formula for the compound and then adding a dot followed by the number of water molecules
present. An example is BaCl2 # 2 H2O. This formula tells us that each formula unit of this
compound contains one barium ion, two chloride ions, and two water molecules. A crystal of
the compound contains many of these units in its crystalline lattice.
In naming hydrates, we first name the compound exclusive of the water and then add
the term hydrate, with the proper prefix representing the number of water molecules in the
formula. For example, BaCl2 # 2 H2O is called barium chloride dihydrate. Hydrates are true
compounds and follow the law of definite composition. The molar mass of BaCl2 # 2 H2O is
244.2 g/mol; it contains 56.22% barium, 29.03% chlorine, and 14.76% water.
Water molecules in hydrates are bonded by electrostatic forces between polar water molecules and the positive or negative ions of the compound. These forces are not as strong as
covalent or ionic chemical bonds. As a result, water of crystallization can be removed by
moderate heating of the compound. A partially dehydrated or completely anhydrous compound
may result. When BaCl2 # 2 H2O is heated, it loses its water at about 100°C:
Key terms
100°C "
BaCl2 # 2 H2O(s) :::
BaCl2(s) + 2 H2O(g)
Richard Megna/Fundamental Photographs
Richard Megna/Fundamental Photographs
When a solution of copper(II) sulfate (CuSO4) is allowed to evaporate, beautiful blue crystals containing 5 moles water per 1 mole CuSO4 are formed (Figure 13.10a). The formula for
(b)
Figure 13.10
(a)
(a) When these blue crystals of
CuSO4 # 5 H2O are dissolved in water, a blue
solution forms. (b) The anhydrous crystals
of CuSO4 are pale green. When water is
added, they immediately change color to
blue CuSO4 # 5 H2O crystals.
hydrate
water of hydration
water of crystallization
296 chapter 13
• Liquids
this hydrate is CuSO4 # 5 H2O; it is called copper(II) sulfate pentahydrate. When CuSO4 # 5 H2O
is heated, water is lost, and a pale green-white powder, anhydrous CuSO4 , is formed:
150°C "
CuSO4 # 5 H2O(s) :::
CuSO4(s) + 5 H2O(g)
When water is added to anhydrous copper(II) sulfate, the foregoing reaction is reversed, and
the compound turns blue again (Figure 13.10b). Because of this outstanding color change,
anhydrous copper(II) sulfate has been used as an indicator to detect small amounts of water.
The formation of the hydrate is noticeably exothermic.
The formula for plaster of paris is (CaSO4)2 # H2O. When mixed with the proper quantity
of water, plaster of paris forms a dihydrate and sets to a hard mass. It is therefore useful for
making patterns for the production of art objects, molds, and surgical casts. The chemical
reaction is
(CaSO4)2 # H2O(s) + 3 H2O(l) ¡ 2 CaSO4 # 2 H2O(s)
Table 13.3 lists a number of common hydrates.
Table 13.3
Selected Hydrates
Hydrate
Name
Hydrate
Name
CaCl2 # 2 H2O
Ba(OH)2 # 8 H2O
MgSO4 # 7 H2O
SnCl2 # 2 H2O
CoCl2 # 6 H2O
calcium chloride dihydrate
barium hydroxide octahydrate
magnesium sulfate heptahydrate
tin(II) chloride dihydrate
cobalt(II) chloride hexahydrate
Na2CO3 # 10 H2O
(NH4)2C2O4 # H2O
NaC2H3O2 # 3 H2O
Na2B4O7 # 10 H2O
Na2S2O3 # 5 H2O
sodium carbonate decahydrate
ammonium oxalate monohydrate
sodium acetate trihydrate
sodium tetraborate decahydrate
sodium thiosulfate pentahydrate
Example 13.8
ENHANCED EXAMPLE
Write the formula for calcium chloride hexahydrate.
Solution
Begin by writing the formula for calcium chloride, CaCl2. Remember the Ca is +2 and
the Cl is -1 so we need to have two Cl atoms in the formula for the charge to be neutral.
Next determine the number of water molecules in the hydrate. The prefix hexa- means 6,
so the formula for the hydrate would be CaCl2 # 6 H2O.
Practice 13.9
Write formulas for
(a) beryllium carbonate tetrahydrate
(b) cadmium permanganate hexahydrate
(c) chromium(III) nitrate nonahydrate
(d) platinum(IV) oxide trihydrate
Example 13.9
Calculate the percent water in CuSO4 # 5 H2O.
Solution
Using the periodic table, find the molar mass of CuSO4. It is 159.6 g/mol. The molar
mass for water is 18.02 g/mol. Since there are 5 mol of water associated with each mol of
CuSO4 in the hydrate, the water has a mass of 5 * 18.02 = 90.10 g. The percent water is
90.10 g
* 100% = 36.08% water
159.6 g + 90.10 g
Practice 13.10
Calculate the percent water in Epsom salts, MgSO4 # 7 H2O
13.7
• Water, a Unique Liquid 297
13.7 Water, a Unique Liquid
Describe the characteristics of water in terms of its structure and list the sources of
drinking water.
Learning objectiv e
Water is our most common natural resource. It covers about 75% of Earth’s surface. Not only
is it found in the oceans and seas, in lakes, rivers, streams, and glacial ice deposits, it is always
present in the atmosphere and in cloud formations.
About 97% of Earth’s water is in the oceans. This saline water contains vast amounts
of dissolved minerals. More than 70 elements have been detected in the mineral content of
seawater. Only four of these—chlorine, sodium, magnesium, and bromine—are now commercially obtained from the sea. The world’s fresh water comprises the other 3%, of which about
two-thirds is locked up in polar ice caps and glaciers. The remaining fresh water is found in
groundwater, lakes, rivers, and the atmosphere.
Water is an essential constituent of all living matter. It is the most abundant compound
in the human body, making up about 70% of total body mass. About 92% of blood plasma is
water; about 80% of muscle tissue is water; and about 60% of a red blood cell is water. Water
is more important than food in the sense that we can survive much longer without food than
without water.
Key tErm
semipermeable membrane
Physical Properties of Water
Water is a colorless, odorless, tasteless liquid with a melting point of 0°C and a boiling point
of 100°C at 1 atm. The heat of fusion of water is 335 J/g (80 cal/g). The heat of vaporization of
water is 2.26 kJ/g (540 cal/g). The values for water for both the heat of fusion and the heat of
vaporization are high compared with those for other substances; this indicates strong attractive
forces between the molecules.
Ice and water exist together in equilibrium at 0°C, as shown in Figure 13.11. When ice
at 0°C melts, it absorbs 335 J/g in changing into a liquid; the temperature remains at 0°C. To
refreeze the water, 335 J/g must be removed from the liquid at 0°C.
In Figure 13.11, both boiling water and steam are shown to have a temperature of 100°C.
It takes 418 J to heat 1 g of water from 0°C to 100°C, but water at its boiling point absorbs
2.26 kJ/g in changing to steam. Although boiling water and steam are both at the same temperature, steam contains considerably more heat per gram and can cause more severe burns
than hot water.
The maximum density of water is 1.000 g/mL at 4°C. Water has the unusual property of
contracting in volume as it is cooled to 4°C and then expanding when cooled from 4°C to 0°C.
Therefore, 1 g of water occupies a volume greater than 1 mL at all temperatures except 4°C.
Although most liquids contract in volume all the way down to the point at which they solidify,
a large increase (about 9%) in volume occurs when water changes from a liquid at 0°C to a
solid (ice) at 0°C. The density of ice at 0°C is 0.917 g/mL, which means that ice, being less
dense than water, will float in water.
0°C
(b)
100°C
Martin Dohrn/Photo Researchers, Inc.
(a)
© Stian Magnus Hatling/iStockphoto
Figure 13.11
(a) Ice and water in equilibrium
at 0°C. When ice melts, 335 J/g
are needed to change it from
solid to liquid. The same amount
is released in freezing. (b) Boiling
water and steam in equilibrium
at 100°C. It takes 2.26 kJ/g to
convert water to steam.
298 chapter 13
• Liquids
>Chemistry in Action
Reverse Osmosis?
If you have ever made a salad and been disappointed to
find that the lettuce has become limp and the dressing has
become watery, you have experienced osmosis. Osmosis is
the process by which water flows through a membrane from a
region of more pure water to a region of less pure water. (See
Section 14.6.) In the case of your salad, lettuce contains lots
of water in its cells. When the lettuce is surrounded by salad
dressing (water with lots of things dissolved in it), the water in
the lettuce leaf will flow through the cell membranes, resulting
in diluted dressing and wilted lettuce.
This same process can be pushed in the reverse direction
as well. Suppose you were to soak a raisin in water. The
water would flow into the raisin to dilute the concentrated
sugar solution inside the raisin, resulting in a plumper raisin.
What would happen if you put the raisin into a box and put
pressure on the sides so that the raisin could not expand?
The water would not be able to enter the raisin. And if you
increased the pressure, some of the water
would even flow out of the raisin. This is reverse
osmosis.
This is the process used in the reverse osmosis
purification of water. A sample of water containing contaminant molecules and ions (such as
seawater) is put into a vessel with a semipermeable membrane. A semipermeable membrane
allows the passage of water (solvent) molecules
through it in either direction but prevents the
passage of larger solute molecules or ions. As
pressure is applied to the water sample, water
molecules will pass through the membrane and
pure water can be isolated.
H20
Purified
water
H20
Contaminants
..
..
Structure of the Water Molecule
..O.. H
.. ..
m
n
96
0.0
105
H
H
O
(a)
(b)
(c)
δ–
..
..
O
H
H
δ+
(d)
H
A single water molecule consists of two hydrogen atoms and one oxygen atom. Each hydrogen atom is bonded to the oxygen atom by a single covalent bond. This bond is formed by
the overlap of the 1s orbital of hydrogen with an unpaired 2p orbital of oxygen. The average
distance between the two nuclei is known as the bond length. The O i H bond length in water
is 0.096 nm. The water molecule is nonlinear and has a bent structure with an angle of about
105° between the two bonds (see Figure 13.12).
Oxygen is the second most electronegative element. As a result, the two covalent OH
bonds in water are polar. If the three atoms in a water molecule were aligned in a linear strucO
H, the two polar bonds would be acting in equal and opposite
ture, such as H
directions and the molecule would be nonpolar. However, water is a highly polar molecule.
It therefore does not have a linear structure. When atoms are bonded in a nonlinear fashion,
the angle formed by the bonds is called the bond angle. In water the HOH bond angle is 105°.
The two polar covalent bonds and the bent structure result in a partial negative charge on the
oxygen atom and a partial positive charge on each hydrogen atom. The polar nature of water
is responsible for many of its properties, including its behavior as a solvent.
Figure 13.12
Diagrams of a water molecule: (a) electron distribution, (b) bond angle and
O i H bond length, (c) molecular structure, and (d) dipole representation.
13.7
• Water, a Unique Liquid 299
One of the greatest challenges facing our planet is finding sufficient potable water for the
global population that now surpasses the 7 billion mark. A U.S. citizen, on average, consumes
about 575 liters of water a day (United Nations Development Program—Human Development
Report, 2006). Thus, it is not difficult to see why water is becoming a very precious resource.
To satisfy the insatiable growing demand, fresh water is being depleted from streams, aquifers,
and reservoirs faster than it can be replenished. Climate change also contributes to the water
problem: Changes in weather patterns are predicted to cause severe drought in many parts of
the globe. Reclamation of wastewater and desalination of seawater are currently being considered to augment the supplies of fresh water.
1. Reclamation of wastewater is actually an excellent source of fresh water. Although it may
be very difficult to get comfortable with the idea of drinking reclaimed wastewater (the
“ick factor” as termed by the reclamation industry), most of the water provided by municipalities is in fact reclaimed water from upstream cities. Many cities are using reclaimed
water for irrigation and for industrial uses. Some cities are beginning to augment their
water supplies using this reclaimed water. As an example, Singapore has been a leader in
the challenge to reclaim wastewater for many uses. NEWater (provided by the Singapore
water treatment department) is water that has been filtered, purified by reverse osmosis
(see Chemistry in Action, Reverse Osmosis), and reused for agricultural and industrial
use. Some of this water is also treated with UV light to further purify it and it is bottled
for drinking. Thirty percent of Singapore’s water is currently reclaimed and plans are to
increase this to 50% by 2060. Orange County in California pumps reclaimed water into a
local aquifer for storage and reuse. San Diego County has a plan to pump reclaimed water
to upstream reservoirs to be mixed with “natural” water to be reused. These are excellent
models for other cities that suffer from a shortage in available fresh water.
2. Desalination of seawater is a good source of fresh water for people living near the
ocean. Unfortunately, desalination is very expensive and it is inefficient to extract fresh
water from the ocean. On average it takes about 4 liters of seawater to make 1 liter of fresh
water. The wastewater, which has a very high salt concentration, may also have an adverse
impact on the environment. Generally, salt water is desalinated using the process of reverse
osmosis in which a sample of salt water is placed in a vessel with a semipermeable membrane and pressure is applied to push the water molecules through the membrane. (See
Figure 13.13.) This method of water purification has been championed by Israel, a nation
with easy access to the sea but limited fresh water. Israel is currently building its fifth desalination plant and will produce over 75% of its water in this way by 2013.
Low temperature distillation and combustion of hydrogen are two other promising technologies being investigated as possible new sources of fresh water.
• In low-temperature distillation, water is subjected to a very low pressure environment.
When the pressure is lowered, the boiling temperature is also lowered. This could possibly result in producing distilled water at low temperature and with a lower energy cost.
• To produce hydrogen gas for combustion, an electric
current is run through a sample of water breaking it
down into its component elements. In regions with
lots of sunshine, solar panels can create electricity
to decompose water. The hydrogen and oxygen gas
can then be collected and recombined later to produce electricity and pure water. This provides both
a source of fresh water and a storage method for energy produced from the sun.
Though these technologies are still in the experimental
stage, hopefully they will become important sources of
fresh water in the future.
Figure 13.13
Catalina Island, California, gets most
of its water from a desalination plant.
© Btrenkel/iStockphoto
Sources of Water for a Thirsty World
Bottled water provides a
convenient source of drinking
water for people on the go.
© Rarpia/iStockphoto
300 chapter 13
• Liquids
C h a p t e r
13 review
13.1 states of matter: a review
• Solids
• Particles close together
• High density
• Incompressible
• Maintain shape
• Liquids
• Particles close together, yet free to move
• Incompressible
• Definite volume, but take the shape of the container
• Gases
• Particles far apart
• Low density
• Compressible
• Take volume and shape of container
13.2 Properties of Liquids
key terms
surface tension
capillary action
meniscus
evaporation
vaporization
sublimation
condensation
vapor pressure
volatile
• The resistance of a liquid to an increase in its surface area is the surface tension of the liquid.
• Capillary action is caused by the cohesive forces within the liquid and adhesive forces between the
liquid and the walls of the container:
• If the forces between the liquid and the container are greater than those within the liquid, the
liquid will climb the walls of the container.
• A meniscus is evidence of cohesive and adhesive forces:
• It is concave if the adhesive forces are stronger than the cohesive forces.
• It is convex if the cohesive forces are stronger than the adhesive forces.
• During evaporation, molecules of greater-than-average kinetic energy escape from the liquid.
• Sublimation is the evaporation of a solid directly to a gas.
• Gaseous molecules return to the liquid state through condensation.
• At equilibrium the rate of evaporation equals the rate of condensation.
• The pressure exerted by the vapor in equilibrium with its liquid in a closed container is the vapor
pressure of the liquid.
• A volatile substance evaporates readily.
13.3 Boiling Point and Melting Point
Key Terms
boiling point
normal boiling point
vapor pressure curve
melting point (or freezing point)
• The boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric
pressure:
• At 1 atm the boiling point is called the normal boiling point for a liquid.
• The temperature at which a solid is in equilibrium with its liquid phase is the freezing point ­or
­melting point.
13.4 Changes of State
Key Terms
heat of fusion
heat of vaporization
Key Terms
intermolecular forces
intramolecular forces
dipole–dipole attractions
hydrogen bond
London dispersion forces
• A graph of the warming of a liquid is called a heating curve:
• Horizontal lines on the heating curve represent changes of state:
• The energy required to change 1 g of solid to liquid at its melting point is the heat of
­fusion.
• The energy required to change 1 g of liquid to gas at its normal boiling point is the heat of
vaporization.
• The energy required to change the phase of a sample (at its melting or boiling point) is
energy = (mass)(heat of fusion (or vaporization))
• The energy required to heat molecules without a phase change is determined by
energy = (mass)(sp ht)(t)
13.5 Intermolecular forces
• Dipole–dipole attractions among polar molecules.
Review Questions 301
• A hydrogen bond is the dipole–dipole attraction between polar molecules containing any of these
types of bonds: F i H, O i H, N i H.
• London dispersion forces cause attractions among nonpolar atoms and molecules.
13.6 Hydrates
• Solids that contain water molecules as part of their crystalline structure are called hydrates.
• Formulas of hydrates are given by writing the formula for the anhydrous compound followed by a
dot and then the number of water molecules present.
• Water molecules in hydrates are bonded by electrostatic forces, which are weaker than covalent or
ionic bonds:
• Water of hydration can be removed by heating the hydrate to form a partially dehydrated compound or the anhydrous compound.
Key Terms
hydrate
water of hydration
water of crystallization
13.7 Water, a Unique Liquid
• Water is our most common resource, covering 75% of the Earth’s surface.
• Water is a colorless, odorless, tasteless liquid with a melting point of 0°C and a boiling point of
100°C at 1 atm.
