Noble
gases
Alkaline
1 earth metals
Halogens 18
1A
8A
metals
1
H
1.008
Alkali metals
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Periodic Table of the Elements
nonmetals
2
13
14
15
16
17
2A
3A
4A
5A
6A
7A
He
4.003
3
4
5
6
7
8
9
10
Li
6.941
Be
9.012
B
10.81
C
12.01
N
14.01
O
16.00
F
19.00
Ne
20.18
13
14
15
16
17
18
Al
26.98
Si
28.09
P
30.97
S
32.07
Cl
35.45
Ar
39.95
11
12
Na
22.99
Mg
24.31
3
4
5
6
7
8
Transition metals
19
20
21
22
23
24
25
26
K
39.10
Ca
40.08
Sc
44.96
Ti
47.88
V
50.94
Cr
52.00
Mn
54.94
Fe
55.85
37
38
39
40
41
42
43
44
Rb
85.47
Sr
87.62
Y
88.91
Zr
91.22
Nb
92.91
Mo
95.94
Tc
(98)
Ru
101.1
55
56
57
72
73
74
75
76
Cs
132.9
Ba
137.3
La*
138.9
Hf
178.5
Ta
180.9
W
183.9
Re
186.2
Os
190.2
9
10
27
28
Co
Ni
58.93 58.69
45
46
Rh
Pd
102.9 106.4
77
78
Ir
Pt
192.2 195.1
11
12
29
30
31
32
33
34
35
36
Cu
63.55
Zn
65.38
Ga
69.72
Ge
72.59
As
74.92
Se
78.96
Br
79.90
Kr
83.80
47
48
49
50
51
52
53
54
Ag
107.9
Cd
112.4
In
114.8
Sn
118.7
Sb
121.8
Te
127.6
I
126.9
Xe
131.3
79
80
81
82
83
84
85
86
Au
197.0
Hg
200.6
Tl
204.4
Pb
207.2
Bi
209.0
Po
(209)
At
(210)
Rn
(222)
87
88
89
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
Fr
(223)
Ra
226
Ac†
(227)
Rf
(261)
Db
(262)
Sg
(263)
Bh
(264)
Hs
(265)
Mt
(268)
Ds
(271)
Rg
(272)
Cn
(285)
Uut
Fl
(289)
Uup
Lv
(293)
Uus
Uuo
62
63
*Lanthanides
†
Actinides
58
59
60
61
Ce
140.1
Pr
140.9
Nd
144.2
Pm
(145)
90
91
92
93
94
95
96
97
98
99
100
Th
232.0
Pa
(231)
U
238.0
Np
(237)
Pu
(244)
Am
(243)
Cm
(247)
Bk
(247)
Cf
(251)
Es
(252)
Fm
(257)
Sm
Eu
150.4 152.0
64
65
66
67
68
69
70
71
Gd
157.3
Tb
158.9
Dy
162.5
Ho
164.9
Er
167.3
Tm
168.9
Yb
173.0
Lu
175.0
101
102
103
Md
(258)
No
(259)
Lr
(260)
Group numbers 1–18 represent the system recommended by the International Union
of Pure and Applied Chemistry.
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Table of Atomic Masses*
Element
Actinium
Aluminum
Americium
Antimony
Argon
Arsenic
Astatine
Barium
Berkelium
Beryllium
Bismuth
Bohrium
Boron
Bromine
Cadmium
Calcium
Californium
Carbon
Cerium
Cesium
Chlorine
Chromium
Cobalt
Copernicium
Copper
Curium
Darmstadtium
Dubnium
Dysprosium
Einsteinium
Erbium
Europium
Fermium
Flerovium
Fluorine
Francium
Gadolinium
Gallium
Symbol
Atomic
Number
Atomic
Mass
Element
Ac
Al
Am
Sb
Ar
As
At
Ba
Bk
Be
Bi
Bh
B
Br
Cd
Ca
Cf
C
Ce
Cs
Cl
Cr
Co
CN
Cu
Cm
Ds
Db
Dy
Es
Er
Eu
Fm
Fl
F
Fr
Gd
Ga
89
13
95
51
18
33
85
56
97
4
83
107
5
35
48
20
98
6
58
55
17
24
27
112
29
96
110
105
66
99
68
63
100
114
9
87
64
31
[227]§
26.98
[243]
121.8
39.95
74.92
[210]
137.3
[247]
9.012
209.0
[264]
10.81
79.90
112.4
40.08
[251]
12.01
140.1
132.90
35.45
52.00
58.93
[285]
63.55
[247]
[271]
[262]
162.5
[252]
167.3
152.0
[257]
[289]
19.00
[223]
157.3
69.72
Germanium
Gold
Hafnium
Hassium
Helium
Holmium
Hydrogen
Indium
Iodine
Iridium
Iron
Krypton
Lanthanum
Lawrencium
Lead
Livermorium
Lithium
Lutetium
Magnesium
Manganese
Meitnerium
Mendelevium
Mercury
Molybdenum
Neodymium
Neon
Neptunium
Nickel
Niobium
Nitrogen
Nobelium
Osmium
Oxygen
Palladium
Phosphorus
Platinum
Plutonium
Polonium
Symbol
Atomic
Number
Atomic
Mass
Element
Ge
Au
Hf
Hs
He
Ho
H
In
I
Ir
Fe
Kr
La
Lr
Pb
Lv
Li
Lu
Mg
Mn
Mt
Md
Hg
Mo
Nd
Ne
Np
Ni
Nb
N
No
Os
O
Pd
P
Pt
Pu
Po
32
79
72
108
2
67
1
49
53
77
26
36
57
103
82
116
3
71
12
25
109
101
80
42
60
10
93
28
41
7
102
76
8
46
15
78
94
84
72.59
197.0
178.5
[265]
4.003
164.9
1.008
114.8
126.9
192.2
55.85
83.80
138.9
[260]
207.2
[293]
6.9419
175.0
24.31
54.94
[268]
[258]
200.6
95.94
144.2
20.18
[237]
58.69
92.91
14.01
[259]
190.2
16.00
106.4
30.97
195.1
[244]
[209]
Potassium
Praseodymium
Promethium
Protactinium
Radium
Radon
Rhenium
Rhodium
Roentgenium
Rubidium
Ruthenium
Rutherfordium
Samarium
Scandium
Seaborgium
Selenium
Silicon
Silver
Sodium
Strontium
Sulfur
Tantalum
Technetium
Tellurium
Terbium
Thallium
Thorium
Thulium
Tin
Titanium
Tungsten
Uranium
Vanadium
Xenon
Ytterbium
Yttrium
Zinc
Zirconium
*The values given here are to four significant figures where possible. §A value given in parentheses denotes the mass of the longest-lived isotope.
Symbol
Atomic
Number
Atomic
Mass
K
Pr
Pm
Pa
Ra
Rn
Re
Rh
Rg
Rb
Ru
Rf
Sm
Sc
Sg
Se
Si
Ag
Na
Sr
S
Ta
Tc
Te
Tb
Tl
Th
Tm
Sn
Ti
W
U
V
Xe
Yb
Y
Zn
Zr
19
59
61
91
88
86
75
45
111
37
44
104
62
21
106
34
14
47
11
38
16
73
43
52
65
81
90
69
50
22
74
92
23
54
70
39
30
40
39.10
140.9
[145]
[231]
226
[222]
186.2
102.9
[272]
85.47
101.1
[261]
150.4
44.96
[263]
78.96
28.09
107.9
22.99
87.62
32.07
180.9
[98]
127.6
158.9
204.4
232.0
168.9
118.7
47.88
183.9
238.0
50.94
131.3
173.0
88.91
65.38
91.22
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Ninth Edition
Steven S. Zumdahl
University of Illinois
Susan A. Zumdahl
University of Illinois
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
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Photo by Dr. Eric Heller
Chemistry
Chemistry, Ninth Edition
Steven S. Zumdahl and Susan A. Zumdahl
Publisher: Mary Finch
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Contents
Chapter 3 Stoichiometry
To the Professor ix
To the Student
xiii
Chapter 1 Chemical Foundations
1.1
1.2
1
Chemistry: An Overview 3
The Scientific Method 5
CHEMICAL CONNECTIONS A Note-able Achievement
1.3
Units of Measurement 8
CHEMICAL CONNECTIONS Critical Units!
1.4
1.5
1.6
1.7
1.8
1.9
1.10
7
9
Uncertainty in Measurement 11
Significant Figures and Calculations 14
Learning to Solve Problems Systematically 18
Dimensional Analysis 18
Temperature 22
Density 26
Classification of Matter 27
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
81
Counting by Weighing 82
Atomic Masses 83
The Mole 85
Molar Mass 90
Learning to Solve Problems 93
Percent Composition of Compounds 94
Determining the Formula of a Compound 96
Chemical Equations 103
Balancing Chemical Equations 105
Stoichiometric Calculations: Amounts of Reactants
and Products 108
CHEMICAL CONNECTIONS High Mountains—Low
Octane 109
3.11 The Concept of Limiting Reactant 114
For Review 124 ∣ Key Terms 124 ∣ Questions and
Exercises 126
For Review 31 ∣ Key Terms 31 ∣ Questions and Exercises 33
Chapter 2 Atoms, Molecules, and Ions
2.1
2.2
2.3
42
The Early History of Chemistry 43
Fundamental Chemical Laws 44
Dalton’s Atomic Theory 47
CHEMICAL CONNECTIONS Berzelius, Selenium, and
Silicon
2.6
2.7
Early Experiments to Characterize the Atom 50
The Modern View of Atomic Structure:
An Introduction 54
Molecules and Ions 55
An Introduction to the Periodic Table 57
CHEMICAL CONNECTIONS Hassium Fits Right In
2.8
Daff/Dreamstime.com
2.4
2.5
48
60
Naming Simple Compounds 60
For Review 71 ∣ Key Terms 71 ∣ Questions and Exercises 72
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iii
iv
Contents
Chapter 4 Types of Chemical Reactions
and Solution Stoichiometry
4.1
4.2
138
Water, the Common Solvent 139
The Nature of Aqueous Solutions: Strong and Weak
Electrolytes 141
CHEMICAL CONNECTIONS Arrhenius: A Man with
4.3
The Composition of Solutions 145
CHEMICAL CONNECTIONS Tiny Laboratories
4.4
4.5
4.6
4.7
4.8
4.9
4.10
152
Types of Chemical Reactions 153
Precipitation Reactions 153
Describing Reactions in Solution 158
Stoichiometry of Precipitation Reactions 160
Acid–Base Reactions 163
Oxidation–Reduction Reactions 170
Balancing Oxidation–Reduction Equations 175
For Review 177 ∣ Key Terms 177 ∣ Questions and
Exercises 179
Chapter 5 Gases
5.1
5.2
5.3
5.4
5.5
Pressure 190
The Gas Laws of Boyle, Charles, and
Avogadro 192
The Ideal Gas Law 198
Gas Stoichiometry 203
Dalton’s Law of Partial Pressures 208
CHEMICAL CONNECTIONS Veggie Gasoline?
and Periodicity
7.1
7.2
210
Bags 211
For Review 230 ∣ Key Terms 230 ∣ Questions and
Exercises 232
245
The Nature of Energy 246
Enthalpy and Calorimetry 252
CHEMICAL CONNECTIONS Nature Has Hot
Plants 256
6.3
6.4
6.5
CHEMICAL CONNECTIONS Farming the Wind
277
282
Chapter 7 Atomic Structure
The Kinetic Molecular Theory of Gases 214
Effusion and Diffusion 222
Real Gases 224
Characteristics of Several Real Gases 226
Chemistry in the Atmosphere 227
Chapter 6 Thermochemistry
New Energy Sources 275
For Review 283 ∣ Key Terms 283 ∣ Questions and
Exercises 285
CHEMICAL CONNECTIONS The Chemistry of Air
6.1
6.2
6.6
189
CHEMICAL CONNECTIONS Separating Gases
5.6
5.7
5.8
5.9
5.10
© Caren Brinkema/Science Faction/Corbis
Solutions 144
Hess’s Law 260
Standard Enthalpies of Formation 264
Present Sources of Energy 271
295
Electromagnetic Radiation 296
The Nature of Matter 298
CHEMICAL CONNECTIONS Fireworks
7.3
7.4
300
The Atomic Spectrum of Hydrogen 305
The Bohr Model 306
CHEMICAL CONNECTIONS 0.035 Femtometer Is a Big
Deal 309
7.5
7.6
7.7
7.8
7.9
7.10
7.11
The Quantum Mechanical Model of the Atom 310
Quantum Numbers 313
Orbital Shapes and Energies 314
Electron Spin and the Pauli Principle 318
Polyelectronic Atoms 318
The History of the Periodic Table 320
The Aufbau Principle and the Periodic Table 322
CHEMICAL CONNECTIONS The Chemistry of
Copernicium 323
7.12 Periodic Trends in Atomic Properties 329
7.13 The Properties of a Group: The Alkali Metals 335
CHEMICAL CONNECTIONS Potassium—Too Much of a
Good Thing Can Kill You
337
For Review 339 ∣ Key Terms 339 ∣ Questions and
Exercises 341
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Contents
Chapter 8 Bonding: General Concepts
351
Types of Chemical Bonds 352
CHEMICAL CONNECTIONS No Lead Pencils
354
Carsten Peter/Speleoresearch Films/National Geographic Stock
8.1
v
8.2
8.3
8.4
8.5
8.6
8.7
8.8
Electronegativity 356
Bond Polarity and Dipole Moments 358
Ions: Electron Configurations and Sizes 361
Energy Effects in Binary Ionic Compounds 365
Partial Ionic Character of Covalent Bonds 369
The Covalent Chemical Bond: A Model 370
Covalent Bond Energies and Chemical
Reactions 373
8.9 The Localized Electron Bonding Model 376
8.10 Lewis Structures 376
8.11 Exceptions to the Octet Rule 380
CHEMICAL CONNECTIONS Nitrogen Under Pressure
381
8.12 Resonance 384
8.13 Molecular Structure: The VSEPR Model 389
Substance? 472
10.6
10.7
10.8
10.9
Communication: Semiochemicals 398
For Review 402 ∣ Key Terms 402 ∣ Questions and
Exercises 404
9.1
9.2
9.3
9.4
9.5
9.6
440
Chapter 11 Properties of Solutions
11.1
11.2
11.3
11.4
Intermolecular Forces 455
The Liquid State 458
An Introduction to Structures and Types of
Solids 459
Solution Composition 511
The Energies of Solution Formation 514
Factors Affecting Solubility 517
The Vapor Pressures of Solutions 521
11.5
Boiling-Point Elevation and Freezing-Point
Depression 527
11.6
11.7
Osmotic Pressure 531
Colligative Properties of Electrolyte Solutions 535
CHEMICAL CONNECTIONS The Drink of Champions—
Water 537
11.8
Colloids 538
CHEMICAL CONNECTIONS Organisms and Ice
463
Formation 539
Structure and Bonding in Metals 465
CHEMICAL CONNECTIONS Closest Packing of M & Ms
510
Tragedy 522
453
CHEMICAL CONNECTIONS Smart Fluids
10.4
For Review 496 ∣ Key Terms 496 ∣ Questions and
Exercises 498
CHEMICAL CONNECTIONS The Lake Nyos
For Review 443 ∣ Key Terms 443 ∣ Questions and
Exercises 444
10.1
10.2
10.3
Pressures: Fooling Mother Nature 494
Photoelectron Spectroscopy (PES) 441
Chapter 10 Liquids and Solids
Molecular Solids 479
Ionic Solids 480
Vapor Pressure and Changes of State 483
Phase Diagrams 491
CHEMICAL CONNECTIONS Making Diamonds at Low
415
Hybridization and the Localized Electron
Model 416
The Molecular Orbital Model 428
Bonding in Homonuclear Diatomic Molecules 431
Bonding in Heteronuclear Diatomic
Molecules 438
Combining the Localized Electron and Molecular
Orbital Models 439
CHEMICAL CONNECTIONS What’s Hot?
Carbon and Silicon: Network Atomic Solids 471
CHEMICAL CONNECTIONS Graphene—Miracle
CHEMICAL CONNECTIONS Chemical Structure and
Chapter 9 Covalent Bonding: Orbitals
10.5
469
For Review 540 ∣ Key Terms 540 ∣ Questions and
Exercises 542
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vi
Contents
13.7
Le Châtelier’s Principle 633
For Review 640 ∣ Key Terms 640 ∣ Questions and
Exercises 642
Chapter 14 Acids and Bases
14.1
14.2
14.3
652
The Nature of Acids and Bases 653
Acid Strength 656
The pH Scale 661
CHEMICAL CONNECTIONS Arnold Beckman, Man of
Science 663
14.4
14.5
14.6
Calculating the pH of Strong Acid Solutions 665
Calculating the pH of Weak Acid Solutions 666
Bases 675
National Cancer Institute/Photo Researchers, Inc.
CHEMICAL CONNECTIONS Amines
14.7
14.8
679
Polyprotic Acids 681
Acid–Base Properties of Salts 686
14.9
The Effect of Structure on Acid–Base
Properties 691
14.10 Acid–Base Properties of Oxides 693
14.11 The Lewis Acid–Base Model 694
14.12 Strategy for Solving Acid–Base Problems:
A Summary 696
For Review 697 ∣ Key Terms 697 ∣ Questions and
Exercises 701
Chapter 12 Chemical Kinetics
12.1
12.2
12.3
12.4
12.5
12.6
12.7
552
Reaction Rates 553
Rate Laws: An Introduction 557
Determining the Form of the Rate Law 559
The Integrated Rate Law 563
Reaction Mechanisms 574
A Model for Chemical Kinetics 577
Catalysis 583
Chapter 15 Acid–Base Equilibria
15.1
15.2
15.3
15.4
15.5
Catalysts 586
Chapter 13 Chemical Equilibrium
13.1
13.2
13.3
13.4
13.5
13.6
606
The Equilibrium Condition 607
The Equilibrium Constant 610
Equilibrium Expressions Involving Pressures 614
Heterogeneous Equilibria 617
Applications of the Equilibrium Constant 618
Solving Equilibrium Problems 628
Solutions of Acids or Bases Containing a
Common Ion 712
Buffered Solutions 715
Buffering Capacity 724
Titrations and pH Curves 727
Acid–Base Indicators 742
For Review 748 ∣ Key Terms 748 ∣ Questions and
Exercises 749
CHEMICAL CONNECTIONS Enzymes: Nature’s
For Review 590 ∣ Key Terms 590 ∣ Questions and
Exercises 592
711
Chapter 16 Solubility and Complex Ion
Equilibria
16.1
758
Solubility Equilibria and the Solubility
Product 759
CHEMICAL CONNECTIONS The Chemistry of
Teeth 763
16.2
16.3
Precipitation and Qualitative Analysis 768
Equilibria Involving Complex Ions 774
For Review 779 ∣ Key Terms 779 ∣ Questions and
Exercises 780
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Contents
Chapter 17 Spontaneity, Entropy,
and Free Energy
17.1
18.8
CHEMICAL CONNECTIONS The Chemistry of Sunken
787
Treasure 868
Spontaneous Processes and Entropy 788
18.9
CHEMICAL CONNECTIONS Entropy: An Organizing
Force?
17.2
17.3
17.4
17.5
17.6
17.7
17.8
17.9
Electrolysis 864
Commercial Electrolytic Processes 868
For Review 874 ∣ Key Terms 874 ∣ Questions and
Exercises 877
794
Entropy and the Second Law of
Thermodynamics 794
The Effect of Temperature on Spontaneity 795
Free Energy 798
Entropy Changes in Chemical Reactions 801
Free Energy and Chemical Reactions 805
The Dependence of Free Energy on Pressure 810
Free Energy and Equilibrium 813
Free Energy and Work 817
Chapter 19 The Nucleus: A Chemist’s
View
19.1
19.2
19.3
890
Nuclear Stability and Radioactive Decay 891
The Kinetics of Radioactive Decay 896
Nuclear Transformations 899
CHEMICAL CONNECTIONS Element 117
19.4
19.5
19.6
For Review 820 ∣ Key Terms 820 ∣ Questions and
Exercises 822
901
Detection and Uses of Radioactivity 902
Thermodynamic Stability of the Nucleus 906
Nuclear Fission and Nuclear Fusion 910
CHEMICAL CONNECTIONS Future Nuclear
Chapter 18 Electrochemistry
18.1
18.2
18.3
18.4
18.5
18.6
Power 912
832
19.7
Balancing Oxidation–Reduction Equations 833
Galvanic Cells 839
Standard Reduction Potentials 842
Cell Potential, Electrical Work, and Free
Energy 849
Dependence of Cell Potential on
Concentration 852
Batteries 858
For Review 917 ∣ Key Terms 917 ∣ Questions and
Exercises 919
Chapter 20 The Representative
Elements
20.1
20.2
20.3
20.4
20.5
20.6
CHEMICAL CONNECTIONS Fuel Cells—Portable
Energy 861
18.7
Effects of Radiation 915
Corrosion 861
926
A Survey of the Representative Elements 927
The Group 1A Elements 932
The Chemistry of Hydrogen 933
The Group 2A Elements 935
The Group 3A Elements 937
The Group 4A Elements 939
CHEMICAL CONNECTIONS Beethoven: Hair Is the
Story 940
20.7
20.8
The Group 5A Elements 941
The Chemistry of Nitrogen 942
CHEMICAL CONNECTIONS Nitrous Oxide: Laughing Gas
NASA/SDO/AIA
That Propels Whipped Cream and Cars 948
20.9
20.10
20.11
20.12
20.13
20.14
The Chemistry of Phosphorus 949
The Group 6A Elements 952
The Chemistry of Oxygen 952
The Chemistry of Sulfur 954
The Group 7A Elements 956
The Group 8A Elements 960
For Review 961 ∣ Key Terms 961 ∣ Questions and
Exercises 964
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vii
viii
Contents
Chapter 21 Transition Metals
and Coordination Chemistry
21.1
21.2
22.4
22.5
972
CHEMICAL CONNECTIONS Wallace Hume
The Transition Metals: A Survey 973
The First-Row Transition Metals 978
Carothers 1045
CHEMICAL CONNECTIONS Super-Slippery Slope
CHEMICAL CONNECTIONS Titanium Dioxide—Miracle
22.6
Coating 980
21.3
21.4
21.6
For Review 1067 ∣ Key Terms 1067 ∣ Questions and
Exercises 1070
990
Appendix 1 Mathematical Procedures
Bonding in Complex Ions: The Localized Electron
Model 992
The Crystal Field Model 994
A1.1
A1.2
A1.3
A1.4
A1.5
CHEMICAL CONNECTIONS Transition Metal Ions Lend
Color to Gems 997
21.7
21.8
The Biological Importance of Coordination
Complexes 1000
Metallurgy and Iron and Steel Production 1004
Exponential Notation A1
Logarithms A4
Graphing Functions A6
Solving Quadratic Equations A7
Uncertainties in Measurements A10
Molecular Model
A13
Appendix 3 Spectral Analysis
Chapter 22 Organic and Biological
A16
Appendix 4 Selected Thermodynamic
1023
Data
Alkanes: Saturated Hydrocarbons 1024
Alkenes and Alkynes 1032
Aromatic Hydrocarbons 1035
A19
Appendix 5 Equilibrium Constants and
Reduction Potentials
A5.1
A5.2
A5.3
A5.4
A5.5
A22
Values of Ka for Some Common Monoprotic
Acids A22
Stepwise Dissociation Constants for Several
Common Polyprotic Acids A23
Values of Kb for Some Common Weak Bases A23
Ksp Values at 258C for Common Ionic Solids A24
Standard Reduction Potentials at 258C (298 K) for
Many Common Half-Reactions A25
Appendix 6 SI Units and Conversion
Chip Clark/Smithsonian Institute
22.1
22.2
22.3
A1
Appendix 2 The Quantitative Kinetic
For Review 1012 ∣ Key Terms 1012 ∣ Questions and
Exercises 1015
Molecules
Natural Polymers 1052
Shade 1059
CHEMICAL CONNECTIONS The Importance of
21.5
1046
CHEMICAL CONNECTIONS Tanning in the
Coordination Compounds 983
Isomerism 987
Being cis
Hydrocarbon Derivatives 1037
Polymers 1044
Factors
Glossary
A26
A27
Answers to Selected Exercises
Index
A39
A71
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To the Professor
Features of Chemistry,
Ninth Edition
Conceptual learning and problem solving are fundamental to
the approach of Chemistry. For the Ninth Edition, we have
extended this emphasis by beginning the problem-solving approach in Chapter 1 (rather than Chapter 3 as in the Eighth
Edition) to assist students as they learn to use dimensional
analysis for unit conversions. Our philosophy is to help students learn to think like chemists so that they can apply the
process of problem solving to all aspects of their lives. We
give students the tools to become critical thinkers: to ask questions, to apply rules and models, and to evaluate the outcome.
It was also our mission to create a media program that embodies this philosophy so that when instructors and students look
online for either study aids or online homework, each resource
supports the goals of the textbook—a strong emphasis on
models, real-world applications, and visual learning.
What’s New
We have made extensive updates to the Ninth Edition to enhance the learning experience for students. Here’s what’s
new:
❯ A new emphasis has been placed on systematic problem
solving in the applications of dimensional analysis.
❯ Critical Thinking questions have been added throughout the
text to emphasize the importance of conceptual learning.
❯ Interactive Examples have been added throughout the text.
These computer-based examples force students to think
through the example step-by-step rather than simply scan
the written example in the text as many students do.
❯ ChemWork problems have been added to the end-ofchapter problems throughout the text. These problems
test students’ understanding of core concepts from each
chapter. Students who solve a particular problem with no
assistance can proceed directly to the answer. However,
students who need help can get assistance through a series of online hints. The online procedure for assisting
students is modeled after the way a teacher would help
with homework problems in his or her office. The hints
are usually in the form of interactive questions that guide
students through the problem-solving process. Students
cannot receive the correct answer from the computer;
rather, it encourages students to continue working though
the hints to arrive at the answer. ChemWork problems
in the text can be worked using the online system or as
pencil-and-paper problems.
❯ New end-of-chapter questions and problems have been
added throughout the text.
❯ The art program has been modified and updated as needed,
and new macro/micro illustrations have been added.
❯ In Chapter 3 the treatment of stoichiometry has been enhanced by the addition of a new section on limiting reactants, which emphasizes calculating the amounts of products that can be obtained from each reactant. Now students
are taught how to select a limiting reactant both by comparing the amounts of reactants present and by calculating
the amounts of products that can be formed by complete
consumption of each reactant.
❯ A section on photoelectron spectroscopy was added to
Chapter 9 (Section 9.6).
Hallmarks of Chemistry
❯ Chemistry contains numerous discussions, illustrations,
and exercises aimed at overcoming misconceptions. It has
become increasingly clear from our own teaching experience that students often struggle with chemistry because
they misunderstand many of the fundamental concepts. In
this text, we have gone to great lengths to provide illustrations and ­explanations aimed at giving students a more
accurate picture of the fundamental ideas of chemistry. In
particular, we have attempted to represent the microscopic
world of chemistry so that students have a picture in their
minds of “what the atoms and molecules are doing.” The
art program along with the animations emphasize this goal.
We have also placed a larger emphasis on the qualitative
understanding of concepts before quantitative problems are
considered. Because using an algorithm to correctly solve
a problem often masks misunderstanding—when students
assume they understand the material because they got the
right “answer”—it is important to probe their understanding in other ways. In this vein, the text includes many Critical Thinking questions throughout the text and a number
of Active Learning Questions at the end of each chapter
that are intended for group discussion. It is our experience
that students often learn the most when they teach each
other. Students are forced to recognize their own lack of
understanding when they try and fail to explain a concept
to another student.
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ix
x
To the Professor
❯ With a strong problem-solving orientation, this text talks
❯ Chemical Connections boxes present applications of
to students about how to approach and solve chemical
problems. We emphasize a thoughtful, logical approach
rather than simply memorizing procedures. In particular,
an innovative method is given for dealing with acid–base
equilibria, the material the typical student finds most difficult and frustrating. The key to this approach involves first
deciding what species are present in solution, then thinking about the chemical properties of these species. This
method provides a general framework for approaching all
types of solution equilibria.
The text contains almost 300 Examples, with more given
in the text discussions, to illustrate general problemsolving strategies. When a specific strategy is presented, it is
summarized in a Problem-Solving Strategy box, and the Example that follows it reinforces the use of the strategy to solve
the problem. In general, we emphasize the use of conceptual
understanding to solve problems rather than an algorithmbased approach. This approach is strongly reinforced by the
inclusion of many Interactive Examples, which encourage
students to thoughtfully consider the example step-by-step.
We have presented a thorough treatment of reactions that
­occur in solution, including acid–base reactions. This material appears in Chapter 4, “Types of Chemical Reactions
and Solution Stoichiometry,” directly after the chapter on
chemical stoichiometry, to emphasize the connection between solution reactions and chemical reactions in general.
The early presentation of this material provides an opportunity to cover some interesting descriptive chemistry and
also supports the lab, which typically involves a great deal
of aqueous chemistry. Chapter 4 also includes oxidation–
reduction reactions and balancing by oxidation state, because a large number of interesting and important chemical
reactions involve redox processes. However, coverage of
oxidation–reduction is optional at this point and depends
on the needs of a specific course.
Descriptive chemistry and chemical principles are thoroughly integrated in this text. Chemical models may appear sterile and confusing without the observations that
stimulated their invention. On the other hand, facts without
organizing principles may seem overwhelming. A combination of observation and models can make chemistry
both interesting and understandable. In the chapter on the
chemistry of the elements, we have used tables and charts
to show how properties and models correlate. Descriptive
chemistry is presented in a variety of ways—as applications of principles in separate sections, in photographs, in
Examples and exercises, in paragraphs, and in Chemical
Connections.
Throughout the book a strong emphasis on models prevails.
Coverage includes how they are constructed, how they are
tested, and what we learn when they inevitably fail. Models are developed naturally, with pertinent observation always presented first to show why a particular model was
invented.
chemistry in various fields and in our daily lives. Margin
notes in the Instructor’s Annotated Edition also highlight
many more Chemical Connections available on the student
website.
❯ We offer end-of-chapter exercises for every type of student
and for every kind of homework assignment: questions
that promote group learning, exercises that reinforce student understanding, and problems that present the ultimate
challenge with increased rigor and by integrating multiple
concepts. We have added biochemistry problems to make
the connection for students in the course who are not chemistry majors.
❯ Judging from the favorable comments of instructors and
students who have used the eighth edition, the text seems to
work very well in a variety of courses. We were especially
pleased that readability was cited as a key strength when
students were asked to assess the text.
❯
❯
❯
❯
Supporting Materials
Please visit www.cengage.com
/chemistry/zumdahl/chemistry9e for
information about student and instructor resources for this text.
Acknowledgments
This book represents the efforts of many talented and dedicated people. We particularly want to thank Mary Finch, Publisher, for her vision and oversight of the project, and Lisa
Lockwood, Executive Editor, whose enthusiasm, powers of
organization, and knowledge of the market have contributed
immensely to the success of this revision. We also greatly appreciate the work of Teresa Trego, Content Project Manager,
who did an outstanding job of managing the production of
this complex project.
We especially appreciate the outstanding and untiring
work of Tom Martin, Developmental Editor. Tom is always
upbeat and has great suggestions. He contributed in many important ways to the successful completion of this edition,
keeping the details in order and managing many different
people with grace and good humor.
We are especially grateful to Tom Hummel, University
of Illinois, Urbana-Champaign, who managed the revision of
the end-of-chapter problems and the solutions manuals.
Tom’s ­extensive experience teaching general chemistry and
his high standards of accuracy and clarity have resulted in
great improvements in the quality of the problems and solutions in this edition. Don DeCoste and Gretchen Adams support us in so many ways it is impossible to list all of them.
Don wrote all of the Critical Thinking questions for this edition. Gretchen constructed all of the online Interactive Examples, created the PowerPoint slides, and worked on many of
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To the Professor
the other media aspects of the program. We are very grateful
to Don and Gretchen for their creativity and their incredible
work ethic and for being such wonderful colleagues.
Special thanks to Kathy Thrush Saginaw, who contributed excellent suggestions for improving the art in the text,
and to Sharon Donahue, who did her usual outstanding job
finding just the right photos for this edition. Also we greatly
appreciate the advice and support of Nicole Hamm, Senior
Marketing Manager.
There are many other people who made important contributions to the success of this edition, including Megan
Greiner at Graphic World; Maria Epes, Art Director; Ellen
Pettengill, Text Designer; Lisa Weber, Senior Media Editor;
and Stephanie VanCamp, Media Editor. Special thanks to
Krista Mastroianni, Assistant Editor, who helped in many different ways.
We are especially thankful to all of the reviewers who
participated in different aspects of the development process,
from reviewing the illustrations and chapters to providing
feedback on the development of new features. We sincerely
appreciate all of these suggestions.
Reviewers
Ninth Edition Reviewers
Kaveh Azimi, Tarrant County College–South
Ron Briggs, Arizona State University
Maureen Burkart, Georgia Perimeter College
Paula Clark, Muhlenberg College
Russell Franks, Stephen F. Austin State University
Judy George, Grossmont College
Roger LeBlanc, University of Miami
Willem Leenstra, University of Vermont
Gary Mort, Lane Community College
Hitish Nathani, St. Philip’s College
Shawn Phillips, Vanderbilt University
Elizabeth Pulliam, Tallahassee Community College
Michael Sommer, University of Wyoming
Clarissa Sorensen-Unruh, Central New Mexico Community
College
William Sweeney, Hunter College, The City University of
New York
Brooke Taylor, Lane Community College
Hongqiu Zhao, Indiana University-Purdue University
Indianapolis
Lin Zhu, Indiana University-Purdue University Indianapolis
AP Reviewers:
Todd Abronowitz, Parish Episcopal High School
Kristen Jones, College Station ISD
xi
Lisa McGaw, Laying the Foundation
Priscilla Tuttle, Eastport-South Manor Junior/Senior High
School
Eighth Edition Reviewers
Yiyan Bai, Houston Community College
David A. Boyajian, Palomar College San Marcos
Carrie Brennan, Austin Peay State University
Alexander Burin, Tulane University
Jerry Burns, Pellissippi State Technical Community College
Stuart Cohen, Horry-Georgetown Technical College
Philip Davis, University of Tennessee at Martin
William M. Davis, The University of Texas at Brownsville
Stephanie Dillon, Florida State University
David Evans, Coastal Carolina University
Leanna Giancarlo, University of Mary Washington
Tracy A. Halmi, Penn State Erie, The Behrend College
Myung Han, Columbus State Community College
Carl Hoeger, University of California, San Diego
Richard Jarman, College of DuPage
Kirk Kawagoe, Fresno City College
Cathie Keenan, Chaffey College
Donald P. Land, University of California, Davis Department
of Chemistry
Craig Martens, University of California, Irvine
Chavonda Mills, Georgia College & State University
John Pollard, University of Arizona
Rene Rodriguez, Idaho State University
Tim Royappa, University of West Florida
Karl Sienerth, Elon University
Brett Simpson, Coastal Carolina University
Alan Stolzenberg, West Virginia University, Morgantown
Paris Svoronos, Queensborough Community College, CUNY
Brooke Taylor, Lane Community College
James Terner, Virgina Commonwealth University
Jackie Thomas, Southwestern College
David W. Thompson, College of William and Mary
Edward Walters, University of New Mexico
Darrin M. York, University of Minnesota
Noel S. Zaugg, Brigham Young University, Idaho
AP Reviewers:
Robert W. Ayton, Jr., Dunnellon High School
David Hostage, The Taft School
Steven Nelson, Addison Trail High School
Connie Su, Adolfo Camarillo High School
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xii
To the Professor
Seventh Edition Reviewers
Dawood Afzal, Truman State
Carol Anderson, University of Connecticut, Avery Point
Jeffrey R. Appling, Clemson University
Dave Blackburn, University of Minnesota
Robert S. Boikess, Rutgers University
Ken Carter, Truman State
Bette Davidowitz, University of Cape Town
Natalie Foster, Lehigh University
Tracy A. Halmi, Penn State Erie, The Behrend College
Carl Hoeger, University of California, San Diego
Ahmad Kabbani, Lebanese American University
Arthur Mar, University of Alberta
Jim McCormick, Truman State
Richard Orwell, Blue Ridge Community College
Jason S. Overby, College of Charleston
Robert D. Pike, The College of William and Mary
Daniel Raferty, Purdue University
Jimmy Rogers, University of Texas, Arlington
Raymond Scott, Mary Washington College
Alan Stolzenberg, West Virginia University, Morgantown
Rashmi Venkateswaran, University of Ottawa
AP Reviewers:
Annis Hapkiewicz, Okemos High School
Tina Ohn-Sabatello, Maine Township HS East
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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To the Student
As you jump into the study of chemistry, we hope that you
will find our text helpful and interesting. Our job is to present
the concepts and ideas of chemistry in a way you can understand. We hope to encourage you in your studies and to help
you learn to solve problems in ways you can apply in all areas
of your professional and personal lives.
Our main goal is to help you learn to become a truly
creative problem solver. Our world badly needs people who
can “think outside the box.” Our focus is to help you learn to
think like a chemist. Why would you want to do that? Chemists are great problem solvers. They use logic, trial and error,
and intuition—along with lots of patience—to work through
complex problems. Chemists make mistakes, as we all do in
our lives. The important thing that a chemist does is to learn
from the mistakes and to try again. This “can do” attitude is
useful in all ­careers.
In this book we develop the concepts in a natural way:
The observations come first and then we develop models to
explain the observed behavior. Models help us to understand
and explain our world. They are central to scientific thinking.
Models are very useful, but they also have limitations, which
we will point out. By understanding the basic concepts in
chemistry we lay the foundation for solving problems.
Our main goal is to help you learn a thoughtful method of
problem solving. True learning is more than memorizing facts.
Truly educated people use their factual knowledge as a starting
point—a basis for creative problem solving. Our strategy for
solving problems is explained first in Section 1.6 and is covered in more details in Section 3.5. To solve a problem we ask
ourselves questions, which help us think through the problem.
We let the problem guide us to the solution. This process can
be applied to all types of problems in all areas of life.
As you study the text, use the Examples and the problemsolving strategies to help you. The strategies are boxed to
highlight them for you, and the Examples show how these
strategies are applied. It is especially important for you to
do the computer-based Interactive Examples that are found
throughout the text. These examples encourage you to think
through the examples step-by-step to help you thoroughly understand the concepts involved.