• The water molecule consists of 2 H atoms and 1 O atom bonded together at a 105° bond angle,
­making the molecule polar.
• Water can be formed in a variety of ways including:
• Reclamation of wastewater
• Reverse osmosis
• Desalination
• Low-temperature distillation
• Hydrogen combustion
Key Term
semipermeable membrane
Review Questions
1. In what state (solid, liquid, or gas) would H2S, H2Se, and H2Te be
at 0°C? (Table 13.2)
2. What property or properties of liquids are similar to solids?
3. What property or properties of liquids are similar to gases?
4. If water were placed in the containers in Figure 13.5, would they all
have the same vapor pressure at the same temperature? Explain.
5. Why doesn’t the vapor pressure of a liquid depend on the amount
of liquid and vapor present?
6. In Figure 13.5, in which case, (a), (b) or (c) will the atmosphere
above the liquid reach a point of saturation?
7. Suppose a solution of ethyl ether and ethyl alcohol is placed in the
closed bottle in Figure 13.5. (Use Figure 13.7 for information on
the substances.)
(a) Are both substances present in the vapor?
(b) If the answer to part (a) is yes, which has more molecules in the
vapor?
8. Explain why rubbing alcohol warmed to body temperature still
feels cold when applied to your skin.
9. The vapor pressure at 20°C for the following substances is
methyl alcohol
acetic acid
benzene
bromine
water
carbon tetrachloride
mercury
toluene
96 torr
11.7 torr
74.7 torr
173 torr
17.5 torr
91 torr
0.0012 torr
23 torr
(a) Arrange these substances in order of increasing rate of
e­ vaporation.
(b) Which substance listed has the highest boiling point? the ­lowest?
10. On the basis of the kinetic-molecular theory, explain why vapor
pressure increases with temperature.
11. The temperature of the water in the pan on the burner (Figure 13.11)
reads 100°C. What is the pressure of the atmosphere?
12. If ethyl alcohol is boiling in a flask and the atmospheric pressure is 543
torr, what is the temperature of the boiling liquid? (Use ­Figure 13.7.)
13. At approximately what temperature would each of the substances
shown in Figure 13.7 boil when the pressure is 30 torr?
14. Use the graph in Figure 13.7 to find the following:
(a) boiling point of water at 500 torr
(b) normal boiling point of ethyl alcohol
(c) boiling point of ethyl ether at 0.50 atm
15. Suggest a method whereby water could be made to boil at 50°C.
16. Explain why a higher temperature is obtained in a pressure cooker
than in an ordinary cooking pot.
17. What is the relationship between vapor pressure and boiling point?
18. The boiling point of ammonia, NH3, is - 33.4°C and that of sulfur
dioxide, SO2, is -10.0°C. Which has the higher vapor pressure at
-40°C?
19. Explain what is occurring physically when a substance is boiling.
20. At what specific temperature will ethyl ether have a vapor pressure
of 760 torr?
21. Compare the potential energy of the three states of water shown in
Figure 13.11.
22. Consider Figure 13.8.
(a) Why is line BC horizontal? What is happening in this interval?
(b) What phases are present in the interval BC?
(c) When heating is continued after point C, another horizontal
line, DE, is reached at a higher temperature. What does this line
represent?
23. Account for the fact that an ice–water mixture remains at 0°C until
all the ice is melted, even though heat is applied to it.
24. Which contains less energy: ice at 0°C or water at 0°C? Explain.
25. Why does a boiling liquid maintain a constant temperature when
heat is continuously being added?
302 chapter 13
• Liquids
26. Define the terms intermolecular forces and intramolecular forces.
27. What kind of covalent bond is formed when atoms do not share
electrons equally?
28. What property of a molecule most affects its ability to form instantaneous dipoles?
29. If water molecules were linear instead of bent, would the heat of
vaporization be higher or lower? Explain.
30. The heat of vaporization for ethyl ether is 351 J/g and that for ethyl
alcohol is 855 J/g. From these data, which of these compounds has
hydrogen bonding? Explain.
31. Would there be more or less H bonding if water molecules were
linear instead of bent? Explain.
32. In which condition are there fewer hydrogen bonds between
­molecules: water at 40°C or water at 80°C? Explain.
33. Which compound
H2NCH2CH2NH2
or
CH3CH2CH2NH2
would you expect to have the higher boiling point? Explain. (Both
compounds have similar molar masses.)
34. Why does water have such a relatively high boiling point?
35. Explain why HF (bp = 19.4°C) has a higher boiling point than
HCl (bp = -85°C) , whereas F2 (bp = -188°C) has a lower boiling point than Cl2 (bp = -34°C).
36. How do we specify 1, 2, 3, 4, 5, 6, 7, and 8 molecules of water in
the formulas of hydrates? (Table 13.3)
37. Diagram a water molecule and point out the negative and positive
ends of the dipole.
38. If the water molecule were linear, with all three atoms in a straight
line rather than in the shape of a V, as shown in Figure 13.12, what
effect would this have on the physical properties of water?
39. List six physical properties of water.
40. What condition is necessary for water to have its maximum ­density?
What is its maximum density?
41. Why does ice float in water? Would ice float in ethyl alcohol
(d = 0.789 g>mL)? Explain.
42. What water temperature would you theoretically expect to find at
the bottom of a very deep lake? Explain.
43. Is the formation of hydrogen and oxygen from water an exothermic
or an endothermic reaction? How do you know?
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com)
All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1. Rank the following molecules in order of increasing polarity:
CH3Cl, CH3F, and CH3Br.
2. Rank the following molecules in order of increasing polarity: H2S,
H2Se, and H2O.
3. Rank the following molecules in order of increasing London
­dispersion forces: CCl4, Cl4, and CBr4.
4. Rank the following molecules in order of increasing London
­dispersion forces: CO2, SO2, and CS2.
5. In which of the following substances would you expect to find
­hydrogen bonding?
(a) C3H7OH
(d) PH3
(b) H2O2
(e) HF
(c) CHCI3
6. In which of the following substances would you expect to find
­hydrogen bonding?
(a) HI
(d) C2H5OH
(b) NH3
(e) H2O
(c) CH2F2
7. For each of the compounds in Question 5 that form hydrogen
bonds, draw a diagram of the two molecules using a dotted line to
indicate where the hydrogen bonding will occur.
8. For each of the compounds in Question 6 that form hydrogen
bonds, draw a diagram of the two molecules using a dotted line to
indicate where the hydrogen bonding will occur.
9. You leave the house wearing a cotton T-shirt and are surprised by
a sudden rainstorm. You notice that the water soaks into the T-shirt
whereas it just beads up on your raincoat. In this example, are the
adhesive forces or cohesive forces stronger between water and your
T-shirt? Between water and your raincoat? Explain.
10. Rainex is a product that causes water to “bead” instead of spread
when sprayed on car windshields. When the water forms these
droplets, are adhesive forces or cohesive forces stronger? Explain.
11. Name these hydrates:
(a) BaBr2 # 2 H2O
(b) AlCl3 # 6 H2O
(c) FePO4 # 4 H2O
12. Name these hydrates:
(a) MgNH4PO4 # 6 H2O
(b) FeSO4 # 7 H2O
(c) SnCl4 # 5 H2O
13. How many moles of compound are in 25.0 g of Na2CO3 # 10 H2O?
14. How many moles of compound are in 25.0 g of Na2B4O7 # 10 H2O?
17. Cobalt(II) chloride hexahydrate, CoCl2 # 6 H2O, is often used as a
humidity indicator. This is due to the fact that the hydrate is a deep
magenta color while the anhydrous form is a pale blue. As the humidity level changes, the color changes as well. What is the mass
percent water in cobalt(II) ­hexahydrate?
18. Gypsum, CaSO4 # 2 H2O, is used to manufacture cement. What is
the mass percent of water in this compound?
19. A hydrated iron chloride compound was found to contain 20.66%
Fe, 39.35% Cl, and 39.99% water. Determine the empirical formula
of this hydrated compound.
20. A hydrated nickel chloride compound was found to contain 24.69%
Ni, 29.83% Cl, and 45.48% water. Determine the empirical formula
of this hydrated compound.
15. Upon heating 125 g MgSO4 # 7 H2O:
(a) how many grams of water can be obtained?
(b) how many grams of anhydrous compound can be obtained?
16. Upon heating 125 g AlCl3 # 6 H2O:
(a) how many grams of water can be obtained?
(b) how many grams of anhydrous compound can be obtained?
Additional Exercises 303
21. How many joules of energy are needed to change 275 g of water
from 15°C to steam at 100.°C?
22. How many joules of energy must be removed from 325 g water at
35°C to form ice at 0°C?
23. Suppose 100. g of ice at 0°C are added to 300. g of water at 25°C.
Is this sufficient ice to lower the temperature of the system to 0°C
and still have ice remaining? Show evidence for your answer.
24. Suppose 35.0 g of steam at 100.°C are added to 300. g of water at
25°C. Is this sufficient steam to heat all the water to 100.°C and still
have steam remaining? Show evidence for your answer.
25. If 75 g of ice at 0.0°C were added to 1.5 L of water at 75°C, what
would be the final temperature of the mixture?
26. If 9560 J of energy were absorbed by 500. g of ice at 0.0°C, what
would be the final temperature?
Additional Exercises
250
250
200
200
150
150
100
100
50
Boiling point
Molar mass
0
50
0
He
Ne
Ar
Kr
Xe
Rn
Molar mass (g/mol)
Boiling point (K)
27. You walk out to your car just after a rain shower and notice that
small droplets of water are scattered on the hood. How was it possible for the water to form those droplets?
28. Which causes a more severe burn: liquid water at 100°C or steam
at 100°C? Why?
29. You have a shallow dish of alcohol set into a tray of water. If you
blow across the tray, the alcohol evaporates, while the water cools
significantly and eventually freezes. Explain why.
30. Regardless of how warm the outside temperature may be, we always feel cool when stepping out of a swimming pool, the ocean,
or a shower. Why is this so?
31. Sketch a heating curve for a substance X whose melting point is
40°C and whose boiling point is 65°C.
(a) Describe what you will observe as a 60.-g sample of X is
warmed from 0°C to 100°C.
(b) If the heat of fusion of X is 80. J/g, the heat of vaporization is
190. J/g, and if 3.5 J are required to warm 1 g of X each degree,
how much energy will be needed to accomplish the change in (a)?
32. For the heating curve of water (see Figure 13.8), why doesn’t the
temperature increase when ice is melting into liquid water and liquid water is changing into steam?
33. Why does the vapor pressure of a liquid increase as the temperature
is increased?
34. Mount McKinley in Alaska is the tallest peak in North America
with a summit at 20,320 ft or 6194 m. It is also known as Denali,
meaning “the high one” in the native Koyukon Athabaskan language. The average atmospheric pressure at the top of Denali is approximately 330 torr. Use Figure 13.7 to determine the approximate
boiling temperature of water on Denali.
35. Explain how anhydrous copper(II) sulfate (CuSO4) can act as an
indicator for moisture.
36. Write formulas for magnesium sulfate heptahydrate and disodium
hydrogen phosphate dodecahydrate.
37. Analyze the graph below and draw conclusions as appropriate. The
boiling point and molar mass are graphed for the noble gases? What
generalizations can you draw from these data?
38. You decide to go to the mountains of Santa Fe, New Mexico (7200
ft elevation), for your spring holiday. One of the important tasks
for you to complete is to boil the eggs to color and hide for the
egg hunt. You are in a hurry and know that eggs boiled for exactly
8 minutes at home at the beach in Santa Barbara, California, come
out perfectly cooked. As everyone is cracking their eggs for breakfast after the egg hunt, you discover to your dismay that your eggs
are not fully cooked. Explain this observation.
39. Look at the drawing on the
right of the water in the glassgraduated cylinder and the
water in the plastic graduated
cylinder. What differences do
you observe in the shape of the
meniscus? Plastics are composed of long-chain hydrocarbons and glass is composed of
silicon oxides. Based on this
information, propose a reason
for the differences in these two
water samples.
40. The flow rate of a liquid is related to the strength of the cohesive
forces between molecules. The
Plastic cylinder
Glass cylinder
stronger the cohesive forces, the
harder it is for the molecules to slide past one another. Honey is a
very thick liquid that pours slowly at ordinary temperatures. A student tried an experiment. She filled five eyedroppers with honey at
different temperatures and then measured the rate of dripping for
each sample (see table below). Graph the rate of dripping ­versus
temperature for the honey and propose a relationship between temperature and flow rate. Propose an explanation for the trend that you
observe.
Temperature (°C)
Number of drops/minute
10
3
20
17
30
52
50
87
70
104
41. You work in a hardware store and notice that whenever you spill
water on the waxed floors it tends to bead up and stay fairly well
confined. When you spill hexane, however, it spreads over the
entire floor and is a big mess to clean up. Based on your knowledge of adhesive and cohesive forces, explain this difference in
­behavior.
304 chapter 13
• Liquids
42. A sample of pure acetic acid is also known as glacial acetic acid.
This name comes from the fact that flasks of acetic acid would often
freeze in the old laboratories of Europe. When acetic acid freezes, it
forms many cracks resembling a glacier and thus the name. A heating curve for a sample of glacial acetic acid is pictured below.
160
140
Temperature (°C)
120
100
80
60
40
20
0
Heat added
(a) Using this graph, determine the melting point and the boiling
point of acetic acid.
(b) Redraw this graph and indicate regions where the solid, liquid,
and gas phases will exist.
(c) Indicate regions where a transition of states is occurring and tell
what phases are present.
43. How many calories are required to change 225 g of ice at 0°C to
steam at 100.°C?
44. The molar heat of vaporization is the number of joules required
to change 1 mol of a substance from liquid to vapor at its boiling
point. What is the molar heat of vaporization of water?
45. The specific heat of zinc is 0.096 cal/g°C. Determine the energy
required to raise the temperature of 250. g of zinc from room temperature (20.0°C) to 150.°C.
46. How many joules of energy would be liberated by condensing
50.0 mol of steam at 100.0°C and allowing the liquid to cool to
30.0°C?
47. How many kilojoules of energy are needed to convert 100. g of ice
at - 10.0°C to water at 20.0°C? (The specific heat of ice at -10.0°C
is 2.01 J/g°C.)
48. What mass of water must be decomposed to produce 25.0 L of
oxygen at STP?
49. Suppose 1.00 mol of water evaporates in 1.00 day. How many water
molecules, on average, leave the liquid each second?
50. Compare the volume occupied by 1.00 mol of liquid water at 0°C
and 1.00 mol of water vapor at STP.
51. A mixture of 80.0 mL of hydrogen and 60.0 mL of oxygen is
­ignited by a spark to form water.
(a) Does any gas remain unreacted? Which one, H2 or O2 ?
(b) What volume of which gas (if any) remains unreacted? (­Assume
the same conditions before and after the reaction.)
52. A student (with slow reflexes) puts his hand in a stream of steam
at 100.°C until 1.5 g of water have condensed. If the water then
cools to room temperature (20.0°C), how many joules have been
absorbed by the student’s hand?
53. Determine which of the following molecules would hydrogenbond with other molecules like it. For those that do not hydrogenbond, explain why. For those that hydrogen-bond, draw a diagram
of two molecules using a dotted line to indicate where the hydrogen
bonding will occur.
(a) Br2
(d) H2O
(c) CH3 i O i CH3
(b) CH3 i O i H
(e) H2S
54. The heat of fusion of a substance is given in units of J/g. The specific heat of a substance is given in units of J/g °C. Why is a temperature factor not needed in the units for heat of fusion?
55. How many joules of energy are required to change 50.0 g Cu from
25.0°C to a liquid at its melting point, 1083°C?
Specific heat of Cu = 0.385 J/g°C
Heat of fusion for Cu = 134 J/g
56. You pour a steaming hot bowl of soup. After one taste, you burn
your tongue and decide to cool it down with some ice. If you add
75 g of ice at 0°C to your soup and the final temperature of the
ice–soup mixture is 87°C, how many joules of energy did the ice
remove from the bowl of soup?
57. Write and balance the chemical equation for the decomposition of
water into its component elements.
58. Use Figure 13.7 to determine the temperature in degrees Fahrenheit
required to purify water by distillation at atmospheric pressures of
500 torr, 300 torr, and 100 torr.
Challenge Exercises
59. You buy a box of borax (Na2B4O7 # 10 H2O) from the corner market
in Phoenix, Arizona, in the middle of the summer. You open up
the box and pour the borax into a weighed beaker. After all, you
do want to be sure that you were not cheated by the manufacturer.
You are distracted from your task and do not get back to weigh the
filled beaker for several days. Upon weighing the beaker you get
the following data from the 5.0-lb box of borax.
Empty beaker
Filled beaker
60. Why does a lake freeze from the top down? What significance does
this have for life on Earth?
61. Suppose 150. g of ice at 0.0°C are added to 0.120 L of water at
45°C. If the mixture is stirred and allowed to cool to 0.0°C, how
many grams of ice remain?
492.5 g
2467.4 g
Were you cheated by the manufacturer? Why or why not?
Answers to Practice Exercises
13.1 Vapor pressure is the pressure exerted by a vapor in equilibrium
with its liquid. The vapor pressure of a liquid increases with
increasing temperature.
13.2 They are the same.
13.3 28°C, 73°C, 93°C
13.4 approximately 95°C
13.5 44.8 kJ
13.6 SO2 and CH3F are polar.
13.7 (a) yes, (b) yes, (c) no
13.8 I2
13.9 (a) BeCO3 # 4 H2O
(b) Cd(MnO4)2 # 6 H2O
13.10 51.17% H2O
(c) Cr(NO3)3 # 9 H2O
(d) PtO2 # 3 H2O
CHAPTER
Photodisc
Brass, a solid solution of zinc and
copper, is used to make musical
instruments and many other
objects.