After you have read and studied each chapter of the
text, you’ll need to practice your problem-solving skills. To
do this we have provided plenty of review questions and
end-of-­chapter exercises. Your instructor may assign these
on ­paper or online; in either case, you’ll want to work with
your fellow students. One of the most effective ways to
learn chemistry is through the exchange of ideas that comes
from helping one another. The online homework assignments will give you instant feedback, and in print, we have
provided ­answers to some of the exercises in the back of
the text. In all cases, your main goal is not just to get the
correct answer but to understand the process for getting the
answer. Memorizing solutions for specific problems is not
a very good way to prepare for an exam (or to solve problems in the real world!).
To become a great problem solver, you’ll need these
skills:
1. Look within the problem for the solution. (Let the problem guide you.)
2. Use the concepts you have learned along with a systematic, logical approach to find the solution.
3. Solve the problem by asking questions and learn to trust
yourself to think it out.
You will make mistakes, but the important thing is to learn
from these errors. The only way to gain confidence is to practice, practice, practice and to use your mistakes to find your
weaknesses. Be patient with yourself and work hard to understand rather than simply memorize.
We hope you’ll have an interesting and successful year
learning to think like a chemist!
Steve and Susan Zumdahl
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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xiii
ter 12
Rules Governing Formal Charge
❯ To calculate the formal charge on an atom:
1. Take the sum of the lone pair electrons and one-half the shared electrons. This is
the number of valence electrons assigned to the atom in the molecule.
Chemistry, Ninth Edition 2. Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom to obtain the formal charge.
A Guide to
❯ The sum of the formal charges of all atoms in a given molecule or ion must equal
the overall charge on that species.
❯ If nonequivalent Lewis structures exist for a species, those with formal charges
closest to zero and with any negative formal charges on the most electronegative
atoms are considered to best describe the bonding in the molecule or ion.
Conceptual Understanding Conceptual learning and problem solving are fundamental to
Example 8.10
Charges
the approach of Chemistry. The text gives students the
tools to Formal
become
critical thinkers: to ask
Give possible Lewis structures for XeO , an explosive compound of xenon. Which
Lewis structure or structures are most appropriate according to the formal charges?
questions, to apply rules and models, and to evaluate the outcome.
Solution
3
For XeO3 (26 valence electrons) we can draw the following possible Lewis structures
(formal charges are indicated in parentheses):
Xe
O
(−1)
O
(+3)
O
O
(−1)
(0)
(−1)
Xe
O
(+2)
Xe
O
O
(−1)
(−1)
(−1)
(+2)
O
O (−1)
Xe
(−1)
O
(+1)
Xe
O
O
(0)
(0)
(0)
O
O
(−1)
(0)
(+1)
Xe
O
O
(0)
(0)
(−1)
O
(0)
(+2)
O
(0)
(−1)
“Before students are ready to figure out complex problems, they need to
master simpler problems in various contortions. This approach works, and
the authors’ presentation of it should have the students buying in.”
O
Xe
O
(+1)
Xe
O
O
(−1)
(0)
O
(0)
O
(0)
(0)
Based on the ideas of formal charge, we would predict that the Lewis structures with
the lower values of formal charge would be most appropriate for describing the
bonding in XeO3.
See Exercises 8.101 and 8.102
—Jerry Burns, Pellissippi State Technical Community College
As a final note, there are a couple of cautions about formal charge to keep in mind.
First, although formal charges are closer to actual atomic charges in molecules than are
oxidation states, formal charges still provide only estimates of charge—they should
not be taken as actual atomic charges. Second, the evaluation of Lewis structures using
formal charge ideas can lead to erroneous predictions. Tests based on experiments
must be used to make the final decisions on the correct description of the bonding in a
molecule or polyatomic ion.
Chemical Kinetics
| The decomposition
g) n 2N2(g) 1 O2(g)
a platinum surface.
] is three times as great
the rate of decomposihe same in both cases
tinum surface can
only a certain number
s a result, this reaction
IBLG: See questions from
The authors’ emphasis on modeling (or chemical theories)
8.13 Molecular Structure: The VSEPR Model
“Molecular Structure: The
VSEPR Model”
The structures of molecules play a very important role in determining their chemical
throughout the text addresses the problem of rote memorization
properties. As we will see later, this is particularly important for biological molecules;
a slight change in the structure of a large biomolecule can completely destroy its useby helping students better understand and appreciate the procfulness to a cell or may even change the cell from a normal one to a cancerous one.
ess of scientific thinking. By stressing the limitations and uses of
scientific models, the authors show students how chemists think
Pt
Pt
and work.
NO
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
2
a
b
11097_Ch08_0351-0414.indd 389
8.13
Critical Thinking
cules, such as methanol (CH3OH). This molecule is represented by the following
Lewis structure:
The text includes a number of open-ended Critical Thinking
questions that emphasize
232
Chapter 5 the
Gases importance of conceptual learning.
These questions7. are
particularly
useful for generating
group
Consider the following velocity distribution curves A
b. If the plots represent the velocity distribution of
and B.
1.0 L of O (g) at temperatures of 273 K versus
discussion.
1273 K, which plot corresponds to each tempera-
Consider the simple reaction aA n products. You run this reaction and wish
H to deterCmine its order. What if you made a graph of reaction rate versus time? Could you use
this to determine
the order? Sketch three plots of rate versus timeHfor the
if
C reaction
O H
O
it is zero, first, or second order. Sketch these plots on the same graph and compare
H them. Defend your answer.
H
H
a
C
2
The molecular structure can be predicted from the arrangement of pairs around the
carbon and oxygen atoms. Note that there are four pairs of electrons around the carbon, for
whichReactions
requires a tetrahedral arrangement [Fig. 8.22(a)]. The oxygen also has four
Integrated Rate Laws
pairs, which requires a tetrahedral arrangement. However, in this case the tetrahedron
with More Than One
Reactant
will be slightly distorted by the space requirements of the lone pairs [Fig. 8.22(b)]. The
H
geometric
arrangement
the molecule
is shown
SoO far we have considered theoverall
integrated
rate laws
for simplefor
reactions
with only
one in Fig. 8.22(c).
reactant. Special techniques are required to deal with more complicated reactions. Let’s
consider the reaction
Let’s
Summary
BrO32 1aq2 1 5Br2 1aq2
1 Review
6H1 1aq2 h
1the
3H2VSEPR
O 1l2 Model
3Br2 1l2of
H
c
Velocity (m/s)
A discussion of the Active Learning Questions can be found online in the Instructor’s Resource Guide and on PowerLecture. The questions
allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the
learning that occurs while students talk to each other about chemical concepts.
Active Learning Questions
Figure 8.22 | The molecular
3Br2 4 5 3Br2 4 0 and 3H1 4 5 3H1 4 0
structure of methanol. (a) The
This
means that
arrangement
of electron
pairsthe
andrate law can be written
atoms around the carbon atom.
Rate 5 k3Br2 4 0 3H1 4 02 3BrO32 4 5 kr 3BrO32 4
(b) The arrangement of bonding and
The VSEPR model is very simple. There are only a few rules to remember, yet the
lone pairs around
oxygen
where,the
since
[Br2atom.
]0 and [H1]0 are constant,
model correctly predicts the molecular structures of most molecules formed from non(c) The molecular structure.
The VSEPR Model—How Well Does It Work?
These questions are designed to be used by groups of students in
class.
1. Consider the following apparatus: a test tube covered with a
nonpermeable elastic membrane inside a container that is
closed with a cork. A syringe goes through the cork.
Syringe
1 2
metallic
of any size can be treated by applying the VSEPR model
40
k3Br2 4 0 3HMolecules
kr 5 elements.
to each appropriate atom (those bonded to at least two other atoms) in the molecule.
Thus we can use this model to predict the structures of molecules with hundreds of
atoms. It does,
however, fail in a few instances. For example, phosphine (PH3), which
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
has a Lewis structure analogous to that of ammonia,
H
P
H
2
H
H
N
B
Let’s Review boxes help students organize their thinking about the
a. If the plots represent the velocity distribution of
1.0 L of He(g) at STP versus 1.0 L of Cl2(g) at
STP, which plot corresponds to each gas? Explain
your reasoning.
D3BrO 2 4
2
1structures,
2
2resonance
with
use any of the structures to predict the
5 k3BrO
Rate 5 2 ❯ For3molecules
3 4 3Br 4 3H 4
Dt
molecular structure.
2 central atom. 23
Sum the
electron where
pairs around
Suppose we run this reaction❯under
conditions
[BrO3the
]0 5 1.0 3 10 M,
2
1
H
❯ In As
counting
pairs, count
each multiple
as a single
[Br
the reaction
proceeds,
[BrO32]bond
decreases
sig-effective pair.
C ]0 5 1.0 M, and [H ]0 5 1.0 M.
2
1
nificantly, but because the Br ion
andarrangement
H ion concentrations
are so largebyinitially,
❯ The
of the pairs is determined
minimizing electron-pair repulsions.
O
relatively little of these two reactants
is consumed.
Thus
[Br2] in
and
[H18.6.
] remain apThese
arrangements
are shown
Table
1
the conditions
where
Br2 ion
Hproximately constant. In other words,
❯ Loneunder
pairs require
more space
thanthe
bonding
pairsand
do. H
Choose an arrangement that
2
ion concentrations are much largergives
thanthe
thelone
BrOpairs
concentration,
we canRecognize
assume that the lone pairs may
as much
room as possible.
3 ion
that throughout the reaction
produce a slight distortion of the structure at angles less than 120 degrees.
H
A
ture? Explain your reasoning. Under which temperature condition would the O2(g) sample behave
most ideally? Explain.
8. Briefly describe two methods one might use to find the
molar mass of a newly synthesized gas for which a
molecular formula was not known.
9. In the van der Waals equation, why is a term added to
the observed pressure and why is a term subtracted
from the container volume to correct for nonideal gas
behavior?
10. Why do real gases not always behave ideally? Under
what conditions does a real gas behave most ideally?
Why?
crucial chemical concepts that they encounter.
The rules for using the VSEPR model to predict molecular structure are as follows:
❯ Determine the Lewis structure(s) for the molecule.
From experimental evidence we know that the rate law is
b
9/6/12 8:24 AM
401
Relative number of molecules
H
Molecular Structure: The VSEPR Model
d. Capillary action of the mercury causes the mercury to go
up the tube.
e. The vacuum that is formed at the top of the tube holds up
the mercury.
Justify your choice, and for the choices you did not pick, explain what is wrong with them. Pictures help!
3. The barometer below shows the level of mercury at a given atmospheric pressure. Fill all the other barometers with mercury
for that same atmospheric pressure. Explain your answer.
Cork
Membrane
H
Hg(l )
H
9/6/12 8:44 AM
would be predicted to have a molecular structure similar to that for NH3, with bond
angles of approximately 107 degrees. However, the bond angles of phosphine are actually 94 degrees. There are ways of explaining this structure, but more rules have to be
added to the model.
This again illustrates the point that simple models are bound to have exceptions. In
introductory chemistry we want to use simple models that fit the majority of cases; we
are willing to accept a few failures rather than complicate the model. The amazing
thing about the VSEPR model is that such a simple model predicts correctly the structures of so many molecules.
The text includes a number of Active Learning Questions at
the end of each chapter that are intended for group discussion,
since students often learn the most when they teach each other.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch08_0351-0414.indd 401
a. As you push down on the syringe, how does the membrane covering the test tube change?
b. You stop pushing the syringe but continue to hold it
down. In a few seconds, what happens to the membrane?
2. Figure 5.2 shows a picture of a barometer. Which of the following statements is the best explanation of how this barometer works?
a. Air pressure outside the tube causes the mercury to move in
the tube until the air pressure inside and outside the tube is
equal.
b. Air pressure inside the tube causes the mercury to move in
the tube until the air pressure inside and outside the tube is
equal.
c. Air pressure outside the tube counterbalances the weight
of the mercury in the tube.
9/6/12 8:24 AM
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch05_0189-0244.indd 232
xiv
4. As you increase the temperature of a gas in a sealed, rigid
container, what happens to the density of the gas? Would the
results be the same if you did the same experiment in a container with a piston at constant pressure? (See Fig. 5.17.)
5. A diagram in a chemistry book shows a magnified view of a
flask of air as follows:
9/6/12 8:27 AM
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Problem Solving This text talks to the student about how to approach and solve chemical
problems, since one of the main goals of general chemistry is to help students become creative problem solvers. The authors emphasize a thoughtful, logical approach rather than simply
memorizing procedures.
“The text gives a meaningful explanation and alternative to memorization. This
approach and the explanation [to the student] of the approach will supply the ‘secret’
of successful problem solving abilities to all students.”
—David Boyajian, Palomar College
3.5 Learning to Solve Problems
93
John Humble/The Image Bank/Getty Images
3.5 Learning to Solve Problems
Pigeonholes can be used for sorting and
classifying objects like mail.
One of the great rewards of studying chemistry is to become a good problem solver.
Being able to solve complex problems is a talent that will serve you well in all walks
of life. It is our purpose in this text to help you learn to solve problems in a flexible,
creative way based on understanding the fundamental ideas of chemistry. We call this
approach conceptual problem solving.
The ultimate goal is to be able to solve new problems (that is, problems you have
not seen before) on your own. In this text we will provide problems and offer solutions
by explaining how to think about the problems. While the answers to these problems
are important, it is perhaps even more important to understand the process—the thinking necessary to get the answer. Although at first we will be solving the problem for
you, do not take a passive role. While studying the solution, it is crucial that you interactively think through the problem with us. Do not skip the discussion and jump to the
answer. Usually, the solution will involve asking a series of questions. Make sure that
you understand each step in the process. This active approach should apply to problems outside of chemistry as well. For example, imagine riding with someone in a car
to an unfamiliar destination. If your goal is simply to have the other person get you to
that destination, you will probably not pay much attention to how to get there (passive), and if you have to find this same place in the future on your own, you probably
will not be able to do it. If, however, your goal is to learn how to get there, you would
pay attention to distances, signs, and turns (active). This is how you should read the
solutions in the text (and the text in general).
While actively studying our solutions to problems is helpful, at some point you will
need to know how to think through these problems on your own. If we help you too
much as you solve a problem, you won’t really learn effectively. If we always “drive,”
you won’t interact as meaningfully with the material. Eventually you need to learn to
drive yourself. We will provide more help at the beginning of the text and less as we
proceed to later chapters.
There are two fundamentally different ways you might use to approach a problem.
One way emphasizes memorization. We might call this the “pigeonholing method.” In
this approach, the first step is to label the problem—to decide in which pigeonhole it
fits. The pigeonholing method requires that we provide you with a set of steps that you
memorize and store in the appropriate slot for each different problem you encounter.
The difficulty with this method is that it requires a new pigeonhole each time a problem is changed by even a small amount.
Consider the driving analogy again. Suppose you have memorized how to drive from
your house to the grocery store. Do you know how to drive back from the grocery store
to your house? Not necessarily. If you have only memorized the directions and do not
understand fundamental principles such as “I traveled north to get to the store, so my
house is south of the store,” you may find yourself stranded. In a more complicated
example, suppose you know how to get from your house to the store (and back) and
from your house to the library (and back). Can you get from the library to the store
without having to go back home? Probably not if you have only memorized directions
and you do not have a “big picture” of where your house, the store, and the library are
relative to one another.
The second approach is conceptual problem solving, in which we help you get the
“big picture”—a real understanding of the situation. This approach to problem solving
looks within the problem for a solution. In this method we assume that the problem is
a new one, and we let the problem guide us as we solve it. In this approach we ask a
series of questions as we proceed and use our knowledge of fundamental principles to
3.7 this
Determining
the Formula
a Compound
99
answer these questions. Learning
approach requires
someofpatience,
but the reward
for learning to solve problems this way is that we become an effective solver of any
new problem that confronts us in daily life or in our work in any field. In summary,
instead of looking outside the problem for a memorized solution, we will look inside
the problem and let the problem help us as we proceed to a solution.
In Chapter 3, “Stoichiometry,” the authors introduce a new section,
Learning to Solve Problems, which emphasizes the importance of problem solving. This new section helps students understand that thinking their
way through a problem produces more long-term, meaningful learning than
simply memorizing steps, which are soon forgotten.
1.8
Figure 1.10 | Normal body
temperature on the Fahrenheit,
Celsius, and Kelvin scales.
Fahrenheit
Celsius
98.6°F
66.6°F
32°F
Example 1.12
?K
?°C
5°C
66.6°F ×
= 37.0°C
9°F
273.15 K
0°C
Temperature Conversions II
Solution
Where are we going?
To show that 40C 40F
What do we know?
❯ The relationship between the Celsius and Fahrenheit scales
How do we get there?
The difference between 32F and 40F is 72F. The difference between 0C and
40C is 40C. The ratio of these is
Examples of substances
whose empirical and molecular
formulas differ. Notice that molecular
formula
5 (empirical
formula)
Unless otherwise
noted, all art on
this page isn,©where
Cengage Learning 2014.
n is an integer.
get there? This more active approach helps students think
their way through the solution to the problem.
72°F
8 3 9°F
9°F
5
5
40°C
8 3 5°C
5°C
as required. Thus 40C is equivalent to 40F.
See Exercise 1.61
17.1
9/6/12 8:46 AM
C6H6 = (CH)6
S8 = (S)8
The tendency to mix is due to the
increased volume available to the particles
of each component of the mixture. For
example, when two liquids are mixed,
the molecules of each liquid have more
available volume and thus more available
positions.
C6H12O6 = (CH2O)6
❯
Obtain the empirical formula.
❯
Compute the mass corresponding to the empirical formula.
Calculate the ratio:
793
portant process of problem solving.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Molar mass
Empirical formula mass
❯
Spontaneous Processes and Entropy
Since, as shown in Example 1.12, 40 on both the Fahrenheit and Celsius scales
Positional
is also verythis
important
in the
formation
of solutions.
Chapter
represents
theentropy
same temperature,
point can
be used
as a reference
pointIn(like
0C
11
sawfor
thata solution
formation
is favored
the natural tendency for substances to
andwe
32F)
relationship
between
the two by
scales:
mix. We can now be more precise. The entropy change associated with the mixing of
Number
of Fahrenheit
degrees AnTincrease
9°F is expected beF 2 12402
two pure substances
is expected
to be positive.
in
entropy
5
5
12402 than
Number
Celsius degrees
TC 2
5°Cfor the separated
cause there are many
moreofmicrostates
for the mixed
condition
condition. This effect is due principally to the increased volume available to a given
T 1 40
9°F
“particle” after mixing occurs. For Fexample,
(1.3)a
5 when two liquids are mixed to form
T
1
40
5°C
C
solution, the molecules of each liquid have more available volume and thus more available
thesame
increase
in positional
entropy
associated
with
represent the
temperature
(but not
the same
number).
Thismixing
equawherepositions.
TF and TCTherefore,
favors
thebeformation
of solutions.
tion can
used to convert
Fahrenheit temperatures to Celsius, and vice versa, and
may be easier to remember than Equations (1.1) and (1.2).
Problem-Solving Strategy boxes focus students’ attention on the very im-
Problem-Solving Strategy
Determining Molecular Formula from Empirical Formula
❯
37.0 + 273.15 K = 310.2 K
One interesting feature of the Celsius and Fahrenheit scales is that 40C and 40F
represent the same temperature, as shown in Fig. 1.9. Verify that this is true.
Chapters 1–6 introduce a series of questions into the inchapter Examples to engage students in the process of problem solving,
Figure 3.6 | such as Where are we going? and How do we
11097_Ch03_0081-0137.indd 93
25
Temperature
Kelvin
Interactive
Example 17.1
The integer from the previous step represents the number of empirical formula
units in one molecule. When the empirical formula subscripts are multiplied by
this integer, the molecular formula results. This procedure is summarized by the
equation:
Molecular formula 5 empirical formula 3
11097_Ch01_0001-0041.indd 25
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
molar mass
empirical formula mass
Positional Entropy
For each of the following pairs, choose the substance with the higher positional en9/25/12
tropy (per mole) at a given temperature.
5:06 PM
a. Solid CO2 and gaseous CO2
b. N2 gas at 1 atm and N2 gas at 1.0 3 1022 atm
Solution
Interactive
Example 3.10
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
a. Since a mole of gaseous CO2 has the greater volume by far, the molecules have
many more available positions than in a mole of solid CO2. Thus gaseous CO2
has the higher positional entropy.
b. A mole of N2 gas at 1 3 1022 atm has a volume 100 times that (at a given
temperature) of a mole of N2 gas at 1 atm. Thus N2 gas at 1 3 1022 atm has
the higher positional entropy.
Determining Empirical and Molecular Formulas I
Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percents):
Interactive Examples engage students in the problem71.65% Cl
24.27% C
4.07% H
See Exercise 17.31
The molar mass is known to be 98.96 g/mol.
Solution
solving process by requiring
them to think through the exWhere are we going?
To find thethan
empirical and
molecular formulas
for the given compound
ample step-by-step rather
simply
scanning
the written
What do we know?
Percent students
of each element
example in the text as many
do.
❯
❯
Molar mass of the compound is 98.96 g/mol
Interactive
Example 17.2
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
What information do we need to find the empirical formula?
❯ Mass of each element in 100.00 g of compound
❯ Moles of each element
Predict the sign of the entropy change for each of the following processes.
a. Solid sugar is added to water to form a solution.
b. Iodine vapor condenses on a cold surface to form crystals.
Solution
a. The sugar molecules become randomly dispersed in the water when the solution
forms and thus have access to a larger volume and a larger number of possible
positions. The positional disorder is increased, and there will be an increase in
entropy. DS is positive, since the final state has a larger entropy than the initial
state, and DS 5 Sfinal 2 Sinitial.
b. Gaseous iodine is forming a solid. This process involves a change from a
relatively large volume to a much smaller volume, which results in lower
positional disorder. For this process DS is negative (the entropy decreases).
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch03_0081-0137.indd 99
Predicting Entropy Changes
9/6/12 8:47 AM
See Exercise 17.32
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
xv
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch17_0787-0831.indd 793
9/6/12 8:57 AM
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
170
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
been emphasizing this approach in dealing with the reactions between ions in solution.
Make it a habit to write down the components of solutions before trying to decide what
reaction(s) might take place as you attempt the end-of-chapter problems involving
titrations.
4.9 Oxidation–Reduction Reactions
We have seen that many important substances are ionic. Sodium chloride, for example,
can be formed by the reaction of elemental sodium and chlorine:
2Na1s2 1 Cl2 1g2 h 2NaCl1s2
Dynamic Art Program Most of the glassware, orbitals, graphs, flowcharts, and molecules
In this reaction, solid sodium, which contains neutral sodium atoms, reacts with chlorine gas, which contains diatomic Cl2 molecules, to form the ionic solid NaCl, which
contains Na1 and Cl2 ions. This process is represented in Fig. 4.19. Reactions like this
one, in which one or more electrons are transferred, are called oxidation–reduction
reactions or redox reactions.
Many important chemical reactions involve oxidation and reduction. Photosynthesis, which stores energy from the sun in plants by converting carbon dioxide and water
to sugar, is a very important oxidation–reduction reaction. In fact, most reactions used
for energy production are redox reactions. In humans, the oxidation of sugars, fats, and
proteins provides the energy necessary for life. Combustion reactions, which provide
have been redrawn to better serve visual learners and enhance the textbook.
Experiment 26: Classification of
Chemical Reactions
IBLG: See questions from “Oxidation
Reduction”
4.3
The Composition of Solutions
149
What information do we need to find volume of blood containing 1.0 mg of NaCl?
❯ Moles of NaCl (in 1.0 mg)
How do we get there?
What are the moles of NaCl (58.44 g/mol)?
1.0 mg NaCl 3
1 g NaCl
1 mol NaCl
3
5 1.7 3 1025 mol NaCl
1000 mg NaCl
58.44 g NaCl
Photos © Cengage Learning. All rights reserved.
What volume of 0.14 M NaCl contains 1.0 mg (1.7 3 1025 mole) of NaCl?
There is some volume, call it V, that when multiplied by the molarity of this solution
will yield 1.7 3 1025 mole of NaCl. That is,
0.14 mol NaCl
The art program emphasizes
that
V 3molecular-level
5 1.7 3 10interactions
mol NaCl
L solution
help students visualize
thefor“micro/macro”
connection.
We want to solve
the volume:
25
V5
j
1.7 3 1025 mol NaCl
5 1.2 3 1024 L solution
0.14 mol NaCl
L solution
Thus 0.12 mL of blood contains 1.7 3 1025 mole of NaCl or 1.0 mg of NaCl.
See Exercises 4.33 and 4.34
Cl−
Na
Na+
Cl−
Na+
A standard solution is a solution whose concentration is accurately known. Standard solutions, often used in chemical analysis, can be prepared as shown in Fig. 4.10
and in Example 4.6.
Na
CCl
l
Cl Cl
2Na(s)
Sodium
Cl2(g)
Chlorine
+
Interactive
Example 4.6
2NaCl(s)
Sodium chloride
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Figure 4.19 | The reaction of solid sodium and gaseous chlorine to form solid sodium chloride.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Solutions of Known Concentration
To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of an aqueous
0.200-M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be
weighed out to make this solution?
Solution
Where are we going?
To find the mass of K2Cr2O7 required for the solution
11097_Ch04_0138-0188.indd 170
9/6/12 8:59 AM
Figure 4.10 | Steps involved in the
preparation of a standard aqueous
solution. (a) Put a weighed amount
of a substance (the solute) into the
volumetric flask, and add a small
quantity of water. (b) Dissolve the
solid in the water by gently swirling
the flask (with the stopper in place).
(c) Add more water (with gentle
swirling) until the level of the solution
just reaches the mark etched on the
neck of the flask. Then mix the
solution thoroughly by inverting the
flask several times.
Realistic drawings of glassware and instrumentation found in
the lab help students make real connections.
Wash
bottle
Volume marker
(calibration mark)
Weighed
amount
of solute
a
b
c
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
8.3
Figure 8.5 | (a) The charge distribution in the water molecule. (b) The
water molecule in an electric field.
(c) The electrostatic potential diagram
of the water molecule.
−
δ+
xvi
Δ−
O
H
H
b
c
+
3δ−
Δ−
N
H
δ+ H
δ+
H
H
δ+
O
a
N
Electrostatic potential maps help students visualize the
distribution of charge in molecules.
H
H
Δ+
−
a
δ−
molecule. (b) The opposed bond
polarities cancel out, and the carbon
dioxide molecule has no dipole
moment. (c) The electrostatic
potential diagram for carbon dioxide.
9/6/12 8:59 AM
H
Δ+
Figure 8.6 | (a) The structure and
Figure 8.7 | (a) The carbon dioxide
+
2δ−
a
charge distribution of the ammonia
molecule. The polarity of the NOH
bonds occurs because nitrogen has a
greater electronegativity than
hydrogen. (b) The dipole moment of
the ammonia molecule oriented in an
electric field. (c) The electrostatic
potential diagram for ammonia.
11097_Ch04_0138-0188.indd 149
H
O
δ+
359
Bond Polarity and Dipole Moments
b
2δ+
C
c
δ−
O
O
b
C
O
c
than the hydrogen atoms, the molecular charge distribution is that shown in Fig. 8.5(a).
Because of this charge distribution, the water molecule behaves in an electric field as if
it had two centers of charge—one positive and one negative—as shown in Fig. 8.5(b).
The water molecule has a dipole moment. The same type of behavior is observed for the
NH3 molecule (Fig. 8.6). Some molecules have polar bonds but do not have a dipole
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
moment. This occurs when the individual bond polarities are arranged in such a way
that they cancel each other out. An example is the CO2 molecule, which is a linear molecule that has the charge distribution shown in Fig. 8.7. In this case the opposing bond
polarities
cancelAll
out,Rights
and the
carbon dioxide
molecule
doesscanned,
not haveora duplicated,
dipole moment.
Copyright 2012 Cengage
Learning.
Reserved.
May not
be copied,
in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
There is no preferential way for this molecule to line up in an electric field. (Try to find
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
a preferred orientation to make sure you understand this concept.)
Real-World Applications Interesting applications of modern chemistry show students the
relevance of chemistry to their world.
I
n doing stoichiometry calculations we assumed that reactions proceed to completion, that is, until one of the reactants runs out. Many reactions do proceed essentially to completion. For such reactions it can be assumed that the reactants are quantitatively converted to products and that the amount of limiting reactant that remains is
negligible. On the other hand, there are many chemical reactions that stop far short of
completion. An example is the dimerization of nitrogen dioxide:
NO2 1g2 1 NO2 1g2 h N2O4 1g2
Each chapter begins
with an engaging introduction that demonstrates how chemistry is
related to everyday life.
The reactant, NO2, is a dark brown gas, and the product, N2O4, is a colorless gas.
When NO2 is placed in an evacuated, sealed glass vessel at 258C, the initial dark
brown color decreases in intensity as it is converted to colorless N2O4. However, even
over a long period of time, the contents of the reaction vessel do not become colorless. Instead, the intensity of the brown color eventually becomes constant, which
means that the concentration of NO2 is no longer changing. This is illustrated on the
molecular level in Fig. 13.1. This observation is a clear indication that the reaction
has stopped short of completion. In fact, the system has reached chemical equilibrium, the state where the concentrations of all reactants and products remain constant with time.
Any chemical reactions carried out in a closed vessel will reach equilibrium. For
some reactions the equilibrium position so favors the products that the reaction appears to have gone to completion. We say that the equilibrium position for such reactions lies far to the right (in the direction of the products). For example, when gaseous
hydrogen and oxygen are mixed in stoichiometric quantities and react to form water
vapor, the reaction proceeds essentially to completion. The amounts of the reactants
that remain when the system reaches equilibrium are so tiny as to be negligible. By
contrast, some reactions occur only to a slight extent. For example, when solid CaO is
placed in a closed vessel at 258C, the decomposition to solid Ca and gaseous O2 is
virtually undetectable. In cases like this, the equilibrium position is said to lie far to the
left (in the direction of the reactants).
In this chapter we will discuss how and why a chemical system comes to equilibrium and the characteristics of equilibrium. In particular, we will discuss how to
calculate the concentrations of the reactants and products present for a given system
at equilibrium.
Chapter 13
Chemical Equilibrium
13.1 The Equilibrium Condition
13.5 Applications of the Equilibrium
Constant
The Characteristics of Chemical
Equilibrium
Reaction Quotient
13.3 Equilibrium Expressions Involving
Pressures
Calculating Equilibrium Pressures and
Concentrations
13.1 The Equilibrium Condition
13.6 Solving Equilibrium Problems
Treating Systems That Have Small
Equilibrium Constants
The Extent of a Reaction
13.2 The Equilibrium Constant
IBLG: See questions from
“The Equilibrium Condition and
the Equilibrium Constant”
13.7 Le Châtelier’s Principle
The Effect of a Change in Concentration
Equilibrium is a dynamic situation.
The Effect of a Change in Pressure
The Effect of a Change in Temperature
13.4 Heterogeneous Equilibria
The equilibrium in a salt water aquarium must be carefully maintained to keep the sea life healthy.
(Borissos/Dreamstime.com)
606
9/6/12 9:06 AM
The Scientific Method
interest boxes cover such topics as the invention of Post-it Notes, farming the
wind, and the use of iron metal to clean up contaminated groundwater. Additional
Chemical Connections are available on the student website.
Photo © Cengage Learning. All rights reserved.
6.6
remarkable stories connected to the
use of these notes. For example, a
Post-it Note was applied to the nose of
a corporate jet, where it was intended
to be read by the plane’s Las Vegas
ground crew. Someone forgot to
remove it, however. The note was still
on the nose of the plane when it
landed in Minneapolis, having survived
a takeoff, a landing, and speeds of
500 miles per hour at temperatures as
low as 2568F. Stories on the 3M Web
site describe how a Post-it Note on the
front door of a home survived the
140-mile-per-hour winds of Hurricane
Hugo and how a foreign official
accepted Post-it Notes in lieu of cash
when a small bribe was needed to cut
through bureaucratic hassles.
Post-it Notes have definitely
changed the way we communicate and
remember things.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch01_0001-0041.indd 7
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
New Energy Sources
277
Chemical connections
Farming the Wind
In the Midwest the wind blows across
fields of corn, soybeans, wheat, and
wind turbines—wind turbines? It turns
out that the wind that seems to blow
almost continuously across the plains is
now becoming the latest cash crop. One
of these new-breed wind farmers is
Daniel Juhl, who recently erected
17 wind turbines on six acres of land
near Woodstock, Minnesota. These
turbines can generate as much as
10 megawatts (MW) of electricity, which
Juhl sells to the local electrical utility.
There is plenty of untapped wind
power in the United States. Wind
mappers rate regions on a scale of
1 to 6 (with 6 being the best) to
indicate the quality of the wind
resource. Wind farms are now being
developed in areas rated from 4 to 6.
The farmers who own the land
welcome the increased income derived
from the wind blowing across their
land. Economists estimate that each
acre devoted to wind turbines can pay
royalties to the farmers of as much as
$8000 per year, or many times the
revenue from growing corn on that
same land. Juhl claims that farmers
who construct the turbines themselves
can realize as much as $20,000 per
year per turbine. Globally, wind
generation of electricity has nearly
quadrupled in the last five years and
is expected to increase by about 60%
per year in the United
States. The economic
feasibility of windgenerated electricity
has greatly improved in
the past 30 years as the
wind turbines have
become more efficient.
Today’s turbines can
produce electricity that
costs about the same as
that from other sources.
The most impressive
thing about wind power
is the magnitude of the
This State Line Wind Project along the Oregon–Washington border
supply. According to the uses approximately 399 wind turbines to create enough electricity to
power some 70,000 households.
American Wind Energy
Association in Washpower 1 million homes if transmission
ington, D.C., the wind-power potential
problems can be solved.
in the United States is comparable or
Another possible scenario for wind
larger than the energy resources under
farms is to use the electrical power
the sands of Saudi Arabia.
generated to decompose water to
The biggest hurdle that must be
produce hydrogen gas that could be
overcome before wind power can
carried to cities by pipelines and used
become a significant electricity
as a fuel. One real benefit of hydrogen
producer in the United States is
is that it produces water as its only
construction of the transmission
combustion product. Thus, it is
infrastructure—the power lines
essentially pollution-free.
needed to move the electricity from
Within a few years, wind power
the rural areas to the cities where
could be a major source of electricity.
most of the power is used. For
There could be a fresh wind blowing
example, the hundreds of turbines
across the energy landscape of the
planned in southwest Minnesota in a
United States in the near future.
development called Buffalo Ridge
could supply enough electricity to
9/6/12 9:02 AM
carbon dioxide. However, even though it appears that hydrogen is a very logical choice
as a major fuel for the future, there are three main problems: the cost of production, storage, and transport.
First let’s look at the production problem. Although hydrogen is very abundant on
the earth, virtually none of it exists as the free gas. Currently, the main source of hydrogen
from the treatment
of natural
gas third
with steam:
part.
Duegastois electronic
rights,
some
party content may be suppressed
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in
from the eBook and/or eChapter(s).
CH4 1g2 1 H2O 1g2 h 3H2 1g2 1 CO 1g2
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves
the right to remove additional content at any time if subsequent rights restrictions require it.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
607
9/6/12 10:57
Chemical Connections describe current applications of chemistry. These special-
A Note-able Achievement
Post-it Notes popped up. One Sunday
Art Fry, a chemical engineer for 3M,
was singing in his church choir when
he became annoyed that the bookmark in his hymnal kept falling out. He
thought to himself that it would be
nice if the bookmark were sticky
enough to stay in place but not so
sticky that it couldn’t be moved.
Luckily, he remembered Silver’s
glue—and the Post-it Note was born.
For the next three years, Fry
worked to overcome the manufacturing obstacles associated with the
product. By 1977 enough Post-it Notes
were being produced to supply 3M’s
corporate headquarters, where the
employees quickly became addicted to
their many uses. Post-it Notes are now
available in 62 colors and 25 shapes.
In the years since the introduction
of Post-it Notes, 3M has heard some
11097_Ch13_0606-0651.indd 607
7
Chemical connections
Post-it Notes, a product of the 3M
Corporation, revolutionized casual
written communications and personal
reminders. Introduced in the United
States in 1980, these sticky-but-nottoo-sticky notes have now found
countless uses in offices, cars, and
homes throughout the world.
The invention of sticky notes
occurred over a period of about 10
years and involved a great deal of
serendipity. The adhesive for Post-it
Notes was discovered by Dr. Spencer
F. Silver of 3M in 1968. Silver found
that when an acrylate polymer
material was made in a particular way,
it formed cross-linked microspheres.
When suspended in a solvent and
sprayed on a sheet of paper, this
substance formed a “sparse monolayer” of adhesive after the solvent
evaporated. Scanning electron
microscope images of the adhesive
show that it has an irregular surface, a
little like the surface of a gravel road.
In contrast, the adhesive on cellophane tape looks smooth and uniform,
like a superhighway. The bumpy
surface of Silver’s adhesive caused it
to be sticky but not so sticky to
produce permanent adhesion,
because the number of contact points
between the binding surfaces was
limited.
When he invented this adhesive,
Silver had no specific ideas for its use,
so he spread the word of his discovery
to his fellow employees at 3M to see if
anyone had an application for it. In
addition, over the next several years
development was carried out to
improve the adhesive’s properties. It
was not until 1974 that the idea for
H2O 1g2 1 CO 1g2 m H2 1g2 1 CO2 1g2
Courtesy, NextEra Energy Resources
1.2
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Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch13_0606-0651.indd 606
Since no changes occur in the concentrations of reactants or products in a reaction
system at equilibrium, it may appear that everything has stopped. However, this is not
the case. On the molecular level, there is frantic activity. Equilibrium is not static but
is a highly dynamic situation. The concept of chemical equilibrium is analogous to the
flow of cars across a bridge connecting two island cities. Suppose the traffic flow on
the bridge is the same in both directions. It is obvious that there is motion, since one
can see the cars traveling back and forth across the bridge, but the number of cars in
each city is not changing because equal numbers of cars are entering and leaving. The
result is no net change in the car population.
To see how this concept applies to chemical reactions, consider the reaction between steam and carbon monoxide in a closed vessel at a high temperature where the
reaction takes place rapidly:
xvii
Comprehensive End-of-Chapter Practice and Review We offer end-of-chapter exercises for every type of student and for every kind of homework assignment.
748
Chapter 15
Acid–Base Equilibria
Each chapter has a For Review section to reinforce key concepts and includes review questions for students to practice
independently.