14
Solutions
M
ost substances we encounter in our daily lives are
mixtures. Often they are homogeneous mixtures,
which are called solutions. Some solutions we
commonly encounter are shampoo, soft drinks, and wine.
Blood plasma is a complex mixture composed of compounds
and ions dissolved in water and proteins suspended in the
solution. These solutions all have water as a main component,
but many common items, such as air, gasoline, and steel, are
also solutions that do not contain water. What are the necessary components of a solution? Why do some substances
dissolve, while others do not? What effect does a dissolved
substance have on the properties of the solution? Answering
these questions is the first step in understanding the solutions
we encounter in our daily lives.
Chapter Outline
14.1 General Properties of Solutions
14.2 Solubility
14.3 Rate of Dissolving Solids
14.4 Concentration of Solutions
14.5 Colligative Properties of Solutions
14.6 Osmosis and Osmotic Pressure
306 chapter 14 • Solutions
14.1 General Properties of Solutions
L earning objec tive
List the properties of a true solution.
Key termS
The term solution is used in chemistry to describe a system in which one or more substances
are homogeneously mixed or dissolved in another substance. A simple solution has two components: a solute and a solvent. The solute is the component that is dissolved or is the least
abundant component in the solution. The solvent is the dissolving agent or the most abundant
component in the solution. For example, when salt is dissolved in water to form a solution,
salt is the solute and water is the solvent. Complex solutions containing more than one solute
and/or more than one solvent are common.
The three states of matter—solid, liquid, and gas—give us nine different types of solutions: solid dissolved in solid, solid dissolved in liquid, solid dissolved in gas, liquid dissolved
in liquid, and so on. Of these, the most common solutions are solid dissolved in liquid, liquid
dissolved in liquid, gas dissolved in liquid, and gas dissolved in gas. Some common types of
solutions are listed in Table 14.1.
A true solution is one in which the particles of dissolved solute are molecular or ionic in
size, generally in the range of 0.1 to 1 nm (10-8 to 10-7 cm). The properties of a true solution
are as follows:
solution
solute
solvent
1. A mixture of two or more components—solute and solvent—is homogeneous and has a
variable composition; that is, the ratio of solute to solvent can be varied.
2. The dissolved solute is molecular or ionic in size.
3. It is either colored or colorless and is usually transparent.
4. The solute remains uniformly distributed throughout the solution and will not settle out
with time.
5. The solute can generally be separated from the solvent by purely physical means (for
example, by evaporation).
Let’s illustrate these properties using water solutions of sugar and of potassium
permanganate.
Example 14.1
Consider two sugar solutions: solution A containing 10 g of sugar added to 100 mL
of water and solution B containing 20 g of sugar added to 100 mL of water. Each is
stirred until all the sugar dissolves. Explain why these solutions are considered to be
true solutions.
Solution
1. All the sugar dissolves in each solution showing we can vary the composition of a sugar
solution.
2. Every portion of each solution has the same sweet taste uniformly (although you should
not taste solutions in the laboratory) because the sugar molecules are uniformly distributed throughout.
3. If the solution is confined so that no water is lost, the solution will look and taste the
same a week or more later.
4. The solutions cannot be separated by filtering.
5. Careful evaporation will separate the solution into its components: sugar and water.
Practice 14.1
Identify the solute and the solvent in each of these solutions:
(a) air
(b) seawater
(c) carbonated water
Table 14.1
307
Richard Megna/Fundamental Photographs
14.2 • Solubility Common Types of Solutions
Phase of solution
Solute
Solvent
Example
Gas
Liquid
Liquid
Liquid
Solid
Solid
gas
gas
liquid
solid
gas
solid
gas
liquid
liquid
liquid
solid
solid
air
soft drinks
antifreeze
salt water
H2 in Pt
brass
To observe the dissolving of potassium permanganate (KMnO4), we drop a few crystals
of it in a beaker of water. Almost at once, the beautiful purple color of dissolved permanganate
ions (MnO4-) appears and streams to the bottom of the beaker as the crystals dissolve. After
a while, the purple color disperses until it’s evenly distributed throughout the solution. This
dispersal demonstrates that molecules and ions move about freely and spontaneously (diffuse)
in a liquid or solution.
Solution permanency is explained in terms of the kinetic-molecular theory (see
Section 12.2). According to this theory, both the solute and solvent particles (molecules or ions)
are in constant random motion. This motion is energetic enough to prevent the solute particles
from settling out under the influence of gravity.
Note the beautiful purple trails of
KMnO4 as the crystals dissolve.
14.2Solubility
The term solubility describes the amount of one substance (solute) that will dissolve in a
specified amount of another substance (solvent) under stated conditions. For example, 36.0 g
of sodium chloride will dissolve in 100 g of water at 20°C. We say then that the solubility
of NaCl in water is 36.0 g/100 g H2O at 20°C. Solubility is often used in a relative way. For
instance, we say that a ­substance is very soluble, moderately soluble, slightly soluble, or insoluble. Although these terms do not accurately indicate how much solute will dissolve, they
are frequently used to describe the solubility of a substance qualitatively.
Two other terms often used to describe solubility are miscible and immiscible. Liquids
that are capable of mixing and forming a homogeneous solution are miscible; those that do
not form solutions or are generally insoluble in each other are immiscible. Methyl alcohol and
water are miscible in each other in all proportions. Oil and water are immiscible, forming two
separate layers when they are mixed, as shown in Figure 14.1.
The general guidelines for the solubility of common ionic compounds (salts) are given in
Figure 14.2. These guidelines have some exceptions, but they provide a solid foundation for
the compounds considered in this course. The solubilities of over 200 compounds are given in
the Solubility Table in Appendix V. Solubility data for thousands of compounds can be found
by consulting standard reference sources.*
The quantitative expression of the amount of dissolved solute in a particular quantity of
solvent is known as the concentration of a solution. Several methods of expressing concentration are described in Section 14.4.
Predicting solubilities is complex and difficult. Many variables, such as size of ions,
charge on ions, interaction between ions, interaction between solute and solvent, and temperature, complicate the problem. Because of the factors involved, the general rules of solubility
given in Figure 14.2 have many exceptions. However, these rules are useful because they do
apply to many of the more common compounds that we encounter in the study of chemistry.
Keep in mind that these are rules, not laws, and are therefore subject to exceptions. Fortunately,
the solubility of a solute is relatively easy to determine experimentally. Now let’s examine the
factors related to solubility.
Learning objectiv e
Key termS
solubility
miscible
immiscible
concentration of a solution
saturated solution
unsaturated solution
supersaturated solution
Many chemists use the term
salt interchangeably with
ionic compound.
Kip Peticolas/Fundamental Photographs
Define solubility and the factors that affect it.
Figure 14.1
*Two commonly used handbooks are Lange’s Handbook of Chemistry, 16th ed. (New York: McGraw-Hill, 2009),
and the CRC Handbook of Chemistry and Physics, 93rd ed. (Cleveland: Chemical Rubber Co., 2012).
An immiscible mixture of oil and
water.
308 chapter 14 • Solutions
Soluble
Insoluble
Na+, K+, NH4+
Nitrates, NO3–
Acetates, C 2 H 3O 2–
Chlorides, C l –
Bromides, Br –
Iodides, I –
Sulfates, SO42–
Ag +, Ca2+ are
slightly soluble
except
Ag+, Hg 22+, Pb2+
except
Ba2+, Sr 2+, Pb2+
NH4+
except
alkali metal cations
Carbonates, CO 2–
3
Phosphates, PO 43–
Hydroxides, OH –
Sulfides, S2–
Figure 14.2
The solubility of various common ions. Substances containing the ions on the
left are generally soluble in cold water, while those substances containing the
ions on the right are insoluble in cold water. The arrows point to the exceptions.
The Nature of the Solute and Solvent
= Water
= Na+
= Cl –
Figure 14.3
Dissolution of sodium chloride in
water. Polar water molecules are
attracted to Na1 and Cl2 ions in
the salt crystal, weakening the
attraction between the ions. As
the attraction between the ions
weakens, the ions move apart
and become surrounded by water
dipoles. The hydrated ions slowly
diffuse away from the crystal to
become dissolved in solution.
The old adage “like dissolves like” has merit, in a general way. Polar or ionic substances tend
to be more compatible with other polar substances. Nonpolar substances tend to be compatible
with other nonpolar substances and less soluble with polar substances. Thus ionic compounds,
which are polar, tend to be much more soluble in water, which is polar, than in solvents such as
ether, hexane, or benzene, which are essentially nonpolar. Sodium chloride, an ionic substance,
is soluble in water, slightly soluble in ethyl alcohol (less polar than water), and insoluble in
ether and benzene. Pentane, C5H12, a nonpolar substance, is only slightly soluble in water but
is very soluble in benzene and ether.
At the molecular level the formation of a solution from two nonpolar substances, such as
hexane and benzene, can be visualized as a process of simple mixing. These nonpolar molecules, having little tendency to either attract or repel one another, easily intermingle to form
a homogeneous solution.
Solution formation between polar substances is much more complex. See, for example,
the process by which sodium chloride dissolves in water (Figure 14.3). Water molecules
are very polar and are attracted to other polar molecules or ions. When salt crystals are
put into water, polar water molecules become attracted to the sodium and chloride ions on
the crystal surfaces and weaken the attraction between Na+ and Cl- ions. The positive end
of the water dipole is attracted to the Cl- ions, and the negative end of the water dipole to
the Na+ ions. The weakened attraction permits the ions to move apart, making room for
more water dipoles. Thus the surface ions are surrounded by water molecules, becoming
hydrated ions, Na+(aq) and Cl-(aq), and slowly diffuse away from the crystals and dissolve
in solution:
H2 O
NaCl(crystal) ¡ Na+(aq) + Cl-(aq)
Examination of the data in Table 14.2 reveals some of the complex questions relating to
solubility.
14.2 • Solubility Table 14.2
309
Solubility of Alkali Metal Halides in Water
Solubility
(g salt/100 g H2O)
Salt
08C
1008C
LiF
LiCl
LiBr
LiI
0.12
63.7
143
151
0.14 (at 358C)
127.5
266
481
NaF
NaCl
NaBr
NaI
4
35.7
79.5
158.7
5
39.8
121
302
KF
KCl
KBr
KI
92.3 (at 188C)
27.6
53.5
127.5
Very soluble
57.6
56.7
102
The Effect of Temperature on Solubility
Temperature affects the solubility of most substances, as shown by the data in Table 14.2.
Most solutes have a limited solubility in a specific solvent at a fixed temperature. For
most solids dissolved in a liquid, an increase in temperature results in increased solubility
(see Figure 14.4). However, no single rule governs the solubility of solids in liquids with
change in temperature. Some solids increase in solubility only slightly with increasing
temperature (see NaCl in Figure 14.4); other solids decrease in solubility with increasing
temperature (see Li2SO4 in Figure 14.4).
On the other hand, the solubility of a gas in water usually decreases with ­increasing temperature (see HCl and SO2 in Figure 14.4). The tiny bubbles that form when water is heated
are due to the decreased solubility of air at higher temperatures. The decreased solubility of
gases at higher temperatures is explained in terms of the kinetic-molecular theory (KMT) by
assuming that, in order to dissolve, the gas molecules must have attraction of some sort with the
molecules of the liquid. An increase in temperature decreases the solubility of the gas because
it increases the kinetic energy (speed) of the gas molecules and thereby decreases their ability
to interact with the liquid molecules.
100
Figure 14.4
NaNO3
90
Solubility of various compounds in
water. Solids are shown in red and
gases are shown in blue.
KNO3
Grams solute per 100 g H2O
80
NH 4 Cl
70
HCl (1atm)
60
KCl
50
40
NaCl
30
Li 2 SO4
CuSO4
20
KClO3
10
SO2 (1 atm)
0
0
10
20
30
40
50
60
Temperature (°C )
70
80
90
100
310 chapter 14 • Solutions
Copyright John Wiley & Sons, Inc.
The Effect of Pressure on Solubility
Small changes in pressure have little effect on the solubility of solids in liquids or liquids in liquids
but have a marked effect on the solubility of gases in liquids. The solubility of a gas in a liquid is
directly proportional to the pressure of that gas above the solution. Thus the amount of a gas dissolved in solution will double if the pressure of that gas over the solution is doubled. For example,
carbonated beverages contain dissolved carbon dioxide under pressures greater than atmospheric
pressure. When a can of carbonated soda is opened, the pressure is immediately reduced to the
atmospheric pressure, and the excess dissolved carbon dioxide bubbles out of the solution.
Saturated, Unsaturated, and Supersaturated Solutions
Pouring root beer into a glass
illustrates the effect of pressure
on solubility. The escaping CO2
produces the foam.
At a specific temperature there is a limit to the amount of solute that will dissolve in a given
amount of solvent. When this limit is reached, the resulting ­solution is said to be saturated. For
example, when we put 40.0 g of KCl into 100 g of H2O at 20°C, we find that 34.0 g of KCl dissolve and 6.0 g of KCl remain undissolved. The solution formed is a saturated solution of KCl.
Two processes are occurring simultaneously in a saturated solution. The solid is dissolving
into solution, and, at the same time, the dissolved solute is crystallizing out of solution. This
may be expressed as
solute (undissolved) m solute (dissolved)
When these two opposing processes are occurring at the same rate, the amount of solute in solution is constant, and a condition of equilibrium is established between dissolved and undissolved
solute. Therefore, a saturated solution contains dissolved solute in equilibrium with undissolved
solute. Thus, any point on any solubility curve (Figure 14.4) represents a saturated solution of
that solute. For example, a solution containing 60 g NH4Cl per 100 g H2O is saturated at 70°C.
It’s important to state the temperature of a saturated solution, because a solution that is
saturated at one temperature may not be saturated at another. If the temperature of a saturated
solution is changed, the equilibrium is disturbed, and the amount of dissolved solute will
change to reestablish equilibrium.
A saturated solution may be either dilute or concentrated, depending on the solubility of
the solute. A saturated solution can be conveniently prepared by dissolving a little more than the
saturated amount of solute at a temperature somewhat higher than room temperature. Then the
amount of solute in solution will be in excess of its solubility at room temperature, and, when
the solution cools, the excess solute will crystallize, leaving the solution saturated. (In this case,
the solute must be more soluble at higher temperatures and must not form a supersaturated solution.) Examples expressing the solubility of saturated solutions at two different temperatures are
given in Table 14.3.
An unsaturated solution contains less solute per unit of volume than does its corresponding saturated solution. In other words, additional solute can be dissolved in an unsaturated solution without altering any other conditions. Consider a solution made by adding 40 g of KCl to
100 g of H2O at 20°C (see Table 14.3). The solution formed will be saturated and will contain
about 6 g of undissolved salt, because the maximum amount of KCl that can dissolve in 100 g
of H2O at 20°C is 34 g. If the solution is now heated and maintained at 50°C, all the salt will
dissolve and even more can be dissolved. The solution at 50°C is unsaturated.
Table 14.3
Saturated Solutions at 208C and 508C
Solubility
(g solute/100 g H2O)
Solute
208C
508C
NaCl
KCl
NaNO3
KClO3
AgNO3
C12H22O11
36.0
34.0
88.0
7.4
222.0
203.9
37.0
42.6
114.0
19.3
455.0
260.4
14.3 • Rate of Dissolving Solids 311
Richard Megna/Fundamental Photographs
In some circumstances, solutions can be prepared that contain more solute than needed for a
saturated solution at a particular temperature. These solutions are said to be supersaturated. However, we must qualify this definition by noting that a supersaturated solution is unstable. Disturbances, such as jarring, stirring, scratching the walls of the container, or dropping in a “seed”
crystal, cause the supersaturation to return to saturation, releasing heat. When a supersaturated solution is disturbed, the excess solute crystallizes out rapidly, returning the solution to a saturated state.
Supersaturated solutions are not easy to prepare but may be made from certain substances
by dissolving, in warm solvent, an amount of solute greater than that needed for a saturated
solution at room temperature. The warm solution is then allowed to cool very slowly. With the
proper solute and careful work, a supersaturated solution will result.
Example 14.2
Will a solution made by adding 2.5 g of CuSO4 to 10 g of H2O be saturated or unsaturated at 20°C?
SOLUTION
We first need to know the solubility of CuSO4 at 20°C. From Figure 14.4, we see that the
solubility of CuSO4 at 20°C is about 21 g per 100 g of H2O. This amount is equivalent to
2.1 g of CuSO4 per 10 g of H2O.
Since 2.5 g per 10 g of H2O is greater than 2.1 g per 10 g of H2O, the solution will
be saturated and 0.4 g of CuSO4 will be undissolved.
The heat released in this hot pack
results from the crystallization of a
supersaturated solution of sodium
acetate.
Practice 14.2
Will a solution made by adding 9.0 g NH4Cl to 20 g of H2O be saturated or unsaturated
at 50°C?
14.3Rate of Dissolving Solids
Describe the factors that affect the rate at which a solid dissolves.
Learning objectiv e
The rate at which a solid dissolves is governed by (1) the size of the solute particles, (2) the
temperature, (3) the concentration of the solution, and (4) agitation or stirring. Let’s look at
each of these conditions:
1. Particle size. A solid can dissolve only at the surface that is in contact with the solvent.
Because the surface-to-volume ratio increases as size decreases, smaller crystals dissolve faster than large ones. For example, if a salt crystal 1 cm on a side (a surface area of
6 cm2) is divided into 1000 cubes, each 0.1 cm on a side, the total surface of the smaller
cubes is 60 cm2—a 10-fold increase in surface area (see Figure 14.5).
Figure 14.5
0.1-cm cube
Surface of this single
cube is 0.06 cm2
Total surface area of
the 1000 smaller cubes is
1000 × 0.06 cm2 = 60 cm2
1 cm
1 cm
Surface area of this cube is
6 × 1 cm2 = 6 cm2
Surface area of crystals. A crystal
1 cm on a side has a surface area
of 6 cm2. Subdivided into 1000
smaller crystals, each 0.1 cm on
a side, the total surface area is
increased to 60 cm2.
312 chapter 14 • Solutions
Figure 14.6
Rate of dissolution of a solid
solute in a solvent. The rate is
maximum at the beginning and
decreases as the concentration
approaches saturation.
Grams solute per 100 g solvent
30
Saturated solution
20
10
0
0
5
10
15
Time (minutes)
20
25
2. Temperature. In most cases, the rate of dissolving of a solid increases with temperature.
This increase is due to kinetic effects. The solvent molecules move more rapidly at higher
temperatures and strike the solid surfaces more often, causing the rate of dissolving to
increase.
3. Concentration of the solution. When the solute and solvent are first mixed, the rate of dissolving is at its maximum. As the concentration of the solution increases and the solution
becomes more nearly saturated with the solute, the rate of dissolving decreases greatly.
The rate of dissolving is graphed in Figure 14.6. Note that about 17 g dissolve in the
first five-minute interval, but only about 1 g dissolves in the fourth five-minute interval.
Although different solutes show different rates, the rate of dissolving ­ always becomes
very slow as the concentration approaches the saturation point.
4. Agitation or stirring. The effect of agitation or stirring is kinetic. When a solid is first put
into water, it comes in contact only with solvent in its immediate vicinity. As the solid
dissolves, the amount of dissolved solute around the solid becomes more and more concentrated, and the rate of dissolving slows down. If the mixture is not stirred, the dissolved
solute ­diffuses very slowly through the solution; weeks may pass before the solid is entirely
dissolved. Stirring distributes the dissolved solute rapidly through the solution, and more
solvent is brought into contact with the solid, causing it to dissolve more rapidly.
Example 14.3
What does it mean to increase the surface area of a solid? Explain how this change causes
an increase in the rate of dissolving.
Solution
Increasing the surface area means breaking up the solid to form smaller particles that
have a larger total surface area. Increasing the surface area speeds up dissolving because
more solid comes in contact with the solute more quickly.
Practice 14.3
Use a molecular explanation to explain why increasing the temperature speeds up the rate
of dissolving a solid in a liquid.
14.4Concentration of Solutions
L earning objec tive
Solve problems involving mass percent, volume percent, molarity, and dilution.
Key termS
Many solids must be put into solution to undergo appreciable chemical reaction. We can
write the equation for the double-displacement reaction between sodium chloride and
silver nitrate:
dilute solution
concentrated solution
parts per million (ppm)
molarity (M)
NaCl + AgNO3 ¡ AgCl + NaNO3
14.4 • Concentration of Solutions But suppose we mix solid NaCl and solid AgNO3 and look for a chemical change. If any reaction occurs, it is slow and virtually undetectable. In fact, the crystalline structures of NaCl and
AgNO3 are so different that we could separate them by tediously picking out each kind of crystal
from the mixture. But if we dissolve the NaCl and AgNO3 separately in water and mix the two
solutions, we observe the immediate formation of a white, curdlike precipitate of silver chloride.
Molecules or ions must collide with one another in order to react. In the foregoing example,
the two solids did not react because the ions were securely locked within their crystal structures.
But when the NaCl and AgNO3 are dissolved, their crystal lattices are broken down and the ions
become mobile. When the two solutions are mixed, the mobile Ag+ and Cl- ions come into contact
and react to form insoluble AgCl, which precipitates out of solution. The soluble Na+ and NO3ions remain mobile in solution but form the crystalline salt NaNO3 when the water is evaporated:
NaCl(aq) + AgNO3(aq) ¡ AgCl(s) + NaNO3(aq)
Na + (aq) + Cl - (aq) + Ag + (aq) + NO -3 (aq) ¡ AgCl(s) + Na + (aq) + NO -3 (aq)
sodium chloride
solution
silver nitrate
solution
silver
chloride
sodium nitrate
in solution
The mixture of the two solutions provides a medium or space in which the Ag+ and Cl- ions
can react. (See Chapter 15 for further discussion of ionic reactions.)
Solutions also function as diluting agents in reactions in which the undiluted reactants
would combine with each other too violently. Moreover, a solution of known concentration
provides a convenient method for delivering a specific amount of reactant.
The concentration of a solution expresses the amount of solute dissolved in a given quantity of solvent or solution. Because reactions are often conducted in solution, it’s important to
understand the methods of expressing concentration and to know how to prepare solutions of
particular concentrations. The concentration of a solution may be expressed qualitatively or
quantitatively. Let’s begin with a look at the qualitative methods of expressing concentration.
Dilute and Concentrated Solutions
When we say that a solution is dilute or concentrated, we are expressing, in a ­ relative way,
the amount of solute present. One gram of a compound and 2 g of a compound in solution
are both dilute solutions when compared with the same volume of a solution containing 20 g
of a compound. Ordinary concentrated hydrochloric acid contains 12 mol of HCl per liter of
solution. In some laboratories, the dilute acid is made by mixing equal volumes of water and
the concentrated acid. In other laboratories the concentrated acid is diluted with two or three
volumes of water, depending on its use. The term dilute solution, then, describes a solution that contains a relatively small amount of dissolved solute. Conversely, a concentrated
solution contains a relatively large amount of dissolved solute.
Mass Percent Solution
The mass percent method expresses the concentration of the solution as the percent of solute
in a given mass of solution. It says that for a given mass of­ solution, a certain percent of that
mass is solute. Suppose we take a bottle from the reagent shelf that reads “sodium hydroxide,
NaOH, 10%.” This statement means that for every 100 g of this solution, 10 g will be NaOH
and 90 g will be water. (Note that this amount of solution is 100 g, not 100 mL.) We could also
make this same concentration of solution by dissolving 2.0 g of NaOH in 18 g of water. Mass
percent concentrations are most generally used for solids dissolved in liquids:
mass percent =
g solute
g solute
* 100 =
* 100
g solute + g solvent
g solution
As instrumentation advances are made in chemistry, our ability to measure the concentration
of dilute solutions is increasing as well. In addition to mass percent, chemists now commonly
use parts per million (ppm):
parts per million =
g solute
* 1,000,000
g solute + g solvent
Note that mass percent is
independent of the formula
for the solute.
313
314 chapter 14 • Solutions
Currently, air and water contaminants, drugs in the human body, and pesticide residues are
measured in parts per million.
Example 14.4
ENHANCED EXAMPLE
What is the mass percent of sodium hydroxide in a solution that is made by dissolving
8.00 g NaOH in 50.0 g H2O?
Solution
Read •
Knowns:
8.00 g NaOH
50.0 g H2O
Solving for: mass percent
Calculate • a
8.00 g NaOH
b = 13.8% NaOH solution
8.00 g NaOH + 50.0 g H2O
Example 14.5
What masses of potassium chloride and water are needed to make 250. g of 5.00% solution?
Solution
Read •
Knowns:
250. g solution
5.00% solution
Solving for: masses KCl and water
Plan •
5.00% of 250. g = (0.0500)(250. g) = 12.5 g KCl (solute)
Calculate • 250. g - 12.5 g = 238 g H2O
Dissolving 12.5 g KCl in 238 g H2O gives a 5.00% solution.
Example 14.6
A 34.0% sulfuric acid solution had a density of 1.25 g/mL. How many grams of H2SO4
are contained in 1.00 L of this solution?
Solution
Read •
Knowns:
d = 1.25 g/mL
V = 1.00 L
34.0% H2SO4 solution
Solving for: mass H2SO4
Plan •
Find the mass of the solution from the density and then use the mass percent
to determine the mass of H2SO4.
Calculate • d = mass/V
mass of solution = a
mass percent = a
g solute =
1.25 g
b (1.00 * 103 mL ) = 1250 g (solution)
mL
g solute
b 100
g solution
(mass percent)(g solution)
100
(34.0)(1250 g)
= 425 g H2SO4
100
Therefore, 1.00 L of 34.0% H2SO4 solution contains 425 g H2SO4.
g solute =
14.4 • Concentration of Solutions Practice 14.4
What is the mass percent of Na2SO4 in a solution made by dissolving 25.0 g Na2SO4 in
225.0 g H2O?
Mass/Volume Percent (m/v)
This method expresses concentration as grams of solute per 100 mL of solution. With this system, a 10.0%-m/v-glucose solution is made by dissolving 10.0 g of glucose in water, diluting to
100 mL, and mixing. The 10.0%-m/v solution could also be made by diluting 20.0 g to 200 mL,
50.0 g to 500 mL, and so on. Of course, any other appropriate dilution ratio may be used:
mass>volume percent =
g solute
* 100
mL solution
Example 14.7
A 3.0% H2O2 solution is commonly used as a topical antiseptic to prevent ­infection.
What volume of this solution will contain 10. g of H2O2?
SOLUTION
First solve the mass/volume percent equation for mL of solution:
mL solution =
(g solute)
(100)
(m>v percent)
mL solution =
(10. g solute)
(100) = 330 mL
(3.0 m>v percent)
Volume Percent
Solutions that are formulated from two liquids are often expressed as volume percent with respect to the solute. The volume percent is the volume of a liquid in 100 mL of solution. The label
on a bottle of ordinary rubbing alcohol reads “isopropyl alcohol, 70% by volume.” Such a solution could be made by mixing 70 mL of alcohol with water to make a total volume of 100 mL,
but we cannot use 30 mL of water, because the two volumes are not necessarily additive:
volume of liquid in question
* 100
total volume of solution
Volume percent is used to express the concentration of alcohol in beverages. Wines generally contain 12% alcohol by volume. This translates into 12 mL of alcohol in each 100 mL
of wine. The beverage industry also uses the concentration unit of proof (twice the volume
percent). Pure ethyl alcohol is 100% and therefore 200 proof. Scotch whiskey is 86 proof, or
43% alcohol by volume.
volume percent =
Molarity
Mass percent solutions do not equate or express the molar masses of the solute in solution. For
example, 1000. g of 10.0% NaOH solution contains 100. g NaOH; 1000. g of 10.0% KOH
solution contains 100. g KOH. In terms of moles of NaOH and KOH, these solutions contain
mol NaOH = (100. g NaOH) a
mol KOH = (100. g KOH) a
1 mol NaOH
b = 2.50 mol NaOH
40.00 g NaOH
1 mol KOH
b = 1.78 mol KOH
56.11 g KOH
From these figures, we see that the two 10.0% solutions do not contain the same number of
moles of NaOH and KOH. As a result, we find that a 10.0% NaOH solution contains more
reactive base than a 10.0% KOH solution.
315
316 chapter 14 • Solutions
1 liter
1 liter
Figure 14.7
Preparation of a 1 M solution.
1 liter
1000 ml
20°C
1000 ml
20°C
1000 ml
20°C
(a)
Add 1 mole of
solute to a 1-liter
volumetric flask
(b)
Dissolve in
solvent
(c)
Add solvent to the
1-liter mark and
mix thoroughly
We need a method of expressing concentration that will easily indicate how many moles
of solute are present per unit volume of solution. For this purpose, the concentration known as
molarity is used in calculations involving chemical reactions.
A 1-molar solution contains 1 mol of solute per liter of solution. For example, to make
a 1-molar solution of sodium hydroxide (NaOH) we dissolve 40.00 g NaOH (1 mol) in
water and dilute the solution with more water to a volume of 1 L. The solution contains
1 mol of the solute in 1 L of solution and is said to be 1 molar in concentration. Figure 14.7
illustrates the preparation of a 1-molar solution. Note that the volume of the solute and the
solvent together is 1 L.
The concentration of a solution can, of course, be varied by using more or less solute or
solvent; but in any case the molarity (M) of a solution is the number of moles of solute per liter
of solution. The abbreviation for molarity is M. The units of molarity are moles per liter. The
expression “2.0 M NaOH” means a 2.0-molar solution of NaOH (2.0 mol, or 80. g, of NaOH
dissolved in water to make 1.0 L of solution):
molarity = M =
number of moles of solute
moles
=
liter of solution
liter
Flasks that are calibrated to contain specific volumes at a particular temperature are used to
prepare solutions of a desired concentration. These volumetric flasks have a calibration mark
on the neck that accurately indicates the measured volume. Molarity is based on a specific
volume of solution and therefore will vary slightly with temperature because volume varies
with temperature (1000 mL H2O at 208C 5 1001 mL at 258C).
Suppose we want to make 500 mL of 1 M solution. This solution can be prepared by
determining the mass of 0.5 mol of the solute and diluting with water in a 500-mL (0.5-L)
volumetric flask. The molarity will be
0.5 mol solute
M =
= 1 molar
0.5 L solution
You can see that it isn’t necessary to have a liter of solution to express molarity. All we need to
know is the number of moles of dissolved solute and the volume of solution. Thus, 0.001 mol
NaOH in 10 mL of solution is 0.1 M:
a
0.001 mol 1000 mL
ba
b = 0.1 M
10 mL
1L
14.4 • Concentration of Solutions When we stop to think that a balance is not calibrated in moles, but in grams, we can incorporate grams into the molarity formula. We do so by using the relationship
grams of solute
molar mass
Substituting this relationship into our expression for molarity, we get
Molarities of concentrated
acids commonly used in the
laboratory:
HCl
HC2H3O2
HNO3
H2SO4
moles =
12 M
17 M
16 M
18 M
g solute
mol
=
L
molar mass solute * L solution
We can now determine the mass of any amount of a solute that has a known formula, dilute it
to any volume, and calculate the molarity of the solution using this formula.
M=
Example 14.8
What is the molarity of a solution containing 1.4 mol of acetic acid (HC2H3O2) in 250. mL
of solution?
Solution
By the unit-conversion method, we note that the concentration given in the problem statement is 1.4 mol per 250. mL (mol/mL). Since molarity 5 mol/L, the needed conversion is
Solution map:
Calculate • a
mol
mol
¡
=M
mL
L
1.4 mol
1000 mL
5.6 mol
ba
b =
= 5.6 M
250. mL
L
L
Example 14.9
What is the molarity of a solution made by dissolving 2.00 g of potassium chlorate in
enough water to make 150. mL of solution?
Solution
mass KClO3 = 2.00 g
volume = 150. mL
molar mass KClO3 = 122.6 g>mol
Knowns:
Solution map:
Calculate • a
g KClO3
g KClO3
mol KClO3
¡
¡
=M
mL
L
L
2.00 g KClO3 1000 mL
1 mol KClO3
0.109 mol
ba
ba
b =
150. mL
L
122.6 g KClO3
L
= 0.109 M KClO3
Example 14.10
How many grams of potassium hydroxide are required to prepare 600. mL of 0.450 M
KOH solution?
SOLUTION
Solution map: milliliters ¡ liters ¡ moles ¡ grams
Plan •
Knowns:
Calculate • (600. mL )a
volume = 600. mL
molar mass KOH =
M =
0.450 mol
L
56.11 g
mol
1L
0.450 mol 56.11 g KOH
ba
ba
b = 15.1 g KOH
1000 mL
L
mol
ENHANCED EXAMPLE
317
318 chapter 14 • Solutions
Practice 14.5
What is the molarity of a solution made by dissolving 7.50 g of magnesium nitrate
[Mg(NO3)2] in enough water to make 25.0 mL of solution?
Practice 14.6
How many grams of sodium chloride are needed to prepare 125 mL of a 0.037 M NaCl solution?
Example 14.11
Calculate the number of moles of nitric acid in 325 mL of 16 M HNO3 solution.
Solution
Read •
Knowns:
Plan •
V 5 325 mL
16 M HNO3 solution
Use the equation moles = liters * M
16 mol HNO3
b = 5.2 mol HNO3
1L
Calculate • moles = (0.325 L )a
Example 14.12
What volume of 0.250 M solution can be prepared from 16.0 g of potassium carbonate?
Solution
Read •
Knowns:
Plan •
mass = 16.0 g K2CO3
M = 0.250 mol>L
Molar mass K2CO3 = 138.2 g>mol
Find the volume that can be prepared from 16.0 g K2CO3.
Solution map: g K2CO3 S mol K2CO3 S L solution
Calculate • (16.0 g K2CO3 )a
1 mol K2CO3
1L
ba
b = 0.463 L
138.2 g K2CO3 0.250 mol K2CO3
= 463 mL
Example 14.13
How many milliliters of 2.00 M HCl will react with 28.0 g NaOH?
Solution
Read •
Knowns:
Plan •
mass = 28.0 g NaOH
M = 2.00 mol>L HCl
Write a balanced equation for the reaction
HCl(aq) + NaOH(aq) ¡ NaCl(aq) + H2O(l)
Find the number of moles NaOH in 28.0 g NaOH.
Solution map: grams NaOH S moles NaOH
(28.0 g NaOH )a
Setup •
1 mole
b = 0.700 mol NaOH
40.00 g NaOH
Solution map: mol NaOH S mol HCl S L HCl S mL HCl
Calculate • (0.700 mol NaOH )a
1 mol HCl
1 L HCl
ba
b = 0.350 L HCl
1 mol NaOH 2.00 mol HCl
= 350. mL HCl
14.4 • Concentration of Solutions Practice 14.7
What volume of 0.035 M AgNO3 can be made from 5.0 g of AgNO3 ?
Practice 14.8
How many milliliters of 0.50 M NaOH are required to react with 25.00 mL of 1.5 M HCl?