For review
Key terms
Buffered solutions
Section 15.1
❯
common ion
common ion effect
❯
❯
Section 15.2
buffered solution
Henderson–Hasselbalch
equation
Section 15.3
Contains a weak acid (HA) and its salt (NaA) or a weak base (B) and its salt (BHCl)
Resists a change in its pH when H1 or OH2 is added
For a buffered solution containing HA and A2
❯ The Henderson–Hasselbalch equation is useful:
buffering capacity
❯
Section 15.4
pH curve (titration curve)
millimole (mmol)
equivalence point
(stoichiometric point)
❯
3A2 4
b
3HA 4
Buffering works because the amounts of HA (which reacts with added OH2) and A2
3A2 4
ratio does not change
3HA 4
significantly when strong acids or bases are added
(which reacts with added H1) are large enough that the
Section 15.5
acid–base indicator
phenolphthalein
pH 5 pKa 1 log a
The capacity of the buffered solution depends on the amounts of HA and A2 present
3A2 4
The most efficient buffering occurs when the
ratio is close to 1
3HA 4
❯
Acid–base titrations
❯
❯
❯
❯
The progress of a titration is represented by plotting the pH of the solution versus the volume
of added titrant; the resulting graph is called a pH curve or titration curve
Strong acid–strong base titrations show a sharp change in pH near the equivalence point
The shape of the pH curve for a strong base–strong acid titration before the equivalence point
is quite different from the shape of the pH curve for a strong base–weak acid titration
❯ The strong base–weak acid pH curve shows the effects of buffering before the equivalence point
❯ For a strong base–weak acid titration, the pH is greater than 7 at the equivalence point
because of the basic properties of A2
Indicators are sometimes used to mark the equivalence point of an acid–base titration
The end point is where the indicator changes color
The goal is to have the end point and the equivalence point be as close as possible
❯
❯
Review questions
Answers to the Review Questions can be found on the Student website.
1. What is meant by the presence of a common ion? How
does the presence of a common ion affect an equilibrium such as
HNO2 1aq2 m H1 1aq2 1 NO22 1aq2
What is an acid–base solution called that contains a
common ion?
2. Define a buffer solution. What makes up a buffer
solution? How do buffers absorb added H1 or OH2
with little pH change?
Is it necessary that the concentrations of the weak
acid and the weak base in a buffered solution be equal?
Explain. What is the pH of a buffer when the weak acid
and conjugate base concentrations are equal?
A buffer generally contains a weak acid and its
weak conjugate base, or a weak base and its weak
conjugate acid, in water. You can solve for the pH by
setting up the equilibrium problem using the Ka reaction
of the weak acid or the Kb reaction of the conjugate
base. Both reactions give the same answer for the pH of
the solution. Explain.
A third method that can be used to solve for the pH
of a buffer solution is the Henderson–Hasselbalch
equation. What is the Henderson–Hasselbalch equation? What assumptions are made when using this
equation?
3. One of the most challenging parts of solving acid–base
problems is writing out the correct equation. When a
A discussion of the Active Learning Questions can be found online in the Instructor’s Resource Guide and on
PowerLecture. The questions allow students to explore their understanding of concepts through discussion and
peer teaching. The real value of these questions is the learning that occurs while students talk to each other
about chemical concepts.
Active Learning Questions
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11097_Ch15_0711-0757.indd 748
9/6/12 9:08 AM
Active Learning Questions are designed to promote discussion among groups of students in class.
These questions are designed to be used by groups of students in
class.
1. Consider two beakers of pure water at different temperatures.
How do their pH values compare? Which is more acidic? more
basic? Explain.
2. Differentiate between the terms strength and concentration as
they apply to acids and bases. When is HCl strong? Weak?
Concentrated? Dilute? Answer the same questions for ammonia. Is the conjugate base of a weak acid a strong base?
3. Sketch two graphs: (a) percent dissociation for weak acid HA
versus the initial concentration of HA ([HA]0) and (b) H1 concentration versus [HA]0. Explain both.
4. Consider a solution prepared by mixing a weak acid HA and
HCl. What are the major species? Explain what is occurring in
solution. How would you calculate the pH? What if you added
NaA to this solution? Then added NaOH?
5. Explain why salts can be acidic, basic, or neutral, and show
examples. Do this without specific numbers.
6. Consider two separate aqueous solutions: one of a weak acid
HA and one of HCl. Assuming you started with 10 molecules
of each:
a. Draw a picture of what each solution looks like at
equilibrium.
b. What are the major species in each beaker?
c. From your pictures, calculate the Ka values of each acid.
d. Order the following from the strongest to the weakest
base: H2O, A2, Cl2. Explain your order.
7. You are asked to calculate the H1 concentration in a solution
of NaOH(aq). Because sodium hydroxide is a base, can we
say there is no H1, since having H1 would imply that the solution is acidic?
8. Consider a solution prepared by mixing a weak acid HA, HCl,
and NaA. Which of the following statements best describes
what happens?
a. The H1 from the HCl reacts completely with the A2 from
the NaA. Then the HA dissociates somewhat.
b. The H1 from the HCl reacts somewhat with the A2 from
the NaA to make HA, while the HA is dissociating. Eventually you have equal amounts of everything.
c. The H1 from the HCl reacts somewhat with the A2 from
the NaA to make HA while the HA is dissociating. Eventually all the reactions have equal rates.
d. The H1 from the HCl reacts completely with the A2 from
the NaA. Then the HA dissociates somewhat until “too
much” H1 and A2 are formed, so the H1 and A2 react to
form HA, and so on. Eventually equilibrium is reached.
Justify your choice, and for choices you did not pick, explain
what is wrong with them.
9. Consider a solution formed by mixing 100.0 mL of 0.10 M
HA (Ka 5 1.0 3 1026), 100.00 mL of 0.10 M NaA, and
100.0 mL of 0.10 M HCl. In calculating the pH for the final
solution, you would make some assumptions about the order
in which various reactions occur to simplify the calculations.
State these assumptions. Does it matter whether the reactions
actually occur in the assumed order? Explain.
For Review
701
10. A certain sodium compound is dissolved in water to liberate
Na1 ions and a certain negative ion. What evidence would you
look for to determine whether the anion is behaving as an acid
or a base? How could you tell whether the anion is a strong
base? Explain how the anion could behave simultaneously as
an acid and a base.
11. Acids and bases can be thought of as chemical opposites (acids are proton donors, and bases are proton acceptors). Therefore, one might think that Ka 5 1yKb. Why isn’t this the case?
What is the relationship between Ka and Kb? Prove it with a
derivation.
12. Consider two solutions of the salts NaX(aq) and NaY(aq) at
equal concentrations. What would you need to know to determine which solution has the higher pH? Explain how you
would decide (perhaps even provide a sample calculation).
13. What is meant by pH? True or false: A strong acid solution
always has a lower pH than a weak acid solution. Explain.
14. Why is the pH of water at 258C equal to 7.00?
15. Can the pH of a solution be negative? Explain.
16. Is the conjugate base of a weak acid a strong base? Explain.
Explain why Cl2 does not affect the pH of an aqueous solution.
17. Match the following pH values: 1, 2, 5, 6, 6.5, 8, 11, 11, and
13 with the following chemicals (of equal concentration):
HBr, NaOH, NaF, NaCN, NH4F, CH3NH3F, HF, HCN, and
NH3. Answer this question without performing calculations.
18. The salt BX, when dissolved in water, produces an acidic solution. Which of the following could be true? (There may be
more than one correct answer.)
a. The acid HX is a weak acid.
b. The acid HX is a strong acid.
c. The cation B1 is a weak acid.
Explain.
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of this book and a
solution appears in the Solutions Guide, as found on PowerLecture.
Questions
19. Anions containing hydrogen (for example, HCO32 and
H2PO42) usually show amphoteric behavior. Write equations
illustrating the amphoterism of these two anions.
20. Which of the following conditions indicate an acidic solution
at 258C?
a. pH 5 3.04
b. [H1] . 1.0 3 1027 M
c. pOH 5 4.51
d. [OH2] 5 3.21 3 10212 M
21. Which of the following conditions indicate a basic solution at
258C?
a. pOH 5 11.21
b. pH 5 9.42
c. [OH2] . [H1]
d. [OH2] . 1.0 3 1027 M
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
11097_Ch14_0652-0710.indd 701
xviii
9/6/12 9:07 AM
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
919
Comprehensive End-of-Chapter Practice and Review
90
problems associated with nuclear reactors? What are
breeder reactors? What are some problems associated
with breeder reactors?
10. The biological effects of a particular source of radiation
depend on several factors. List some of these factors.
Even though 85Kr and 90Sr are both b-particle emitters,
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of this book and a
solution appears in the Solutions Guide, as found on PowerLecture.
1. When nuclei undergo nuclear transformations, g rays of characteristic frequencies are observed. How does this fact, along with
other information in the chapter on nuclear stability, suggest
that a quantum mechanical model may apply to the nucleus?
2. There is a trend in the United States toward using coal-fired
power plants to generate electricity rather than building new
nuclear fission power plants. Is the use of coal-fired power
plants without risk? Make a list of the risks to society from the
use of each type of power plant.
3. Which type of radioactive decay has the net effect of changing
a neutron into a proton? Which type of decay has the net effect
of turning a proton into a neutron?
4. Consider the following graph of binding energy per nucleon as
a function of mass number.
Binding energy per nucleon (MeV)
16
12
8
C
56
O
34
Fe
84
Kr
119
Sn
205
S
Tl
14
N
4
7
235
U
238
U
He
6
7
Li
6
Li
5
4
3
2
1
0
decays to 176Hf, was used to estimate this age. The half-life of
176
Lu is 37 billion years. How are ratios of 176Lu to 176Hf utilized to date very old rocks?
7. Why are the observed energy changes for nuclear processes so
much larger than the energy changes for chemical and physical processes?
8. Natural uranium is mostly nonfissionable 238U; it contains
only about 0.7% of fissionable 235U. For uranium to be useful
as a nuclear fuel, the relative amount of 235U must be increased
to about 3%. This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of 238UF6(g) and 235UF6(g). The fluoride mixture is then enriched through a multistage diffusion
process to produce a 3% 235U nuclear fuel. The diffusion process utilizes Graham’s law of effusion (see Chapter 5, Section
5.7). Explain how Graham’s law of effusion allows natural
Exercises
uranium to be enriched by the gaseous
diffusion process.
9. Much of the research on controlled
fusion
focuses
on the
prob- are paired.
In this
section
similar
exercises
lem of how to contain the reacting material. Magnetic fields
Localized
Electron
Model
appear to be the most promisingThe
mode
of containment.
Why
is and Hybrid Orbitals
containment such a problem? Why must one resort to mag17. Use the localized electron model to describe the bonding in
netic fields for containment?
H2O.
10. A recent study concluded that any amount of radiation expo18. Use the localized electron model to describe the bonding in
sure can cause biological damage. Explain the differences beCCl4.
tween the two models of radiation damage, the linear model
19. Use the localized electron model to describe the bonding in
and the threshold model.
H2CO (carbon is the central atom).
20. Use the localized electron model to describe the bonding in
Exercises
C2H2 (exists as HCCH).
In this section similar exercises are paired.
21. The space-filling models of ethane and ethanol are shown
Questions are homework problems directed at concepts
Questions
9
the dangers associated with the decay of Sr are much
greater than those linked to 85Kr. Why? Although g rays
are far more penetrating than a particles, the latter are
more likely to cause damage to an organism. Why?
Which type of radiation is more effective at promoting
the ionization of biomolecules?
3
H
3
within the chapter and in general don’t require calculation.
For Review
30. For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid
orbitals for sulfur.
a. SO2
b. SO3
2−
c.
O
S2O32−
H
20 40 60 80 100 120 140 160 180 200 220 240 260
Mass number (A)
a. What does this graph tell us about the relative half-lives
of the nuclides? Explain your answer.
b. Which nuclide shown is the most thermodynamically stable? Which is the least thermodynamically stable?
c. What does this graph tell us about which nuclides
undergo fusion and which undergo fission to become
more stable? Support your answer.
5. What are transuranium elements and how are they synthesized?
6. Scientists have estimated that the earth’s crust was formed
4.3 billion years ago. The radioactive nuclide 176Lu, which
11. Write an equation describing the radioactive decay of each of
C
the following nuclides. (The particle produced is shown in paH
rentheses, except for electron capture, where an electron is a
Ethane
Ethanol
reactant.)
O
(C2H6)
(C2H5OH)
a. 31H (b)
Use the localized electron model to describe the bonding in
b. 83Li (b followed by a)
ethane and ethanol.
c. 74Be (electron capture)
22. The space-filling models of hydrogen cyanide and phosgene
d. 85B (positron)
shown
below. supply
12. In each of the following radioactive are
decay
processes,
the missing particle.
a.
60
Co S 60Ni 1 ?
b.
97
Tc 1 ? S 97Mo
c.
99
d.
239
C
H
Hydrogen cyanide
(HCN)
99
Tc S Ru 1 ?
Pu S 235U 1 ?
d.
11097_Ch19_0890-0925.indd 919
ChemWork Problems
These multiconcept problems (and additional ones) are found interactively online with the same type of assistance a student would get
from an instructor.
95. Which of the following reactions (or processes) are expected
to have a negative value for DS8?
a. SiF6 1aq2 1 H2 1g2 h 2HF1g2 1 SiF4 1g2
b. 4Al1s2 1 3O2 1g2 h 2Al2O3 1s2
c. CO 1g2 1 Cl2 1g2 h COCl2 1g2
d. C2H4 1g2 1 H2O 1l2 h C2H5OH 1l2
e. H2O 1s2 h H2O 1l2
96. For rubidium DH8vap 5 69.0 kJ/mol at 6868C, its boiling point.
Calculate DS8, q, w, and DE for the vaporization of 1.00 mole
of rubidium at 6868C and 1.00 atm pressure.
97. Given the thermodynamic data below, calculate DS and DSsurr
for the following reaction at 258C and 1 atm:
XeF6 1g2 h XeF4 1s2 1 F2 1g2
XeF6(g)
XeF4(s)
F2(g)
DH8f (kJ/mol)
S8 (J/K ? mol)
2294
2251
0
300.
146
203
98. Consider the reaction:
O
S
O
O
2−
S2O8
O
S
e.
f.
g.
h.
i.
j.
k.
2−
O
O
S
O
O
O
O
SO322
SO422
SF2
SF4
SF6
F3SOSF
SF51
31. Why must all six atoms in C2H4 lie in the same plane?
32. The allene molecule has the following Lewis structure:
H
H
N
C
Cl
C
C
H
H
Use the localized electron model to describe the bonding in
hydrogen cyanide and phosgene.
Must all hydrogen atoms lie the same plane? If not, what is
their spatial relationship? Explain.
23. Give the expected hybridization of the central atom for the
molecules or ions in Exercises 83 and 89 from Chapter 8.
24. Give the expected hybridization of the central atom for the
molecules or ions in Exercises 84 and 90 from Chapter 8.
33. Indigo is the dye used in coloring blue jeans. The term navy
blue is derived from the use of indigo to dye British naval
uniforms in the eighteenth century. The structure of the indigo
molecule is
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
There are numerous Exercises to reinforce students’ understanding of each section. These problems are paired and organized by topic so that
instructors can review them in class and assign
them for homework.
Phosgene
(COCl2)
S
O
Radioactive Decay and Nuclear Transformations
below.
He
2
9/6/12 9:01 AM
25. Give the expected hybridization of the central atom for the
molecules or ions in Exercise 87 from Chapter 8.
26. Give the expected hybridization of the central atom for the
molecules in Exercise 88 from Chapter 8.
27. Give the expected hybridization of the central atom for the
molecules in Exercises 113 and 114 from Chapter 8.
28. Give the expected hybridization of the central atom for the
molecules in Exercises 115 and 116 from Chapter 8.
29. For each of the following molecules, write the Lewis
structure(s), predict the molecular structure (including bond
angles), give the expected hybrid orbitals on the central atom,
and predict the overall polarity.
a. CF4
e. BeH2
i. KrF4
b. NF
f. TeF4
j. SeF6
For3 Review
829
c. OF2
g. AsF5
k. IF5
d. BF
h. KrF2
l. IF3
e. When DG8 for this reaction is negative,3 then Kp is greater
than 1.00.
102. The equilibrium constant for aUnless
certain
reaction
increases
by isa© Cengage Learning 2014.
otherwise
noted, all
art on this page
factor of 6.67 when the temperature is increased from 300.0 K
to 350.0 K. Calculate the standard change in enthalpy (DH8)
for this reaction (assuming DH8 is temperature-independent).
Challenge Problems
11097_Ch09_0415-0452.indd
H
H
C
H
C
C
C
C
C
H
O
H
C
N
C
C
N
C
H
O
H
C
C
C
C
H
C
C
H
H
a. How many s bonds and p bonds exist in the molecule?
b. What hybrid orbitals are used by the carbon atoms in the
indigo molecule?
34. Urea, a compound formed in the liver, is one of the ways humans excrete nitrogen. The Lewis structure for urea is
H
H
O
H
N
C
N
H
Using hybrid orbitals for carbon, nitrogen, and oxygen, determine which orbitals overlap to form the various bonds in urea.
445
9/6/12 9:08 AM
103. Consider two perfectly insulated vessels. Vessel 1 initially
contains an ice cube at 08C and water at 08C. Vessel 2 initially
contains an ice cube at 08C and a saltwater solution at 08C.
Consider the process H2O 1s2 S H2O 1l2 .
a. Determine the sign of DS, DSsurr, and DSuniv for the process in vessel 1.
b. Determine the sign of DS, DSsurr, and DSuniv for the process in vessel 2.
(Hint: Think about the effect that a salt has on the freezing
point of a solvent.)
104. Liquid water at 258C is introduced into an evacuated, insulated
vessel. Identify the signs of the following thermodynamic functions for the process that occurs: DH, DS, DTwater, DSsurr, DSuniv.
105. Using data from Appendix 4, calculate DH8, DG8, and K (at
298 K) for the production of ozone from oxygen:
New ChemWork end-of-chapter problems are now included, with many additional problems available to assign online for more practice.
3O2 1g2 m 2O3 1g2
At 30 km above the surface of the earth, the temperature
is about 230. K and the partial pressure of oxygen is about
Unless otherwise noted, all art on this page is © Cengage Learning 2014. 1.0 3 1023 atm. Estimate the partial pressure of ozone in equifor which DH is 2233 kJ and DS is 2424 J/K.
librium with oxygen at 30 km above the earth’s surface. Is it
a. Calculate the free energy change for the reaction (DG) at
reasonable to assume that the equilibrium between oxygen
393 K.
and ozone is maintained under these conditions? Explain.
b. Assuming
DHAll
andRights
DS do not
depend on
temperature,
at
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Learning.
Reserved.
May
not be copied,
scanned,
or duplicated,
whole orbyin apart.
Due to electronic
some third party content may be suppressed from the eBook and/or eChapter(s).
106. Entropy
can be in
calculated
relationship
proposed rights,
by
what temperatures is this reaction spontaneous?
Ludwig
Boltzmann:
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learning
experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
99. The following reaction occurs in pure water:
H2S1g2 1 SO2 1g2 h 3S1g2 1 2H2O 1l2
445
xix
Wealth of End-of-Chapter Problems The text offers an unparalleled variety of end-ofchapter content with problems that increase in rigor and integrate multiple concepts.
For Review
10 ) and the over(1.0 3 1013), calcuant for the following
21
98. The Hg
219
1aq2 1 2OH2 1aq2
ant you calculated
mol/L) of Cu(OH)2
ntration of OH2 is
s in each of the fol-
4 say that Ba(OH)2,
soluble hydroxides.
f each of these mar-
ormation, we ran the
then let some of the
ibrium. To see why,
Equilibrium
entration (mol/L)
5 3.75 3 1023 2 y
6.25 3 1022 2 2y
nes) are found intera student would get
)2(s) is 1.3 3 10232
lt. Ignore any poten-
east soluble to most
the ions with water.
ate the solubility of
21
785
ion forms complex ions with I as follows:
For Review
2
Hg21 1aq2 1 I2 1aq2 m HgI1 1aq2
K1 5 1.0 3 108
HgI1 1aq2 1 I2 1aq2 m HgI2 1aq2
K2 5 1.0 3 105
2
2
HgI2 1aq2 1 I 1aq2 m HgI3 1aq2
K3 5 1.0 3 109
HgI32 1aq2 1 I2 1aq2 m HgI422 1aq2
K4 5 1.0 3 108
A solution is prepared by dissolving 0.088 mole of Hg(NO3)2
and 5.00 mole of NaI in enough water to make 1.0 L of
solution.
a. Calculate the equilibrium concentration of [HgI422].
b. Calculate the equilibrium concentration of [I2].
c. Calculate the equilibrium concentration of [Hg21].
a. What­fraction­of­the­moles­of­NaCl­in­this­solution­exist­
as­ion­pairs?
b. Calculate­the­freezing­point­that­would­be­observed­for­
this­solution.
123. The­ vapor­ in­ equilibrium­ with­ a­ pentane–hexane­ solution­ at­
258C­ has­ a­ mole­ fraction­ of­ pentane­ equal­ to­ 0.15­ at­ 258C.­
What­is­the­mole­fraction­of­pentane­in­the­solution?­(See­Exercise­57­for­the­vapor­pressures­of­the­pure­liquids.)
124. A­forensic­chemist­is­given­a­white­solid­that­is­suspected­of­
being­pure­cocaine­(C17H21NO4,­molar­mass­5­303.35­g/mol).­
She­ dissolves­ 1.22­ 6­ 0.01­ g­ of­ the­ solid­ in­ 15.60­ 6­ 0.01­ g­
benzene.­The­freezing­point­is­lowered­by­1.32­6­0.048C.
Challenge Problems
a. What­is­the­molar­mass­of­the­substance?­Assuming­that­
99. The copper(I) ion forms a complex ion with CN2 according to
the­percent­uncertainty­in­the­calculated­molar­mass­is­the­
the following equation:
same­as­the­percent­uncertainty­in­the­temperature­
1
2
11
22
K 5 1.0 3 10
Cu 1aq2 1 3CN 1aq2 m Cu 1CN2 3 1aq2
change,­calculate­the­uncertainty­in­the­molar­mass.
b. Could­the­chemist­unequivocally­state­that­the­substance­
a. Calculate the solubility of CuBr(s) (Ksp 5 1.0 3 1025) in
is­cocaine?­For­example,­is­the­uncertainty­small­enough­
1.0 L of 1.0 M NaCN.
to­distinguish­cocaine­from­codeine­(C18H21NO3,­molar­
b. Calculate the concentration
of
Br2 at equilibrium.
188
Chapter
4 Types of Chemical Reactions and Solution Stoichiometry
mass­5­299.36­g/mol)?
c. Calculate the concentration of CN2 at equilibrium.
c. Assuming­that­the­absolute­uncertainties­in­the­measure100. Consider a solution made by
mixing
500.0 mL of 4.0
M NH3the molarity of the original
here. If the percent yield of the reaction
was 88.0%, what mass
HCl
for 1
neutralization.
Calculate
ments­of­temperature­and­mass­remain­unchanged,­how­
and 500.0 mL of 0.40 M AgNO
reacts
with NH3 to form
3. Ag
of chromium(III) chromate wascould­the­chemist­improve­the­precision­of­her­results?
isolated?
sample
of H
2SO4. Sulfuric acid has two acidic hydrogens.
AgNH31 and Ag(NH3)21:
142. The vanadium in a sample
of ore is converted to VO21. The
136. A 6.50-g sample of a diprotic acid requires 137.5 mL of a
125. A­1.60-g­sample­of­a­mixture­of­naphthalene­(C
10H8)­and­an1
3
1
2
1aq2NaOH solution
Ag 1aq2 1 NH3 1aq2 m AgNH
K1 5 2.1
10
0.7503 M
for3complete
neutralization. DeterVO21 ion is subsequently titrated
with 14
MnO
thracene­(C
H10)­is­dissolved­in­20.0­g­benzene­(C
4 in acidic solu6H6).­The­
tion to form V(OH)41 and manganese(II)
ion. The unbalanced
freezing­point­of­the­solution­is­2.818C.­What­is­the­composi1NH3the
2 21molar
1aq2 mass
AgNH31 1aq2 1 NH3 1aq2 m Agmine
K2 of
5 the
8.2 acid.
3 103
titration reaction is
tion­as­mass­percent­of­the­sample­mixture?­The­freezing­point­
137. Citric
which
can be obtained from lemon juice, has the
Determine the concentration
of allacid,
species
in solution.
molecular formula C6H8O7. A 0.250-g sample of citric acid
f­is­5.128C­?­kg/mol.
MnO42 1aq2 1 VO21 1aq2 1of­benzene­is­5.518C­and­K
H2O 1l2 h
101. a. Calculate the molar solubility of AgBr in pure water. Ksp
21
1 ­and­NaCl.­When­0.5000­g­of­
1
dissolved in 25.0 mL of water requires 37.2 mL of 0.105 M
126.
A­solid­mixture­contains­MgCl
V 1OH2
2
4 1aq2 1 Mn 1aq2 1 H 1aq2
for AgBr is 5.0 3 10213.
NaOH for complete neutralization. What number of acidic hythis­solid­is­dissolved­in­enough­water­to­form­1.000­L­of­soluTo titrate the solution, 26.45 mL of 0.02250 M MnO42 was
b. Calculate the molar solubility
of AgBr
in 3.0 Mdoes
NH3.citric
The acid have?
drogens
per molecule
tion,­the­osmotic­pressure­at­25.08C­is­observed­to­be­0.3950­
required.
If
the
mass
percent
of
vanadium
in
the
ore
was
overall formation constant for Ag(NH3)21 is 1.7 3 107,
atm.­What­is­the­mass­percent­of­MgCl2­in­the­solid?­(Assume­
138. A stream flows at a rate of 5.00 3 104 liters per second (L/s)
58.1%, what was the mass of the ore sample? Hint: Balance
that is,
ideal­­behavior­for­the­solution.)
upstream of a manufacturing plant. The plant discharges
the titration reaction by the oxidation states method.
31 1aq2
3 10
of water
that3contains
Ag1 1aq2 1 2NH3 1aq2 h 3.50
Ag 1NH
K 5 1.7
107. 65.0 ppm HCl into the
127. Formic­acid­(HCO2H)­is­a­monoprotic­acid­that­ionizes­only­
32 2 L/s
143. The unknown acid H2X can be neutralized completely by
stream.
(See
Exercise
121
for
definitions.)
partially­in­aqueous­solutions.­A­0.10-M­formic­acid­solution­
c. Compare the calculated solubilities from parts a and b.
OH2 according to the following (unbalanced) equation:
is­4.2%­ionized.­Assuming­that­the­molarity­and­molality­of­
Explain any differences.a. Calculate the stream’s total flow rate downstream from
H2X 1aq2 1 OH2 1aq2the­solution­are­the­same,­calculate­the­freezing­point­and­the­
h X22 1aq2 1 H2O 1l2
this
plant.
d. What mass of AgBr will dissolve in 250.0 mL of 3.0 M
boiling­point­of­0.10­M­formic­acid.
The ion formed as a product,
X22, was shown to have 36 total
b. Calculate the concentration of HCl in ppm downstream
NH3?
128. You­have­a­solution­of­two­volatile­liquids,­A­and­B­(assume­
electrons. What is element
X? Propose a name for H2X. To
fromhave
this on
plant.
e. What effect does adding HNO
the solubilities
3
ideal­behavior).­Pure­liquid­A­has­a­vapor­pressure­of­350.0­torr­
completely neutralize a sample
of H2X, 35.6 mL of 0.175 M
calculated in parts a andc.b? Further downstream, another manufacturing plant diverts
4
pure­
liquid­
has­of
a­ vapor­
of­ 100.0­ torr­ at­ the­
OH2 solution was required.and­
What
was
the B­
mass
the H2pressure­
X
1.80
3
10
L/s
of
water
from
21the stream for its own use.
102. Calculate the equilibrium concentrations of NH3, Cu ,
sample used?
This) plant
must first neutralize
21
21
Cu(NH3)21, Cu(NH3)221, Cu(NH
in a the acid and does so by
3 3 , and Cu(NH3)4
addingmL
lime:
solution prepared by mixing 500.0
of 3.00 M NH with
Challenge Problems take students one step
further and challenge them more rigorously than
the Additional Exercises.
Integrative Problems combine concepts from
multiple chapters.
3
CaO
2H1 1aq2equilib500.0 mL of 2.00 3 1023 M Cu(NO
The1stepwise
h Ca21 1aq2 1 H2O 1l2
3)2. 1s2
ria are
What mass of CaO is consumed in an 8.00-h work day by
this CuNH
plant? 21 1aq2
Cu21 1aq2 1 NH3 1aq2 m
3
d. The original stream
water3contained
10.2 ppm Ca21.
K1 5 1.86
104
Although no calcium
was in the waste water from the first
21
21
1aq2
1aq2
1NH
CuNH3
1 NH3
m Cu
32 2 1aq2
plant, the waste
water of the second
plant
contains Ca21
K2 5 3.88 3 103
from the neutralization
process. If 90.0% of the water
2 321 1aq2plant is returned to the stream, calcuCu 1NH32 221 1aq2 1 NH3 1aq2 m
usedCu
by1NH
the3second
21 3
late the concentration
of 3
Ca10
in ppm downstream of the
K3 5 1.00
second
plant.2 21 1aq2
Cu 1NH32 321 1aq2 1 NH3 1aq2 m
Cu 1NH
3 4
2
139. It took 25.06 60.05
mL
of
a
sodium
hydroxide solution to tiK4 5 1.55 3 10
trate a 0.4016-g sample of KHP (see Exercise 77). Calculate
the concentration and uncertainty in the concentration of the
sodium hydroxide solution. (See Appendix 1.5.) Neglect any
uncertainty in the mass.
022-M KIO3 solution
ue for Pb(IO3)2(s).
) is added to 50.0 mL
at equilibrium in the
7 3 1028.]
014.
Integrative Problems
These problems require the integration
of multiple
concepts to find
9/6/12
9:11 AM
the solutions.
140. Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of
ethylene or propylene in the presence of a catalytic transition
metal compound. It is composed solely of C, F, and B; it is
42.23% C and 55.66% F by mass.
a. What is the empirical formula of BARF?
b. A 2.251-g sample of BARF dissolved in 347.0 mL of
solution produces a 0.01267-M solution. What is the
molecular formula of BARF?
141. In a 1-L beaker, 203 mL of 0.307 M ammonium chromate was
mixed with 137 mL of 0.269 M chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write
the balanced chemical equation for the reaction occurring
551
t­emperature­ of­ the­ solution.­The­ vapor­ at­ equilibrium­ above­
the­solution­has­double­the­mole­fraction­of­substance­A­that­
the­solution­does.­What­is­the­mole­fraction­of­liquid­A­in­the­
solution?
129. In­some­regions­of­the­southwest­United­States,­the­water­is­
very­hard.­For­example,­in­Las­Cruces,­New­Mexico,­the­tap­
water­contains­about­560­mg­of­dissolved­solids­per­milliliter.­
Reverse­ osmosis­ units­ are­ marketed­ in­ this­ area­ to­ soften­­
water.­A­­typical­unit­exerts­a­pressure­of­8.0­atm­and­can­produce­45­L­­­water­per­day.
a. Assuming­all­of­the­dissolved­solids­are­MgCO3­and­
assuming­a­temperature­of­278C,­what­total­volume­of­
water­must­be­processed­to­produce­45­L­pure­water?
b. Would­the­same­system­work­for­purifying­seawater?­
(Assume­seawater­is­0.60­M­NaCl.)
Integrative Problems
These­problems­require­the­integration­of­multiple­concepts­to­find­
the­solutions.
130. Creatinine,­C4H7N3O,­is­a­by-product­of­muscle­metabolism,­
and­ creatinine­ levels­ in­ the­ body­ are­ known­ to­ be­ a­ fairly­
­reliable­indicator­of­kidney­function.­The­normal­level­of­creatinine­ in­ the­ blood­ for­ adults­ is­ approximately­ 1.0­ mg­ per­
deciliter­(dL)­of­blood.­If­the­density­of­blood­is­1.025­g/mL,­
calculate­the­molality­of­a­normal­creatinine­level­in­a­10.0-mL­
blood­sample.­What­is­the­osmotic­pressure­of­this­solution­at­
25.08C?
131. An­ aqueous­ solution­ containing­ 0.250­ mole­ of­ Q,­ a­ strong­
electrolyte,­in­5.00­3­102­g­water­freezes­at­22.798C.­What­is­
the­van’t­Hoff­factor­for­Q?­The­molal­freezing-point­depression­constant­for­water­is­1.868C­?­kg/mol.­What­is­the­formula­
of­Q­if­it­is­38.68%­chlorine­by­mass­and­there­are­twice­as­
many­anions­as­cations­in­one­formula­unit­of­Q?
132. Anthraquinone­contains­only­carbon,­hydrogen,­and­oxygen.­
When­ 4.80­ mg­ anthraquinone­ is­ burned,­ 14.2­ mg­ CO2­ and­
1.65­mg­H2O­are­produced.­The­freezing­point­of­camphor­is­
lowered­by­22.38C­when­1.32­g­anthraquinone­is­dissolved­in­
11.4­g­camphor.­Determine­the­empirical­and­molecular­formulas­of­anthraquinone.
Marathon Problems
These problems are designed to incorporate several concepts and
techniques into one situation.
144. Three students were asked to find the identity of the metal in a
particular sulfate salt. They dissolved a 0.1472-g sample of the
salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate
had been filtered and dried, it weighed 0.2327 g.
Each student analyzed the data independently and came to
different conclusions. Pat decided that the metal was titanium.
Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt?
Look for information on the sulfates of gallium, sodium, and
titanium in this text and reference books such as the CRC
Handbook of Chemistry and Physics. What further tests would
you suggest to determine
student
correct?
Unlesswhich
otherwise
noted, allis
artmost
on thislikely
page is ©
Cengage Learning 2014.
145. You have two 500.0-mL aqueous solutions. Solution A is a
solution of a metal nitrate that is 8.246% nitrogen by mass.
The ionic compound in solution B consists of potassium, chromium, and oxygen; chromium has an oxidation state of 16
and there are 2 potassiums and 1 chromium in the formula.
11097_Ch11_0510-0551.indd 551
The masses of the solutes in each of the solutions are the same.
When the solutions are added together, a blood-red precipitate
forms. After the reaction has gone to completion, you dry the
solid and find that it has a mass of 331.8 g.
a. Identify the ionic compounds in solution A and solution B.
b. Identify the blood-red precipitate.
c. Calculate the concentration (molarity) of all ions in the
original solutions.
d. Calculate the concentration (molarity) of all ions in the
final solution.
Marathon Problems also combine concepts from multiple chapters; they are the
most challenging problems in the end-ofchapter material.
9/6/12 9:10 AM
Marathon Problems can be used in class by groups of students to
help facilitate problem-solving skills.
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11097_Ch04_0138-0188.indd 188
9/6/12 9:00 AM
“The end-of-chapter content helps students identify and review the central concepts.
There is an impressive range of problems that are well graded by difficulty.”
—Alan M. Stolzenberg, West Virginia University
xx
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
About the Authors
Steven S. Zumdahl earned a B.S. in Chemistry from
Wheaton College (IL) and a Ph.D. from the University of
Illinois, Urbana-Champaign. He has been a faculty member
at the University of Colorado–Boulder, Parkland College
(IL), and the University of Illinois at Urbana-Champaign
(UIUC), where he is Professor Emeritus. He has received
numerous awards, including the National Catalyst Award for
Excellence in Chemical Education, the University of Illinois
Teaching Award, the UIUC Liberal Arts and Sciences Award
for Excellence in Teaching, UIUC Liberal Arts and Sciences Advising Award, and the School of Chemical Sciences
Teaching award (five times). He is the author of several
chemistry textbooks. In his leisure time he enjoys traveling
and collecting classic cars.
Susan A. Zumdahl earned a B.S. and M.A. 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 at Urbana-Champaign, 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 programs for science teachers from
elementary through college levels. These programs encourage and support active learning and creative techniques for
teaching science. For several years she was director of
an Institute for Chemical Education (ICE) field center in
Southern California, and she has authored several chemistry textbooks. Susan spearheaded the development of a
sophisticated web-based electronic homework system for
teaching chemistry. She enjoys traveling, classic cars, and
gardening in her spare time—when she is not playing with
her grandchildren.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xxi
Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chapter 1
Chemical Foundations
1.1
Chemistry: An Overview
Science: A Process for Understanding
Nature and Its Changes
1.2
1.5
The Scientific Method
Scientific Models
1.3
1.4
1.6
Units of Measurement
Uncertainty in Measurement
1.7
Dimensional Analysis
Precision and Accuracy
1.8
Temperature
S ignificant Figures and
Calculations
1.9
Density
Learning to Solve Problems
Systematically
1.10 Classification of Matter
A high-performance race car uses chemistry for its structure, tires, and fuel. (© Maria Green/Alamy)
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1
W
hen you start your car, do you think about chemistry? Probably not, but you
should. The power to start your car is furnished by a lead storage battery.
How does this battery work, and what does it contain? When a battery goes dead, what
does that mean? If you use a friend’s car to “jump-start” your car, did you know that
your battery could explode? How can you avoid such an unpleasant possibility? What
is in the gasoline that you put in your tank, and how does it furnish energy to your car
so that you can drive it to school? What is the vapor that comes out of the exhaust pipe,
and why does it cause air pollution? Your car’s air conditioner might have a substance
in it that is leading to the destruction of the ozone layer in the upper atmosphere. What
are we doing about that? And why is the ozone layer important anyway?
All of these questions can be answered by understanding some chemistry. In fact,
we’ll consider the answers to all of these questions in this text.
Chemistry is around you all the time. You are able to read and understand this sentence because chemical reactions are occurring in your brain. The food you ate for
breakfast or lunch is now furnishing energy through chemical reactions. Trees and
grass grow because of chemical changes.
Chemistry also crops up in some unexpected places. When archaeologist Luis
­Alvarez was studying in college, he probably didn’t realize that the chemical elements
iridium and niobium would make him very famous when they helped him solve the
problem of the disappearing dinosaurs. For decades scientists had wrestled with the
mystery of why the dinosaurs, after ruling the earth for millions of years, suddenly
became extinct 65 million years ago. In studying core samples of rocks dating back to
that period, Alvarez and his coworkers recognized unusual levels of iridium and niobium in these samples—levels much more characteristic of extraterrestrial bodies than
of the earth. Based on these ­observations, Alvarez hypothesized that a large meteor hit
the earth 65 million years ago, changing atmospheric conditions so much that the
dinosaurs’ food couldn’t grow, and they died—almost instantly in the geologic
timeframe.