We’ve now examined several ways to measure concentration of solutions quantitatively.
A summary of these concentration units is found in Table 14.4.
Table 14.4
Concentration Units for Solutions
Units
Symbol
Definition
Mass percent
% m/m
mass solute
* 100
mass solution
Parts per million
ppm
mass solute
* 1,000,000
mass solution
Mass/volume percent
% m/v
mass solute
* 100
mL solution
Volume percent
% v/v
mL solute
* 100
mL solution
Molarity
M
moles solute
L solution
Molality
m
moles solute
kg solvent
Molality is covered in
Section 14.5.
Dilution Problems
Chemists often find it necessary to dilute solutions from one concentration to another by adding more solvent to the solution. If a solution is diluted by adding pure solvent, the volume
of the solution increases, but the number of moles of solute in the solution remains the same.
Thus the moles/liter (molarity) of the solution decreases. Always read a problem carefully to
distinguish ­between (1) how much solvent must be added to dilute a solution to a particular
concentration and (2) to what volume a solution must be diluted to prepare a solution of a
particular concentration.
Example 14.14
Calculate the molarity of a sodium hydroxide solution that is prepared by mixing 100. mL
of 0.20 M NaOH with 150. mL of water. Assume that the volumes are additive.
Solution
Read •
Knowns:
Plan •
V1 = 100. mL NaOH
V2 = 150. mL water
Vsolution = 250. mL
Minitial = 0.20 M
In the dilution, the moles NaOH remain the same; the molarity and volume
change.
1. Calculate the moles of NaOH in the original solution.
0.20 mol NaOH
b = 0.020 mol NaOH
1 L
2. Divide the mol NaOH by the final volume of the solution to find the new
molarity.
mol = (0.100 L )a
ENHANCED EXAMPLE
319
320 chapter 14 • Solutions
Calculate • M =
Check •
0.020 mol NaOH
= 0.080 M NaOH
0.250 L
If we double the volume of the solution, the concentration is half. Therefore
the concentration of the new solution here should be less than 0.10 M.
Practice 14.9
Calculate the molarity of a solution prepared by diluting 125 mL of 0.400 M K2Cr2O7
with 875 mL of water.
Example 14.15
How many grams of silver chloride will be precipitated by adding sufficient silver nitrate
to react with 1500. mL of 0.400 M barium chloride solution?
2 AgNO3(aq) + BaCl2(aq) ¡ 2 AgCl(s) + Ba(NO3)2(aq)
Solution
Read •
Knowns: V = 1500. mL BaCl2
M = 0.400 M BaCl2
Plan •
Solve as a stoichiometry problem.
Determine the mol BaCl2 in 1500. mL of 0.400 M solution.
mol = M * V = (1.500 L )a
0.400 mol BaCl2
b = 0.600 mol BaCl2
L
Solution map: moles BaCl2 S moles AgCl S grams AgCl
Calculate • (0.600 mol BaCl2 )a
2 mol AgCl
143.4 g AgCl
ba
b = 172 g AgCl
1 mol BaCl2
mol AgCl
Practice 14.10
How many grams of lead(II) iodide will be precipitated by adding sufficient Pb(NO3)2 to
react with 750 mL of 0.250 M KI solution?
2 KI(aq) + Pb(NO3)2(aq) ¡ PbI2(s) + 2 KNO3(aq)
14.5Colligative Properties of Solutions
Learning objective
Use the concept of colligative properties to calculate molality, freezing point, boiling
point, freezing point depression, and boiling point elevation of various solutions.
KEY TERMS
Two solutions—one containing 1 mol (60.06 g) of urea (NH2CONH2) and the other containing
1 mol (342.3 g) of sucrose (C12H22O11) each in 1 kg of water—both have a freezing point of
-1.86°C, not 0°C as for pure water. Urea and sucrose are distinct substances, yet they lower
the freezing point of the water by the same amount. The only thing apparently common to
these two solutions is that each contains 1 mol (6.022 * 1023 molecules) of solute and 1 kg
of solvent. In fact, when we dissolve 1 mol of any nonionizable solute in 1 kg of water, the
freezing point of the resulting solution is -1.86°C.
These results lead us to conclude that the freezing point depression for a solution
containing 6.022 * 1023 solute molecules (particles) and 1 kg of water is a constant,
namely, 1.86°C. Freezing point depression is a general property of solutions. Furthermore, the amount by which the freezing point is depressed is the same for all solutions
colligative properties
molality (m)
14.5 • Colligative Properties of Solutions Table 14.5
Freezing Point Depression and Boiling Point Elevation Constants of Selected Solvents
Freezing point
Solvent
Water
Acetic acid
Benzene
Camphor
321
Freezing point
depression constant, Kf
Boiling point
C kg solvent
of pure solvent
of
pure solvent
a
b
(8C)
mol solute
(8C)
0.00
16.6
5.5
178
1.86
3.90
5.1
40
100.0
118.5
80.1
208.2
Boiling point
elevation constant, Kb
C kg solvent
a
b
mol solute
0.512
3.07
2.53
5.95
made with a given solvent; that is, each solvent shows a characteristic freezing point
depression constant. Freezing point depression constants for several solvents are given
in Table 14.5.
The solution formed by the addition of a nonvolatile solute to a solvent has a lower freezing point, a higher boiling point, and a lower vapor pressure than that of the pure solvent. These
effects are related and are known as colligative properties. The colligative properties are
properties that depend only on the number of solute particles in a solution, not on the nature of
those particles. Freezing point depression, boiling point elevation, and vapor pressure lowering
are colligative properties of solutions.
The colligative properties of a solution can be considered in terms of vapor pressure. The
vapor pressure of a pure liquid depends on the tendency of molecules to escape from its surface. If 10% of the molecules in a solution are nonvolatile solute molecules, the vapor pressure
of the solution is 10% lower than that of the pure solvent. The vapor pressure is lower because
the surface of the solution contains 10% nonvolatile molecules and 90% of the volatile solvent
molecules. A liquid boils when its vapor pressure equals the pressure of the atmosphere. We
can thus see that the solution just described as having a lower vapor pressure will have a higher
boiling point than the pure solvent. The solution with a lowered vapor pressure doesn’t boil
until it has been heated above the boiling point of the solvent (see Figure 14.8a). Each solvent has its own characteristic boiling point elevation constant (Table 14.5). The boiling point
elevation constant is based on a solution that contains 1 mol of solute particles per kilogram
of solvent. For example, the boiling point elevation constant for a solution containing 1 mol
of solute particles per kilogram of water is 0.512°C, which means that this water solution will
boil at 100.512°C.
The freezing behavior of a solution can also be considered in terms of lowered vapor pressure. Figure 14.8b shows the vapor pressure relationships of ice, water, and a solution containing 1 mol of solute per kilogram of water. The freezing point of water is at the intersection of
the liquid and solid vapor pressure curves (i.e., at the point where water and ice have the same
Figure 14.8
Pure water
Vapor pressure (torr)
Vapor pressure (torr)
760
Pure water
Solution
4.58
Solution
100 100.512
Temperature (°C)
(a)
–1.86 0
Temperature (°C)
(b)
Vapor pressure curves of pure
water and water solutions,
showing (a) boiling point elevation
and (b) freezing point depression
effects (concentration: 1 mol
solute/1 kg water).
chapter 14 • Solutions
© Brian H. Thomas/Alamy Limited
Engine coolant is one application of
colligative properties. The addition
of coolant to the water in a radiator
raises its boiling point and lowers
its freezing point.
vapor pressure). Because the vapor pressure of the liquid is lowered by the solute, the vapor
pressure curve of the solution does not intersect the vapor pressure curve of the solid until the
solution has been cooled below the freezing point of pure water. So the solution must be cooled
below 0°C in order for it to freeze.
The foregoing discussion dealing with freezing point depressions is restricted to unionized substances. The discussion of boiling point elevations is restricted to nonvolatile and
un-ionized substances. The colligative properties of ionized substances are not under consideration at this point; we will discuss them in Chapter 15.
Some practical applications involving colligative properties are (1) use of salt–ice mixtures to provide low freezing temperatures for homemade ice cream, (2) use of sodium chloride
or calcium chloride to melt ice from streets, and (3) use of ethylene glycol–water mixtures
as antifreeze in automobile radiators (ethylene glycol also raises the boiling point of radiator
fluid, thus ­allowing the engine to operate at a higher temperature).
Sodium chloride or calcium
chloride is used to melt ice on
snowy streets and highways.
J Morrill Photo/Photolibrary/Getty Images, Inc.
322 Both the freezing point depression and the boiling point elevation are directly proportional
to the number of moles of solute per kilogram of solvent. When we deal with the colligative
properties of solutions, another concentration expression, molality, is used. The molality (m)
of a solution is the number of moles of solute per kilogram of solvent:
m=
mol solute
kg solvent
Note that a lowercase m is used for molality concentrations and a capital M for molarity.
The difference between molality and molarity is that molality refers to moles of solute per
kilogram of solvent, whereas molarity refers to moles of solute per liter of solution. For
un-ionized substances, the colligative properties of a solution are directly proportional to
its molality.
Molality is independent of volume. It is a mass-to-mass relationship of solute to solvent
and allows for experiments, such as freezing point depression and boiling point elevation, to
be conducted at variable temperatures.
The following equations are used in calculations involving colligative properties and molality:
tf = mK f
m
tf
tb
Kf
Kb
=
=
=
=
=
tb = mKb
m=
mol solute
kg solvent
molality; mol solute/kg solvent
freezing point depression; °C
boiling point elevation; °C
freezing point depression constant; °C kg solvent/mol solute
boiling point elevation constant; °C kg solvent/mol solute
14.5 • Colligative Properties of Solutions Example 14.16
What is the molality (m) of a solution prepared by dissolving 2.70 g CH3OH in 25.0 g H2O?
SOLUTION
Since m =
mol solute
, the conversion is
kg solvent
2.70 g CH3OH
mol CH3OH
mol CH3OH
¡
¡
25.0 g H2O
25.0 g H2O
1 kg H2O
The molar mass of CH3OH is (12.01 + 4.032 + 16.00), or 32.04 g/mol:
2.70 g CH3OH
1 mol CH3OH
3.37 mol CH3OH
1000 g H2O
ba
ba
b =
25.0 g H2O
32.04 g CH3OH
1 kg H2O
1 kg H2O
The molality is 3.37 m.
a
Practice 14.11
What is the molality of a solution prepared by dissolving 150.0 g C6H12O6 in 600.0 g
H2O?
Example 14.17
A solution is made by dissolving 100. g of ethylene glycol (C2H6O2) in 200. g of water.
What is the freezing point of this solution?
SOLUTION
To calculate the freezing point of the solution, we first need to calculate tf , the change
in freezing point. Use the equation
tf = mKf =
mol solute
* Kf
kg solvent
K f (for water):
1.86°C kg solvent
(from Table 14.5)
mol solute
mol solute: (100. g C2H6O2 )a
kg solvent: (200. g H2O) a
tf = ¢
1 mol C2H6O2
b = 1.61 mol C2H6O2
62.07 g C2H6O2
1 kg
b = 0.200 kg H2O
1000 g
1.61 mol C2H6O2 1.86°C kg H2O
≤¢
≤ = 15.0°C
0.200 kg H2O
1 mol C2H6O2
The freezing point depression, 15.0°C, must be subtracted from 0°C, the freezing point of
the pure solvent (water):
freezing point of solution = freezing point of solvent - tf
= 0.0°C - 15.0°C = -15.0°C
Therefore, the freezing point of the solution is -15.0°C. This solution will protect an
automobile radiator down to -15.0°C (5°F).
Example 14.18
A solution made by dissolving 4.71 g of a compound of unknown molar mass in
100.0 g of water has a freezing point of -1.46°C. What is the molar mass of the
compound?
ENHANCED EXAMPLE
323
324 chapter 14 • Solutions
SOLUTION
First substitute the data in tf = mKf and solve for m:
tf = +1.46 (since the solvent, water, freezes at 0°C)
Kf =
1.86°C kg H2O
mol solute
1.86°C kg H2O
mol solute
1.46° C * mol solute 0.785 mol solute
m=
=
1.86° C * kg H2O
kg H2O
1.46°C = mKf = m *
Now convert the data, 4.71 g solute/100.0 g H2O, to g/mol:
¢
4.71 g solute 1000 g H2O
1 kg H2O
≤¢
≤¢
≤ = 60.0 g>mol
100.0 g H2O
1 kg H2O
0.785 mol solute
The molar mass of the compound is 60.0 g/mol.
Practice 14.12
What is the freezing point of the solution in Practice Exercise 14.11? What is the boiling
point?
>Chemistry in action
The Scoop on Ice Cream
Ice cream is mainly composed of water (from milk and
cream), milk solids, milk fats, and, frequently, various sweeteners (corn syrup or sugar), flavorings, emulsifiers, and
stabilizers. But that smooth, creamy, rich flavor and texture
are the result of the chemistry of the mixing and freezing process. The rich, smooth texture of great ice cream
results from the milk fat. By law, if the carton is labeled “ice
cream,” it must contain a minimum of 10% milk fat. That
carton of ice cream also contains 20%–50% air whipped
into the ingredients during the initial mixing process.
typical serving temperature (216°C for most ice cream),
only about 72% of the water in the ice cream is frozen.
The unfrozen solution keeps the ice cream “scoopable.”
One last factor that affects the quality of ice cream is the
size of the ice crystals. For very smooth ice cream, tiny
crystals are needed. To produce these, the ice cream must
freeze very slowly. Large crystals give a coarse, grainy
texture. Now, as you savor that premium ice cream cone,
you’ll know just how colligative properties and the chemistry of freezing helped make it so delicious!
Once the mixture is fully whipped, it is cooled to begin
the freezing process. But ice cream does not freeze at
0°C even though it is 55%–64% water. The freezing point
is depressed as a colligative property of the ice cream
solution. Once the freezing of the water begins, the
concentration of the solution increases, which continues
to depress the freezing point. Goff tells us that even at a
© DNY59/iStockphoto
H. Douglas Goff, a professor of food science and an ice
cream expert from Ontario, Canada, says, “There are no
real chemical reactions that take place when you make
ice cream, but that doesn’t mean that there isn’t plenty of
chemistry.” The structure of ice cream contributes greatly
to its taste. Tiny air bubbles are formed in the initial whipping process. These bubbles are distributed through a
network of fat globules and liquid water. The milk fat has
surface proteins on the globules in the milk to keep the
fat dissolved in solution. Ice cream manufacturers destabilize these globules using emulsifiers (such as egg yolks,
mono- or diglycerides) and let them come together in
larger networks sort of like grape clusters.
14.6 • Osmosis and Osmotic Pressure 325
14.6Osmosis and Osmotic Pressure
Discuss osmosis and osmotic pressure and their importance in living systems.
Learning objectiv e
Key terms
semipermeable membrane
osmosis
osmotic pressure
Dennis Kunkel/Phototake
Dennis Kunkel/Phototake
Dennis Kunkel/Phototake
When red blood cells are put into distilled water, they gradually swell and in time may
burst. If red blood cells are put in a 5% urea (or a 5% salt) solution, they gradually
shrink and take on a wrinkled appearance. The cells behave in this fashion because they
are enclosed in semipermeable membranes. A semipermeable membrane allows the
passage of water (solvent) molecules through it in either direction but prevents the passage of larger solute molecules or ions. When two solutions of different concentrations
(or water and a water solution) are separated by a semipermeable membrane, water diffuses through the membrane from the solution of lower concentration into the solution
of higher concentration. The diffusion of water, either from a dilute solution or from
pure water, through a semipermeable membrane into a solution of higher concentration
is called osmosis.
A 0.90% (0.15 M) sodium chloride solution is known as a physiological saline solution because it is isotonic with blood plasma; that is, it has the same concentration of NaCl
as blood plasma. Because each mole of NaCl yields about 2 mol of ions when in solution,
the solute particle concentration in physiological saline solution is nearly 0.30 M. A 5%
glucose solution (0.28 M) is also approximately isotonic with blood plasma. Blood cells
neither swell nor shrink in an isotonic solution. The cells described in the preceding paragraph swell in water because water is hypotonic to cell plasma. The cells shrink in 5% urea
solution because the urea solution is hypertonic to the cell plasma. To prevent possible
injury to blood cells by osmosis, fluids for intravenous use are usually made up at approximately isotonic concentration.
All solutions exhibit osmotic pressure, which is another colligative property. Osmotic
pressure is a pressure difference between the system and atmospheric pressure. The osmotic
pressure of a system can be measured by applying enough pressure to stop the flow of water
due to osmosis in the system. The difference between the applied pressure and atmospheric
pressure is the osmotic pressure. When pressure greater than the osmotic pressure is applied to a system, the flow of water can be reversed from that of osmosis. This process can be
used to obtain useful drinking water from seawater and is known as reverse osmosis. Osmotic
pressure is dependent only on the concentration of the solute particles and is independent of
Human red blood cells. Left: In an isotonic solution, the concentration is the same inside and outside the cell (0.9% saline).
Center: In a hypertonic solution (1.6% saline), water leaves the cells, causing them to crenate (shrink). Right: In a hypotonic
solution (0.2% saline), the cells swell as water moves into the cell center. Magnification is 260,0003.