Chemistry is also important to historians. Did you realize that lead poisoning probably was a significant contributing factor to the decline of the Roman Empire? The
Romans had high exposure to lead from lead-glazed pottery, lead water pipes, and a
sweetening syrup called sapa that was prepared by boiling down grape juice in leadlined vessels. It turns out that one reason for sapa’s sweetness was lead acetate (“sugar
of lead”), which formed as the juice was cooked down. Lead poisoning, with its symptoms of lethargy and mental malfunctions, certainly could have contributed to the demise of the Roman society.
Chemistry is also apparently very important in determining a person’s behavior.
Various studies have shown that many personality disorders can be linked directly
to imbalances of trace elements in the body. For example, studies on the inmates at
State­ville Prison in Illinois have linked low cobalt levels with violent behavior. Lithium salts have been shown to be very effective in controlling the effects of manicdepressive disease, and you’ve probably at some time in your life felt a special “chemistry” for another person. Studies suggest there is literally chemistry going on between
two people who are attracted to each other. “Falling in love” apparently causes changes
in the chemistry of the brain; chemicals are produced that give that “high” associated
with a new relationship. Unfortunately, these chemical effects seem to wear off over
time, even if the relationship persists and grows.
The importance of chemistry in the interactions of people should not really surprise
us. We know that insects communicate by emitting and receiving chemical signals
via molecules called pheromones. For example, ants have a very complicated set of
chemical signals to signify food sources, danger, and so forth. Also, various female sex
2
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1.1
3
Chemistry: An Overview
attractants have been isolated and used to lure males into traps to control insect populations. It would not be surprising if humans also emitted chemical signals that we
were not aware of on a conscious level. Thus chemistry is pretty interesting and pretty
­important. The main goal of this text is to help you understand the concepts of chemistry so that you can better appreciate the world around you and can be more effective
in ­whatever career you choose.
1.1 Chemistry: An Overview
Lawrence Berkeley National Laboratory/MCT
Lawrence Livermore Laboratory/Science Photo Library/Photo
Researchers, Inc.
Since the time of the ancient Greeks, people have wondered about the answer to the
question: What is matter made of? For a long time, humans have believed that matter is
composed of atoms, and in the previous three centuries, we have collected much indirect evidence to support this belief. Very recently, something exciting has happened—
for the first time we can “see” individual atoms. Of course, we cannot see atoms with
the naked eye; we must use a special microscope called a scanning tunneling microscope (STM). Although we will not consider the details of its operation here, the STM
uses an electron current from a tiny needle to probe the surface of a substance. The
STM pictures of ­several substances are shown in Fig. 1.1. Notice how the atoms are
connected to one another by “bridges,” which, as we will see, represent the electrons
that interconnect atoms.
So, at this point, we are fairly sure that matter consists of individual atoms. The
­nature of these atoms is quite complex, and the components of atoms don’t behave
much like the objects we see in the world of our experience. We call this world the
macroscopic world—the world of cars, tables, baseballs, rocks, oceans, and so forth.
One of the main jobs of a scientist is to delve into the macroscopic world and discover
its “parts.” For ­example, when you view a beach from a distance, it looks like a continuous solid substance. As you get closer, you see that the beach is really made up of
individual grains of sand. As we examine these grains of sand, we find that they are
composed of silicon and oxygen atoms connected to each other to form intricate shapes
(Fig. 1.2). One of the main challenges of chemistry is to understand the connection
between the macroscopic world that we experience and the microscopic world of
atoms and molecules. To truly understand chemistry, you must learn to think on the
atomic level. We will spend much time in this text helping you learn to do that.
Figure 1.1 | Scanning tunneling
microscope images.
An image showing the individual carbon
atoms in a sheet of graphene.
Scanning tunneling microscope image
of DNA.
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4
Chapter 1
Chemical Foundations
Figure 1.2 | Sand on a beach looks
Chuck Place. Inset: Jeremy Burgess/SPL/Photo Researchers, Inc.
uniform from a distance, but up close
the irregular sand grains are visible,
and each grain is composed of tiny
atoms.
O
Si
Critical Thinking
The scanning tunneling microscope allows us to “see” atoms. What if you were sent
back in time before the invention of the scanning tunneling microscope? What
evidence could you give to support the theory that all matter is made of atoms and
molecules?
One of the amazing things about our universe is that the tremendous variety of
substances we find there results from only about 100 different kinds of atoms. You can
think of these approximately 100 atoms as the letters in an alphabet from which all the
“words” in the universe are made. It is the way the atoms are organized in a given
substance that determines the properties of that substance. For example, water, one of
the most common and important substances on the earth, is composed of two types of
atoms: hydrogen and ­oxygen. Two hydrogen atoms and one oxygen atom are bound
together to form the water molecule:
oxygen atom
water molecule
hydrogen atom
When an electric current passes through it, water is decomposed to hydrogen and oxygen. These chemical elements themselves exist naturally as diatomic (two-atom)
molecules:
oxygen molecule
written O2
hydrogen molecule
written H2
We can represent the decomposition of water to its component elements, hydrogen and
oxygen, as follows:
two water
molecules
written 2H2O
electric
current
one oxygen molecule
written O2
two hydrogen molecules
written 2H2
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1.2
The Scientific Method
5
Notice that it takes two molecules of water to furnish the right number of oxygen and
hydrogen atoms to allow for the formation of the two-atom molecules. This reaction
­explains why the battery in your car can explode if you jump-start it improperly. When
you hook up the jumper cables, current flows through the dead battery, which contains
water (and other things), and causes hydrogen and oxygen to form by decomposition of
some of the water. A spark can cause this accumulated hydrogen and oxygen to explode,
forming water again.
O2
spark
2H2O
2H2
This example illustrates two of the fundamental concepts of chemistry: (1) Matter is
composed of various types of atoms, and (2) one substance changes to another by reorganizing the way the atoms are attached to each other.
These are core ideas of chemistry, and we will have much more to say about them.
Science: A Process for Understanding
Nature and Its Changes
How do you tackle the problems that confront you in real life? Think about your trip
to school. If you live in a city, traffic is undoubtedly a problem you confront daily. How
do you decide the best way to drive to school? If you are new in town, you first get a
map and look at the possible ways to make the trip. Then you might collect information about the advantages and disadvantages of various routes from people who know
the area. Based on this information, you probably try to predict the best route. However, you can find the best route only by trying several of them and comparing the results. After a few experiments with the various possibilities, you probably will be able
to select the best way. What you are doing in solving this everyday problem is applying
the same process that scientists use to study nature. The first thing you did was collect
relevant data. Then you made a prediction, and then you tested it by trying it out. This
process contains the fundamental elements of science.
1. Making observations (collecting data)
2. Suggesting a possible explanation (formulating a hypothesis)
3. Doing experiments to test the possible explanation (testing the hypothesis)
Scientists call this process the scientific method. We will discuss it in more detail in the
next section. One of life’s most important activities is solving problems—not “plug and
chug” exercises, but real problems—problems that have new facets to them, that involve
things you may have never confronted before. The more creative you are at solving these
problems, the more effective you will be in your career and your personal life. Part of the
reason for learning chemistry, therefore, is to become a better problem solver. Chemists
are usually excellent problem solvers because to master chemistry, you have to master
the scientific approach. Chemical problems are frequently very complicated—there is
usually no neat and tidy solution. Often it is difficult to know where to begin.
1.2 The Scientific Method
IBLG: See questions from “Chemistry:
An Overview and the Scientific Method”
Science is a framework for gaining and organizing knowledge. Science is not simply a
set of facts but also a plan of action—a procedure for processing and understanding
­certain types of information. Scientific thinking is useful in all aspects of life, but in
this text we will use it to understand how the chemical world operates. As we said in
our ­previous discussion, the process that lies at the center of scientific inquiry is called
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6
Chapter 1
Chemical Foundations
the scientific method. There are actually many scientific methods, depending on the
nature of the ­specific problem under study and the particular investigator involved.
However, it is useful to consider the following general framework for a generic scientific method (Fig. 1.3):
Observation
Hypothesis
Experiment
Steps in the Scientific Method
Theory (model)
1. Making observations. Observations may be qualitative (the sky is blue; water is a
liquid) or quantitative (water boils at 1008C; a certain chemistry book weighs 2 kg).
A qualitative observation does not involve a number. A quantitative observation
(called a measurement) involves both a number and a unit.
2. Formulating hypotheses. A hypothesis is a possible explanation for an observation.
Theory
modified
as needed
Prediction
3. Performing experiments. An experiment is carried out to test a hypothesis. This involves gathering new information that enables a scientist to decide whether the
hypothesis is valid—that is, whether it is supported by the new information
learned from the experiment. Experiments always produce new observations, and
this brings the process back to the beginning again.
Experiment
Figure 1.3 | The fundamental steps
of the scientific method.
To understand a given phenomenon, these steps are repeated many times, gradually
accumulating the knowledge necessary to provide a possible explanation of the
phenomenon.
Scientific Models
Observation
Hypothesis
Experiment
Theory
(model)
Theory
modified
as needed
Law
Prediction
Experiment
Figure 1.4 | The various parts of the
scientific method.
Once a set of hypotheses that agrees with the various observations is obtained, the hypotheses are assembled into a theory. A theory, which is often called a model, is a set
of tested hypotheses that gives an overall explanation of some natural phenomenon.
It is very important to distinguish between observations and theories. An observation is something that is witnessed and can be recorded. A theory is an interpretation—
a possible explanation of why nature behaves in a particular way. Theories inevitably
change as more information becomes available. For example, the motions of the sun
and stars have remained virtually the same over the thousands of years during which
humans have been observing them, but our explanations—our theories—for these motions have changed greatly since ancient times.
The point is that scientists do not stop asking questions just because a given theory
seems to account satisfactorily for some aspect of natural behavior. They continue doing experiments to refine or replace the existing theories. This is generally done by using the currently accepted theory to make a prediction and then performing an experiment (making a new observation) to see whether the results bear out this prediction.
Always remember that theories (models) are human inventions. They represent attempts to explain observed natural behavior in terms of human experiences. A theory
is actually an educated guess. We must continue to do experiments and to refine our
theories (making them consistent with new knowledge) if we hope to approach a more
complete understanding of nature.
As scientists observe nature, they often see that the same observation applies to
many different systems. For example, studies of innumerable chemical changes have
shown that the total observed mass of the materials involved is the same before and
after the change. Such generally observed behavior is formulated into a statement
called a natural law. For example, the observation that the total mass of materials is
not affected by a chemical change in those materials is called the law of conservation
of mass.
Note the difference between a natural law and a theory. A natural law is a summary of
observed (measurable) behavior, whereas a theory is an explanation of behavior. A law
summarizes what happens; a theory (model) is an attempt to explain why it happens.
In this section we have described the scientific method as it might ideally be applied
(Fig. 1.4). However, it is important to remember that science does not always progress
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1.2
The Scientific Method
Chemical connections
Post-it Notes, a product of the 3M
Corporation, revolutionized casual
written communications and personal
reminders. Introduced in the United
States in 1980, these sticky-but-nottoo-sticky notes have now found
countless uses in offices, cars, and
homes throughout the world.
The invention of sticky notes
occurred over a period of about 10
years and involved a great deal of
serendipity. The adhesive for Post-it
Notes was discovered by Dr. Spencer
F. Silver of 3M in 1968. Silver found
that when an acrylate polymer
material was made in a particular way,
it formed cross-linked microspheres.
When suspended in a solvent and
sprayed on a sheet of paper, this
substance formed a “sparse monolayer” of adhesive after the solvent
evaporated. Scanning electron
microscope images of the adhesive
show that it has an irregular surface, a
little like the surface of a gravel road.
In contrast, the adhesive on cellophane tape looks smooth and uniform,
like a superhighway. The bumpy
surface of Silver’s adhesive caused it
to be sticky but not so sticky to
produce permanent adhesion,
because the number of contact points
between the binding surfaces was
limited.
When he invented this adhesive,
Silver had no specific ideas for its use,
so he spread the word of his discovery
to his fellow employees at 3M to see if
anyone had an application for it. In
addition, over the next several years
development was carried out to
improve the adhesive’s properties. It
was not until 1974 that the idea for
Photo © Cengage Learning. All rights reserved.
A Note-able Achievement
Post-it Notes popped up. One Sunday
Art Fry, a chemical engineer for 3M,
was singing in his church choir when
he became annoyed that the bookmark in his hymnal kept falling out. He
thought to himself that it would be
nice if the bookmark were sticky
enough to stay in place but not so
sticky that it couldn’t be moved.
Luckily, he remembered Silver’s
glue—and the Post-it Note was born.
For the next three years, Fry
worked to overcome the manufacturing obstacles associated with the
product. By 1977 enough Post-it Notes
were being produced to supply 3M’s
corporate headquarters, where the
employees quickly became addicted to
their many uses. Post-it Notes are now
available in 62 colors and 25 shapes.
In the years since the introduction
of Post-it Notes, 3M has heard some
remarkable stories connected to the
use of these notes. For example, a
Post-it Note was applied to the nose of
a corporate jet, where it was intended
to be read by the plane’s Las Vegas
ground crew. Someone forgot to
remove it, however. The note was still
on the nose of the plane when it
landed in Minneapolis, having survived
a takeoff, a landing, and speeds of
500 miles per hour at temperatures as
low as 2568F. Stories on the 3M Web
site describe how a Post-it Note on the
front door of a home survived the
140-mile-per-hour winds of Hurricane
Hugo and how a foreign official
accepted Post-it Notes in lieu of cash
when a small bribe was needed to cut
through bureaucratic hassles.
Post-it Notes have definitely
changed the way we communicate and
remember things.
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7
Chapter 1
Chemical Foundations
© Devonshire Collection/Reproduced by permission of Chatsworth Settlement Trustees/The Bridgeman
Art Library
8
Robert Boyle (1627–1691) was born in Ireland. He became especially interested in experiments
­involving air and developed an air pump with which he produced evacuated cylinders. He used
these cylinders to show that a feather and a lump of lead fall at the same rate in the absence of
air resistance and that sound cannot be produced in a vacuum. His most famous experiments
­involved careful measurements of the volume of a gas as a function of pressure. In his book Boyle
urged that the ancient view of elements as mystical substances should be abandoned and that an
element should instead be defined as anything that cannot be broken down into simpler
substances. This concept was an important step in the development of m
­ odern chemistry.
smoothly and efficiently. For one thing, hypotheses and observations are not totally
independent of each other, as we have assumed in the description of the idealized scientific method. The coupling of observations and hypotheses occurs because once we
begin to proceed down a given theoretical path, our hypotheses are unavoidably
couched in the language of that theory. In other words, we tend to see what we expect
to see and often fail to notice things that we do not expect. Thus the theory we are testing helps us ­because it focuses our questions. However, at the same time, this focusing
process may limit our ability to see other possible explanations.
It is also important to keep in mind that scientists are human. They have prejudices;
they misinterpret data; they become emotionally attached to their theories and thus
lose objectivity; and they play politics. Science is affected by profit motives, budgets,
fads, wars, and religious beliefs. Galileo, for example, was forced to recant his astronomical observations in the face of strong religious resistance. Lavoisier, the father of
modern chemistry, was beheaded because of his political affiliations. Great progress in
the chemistry of nitrogen fertilizers resulted from the desire to produce explosives to
fight wars. The progress of science is often affected more by the frailties of humans
and their ­institutions than by the limitations of scientific measuring devices. The scientific methods are only as effective as the humans using them. They do not automatically lead to progress.
Critical Thinking
What if everyone in the government used the scientific method to analyze and solve
society’s problems, and politics were never involved in the solutions? How would this
be different from the present situation, and would it be better or worse?
1.3 Units of Measurement
IBLG: See questions from “Uncertainty,
Measurement, and Calculations”
Making observations is fundamental to all science. A quantitative observation, or measurement, always consists of two parts: a number and a scale (called a unit). Both parts
must be present for the measurement to be meaningful.
In this textbook we will use measurements of mass, length, time, temperature, electric
current, and the amount of a substance, among others. Scientists recognized long ago
that standard systems of units had to be adopted if measurements were to be useful. If
every scientist had a different set of units, complete chaos would result. Unfortunately,
different standards were adopted in different parts of the world. The two major systems
are the English system used in the United States and the metric system used by most of
the rest of the industrialized world. This duality causes a good deal of trouble; for example, parts as simple as bolts are not interchangeable between machines built using the
two systems. As a result, the United States has begun to adopt the metric system.
Most scientists in all countries have used the metric system for many years. In 1960,
an international agreement set up a system of units called the International ­System (le
Système International in French), or the SI system. This system is based on the metric
system and units derived from the metric system. The fundamental SI units are listed in
Table 1.1. We will discuss how to manipulate these units later in this chapter.
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1.3
Units of Measurement
9
Chemical connections
How important are conversions
from one unit to another? If you ask
the National Aeronautics and Space
Administration (NASA), very important! In 1999, NASA lost a $125 million
Mars Climate Orbiter because of a
failure to convert from English to
­metric units.
The problem arose because two
teams working on the Mars mission
were using different sets of units.
NASA’s scientists at the Jet Propulsion
Laboratory in Pasadena, California,
assumed that the thrust data for the
rockets on the Orbiter they received
from Lockheed Martin Astronautics in
Denver, which built the spacecraft,
were in metric units. In reality, the
units were English. As a result, the
Orbiter dipped 100 km lower into the
Mars atmosphere than planned, and
the friction from the atmosphere
caused the craft to burn up.
NASA’s mistake refueled the
controversy over whether Congress
should require the United States to
NASA
Critical Units!
Artist’s conception of the lost Mars Climate Orbiter.
switch to the metric system. About
95% of the world now uses the ­metric
system, and the United States is slowly
switching from English to metric. For
example, the automobile industry has
adopted metric fasteners, and we buy
our soda in 2-L bottles.
Units can be very important. In
fact, they can mean the difference
between life and death on some
occasions. In 1983, for example, a
Canadian jetliner almost ran out of
fuel when someone pumped 22,300 lb
of fuel into the ­aircraft instead of
22,300 kg. Remember to watch your
units!
Photo © Cengage Learning. All rights reserved.
Because the fundamental units are not always convenient (expressing the mass of a
pin in kilograms is awkward), prefixes are used to change the size of the unit. These
are listed in Table 1.2. Some common objects and their measurements in SI units are
listed in Table 1.3.
One physical quantity that is very important in chemistry is volume, which is not a
fundamental SI unit but is derived from length. A cube that measures 1 meter (m) on
Soda is commonly sold in 2-L ­bottles—
an example of the use of SI units in
everyday life.
Table 1.1 | Fundamental SI Units
Physical Quantity
Mass
Length
Time
Temperature
Electric current
Amount of substance
Luminous intensity
Name of Unit
Abbreviation
kilogram
meter
second
kelvin
ampere
mole
candela
kg
m
s
K
A
mol
cd
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10
Chapter 1
Chemical Foundations
Table 1.2 | Prefixes Used in the SI System (The most commonly encountered
are shown in blue.)
Prefix
Symbol
Meaning
Exponential
Notation*
exa
peta
tera
giga
mega
kilo
hecto
deka
—
deci
centi
milli
micro
nano
pico
femto
atto
E
P
T
G
M
k
h
da
—
d
c
m
m
n
p
f
a
1,000,000,000,000,000,000
1,000,000,000,000,000
1,000,000,000,000
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
0.000000000000000001
1018
1015
1012
109
106
103
102
101
100
1021
1022
1023
1026
1029
10212
10215
10218
*See Appendix 1.1 if you need a review of exponential notation.
Table 1.3 | Some Examples of
Commonly Used Units
Length
dime is 1 mm thick.
A
A quarter is 2.5 cm in
diameter.
The average height of
an adult man is 1.8 m.
Mass
nickel has a mass of
A
about 5 g.
A 120-lb person has a
mass of about 55 kg.
Volume
12-oz can of soda
A
has a volume of about
360 mL.
each edge is represented in Fig. 1.5. This cube has a volume of (1 m)3 5 1 m3. Recognizing that there are 10 decimeters (dm) in a meter, the volume of this cube is (1 m)3
5 (10 dm)3 5 1000 dm3. A cubic decimeter, that is, (1 dm)3, is commonly called a liter
(L), which is a unit of volume slightly larger than a quart. As shown in Fig. 1.5, 1000 L
is contained in a cube with a volume of 1 cubic meter. Similarly, since 1 decimeter
equals 10 centimeters (cm), the liter can be divided into 1000 cubes, each with a volume of 1 cubic ­centimeter:
1 L 5 11 dm2 3 5 110 cm2 3 5 1000 cm3
1 m3
1 dm3 = 1 L
Figure 1.5 | The largest cube has
sides 1 m in length and a volume of
1 m3. The middle-sized cube has sides
1 dm in length and a volume of 1 dm3,
or 1 L. The smallest cube has sides
1 cm in length and a volume of 1 cm3,
or 1 mL.
1 cm
1 cm
1 cm3 = 1 mL
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1.4
Uncertainty in Measurement
11
Figure 1.6 | Common types of
laboratory equipment used to
measure liquid volume.
mL
100
Calibration
mark indicates
25-mL volume
mL
0
1
2
3
4
90
80
70
60
50
40
30
20
10
100-mL
graduated cylinder
Valve
(stopcock)
controls the
liquid flow
25-mL pipet
Calibration
mark indicates
250-mL volume
44
45
46
47
48
49
50
50-mL buret
250-mL
volumetric flask
Also, since 1 cm3 5 1 milliliter (mL),
1 L 5 1000 cm3 5 1000 mL
Experiment 1: The Determination
of Mass
Experiment 2: The Use of Volumetric
Glassware
Thus 1 liter contains 1000 cubic centimeters, or 1000 milliliters.
Chemical laboratory work frequently requires measurement of the volumes of
­liquids. Several devices for the accurate determination of liquid volume are shown in
Fig. 1.6.
An important point concerning measurements is the relationship between mass and
weight. Although these terms are sometimes used interchangeably, they are not the
same. Mass is a measure of the resistance of an object to a change in its state of motion. Mass is measured by the force necessary to give an object a certain acceleration.
On the earth we use the force that gravity exerts on an object to measure its mass. We
call this force the object’s weight. Since weight is the response of mass to gravity, it
varies with the strength of the gravitational field. Therefore, your body mass is the
same on the earth and on the moon, but your weight would be much less on the moon
than on the earth because of the moon’s smaller gravitational field.
Because weighing something on a chemical balance involves comparing the mass
of that object to a standard mass, the terms weight and mass are sometimes used interchangeably, although this is incorrect.
Critical Thinking
What if you were not allowed to use units for one day? How would this affect your life
for that day?
1.4 Uncertainty in Measurement
The number associated with a measurement is obtained using some measuring device.
For example, consider the measurement of the volume of a liquid using a buret (shown
in Fig. 1.7 with the scale greatly magnified). Notice that the meniscus of the liquid
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12
mL
Chapter 1
Chemical Foundations
20
20
21
22
23
24
occurs at about 20.15 mL. This means that about 20.15 mL of liquid has been delivered
from the buret (if the initial position of the liquid meniscus was 0.00 mL). Note that
we must estimate the last number of the volume reading by interpolating between the
0.1-mL marks. Since the last number is estimated, its value may be different if another
person makes the same measurement. If five different people read the same volume,
the results might be as follows:
Person
Results of Measurement
1
2
3
4
5
20.15 mL
20.14 mL
20.16 mL
20.17 mL
20.16 mL
25
Figure 1.7 | Measurement of
volume using a buret. The volume is
read at the bottom of the liquid curve
(called the meniscus).
A measurement always has some degree
of uncertainty.
These results show that the first three numbers (20.1) remain the same regardless of
who makes the measurement; these are called certain digits. However, the digit to the
right of the 1 must be estimated and therefore varies; it is called an uncertain digit. We
customarily report a measurement by recording all the certain digits plus the first uncertain digit. In our example it would not make any sense to try to record the volume
to thousandths of a milliliter because the value for hundredths of a milliliter must be
estimated when ­using the buret.
It is very important to realize that a measurement always has some degree of uncertainty. The uncertainty of a measurement depends on the precision of the measuring
device. For example, using a bathroom scale, you might estimate the mass of a grapefruit to be approximately 1.5 lb. Weighing the same grapefruit on a highly precise
balance might produce a result of 1.476 lb. In the first case, the uncertainty occurs in
the tenths of a pound place; in the second case, the uncertainty occurs in the thousandths of a pound place. Suppose we weigh two similar grapefruits on the two devices and obtain the following results:
Grapefruit 1
Grapefruit 2
Uncertainty in measurement is discussed
in more detail in Appendix 1.5.
Example 1.1
Bathroom Scale
Balance
1.5 lb
1.5 lb
1.476 lb
1.518 lb
Do the two grapefruits have the same mass? The answer depends on which set of results you consider. Thus a conclusion based on a series of measurements depends on
the certainty of those measurements. For this reason, it is important to indicate the
uncertainty in any measurement. This is done by always recording the certain digits
and the first uncertain digit (the estimated number). These numbers are called the significant figures of a measurement.
The convention of significant figures automatically indicates something about the
uncertainty in a measurement. The uncertainty in the last number (the estimated number) is usually assumed to be 61 unless otherwise indicated. For example, the measurement 1.86 kg can be taken to mean 1.86 6 0.01 kg.
Uncertainty in Measurement
In analyzing a sample of polluted water, a chemist measured out a 25.00-mL water
sample with a pipet (see Fig. 1.6). At another point in the analysis, the chemist used a
graduated cylinder (see Fig. 1.6) to measure 25 mL of a solution. What is the difference between the measurements 25.00 mL and 25 mL?
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1.4
Uncertainty in Measurement
13
Solution
Even though the two volume measurements appear to be equal, they really convey different information. The quantity 25 mL means that the volume is between 24 mL and
26 mL, whereas the quantity 25.00 mL means that the volume is between 24.99 mL
and 25.01 mL. The pipet measures volume with much greater precision than does the
graduated cylinder.
See Exercise 1.33
When making a measurement, it is important to record the results to the appropriate
number of significant figures. For example, if a certain buret can be read to 60.01 mL,
you should record a reading of twenty-five milliliters as 25.00 mL, not 25 mL. This
way at some later time when you are using your results to do calculations, the uncertainty in the measurement will be known to you.
Precision and Accuracy
Two terms often used to describe the reliability of measurements are precision and
accuracy. Although these words are frequently used interchangeably in everyday life,
they have different meanings in the scientific context. Accuracy refers to the agreement of a particular value with the true value. Precision refers to the degree of agreement among several measurements of the same quantity. Precision reflects the reproducibility of a given type of measurement. The difference between these terms is
illustrated by the results of three different dart throws shown in Fig. 1.8.
Two different types of errors are illustrated in Fig. 1.8. A random error (also called
an indeterminate error) means that a measurement has an equal probability of being
high or low. This type of error occurs in estimating the value of the last digit of a measurement. The second type of error is called systematic error (or determinate error).
This type of error occurs in the same direction each time; it is either always high or
always low. Fig. 1.8(a) indicates large random errors (poor technique). Fig. 1.8(b) indicates small random errors but a large systematic error, and Fig. 1.8(c) indicates small
random errors and no systematic error.
In quantitative work, precision is often used as an indication of accuracy; we assume that the average of a series of precise measurements (which should “average
out” the random errors because of their equal probability of being high or low) is
accurate, or close to the “true” value. However, this assumption is valid only if
a
Neither accurate nor precise.
b
Precise but not accurate.
c
Accurate and precise.
Figure 1.8 | The results of several dart throws show the difference between precise and accurate.
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14
Chapter 1
Chemical Foundations
systematic errors are absent. Suppose we weigh a piece of brass five times on a very
precise balance and obtain the following results:
Weighing
Result
1
2
3
4
5
2.486 g
2.487 g
2.485 g
2.484 g
2.488 g
Normally, we would assume that the true mass of the piece of brass is very close to
2.486 g, which is the average of the five results:
2.486 g 1 2.487 g 1 2.485 g 1 2.484 g 1 2.488 g
5 2.486 g
5
However, if the balance has a defect causing it to give a result that is consistently
1.000 g too high (a systematic error of 11.000 g), then the measured value of 2.486 g
would be seriously in error. The point here is that high precision among several measurements is an indication of accuracy only if systematic errors are absent.
Example 1.2
Precision and Accuracy
To check the accuracy of a graduated cylinder, a student filled the cylinder to the
25-mL mark using water delivered from a buret (see Fig. 1.6) and then read the volume
delivered. Following are the results of five trials:
Trial
Volume Shown by
Graduated Cylinder
Volume Shown
by the Buret
1
2
3
4
5
25 mL
25 mL
25 mL
25 mL
25 mL
26.54 mL
26.51 mL
26.60 mL
26.49 mL
26.57 mL
Average
25 mL
26.54 mL
Is the graduated cylinder accurate?
Solution
Precision is an indication of accuracy only
if there are no systematic errors.
The results of the trials show very good precision (for a graduated cylinder). The student has good technique. However, note that the average value measured using the
buret is significantly different from 25 mL. Thus this graduated cylinder is not very
accurate. It produces a systematic error (in this case, the indicated result is low for each
measurement).
See Question 1.11
1.5 Significant Figures and Calculations
Calculating the final result for an experiment usually involves adding, subtracting,
multiplying, or dividing the results of various types of measurements. Since it is very
important that the uncertainty in the final result is known correctly, we have developed
rules for counting the significant figures in each number and for determining the correct number of significant figures in the final result.
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1.5 Significant Figures and Calculations
15
Rules for Counting Significant Figures
1. Nonzero integers. Nonzero integers always count as significant figures.
2. Zeros. There are three classes of zeros:
Leading zeros are never significant figures.
a. L eading zeros are zeros that precede all the nonzero digits. These do not count as
significant figures. In the number 0.0025, the three zeros simply indicate the
position of the decimal point. This number has only two significant figures.
Captive zeros are always significant
figures.
b. C
aptive zeros are zeros between nonzero digits. These always count as significant figures. The number 1.008 has four significant figures.
Trailing zeros are sometimes significant
figures.
c. T railing zeros are zeros at the right end of the number. They are significant only
if the number contains a decimal point. The number 100 has only one significant figure, whereas the number 1.00 3 102 has three significant figures. The
number one hundred written as 100. also has three significant figures.
Exact numbers never limit the number of
significant figures in a calculation.
Exponential notation is reviewed in
Appendix 1.1.
Interactive
Example 1.3
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3. Exact numbers. Many times calculations involve numbers that were not obtained
using measuring devices but were determined by counting: 10 experiments,
3 apples, 8 molecules. Such numbers are called exact numbers. They can be assumed to have an infinite number of significant figures. Other examples of exact
numbers are the 2 in 2pr (the circumference of a circle) and the 4 and the 3 in
4
3
3 pr (the volume of a sphere). Exact numbers also can arise from definitions. For
example, 1 inch is defined as exactly 2.54 centimeters. Thus, in the statement
1 in 5 2.54 cm, neither the 2.54 nor the 1 limits the number of significant figures
when used in a calculation.
Note that the number 1.00 3 102 above is written in exponential notation. This
type of notation has at least two advantages: The number of significant figures can be
easily ­indicated, and fewer zeros are needed to write a very large or very small number.
For example, the number 0.000060 is much more conveniently represented as
6.0 3 1025 (the number has two significant figures).
Significant Figures
Give the number of significant figures for each of the following results.
a. A student’s extraction procedure on tea yields 0.0105 g of caffeine.
b. A chemist records a mass of 0.050080 g in an analysis.
c. In an experiment a span of time is determined to be 8.050 3 1023 s.
Solution
a.The number contains three significant figures. The zeros to the left of the 1 are
leading zeros and are not significant, but the remaining zero (a captive zero) is
significant.
b.The number contains five significant figures. The leading zeros (to the left of the
5) are not significant. The captive zeros between the 5 and the 8 are significant,
and the trailing zero to the right of the 8 is significant because the number
contains a decimal point.
c. This number has four significant figures. Both zeros are significant.
See Exercises 1.27 through 1.30
To this point we have learned to count the significant figures in a given number.
Next, we must consider how uncertainty accumulates as calculations are carried out.
The detailed analysis of the accumulation of uncertainties depends on the type of
calculation involved and can be complex. However, in this textbook we will use the
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16
Chapter 1
Chemical Foundations
following simple rules that have been developed for determining the appropriate number of significant figures in the result of a calculation.
Rules for Significant Figures in Mathematical Operations
1. For multiplication or division, the number of significant figures in the result is the
same as the number in the least precise measurement used in the calculation. For
example, consider the calculation
Corrected
4.56 3 1.4 5 6.38 8888888n 6.4
h
Limiting term has
two significant
figures
h
Two significant
figures
T he product should have only two significant figures, since 1.4 has two significant
figures.
2. For addition or subtraction, the result has the same number of decimal places as the
least precise measurement used in the calculation. For example, consider the sum
Although these simple rules work
well for most cases, they can give
misleading results in certain cases. For
more information, see L. M. Schwartz,
“Propagation of Significant Figures,”
J. Chem. Ed. 62 (1985): 693; and H.
Bradford Thompson, “Is 88C Equal to
508F?” J. Chem. Ed. 68 (1991): 400.
12.11
18.0
m Limiting term has one decimal place
1.013
Corrected
31.123 8888888n
31.1
h
One decimal place
The correct result is 31.1, since 18.0 has only one decimal place.
Note that for multiplication and division, significant figures are counted. For addition and subtraction, the decimal places are counted.
In most calculations you will need to round numbers to obtain the correct number
of significant figures. The following rules should be applied when rounding.
Rules for Rounding
1. In a series of calculations, carry the extra digits through to the final result, then
round.
2. If the digit to be removed
a. is less than 5, the preceding digit stays the same. For example, 1.33 rounds
to 1.3.
Rule 2 is consistent with the operation of
electronic calculators.
b. is equal to or greater than 5, the preceding digit is increased by 1. For example,
1.36 rounds to 1.4.
Although rounding is generally straightforward, one point requires special emphasis. As an illustration, suppose that the number 4.348 needs to be rounded to two significant figures. In doing this, we look only at the first number to the right of the 3:
4.348
h
Look at this number to
round to two significant figures.
Do not round sequentially. The number
6.8347 rounded to three significant figures
is 6.83, not 6.84.
The number is rounded to 4.3 because 4 is less than 5. It is incorrect to round sequentially. For example, do not round the 4 to 5 to give 4.35 and then round the 3 to 4 to
give 4.4.
When rounding, use only the first number to the right of the last significant figure.
It is important to note that Rule 1 above usually will not be followed in the examples in this text because we want to show the correct number of significant figures in
each step of a problem. This same practice is followed for the detailed solutions given
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1.5 Significant Figures and Calculations
17
in the Solutions Guide. However, when you are doing problems, you should carry extra
digits throughout a series of calculations and round to the correct number of significant
figures only at the end. This is the practice you should follow. The fact that your rounding procedures are different from those used in this text must be taken into account
when you check your answer with the one given at the end of the book or in the Solutions Guide. Your answer (based on rounding only at the end of a calculation) may
differ in the last place from that given here as the “correct” answer because we have
rounded after each step. To help you understand the difference between these rounding
procedures, we will consider them further in Example 1.4.
Interactive
Example 1.4
Significant Figures in Mathematical Operations
Carry out the following mathematical operations, and give each result with the correct
number of significant figures.
a. 1.05 3 1023 4 6.135
b. 21 2 13.8
c.As part of a lab assignment to determine the value of the gas constant (R), a
student measured the pressure (P), volume (V ), and temperature (T) for a sample
of gas, where
PV
R5
T
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in OWL.
The following values were obtained: P 5 2.560, T 5 275.15, and V 5 8.8. (Gases will
be discussed in detail in Chapter 5; we will not be concerned at this time about the
units for these quantities.) Calculate R to the correct number of significant figures.
Solution
a.The result is 1.71 3 1024, which has three significant figures because the term
with the least precision (1.05 3 1023) has three significant figures.
b.The result is 7 with no decimal point because the number with the least number
of decimal places (21) has none.
12.5602 18.82
PV
5
c. R 5
T
275.15
The correct procedure for obtaining the final result can be represented as follows:
12.5602 18.82
22.528
5
5 0.0818753
275.15
275.15
5 0.082 5 8.2 3 1022 5 R
The final result must be rounded to two significant figures because 8.8 (the least precise measurement) has two significant figures. To show the effects of rounding at intermediate steps, we will carry out the calculation as follows:
Photo © Cengage Learning
Rounded to two
significant figures
g
This number must be rounded to two
significant figures.
12.5602 18.82
22.528
23
5
5
275.15
275.15
275.15
Now we proceed with the next calculation:
23
5 0.0835908
275.15
Rounded to two significant figures, this result is
0.084 5 8.4 3 1022
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18
Chapter 1
Chemical Foundations
Note that intermediate rounding gives a significantly different result than that obtained
by rounding only at the end. Again, we must reemphasize that in your calculations you
should round only at the end. However, because rounding is carried out at intermediate
steps in this text (to always show the correct number of significant figures), the final
answer given in the text may differ slightly from the one you obtain (rounding only at
the end).
See Exercises 1.35 through 1.38
There is a useful lesson to be learned from Part c of Example 1.4. The student measured the pressure and temperature to greater precision than the volume. A more precise value of R (one with more significant figures) could have been obtained if a more
precise measurement of V had been made. As it is, the efforts expended to measure
P and T very precisely were wasted. Remember that a series of measurements to obtain
some final result should all be done to about the same precision.
1.6 Learning to Solve Problems
Systematically
One of the main activities in learning chemistry is solving various types of problems.
The best way to approach a problem, whether it is a chemistry problem or one from
your daily life, is to ask questions such as the following:
1. What is my goal? Or you might phrase it as: Where am I going?
2. Where am I starting? Or you might phrase it as: What do I know?
3. How do I proceed from where I start to where I want to go? Or you might say:
How do I get there?
We will use these ideas as we consider unit conversions in this chapter. Then we will
have much more to say about problem solving in Chapter 3, where we will start to
consider more complex problems.