326 chapter 14 • Solutions
Rising
solution
level
Cross section on
molecular level
Thistle
tube
Semipermeable
membrane
Sugar
solution
Water
Sugar molecule
Water molecule
Semipermeable
membrane (cellophane)
Figure 14.9
Laboratory demonstration of osmosis: As a result of osmosis, water passes through the
membrane, causing the solution to rise in the thistle tube.
their nature. The osmotic pressure of a solution can be measured by determining the amount
of counterpressure needed to prevent osmosis; this pressure can be very large. The osmotic
pressure of a solution containing 1 mol of solute particles in 1 kg of water is about 22.4 atm,
which is about the same as the pressure exerted by 1 mol of a gas confined in a volume of
1 L at 0°C.
Osmosis has a role in many biological processes, and semipermeable membranes occur
commonly in living organisms. An example is the roots of plants, which are covered with
tiny structures called root hairs; soil water enters the plant by osmosis, passing through the
semipermeable membranes covering the root hairs. Artificial or synthetic membranes can
also be made.
Osmosis can be demonstrated with the simple laboratory setup shown in Figure 14.9.
As a result of osmotic pressure, water passes through the cellophane membrane into the
thistle tube, causing the solution level to rise. In osmosis, the net transfer of water is
always from a less concentrated to a more concentrated solution; that is, the effect is toward
equalization of the concentration on both sides of the membrane. Note that the effective
movement of water in osmosis is always from the region of higher water concentration to
the region of lower water concentration.
Osmosis can be explained by assuming that a semipermeable membrane has passages
that permit water molecules and other small molecules to pass in either direction. Both
sides of the membrane are constantly being struck by water molecules in random motion.
The number of water molecules crossing the membrane is proportional to the number of
water molecule-to-membrane impacts per unit of time. Because the solute molecules or
ions reduce the concentration of water, there are more water molecules and thus more water
molecule impacts on the side with the lower solute concentration (more dilute solution).
The greater number of water molecule-to-membrane impacts on the dilute side thus causes
a net transfer of water to the more concentrated solution. Again, note that the overall process involves the net transfer, by diffusion through the membrane, of water molecules from
a region of higher water concentration (dilute solution) to one of lower water concentration
(more concentrated solution).
This is a simplified picture of osmosis. No one has ever seen the hypothetical passages
that allow water molecules and other small molecules or ions to pass through them. Alternative explanations have been proposed, but our discussion has been confined to water solutions.
Osmotic pressure is a general colligative property, however, and is known to occur in nonaqueous systems.
Review 327
C hapter
14 review
14.1 General Properties of Solutions
• A solution is a homogeneous mixture of two or more substances:
• Consists of solvent—the dissolving agent—and solute—the component(s) dissolved in the solvent
• Is a homogeneous mixture
• Contains molecular or ionic particles
• Can be colored or colorless
• Can be separated into solute and solvent by a physical separation process
Key Terms
solution
solute
solvent
14.2 Solubility
• Solubility describes the amount of solute that will dissolve in a specified amount of solvent.
• Solubility can also be qualitative.
• General guidelines for ionic solubility are:
Soluble
Insoluble
Na+, K+, NH4+
Nitrates, NO3–
Acetates, C 2 H 3O 2–
Chlorides, C l –
Bromides, Br –
Iodides, I –
Sulfates, SO42–
Ag +, Ca2+ are
slightly soluble
NH4+
alkali metal cations
except
Ag+, Hg 22+, Pb2+
except
Ba2+, Sr 2+, Pb2+
except
Carbonates, CO 2–
3
Phosphates, PO 43–
Hydroxides, OH –
Sulfides, S2–
• Liquids can also be classified as miscible (soluble in each other) or immiscible (not soluble in each
other).
• The concentration of a solution is the quantitative measurement of the amount of solute that is dissolved in a solution.
• Like tends to dissolve like is a general rule for solvents and solutes.
• As temperature increases:
• Solubility of a solid in a liquid tends to increase.
• Solubility of a gas in a liquid tends to decrease.
• As pressure increases:
• Solubility of a solid in a liquid remains constant.
• Solubility of a gas in a liquid tends to increase.
• At a specific temperature, the amount of solute that can dissolve in a solvent has a limit:
• Unsaturated solutions contain less solute than the limit.
• Saturated solutions contain dissolved solute at the limit.
• Supersaturated solutions contain more solute than the limit and are therefore unstable:
• If disturbed, the excess solute will precipitate out of solution.
Key Terms
solubility
miscible
immiscible
concentration of a solution
saturated solution
unsaturated solution
supersaturated solution
328 chapter 14 • Solutions
14.3 Rate of Dissolving Solids
• The rate at which a solute dissolves is determined by these factors:
• Particle size
• Temperature
• Concentration of solution
• Agitation
14.4 Concentration of Solutions
Key Terms
dilute solution
concentrated solution
parts per million (ppm)
molarity (M)
• Molecules or ions must collide in order to react.
• Solutions provide a medium for the molecules or ions to collide.
• Concentrations can be measured in many ways:
Mass percent
% m/m
mass solute
* 100
mass solution
Parts per million
ppm
mass solute
* 1,000,000
mass solution
Mass/volume percent
% m/v
mass solute
* 100
mL solution
Volume percent
% v/v
mL solute
* 100
mL solution
Molarity
M
moles solute
L solution
Molality
m
moles solute
kg solvent
• Dilution of solutions requires the addition of more solvent to an existing solution:
• The number of moles in the diluted solution is the same as that in the original solution.
• M1V1 = M2V2.
14.5 Colligative Properties of Solutions
Key Terms
colligative properties
molality (m)
• Properties of a solution that depend only on the number of solute particles in solution are called colligative properties:
• Freezing point depression
tf = mK f
• Boiling point elevation
tb = mKb
• Osmotic pressure
• Molality is used in working with colligative properties.
14.6 Osmosis and Osmotic Pressure
Key Terms
semipermeable membrane
osmosis
osmotic pressure
• Osmosis is the diffusion of water through a semipermeable membrane:
• Occurs from dilute solution to a solution of higher concentration
• Osmosis results in osmotic pressure, which is a colligative property of a solution.
R e v i e w Q u e st i o n s
1. What is a true solution?
2. Name and distinguish between the two components of a
solution.
3. Is it always apparent in a solution which component is the solute,
for example, in a solution of a liquid in a liquid?
4. Explain why the solute does not settle out of a solution.
5. Is it possible to have one solid dissolved in another? Explain.
6. An aqueous solution of KCl is colorless, KMnO4 is purple, and
K2Cr2O7 is orange. What color would you expect of an aqueous
solution of Na2Cr2O7 ? Explain.
7. Why is air considered to be a solution?
8. Sketch the orientation of water molecules (a) about a single sodium
ion and (b) about a single chloride ion in solution.
9. Refer to Table 14.2 and estimate the number of grams of potassium
bromide that would dissolve in 50 g water at 08C.
10. Refer to Figure 14.4 to determine the solubility of each of these
substances at 258C:
(a) ammonium chloride
(b) copper(II) sulfate
(c) sodium nitrate
Paired Exercises 329
11. As you go down Group 1A from lithium to potassium, what is the
trend in solubilities of the chlorides and bromides of these metals?
(Table 14.2)
12. State the solubility, in grams of solute per 100 g of H2O, of:
(Figure 14.4)
(a) NH4Cl at 35°C
(c) SO2 gas at 30°C
(b) CuSO4 at 60°C
(d) NaNO3 at 15°C
13. Which solid substance in Figure 14.4 shows an overall decrease in
solubility with an increase in temperature?
14. At a temperature of 45°C, will a solution made with 6.5 g KNO3
in 15.0 g water be saturated or unsaturated? What will be the mass
percent of solute in the solution? (Figure 14.4)
15. If 40. g Li2SO4 are added to 75.0 g water at 40°C, will the solution be saturated or unsaturated? What will be the mass percent of
solute in the solution? (Figure 14.4)
16. Explain how a supersaturated solution of NaC2H3O2 can be prepared and proven to be supersaturated.
17. Explain why hexane will dissolve benzene but will not dissolve
sodium chloride.
18. Some drinks like tea are consumed either hot or cold, whereas others like Coca-Cola are drunk only cold. Why?
19. What is the effect of pressure on the solubility of gases in liquids?
of solids in liquids?
20. In a saturated solution containing undissolved solute, solute is continuously dissolving, but the concentration of the solution remains
unchanged. Explain.
21. Champagne is usually cooled in a refrigerator prior to opening. It’s
also opened very carefully. What would happen if a warm bottle of
champagne were shaken and opened quickly and forcefully?
22. What would be the total surface area if the 1-cm cube in Figure
14.5 were cut into cubes 0.01 cm on a side?
23. When a solid solute is put into a solvent, does the rate of dissolving increase or decrease as the dissolution proceeds? Explain.
(Figure 14.6)
24. In which will a teaspoonful of sugar dissolve more rapidly, 200 mL
of iced tea or 200 mL of hot coffee? Explain in terms of the KMT.
25. Why do smaller particles dissolve faster than large ones?
26. Explain why there is no apparent reaction when crystals of AgNO3
and NaCl are mixed, but a reaction is apparent immediately when
solutions of AgNO3 and NaCl are mixed.
27. What do we mean when we say that concentrated nitric acid
(HNO3) is 16 molar?
28. Will 1 L of 1 M NaCl contain more chloride ions than 0.5 L of 1 M
MgCl2 ? Explain.
29. Describe how you would prepare 750 mL of 5.0 M NaCl solution.
30. Arrange the following bases (in descending order) according to the
volume of each that will react with 1 L of 1 M HCl:
(a) 1 M NaOH
(b) 1.5 M Ca(OH)2
(c) 2 M KOH
(d) 0.6 M Ba(OH)2
31. Explain in terms of vapor pressure why the boiling point of a solution
containing a nonvolatile solute is higher than that of the pure solvent.
32. Explain why the freezing point of a solution is lower than the freezing point of the pure solvent.
33. When water and ice are mixed, the temperature of the mixture is
0°C. But if methyl alcohol and ice are mixed, a temperature of
-10°C is readily attained. Explain why the two mixtures show
such different temperature behavior.
34. Which would be more effective in lowering the freezing point of
500. g of water?
(a) 100. g of sucrose, C12H22O11 , or 100. g of ethyl alcohol,
C2H5OH
(b) 100. g of sucrose or 20.0 g of ethyl alcohol
(c) 20.0 g of ethyl alcohol or 20.0 g of methyl alcohol, CH3OH.
35. What is the difference between molarity and molality?
36. Is the molarity of a 5 m aqueous solution of NaCl greater or less than
5 M? Explain.
37. Why do salt trucks distribute salt over icy roads in the winter?
38. Assume that the thistle tube in Figure 14.9 contains 1.0 M sugar
solution and that the water in the beaker has just been replaced by
a 2.0 M solution of urea. Would the solution level in the thistle tube
continue to rise, remain constant, or fall? Explain.
39. Explain in terms of the KMT how a semipermeable membrane functions when placed between pure water and a 10% sugar solution.
40. Which has the higher osmotic pressure, a solution containing 100 g
of urea (NH2CONH2) in 1 kg H2O or a solution containing 150 g
of glucose, C6H12O6 , in 1 kg H2O?
41. Explain why a lettuce leaf in contact with salad dressing containing
salt and vinegar soon becomes wilted and limp whereas another
lettuce leaf in contact with plain water remains crisp.
42. A group of shipwreck survivors floated for several days on a life
raft before being rescued. Those who had drunk some seawater
were found to be suffering the most from dehydration. Explain.
Most of the exercises in this chapter are available for assignment via the online homework management program, WileyPLUS (www.wileyplus.com).
All exercises with blue numbers have answers in Appendix VI.
Pa i r e d E x e r c i s e s
1. Of the following substances, which ones are generally soluble in
water? (See Figure 14.2 or Appendix V.)
(a) AgCl
(d) NaOH
(b) K2SO4
(e) PbI2
(c) Na3PO4
(f) SnCO3
2. Of the following substances, which ones are generally soluble in
water? (See Figure 14.2 or Appendix V.)
(a) Ba3(PO4)2
(d) NH4C2H3O2
(b) Cu(NO3)2
(e) MgO
(c) Fe(OH)3
(f) AgNO3
3. Calculate the mass percent of the following solutions:
(a) 15.0 g KCl 1 100.0 g H2O
(b) 2.50 g Na3PO4 1 10.0 g H2O
(c) 0.20 mol NH4C2H3O2 1 125 g H2O
(d) 1.50 mol NaOH in 33.0 mol H2O
4. Calculate the mass percent of the following solutions:
(a) 25.0 g NaNO3 in 125.0 g H2O
(b) 1.25 g CaCl2 in 35.0 g H2O
(c) 0.75 mol K2CrO4 in 225 g H2O
(d) 1.20 mol H2SO4 in 72.5 mol H2O
5. A bleaching solution requires 5.23 g of sodium hypochlorite. How
many grams of a 21.5% by mass solution of sodium hypochorite
should be used?
6. A paint requires 42.8 g of iron(III) oxide to give it the right yellow
tint. How many grams of a 30.0% by mass solution of iron(III)
oxide should be used?
330 chapter 14 • Solutions
7. In 25 g of a 7.5% by mass solution of CaSO4
(a) how many grams of solute are present?
(b) how many grams of solvent are present?
8. In 75 g of a 12.0% by mass solution of BaCl2
(a) how many grams of solute are present?
(b) how many grams of solvent are present?
9. Determine the mass/volume percent of a solution made by dissolving:
(a) 15.0 g of C2H5OH (ethanol) in water to make 150.0 mL of
solution
(b) 25.2 g of NaCl in water to make 125.5 mL of solution
10. Determine the mass/volume percent of a solution made by dissolving:
(a) 175.2 g of table sugar, C12H22O11, in water to make 275.5 mL
of solution
(b) 35.5 g of CH3OH (methanol) in water to make 75.0 mL of solution
11. Determine the volume percent of a solution made by dissolving:
(a) 50.0 mL of hexanol in enough ethanol to make 125 mL of
solution
(b) 2.0 mL of ethanol in enough methanol to make 15.0 mL of
solution
(c) 15.0 mL of acetone in enough hexane to make 325 mL of solution
12. Determine the volume percent of a solution made by dissolving:
(a) 37.5 mL of butanol in enough ethanol to make 275 mL of
solution
(b) 4.0 mL of methanol in enough water to make 25.0 mL of solution
(c) 45.0 mL of isoamyl alcohol in enough acetone to make 750. mL
of solution
13. Calculate the molarity of the following solutions:
(a) 0.25 mol of solute in 75.0 mL of solution
(b) 1.75 mol of KBr in 0.75 L of solution
(c) 35.0 g of NaC2H3O2 in 1.25 L of solution
(d) 75 g of CuSO4 ? 5 H2O in 1.0 L of solution
14. Calculate the molarity of the following solutions:
(a) 0.50 mol of solute in 125 mL of solution
(b) 2.25 mol of CaCl2 in 1.50 L of solution
(c) 275 g C6H12O6 in 775 mL of solution
(d) 125 g MgSO4 ? 7 H2O in 2.50 L of solution
15. Calculate the number of moles of solute in each of the following
solutions:
(a) 1.5 L of 1.20 M H2SO4
(b) 25.0 mL of 0.0015 M BaCl2
(c) 125 mL of 0.35 M K3PO4
16. Calculate the number of moles of solute in each of the following
solutions:
(a) 0.75 L of 1.50 M HNO3
(b) 10.0 mL of 0.75 M NaClO3
(c) 175 mL of 0.50 M LiBr
17. Calculate the grams of solute in each of the following solutions:
(a) 2.5 L of 0.75 M K2CrO4
(b) 75.2 mL of 0.050 M HC2H3O2
(c) 250 mL of 16 M HNO3
18. Calculate the grams of solute in each of the following solutions:
(a) 1.20 L of 18 M H2SO4
(b) 27.5 mL of 1.50 M KMnO4
(c) 120 mL of 0.025 M Fe2(SO4)3
19. How many milliliters of 0.750 M H3PO4 will contain the following?
(a) 0.15 mol H3PO4
(b) 35.5 g H3PO4
(c) 7.34 3 1022 molecules of H3PO4
20. How many milliliters of 0.250 M NH4Cl will contain the following?
(a) 0.85 mol NH4Cl
(b) 25.2 g NH4Cl
(c) 2.06 3 1020 formula units of NH4Cl
21. What will be the molarity of the resulting solutions made by mixing
the following? Assume that volumes are additive.
(a) 125 mL of 5.0 M H3PO4 with 775 mL of H2O
(b) 250 mL of 0.25 M Na2SO4 with 750 mL of H2O
(c) 75 mL of 0.50 M HNO3 with 75 mL of 1.5 M HNO3
22. What will be the molarity of the resulting solutions made by mixing the following? Assume that volumes are additive.
(a) 175 mL of 3.0 M H2SO4 with 275 mL of H2O
(b) 350 mL of 0.10 M CuSO4 with 150 mL of H2O
(c) 50.0 mL of 0.250 M HCl with 25.0 mL of 0.500 M HCl
23. Calculate the volume of concentrated reagent required to prepare
the diluted solutions indicated:
(a) 15 M H3PO4 to prepare 750 mL of 3.0 M H3PO4
(b) 16 M HNO3 to prepare 250 mL of 0.50 M HNO3
24. Calculate the volume of concentrated reagent required to prepare
the diluted solutions indicated:
(a) 18 M H2SO4 to prepare 225 mL of 2.0 M H2SO4
(b) 15 M NH3 to prepare 75 mL of 1.0 M NH3
25. Calculate the molarity of the solutions made by mixing 125 mL of
6.0 M HC2H3O2 with the following:
(a) 525 mL of H2O
(b) 175 mL of 1.5 M HC2H3O2
26. Calculate the molarity of the solutions made by mixing 175 mL of
3.0 M HCl with the following:
(a) 250 mL of H2O
(b) 115 mL of 6.0 M HCl
27. Use the equation to calculate the following:
28. Use the equation to calculate the following:
3 Ca(NO3)2(aq) + 2 Na3PO4(aq) ¡
Ca3(PO4)2(s) + 6 NaNO3(aq)
(a) the moles of Ca3(PO4)2 produced from 2.7 mol Na3PO4
(b) the moles of NaNO3 produced from 0.75 mol Ca(NO3)2
(c) the moles of Na3PO4 required to react with 1.45 L of 0.225 M
Ca(NO3)2
(d) the grams of Ca3(PO4)2 that can be obtained from 125 mL of
0.500 M Ca(NO3)2
(e) the volume of 0.25 M Na3PO4 needed to react with 15.0 mL of
0.50 M Ca(NO3)2
(f) the molarity (M) of the Ca(NO3)2 solution when 50.0 mL react
with 50.0 mL of 2.0 M Na3PO4
2 NaOH(aq) + H2SO4(aq) ¡ Na2SO4(aq) + 2 H2O(l)
(a) the moles of Na2SO4 produced from 3.6 mol H2SO4
(b) the moles of H2O produced from 0.025 mol NaOH
(c) the moles of NaOH required to react with 2.50 L of 0.125 M
H2SO4
(d) the grams of Na2SO4 that can be obtained from 25 mL of 0.050 M
NaOH
(e) the volume of 0.250 M H2SO4 needed to react with 25.5 mL of
0.750 M NaOH
(f) the molarity (M) of the NaOH solution when 48.20 mL react
with 35.72 mL of 0.125 M H2SO4
Paired Exercises 331
29. Use the equation to calculate the following:
30. Use the equation to calculate the following:
2 KMnO4(aq) + 16 HCl(aq) ¡
2 MnCl2(aq) + 5 Cl2(g) + 8 H2O(l) + 2 KCl(aq)
K2CO3(aq) + 2 HC2H3O2(aq) ¡
2 KC2H3O2(aq) + H2O(l) + CO2(g)
(a) the moles of H2O that can be obtained from 15.0 mL of 0.250 M
HCl
(b) the volume of 0.150 M KMnO4 needed to produce 1.85 mol
MnCl2
(c) the volume of 2.50 M HCl needed to produce 125 mL of 0.525 M
KCl
(d) the molarity (M) of the HCl solution when 22.20 mL react with
15.60 mL of 0.250 M KMnO4
(e) the liters of Cl2 gas at STP produced by the reaction of 125 mL
of 2.5 M HCl
(f) the liters of Cl2 gas at STP produced by the reaction of 15.0 mL
of 0.750 M HCl and 12.0 mL of 0.550 M KMnO4
(a) the moles of H2O that can be obtained from 25.0 mL of 0.150 M
HC2H3O2
(b) the volume of 0.210 M K2CO3 needed to produce 17.5 mol
KC2H3O2
(c) the volume of 1.25 M HC2H3O2 needed to react with 75.2 mL
0.750 M K2CO3
(d) the molarity (M) of the HC2H3O2 solution when 10.15 mL react
with 18.50 mL of 0.250 M K2CO3
(e) the liters of CO2 gas at STP produced by the reaction of 105 mL
of 1.5 M HC2H3O2
(f) the liters of CO2 gas at STP produced by the reaction of 25.0 mL
of 0.350 M K2CO3 and 25.0 mL of 0.250 M HC2H3O2
31. Calculate the molality of each of the following solutions:
(a) 2.0 mol HCl in 175 g water
(b) 14.5 g C12H22O11 in 550.0 g water
(c) 25.2 mL methanol, CH3OH (d = 0.791 g>mL) in 595 g ethanol,
CH3CH2OH
32. Calculate the molality of each of the following solutions:
(a) 1.25 mol CaCl2 in 750.0 g water
(b) 2.5 g C6H12O6 in 525 g water
(c) 17.5 mL 2-propanol, (CH3)2CHOH (d = 0.785 g>mL) in
35.5 mL H2O (d = 1.00 g>mL)
33. What is the (a) molality, (b) freezing point, and (c) boiling point
of a solution containing 2.68 g of naphthalene (C10H8) in 38.4 g of
benzene (C6H6)?
34. What is the (a) molality, (b) freezing point, and (c) boiling point of a
solution containing 100.0 g of ethylene glycol (C2H6O2) in 150.0 g
of water?
35. The freezing point of a solution of 8.00 g of an unknown compound
dissolved in 60.0 g of acetic acid is 13.2°C. Calculate the molar
mass of the compound.
36. What is the molar mass of a compound if 4.80 g of the compound dissolved in 22.0 g of H2O given a solution that freezes at
-2.50°C?
37. Identify which of the following substances are examples of true
solutions.
(a) jasmine tea
(b) chromium metal
(c) muddy water
(d) gasoline
38. Identify which of the following substances are examples of true
solutions.
(a) red paint
(b) Concord grape juice
(c) oil and vinegar salad dressing
(d) stainless steel
39. For those substances in Exercise 37 that are not true solutions,
explain why.
40. For those substances in Exercise 38 that are not true solutions,
explain why.
41. For each pairing below, predict which will dissolve faster and
explain why.
(a) a teaspoon of sugar or a large crystal of sugar dissolving in hot
coffee
(b) 15.0 grams of copper(II) sulfate in 100 mL of water or in 100 mL
of a 15% copper(II) sulfate solution
(c) a packet of artificial sweetener dissolving in a cup of iced tea or
a cup of hot tea
(d) 20 grams of silver nitrate in 1.5 L of water sitting on a table or
in 1.5 L of water sloshing around in a speeding car
42. For each pairing below, predict which will dissolve faster and
explain why.
(a) 0.424 g of the amino acid phenylalanine in 50.0 g of isopropyl
alcohol at 258C or in 50.0 g of isopropyl alcohol at 758C
(b) a 1.42-g crystal of sodium acetate in 300 g of 378C water or 1.42 g
of powdered sodium acetate in 300 g of 378C water
(c) a 500-g carton of table salt (NaCl) in 5.0 L of water at 358C
or a 500-g salt lick (NaCl) for livestock in 5.0 L of water at
218C
(d) 25 mg of acetaminophen in 100 mL of infant pain medication
containing 160 mg of acetaminophen per 15 mL or in 100 mL
of adult pain medication containing 320 mg of acetaminophen
per 15 mL
43. Law enforcement uses a quick and easy test for the presence of
the illicit drug PCP, reacting it with potassium iodide. The PCP
will form a crystalline solid with a long branching needle-like
structure with Kl. What is the molarity of Kl in a stock solution prepared by dissolving 396.1 g of Kl to a total volume
of 750.0 mL?
44. Drug enforcement agents have a variety of chemical tests they
can use to identify crystalline substances they may find at a crime
scene. Heroin can be detected by treating a sample with a solution
of mercury(II) chloride. The heroin will form a rosette of needleshaped crystals. What is the molarity of HgCl2 in a stock solution
prepared by dissolving 74.15 g of HgCl2 to a final volume of
250.0 mL?
332 chapter 14 • Solutions
Additional Exercises
45. What happens to salt (NaCl) crystals when they are dissolved in
water?
46. What happens to sugar molecules (C12H22O11) when they are dissolved in water?
47. Why do sugar and salt behave differently when dissolved in
water?
48. Why don’t blood cells shrink or swell in an isotonic sodium chloride solution (0.9% saline)?
49. In the picture of dissolving KMnO4 found in Section 14.1, the compound is forming purple streaks as it dissolves. Why?
50. In Figure 14.4, observe the line for KNO3 . Explain why it slopes
up from left to right. How does the slope compare to the slopes of
the other substances? What does this mean?
51. An IV bag contains 9.0 g of sodium chloride per liter of solution.
What is the molarity of sodium chloride in this solution?
52. How many grams of solution, 10.0% NaOH by mass, are required
to neutralize 250.0 g of a 1.0 m solution of HCl?
53. A sugar syrup solution contains 15.0% sugar, C12H22O11 , by mass
and has a density of 1.06 g/mL.
(a) How many grams of sugar are in 1.0 L of this syrup?
(b) What is the molarity of this solution?
(c) What is the molality of this solution?
54. A solution of 3.84 g C4H2N (empirical formula) in 250.0 g of benzene depresses the freezing point of benzene 0.614°C. What is the
molecular formula for the compound?
55. Hydrochloric acid (HCl) is sold as a concentrated aqueous solution
(12.0 mol/L). If the density of the solution is 1.18 g/mL, determine
the molality of the solution.
56. How many grams of KNO3 are needed to make 450 mL of a solution that is to contain 5.5 mg/mL of potassium ion? Calculate the
molarity of the solution.
57. Witch hazel solution, an astringent for skin, contains 14% ethyl
alcohol, C2H5OH, by volume. How many mL of ethyl alcohol are
contained in a 16 fluid ounce bottle of witch hazel?
58. Given a solution containing 16.10 g C2H6O2 in 82.0 g H2O that has
a boiling point of 101.62°C, verify that the boiling point elevation
constant Kf for water is 0.512°C kg H2O>mole solute.
59. Physiological saline (NaCl) solutions used in intravenous injections have a concentration of 0.90% NaCl (mass/volume).
(a) How many grams of NaCl are needed to prepare 500.0 mL of
this solution?
(b) How much water must evaporate from this solution to give a
solution that is 9.0% NaCl (mass/volume)?
60. A solution is made from 50.0 g KNO3 and 175 g H2O. How many
grams of water must evaporate to give a saturated solution of KNO3
in water at 20°C? (See Figure 14.4.)
61. How many liters of a 0.25% (v/v) oil of wintergreen solution can
you prepare if you have only 7.35 mL of oil of wintergreen on
hand?
62. At 20°C, an aqueous solution of HNO3 that is 35.0% HNO3 by
mass has a density of 1.21 g/mL.
(a) How many grams of HNO3 are present in 1.00 L of this solution?
(b) What volume of this solution will contain 500. g HNO3 ?
63. What is the molarity of a phosphoric acid solution if the solution is
85% by mass H3PO4 and has a density of 1.7 g/mL?
64. To what volume must a solution of 80.0 g H2SO4 in 500.0 mL of
solution be diluted to give a 0.10 M solution?
65. How many pounds of glycerol, C3H8O3, are present in 30.0 gallons
of a 4.28 M solution?
66. (a) How many moles of hydrogen will be liberated from 200.0 mL
of 3.00 M HCl reacting with an excess of magnesium? The
equation is
Mg(s) + 2 HCl(aq) ¡ MgCl2(aq) + H2(g)
(b) How many liters of hydrogen gas (H2) measured at 27°C and
720 torr will be obtained?
(Hint: Use the ideal gas law.)
67. Which will be more effective in neutralizing stomach acid, HCl:
a tablet containing 1.20 g Mg(OH)2 or a tablet containing 1.00 g
Al(OH)3 ? Show evidence for your answer.
68. Which would be more effective as an antifreeze in an automobile
radiator? A solution containing
(a) 10 kg of methyl alcohol (CH3OH) or 10 kg of ethyl alcohol
(C2H5OH)?
(b) 10 m solution of methyl alcohol or 10 m solution of ethyl alcohol?
69. Automobile battery acid is 38% H2SO4 and has a density of 1.29 g/mL.
Calculate the molality and the molarity of this solution.
70. What is the (a) molality and (b) boiling point of an aqueous sugar,
C12H22O11 , solution that freezes at -5.4°C?
71. A solution of 6.20 g C2H6O2 in water has a freezing point of
-0.372°C. How many grams of H2O are in the solution?
72. What (a) mass and (b) volume of ethylene glycol (C2H6O2 ,
density = 1.11 g>mL) should be added to 12.0 L of water in an
automobile radiator to protect it from freezing at - 20°C? (c) To
what temperature Fahrenheit will the radiator be protected?
73. If 150 mL of 0.055 M HNO3 are needed to completely neutralize
1.48 g of an impure sample of sodium hydrogen carbonate (baking
soda), what percent of the sample is baking soda?
74. (a) How much water must be added to concentrated sulfuric acid
(H2SO4) (17.8 M) to prepare 8.4 L of 1.5 M sulfuric acid solution?
(b) How many moles of H2SO4 are in each milliliter of the original
concentrate?
(c) How many moles are in each milliliter of the diluted solution?
75. How would you prepare a 6.00 M HNO3 solution if only 3.00 M
and 12.0 M solutions of the acid are available for mixing?
76. A 20.0-mL portion of an HBr solution of unknown strength is diluted
to exactly 240 mL. If 100.0 mL of this diluted solution requires
88.4 mL of 0.37 M NaOH to achieve complete neutralization, what
was the strength of the original HBr solution?
77. When 80.5 mL of 0.642 M Ba(NO3)2 are mixed with 44.5 mL of
0.743 M KOH, a precipitate of Ba(OH)2 forms. How many grams
of Ba(OH)2 do you expect?
78. A 0.25 M solution of lithium carbonate (Li2CO3), a drug used to
treat manic depression, is prepared.
(a) How many moles of Li2CO3 are present in 45.8 mL of the
solution?
(b) How many grams of Li2CO3 are in 750 mL of the same solution?
(c) How many milliliters of the solution would be needed to supply
6.0 g of the solute?
(d) If the solution has a density of 1.22 g/mL, what is its mass percent?
79. Forensic chemists use Super Glue® to help them see fingerprints left
behind at crime scenes. If cyanoacrylate, the main component of Super
Glue®, is vaporized, it will be attracted to the oils left behind after someone touches an object. Forensic investigators will then rinse the print
with a solution of europium chloride hexahydrate, EuCl3 # 6 H2O, which
fluoresces when exposed to ultraviolet light. If 0.625 g of europium
chloride hexahydrate is dissolved in enough water to make 500.0 mL
of solution, what is the molarity of the resulting solution?
Answers to Practice Exercises 333
80. The pictures below represent solutions of sodium phosphate, potassium chloride, and sucrose dissolved in water. Identify each representation
below:
(a)
(b)
(c)
Represents H2O molecules
81. Dark streaks have been observed on the surface of Mars. These
streaks sometimes seem to run or flow, and other times they appear
static. It has been hypothesized that these streaks are salt (sodium
chloride) water rivers that are sometimes melted, allowing them
to flow, and at other times these rivers are frozen. If the salt water
mixtures melt when the temperature rises to 212.78C, what is the
molality of salt in the rivers? (Hint: The freezing point is determined by the molality of ions. How will the molality of sodium
chloride be related to the molality of the ions?)
82. Snails are very sensitive to the concentration of salt in their environment. Because their “skin” is a semipermeable membrane,
water can easily flow through this membrane. If you spill salt onto
the “skin” of a snail, water from the inside of the body will flow out
very quickly. Explain this phenomenon in terms of the chemistry
described in this chapter.
83. Eggplant is often considered to be very bitter. One of the secrets chefs use to remove the bitterness is to coat the surface
of the eggplant with salt. As the eggplant sits, it becomes wet
and when the wetness is rinsed or wiped away the bitterness is
gone. Explain what is happening to improve the flavor of the
eggplant.
84. To make taffy a sugar water solution is heated until the boiling
point is 1278C. What is the concentration (m) of sugar in taffy?
Challenge Exercises
85. When solutions of hydrochloric acid and sodium sulfite react, a
salt, water, and sulfur dioxide gas are produced. How many liters
of sulfur dioxide gas at 775 torr and 22°C can be produced when
125 mL of 2.50 M hydrochloric acid react with 75.0 mL of 1.75 M
sodium sulfite?
86. Consider a saturated solution at 20°C made from 5.549 moles of
water and an unknown solute. You determine the mass of the container containing the solution to be 563 g. The mass of the empty
container is 375 g. Identify the solute.
Answers to Practice Exercises
14.1 (a) solute, oxygen, CO2 other gases; solvent, nitrogen (b) solute, salt; solvent, water (c) solute CO2; solvent, water
14.7 0.84 L (840 mL)
14.2 unsaturated
14.8 75 mL NaOH solution
14.3 The rate of dissolving increases as the temperature is increased
because the molecules of solvent are moving faster and strike
the surfaces of the solute more frequently.
14.9 5.00 * 10-2 M
14.10 43 g
14.4 10.0% Na2SO4 solution
14.11 1.387 m
14.5 2.02 M
14.12 freezing point = - 2.58°C,
boiling point = 100.71°C
14.6 0.27 g NaCl
PUTTING IT TOGETHER:
Review for Chapters 12–14
Answers for Putting It Together Reviews are found in Appendix VII.
Multiple Choice
Choose the correct answer to each of the following.
1. Which of these statements is not one of the principal assumptions
of the kinetic-molecular theory for an ideal gas?
(a) All collisions of gaseous molecules are perfectly elastic.
(b) A mole of any gas occupies 22.4 L at STP.
(c) Gas molecules have no attraction for one another.
(d) The average kinetic energy for molecules is the same for all
gases at the same temperature.
2. Which of the following is not equal to 1.00 atm?
(a) 760. cm Hg
(c) 760. mm Hg
(b) 29.9 in. Hg
(d) 760. torr
3. If the pressure on 45 mL of gas is changed from 600. torr to
800. torr, the new volume will be
(a) 60 mL
(b) 34 mL (c) 0.045 L
(d) 22.4 L
4. The volume of a gas is 300. mL at 740. torr and 25°C. If the pressure remains constant and the temperature is raised to 100.°C, the
new volume will be
(a) 240. mL (b) 1.20 L (c) 376 mL (d) 75.0 mL
5. The volume of a dry gas is 4.00 L at 15.0°C and 745 torr. What
volume will the gas occupy at 40.0°C and 700. torr?
(a) 4.63 L
(b) 3.46 L (c) 3.92 L
(d) 4.08 L
6. A sample of Cl2 occupies 8.50 L at 80.0°C and 740. mm Hg. What
volume will the Cl2 occupy at STP?