1.7 Dimensional Analysis
Table 1.4 | English–Metric
Equivalents
Length
1 m 5 1.094 yd
2.54 cm 5 1 in
Mass
1 kg 5 2.205 lb
453.6 g 5 1 lb
Volume
1 L 5 1.06 qt
1 ft3 5 28.32 L
It is often necessary to convert a given result from one system of units to another. The
best way to do this is by a method called the unit factor method or, more commonly,
dimensional analysis. To illustrate the use of this method, we will consider several
unit conversions. Some equivalents in the English and metric systems are listed in
Table 1.4. A more complete list of conversion factors given to more significant figures
appears in Appendix 6.
Consider a pin measuring 2.85 cm in length. What is its length in inches? To accomplish this conversion, we must use the equivalence statement
2.54 cm 5 1 in
If we divide both sides of this equation by 2.54 cm, we get
15
1 in
2.54 cm
This expression is called a unit factor. Since 1 inch and 2.54 cm are exactly equivalent,
multiplying any expression by this unit factor will not change its value.
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1.7
Dimensional Analysis
19
The pin has a length of 2.85 cm. Multiplying this length by the appropriate unit
factor gives
2.85 cm 3
1 in
2.85
5
in 5 1.12 in
2.54 cm
2.54
Note that the centimeter units cancel to give inches for the result. This is exactly what
we wanted to accomplish. Note also that the result has three significant figures, as required by the number 2.85. Recall that the 1 and 2.54 in the conversion factor are exact
numbers by definition.
Interactive
Example 1.5
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Unit Conversions I
A pencil is 7.00 in long. What is its length in centimeters?
Solution
Where are we going?
To convert the length of the pencil from inches to centimeters
What do we know?
❯ The pencil is 7.00 in long.
How do we get there?
Since we want to convert from inches to centimeters, we need the equivalence state2.54 cm
ment 2.54 cm 5 1 in. The correct unit factor in this case is
:
1 in
7.00 in 3
2.54 cm
5 17.002 12.542 cm 5 17.8 cm
1 in
Here the inch units cancel, leaving centimeters, as requested.
See Exercises 1.41 and 1.42
Note that two unit factors can be derived from each equivalence statement. For example, from the equivalence statement 2.54 cm 5 1 in, the two unit factors are
2.54 cm
1 in
Consider the direction of the required
change to select the correct unit factor.
and
1 in
2.54 cm
How do you choose which one to use in a given situation? Simply look at the direction
of the required change. To change from inches to centimeters, the inches must cancel.
Thus the factor 2.54 cm/1 in is used. To change from centimeters to inches, centimeters must cancel, and the factor 1 in/2.54 cm is appropriate.
Problem-Solving Strategy
Converting from One Unit to Another
❯
❯
❯
To convert from one unit to another, use the equivalence statement that
relates the two units.
Derive the appropriate unit factor by looking at the direction of the required
change (to cancel the unwanted units).
Multiply the quantity to be converted by the unit factor to give the quantity
with the desired units.
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20
Chapter 1
Chemical Foundations
Interactive
Example 1.6
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in OWL.
Unit Conversions II
You want to order a bicycle with a 25.5-in frame, but the sizes in the catalog are given
only in centimeters. What size should you order?
Solution
Where are we going?
To convert from inches to centimeters
What do we know?
❯ The size needed is 25.5 in.
How do we get there?
Since we want to convert from inches to centimeters, we need the equivalence state2.54 cm
ment 2.54 cm 5 1 in. The correct unit factor in this case is
:
1 in
25.5 in 3
2.54 cm
5 64.8 cm
1 in
See Exercises 1.41 and 1.42
To ensure that the conversion procedure is clear, a multistep problem is considered
in Example 1.7.
Interactive
Example 1.7
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Unit Conversions III
A student has entered a 10.0-km run. How long is the run in miles?
Solution
Where are we going?
To convert from kilometers to miles
What do we know?
❯ The run is 10.00 km long.
How do we get there?
This conversion can be accomplished in several different ways. Since we have the
equivalence statement 1 m 5 1.094 yd, we will proceed by a path that uses this fact.
Before we start any calculations, let us consider our strategy. We have kilometers,
which we want to change to miles. We can do this by the following route:
kilometers
meters
yards
miles
To proceed in this way, we need the following equivalence statements:
1 km 5 1000 m
1 m 5 1.094 yd
1760 yd 5 1 mi
To make sure the process is clear, we will proceed step by step:
Kilometers to Meters
10.0 km 3
1000 m
5 1.00 3 104 m
1 km
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1.7
Dimensional Analysis
21
Meters to Yards
1.00 3 104 m 3
1.094 yd
5 1.094 3 104 yd
1m
Note that we should have only three significant figures in the result. However, since
this is an intermediate result, we will carry the extra digit. Remember, round off only
the final result.
In the text we round to the correct
number of significant figures after each
step to show the correct significant figures
for each calculation. However, since you
use a calculator and combine steps on it,
you should round only at the end.
Yards to Miles
1.094 3 104 yd 3
1 mi
5 6.216 mi
1760 yd
Note in this case that 1 mi equals exactly 1760 yd by designation. Thus 1760 is an
e­ xact number.
Since the distance was originally given as 10.0 km, the result can have only three
significant figures and should be rounded to 6.22 mi. Thus
10.0 km 5 6.22 mi
Alternatively, we can combine the steps:
10.0 km 3
1000 m
1.094 yd
1 mi
3
3
5 6.22 mi
1 km
1m
1760 yd
See Exercises 1.41 and 1.42
In using dimensional analysis, your verification that everything has been done correctly is that you end up with the correct units. In doing chemistry problems, you
should always include the units for the quantities used. Always check to see that the
units cancel to give the correct units for the final result. This provides a very valuable
check, especially for complicated problems.
Study the procedures for unit conversions in the following examples.
Interactive
Example 1.8
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in OWL.
Unit Conversions IV
The speed limit on many highways in the United States is 55 mi/h. What number
would be posted in kilometers per hour?
Solution
Where are we going?
To convert the speed limit from 55 miles per hour to kilometers per hour
What do we know?
❯ The speed limit is 55 mi/h.
How do we get there?
We use the following unit factors to make the required conversion:











Result obtained by
rounding only at the
end of the calculation
888n
55 mi
1760 yd
1m
1 km
3
3
3
5 88 km/h
h
1 mi
1.094 yd
1000 m
Note that all units cancel except the desired kilometers per hour.
See Exercises 1.49 through 1.51
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22
Chapter 1
Chemical Foundations
Interactive
Example 1.9
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in OWL.
Unit Conversions V
A Japanese car is advertised as having a gas mileage of 15 km/L. Convert this rating to
miles per gallon.
Solution
Where are we going?
To convert gas mileage from 15 kilometers per liter to miles per gallon
What do we know?
❯ The gas mileage is 15 km/L.
How do we get there?
We use the following unit factors to make the required conversion:











Result obtained by
rounding only at the
end of the calculation
n
15 km
1000 m
1.094 yd
1 mi
1L
4 qt
3
3
3
3
3
5 35 mi/gal
L
1 km
1m
1760 yd
1.06 qt
1 gal
See Exercise 1.52
Interactive
Example 1.10
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in OWL.
Unit Conversions VI
The latest model Corvette has an engine with a displacement of 6.20 L. What is the
displacement in units of cubic inches?
Solution
Where are we going?
To convert the engine displacement from liters to cubic inches
What do we know?
❯ The displacement is 6.20 L.
How do we get there?
We use the following unit factors to make the required conversion:
6.20 L 3
112 in2 3
1 ft3
3
5 378 in3
11 ft2 3
28.32 L
Note that the unit factor for conversion of feet to inches must be cubed to accommodate the conversion of ft3 to in3.
See Exercise 1.56
1.8 Temperature
IBLG: See questions from “Temperature
and Density”
Three systems for measuring temperature are widely used: the Celsius scale, the Kelvin scale, and the Fahrenheit scale. The first two temperature systems are used in the
physical sciences, and the third is used in many of the engineering sciences. Our purpose here is to define the three temperature scales and show how conversions from one
scale to another can be performed. Although these conversions can be carried out rouUnless otherwise noted, all art on this page is © Cengage Learning 2014.
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1.8
Experiment 4: The Determination of
Boiling Point
Temperature
23
tinely on most calculators, we will consider the process in some detail here to illustrate
methods of ­problem solving.
The three temperature scales are defined and compared in Fig. 1.9. Note that the
size of the temperature unit (the degree) is the same for the Kelvin and Celsius scales.
The fundamental difference between these two temperature scales is their zero points.
­Conversion between these two scales simply requires an adjustment for the different
zero points.
Temperature 1Kelvin2 5 temperature 1Celsius2 1 273.15
TK 5 TC 1 273.15
or
Temperature 1Celsius2 5 temperature 1Kelvin2 2 273.15
TC 5 TK 2 273.15
For example, to convert 300.00 K to the Celsius scale, we do the following
calculation:
300.00 2 273.15 5 26.85°C
Note that in expressing temperature in Celsius units, the designation 8C is used. The
degree symbol is not used when writing temperature in terms of the Kelvin scale. The
unit of temperature on this scale is called a kelvin and is symbolized by the letter K.
Converting between the Fahrenheit and Celsius scales is somewhat more complicated because both the degree sizes and the zero points are different. Thus we need to
consider two adjustments: one for degree size and one for the zero point. First, we
must account for the difference in degree size. This can be done by reconsidering Fig.
1.9. Notice that since 2128F 5 1008C and 328F 5 08C,
212 2 32 5 180 Fahrenheit degrees 5 100 2 0 5 100 Celsius degrees
Thus 1808 on the Fahrenheit scale is equivalent to 1008 on the Celsius scale, and the
unit factor is
180°F
9°F
or 100°C
5°C
or the reciprocal, depending on the direction in which we need to go.
Next, we must consider the different zero points. Since 328F 5 08C, we obtain
the corresponding Celsius temperature by first subtracting 32 from the Fahrenheit
Experiment 5: The Determination of
Melting Point
Fahrenheit
Boiling
point
of water
212°F
Kelvin
373.15 K
100°C
100
Celsius
degrees
180
Fahrenheit
degrees
Freezing
point
of water
Celsius
100
kelvins
32°F
0°C
273.15 K
−40°F
−40°C
233.15 K
Figure 1.9 | The three major
temperature scales.
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24
Chapter 1
Chemical Foundations
temperature to account for the different zero points. Then the unit factor is applied to
adjust for the difference in the degree size. This process is summarized by the
equation
1TF 2 32°F2
5°C
5 TC
9°F
(1.1)
where TF and TC represent a given temperature on the Fahrenheit and Celsius scales,
­respectively. In the opposite conversion, we first correct for degree size and then
correct for the different zero point. This process can be summarized in the following ­general equation:
TF 5 TC 3
9°F
1 32°F
5°C
(1.2)
Equations (1.1) and (1.2) are really the same equation in different forms. See if you
can obtain Equation (1.2) by starting with Equation (1.1) and rearranging.
At this point it is worthwhile to weigh the two alternatives for learning to do temperature conversions: You can simply memorize the equations, or you can take the
time to learn the differences between the temperature scales and to understand the
processes involved in converting from one scale to another. The latter approach may
take a little more effort, but the understanding you gain will stick with you much longer than the memorized formulas. This choice also will apply to many of the other
chemical concepts. Try to think things through!
Understand the process of converting
from one temperature scale to another; do
not simply memorize the equations.
Interactive
Example 1.11
Temperature Conversions I
Normal body temperature is 98.68F. Convert this temperature to the Celsius and
Kelvin scales.
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in OWL.
Solution
Where are we going?
To convert the body temperature from degrees Fahrenheit to degrees Celsius and to
kelvins.
Thinkstock/Getty Images
What do we know?
❯ The body temperature is 98.6°F.
A nurse taking the temperature of a
patient.
How do we get there?
Rather than simply using the formulas to solve this problem, we will proceed by thinking it through. The situation is diagramed in Fig. 1.10. First, we want to convert 98.68F
to the Celsius scale. The number of Fahrenheit degrees between 32.08F and 98.68F is
66.68F. We must convert this difference to Celsius degrees:
66.6°F 3
5°C
5 37.0°C
9°F
Thus 98.68F corresponds to 37.08C.
Now we can convert to the Kelvin scale:
TK 5 TC 1 273.15 5 37.0 1 273.15 5 310.2 K
Note that the final answer has only one decimal place (37.0 is limiting).
See Exercises 1.57, 1.59, and 1.60
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1.8
Figure 1.10 | Normal body
Fahrenheit
Celsius
25
Temperature
Kelvin
temperature on the Fahrenheit,
Celsius, and Kelvin scales.
98.6°F
66.6°F
32°F
Example 1.12
?K
?°C
66.6°F ×
5°C
= 37.0°C
9°F
37.0 + 273.15 K = 310.2 K
273.15 K
0°C
Temperature Conversions II
One interesting feature of the Celsius and Fahrenheit scales is that 2408C and 2408F
­represent the same temperature, as shown in Fig. 1.9. Verify that this is true.
Solution
Where are we going?
To show that 2408C 5 2408F
What do we know?
❯ The relationship between the Celsius and Fahrenheit scales
How do we get there?
The difference between 328F and 2408F is 728F. The difference between 08C and
2408C is 408C. The ratio of these is
72°F
8 3 9°F
9°F
5
5
40°C
8 3 5°C
5°C
as required. Thus 2408C is equivalent to 2408F.
See Exercise 1.61
Since, as shown in Example 1.12, 2408 on both the Fahrenheit and Celsius scales
represents the same temperature, this point can be used as a reference point (like 08C
and 328F) for a relationship between the two scales:
Number of Fahrenheit degrees
T 2 12402
9°F
5 F
5
Number of Celsius degrees
TC 2 12402
5°C
TF 1 40
9°F
5
TC 1 40
5°C
(1.3)
where TF and TC represent the same temperature (but not the same number). This equation can be used to convert Fahrenheit temperatures to Celsius, and vice versa, and
may be easier to remember than Equations (1.1) and (1.2).
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26
Chapter 1
Chemical Foundations
Interactive
Example 1.13
Temperature Conversions III
Liquid nitrogen, which is often used as a coolant for low-temperature experiments, has
a boiling point of 77 K. What is this temperature on the Fahrenheit scale?
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Solution
Richard Megna/Fundamental Photographs © Cengage Learning
Where are we going?
To convert 77 K to the Fahrenheit scale
What do we know?
❯ The relationship between the Kelvin and Fahrenheit scales
How do we get there?
We will first convert 77 K to the Celsius scale:
TC 5 TK 2 273.15 5 77 2 273.15 5 2196°C
To convert to the Fahrenheit scale, we will use Equation (1.3):
TF 1 40
9°F
5
TC 1 40
5°C
TF 1 40
T 1 40
9°F
5 F
5
2196°C 1 40
2156°C
5°C
9°F
12156°C2 5 2281°F
TF 1 40 5
5°C
TF 5 2281°F 2 40 5 2321°F
Liquid nitrogen is so cold that water
condenses out of the surrounding air,
forming a cloud as the nitrogen is poured.
See Exercises 1.57, 1.59, and 1.60
1.9 Density
A property of matter that is often used by chemists as an “identification tag” for a
substance is density, the mass of substance per unit volume of the substance:
Density 5
mass
volume
The density of a liquid can be determined easily by weighing an accurately known
volume of liquid. This procedure is illustrated in Example 1.14.
Interactive
Example 1.14
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Experiment 3: Density Determinations
Determining Density
A chemist, trying to identify an unknown liquid, finds that 25.00 cm3 of the substance
has a mass of 19.625 g at 208C. The following are the names and densities of the compounds that might be the liquid:
Compound
Chloroform
Diethyl ether
Ethanol
Isopropyl alcohol
Toluene
Density in g/cm3 at 208C
1.492
0.714
0.789
0.785
0.867
Which of these compounds is the most likely to be the unknown liquid?
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1.10 Classification of Matter
27
Solution
Where are we going?
To calculate the density of the unknown liquid
What do we know?
❯ The mass of a given volume of the liquid.
There are two ways of indicating units that
occur in the denominator. For example, we
can write g/cm3 or g cm23. Although we
will use the former system here, the other
system is widely used.
How do we get there?
To identify the unknown substance, we must determine its density. This can be done
by using the definition of density:
Density 5
mass
19.625 g
5
5 0.7850 g/cm3
volume
25.00 cm3
This density corresponds exactly to that of isopropyl alcohol, which therefore most
likely is the unknown liquid. However, note that the density of ethanol is also very
close. To be sure that the compound is isopropyl alcohol, we should run several more
density experiments. (In the modern laboratory, many other types of tests could be
done to distinguish between these two liquids.)
See Exercises 1.67 and 1.68
Besides being a tool for the identification of substances, density has many other
uses. For example, the liquid in your car’s lead storage battery (a solution of sulfuric
acid) changes density because the sulfuric acid is consumed as the battery discharges.
In a fully charged battery, the density of the solution is about 1.30 g/cm3. If the density
falls below 1.20 g/cm3, the battery will have to be recharged. Density measurement is
also used to determine the amount of antifreeze, and thus the level of protection against
freezing, in the cooling system of a car.
The densities of various common substances are given in Table 1.5.
Table 1.5 | Densities of Various Common Substances* at 208C
Substance
Oxygen
Hydrogen
Ethanol
Benzene
Water
Magnesium
Salt (sodium chloride)
Aluminum
Iron
Copper
Silver
Lead
Mercury
Gold
Physical State
Density (g/cm3)
Gas
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
Solid
Solid
Solid
Solid
Liquid
Solid
0.00133
0.000084
0.789
0.880
0.9982
1.74
2.16
2.70
7.87
8.96
10.5
11.34
13.6
19.32
*At 1 atmosphere pressure.
IBLG: See questions from
“Classification of Matter”
1.10 Classification of Matter
Before we can hope to understand the changes we see going on around us—the growth
of plants, the rusting of steel, the aging of people, the acidification of rain—we must
find out how matter is organized. Matter, best defined as anything occupying space
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28
Chapter 1
Chemical Foundations
PowerLecture:
Comparison of a Compound and
a Mixture
Comparison of a Solution and a
Mixture
Homogeneous Mixtures: Air and
Brass
and having mass, is the material of the universe. Matter is complex and has many levels of organization. In this section we will introduce basic ideas about the structure of
matter and its behavior.
We will start by considering the definitions of the fundamental properties of matter.
Matter exists in three states: solid, liquid, and gas. A solid is rigid; it has a fixed volume and shape. A liquid has a definite volume but no specific shape; it assumes the
shape of its container. A gas has no fixed volume or shape; it takes on the shape and
volume of its container. In contrast to liquids and solids, which are only slightly compressible, gases are highly compressible; it is relatively easy to decrease the volume of
a gas. Molecular-level pictures of the three states of water are given in Fig. 1.11. The
different properties of ice, liquid water, and steam are determined by the different arrangements of the molecules in these substances. Table 1.5 gives the states of some
common substances at 208C and 1 atmosphere pressure.
Most of the matter around us consists of mixtures of pure substances. Wood, gasoline, wine, soil, and air all are mixtures. The main characteristic of a mixture is that it
has variable composition. For example, wood is a mixture of many substances, the
proportions of which vary depending on the type of wood and where it grows. Mixtures can be classified as homogeneous (having visibly indistinguishable parts) or
heterogeneous (having visibly distinguishable parts).
A homogeneous mixture is called a solution. Air is a solution consisting of a mixture of gases. Wine is a complex liquid solution. Brass is a solid solution of copper and
zinc. Sand in water and iced tea with ice cubes are examples of heterogeneous mixtures. Heterogeneous mixtures usually can be separated into two or more homogeneous mixtures or pure substances (for example, the ice cubes can be separated from
the tea).
Mixtures can be separated into pure substances by physical methods. A pure substance is one with constant composition. Water is a good illustration of these ideas. As
we will discuss in detail later, pure water is composed solely of H2O molecules, but the
water found in nature (groundwater or the water in a lake or ocean) is really a mixture.
Experiment 8: Resolution of Mixtures 1:
Filtration and Distillation
Experiment 9: Resolution of Mixtures 2:
Paper Chromatography
Experiment 10: Resolution of Mixtures
3: Thin-Layer Chromatography
Figure 1.11 | The three states of
water (where red spheres represent
oxygen atoms and blue spheres
represent hydrogen atoms).
Ice
Solid: The water molecules are
locked into rigid positions
and are close together.
Water
Liquid: The water molecules
are still close together but can
move around to some extent.
Steam
Gas: The water molecules are
far apart and move randomly.
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1.10 Classification of Matter
The term volatile refers to the ease with
which a substance can be changed to its
vapor.
PowerLecture:
Structure of a Solid
Structure of a Liquid
Structure of a Gas
29
Seawater, for example, contains large amounts of dissolved minerals. Boiling seawater
produces steam, which can be condensed to pure water, leaving the minerals behind as
solids. The dissolved minerals in seawater also can be separated out by freezing the
mixture, since pure water freezes out. The processes of boiling and freezing are physical changes. When water freezes or boils, it changes its state but remains water; it is
still composed of H2O molecules. A physical change is a change in the form of a substance, not in its chemical composition. A physical change can be used to separate a
mixture into pure compounds, but it will not break compounds into elements.
One of the most important methods for separating the components of a mixture is
distillation, a process that depends on differences in the volatility (how readily substances become gases) of the components. In simple distillation, a mixture is heated in
a device such as that shown in Fig. 1.12. The most volatile component vaporizes at the
lowest temperature, and the vapor passes through a cooled tube (a condenser), where
it condenses back into its liquid state.
The simple, one-stage distillation apparatus shown in Fig. 1.12 works very well
when only one component of the mixture is volatile. For example, a mixture of water
and sand is easily separated by boiling off the water. Water containing dissolved minerals behaves in much the same way. As the water is boiled off, the minerals remain
behind as nonvolatile solids. Simple distillation of seawater using the sun as the heat
source is an excellent way to desalinate (remove the minerals from) seawater.
However, when a mixture contains several volatile components, the one-step distillation does not give a pure substance in the receiving flask, and more elaborate methods are required.
Another method of separation is simple filtration, which is used when a mixture
consists of a solid and a liquid. The mixture is poured onto a mesh, such as filter paper,
which passes the liquid and leaves the solid behind.
A third method of separation is chromatography. Chromatography is the general
name applied to a series of methods that use a system with two phases (states) of matter: a mobile phase and a stationary phase. The stationary phase is a solid, and the
mobile phase is either a liquid or a gas. The separation process occurs because the
Thermometer
Distilling
flask
Vapors
Condenser
Water out
Cool
water in
Figure 1.12 | Simple laboratory
distillation apparatus. Cool water
circulates through the outer portion
of the condenser, causing vapors from
the distilling flask to condense into a
liquid. The nonvolatile component of
the mixture remains in the distilling
flask.
Burner
Distillate
Receiving
flask
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30
Chapter 1
Chemical Foundations
Photos © Charles D. Winters
Figure 1.13 | Paper chromatography of ink. (a) A dot of the mixture to
be separated is placed at one end of a
sheet of porous paper. (b) The paper
acts as a wick to draw up the liquid.
Kristen Brochmann/Fundamental Photographs
a
The element mercury (top left) combines
with the element iodine (top right) to
form the compound mercuric iodide
(bottom). This is an example of a
chemical change.
b
components of the mixture have different affinities for the two phases and thus move
through the system at different rates. A component with a high affinity for the mobile
phase moves relatively quickly through the chromatographic system, whereas one
with a high affinity for the solid phase moves more slowly.
One simple type of chromatography, paper chromatography, uses a strip of
porous paper, such as filter paper, for the stationary phase. A drop of the mixture to
be separated is placed on the paper, which is then dipped into a liquid (the mobile
phase) that travels up the paper as though it were a wick (Fig. 1.13). This method of
separating a mixture is often used by biochemists, who study the chemistry of living
systems.
It should be noted that when a mixture is separated, the absolute purity of the separated components is an ideal. Because water, for example, inevitably comes into contact with other materials when it is synthesized or separated from a mixture, it is never
absolutely pure. With great care, however, substances can be obtained in very nearly
pure form.
Pure substances are either compounds (combinations of elements) or free elements.
A compound is a substance with constant composition that can be broken down into
elements by chemical processes. An example of a chemical process is the electrolysis
of water, in which an electric current is passed through water to break it down into the
free elements hydrogen and oxygen. This process produces a chemical change because
the water molecules have been broken down. The water is gone, and in its place we
have the free elements hydrogen and oxygen. A chemical change is one in which a
given substance becomes a new substance or substances with different properties and
different composition. Elements are substances that cannot be decomposed into simpler substances by chemical or physical means.
We have seen that the matter around us has various levels of organization. The most
fundamental substances we have discussed so far are elements. As we will see in later
chapters, elements also have structure: They are composed of atoms, which in turn are
composed of nuclei and electrons. Even the nucleus has structure: It is composed of
protons and neutrons. And even these can be broken down further, into elementary
particles called quarks. However, we need not concern ourselves with such details at
this point. Fig. 1.14 summarizes our discussion of the organization of matter.
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For Review
Figure 1.14 | The organization of
Matter
matter.
Variable composition?
Yes
No
Mixtures
Pure substances
Visibly
distinguishable parts?
Contains various
types of atoms?
No
Yes
Heterogeneous
Homogeneous
No
Elements
Chemical
methods
Yes
Compounds
Atoms
For review
Key terms
Scientific method
Section 1.2
❯
scientific method
measurement
hypothesis
theory
model
natural law
law of conservation of mass
Section 1.3
SI system
mass
weight
❯
❯
Models (theories) are explanations of why nature behaves in a
particular way.
❯
❯
❯
❯
uncertainty
significant figures
accuracy
precision
random error
systematic error
❯
exponential notation
Section 1.9
density
Measurements consist of a number and a unit.
Measurements involve some uncertainty.
Uncertainty is indicated by the use of significant figures.
❯ Rules to determine significant figures
❯ Calculations using significant figures
Preferred system is the SI system.
Temperature conversions
❯
TK 5 TC 1 273.15
❯
TC 5 1TF 2 32°F2 a
Section 1.7
unit factor method
dimensional analysis
They are subject to modification over time and sometimes fail.
Quantitative observations are called measurements.
Section 1.4
Section 1.5
Make observations
Formulate hypotheses
Perform experiments
❯
TF 5 T C a
5°C
b
9°F
9°F
b 1 32°F
5°C
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
31
32
Chapter 1
Chemical Foundations
Key terms
Density
Section 1.10
❯
matter
states (of matter)
homogeneous mixture
heterogeneous mixture
solution
pure substance
physical change
distillation
filtration
chromatography
paper chromatography
compound
chemical change
element
Review questions
Density 5
mass
volume
Matter can exist in three states:
❯
❯
❯
Solid
Liquid
Gas
Mixtures can be separated by methods involving only physical changes:
❯
❯
❯
Distillation
Filtration
Chromatography
Compounds can be decomposed to elements only through chemical changes.
Answers to the Review Questions can be found on the Student website (accessible from www.cengagebrain.com).
1. Define and explain the differences between the following terms.
a. law and theory
b. theory and experiment
c. qualitative and quantitative
d. hypothesis and theory
2. Is the scientific method suitable for solving problems
only in the sciences? Explain.
3. Which of the following statements could be tested by
quantitative measurement?
a. Ty Cobb was a better hitter than Pete Rose.
b. Ivory soap is 99.44% pure.
c. Rolaids consumes 47 times its weight in excess
stomach acid.
4. For each of the following pieces of glassware, provide a
sample measurement and discuss the number of
significant figures and uncertainty.
5
5. A student performed an analysis of a sample for its
calcium content and got the following results:
14.92% 14.91% 14.88% 14.91%
6.
7.
8.
9.
10.
The actual amount of calcium in the sample is 15.70%.
What conclusions can you draw about the accuracy and
precision of these results?
Compare and contrast the multiplication/division
significant figure rule to the significant figure rule
applied for addition/subtraction in mathematical
operations.
Explain how density can be used as a conversion factor
to convert the volume of an object to the mass of the
object, and vice versa.
On which temperature scale (8F, 8C, or K) does 1 degree
represent the smallest change in temperature?
Distinguish between physical changes and chemical
changes.
Why is the separation of mixtures into pure or relatively
pure substances so important when performing a
chemical analysis?
11
4
3
2
30
1
10
a
b
20
10
c
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A discussion of the Active Learning ­Questions can be found online in the ­Instructor’s Resource Guide and on
PowerLecture. The questions allow students to explore their understanding of concepts through discussion and
peer teaching. The real value of these questions is the learning that occurs while students talk to each other
about chemical concepts.
Active Learning Questions
These questions are designed to be used by groups of students in
class.
1. a. There are 365 days per year, 24 hours per day, 12 months
per year, and 60 minutes per hour. Use these data to determine how many minutes are in a month.
b. Now use the following data to calculate the number of
minutes in a month: 24 hours per day, 60 minutes per
hour, 7 days per week, and 4 weeks per month.
c. Why are these answers different? Which (if any) is more
correct? Why?
2. You go to a convenience store to buy candy and find the owner
to be rather odd. He allows you to buy pieces in multiples of
four, and to buy four, you need $0.23. He only allows you to
do this by using 3 pennies and 2 dimes. You have a bunch of
pennies and dimes, and instead of counting them, you decide
to weigh them. You have 636.3 g of pennies, and each penny
weighs 3.03 g. Each dime weighs 2.29 g. Each piece of candy
weighs 10.23 g.
a. How many pennies do you have?
b. How many dimes do you need to buy as much candy as
possible?
c. How much should all these dimes weigh?
d. How many pieces of candy could you buy? (number of
dimes from part b)
e. How much would this candy weigh?
f. How many pieces of candy could you buy with twice as
many dimes?
3. When a marble is dropped into a beaker of water, it sinks to
the bottom. Which of the following is the best explanation?
a. The surface area of the marble is not large enough to be
held up by the surface tension of the water.
b. The mass of the marble is greater than that of the water.
c. The marble weighs more than an equivalent volume of the
water.
d. The force from dropping the marble breaks the surface
tension of the water.
e. The marble has greater mass and volume than the water.
Justify your choice, and for choices you did not pick, explain
what is wrong about them.
4. You have two beakers, one filled to the 100-mL mark with
sugar (the sugar has a mass of 180.0 g) and the other filled to
the 100-mL mark with water (the water has a mass of 100.0 g).
You pour all the sugar and all the water together in a bigger
beaker and stir until the sugar is completely dissolved.
a. Which of the following is true about the mass of the
solution? Explain.
i. It is much greater than 280.0 g.
ii. It is somewhat greater than 280.0 g.
iii. It is exactly 280.0 g.
iv. It is somewhat less than 280.0 g.
v. It is much less than 280.0 g.
b. Which of the following is true about the volume of the
solution? Explain.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
For Review
33
i. It is much greater than 200.0 mL.
ii. It is somewhat greater than 200.0 mL.
iii. It is exactly 200.0 mL.
iv. It is somewhat less than 200.0 mL.
v. It is much less than 200.0 mL.
You may have noticed that when water boils, you can see bubbles that rise to the surface of the water.
a. What is inside these bubbles?
i. air
ii. hydrogen and oxygen gas
iii. oxygen gas
iv. water vapor
v. carbon dioxide gas
b. Is the boiling of water a chemical or physical change?
Explain.
If you place a glass rod over a burning candle, the glass appears to turn black. What is happening to each of the following
(physical change, chemical change, both, or neither) as the
candle burns? Explain each answer.
a. the wax b. the wick c. the glass rod
Which characteristics of a solid, a liquid, and a gas are exhibited by each of the following substances? How would you
classify each substance?
a. a bowl of pudding b. a bucketful of sand
Sketch a magnified view (showing atoms/molecules) of each
of the following and explain:
a. a heterogeneous mixture of two different compounds
b. a homogeneous mixture of an element and a compound
Paracelsus, a sixteenth-century alchemist and healer, adopted
as his slogan: “The patients are your textbook, the sickbed is
your study.” Is this view consistent with using the scientific
method?
What is wrong with the following statement?
“The results of the experiment do not agree with the theory.
Something must be wrong with the experiment.”
Why is it incorrect to say that the results of a measurement
were accurate but not precise?
What data would you need to estimate the money you would
spend on gasoline to drive your car from New York to Chicago? Provide estimates of values and a sample calculation.
Sketch two pieces of glassware: one that can measure volume
to the thousandths place and one that can measure volume
only to the ones place.
You have a 1.0-cm3 sample of lead and a 1.0-cm3 sample of
glass. You drop each in separate beakers of water. How do the
volumes of water displaced by each sample compare?
Explain.
Consider the addition of 15.4 to 28. What would a mathematician say the answer is? What would a scientist say? Justify the
scientist’s answer, not merely citing the rule, but explaining it.
Consider multiplying 26.2 by 16.43. What would a mathematician say the answer is? What would a scientist say? Justify
the scientist’s answer, not merely citing the rule, but explaining it.
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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34
Chapter 1
Chemical Foundations
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of this book and a
solution appears in the Solutions Guide, as found on PowerLecture.
Questions
17. The difference between a law and a theory is the difference
between what and why. Explain.
18. The scientific method is a dynamic process. What does this
mean?
19. Explain the fundamental steps of the scientific method.
20. What is the difference between random error and systematic
error?
21. A measurement is a quantitative observation involving both a
number and a unit. What is a qualitative observation? What are
the SI units for mass, length, and volume? What is the assumed uncertainty in a number (unless stated otherwise)? The
uncertainty of a measurement depends on the precision of the
measuring device. Explain.
22. To determine the volume of a cube, a student measured one of
the dimensions of the cube several times. If the true dimension
of the cube is 10.62 cm, give an example of four sets of measurements that would illustrate the following.
a. imprecise and inaccurate data
b. precise but inaccurate data
c. precise and accurate data
Give a possible explanation as to why data can be imprecise or inaccurate. What is wrong with saying a set of measurements is imprecise but accurate?
23. What are significant figures? Show how to indicate the number one thousand to 1 significant figure, 2 significant figures,
3 significant figures, and 4 significant figures. Why is the answer, to the correct number of significant figures, not 1.0 for
the following calculation?
1.5 2 1.0
5
0.50
24. A cold front moves through and the temperature drops by
20 degrees. In which temperature scale would this 20 degree
change represent the largest change in temperature?
25. When the temperature in degrees Fahrenheit (TF) is plotted vs.
the temperature in degrees Celsius (TC), a straight-line plot
results. A straight-line plot also results when TC is plotted vs.
TK (the temperature in kelvins). Reference Appendix A1.3 and
determine the slope and y-intercept of each of these two plots.
26. Give four examples illustrating each of the following terms.
a. homogeneous mixture
d. element
b. heterogeneous mixture
e. physical change
c. compound
f. chemical change
Exercises
In this section similar exercises are paired.
Significant Figures and Unit Conversions
c. We can use the equation
°F 5 95°C 1 32
to convert from Celsius to Fahrenheit temperature. Are the
numbers 95 and 32 exact or inexact?
d. p 5 3.1415927.
28. Indicate the number of significant figures in each of the
following:
a. This book contains more than 1000 pages.
b. A mile is about 5300 ft.
c. A liter is equivalent to 1.059 qt.
d. The population of the United States is approaching
3.1 3 102 million.
e. A kilogram is 1000 g.
f. The Boeing 747 cruises at around 600 mi/h.
29. How many significant figures are there in each of the following values?
a. 6.07 3 10215
e. 463.8052
b. 0.003840
f. 300
c. 17.00
g. 301
d. 8 3 108
h. 300.
30. How many significant figures are in each of the following?
a. 100
e. 0.0048
b. 1.0 3 102
f. 0.00480
c. 1.00 3 103
g. 4.80 3 1023
d. 100.
h. 4.800 3 1023
31. Round off each of the following numbers to the indicated
number of significant digits, and write the answer in standard
scientific notation.
a. 0.00034159 to three digits
b. 103.351 3 102 to four digits
c. 17.9915 to five digits
d. 3.365 3 105 to three digits
32. Use exponential notation to express the number 385,500 to
a. one significant figure.
b. two significant figures.
c. three significant figures.
d. five significant figures.
33. You have liquid in each graduated cylinder shown:
mL
5
mL
1
4
3
0.5
2
1
27. Which of the following are exact numbers?
a. There are 100 cm in 1 m.
b. One meter equals 1.094 yards.
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
You then add both samples to a beaker. How would you write
the number describing the total volume? What limits the precision of this number?
34. The beakers shown below have different precisions.
34
50
32.9
33
40
32.8
32
30
32.7
a. L
abel the amount of water in each of the three beakers to
the correct number of significant figures.
b. Is it possible for each of the three beakers to contain the
exact same amount of water? If no, why not? If yes, did
you report the volumes as the same in part a? Explain.
c. Suppose you pour the water from these three beakers into
one container. What should be the volume in the container
reported to the correct number of significant figures?
35. Evaluate each of the following, and write the answer to the
appropriate number of significant figures.
a. 212.2 1 26.7 1 402.09
b. 1.0028 1 0.221 1 0.10337
c. 52.331 1 26.01 2 0.9981
d. 2.01 3 102 1 3.014 3 103
e. 7.255 2 6.8350
36. Perform the following mathematical operations, and express
each result to the correct number of significant figures.
0.102 3 0.0821 3 273
a.
1.01
b. 0.14 3 6.022 3 1023
c. 4.0 3 104 3 5.021 3 1023 3 7.34993 3 102
2.00 3 106
d.
3.00 3 1027
37. Perform the following mathematical operations, and express
the result to the correct number of significant figures.
2.526
0.470
80.705
a.
1
1
3.1
0.623
0.4326
b. (6.404 3 2.91)y(18.7 2 17.1)
c. 6.071 3 1025 2 8.2 3 1026 2 0.521 3 1024
d. (3.8 3 10212 1 4.0 3 10213)y(4 3 1012 1 6.3 3 1013)
9.5 1 4.1 1 2.8 1 3.175
e.
4
(Assume that this operation is taking the average of four
numbers. Thus 4 in the denominator is exact.)
8.925 2 8.905
f.
3 100
8.925
(This type of calculation is done many times in calculating a
percentage error. Assume that this example is such a calcu­
lation; thus 100 can be considered to be an exact number.)
38. Perform the following mathematical operations, and express
the result to the correct number of significant figures.
a. 6.022 3 1023 3 1.05 3 102
6.6262 3 10234 3 2.998 3 108
b.
2.54 3 1029
35
c. 1.285 3 1022 1 1.24 3 1023 1 1.879 3 1021
11.00866 2 1.007282
d.
6.02205 3 1023
9.875 3 102 2 9.795 3 102
e.
3 100 1100 is exact2
9.875 3 102
9.42 3 102 1 8.234 3 102 1 1.625 3 103
13 is exact2
f.
3
39. Perform each of the following conversions.
a. 8.43 cm to millimeters
b. 2.41 3 102 cm to meters
c. 294.5 nm to centimeters
d. 1.445 3 104 m to kilometers
e. 235.3 m to millimeters
f. 903.3 nm to micrometers
40. a. How many kilograms are in 1 teragram?
b. How many nanometers are in 6.50 3 102 terameters?
c. How many kilograms are in 25 femtograms?
d. How many liters are in 8.0 cubic decimeters?
e. How many microliters are in 1 milliliter?
f. How many picograms are in 1 microgram?
41. Perform the following unit conversions.
a. Congratulations! You and your spouse are the proud
parents of a new baby, born while you are studying in a
country that uses the metric system. The nurse has
informed you that the baby weighs 3.91 kg and measures
51.4 cm. Convert your baby’s weight to pounds and
ounces and her length to inches (rounded to the nearest
quarter inch).
b. The circumference of the earth is 25,000 mi at the
equator. What is the circumference in kilometers? in
meters?
c. A rectangular solid measures 1.0 m by 5.6 cm by 2.1 dm.
Express its volume in cubic meters, liters, cubic inches,
and cubic feet.
42. Perform the following unit conversions.
a. 908 oz to kilograms
b. 12.8 L to gallons
c. 125 mL to quarts
d. 2.89 gal to milliliters
e. 4.48 lb to grams
f. 550 mL to quarts
43. Use the following exact conversion factors to perform the
stated calculations:
512 yd 5 1 rod
40 rods 5 1 furlong
8 furlongs 5 1 mile
a. T
he Kentucky Derby race is 1.25 miles. How long is the
race in rods, furlongs, meters, and kilometers?
b. A marathon race is 26 miles, 385 yards. What is this
distance in rods, furlongs, meters, and kilometers?
44. Although the preferred SI unit of area is the square meter, land
is often measured in the metric system in hectares (ha). One
hectare is equal to 10,000 m2. In the English system, land is
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
36
Chapter 1
Chemical Foundations
often measured in acres (1 acre 5 160 rod2). Use the exact
conversions and those given in Exercise 43 to calculate the
following.
a. 1 ha 5 ________ km2
b. The area of a 5.5-acre plot of land in hectares, square
meters, and square kilometers
c. A lot with dimensions 120 ft by 75 ft is to be sold for
$6500. What is the price per acre? What is the price per
hectare?
45. Precious metals and gems are measured in troy weights in the
English system:
24 grains 5 1 pennyweight 1exact2
20 pennyweight 5 1 troy ounce 1exact2
12 troy ounces 5 1 troy pound 1exact2
1 grain 5 0.0648 g
1 carat 5 0.200 g
a. T
he most common English unit of mass is the pound
avoirdupois. What is 1 troy pound in kilograms and in
pounds?
b. What is the mass of a troy ounce of gold in grams and in
carats?
c. The density of gold is 19.3 g/cm3. What is the volume of a
troy pound of gold?
46. Apothecaries (druggists) use the following set of measures in
the English system:
20 grains ap
3 scruples
8 dram ap
1 dram ap
5 1 scruple 1exact2
5 1 dram ap 1exact2
5 1 oz ap 1exact2
5 3.888 g
a. I s an apothecary grain the same as a troy grain? (See
Exercise 45.)
b. 1 oz ap 5 ________ oz troy.
c. An aspirin tablet contains 5.00 3 102 mg of active
ingredient. What mass in grains ap of active ingredient
does it contain? What mass in scruples?
d. What is the mass of 1 scruple in grams?
47. For a pharmacist dispensing pills or capsules, it is often easier
to weigh the medication to be dispensed than to count the individual pills. If a single antibiotic capsule weighs 0.65 g, and
a pharmacist weighs out 15.6 g of capsules, how many capsules have been dispensed?
48. A children’s pain relief elixir contains 80. mg acetaminophen
per 0.50 teaspoon. The dosage recommended for a child who
weighs between 24 and 35 lb is 1.5 teaspoons. What is the range
of ­acetaminophen dosages, expressed in mg acetaminophen/kg
body weight, for children who weigh between 24 and 35 lb?
49. Science fiction often uses nautical analogies to describe space
travel. If the starship U.S.S. Enterprise is traveling at warp
­factor 1.71, what is its speed in knots and in miles per hour?
(Warp 1.71 5 5.00 times the speed of light; speed of light 5
3.00 3 108 m/s; 1 knot 5 2030 yd/h.)
50. The world record for the hundred meter dash is 9.58 s. What is
the corresponding average speed in units of m/s, km/h, ft/s,
and mi/h? At this speed, how long would it take to run 1.00 3
102 yards?
51. Would a car traveling at a constant speed of 65 km/h violate a
40 mi/h speed limit?
52. You pass a road sign saying “New York 112 km.” If you drive
at a constant speed of 65 mi/h, how long should it take you to
reach New York? If your car gets 28 miles to the gallon, how
many liters of gasoline are necessary to travel 112 km?
53. You are in Paris, and you want to buy some peaches for lunch.
The sign in the fruit stand indicates that peaches cost 2.45 euros per kilogram. Given that 1 euro is equivalent to approximately $1.32, calculate what a pound of peaches will cost in
dollars.
54. In recent years, there has been a large push for an increase in
the use of renewable resources to produce the energy we need
to power our vehicles. One of the newer fuels that has become
more widely available is E85, a mixture of 85% ethanol and
15% gasoline. Despite being more environmentally friendly,
one of the potential drawbacks of E85 fuel is that it produces
less energy than conventional gasoline. Assume a car gets
28.0 mi/gal using gasoline at $3.50/gal and 22.5 mi/gal using
E85 at $2.85/gal. How much will it cost to drive 500. miles
using each fuel?
55. Mercury poisoning is a debilitating disease that is often fatal.
In the human body, mercury reacts with essential enzymes
leading to irreversible inactivity of these enzymes. If the
amount of mercury in a polluted lake is 0.4 mg Hg/mL, what
is the total mass in kilograms of mercury in the lake? (The lake
has a surface area of 100 mi2 and an average depth of 20 ft.)
56. Carbon monoxide (CO) detectors sound an alarm when peak
levels of carbon monoxide reach 100 parts per million (ppm).
This level roughly corresponds to a composition of air that
contains 400,000 mg carbon monoxide per cubic meter of air
(400,000 mg/m3). Assuming the dimensions of a room are
18 ft 3 12 ft 3 8 ft, estimate the mass of carbon monoxide in
the room that would register 100 ppm on a carbon monoxide
detector.
Temperature
57. Convert the following Fahrenheit temperatures to the Celsius
and Kelvin scales.
a. 24598F, an extremely low temperature
b. 240.8F, the answer to a trivia question
c. 688F, room temperature
d. 7 3 107 8F, temperature required to initiate fusion
reactions in the sun
58. A thermometer gives a reading of 96.18F 6 0.28F. What is the
temperature in 8C? What is the uncertainty?
59. Convert the following Celsius temperatures to Kelvin and to
Fahrenheit degrees.
a. the temperature of someone with a fever, 39.28C
b. a cold wintery day, 2258C
c. the lowest possible temperature, 22738C
d. the melting-point temperature of sodium chloride, 8018C
60. Convert the following Kelvin temperatures to Celsius and
Fahrenheit degrees.
a. the temperature that registers the same value on both the
Fahrenheit and Celsius scales, 233 K
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
b. the boiling point of helium, 4 K
c. the temperature at which many chemical quantities are
determined, 298 K
d. the melting point of tungsten, 3680 K
61. At what temperature is the temperature in degrees Fahrenheit
equal to twice the temperature in degrees Celsius?
62. The average daytime temperatures on the earth and Jupiter are
728F and 313 K, respectively. Calculate the difference in temperature, in 8C, between these two planets.
63. Use the figure below to answer the following questions.
130°C
−10°C
50°X
0°X
a. D
erive the relationship between 8C and 8X.
b. If the temperature outside is 22.08C, what is the temperature in units of 8X?
c. Convert 58.08X to units of 8C, K, and 8F.
64. Ethylene glycol is the main component in automobile antifreeze. To monitor the temperature of an auto cooling system,
you intend to use a meter that reads from 0 to 100. You devise
a new temperature scale based on the approximate melting and
boiling points of a typical antifreeze solution (2458C and
1158C). You wish these points to correspond to 08A and
1008A, respectively.
a. Derive an expression for converting between 8A and 8C.
b. Derive an expression for converting between 8F and 8A.
c. At what temperature would your thermometer and a
Celsius thermometer give the same numerical reading?
d. Your thermometer reads 868A. What is the temperature in
8C and in 8F?
e. What is a temperature of 458C in 8A?
Density
65. A material will float on the surface of a liquid if the material
has a density less than that of the liquid. Given that the density
of water is approximately 1.0 g/mL, will a block of material
having a volume of 1.2 3 104 in3 and weighing 350 lb float or
sink when placed in a reservoir of water?
66. For a material to float on the surface of water, the material must
have a density less than that of water (1.0 g/mL) and must not
react with the water or dissolve in it. A spherical ball has a radius of 0.50 cm and weighs 2.0 g. Will this ball float or sink
when placed in water? (Note: Volume of a sphere 5 43pr 3.)
37
67. A star is estimated to have a mass of 2 3 1036 kg. Assuming it
to be a sphere of average radius 7.0 3 105 km, calculate the
average density of the star in units of grams per cubic
centimeter.
68. A rectangular block has dimensions 2.9 cm 3 3.5 cm 3
10.0 cm. The mass of the block is 615.0 g. What are the volume and density of the block?
69. Diamonds are measured in carats, and 1 carat 5 0.200 g. The
density of diamond is 3.51 g/cm3.
a. What is the volume of a 5.0-carat diamond?
b. What is the mass in carats of a diamond measuring 2.8 mL?
70. Ethanol and benzene dissolve in each other. When 100. mL of
ethanol is dissolved in 1.00 L of benzene, what is the mass of
the mixture? (See Table 1.5.)
71. A sample containing 33.42 g of metal pellets is poured into a
graduated cylinder initially containing 12.7 mL of water, causing the water level in the cylinder to rise to 21.6 mL. Calculate
the density of the metal.
72. The density of pure silver is 10.5 g/cm3 at 208C. If 5.25 g of
pure silver pellets is added to a graduated cylinder containing
11.2 mL of water, to what volume level will the water in the
cylinder rise?
73. In each of the following pairs, which has the greater mass?
(See Table 1.5.)
a. 1.0 kg of feathers or 1.0 kg of lead
b. 1.0 mL of mercury or 1.0 mL of water
c. 19.3 mL of water or 1.00 mL of gold
d. 75 mL of copper or 1.0 L of benzene
74. a. Calculate the mass of ethanol in 1.50 qt of ethanol. (See
Table 1.5.)
b. Calculate the mass of mercury in 3.5 in3 of mercury. (See
Table 1.5.)
75. In each of the following pairs, which has the greater volume?
a. 1.0 kg of feathers or 1.0 kg of lead
b. 100 g of gold or 100 g of water
c. 1.0 L of copper or 1.0 L of mercury
76. Using Table 1.5, calculate the volume of 25.0 g of each of the
following substances at 1 atm.
a. hydrogen gas
b. water
c. iron
Chapter 5 discusses the properties of gases. One property
unique to gases is that they contain mostly empty space. Explain using the results of your calculations.
77. The density of osmium (the densest metal) is 22.57 g/cm3. If a
1.00-kg rectangular block of osmium has two dimensions of
4.00 cm 3 4.00 cm, calculate the third dimension of the block.
78. A copper wire (density 5 8.96 g/cm3) has a diameter of
0.25 mm. If a sample of this copper wire has a mass of 22 g,
how long is the wire?
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38
Chapter 1
Chemical Foundations
Classification and Separation of Matter
79. Match each description below with the following microscopic
pictures. More than one picture may fit each description. A
picture may be used more than once or not used at all.
i
ii
iii
iv
v
vi
a. a gaseous compound
b. a mixture of two gaseous elements
c. a solid element
d. a mixture of a gaseous element and a gaseous compound
80. Define the following terms: solid, liquid, gas, pure substance,
element, compound, homogeneous mixture, heterogeneous
mixture, solution, chemical change, physical change.
81. What is the difference between homogeneous and heterogeneous matter? Classify each of the following as homogeneous
or heterogeneous.
a. a door
d. the water you drink
b. the air you breathe
e. salsa
c. a cup of coffee (black)
f. your lab partner
82. Classify the following mixtures as homogeneous or hetero­
geneous.
a. potting soil
d. window glass
b. white wine
e. granite
c. your sock drawer
83. Classify each of the following as a mixture or a pure
substance.
a. water
f. uranium
b. blood
g. wine
c. the oceans
h. leather
d. iron
i. table salt
e. brass
Of the pure substances, which are elements and which are
­compounds?
84. Suppose a teaspoon of magnesium filings and a teaspoon of
powdered sulfur are placed together in a metal beaker. Would
this constitute a mixture or a pure substance? Suppose the
magnesium filings and sulfur are heated so that they react with
each other, forming magnesium sulfide. Would this still be a
“mixture”? Why or why not?
85. If a piece of hard, white blackboard chalk is heated strongly in
a flame, the mass of the piece of chalk will decrease, and eventually the chalk will crumble into a fine white dust. Does this
change suggest that the chalk is composed of an element or a
compound?
86. During a very cold winter, the temperature may remain below
freezing for extended periods. However, fallen snow can still
disappear, even though it cannot melt. This is possible because
a solid can vaporize directly, without passing through the liquid state. Is this process (sublimation) a physical or a chemical
change?
87. Classify the following as physical or chemical changes.
a. Moth balls gradually vaporize in a closet.
b. Hydrofluoric acid attacks glass and is used to etch
calibration marks on glass laboratory utensils.
c. A French chef making a sauce with brandy is able to boil
off the alcohol from the brandy, leaving just the brandy
flavoring.
d. Chemistry majors sometimes get holes in the cotton jeans
they wear to lab because of acid spills.
88. The properties of a mixture are typically averages of the properties of its components. The properties of a compound may
differ dramatically from the properties of the elements that
combine to produce the compound. For each process described below, state whether the material being discussed is
most likely a mixture or a compound, and state whether the
process is a chemical change or a physical change.
a. An orange liquid is distilled, resulting in the collection of
a yellow liquid and a red solid.
b. A colorless, crystalline solid is decomposed, yielding a
pale yellow-green gas and a soft, shiny metal.
c. A cup of tea becomes sweeter as sugar is added to it.
Additional Exercises
89. Lipitor, a pharmaceutical drug that has been shown to lower
“bad” cholesterol levels while raising “good” cholesterol levels in patients taking the drug, had over $11 billion in sales in
2006. Assuming one 2.5-g pill contains 4.0% of the active ingredient by mass, what mass in kg of active ingredient is present in one bottle of 100 pills?
90. In Shakespeare’s Richard III, the First Murderer says:
“Take that, and that! [Stabs Clarence]
If that is not enough, I’ll drown you in a malmsey butt within!”
Given that 1 butt 5 126 gal, in how many liters of malmsey (a
foul brew similar to mead) was the unfortunate Clarence about
to be drowned?
91. The contents of one 40. lb bag of topsoil will cover 10. square
feet of ground to a depth of 1.0 inch. What number of bags is
needed to cover a plot that measures 200. by 300. m to a depth
of 4.0 cm?
92. In the opening scenes of the movie Raiders of the Lost Ark,
­Indiana Jones tries to remove a gold idol from a booby-trapped
pedestal. He replaces the idol with a bag of sand of approximately equal volume. (Density of gold 5 19.32 g/cm3; density
of sand < 2 g/cm3.)
a. Did he have a reasonable chance of not activating the
mass-sensitive booby trap?
b. In a later scene, he and an unscrupulous guide play catch
with the idol. Assume that the volume of the idol is about
1.0 L. If it were solid gold, what mass would the idol
have? Is playing catch with it plausible?
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For Review
93. A parsec is an astronomical unit of distance where 1 parsec 5
3.26 light years (1 light year equals the distance traveled by
light in one year). If the speed of light is 186,000 mi/s, calculate the distance in meters of an object that travels 9.6 parsecs.
94. You are driving 65 mi/h and take your eyes off the road for
“just a second.” What distance (in feet) do you travel in this
time?
95. This year, like many past years, you begin to feel very sleepy
­after eating a large helping of Thanksgiving turkey. Some
people attribute this sleepiness to the presence of the amino
acid tryptophan in turkey. Tryptophan can be used by the body
to produce serotonin, which can calm the brain’s activity and
help to bring on sleep.
a. What mass in grams of tryptophan is in a 0.25-lb serving
of turkey? (Assume tryptophan accounts for 1.0% of the
turkey mass.)
b. What mass in grams of tryptophan is in 0.25 quart of
milk? (Assume tryptophan accounts for 2.0% of milk by
mass and that the density of milk is 1.04 kg/L.)
96. Which of the following are chemical changes? Which are
physical changes?
a. the cutting of food
b. interaction of food with saliva and digestive enzymes
c. proteins being broken down into amino acids
d. complex sugars being broken down into simple sugars
e. making maple syrup by heating maple sap to remove
water through evaporation
f. DNA unwinding
97. A column of liquid is found to expand linearly on heating. Assume the column rises 5.25 cm for a 10.08F rise in temperature. If the initial temperature of the liquid is 98.68F, what will
the final temperature be in 8C if the liquid has expanded by
18.5 cm?
98. A 25.00-g sample of a solid is placed in a graduated cylinder,
and then the cylinder is filled to the 50.0-mL mark with benzene. The mass of benzene and solid together is 58.80 g. Assuming that the solid is insoluble in benzene and that the density of benzene is 0.880 g/cm3, calculate the density of the
solid.
99. For each of the following, decide which block is more dense:
the orange block, the blue block, or it cannot be determined.
Explain your answers.
a
b
c
39
d
100. According to the Official Rules of Baseball, a baseball must
have a circumference not more than 9.25 in or less than
9.00 in and a mass not more than 5.25 oz or less than 5.00 oz.
What range of densities can a baseball be expected to have?
Express this range as a single number with an accompanying
uncertainty limit.
101. The density of an irregularly shaped object was determined as
follows. The mass of the object was found to be 28.90 g 6
0.03 g. A graduated cylinder was partially filled with water.
The reading of the level of the water was 6.4 cm3 6 0.1 cm3.
The object was dropped in the cylinder, and the level of the
water rose to 9.8 cm3 6 0.1 cm3. What is the density of the
object with appropriate error limits? (See Appendix 1.5.)
102. The chemist in Example 1.14 did some further experiments.
She found that the pipet used to measure the volume of the
liquid is accurate to 60.03 cm3. The mass measurement is accurate to 60.002 g. Are these measurements sufficiently precise for the chemist to distinguish between isopropyl alcohol
and ethanol?
ChemWork Problems
These multiconcept problems (and additional ones) are found interactively online with the same type of assistance a student would get
from an instructor.
103. The longest river in the world is the Nile River with a length
of 4,145 mi. How long is the Nile in cable lengths, meters, and
nautical miles?
Use these exact conversions to help solve the problem:
6 ft 5 1 fathom
100 fathoms 5 1 cable length
10 cable lengths 5 1 nautical mile
3 nautical miles 5 1 league
104. Secretariat is known as the horse with the fastest run in the
Kentucky Derby. If Secretariat’s record 1.25-mi run lasted
1 minute 59.2 seconds, what was his average speed in m/s?
105. The hottest temperature recorded in the United States is 1348F
in Greenland Ranch, CA. The melting point of phosphorus
is 448C. At this temperature, would phosphorus be a liquid or
a solid?
106. The radius of a neon atom is 69 pm, and its mass is 3.35 3
10223 g. What is the density of the atom in grams per cubic
centimeter (g/cm3)? Assume the nucleus is a sphere with volume 5 43 pr 3.
107. Which of the following statements is(are) true?
a. A spoonful of sugar is a mixture.
b. Only elements are pure substances.
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Chapter 1
Chemical Foundations
c. Air is a mixture of gases.
d. Gasoline is a pure substance.
e. Compounds can be broken down only by chemical means.
108. Which of the following describes a chemical property?
a. The density of iron is 7.87 g/cm3.
b. A platinum wire glows red when heated.
c. An iron bar rusts.
d. Aluminum is a silver-colored metal.
Challenge Problems
109. A rule of thumb in designing experiments is to avoid using a
­result that is the small difference between two large measured
quantities. In terms of uncertainties in measurement, why is
this good advice?
110. Draw a picture showing the markings (graduations) on glassware that would allow you to make each of the following volume measurements of water, and explain your answers (the
numbers given are as precise as possible).
a. 128.7 mL b. 18 mL c. 23.45 mL
If you made these measurements for three samples of water and
then poured all of the water together in one container, what total
volume of water should you report? Support your answer.
111. Many times errors are expressed in terms of percentage. The
percent error is the absolute value of the difference of the true
value and the experimental value, divided by the true value,
and multiplied by 100.
0 true value 2 experimental value 0
Percent error 5
3 100
true value
Calculate the percent error for the following measurements.
a. The density of an aluminum block determined in an
experiment was 2.64 g/cm3. (True value 2.70 g/cm3.)
b. The experimental determination of iron in iron ore was
16.48%. (True value 16.12%.)
c. A balance measured the mass of a 1.000-g standard as
0.9981 g.
112. A person weighed 15 pennies on a balance and recorded the
following masses:
3.112 g
2.467 g
3.129 g
3.053 g
3.081 g
3.109 g
3.079 g
2.545 g
3.054 g
3.131 g
3.059 g
2.518 g
3.050 g
3.072 g
3.064 g
Curious about the results, he looked at the dates on each penny.
Two of the light pennies were minted in 1983 and one in 1982.
The dates on the 12 heavier pennies ranged from 1970 to 1982.
Two of the 12 heavier pennies were minted in 1982.
a. Do you think the Bureau of the Mint changed the way it
made pennies? Explain.
b. The person calculated the average mass of the 12 heavy
pennies. He expressed this average as 3.0828 g 6 0.0482 g.
What is wrong with the numbers in this result, and how
should the value be expressed?
113. On October 21, 1982, the Bureau of the Mint changed the
composition of pennies (see Exercise 112). Instead of an alloy
of 95% Cu and 5% Zn by mass, a core of 99.2% Zn and 0.8%
Cu with a thin shell of copper was adopted. The overall composition of the new penny was 97.6% Zn and 2.4% Cu by
mass. Does this account for the difference in mass among the
pennies in Exercise 112? Assume the volume of the individual
metals that make up each penny can be added together to give
the overall volume of the penny, and assume each penny is
the same size. (Density of Cu 5 8.96 g/cm3; density of Zn 5
7.14 g/cm3.)
114. As part of a science project, you study traffic patterns in your
city at an intersection in the middle of downtown. You set up a
device that counts the cars passing through this intersection
for a 24-hr period during a weekday. The graph of hourly traffic looks like this.
60
Number of Cars
40
50
40
30
20
10
0
12 A.M.
6 A.M.
noon
6 P.M.
Time
a. A
t what time(s) does the highest number of cars pass
through the intersection?
b. At what time(s) does the lowest number of cars pass
through the intersection?
c. Briefly describe the trend in numbers of cars over the
course of the day.
d. Provide a hypothesis explaining the trend in numbers of
cars over the course of the day.
e. Provide a possible experiment that could test your
hypothesis.
115. Sterling silver is a solid solution of silver and copper. If a
piece of a sterling silver necklace has a mass of 105.0 g and a
volume of 10.12 mL, calculate the mass percent of copper in
the piece of necklace. Assume that the volume of silver present plus the volume of copper present equals the total volume.
Refer to Table 1.5.
mass of copper
Mass percent of copper 5
3 100
total mass
116. Make molecular-level (microscopic) drawings for each of the
following.
a. Show the differences between a gaseous mixture that is a
homogeneous mixture of two different compounds, and a
gaseous mixture that is a homogeneous mixture of a
compound and an element.
b. Show the differences among a gaseous element, a liquid
element, and a solid element.
117. Confronted with the box shown in the diagram, you wish to
discover something about its internal workings. You have no
tools and cannot open the box. You pull on rope B, and it
moves rather freely. When you pull on rope A, rope C appears
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For Review
to be pulled slightly into the box. When you pull on rope C,
rope A almost disappears into the box.*
A
B
C
Mass of cylinder plus wet sand
Mass of cylinder plus dry sand
Mass of empty cylinder
Volume of dry sand
Volume of sand plus methanol
Volume of methanol
41
45.2613 g
37.3488 g
22.8317 g
10.0 mL
17.6 mL
10.00 mL
Integrative Problems
a. B
ased on these observations, construct a model for the
interior mechanism of the box.
b. What further experiments could you do to refine your
model?
118. An experiment was performed in which an empty 100-mL
graduated cylinder was weighed. It was weighed once again
after it had been filled to the 10.0-mL mark with dry sand. A
10-mL pipet was used to transfer 10.00 mL of methanol to the
cylinder. The sand–methanol mixture was stirred until bubbles
no longer emerged from the mixture and the sand looked uniformly wet. The cylinder was then weighed again. Use the
data obtained from this experiment (and displayed at the end
of this problem) to find the density of the dry sand, the density
of methanol, and the density of sand particles. Does the bubbling that occurs when the methanol is added to the dry sand
indicate that the sand and methanol are reacting?
*From Yoder, Suydam, and Snavely, Chemistry (New York: Harcourt
Brace Jovanovich, 1975), pp. 9–11.
These problems require the integration of multiple concepts to find
the solutions.
119. The U.S. trade deficit at the beginning of 2005 was
$475,000,000. If the wealthiest 1.00% of the U.S. population
(297,000,000) contributed an equal amount of money to bring
the trade deficit to $0, how many dollars would each person
contribute? If one of these people were to pay his or her share
in nickels only, how many nickels are needed? Another person
living abroad at the time decides to pay in pounds sterling (£).
How many pounds sterling does this person contribute (assume a conversion rate of 1 £ 5 $1.869)?
120. The density of osmium is reported by one source to be 22610
kg/m3. What is this density in g/cm3? What is the mass of a
block of osmium measuring 10.0 cm 3 8.0 cm 3 9.0 cm?
121. At the Amundsen-Scott South Pole base station in Antarctica,
when the temperature is 2100.08F, researchers who live there
can join the “300 Club” by stepping into a sauna heated to
200.08F then quickly running outside and around the pole that
marks the South Pole. What are these temperatures in 8C?
What are these temperatures in K? If you measured the temperatures only in 8C and K, can you become a member of the
“300 Club” (that is, is there a 300.-degree difference between
the temperature extremes when measured in 8C and K)?
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Chapter 2
Atoms, Molecules, and Ions
2.1
The Early History of Chemistry
2.2
Fundamental Chemical Laws
2.3
Dalton’s Atomic Theory
2.6
Molecules and Ions
Formulas from Names
arly Experiments to Characterize
E
the Atom
2.7
n Introduction to the Periodic
A
Table
Ionic Compounds with Polyatomic Ions
2.4
The Electron
Radioactivity
2.5
T he Modern View of Atomic
Structure: An Introduction
2.8
Naming Simple Compounds
Binary Ionic Compounds (Type I)
Binary Ionic Compounds (Type II)
Binary Covalent Compounds (Type III)
Acids
The Nuclear Atom
Polarized light micrograph of crystals of tartaric acid. (Sinclair Stammers/Photo Researchers, Inc.)
42
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W
here does one start in learning chemistry? Clearly we must consider some
essential vocabulary and something about the origins of the science before
we can proceed very far. Thus, while Chapter 1 provided background on the fundamental ideas and ­procedures of science in general, Chapter 2 covers the specific chemical background necessary for understanding the material in the next few chapters. The
coverage of these topics is necessarily brief at this point. We will develop these ideas
more fully as it becomes appropriate to do so. A major goal of this chapter is to present
the systems for naming chemical compounds to provide you with the vocabulary necessary to understand this book and to pursue your laboratory studies.
Because chemistry is concerned first and foremost with chemical changes, we will
proceed as quickly as possible to a study of chemical reactions (Chapters 3 and 4).
However, before we can discuss reactions, we must consider some fundamental ideas
about atoms and how they combine.
IBLG: See questions
from “Development of
Atomic Theory”
2.1 The Early History of Chemistry
Chemistry has been important since ancient times. The processing of natural ores to
­produce metals for ornaments and weapons and the use of embalming fluids are just
two applications of chemical phenomena that were utilized prior to 1000 b.c.
The Greeks were the first to try to explain why chemical changes occur. By about
400 b.c. they had proposed that all matter was composed of four fundamental substances: fire, earth, water, and air. The Greeks also considered the question of whether
matter is continuous, and thus infinitely divisible into smaller pieces, or composed of
small, indivisible particles. Supporters of the latter position were Demokritos* of
Abdera (c. 460–c. 370 b.c.) and Leucippos, who used the term atomos (which later
became atoms) to describe these ultimate particles. However, because the Greeks had
no experiments to test their ideas, no definitive conclusion could be reached about the
divisibility of matter.
The next 2000 years of chemical history were dominated by a pseudoscience called
alchemy. Some alchemists were mystics and fakes who were obsessed with the idea of
turning cheap metals into gold. However, many alchemists were serious scientists, and
this period saw important advances: The alchemists discovered several elements and
learned to prepare the mineral acids.
The foundations of modern chemistry were laid in the sixteenth century with the
­development of systematic metallurgy (extraction of metals from ores) by a German,
Georg Bauer (1494–1555), and the medicinal application of minerals by a Swiss
­alchemist/physician known as Paracelsus (full name: Philippus Theophrastus Bombastus von Hohenheim [1493–1541]).
The first “chemist” to perform truly quantitative experiments was Robert Boyle
(1627–1691), who carefully measured the relationship between the pressure and volume
of air. When Boyle published his book The Skeptical Chymist in 1661, the quantitative
sciences of physics and chemistry were born. In addition to his results on the quantitative behavior of gases, Boyle’s other major contribution to chemistry consisted of his
ideas about the chemical elements. Boyle held no preconceived notion about the number of elements. In his view, a substance was an element unless it could be broken
down into two or more simpler substances. As Boyle’s experimental definition of an
element became generally accepted, the list of known elements began to grow, and the
Greek system of four elements finally died. Although Boyle was an excellent scientist,
*Democritus is an alternate spelling.
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43
44
Chapter 2
Atoms, Molecules, and Ions
Figure 2.1 | The Priestley Medal is the highest honor given by the American Chemical
Roald Hoffmann
Society. It is named for Joseph Priestley, who was born in England on March 13, 1733. He
performed many important scientific experiments, among them the discovery that a gas
later identified as carbon dioxide could be dissolved in water to produce seltzer. Also, as a
result of meeting Benjamin Franklin in London in 1766, Priestley became interested in
electricity and was the first to observe that graphite was an electrical conductor. However,
his greatest discovery occurred in 1774 when he isolated oxygen by heating mercuric oxide.
Because of his nonconformist political views, Priestley was forced to leave England. He
died in the United States in 1804.
he was not always right. For example, he clung to the alchemists’ views that metals
were not true elements and that a way would eventually be found to change one metal
into another.
The phenomenon of combustion evoked intense interest in the seventeenth and
eighteenth centuries. The German chemist Georg Stahl (1660–1734) suggested that a
substance he called “phlogiston” flowed out of the burning material. Stahl postulated
that a substance burning in a closed container eventually stopped burning because the
air in the container became saturated with phlogiston. Oxygen gas, discovered by
Joseph Priestley (1733–1804),* an English clergyman and scientist (Fig. 2.1), was
found to support vigorous combustion and was thus supposed to be low in phlogiston.
In fact, oxygen was originally called “dephlogisticated air.”
2.2 Fundamental Chemical Laws
Experiment 14: Composition 1:
Percentage Composition and Empirical
Formula of Magnesium Oxide
Oxygen is from the French oxygène,
meaning “generator of acid,” because it
was initially considered to be an integral
part of all acids.
Experiment 15: Composition 15:
Percentage Water in a Hydrate
By the late eighteenth century, combustion had been studied extensively; the gases
carbon dioxide, nitrogen, hydrogen, and oxygen had been discovered; and the list of
elements continued to grow. However, it was Antoine Lavoisier (1743–1794), a French
chemist (Fig. 2.2), who finally explained the true nature of combustion, thus clearing
the way for the tremendous progress that was made near the end of the eighteenth
century. Lavoisier, like Boyle, regarded measurement as the essential operation of
chemistry. His experiments, in which he carefully weighed the reactants and products
of various reactions, suggested that mass is neither created nor destroyed. Lavoisier’s
verification of this law of conservation of mass was the basis for the developments in
chemistry in the nineteenth century. Mass is neither created nor destroyed in a chemical reaction.
Lavoisier’s quantitative experiments showed that combustion involved oxygen
(which Lavoisier named), not phlogiston. He also discovered that life was supported
by a process that also involved oxygen and was similar in many ways to combustion.
In 1789 Lavoisier published the first modern chemistry textbook, Elementary Treatise
on Chemistry, in which he presented a unified picture of the chemical knowledge assembled up to that time. ­Unfortunately, in the same year the text was published, the
French Revolution broke out. Lavoisier, who had been associated with collecting taxes
for the government, was executed on the guillotine as an enemy of the people in 1794.
After 1800, chemistry was dominated by scientists who, following Lavoisier’s lead,
performed careful weighing experiments to study the course of chemical reactions and
to determine the composition of various chemical compounds. One of these chemists,
a Frenchman, Joseph Proust (1754–1826), showed that a given compound always contains exactly the same proportion of elements by mass. For example, Proust found that
the substance copper carbonate is always 5.3 parts copper to 4 parts oxygen to 1 part
*Oxygen gas was actually first observed by the Swedish chemist Karl W. Scheele (1742–1786), but
because his results were published after Priestley’s, the latter is commonly credited with the discovery
of oxygen.
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2.2
Fundamental Chemical Laws
45
The Metropolitan Museum of Art, New York. Image copyright © The Metropolitan Museum of Art/Art Resource, NY
Figure 2.2 | Antoine Lavoisier was
born in Paris on August 26, 1743.
Although Lavoisier’s ­father wanted his
son to follow him into the legal
profession, young Lavoisier was
fascinated by science. From the
beginning of his scientific career,
Lavoisier recognized the importance
of accurate measurements. His careful
weighings showed that mass is
conserved in chemical reactions and
that combustion involves reaction
with oxygen. Also, he wrote the first
modern chemistry textbook. It is not
surprising that Lavoisier is often
called the father of modern chemistry.
To help support his scientific work,
Lavoisier invested in a private
tax-collecting firm and married the
daughter of one of the company
executives. His connection to the tax
collectors proved fatal, for radical
French revolutionaries demanded his
execution, which occurred on the
guillotine on May 8, 1794.
carbon (by mass). The principle of the constant composition of compounds, originally
called “Proust’s law,” is now known as the law of definite proportion. A given compound always contains exactly the same proportion of elements by mass.
Proust’s discovery stimulated John Dalton (1766–1844), an English schoolteacher
(Fig. 2.3), to think about atoms as the particles that might compose elements. Dalton
reasoned that if elements were composed of tiny individual particles, a given compound should always contain the same combination of these atoms. This concept explained why the same relative masses of elements were always found in a given
compound.
But Dalton discovered another principle that convinced him even more of the existence of atoms. He noted, for example, that carbon and oxygen form two different
compounds that contain different relative amounts of carbon and oxygen, as shown by
the ­following data:
Mass of Oxygen That Combines
with 1 g of Carbon
Compound I
Compound II
1.33 g
2.66 g
Manchester Literary and Philosophical Society
Dalton noted that compound II contains twice as much oxygen per gram of carbon as
compound I, a fact that could easily be explained in terms of atoms. Compound I might
Figure 2.3 | John Dalton (1766–1844), an Englishman, began teaching at a Quaker school
when he was 12. His fascination with science included an intense interest in meteorology,
which led to an interest in the gases of the air and their ultimate components, atoms.
Dalton is best known for his atomic theory, in which he postulated that the fundamental
differences among atoms are their masses. He was the first to prepare a table of relative
atomic weights.
Dalton was a humble man with several apparent handicaps: He was not articulate and
he was color-blind, a terrible problem for a chemist. Despite these disadvantages, he helped
to revolutionize the science of chemistry.
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46
Chapter 2
Atoms, Molecules, and Ions
be CO, and compound II might be CO2.* This principle, which was found to apply to
compounds of other elements as well, became known as the law of multiple proportions: When two elements form a series of compounds, the ratios of the masses of the
second element that combine with 1 g of the first element can always be reduced to
small whole numbers.
To make sure the significance of this observation is clear, in Example 2.1 we will
consider data for a series of compounds consisting of nitrogen and oxygen.
Example 2.1
Illustrating the Law of Multiple Proportions
The following data were collected for several compounds of nitrogen and oxygen:
Mass of Nitrogen That Combines
with 1 g of Oxygen
Compound A
Compound B
Compound C
1.750 g
0.8750 g
0.4375 g
Show how these data illustrate the law of multiple proportions.
Solution
For the law of multiple proportions to hold, the ratios of the masses of nitrogen combining with 1 g of oxygen in each pair of compounds should be small whole numbers.
We therefore compute the ratios as follows:
A
1.750
2
5
5
B
0.8750
1
B
0.8750
2
5
5
C
0.4375
1
A
1.750
4
5
5
C
0.4375
1
These results support the law of multiple proportions.
See Exercises 2.37 and 2.38
The significance of the data in Example 2.1 is that compound A contains twice as
much nitrogen (N) per gram of oxygen (O) as does compound B and that compound B
contains twice as much nitrogen per gram of oxygen as does compound C.
These data can be explained readily if the substances are composed of molecules
made up of nitrogen atoms and oxygen atoms. For example, one set of possibilities for
compounds A, B, and C is
B:
A:
N
O
=
2
1
C:
N
O
=
1
1
N
O
=
1
2
Now we can see that compound A contains two atoms of N for every atom of O,
whereas compound B contains one atom of N per atom of O. That is, compound A
contains twice as much nitrogen per given amount of oxygen as does compound B.
Similarly, since compound B contains one N per O and compound C contains one N
*Subscripts are used to show the numbers of atoms present. The number 1 is understood (not written). The symbols for the elements and the writing of chemical formulas will be illustrated further
in Sections 2.6 and 2.7.