(a) 10.7 L
(b) 6.75 L (c) 11.3 L
(d) 6.40 L
7. What volume will 8.00 g O2 occupy at 45°C and 2.00 atm?
(a) 0.462 L (b) 104 L
(c) 9.62 L
(d) 3.26 L
8. The density of NH3 gas at STP is
(a) 0.760 g/mL
(c) 1.32 g/mL
(b) 0.760 g/L
(d) 1.32 g/L
9. Which temperature is equivalent to 22°C?
(a) 295 K
(b) 251 K
(c) 191 K
(d) 234°F
10. Measured at 65°C and 500. torr, the mass of 3.21 L of a gas is 3.5 g.
The molar mass of this gas is
(a) 21 g/mole
(c) 24 g/mole
(b) 46 g/mole
(d) 130 g/mole
11. Box A contains O2 (molar mass = 32.0) at a pressure of 200 torr.
Box B, which is identical to box A in volume, contains twice as
many molecules of CH4 (molar mass = 16.0) as the molecules of
O2 in box A. The temperatures of the gases are identical. The pressure in box B is
(a) 100 torr
(c) 400 torr
(b) 200 torr
(d) 800 torr
12. A 300.-mL sample of oxygen (O2) is collected over water at 23°C
and 725 torr. If the vapor pressure of water at 23°C is 21.0 torr, the
volume of dry O2 at STP is
(a) 256 mL (b) 351 mL (c) 341 mL (d) 264 mL
334
13. A tank containing 0.01 mol of neon and 0.04 mol of helium shows
a pressure of 1 atm. What is the partial pressure of neon in the
tank?
(a) 0.8 atm (b) 0.01 atm (c) 0.2 atm
(d) 0.5 atm
14. How many liters of NO2 (at STP) can be produced from 25.0 g Cu
reacting with concentrated nitric acid?
Cu(s) + 4 HNO3(aq) ¡ Cu(NO3)2(aq) + 2 H2O(l) + 2 NO2(g)
(a) 4.41 L
(b) 8.82 L
(c) 17.6 L
(d) 44.8 L
15. How many liters of butane vapor are required to produce 2.0 L CO2
at STP?
2 C4H10(g) + 13 O2(g) ¡ 8 CO2(g) + 10 H2O(g)
butane
(a) 2.0 L
(b) 4.0 L
(c) 0.80 L
(d) 0.50 L
16. What volume of CO2 (at STP) can be produced when 15.0 g C2H6
and 50.0 g O2 are reacted?
2 C2H6(g) + 7 O2(g) ¡ 4 CO2(g) + 6 H2O(g)
(a) 20.0 L
(b) 22.4 L
(c) 35.0 L
(d) 5.6 L
17. Which of these gases has the highest density at STP?
(b) NO2
(c) Cl2
(d) SO2
(a) N2O
18. What is the density of CO2 at 25°C and 0.954 atm?
(a) 1.72 g/L
(c) 0.985 g/L
(b) 2.04 g/L
(d) 1.52 g/L
19. How many molecules are present in 0.025 mol of H2 gas?
(a) 1.5 * 1022 molecules (c) 2.40 * 1025 molecules
(b) 3.37 * 1023 molecules (d) 1.50 * 1022 molecules
20. 5.60 L of a gas at STP have a mass of 13.0 g. What is the molar
mass of the gas?
(a) 33.2 g/mol
(c) 66.4 g/mol
(b) 26.0 g/mol
(d) 52.0 g/mol
21. The heat of fusion of ice at 0°C is
(a) 4.184 J/g
(c) 2.26 kJ/g
(b) 335 J/g
(d) 2.26 kJ/mol
22. The heat of vaporization of water is
(a) 4.184 J/g
(c) 2.26 kJ/g
(b) 335 J/g
(d) 2.26 kJ/mol
23. The specific heat of water is
(a) 4.184 J/g°C
(c) 2.26 kJ/g°C
(b) 335 J/g°C
(d) 18 J/g°C
24. The density of water at 4°C is
(a) 1.0 g/mL
(c) 18.0 g/mL
(b) 80 g/mL
(d) 14.7 lb>in.3
25. SO2 can be properly classified as a(n)
(a) basic anhydride
(c) anhydrous salt
(b) hydrate
(d) acid anhydride
Putting It Together 335
26. When compared to H2S, H2Se, and H2Te, water is found to have
the highest boiling point because it
(a) has the lowest molar mass
(b) is the smallest molecule
(c) has the highest bonding
(d) forms hydrogen bonds better than the others
27. In which of the following molecules will hydrogen bonding be
important?
(a) H F
(c) H Br
H
H
(b) S
(d) H
H
H
C
H
O
C
H
H
28. Which of the following is an incorrect equation?
(a) H2SO4 + 2 NaOH ¡ Na2SO4 + 2 H2O
(b) C2H6 + O2 ¡ 2 CO2 + 3 H2
electrolysis
(c) 2 H2O ::::" 2 H2 + O2
H2SO4
(d) Ca + 2 H2O ¡ H2 + Ca(OH)2
29. Which of the following is not an example of an intermolecular
force?
(a) London dispersion forces
(b) hydrogen bonds
(c) covalent bonds
(d) dipole–dipole attractions
30. How many kilojoules are required to change 85 g of water at 25°C
to steam at 100.°C?
(a) 219 kJ
(b) 27 kJ
(c) 590 kJ
(d) 192 kJ
31. A chunk of 0°C ice, mass 145 g, is dropped into 75 g of water at
62°C. The heat of fusion of ice is 335 J/g. The result, after thermal
equilibrium is attained, will be
(a) 87 g ice and 133 g liquid water, all at 0°C
(b) 58 g ice and 162 g liquid water, all at 0°C
(c) 220 g water at 7°C
(d) 220 g water at 17°C
32. The formula for iron(II) sulfate heptahydrate is
(a) Fe2SO4 # 7 H2O
(c) FeSO4 # 7 H2O
#
(b) Fe(SO4)2 6 H2O
(d) Fe2(SO4)3 # 7 H2O
33. The process by which a solid changes directly to a vapor is called
(a) vaporization
(c) sublimation
(b) evaporation
(d) condensation
34. Hydrogen bonding
(a) occurs only between water molecules
(b) is stronger than covalent bonding
(c) can occur between NH3 and H2O
(d) results from strong attractive forces in ionic compounds
35. A liquid boils when
(a) the vapor pressure of the liquid equals the external pressure
above the liquid
(b) the heat of vaporization exceeds the vapor pressure
(c) the vapor pressure equals 1 atm
(d) the normal boiling temperature is reached
36. Consider two beakers, one containing 50 mL of liquid A and the
other 50 mL of liquid B. The boiling point of A is 90°C and that of
B is 72°C. Which of these statements is correct?
(a) A will evaporate faster than B.
(b) B will evaporate faster than A.
(c) Both A and B evaporate at the same rate.
(d) Insufficient data to answer the question.
37. 95.0 g of 0.0°C ice are added to exactly 100. g of water at 60.0°C.
When the temperature of the mixture first reaches 0.0°C, the mass
of ice still present is
(a) 0.0 g (b) 20.0 g (c) 10.0 g (d) 75.0 g
38. Which of the following is not a general property of solutions?
(a) a homogeneous mixture of two or more substances
(b) variable composition
(c) dissolved solute breaks down to individual molecules
(d) the same chemical composition, the same chemical properties,
and the same physical properties in every part
39. If NaCl is soluble in water to the extent of 36.0 g NaCl>100 g H2O
at 20°C, then a solution at 20°C containing 45 g NaCl>150 g H2O
would be
(a) dilute
(c) supersaturated
(b) saturated
(d) unsaturated
40. If 5.00 g NaCl are dissolved in 25.0 g of water, the percent of NaCl
by mass is
(a) 16.7%
(c) 0.20%
(b) 20.0%
(d) no correct answer given
41. How many grams of 9.0% AgNO3 solution will contain 5.3 g
AgNO3 ?
(a) 47.7 g
(c) 59 g
(b) 0.58 g
(d) no correct answer given
42. The molarity of a solution containing 2.5 mol of acetic acid
(HC2H3O2) in 400. mL of solution is
(a) 0.063 M (b) 1.0 M
(c) 0.103 M (d) 6.3 M
43. What volume of 0.300 M KCl will contain 15.3 g KCl?
(a) 1.46 L
(b) 683 mL (c) 61.5 mL (d) 4.60 L
44. What mass of BaCl2 will be required to prepare 200. mL of 0.150
M solution?
(a) 0.750 g (b) 156 g (c) 6.25 g (d) 31.2 g
Problems 45–47 relate to the reaction
CaCO3 + 2 HCl ¡ CaCl2 + H2O + CO2
45. What volume of 6.0 M HCl will be needed to react with 0.350 mol
of CaCO3 ?
(a) 42.0 mL (b) 1.17 L (c) 117 mL (d) 583 mL
46. If 400. mL of 2.0 M HCl react with excess CaCO3 , the volume of
CO2 produced, measured at STP, is
(a) 18 L
(b) 5.6 L
(c) 9.0 L
(d) 56 L
47. If 5.3 g CaCl2 are produced in the reaction, what is the molarity of
the HCl used if 25 mL of it reacted with excess CaCO3 ?
(a) 3.8 M
(b) 0.19 M (c) 0.38 M
(d) 0.42 M
48. If 20.0 g of the nonelectrolyte urea (CO(NH2)2) is dissolved in 25.0 g
of water, the freezing point of the solution will be
(a) -2.47°C (b) -1.40°C (c) -24.7°C (d) - 3.72°C
49. When 256 g of a nonvolatile, nonelectrolyte unknown were dissolved in 500. g H2O, the freezing point was found to be - 2.79°C.
The molar mass of the unknown solute is
(a) 357
(b) 62.0
(c) 768
(d) 341
50. How many milliliters of 6.0 M H2SO4 must you use to prepare 500. mL
of 0.20 M sulfuric acid solution?
(a) 30
(b) 17
(c) 12
(d) 100
51. How many milliliters of water must be added to 200. mL of 1.40 M
HCl to make a solution that is 0.500 M HCl?
(a) 360. mL (b) 560. mL (c) 140. mL (d) 280. mL
336 Putting It Together
52. Which procedure is most likely to increase the solubility of most
solids in liquids?
(a) stirring
(c) heating the solution
(b) pulverizing the solid
(d) increasing the pressure
53. The addition of a crystal of NaClO3 to a solution of NaClO3 causes
additional crystals to precipitate. The original solution was
(a) unsaturated
(c) saturated
(b) dilute
(d) supersaturated
54. Which of these anions will not form a precipitate with silver ions,
Ag+?
(a) Cl(b) NO3(c) Br(d) CO23
55. Which of these salts are considered to be soluble in water?
(a) BaSO4 (b) NH4Cl (c) AgI
(d) PbS
56. A solution of ethyl alcohol and benzene is 40% alcohol by volume.
Which statement is correct?
(a) The solution contains 40 mL of alcohol in 100 mL of solution.
(b) The solution contains 60 mL of benzene in 100 mL of solution.
(c) The solution contains 40 mL of alcohol in 100 g of solution.
(d) The solution is made by dissolving 40 mL of alcohol in 60 mL
of benzene.
57. Which of the following is not a colligative property?
(a) boiling point elevation
(c) osmotic pressure
(b) freezing point depression (d) surface tension
58. When a solute is dissolved in a solvent
(a) the freezing point of the solution increases
(b) the vapor pressure of the solution increases
(c) the boiling point of the solution increases
(d) the concentration of the solvent increases
59. Which of the following solutions will have the lowest freezing point
where X is any element or nonelectrolytic compound?
(a) 1.0 mol X in 1 kg H2O (c) 1.2 mol X in 1 kg H2O
(b) 2.0 mol X in 1 kg H2O (d) 0.80 mol X in 1 kg H2O
60. In the process of osmosis, water passes through a semipermeable
membrane
(a) from a more concentrated solution to a dilute solution
(b) from a dilute solution to a more concentrated solution
(c) in order to remove a solute from a solution
(d) so that a sugar solution can become sweeter
Free Response Questions
Answer each of the following. Be sure to include your work and explanations in a clear, logical form.
1. Which solution should have a higher boiling point: 215 mL of a
10.0% (m/v) aqueous KCl solution or 224 mL of a 1.10 M aqueous
NaCl solution?
2. A glass containing 345 mL of a soft drink (a carbonated beverage)
was left sitting out on a kitchen counter. If the CO2 released at
room temperature (25°C) and pressure (1 atm) occupies 1.40 L, at
a minimum, what is the concentration (in ppm) of the CO2 in the
original soft drink? (Assume the density of the original soft drink
is 0.965 g/mL.)
3. Dina and Murphy were trying to react 100. mL of a 0.10 M HCl
solution with KOH. The procedure called for a 10% KOH solution.
Dina made a 10% mass/volume solution, while Murphy made a 10%
by mass solution. (Assume there is no volume change upon dissolving KOH.) Which solution required less volume to fully react with
100. mL of the HCl solution?
4. A flask containing 825 mL of a solution containing 0.355 mol of
CuSO4 was left open overnight. The next morning the flask only
contained 755 mL of solution.
(a) What is the concentration (molarity) of the CuSO4 solution remaining in the flask?
(b) Which of the pathways shown below best represents the evaporation of water, and why are the others wrong?
H
H
H
H
O
O
H
H
H
O
(1)
(2)
�
�
H (g) + OH (g)
O
H
O(g)
(3)
O
H
H(g) + O
H
H(g)
+
O
H
H(g)
5. Three students at Jamston High—Zack, Gaye, and Lamont—each
had the opportunity to travel over spring break. As part of a project,
each of them measured the boiling point of pure water at their vacation spot. Zack found a boiling point of 93.9°C, Gaye measured
101.1°C, and Lamont read 100.°C. Which student most likely went
snow skiing near Ely, Nevada, and which student most likely went
water skiing in Honolulu? From the boiling point information, what
can you surmise about the Dead Sea region, the location of the third
student’s vacation? Explain.
6. Why does a change in pressure of a gas significantly affect its volume, whereas a change in pressure on a solid or liquid has negligible
effect on their respective volumes? If the accompanying picture represents a liquid at the molecular level, draw what you might expect
a solid and a gas to look like.
7. (a) If you filled up three balloons with equal volumes of hydrogen,
argon, and carbon dioxide gas, all at the same temperature and
pressure, which balloon would weigh the most? the least? Explain.
(b) If you filled up three balloons with equal masses of nitrogen,
oxygen, and neon, all to the same volume at the same temperature, which would have the lowest pressure?
8. Ray ran a double-displacement reaction using 0.050 mol CuCl2 and
0.10 mol AgNO3 . The resulting white precipitate was removed by
filtration. The filtrate was accidentally left open on the lab bench
for over a week, and when Ray returned the flask contained solid,
blue crystals. Ray weighed the crystals and found they had a mass of
14.775 g. Was Ray expecting this number? If not, what did he expect?
9. Why is it often advantageous or even necessary to run reactions in
solution rather than mixing two solids? Would you expect reactions
run in the gas phase to be more similar to solutions or to solids?
Why?
10. A solution of 5.36 g of a molecular compound dissolved in 76.8 g
benzene (C6H6) has a boiling point of 81.48°C. What is the molar
mass of the compound?
Boiling point for benzene = 80.1°C
Kb for benzene = 2.53°C kg solvent/mol solute
C h a p te r
15
Acids, Bases, and Salts
A
cids are important chemicals. They are used in cooking
to produce the surprise of tartness (from lemons) and
to release CO2 bubbles from leavening agents in baking.
Vitamin C is an acid that is an essential nutrient in our diet. Our
stomachs release acid to aid in digestion. Excess stomach acid
can produce heartburn and indigestion. Bacteria in our mouths
produce acids that can dissolve tooth enamel to form cavities.
In our recreational activities we are concerned about acidity
levels in swimming pools and spas. Acids are essential in the
manufacture of detergents, plastics, and storage batteries. The
acid–base properties of substances are found in all areas of our
lives. In this chapter we consider the properties of acids, bases,
and salts.
Chapter Outline
15.1 Acids and Bases
15.2 Reactions of Acids and Bases
15.3 Salts
15.4 Electrolytes and Nonelectrolytes
15.5 Introduction to pH
15.6 Neutralization
15.7 Writing Net Ionic Equations
15.8 Acid Rain
Jan Rihak/iStockphoto
Lemons and limes are examples
of food that contains acidic
solutions.
338 chapter 15
• Acids, Bases, and Salts
15.1 Acids and Bases
L earning objective
Compare the definitions of acids and bases, including Arrhenius, Brønsted–Lowry,
and Lewis acids and bases.
Key term
The word acid is derived from the Latin acidus, meaning “sour” or “tart,” and is also related to
the Latin word acetum, meaning “vinegar.” Vinegar has been around since antiquity as a product of the fermentation of wine and apple cider. The sour constituent of vinegar is acetic acid
(HC2H3O2). Characteristic properties commonly associated with acids include the following:
hydronium ion
1. sour taste
2. the ability to change the color of litmus, a vegetable dye, from blue to red
3. the ability to react with
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• metals such as zinc and magnesium to produce hydrogen gas
• hydroxide bases to produce water and an ionic compound (salt)
• carbonates to produce carbon dioxide
These properties are due to the hydrogen ions, H+, released by acids in a water solution.
Classically, a base is a substance capable of liberating hydroxide ions, OH- , in water solution. Hydroxides of the alkali metals (Group 1A) and alkaline earth metals (Group 2A), such
as LiOH , NaOH , KOH , Mg(OH)2 , Ca(OH)2 , and Ba