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2.3 Dalton’s Atomic Theory
47
per two Os, the nitrogen content of compound C per given amount of oxygen is half
that of compound B.
Another set of compounds that fits the data in Example 2.1 is
B:
A:
N
O
N
O
1
1
=
C:
=
N
O
1
2
=
1
4
N
O
=
Verify for yourself that these compounds satisfy the requirements.
Still another set that works is
B:
A:
N
O
=
4
2
C:
N
O
=
2
2
2
4
See if you can come up with still another set of compounds that satisfies the data in
­Example 2.1. How many more possibilities are there?
In fact, an infinite number of other possibilities exists. Dalton could not deduce
absolute formulas from the available data on relative masses. However, the data on the
­composition of compounds in terms of the relative masses of the elements supported
his hypothesis that each element consisted of a certain type of atom and that compounds were formed from specific combinations of atoms.
2.3 Dalton’s Atomic Theory
In 1808 Dalton published A New System of Chemical Philosophy, in which he presented his theory of atoms:
Dalton’s Atomic Theory
These statements are a modern
paraphrase of Dalton’s ideas.
1. Each element is made up of tiny particles called atoms.
2. The atoms of a given element are identical; the atoms of different elements are different in some fundamental way or ways.
3. Chemical compounds are formed when atoms of different elements combine with
each other. A given compound always has the same relative numbers and types of
atoms.
4. Chemical reactions involve reorganization of the atoms—changes in the way they
are bound together. The atoms themselves are not changed in a chemical reaction.
It is instructive to consider Dalton’s reasoning on the relative masses of the atoms of
the various elements. In Dalton’s time water was known to be composed of the elements
hydrogen and oxygen, with 8 g of oxygen present for every 1 g of hydrogen. If the formula for water were OH, an oxygen atom would have to have 8 times the mass of a
hydrogen atom. However, if the formula for water were H2O (two atoms of hydrogen
for every oxygen atom), this would mean that each atom of oxygen is 16 times as massive as each atom of hydrogen (since the ratio of the mass of one oxygen to that of two
hydrogens is 8 to 1). Because the formula for water was not then known, Dalton could
not specify the relative masses of oxygen and hydrogen unambiguously. To solve the
­problem, Dalton made a fundamental assumption: He decided that nature would be as
simple as possible. This assumption led him to conclude that the formula for water
should be OH. He thus assigned hydrogen a mass of 1 and oxygen a mass of 8.
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48
Chapter 2
Atoms, Molecules, and Ions
Chemical connections
Berzelius, Selenium, and Silicon
Jöns Jakob Berzelius was probably the
best experimental chemist of his
generation and, given the crudeness of
his laboratory equipment, maybe the
best of all time. Unlike Lavoisier, who
Comparison of Several of Berzelius’s Atomic
Masses with the Modern Values
Atomic Mass
Element
Berzelius’s
Value
Current
Value
Chlorine
Copper
Hydrogen
Lead
Nitrogen
Oxygen
Potassium
Silver
Sulfur
35.41
63.00
1.00
207.12
14.05
16.00
39.19
108.12
32.18
35.45
63.55
1.01
207.2
14.01
16.00
39.10
107.87
32.07
could ­afford to buy the best laboratory
equipment available, Berzelius worked
with minimal equipment in very plain
­surroundings. One of Berzelius’s
students described the Swedish
chemist’s workplace: “The laboratory
consisted of two ordinary rooms with
the very simplest arrangements; there
were neither furnaces nor hoods,
neither water system nor gas. Against
the walls stood some closets with the
chemicals, in the middle the ­mercury
trough and the blast lamp table.
­Beside this was the sink consisting of a
stone water holder with a stopcock
and a pot standing under it. [Next
door in the kitchen] stood a small
heating furnace.”
In these simple facilities, Berzelius
performed more than 2000 experiments over a 10-year period to
determine accurate atomic masses for
the 50 elements then known. His
success can be seen from the data in
the table at left. These remarkably
accurate values attest to his experimental skills and patience.
Besides his table of atomic masses,
Berzelius made many other major
contributions to chemistry. The most
important of these was the invention
of a simple set of symbols for the
elements along with a system for
writing the formulas of compounds to
replace the awkward symbolic
representations of the alchemists.
Although some chemists, including
Dalton, objected to the new system, it
was gradually adopted and forms the
basis of the system we use today.
In addition to these accomplishments, Berzelius discovered the
elements cerium, thorium, selenium,
and ­silicon. Of these elements,
The Granger Collection, New York
Using similar reasoning for other compounds, Dalton prepared the first table of
atomic masses (sometimes called atomic weights by chemists, since mass is often
determined by comparison to a standard mass—a process called weighing). Many of
the masses were later proved to be wrong because of Dalton’s incorrect assumptions
about the formulas of certain compounds, but the construction of a table of masses
was an important step forward.
Although not recognized as such for many years, the keys to determining absolute
formulas for compounds were provided in the experimental work of the French chemist Joseph Gay-Lussac (1778–1850) and by the hypothesis of an Italian chemist named
Amadeo Avogadro (1776–1856). In 1809 Gay-Lussac performed experiments in
which he measured (under the same conditions of temperature and pressure) the volumes of gases that reacted with each other. For example, Gay-Lussac found that
Joseph Louis Gay-Lussac, a French physicist and chemist, was remarkably versatile. Although he
is now known primarily for his studies on the combining of volumes of gases, Gay-Lussac was
instrumental in the studies of many of the other properties of gases. Some of Gay-Lussac’s
motivation to learn about gases arose from his passion for ballooning. In fact, he made ascents
to heights of over 4 miles to collect air samples, setting altitude records that stood for about
50 years. Gay-Lussac also was the codiscoverer of boron and the developer of a process for
manufacturing sulfuric acid. As chief assayer of the French mint, Gay-Lussac developed many
techniques for chemical analysis and invented many types of glassware now used routinely in
labs. Gay-Lussac spent his last 20 years as a lawmaker in the French government.
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2.3 Dalton’s Atomic Theory
Substance
Alchemists’ Symbol
Silver
Lead
Tin
Platinum
Sulfuric acid
Alcohol
Sea salt
selenium and silicon are particularly
important in today’s world. Berzelius
discovered ­selenium in 1817 in
connection with his studies of sulfuric
acid. For years selenium’s toxicity has
been known, but only ­recently have we
become aware that it may have a
positive effect on human health.
Studies have shown that trace
amounts of selenium in the diet may
protect people from heart disease and
cancer. One study based on data from
27 countries showed an inverse
relationship between the cancer death
rate and the selenium content of soil
in a particular region (low cancer
death rate in areas with high selenium
content). Another research paper
reported an inverse relationship
between the selenium content of the
blood and the incidence of breast
cancer in women. A study reported in
1998 used the toenail clippings of
33,737 men to show that selenium
seems to protect against prostate
cancer. Selenium is also found in the
heart muscle and may play an
important role in proper heart
function. Because of these and other
studies, selenium’s reputation has
improved, and many scientists are now
studying its function in the human
body.
Silicon is the second most abundant element in the earth’s crust,
exceeded only by oxygen. As we will
see in Chapter 10, compounds
involving silicon bonded to oxygen
make up most of the earth’s sand, rock,
and soil. Berzelius prepared silicon in
its pure form in 1824 by heating silicon
tetrafluoride (SiF4) with potassium
metal. Today, silicon forms the basis
for the modern microelectronics
industry centered near San Francisco
in a place that has come to be known
as “Silicon Valley.” The technology of
the silicon chip (see figure) with its
printed ­circuits has transformed
computers from room-sized monsters
with thousands of unreliable vacuum
tubes to desktop and notebook-sized
units with trouble-free “solid-state”
circuitry.
Courtesy IBM
The Alchemists’ Symbols for Some
Common Elements and Compounds
49
A chip capable of transmitting 4,000,000
simultaneous phone conversations.
2 volumes of hydrogen react with 1 volume of oxygen to form 2 volumes of gaseous
water and that 1 volume of hydrogen reacts with 1 volume of chlorine to form 2 volumes of hydrogen chloride. These results are represented schematically in Fig. 2.4.
In 1811 Avogadro interpreted these results by proposing that at the same temperature and pressure, equal volumes of different gases contain the same number of particles. This assumption (called Avogadro’s hypothesis) makes sense if the distances
between the particles in a gas are very great compared with the sizes of the particles.
Under these conditions, the volume of a gas is determined by the number of molecules
present, not by the size of the individual particles.
+
2 volumes hydrogen
to form 2 volumes gaseous water
+
Figure 2.4 | A representation of
some of Gay-Lussac’s experimental
results on combining gas volumes.
combines with 1 volume oxygen
1 volume hydrogen
combines with 1 volume chlorine to form 2 volumes hydrogen chloride
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50
Chapter 2
Atoms, Molecules, and Ions
Figure 2.5 | A representation of
combining gases at the molecular
level. The spheres ­represent atoms in
the molecules.
H H
H H
H H
+
+
O
Cl
O
Cl
H
H
O
Cl
H
H
H
O
H
Cl
If Avogadro’s hypothesis is correct, Gay-Lussac’s result,
2 volumes of hydrogen react with 1 volume of oxygen 8n 2 volumes of water vapor
can be expressed as follows:
2 molecules* of hydrogen react with 1 molecule of oxygen 8n 2 molecules of water
These observations can best be explained by assuming that gaseous hydrogen, oxygen,
and chlorine are all composed of diatomic (two-atom) molecules: H2, O2, and Cl2,
­respectively. Gay-Lussac’s results can then be represented as shown in Fig. 2.5. (Note
that this reasoning suggests that the formula for water is H2O, not OH as Dalton
believed.)
Unfortunately, Avogadro’s interpretations were not accepted by most chemists, and
a half-century of confusion followed, in which many different assumptions were made
about formulas and atomic masses.
During the nineteenth century, painstaking measurements were made of the masses
of various elements that combined to form compounds. From these experiments a list
of relative atomic masses could be determined. One of the chemists involved in contributing to this list was a Swede named Jöns Jakob Berzelius (1779–1848), who discovered the elements cerium, selenium, silicon, and thorium and developed the modern symbols for the elements used in writing the formulas of compounds.
There are seven elements that occur as
diatomic molecules:
H2, N2, O2, F2, Cl2, Br2, I2
IBLG: See questions from
“The Nature of the Atom”
2.4Early Experiments to Characterize
the Atom
On the basis of the work of Dalton, Gay-Lussac, Avogadro, and others, chemistry was
beginning to make sense. The concept of atoms was clearly a good idea. Inevitably,
scientists began to wonder about the nature of the atom. What is an atom made of, and
how do the atoms of the various elements differ?
The Electron
The Cavendish Laboratory
The first important experiments that led to an understanding of the composition of the
atom were done by the English physicist J. J. Thomson (Fig. 2.6), who studied electrical discharges in partially evacuated tubes called cathode-ray tubes (Fig. 2.7) during
*A molecule is a collection of atoms (see Section 2.6).
Figure 2.6 | J. J. Thomson (1856–1940) was an ­English physicist at Cambridge ­University.
He received the Nobel Prize in physics in 1906.
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2.4
51
Early Experiments to Characterize the Atom
Richard Megna/Fundamental Photographs
© Cengage Learning
Source of electrical potential
Stream of negative
particles (electrons)
(–)
Metal
electrode
(+)
Partially
evacuated
glass tube
Metal
electrode
Figure 2.7 | A cathode-ray tube. The fast-moving electrons excite the gas in the tube, causing a glow between the
electrodes. The green color in the photo is due to the response of the screen (coated with zinc sulfide) to the electron
beam.
the period from 1898 to 1903. Thomson found that when high voltage was applied to
the tube, a “ray” he called a cathode ray (because it emanated from the negative electrode, or cathode) was produced. Because this ray was produced at the negative electrode and was repelled by the negative pole of an applied electric field (Fig. 2.8),
Thomson postulated that the ray was a stream of negatively charged particles, now
called electrons. From experiments in which he measured the deflection of the beam
of electrons in a magnetic field, Thomson determined the charge-to-mass ratio of an
electron:
StockFood/Getty Images
e
5 21.76 3 108 C/g
m
A classic English plum pudding in which
the raisins represent the distribution of
electrons in the atom.
where e represents the charge on the electron in coulombs (C) and m represents the electron mass in grams.
One of Thomson’s primary goals in his cathode-ray tube experiments was to gain
an understanding of the structure of the atom. He reasoned that since electrons could
be produced from electrodes made of various types of metals, all atoms must contain
electrons. Since atoms were known to be electrically neutral, Thomson further assumed that atoms also must contain some positive charge. Thomson postulated that an
atom consisted of a diffuse cloud of positive charge with the negative electrons embedded randomly in it. This model, shown in Fig. 2.9, is often called the plum pudding
model because the electrons are like raisins dispersed in a pudding (the positive charge
cloud), as in plum pudding, a favorite English dessert.
Spherical cloud of
positive charge
Applied
electric field
(+)
Electrons
(–)
Metal
electrode
(+)
(–)
Metal
electrode
Figure 2.8 | Deflection of cathode rays by an applied electric
Figure 2.9 | The plum pudding
field.
model of the atom.
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52
Chapter 2
Atoms, Molecules, and Ions
Atomizer to
produce oil
droplets
Oil spray
(+)
Electrically
charged plates
The Granger Collection, New York
Microscope
X rays produce
charges on the
oil drops
(–)
a
b
Figure 2.10 | (a) A schematic representation of the apparatus Millikan used to determine the charge on the
electron. The fall of charged oil droplets due to gravity can be halted by adjusting the voltage across the two
plates. This voltage and the mass of the oil drop can then be used to calculate the charge on the oil drop.
Millikan’s experiments showed that the charge on an oil drop is always a whole-number multiple of the electron
charge. (b) Robert Millikan using his apparatus.
In 1909 Robert Millikan (1868–1953), working at the University of Chicago, performed very clever experiments involving charged oil drops. These experiments allowed him to determine the magnitude of the electron charge (Fig. 2.10). With this
value and the charge-to-mass ratio determined by Thomson, Millikan was able to calculate the mass of the electron as 9.11 3 10231 kg.
PowerLectures:
Cathode-Ray Tube
Millikan’s Oil-Drop Experiment
Radioactivity
In the late nineteenth century, scientists discovered that certain elements produce highenergy radiation. For example, in 1896 the French scientist Henri Becquerel found
accidentally that a piece of a mineral containing uranium could produce its image on
a photographic plate in the absence of light. He attributed this phenomenon to a spontaneous emission of radiation by the uranium, which he called radioactivity. Studies
in the early twentieth century demonstrated three types of radioactive emission:
gamma (g) rays, beta (b) particles, and alpha (a) particles. A g ray is high-energy
“light”; a b particle is a high-speed electron; and an a particle has a 21 charge, that is,
a charge twice that of the electron and with the opposite sign. The mass of an a particle
is 7300 times that of the electron. More modes of radioactivity are now known, and we
will discuss them in Chapter 19. Here we will consider only a particles because they
were used in some crucial early experiments.
The Nuclear Atom
Topham Picture Library/The Image Works
In 1911 Ernest Rutherford (Fig. 2.11), who performed many of the pioneering experiments to explore radioactivity, carried out an experiment to test Thomson’s plum pudding model. The experiment involved directing a particles at a thin sheet of metal foil,
Figure 2.11 | Ernest Rutherford (1871–1937) was born on a farm in New Zealand. In 1895 he
placed second in a scholarship competition to attend Cambridge University but was awarded
the scholarship when the winner decided to stay home and get married. As a scientist in
England, ­Rutherford did much of the early work on characterizing radioactivity. He named
the a and b particles and the g ray and coined the term half-life to describe an important
attribute of radioactive elements. His experiments on the behavior of a particles striking
thin metal foils led him to postulate the nuclear atom. He also invented the name proton for
the nucleus of the hydrogen atom. He received the Nobel Prize in chemistry in 1908.
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2.4
Early Experiments to Characterize the Atom
Figure 2.12 | Rutherford’s experi-
Some α particles
are scattered.
ment on a-particle bombardment of
metal foil.
PowerLecture: Gold Foil Experiment
Source of
α particles
53
Most particles
pass straight
through foil.
Beam of
α particles
Screen to detect
scattered α particles
Thin
metal foil
as illustrated in Fig. 2.12. Rutherford reasoned that if Thomson’s model were accurate,
the massive a particles should crash through the thin foil like cannonballs through
gauze, as shown in Fig. 2.13(a). He expected the a particles to travel through the foil
with, at the most, very minor deflections in their paths. The results of the experiment
were very ­different from those Rutherford anticipated. Although most of the a particles passed straight through, many of the particles were deflected at large angles, as
shown in Fig. 2.13(b), and some were reflected, never hitting the detector. This outcome was a great surprise to Rutherford. (He wrote that this result was comparable
with shooting a howitzer at a piece of paper and having the shell reflected back.)
Rutherford knew from these results that the plum pudding model for the atom could
not be correct. The large deflections of the a particles could be caused only by a center
of concentrated positive charge that contains most of the atom’s mass, as illustrated in
Fig. 2.13(b). Most of the a particles pass directly through the foil because the atom is
mostly open space. The deflected a particles are those that had a “close encounter”
with the massive positive center of the atom, and the few reflected a particles are those
that made a “direct hit” on the much more massive positive center.
In Rutherford’s mind these results could be explained only in terms of a nuclear
atom—an atom with a dense center of positive charge (the nucleus) with electrons moving around the nucleus at a distance that is large relative to the nuclear radius.
Critical Thinking
You have learned about three different models of the atom: Dalton’s model, Thomson’s model, and Rutherford’s model. What if Dalton was correct? What would
Rutherford have expected from his experiments with gold foil? What if Thomson was
correct? What would Rutherford have expected from his experiments with gold foil?
Electrons scattered
throughout
–
–
Diffuse
positive
charge
–
–
–
–
–
–
–
–
–
n+
–
–
–
–
Figure 2.13 | Rutherford’s
experiment.
–
–
–
–
a
The expected results of the metal foil
experiment if Thomson’s model were correct.
–
b
Actual results.
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54
Chapter 2
Atoms, Molecules, and Ions
2.5The Modern View of Atomic Structure:
An Introduction
The forces that bind the positively
charged protons in the nucleus will be
discussed in Chapter 19.
Photo © Cengage Learning. All rights reserved.
The chemistry of an atom arises from its
electrons.
If the atomic nucleus were the size of
this ball bearing, a typical atom would
be the size of this stadium.
In the years since Thomson and Rutherford, a great deal has been learned about atomic
structure. Because much of this material will be covered in detail in later chapters,
only an introduction will be given here. The simplest view of the atom is that it consists of a tiny nucleus (with a diameter of about 10213 cm) and electrons that move
about the ­nucleus at an average distance of about 1028 cm from it (Fig. 2.14).
As we will see later, the chemistry of an atom mainly results from its electrons. For
this reason, chemists can be satisfied with a relatively crude nuclear model. The nucleus is assumed to contain protons, which have a positive charge equal in magnitude
to the electron’s negative charge, and neutrons, which have virtually the same mass as
a proton but no charge. The masses and charges of the electron, proton, and neutron are
shown in Table 2.1.
Two striking things about the nucleus are its small size compared with the overall
size of the atom and its extremely high density. The tiny nucleus accounts for almost all
the atom’s mass. Its great density is dramatically demonstrated by the fact that a piece
of nuclear material about the size of a pea would have a mass of 250 million tons!
An important question to consider at this point is, “If all atoms are composed of these
same components, why do different atoms have different chemical properties?” The answer to this question lies in the number and the arrangement of the electrons. The electrons constitute most of the atomic volume and thus are the parts that “intermingle”
when atoms combine to form molecules. Therefore, the number of electrons possessed
by a given atom greatly affects its ability to interact with other atoms. As a result, the
atoms of different elements, which have different numbers of protons and electrons,
show different chemical behavior.
A sodium atom has 11 protons in its nucleus. Since atoms have no net charge, the
number of electrons must equal the number of protons. Therefore, a sodium atom has
11 electrons moving around its nucleus. It is always true that a sodium atom has
11 protons and 11 electrons. However, each sodium atom also has neutrons in its nucleus, and different types of sodium atoms exist that have different numbers of neutrons. For example, consider the sodium atoms represented in Fig. 2.15. These two
atoms are isotopes, or atoms with the same number of protons but different numbers of
neutrons. Note that the symbol for one particular type of sodium atom is written
Mass number
Mass number 88n
A
X m Element symbol
8nZ
8
Atomic number
Nucleus
88n
n
Atomic number­­ 88
23
11Na
m Element symbol
where the atomic number Z (number of protons) is written as a subscript, and the mass
number A (the total number of protons and neutrons) is written as a superscript. (The
­particular atom represented here is called “sodium twenty-three.” It has 11 electrons,
11 protons, and 12 neutrons.) Because the chemistry of an atom is due to its electrons,
isotopes show almost identical chemical properties. In nature most elements contain mixtures of isotopes.
Table 2.1 | The Mass and Charge of the Electron,
Proton, and Neutron
~10−13 cm
~2 × 10−8 cm
Figure 2.14 | A nuclear atom
viewed in cross section. Note that this
drawing is not to scale.
Particle
Mass
Charge*
Electron
Proton
Neutron
9.109 3 10231 kg
1.673 3 10227 kg
1.675 3 10227 kg
12
11
None
*The magnitude of the charge of the electron and the
proton is 1.60 3 10219 C.
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2.6
Figure 2.15 | Two isotopes of
Nucleus
sodium. Both have 11 protons and
11 electrons, but they ­differ in the
number of neutrons in their nuclei.
55
Nucleus
11 protons
12 neutrons
11 protons
13 neutrons
11 electrons
23
11
Molecules and Ions
11 electrons
24
11
Na
Na
Critical Thinking
The average diameter of an atom is 2 3 10210 m. What if the average diameter of an
atom were 1 cm? How tall would you be?
Interactive
Example 2.2
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Writing the Symbols for Atoms
Write the symbol for the atom that has an atomic number of 9 and a mass number of
19. How many electrons and how many neutrons does this atom have?
Solution
The atomic number 9 means the atom has 9 protons. This element is called fluorine,
symbolized by F. The atom is represented as
19
9F
and is called fluorine nineteen. Since the atom has 9 protons, it also must have 9 electrons to achieve electrical neutrality. The mass number gives the total number of protons and neutrons, which means that this atom has 10 neutrons.
See Exercises 2.59 through 2.62
2.6 Molecules and Ions
PowerLecture: Covalent Bonding
From a chemist’s viewpoint, the most interesting characteristic of an atom is its ability to
combine with other atoms to form compounds. It was John Dalton who first recognized
that chemical compounds are collections of atoms, but he could not determine the structure of atoms or their means for binding to each other. During the twentieth century, we
learned that atoms have electrons and that these electrons participate in bonding one
atom to another. We will discuss bonding thoroughly in Chapters 8 and 9; here, we will
introduce some simple bonding ideas that will be useful in the next few chapters.
The forces that hold atoms together in compounds are called chemical bonds.
One way that atoms can form bonds is by sharing electrons. These bonds are called
covalent bonds, and the resulting collection of atoms is called a molecule. Molecules can be represented in several different ways. The simplest method is the
chemical formula, in which the symbols for the elements are used to indicate the
types of atoms present and subscripts are used to indicate the relative numbers of atoms. For example, the formula for carbon dioxide is CO2, meaning that each molecule
contains 1 atom of carbon and 2 atoms of oxygen.
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Chapter 2
Atoms, Molecules, and Ions
Figure 2.16 | (a) The structural
formula for methane. (b) Space-filling
model of methane. This type of model
shows both the relative sizes of the
atoms in the molecule and their spatial
relationships. (c) Ball-and-stick model
of methane.
H
Photos Ken O’Donoghue © Cengage Learning
56
C
H
H
H
Methane
a
b
c
Examples of molecules that contain covalent bonds are hydrogen (H2), water (H2O),
oxygen (O2), ammonia (NH3), and methane (CH4). More information about a molecule
is given by its structural formula, in which the individual bonds are shown (indicated
by lines). Structural formulas may or may not indicate the actual shape of the molecule. For example, water might be represented as
H O H
N
H H H
Ammonia
or
O
H
H
The structure on the right shows the actual shape of the water molecule. Scientists
know from experimental evidence that the molecule looks like this. (We will study the
shapes of molecules further in Chapter 8.)
The structural formula for ammonia is shown in the margin at left. Note that atoms
connected to the central atom by dashed lines are behind the plane of the paper, and
atoms connected to the central atom by wedges are in front of the plane of the paper.
In a compound composed of molecules, the individual molecules move around as
independent units. For example, a molecule of methane gas can be represented in
several ways. The structural formula for methane (CH4) is shown in Fig. 2.16(a). The
space-filling model of methane, which shows the relative sizes of the atoms as well
as their relative orientation in the molecule, is given in Fig. 2.16(b). Ball-and-stick
models are also used to represent molecules. The ball-and-stick structure of methane
is shown in Fig. 2.16(c).
A second type of chemical bond results from attractions among ions. An ion is an
atom or group of atoms that has a net positive or negative charge. The best-known
ionic compound is common table salt, or sodium chloride, which forms when neutral
chlorine and sodium react.
To see how the ions are formed, consider what happens when an electron is transferred
from a sodium atom to a chlorine atom (the neutrons in the nuclei will be ignored):
Neutral sodium
atom (Na)
Sodium ion
(Na+)
PowerLecture: Determining Formulas
for Ionic ­Compounds
11+
Minus 1 electron
11+
10 electrons
11 electrons
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2.7
Na1 is usually called the sodium ion rather
than the sodium cation. Also Cl2 is called
the chloride ion rather than the chloride
anion. In general, when a specific ion
is referred to, the word ion rather than
cation or anion is used.
An Introduction to the Periodic Table
57
With one electron stripped off, the sodium, with its 11 protons and only 10 electrons,
now has a net 11 charge—it has become a positive ion. A positive ion is called a
cation. The sodium ion is written as Na1, and the process can be represented in shorthand form as
Na h Na 1 1 e 2
If an electron is added to chlorine,
Chloride ion
(Cl−)
Neutral chlorine
atom (Cl)
17+
Plus 1 electron
17+
17 electrons
18 electrons
the 18 electrons produce a net 12 charge; the chlorine has become an ion with a negative charge—an anion. The chloride ion is written as Cl2, and the process is represented as
Cl 1 e 2 h Cl 2
Because anions and cations have opposite charges, they attract each other. This
force of attraction between oppositely charged ions is called ionic bonding. As illustrated in Fig. 2.17, sodium metal and chlorine gas (a green gas composed of
Cl2 molecules) react to form solid sodium chloride, which contains many Na1 and
Cl2 ions packed together and forms the beautiful colorless cubic crystals.
A solid consisting of oppositely charged ions is called an ionic solid. Ionic solids
can consist of simple ions, as in sodium chloride, or of polyatomic (many atom) ions,
as in ammonium nitrate (NH4NO3), which contains ammonium ions (NH41) and nitrate ions (NO32). The ball-and-stick models of these ions are shown in Fig. 2.18.
2.7 An Introduction to the Periodic Table
Experiment 25: Properties of
Representative Elements
In a room where chemistry is taught or practiced, a chart called the periodic table is
almost certain to be found hanging on the wall. This chart shows all the known elements and gives a good deal of information about each. As our study of chemistry
progresses, the usefulness of the periodic table will become more obvious. This section will simply introduce it to you.
A simplified version of the periodic table is shown in Fig. 2.19. The letters in the
boxes are the symbols for the elements; these abbreviations are based on the current
­element names or the original names (Table 2.2). The number shown above each
symbol is the atomic number (number of protons) for that element. For example, carbon
(C) has atomic number 6, and lead (Pb) has atomic number 82. Most of the elements
are metals. Metals have characteristic physical properties such as efficient conduction
of heat and electricity, malleability (they can be hammered into thin sheets), ductility
(they can be pulled into wires), and (often) a lustrous appearance. Chemically, metals
tend to lose electrons to form positive ions. For example, copper is a typical metal. It
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Charles D. Winters/Photo Researchers, Inc.
Photo © Cengage Learning. All rights reserved.
Atoms, Molecules, and Ions
Photo © Cengage Learning. All rights reserved.
Chapter 2
Photo © Cengage Learning. All rights reserved.
58
Cl−
Na+
Cl−
Na+
Na
Na
Cl Cl
Figure 2.17 | Sodium metal (which is so soft it can be cut with a knife and which consists of individual sodium atoms) reacts with
chlorine gas (which contains Cl2 molecules) to form solid sodium chloride (which contains Na1 and Cl2 ions packed together).
Figure 2.18 | Ball-and-stick models
of the ammonium ion (NH41) and the
nitrate ion (NO32). These ions are
each held together by covalent bonds.
is lustrous (although it tarnishes readily); it is an excellent conductor of electricity (it
is widely used in electrical wires); and it is readily formed into various shapes, such as
pipes for water systems. Copper is also found in many salts, such as the beautiful blue
copper sulfate, in which copper is present as Cu21 ions. Copper is a member of the
transition metals—the metals shown in the center of the periodic table.
The relatively few nonmetals appear in the upper-right corner of the table (to the
right of the heavy line in Fig. 2.19), except hydrogen, a nonmetal that resides in the
Table 2.2 | The Symbols for the Elements That Are
PowerLecture: Comparison of a
Molecular Compound and an Ionic
Compound
Based on the Original Names
Current Name
Original Name
Symbol
Antimony
Copper
Iron
Lead
Mercury
Potassium
Silver
Sodium
Tin
Tungsten
Stibium
Cuprum
Ferrum
Plumbum
Hydrargyrum
Kalium
Argentum
Natrium
Stannum
Wolfram
Sb
Cu
Fe
Pb
Hg
K
Ag
Na
Sn
W
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2.7
Noble
gases
Alkaline
1 earth metals
Halogens 18
1A
1
Alkali metals
H
59
An Introduction to the Periodic Table
8A
2
13
14
15
16
17
2A
3A
4A
5A
6A
7A
2
He
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
3
4
5
6
7
8
Transition metals
9
10
11
12
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
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
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
87
88
89
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
Fr
Ra
Ac†
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Uut
Fl
Uup
Lv
Uus
Uuo
*Lanthanides
†
Actinides
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Figure 2.19 | The periodic table.
Metals tend to form positive ions;
­nonmetals tend to form negative ions.
Photo © Cengage Learning. All rights reserved.
Elements in the same vertical column in
the periodic table form a group (or family)
and generally have similar properties.
Samples of chlorine gas, liquid bromine,
and solid iodine.
u­ pper-left corner. The nonmetals lack the physical properties that characterize the metals. Chemically, they tend to gain electrons in reactions with metals to form negative
ions. Nonmetals often bond to each other by forming covalent bonds. For example,
chlorine is a typical nonmetal. Under normal conditions it exists as Cl2 molecules; it
reacts with metals to form salts containing Cl2 ions (NaCl, for example); and it forms
covalent bonds with nonmetals (for example, hydrogen chloride gas, HCl).
The periodic table is arranged so that elements in the same vertical columns (called
groups or families) have similar chemical properties. For example, all of the alkali
metals, members of Group 1A—lithium (Li), sodium (Na), potassium (K), rubidium
(Rb), cesium (Cs), and francium (Fr)—are very active elements that readily form
ions with a 11 charge when they react with nonmetals. The members of Group 2A—
beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra)—are called the alkaline earth metals. They all form ions with a 21 charge
when they react with nonmetals. The halogens, the members of Group 7A—fluorine
(F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At)—all form diatomic molecules. Fluorine, chlorine, bromine, and iodine all react with metals to form salts
containing ions with a 12 charge (F2, Cl2, Br2, and I2). The members of Group 8A—
helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn)—
are known as the noble gases. They all ­exist under normal conditions as monatomic
(single-atom) gases and have little chemical reactivity.
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60
Chapter 2
Atoms, Molecules, and Ions
Chemical connections
Hassium Fits Right In
Hassium, element 108, does not exist
in ­nature but must be made in a
particle accelerator. It was first created
in 1984 and can be made by shooting
magnesium-26 (26
1 2Mg) atoms at
curium-248 (248
96Cm) atoms. The
collisions between these atoms
produce some hassium-265 (265
108Hs)
atoms. The position of hassium in the
periodic table (see Fig. 2.19) in the
vertical column containing iron,
ruthenium, and osmium suggests that
hassium should have chemical
properties similar to these metals.
Another format of the periodic table will
be discussed in Section 7.11.
IBLG: See questions from
“Nomenclature” and
“Naming Compounds”
However, it is not easy to test this
prediction—only a few atoms of
hassium can be made at a given time
and they last for only about 9 seconds.
Imagine having to get your next lab
experiment done in 9 seconds!
Amazingly, a team of chemists from
the Lawrence ­Berkeley National
Laboratory in California, the Paul
Scherrer Institute and the University of
Bern in Switzerland, and the Institute
of Nuclear Chemistry in Germany have
done experiments to characterize the
chemical behavior of hassium. For
example, they have observed that
hassium atoms react with oxygen to
form a hassium oxide compound of the
type expected from its position on the
periodic table. The team has also
measured other properties of hassium,
including the energy released as it
undergoes nuclear decay to another
atom.
This work would have surely
pleased Dmitri Mendeleev (see Fig.
7.24), who originally developed the
periodic table and showed its power to
predict chemical properties.
Note from Fig. 2.19 that alternate sets of symbols are used to denote the groups.
The symbols 1A through 8A are the traditional designations, whereas the numbers 1
to 18 have been suggested recently. In this text the 1A to 8A designations will be used.
The horizontal rows of elements in the periodic table are called periods. Horizontal
row 1 is called the first period (it contains H and He); row 2 is called the second period
(elements Li through Ne); and so on.
We will learn much more about the periodic table as we continue with our study of
chemistry. Meanwhile, when an element is introduced in this text, you should always
note its position on the periodic table.
2.8 Naming Simple Compounds
When chemistry was an infant science, there was no system for naming compounds.
Names such as sugar of lead, blue vitrol, quicklime, Epsom salts, milk of magnesia,
gypsum, and laughing gas were coined by early chemists. Such names are called common names. As chemistry grew, it became clear that using common names for compounds would lead to unacceptable chaos. Nearly 5 million chemical compounds are
currently known. Memorizing common names for these compounds would be an impossible task.
The solution, of course, is to adopt a system for naming compounds in which the
name tells something about the composition of the compound. After learning the system, a chemist given a formula should be able to name the compound or, given a name,
should be able to construct the compound’s formula. In this section we will specify the
most important rules for naming compounds other than organic compounds (those
based on chains of carbon atoms).
We will begin with the systems for naming inorganic binary compounds—
compounds composed of two elements—which we classify into various types for
easier recognition. We will consider both ionic and covalent compounds.
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2.8 Naming Simple Compounds
61
Binary Ionic Compounds (Type I)
Binary ionic compounds contain a positive ion (cation) always written first in the formula and a negative ion (anion). In naming these compounds, the following rules apply:
Naming Type I Binary Compounds
1. The cation is always named first and the anion second.
A monatomic cation has the same name as
its parent element.
2. A monatomic (meaning “one-atom”) cation takes its name from the name of the
element. For example, Na1 is called sodium in the names of compounds containing this ion.
3. A monatomic anion is named by taking the root of the element name and adding
-ide. Thus the Cl2 ion is called chloride.
Some common monatomic cations and anions and their names are given in Table 2.3.
The rules for naming binary ionic compounds are illustrated by the following
examples:
In formulas of ionic compounds, simple
ions are represented by the element
symbol: Cl means Cl2, Na means Na1, and
so on.
Interactive
Example 2.3
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in OWL.
Compound
Ions Present
NaCl Na1, Cl2
KI K1, I2
CaS Ca21, S22
Li3N Li1, N32
CsBr Cs1, Br2
MgO Mg21, O22
Name
Sodium chloride
Potassium iodide
Calcium sulfide
Lithium nitride
Cesium bromide
Magnesium oxide
Naming Type I Binary Compounds
Name each binary compound.
a. CsF b. AlCl3 c. LiH
Solution
a. CsF is cesium fluoride.
b. AlCl3 is aluminum chloride.
c. LiH is lithium hydride.
Notice that, in each case, the cation is named first and then the anion is named.
See Exercise 2.71
Table 2.3 | Common Monatomic Cations and Anions
Cation
Name
Anion
Name
H
Li1
Na1
K1
Cs1
Be21
Mg21
Ca21
Ba21
Al31
Hydrogen
Lithium
Sodium
Potassium
Cesium
Beryllium
Magnesium
Calcium
Barium
Aluminum
H
F2
Cl2
Br2
I2
O22
S22
N32
P32
Hydride
Fluoride
Chloride
Bromide
Iodide
Oxide
Sulfide
Nitride
Phosphide
1
2
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62
Chapter 2
Atoms, Molecules, and Ions
Formulas from Names
Table 2.4 | Common Type II Cations
Ion
31
Fe
Fe21
Cu21
Cu1
Co31
Co21
Sn41
Sn21
Pb41
Pb21
Hg21
Hg221*
Ag1
Zn21
Cd21
Systematic Name
Iron(III)
Iron(II)
Copper(II)
Copper(I)
Cobalt(III)
Cobalt(II)
Tin(IV)
Tin(II)
Lead(IV)
Lead(II)
Mercury(II)
Mercury(I)
Silver†
Zinc†
Cadmium†
*Note that mercury(I) ions always occur
bound together to form Hg221 ions.
†
Although these are transition metals, they
form only one type of ion, and a Roman
numeral is not used.
Interactive
Example 2.4
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in OWL.
So far we have started with the chemical formula of a compound and decided on its
systematic name. The reverse process is also important. For example, given the name
calcium chloride, we can write the formula as CaCl2 because we know that calcium
forms only Ca21 ions and that, since chloride is Cl2, two of these anions will be required to give a neutral compound.
Binary Ionic Compounds (Type II)
In the binary ionic compounds considered earlier (Type I), the metal present forms
only a single type of cation. That is, sodium forms only Na1, calcium forms only Ca21,
and so on. However, as we will see in more detail later in the text, there are many metals that form more than one type of positive ion and thus form more than one type of
ionic compound with a given anion. For example, the compound FeCl2 contains Fe21
ions, and the compound FeCl3 contains Fe31 ions. In a case such as this, the charge on
the metal ion must be specified. The systematic names for these two iron compounds
are iron(II) chloride and iron(III) chloride, respectively, where the Roman numeral
indicates the charge of the cation.
Another system for naming these ionic compounds that is seen in the older literature was used for metals that form only two ions. The ion with the higher charge has a
name ending in -ic, and the one with the lower charge has a name ending in -ous. In
this system, for example, Fe31 is called the ferric ion, and Fe21 is called the ferrous
ion. The names for FeCl3 and FeCl2 are then ferric chloride and ferrous chloride, respectively. In this text we will use the system that employs Roman numerals. Table 2.4
lists the systematic names for many common type II cations.
Formulas from Names for Type I Binary Compounds
Given the following systematic names, write the formula for each compound:
a. potassium iodide
b. calcium oxide
c. gallium bromide
Solution
Name
Formula
Comments
a. potassium iodide
KI
Contains K1 and I2
b. calcium oxide
CaO
Contains Ca21 and O22
c. gallium bromide
GaBr3
Contains Ga31 and Br2
Must have 3Br2 to balance charge of Ga31
See Exercise 2.71
Interactive
Example 2.5
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.com to try this Interactive Example
in OWL.
Naming Type II Binary Compounds
1. Give the systematic name for each of the following compounds:
a. CuCl b. HgO c. Fe2O3
2. Given the following systematic names, write the formula for each compound:
a. Manganese(IV) oxide
b. Lead(II) chloride
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2.8 Naming Simple Compounds
Type II binary ionic compounds contain a
metal that can form more than one type
of cation.
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reserved.
A compound must be electrically neutral.
Mercury(II) oxide.
63
Solution
All of these compounds include a metal that can form more than one type of cation.
Thus we must first determine the charge on each cation. This can be done by recognizing that a compound must be electrically neutral; that is, the positive and negative
charges must exactly balance.
1.
Formula
Name
Comments
a. CuCl
Copper(I) chlorideBecause the anion is Cl2, the cation must be
Cu1 (for charge balance), which requires a
Roman numeral I.
b. HgO
Mercury(II) oxideBecause the anion is O22, the cation must be
Hg21 [mercury(II)].
c. Fe2O3
Iron(III) oxideThe three O22 ions carry a total charge of 62,
so two Fe31 ions [iron(III)] are needed to give
a 61 charge.
2.
Name
Formula
Comments
a. Manganese(IV) oxide
MnO2Two O22 ions (total charge 42) are
required by the Mn41 ion
[manganese(IV)].
b. Lead(II) chloride
PbCl2Two Cl2 ions are required by the Pb21
ion [lead(II)] for charge balance.
See Exercise 2.72
A compound containing a transition metal
usually requires a Roman numeral in its
name.
Note that the use of a Roman numeral in a systematic name is required only in cases
where more than one ionic compound forms between a given pair of elements. This
case most commonly occurs for compounds containing transition metals, which often
form more than one cation. Elements that form only one cation do not need to be identified by a Roman numeral. Common metals that do not require Roman numerals are
the Group 1A elements, which form only 11 ions; the Group 2A elements, which
form only 21 ions; and aluminum, which forms only Al31. The element silver deserves special mention at this point. In virtually all its compounds, silver is found as
the Ag1 ion. Therefore, although silver is a transition metal (and can potentially form
ions other than Ag1), silver compounds are usually named without a Roman numeral.
Thus AgCl is typically called silver chloride rather than silver(I) chloride, although the
latter name is technically correct. Also, a Roman numeral is not used for zinc compounds, since zinc forms only the Zn21 ion.
As shown in Example 2.5, when a metal ion is present that forms more than one
type of cation, the charge on the metal ion must be determined by balancing the positive and negative charges of the compound. To do this you must be able to recognize
the common cations and anions and know their charges (see Tables 2.3 and 2.5). The
procedure for naming binary ionic compounds is summarized in Fig. 2.20.
Critical Thinking
We can use the periodic table to tell us something about the stable ions formed by
many atoms. For example, the atoms in column 1 always form 11 ions. The transition
metals, however, can form more than one type of stable ion. What if each transition
metal ion had only one possible charge? How would the naming of compounds be
different?
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64
Chapter 2
Atoms, Molecules, and Ions
Figure 2.20 | Flowchart for naming
binary ionic compounds.
Does the compound contain
Type I or Type II cations?
Type I
Name the cation using
the element name.
Type II
Using the principle of charge
balance, determine the cation charge.
Include in the cation name a Roman
numeral indicating the charge.
Interactive
Example 2.6
Naming Binary Compounds
1. Give the systematic name for each of the following compounds:
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in OWL.
a. CoBr2 b. CaCl2 c. Al2O3
Richard Megna/Fundamental Photographs © Cengage Learning
2. Given the following systematic names, write the formula for each compound:
Various chromium compounds dissolved
in water. From left to right: CrCl2,
K2Cr2O7, Cr(NO3)3, CrCl3, K2CrO4.
a. Chromium(III) chloride
b. Gallium iodide
Solution
1.
Formula
a. CoBr2
b. CaCl2
c. Al2O3
Name
Comments
Cobalt(II) bromideCobalt is a transition metal; the compound
name must have a Roman numeral. The two
Br2 ions must be balanced by a Co21 ion.
Calcium chlorideCalcium, an alkaline earth metal, forms only
the Ca21 ion. A Roman numeral is not
necessary.
Aluminum oxideAluminum forms only the Al31 ion. A Roman ­
numeral is not necessary.
2.
Name
a. Chromium(III) chloride
b. Gallium iodide
Formula
Comments
CrCl3Chromium(III) indicates that Cr31 is
present, so 3 Cl2 ions are needed for
charge balance.
GaI3Gallium always forms 31 ions, so
3 I2 ions are required for charge
balance.
See Exercises 2.73 and 2.74
The common Type I and Type II ions are summarized in Fig. 2.21. Also shown in
Fig. 2.21 are the common monatomic ions.
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2.8 Naming Simple Compounds
Figure 2.21 | The common cations
1A
and anions.
65
8A
2A
3A
4A
+
Li
5A
6A
7A
3−
2−
O
F−
S2−
Cl−
N
Al3+
Na+ Mg2+
K+ Ca2+
Cu+ Zn2+ Ga3+
Cu2+
Cr2+ Mn2+ Fe2+ Co2+
Cr3+ Mn3+ Fe3+ Co3+
Rb+ Sr2+
Ag+ Cd2+
Hg22+
Cs+ Ba2+
Hg2+
Common Type I cations
Common Type II cations
Br−
Sn2+
Sn4+
Pb2+
Pb4+
I−
Common monatomic anions
Ionic Compounds with Polyatomic Ions
Polyatomic ion formulas must be
memorized.
We have not yet considered ionic compounds that contain polyatomic ions. For example, the compound ammonium nitrate, NH4NO3, contains the polyatomic ions
NH41 and NO32. Polyatomic ions are assigned special names that must be memorized
to name the compounds containing them. The most important polyatomic ions and
their names are listed in Table 2.5.
Note in Table 2.5 that several series of anions contain an atom of a given element
and different numbers of oxygen atoms. These anions are called oxyanions. When
there are two members in such a series, the name of the one with the smaller number
of oxygen atoms ends in -ite and the name of the one with the larger number ends in
-ate—for example, sulfite (SO322) and sulfate (SO422). When more than two oxy­
anions make up a series, hypo- (less than) and per- (more than) are used as prefixes to
name the members of the series with the fewest and the most oxygen atoms, respectively. The best example involves the oxyanions containing chlorine, as shown in
Table 2.5.
Table 2.5 | Common Polyatomic Ions
Ion
Name
21
Hg2
NH41
NO22
NO32
SO322
SO422
HSO42
OH2
CN2
PO432
HPO422
H2PO42
Mercury(I)
Ammonium
Nitrite
Nitrate
Sulfite
Sulfate
Hydrogen sulfate
(bisulfate is a widely
used common name)
Hydroxide
Cyanide
Phosphate
Hydrogen phosphate
Dihydrogen phosphate
Ion
Name
NCS or SCN
CO322
HCO32
2
2
ClO2 or OCl2
ClO22
ClO32
ClO42
C2H3O22
MnO42
Cr2O722
CrO422
O222
C2O422
S2O322
Thiocyanate
Carbonate
Hydrogen carbonate
(bicarbonate is a widely
used common name)
Hypochlorite
Chlorite
Chlorate
Perchlorate
Acetate
Permanganate
Dichromate
Chromate
Peroxide
Oxalate
Thiosulfate
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66
Chapter 2
Atoms, Molecules, and Ions
Interactive
Example 2.7
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in OWL.
Naming Compounds Containing Polyatomic Ions
1. Give the systematic name for each of the following compounds:
a. Na2SO4
d. Mn(OH)2
b. KH2PO4
e. Na2SO3
c. Fe(NO3)3
f. Na2CO3
2. Given the following systematic names, write the formula for each compound:
a. Sodium hydrogen carbonate
b. Cesium perchlorate
c. Sodium hypochlorite
d. Sodium selenate
e. Potassium bromate
Solution
1.
Formula
Name
Comments
a. Na2SO4
Sodium sulfate
b. KH2PO4
Potassium dihydrogen
phosphate
c. Fe(NO3)3
Iron(III) nitrateTransition metal—name must
contain a Roman numeral. The
Fe31 ion balances three NO32 ions.
d. Mn(OH)2
Manganese(II) hydroxideTransition metal—name must
contain a Roman numeral.
The Mn21 ion balances three
OH2 ions.
e. Na2SO3
Sodium sulfite
f. Na2CO3
Sodium carbonate
2.
Name
a. Sodium hydrogen
carbonate
b. Cesium perchlorate
c. Sodium hypochlorite
d. Sodium selenate
e. Potassium bromate
Formula
NaHCO3
Comments
Often called sodium bicarbonate.
CsClO4
NaOCl
Na2SeO4Atoms in the same group, like sulfur and
selenium, often form similar ions that
are named similarly. Thus SeO422 is
selenate, like SO42– (sulfate).
KBrO3As above, BrO32 is bromate, like ClO32
(chlorate).
See Exercises 2.75 and 2.76
Binary Covalent Compounds (Type III)
In binary covalent compounds, the element
names follow the same rules as for binary
ionic compounds.
Binary covalent compounds are formed between two nonmetals. Although these
compounds do not contain ions, they are named very similarly to binary ionic
compounds.
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2.8 Naming Simple Compounds
Table 2.6 | Prefixes Used to
Indicate Number
in Chemical Names
Prefix
Number Indicated
monoditritetrapentahexaheptaoctanonadeca-
1
2
3
4
5
6
7
8
9
10
67
Naming Binary Covalent Compounds
1. The first element in the formula is named first, using the full element name.
2. The second element is named as if it were an anion.
3. Prefixes are used to denote the numbers of atoms present. These prefixes are
given in Table 2.6.
4. The prefix mono- is never used for naming the first element. For example, CO is
called carbon monoxide, not monocarbon monoxide.
To see how these rules apply, we will now consider the names of the several covalent compounds formed by nitrogen and oxygen:
Compound Systematic Name
N2O
NO
NO2
N2O3
N2O4
N2O5
Common Name
Dinitrogen monoxide Nitrous oxide
Nitrogen monoxide Nitric oxide
Nitrogen dioxide
Dinitrogen trioxide
Dinitrogen tetroxide
Dinitrogen pentoxide
Notice from the preceding examples that to avoid awkward pronunciations, we often drop the final o or a of the prefix when the element begins with a vowel. For example, N2O4 is called dinitrogen tetroxide, not dinitrogen tetraoxide, and CO is called
carbon monoxide, not carbon monooxide.
Some compounds are always referred to by their common names. Three examples
are water, ammonia, and hydrogen peroxide. The systematic names for H2O, NH3, and
H2O2 are never used.
Interactive
Example 2.8
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in OWL.
Naming Type III Binary Compounds
1. Name each of the following compounds:
a. PCl5 b. PCl3 c. SO2
2. From the following systematic names, write the formula for each compound:
a. Sulfur hexafluoride
b. Sulfur trioxide
c. Carbon dioxide
Solution
1.
Formula
a. PCl5
b. PCl3
c. SO2
Name
Phosphorus pentachloride
Phosphorus trichloride
Sulfur dioxide
2.
Name
a. Sulfur hexafluoride
b. Sulfur trioxide
c. Carbon dioxide
Formula
SF6
SO3
CO2
See Exercises 2.77 and 2.78
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68
Chapter 2
Atoms, Molecules, and Ions
Figure 2.22 | A flowchart for
naming binary compounds.
Binary compound?
Yes
Metal present?
No
Yes
Type III:
Use prefixes.
Does the metal form
more than one cation?
No
Yes
Type II:
Determine the charge
of the cation; use a Roman
numeral after the element
name for the cation.
Type I:
Use the element
name for the cation.
Figure 2.23 | Overall strategy for
naming chemical compounds.
Binary compound?
No
Use the strategy
summarized
in Figure 2.22.
Polyatomic ion
or ions present?
No
This is a compound
for which naming
procedures have not
yet been considered.
Yes
Yes
Name the compound
using procedures similar
to those for naming
binary ionic compounds.
The rules for naming binary compounds are summarized in Fig. 2.22. Prefixes
to indicate the number of atoms are used only in Type III binary compounds (those
containing two nonmetals). An overall strategy for naming compounds is given in
Fig. 2.23.
Interactive
Example 2.9
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in OWL.
Naming Various Types of Compounds
1. Give the systematic name for each of the following compounds:
a. P4O10 b. Nb2O5 c. Li2O2 d. Ti(NO3)4
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2.8 Naming Simple Compounds
69
2. Given the following systematic names, write the formula for each compound:
a. Vanadium(V) fluoride
b. Dioxygen difluoride
c. Rubidium peroxide
d. Gallium oxide
Solution
1.
Compound
Name
Comments
a. P4O10
Tetraphosphorus
Binary covalent compound (Type III), so
decaoxide prefixes are used. The a in deca- is
sometimes dropped.
b. Nb2O5
Niobium(V) oxideType II binary compound containing Nb51
and O22 ions. Niobium is a transition
metal and requires a Roman ­numeral.
c. Li2O2
Lithium peroxideType I binary compound containing the
Li1 and O222 (peroxide) ions.
d. Ti(NO3)4 Titanium(IV) nitrateNot a binary compound. Contains the Ti41
and NO32 ions. Titanium is a transition
metal and requires a Roman numeral.
2.
Name
Chemical Formula
Comments
a. Vanadium(V)
VF5
The compound contains V51
fluoride ions and requires five F2 ions
for charge balance.
b. Dioxygen difluoride
O2F2The prefix di- indicates two of
each atom.
c. Rubidium peroxide
Rb2O2Because rubidium is in Group
1A, it forms only 11 ions.
Thus two Rb1 ions are needed
to balance the 22 charge on
the peroxide ion (O222).
d. Gallium oxide
Ga2O3Because gallium is in Group
3A, like aluminum, it forms
only 31 ions. Two Ga31 ions
are required to balance the
charge on three O22 ions.
See Exercises 2.79, 2.83, and 2.84
Acids
Acids can be recognized by the hydrogen
that appears first in the formula.
When dissolved in water, certain molecules produce a solution containing free H1 ions
(protons). These substances, acids, will be discussed in detail in Chapters 4, 14, and
15. Here we will simply present the rules for naming acids.
An acid is a molecule in which one or more H1 ions are attached to an anion. The
rules for naming acids depend on whether the anion contains oxygen. If the name
of the anion ends in -ide, the acid is named with the prefix hydro- and the suffix -ic.
For example, when gaseous HCl is dissolved in water, it forms hydrochloric acid.
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70
Chapter 2
Atoms, Molecules, and Ions
Table 2.7 | Names of Acids* That
Do Not Contain Oxygen
Acid
Name
HF
HCl
HBr
HI
HCN
H2S
Hydrofluoric acid
Hydrochloric acid
Hydrobromic acid
Hydroiodic acid
Hydrocyanic acid
Hydrosulfuric acid
*Note that these acids are aqueous
solutions containing these substances.
Table 2.8 | Names of Some
Oxygen-Containing
Acids
Acid
Name
HNO3
HNO2
H2SO4
H2SO3
H3PO4
HC2H3O2
Nitric acid
Nitrous acid
Sulfuric acid
Sulfurous acid
Phosphoric acid
Acetic acid
Similarly, HCN and H2S dissolved in water are called hydrocyanic and hydrosulfuric acids, respectively.
When the anion contains oxygen, the acidic name is formed from the root name of
the anion with a suffix of -ic or -ous, depending on the name of the anion.
1. If the anion name ends in -ate, the suffix -ic is added to the root name. For example, H2SO4 contains the sulfate anion (SO422) and is called sulfuric acid; H3PO4
contains the phosphate anion (PO432) and is called phosphoric acid; and HC2H3O2
contains the acetate ion (C2H3O22) and is called acetic acid.
2. If the anion has an -ite ending, the -ite is replaced by -ous. For example, H2SO3,
which contains sulfite (SO322), is named sulfurous acid; and HNO2, which contains nitrite (NO22), is named nitrous acid.
The application of these rules can be seen in the names of the acids of the oxyanions
of chlorine:
Acid
Anion
Name
HClO4
HClO3
HClO2
HClO
Perchlorate
Chlorate
Chlorite
Hypochlorite
Perchloric acid
Chloric acid
Chlorous acid
Hypochlorous acid
The names of the most important acids are given in Tables 2.7 and 2.8. An overall
strategy for naming acids is shown in Fig. 2.24.
Critical Thinking
In this chapter, you have learned a systematic way to name chemical compounds.
What if all compounds had only common names? What problems would this cause?
Does the anion contain oxygen?
No
hydro+ anion root
+ -ic
hydro(anion root)ic acid
Yes
Check the ending
of the anion.
-ite
Figure 2.24 | A flowchart for
naming acids. An acid is best considered as one or more H1 ions attached
to an anion.
anion or element root
+ -ous
(root)ous acid
-ate
anion or element root
+ -ic
(root)ic acid
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For Review
71
For review
Key terms
Fundamental laws
Section 2.2
❯
law of conservation of mass
law of definite proportion
law of multiple proportions
Section 2.3
atomic masses
atomic weights
Avogadro’s hypothesis
Section 2.4
❯
❯
Dalton’s atomic theory
❯
❯
❯
cathode-ray tubes
electrons
radioactivity
nuclear atom
nucleus
❯
Section 2.5
❯
proton
neutron
isotopes
atomic number
mass number
❯
❯
❯
Thomson model
Millikan experiment
Rutherford experiment
Nuclear model
Atomic structure
❯
Small, dense nucleus contains protons and neutrons.
Protons—positive charge
❯ Neutrons—no charge
Electrons reside outside the nucleus in the relatively large remaining atomic volume.
❯ Electrons—negative charge, small mass (1y1840 of proton)
Isotopes have the same atomic number but different mass numbers.
❯
❯
❯
Atoms combine to form molecules by sharing electrons to form covalent
bonds.
❯
❯
Section 2.7
periodic table
metal
nonmetal
group (family)
alkali metals
alkaline earth metals
halogens
noble gases
period
All elements are composed of atoms.
All atoms of a given element are identical.
Chemical compounds are formed when atoms combine.
Atoms are not changed in chemical reactions, but the way they are bound together changes.
Early atomic experiments and models
Section 2.6
chemical bond
covalent bond
molecule
chemical formula
structural formula
space-filling model
ball-and-stick model
ion
cation
anion
ionic bond
ionic solid
polyatomic ion
Conservation of mass
Definite proportion
Multiple proportions
Molecules are described by chemical formulas.
Chemical formulas show number and type of atoms.
❯ Structural formula
❯ Ball-and-stick model
❯ Space-filling model
Formation of ions
❯
❯
❯
Cation—formed by loss of an electron, positive charge
Anion—formed by gain of an electron, negative charge
Ionic bonds—formed by interaction of cations and anions
The periodic table organizes elements in order of increasing atomic number.
❯
❯
❯
Elements with similar properties are in columns, or groups.
Metals are in the majority and tend to form cations.
Nonmetals tend to form anions.
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72
Chapter 2
Atoms, Molecules, and Ions
Key terms
Compounds are named using a system of rules depending on the type of
compound.
Section 2.8
binary compounds
binary ionic compounds
oxyanions
binary covalent compounds
acid
❯
❯
Review questions
Binary compounds
Type I—contain a metal that always forms the same cation
❯ Type II—contain a metal that can form more than one cation
❯ Type III—contain two nonmetals
Compounds containing a polyatomic ion
❯
Answers to the Review Questions can be found on the Student website (accessible from www.cengagebrain.com).
1. Use Dalton’s atomic theory to account for each of the
following.
a. the law of conservation of mass
b. the law of definite proportion
c. the law of multiple proportions
2. What evidence led to the conclusion that cathode rays
had a negative charge?
3. What discoveries were made by J. J. Thomson, Henri
Becquerel, and Lord Rutherford? How did Dalton’s
model of the atom have to be modified to account for
these discoveries?
4. Consider Ernest Rutherford’s a-particle bombardment
experiment illustrated in Fig. 2.12. How did the
results of this experiment lead Rutherford away from
the plum pudding model of the atom to propose the
nuclear model of the atom?
5. Do the proton and the neutron have exactly the same
mass? How do the masses of the proton and neutron
compare to the mass of the electron? Which particles
make the greatest contribution to the mass of an atom?
Which particles make the greatest contribution to the
chemical properties of an atom?
6. What is the distinction between atomic number and
mass number? Between mass number and atomic mass?
7. Distinguish between the terms family and period in
connection with the periodic table. For which of these
terms is the term group also used?
8. The compounds AlCl3, CrCl3, and ICl3 have similar
formulas, yet each follows a different set of rules to
name it. Name these compounds, and then compare and
contrast the nomenclature rules used in each case.
9. When metals react with nonmetals, an ionic compound
generally results. What is the predicted general formula
for the compound formed between an alkali metal and
sulfur? Between an alkaline earth metal and nitrogen?
Between aluminum and a halogen?
10. How would you name HBrO4, KIO3, NaBrO2, and
HIO? Refer to Table 2.5 and the acid nomenclature
discussion in the text.
A discussion of the Active Learning ­Questions can be found online in the ­Instructor’s Resource Guide and on PowerLecture. The questions
allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the
learning that occurs while students talk to each other about chemical concepts.
Active Learning Questions
These questions are designed to be used by groups of students in
class.
1. Which of the following is true about an individual atom?
Explain.
a. An individual atom should be considered to be a solid.
b. An individual atom should be considered to be a liquid.
c. An individual atom should be considered to be a gas.
d. The state of the atom depends on which element it is.
e. An individual atom cannot be considered to be a solid,
liquid, or gas.
Justify your choice, and for choices you did not pick, explain
what is wrong with them.
2. How would you go about finding the number of “chalk molecules” it takes to write your name on the board? Provide
an explanation of all you would need to do and a sample
c­ alculation.
3. These questions concern the work of J. J. Thomson.
a. From Thomson’s work, which particles do you think he
would feel are most important for the formation of
compounds (chemical changes) and why?
b. Of the remaining two subatomic particles, which do you
place second in importance for forming compounds and
why?
c. Propose three models that explain Thomson’s findings
and evaluate them. To be complete you should include
Thomson’s findings.
4. Heat is applied to an ice cube in a closed container until only
steam is present. Draw a representation of this process, assuming you can see it at an extremely high level of magnification.
What happens to the size of the molecules? What happens to
the total mass of the sample?
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For Review
5. You have a chemical in a sealed glass container filled with air.
The setup is sitting on a balance as shown below. The chemical is ignited by means of a magnifying glass focusing sunlight on the reactant. After the chemical has completely
burned, which of the following is true? Explain your answer.
250.0 g
a.
b.
c.
d.
6.
7.
8.
9.
10.
11.
12.
13.
The balance will read less than 250.0 g.
The balance will read 250.0 g.
The balance will read greater than 250.0 g.
Cannot be determined without knowing the identity of the
chemical.
The formula of water is H2O. Which of the following is indicated by this formula? Explain your answer.
a. The mass of hydrogen is twice that of oxygen in each
­molecule.
b. There are two hydrogen atoms and one oxygen atom per
water molecule.
c. The mass of oxygen is twice that of hydrogen in each
molecule.
d. There are two oxygen atoms and one hydrogen atom per
water molecule.
You may have noticed that when water boils, you can see bubbles that rise to the surface of the water. Which of the following is inside these bubbles? Explain.
a. air
b. hydrogen and oxygen gas
c. oxygen gas
d. water vapor
e. carbon dioxide gas
One of the best indications of a useful theory is that it raises
more questions for further experimentation than it originally
answered. Does this apply to Dalton’s atomic theory? Give
examples.
Dalton assumed that all atoms of the same element were identical in all their properties. Explain why this assumption is not
valid.
Evaluate each of the following as an acceptable name for
water:
a. dihydrogen oxide
c. hydrogen hydroxide
b. hydroxide hydride
d. oxygen dihydride
Why do we call Ba(NO3)2 barium nitrate, but we call Fe(NO3)2
iron(II) nitrate?
Why is calcium dichloride not the correct systematic name for
CaCl2?
The common name for NH3 is ammonia. What would be the
systematic name for NH3? Support your answer.
73
14. Which (if any) of the following can be determined by knowing
the number of protons in a neutral element? Explain your
answer.
a. the number of neutrons in the neutral element
b. the number of electrons in the neutral element
c. the name of the element
15. Which of the following explain how an ion is formed? Explain
your answer.
a. adding or subtracting protons to/from an atom
b. adding or subtracting neutrons to/from an atom
c. adding or subtracting electrons to/from an atom
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of this book and a
solution appears in the Solutions Guide, as found on PowerLecture.
Questions
16. What refinements had to be made in Dalton’s atomic theory to
account for Gay-Lussac’s results on the combining volumes of
gases?
17. When hydrogen is burned in oxygen to form water, the composition of water formed does not depend on the amount of oxygen
reacted. Interpret this in terms of the law of definite proportion.
18. The two most reactive families of elements are the halogens
and the alkali metals. How do they differ in their reactivities?
19. Explain the law of conservation of mass, the law of definite
proportion, and the law of multiple proportions.
20. Section 2.3 describes the postulates of Dalton’s atomic theory.
With some modifications, these postulates hold up very well
regarding how we view elements, compounds, and chemical
reactions today. Answer the following questions concerning
Dalton’s atomic theory and the modifications made today.
a. The atom can be broken down into smaller parts. What
are the smaller parts?
b. How are atoms of hydrogen identical to each other, and
how can they be different from each other?
c. How are atoms of hydrogen different from atoms of
helium? How can H atoms be similar to He atoms?
d. How is water different from hydrogen peroxide (H2O2)
even though both compounds are composed of only
hydrogen and oxygen?
e. What happens in a chemical reaction, and why is mass
conserved in a chemical reaction?
21. The contributions of J. J. Thomson and Ernest Rutherford led
the way to today’s understanding of the structure of the atom.
What were their contributions?
22. What is the modern view of the structure of the atom?
23. The number of protons in an atom determines the identity of
the atom. What does the number and arrangement of the electrons in an atom determine? What does the number of neutrons in an atom determine?
24. If the volume of a proton were similar to the volume of an
electron, how will the densities of these two particles compare
to each other?
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74
Chapter 2
Atoms, Molecules, and Ions
25. For lighter, stable isotopes, the ratio of the mass number to the
atomic number is close to a certain value. What is the value?
What happens to the value of the mass number to atomic number ratio as stable isotopes become heavier?
26. List some characteristic properties that distinguish the metallic elements from the nonmetallic elements.
27. Consider the elements of Group 4A (the “carbon family”):
C, Si, Ge, Sn, and Pb. What is the trend in metallic character
as one goes down this group? What is the trend in metallic
character going from left to right across a period in the periodic table?
28. Distinguish between the following terms.
a. molecule versus ion
b. covalent bonding versus ionic bonding
c. molecule versus compound
d. anion versus cation
29. Label the type of bonding for each of the following.
a.
b.
30. The vitamin niacin (nicotinic acid, C6H5NO2) can be isolated
from a variety of natural sources such as liver, yeast, milk, and
whole grain. It also can be synthesized from commercially
available materials. From a nutritional point of view, which
source of nicotinic acid is best for use in a multivitamin tablet?
Why?
31. Which of the following statements is(are) true? For the false
statements, correct them.
a. Most of the known elements are metals.
b. Element 118 should be a nonmetal.
c. Hydrogen has mostly metallic properties.
d. A family of elements is also known as a period of
elements.
e. When an alkaline earth metal, A, reacts with a halogen,
X, the formula of the covalent compound formed should
be A2X.
32. Each of the following compounds has three possible names
listed for it. For each compound, what is the correct name and
why aren’t the other names used?
a. N2O: nitrogen oxide, nitrogen(I) oxide, dinitrogen
monoxide
b. Cu2O: copper oxide, copper(I) oxide, dicopper monoxide
c. Li2O: lithium oxide, lithium(I) oxide, dilithium monoxide
Exercises
In this section similar exercises are paired.
Development of the Atomic Theory
33. When mixtures of gaseous H2 and gaseous Cl2 react, a product
forms that has the same properties regardless of the relative
amounts of H2 and Cl2 used.
a. How is this result interpreted in terms of the law of
definite proportion?
b. When a volume of H2 reacts with an equal volume of Cl2
at the same temperature and pressure, what volume of
product having the formula HCl is formed?
34. Observations of the reaction between nitrogen gas and hydrogen gas show us that 1 volume of nitrogen reacts with 3 volumes of hydrogen to make 2 volumes of gaseous product, as
shown below:
N N
+
HH HH HH
Determine the formula of the product and justify your answer.
35. A sample of chloroform is found to contain 12.0 g of carbon,
106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of chloroform is found to contain 30.0 g of carbon, what is
the total mass of chloroform in the second sample?
36. A sample of H2SO4 contains 2.02 g of hydrogen, 32.07 g of
sulfur, and 64.00 g of oxygen. How many grams of sulfur and
grams of oxygen are present in a second sample of H2SO4 containing 7.27 g of hydrogen?
37. Hydrazine, ammonia, and hydrogen azide all contain only nitrogen and hydrogen. The mass of hydrogen that combines
with 1.00 g of nitrogen for each compound is 1.44 3 1021 g,
2.16 3 1021 g, and 2.40 3 1022 g, respectively. Show how
these data illustrate the law of multiple proportions.
38. Consider 100.0-g samples of two different compounds consisting only of carbon and oxygen. One compound contains
27.2 g of carbon and the other has 42.9 g of carbon. How can
these data support the law of multiple proportions if 42.9 is not
a multiple of 27.2? Show that these data support the law of
multiple proportions.
39. The three most stable oxides of carbon are carbon monoxide
(CO), carbon dioxide (CO2), and carbon suboxide (C3O2). The
molecules can be represented as
Explain how these molecules illustrate the law of multiple
proportions.
40. Two elements, R and Q, combine to form two binary compounds. In the first compound, 14.0 g of R combines with
3.00 g of Q. In the second compound, 7.00 g of R combines
with 4.50 g of Q. Show that these data are in accord with the
law of multiple proportions. If the formula of the second compound is RQ, what is the formula of the first compound?
41. In Section 1.1 of the text, the concept of a chemical reaction
was introduced with the example of the decomposition of water, represented as follows:
two water
molecules
written 2H2O
one oxygen molecule
written O2
electric
current
two hydrogen molecules
written 2H2
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
Use ideas from Dalton’s atomic theory to explain how the above
representation illustrates the law of conservation of mass.
42. In a combustion reaction, 46.0 g of ethanol reacts with 96.0 g
of oxygen to produce water and carbon dioxide. If 54.0 g of
water is produced, what mass of carbon dioxide is produced?
43. Early tables of atomic weights (masses) were generated by
mea­s­uring the mass of a substance that reacts with 1.00 g of
oxygen. Given the following data and taking the atomic mass
of hydrogen as 1.00, generate a table of relative atomic masses
for oxygen, sodium, and magnesium.
Element
Mass That Combines
with 1.00 g Oxygen
Assumed Formula
0.126 g
2.875 g
1.500 g
HO
NaO
MgO
Hydrogen
Sodium
Magnesium
How do your values compare with those in the periodic table?
How do you account for any differences?
44. Indium oxide contains 4.784 g of indium for every 1.000 g of
oxygen. In 1869, when Mendeleev first presented his version
of the periodic table, he proposed the formula In2O3 for indium oxide. Before that time it was thought that the formula
was InO. What values for the atomic mass of indium are obtained using these two formulas? Assume that oxygen has an
atomic mass of 16.00.
The Nature of the Atom
45. From the information in this chapter on the mass of the proton,
the mass of the electron, and the sizes of the nucleus and the
atom, calculate the densities of a hydrogen nucleus and a hydrogen atom.
46. If you wanted to make an accurate scale model of the hydrogen atom and decided that the nucleus would have a diameter
of 1 mm, what would be the diameter of the entire model?
47. In an experiment it was found that the total charge on an oil
drop was 5.93 3 10218 C. How many negative charges does
the drop contain?
48. A chemist in a galaxy far, far away performed the Millikan oil
drop experiment and got the following results for the charges
on various drops. Use these data to calculate the charge of the
electron in zirkombs.
2.56 3 10212 zirkombs 7.68 3 10212 zirkombs
3.84 3 10212 zirkombs 6.40 3 10213 zirkombs
49. What are the symbols of the following metals: sodium, radium, iron, gold, manganese, lead?
50. What are the symbols of the following nonmetals: fluorine,
chlorine, bromine, sulfur, oxygen, phosphorus?
51. Give the names of the metals that correspond to the following
symbols: Sn, Pt, Hg, Mg, K, Ag.
52. Give the names of the nonmetals that correspond to the following symbols: As, I, Xe, He, C, Si.
75
53. a. Classify the following elements as metals or nonmetals:
Mg
Ti
Au
Bi
Si
Ge
B
At
Rn
Eu
Am
Br
b. The distinction between metals and nonmetals is really
not a clear one. Some elements, called metalloids, are
intermediate in their properties. Which of these elements
would you reclassify as metalloids? What other elements
in the periodic table would you expect to be metalloids?
54. a. L
ist the noble gas elements. Which of the noble gases has
only radioactive isotopes? (This situation is indicated on
most periodic tables by parentheses around the mass of the
element. See inside front cover.)
b. Which lanthanide element has only radioactive isotopes?
55. For each of the following sets of elements, label each as either
noble gases, halogens, alkali metals, alkaline earth metals, or
transition metals.
a. Ti, Fe, Ag
b. Mg, Sr, Ba
c. Li, K, Rb
d. Ne, Kr, Xe
e. F, Br, I
56. Identify the elements that correspond to the following atomic
numbers. Label each as either a noble gas, a halogen, an alkali
metal, an alkaline earth metal, a transition metal, a lanthanide
metal, or an actinide metal.
a. 17
e. 2
b. 4
f. 92
c. 63
g. 55
d. 72
57. Write the atomic symbol ( AZX) for each of the following
isotopes.
a. Z 5 8, number of neutrons 5 9
b. the isotope of chlorine in which A 5 37
c. Z 5 27, A 5 60
d. number of protons 5 26, number of neutrons 5 31
e. the isotope of I with a mass number of 131
f. Z 5 3, number of neutrons 5 4
58. Write the atomic symbol ( AZX) for each of the isotopes described below.
a. number of protons 5 27, number of neutrons 5 31
b. the isotope of boron with mass number 10
c. Z 5 12, A 5 23
d. atomic number 53, number of neutrons 5 79
e. Z 5 20, number of neutrons 5 27
f. number of protons 5 29, mass number 65
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76
Chapter 2
Atoms, Molecules, and Ions
59. Write the symbol of each atom using the ZAX format.
Nucleus
11 protons
12 neutrons
a.
10 electrons
Nucleus
9 protons
10 neutrons
63. For each of the following ions, indicate the number of protons
and electrons the ion contains.
a. Ba21
e. Co31
21
b. Zn
f. Te22
32
c. N
g. Br2
d. Rb1
64. How many protons, neutrons, and electrons are in each of the
following atoms or ions?
31
22
a. 24
d. 59
g. 79
12Mg
27Co
34Se
24
21
59
63
b. 12Mg
e. 27Co
h. 28Ni
21
21
c. 59
f. 79
i. 59
27Co
34Se
28Ni
65. What is the symbol for an ion with 63 protons, 60 electrons,
and 88 neutrons? If an ion contains 50 protons, 68 neutrons,
and 48 electrons, what is its symbol?
66. What is the symbol of an ion with 16 protons, 18 neutrons, and
18 electrons? What is the symbol for an ion that has 16 protons, 16 neutrons, and 18 electrons?
67. Complete the following table:
b.
11 electrons
Symbol
Number of
Neutrons
in Nucleus
Number of
Electrons
Net
Charge
238
92U
Nucleus
8 protons
8 neutrons
Number of
Protons in
Nucleus
20
20
23
28
20
35
44
36
15
16
21
89
39Y
32
68. Complete the following table:
c.
8 electrons
60. For carbon-14 and carbon-12, how many protons and neutrons
are in each nucleus? Assuming neutral atoms, how many electrons are present in an atom of carbon-14 and in an atom of
carbon-12?
61. How many protons and neutrons are in the nucleus of each of
the following atoms? In a neutral atom of each element, how
many electrons are present?
a. 79Br
d. 133Cs
81
b. Br
e. 3H
c. 239Pu
f. 56Fe
62. What number of protons and neutrons are contained in the
nucleus of each of the following atoms? Assuming each atom
is uncharged, what number of electrons are present?
a. 235
d. 208
92U
82Pb
27
86
b. 13Al
e. 37Rb
c. 57
f. 41
26Fe
20Ca
Symbol
Number of
Protons in
Nucleus
Number of
Neutrons
in Nucleus
26
33
85
125
86
13
14
10
76
54
Number of
Electrons
Net
Charge
53 21
26Fe
31
22
69. Would you expect each of the following atoms to gain or lose
electrons when forming ions? What ion is the most likely in
each case?
a. Ra
c. P
e. Br
b. In
d. Te
f. Rb
70. For each of the following atomic numbers, use the periodic
table to write the formula (including the charge) for the simple
ion that the element is most likely to form in ionic
compounds.
a. 13
c. 56
e. 87
b. 34
d. 7
f. 35
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
Nomenclature
71. Name the compounds in parts a–d and write the formulas for
the compounds in parts e–h.
a. NaBr
e. strontium fluoride
b. Rb2O
f. aluminum selenide
c. CaS
g. potassium nitride
d. AlI3
h. magnesium phosphide
72. Name the compounds in parts a–d and write the formulas for
the compounds in parts e–h.
a. Hg2O
e. tin(II) nitride
b. FeBr3
f. cobalt(III) iodide
c. CoS
g. mercury(II) oxide
d. TiCl4
h. chromium(VI) sulfide
73. Name each of the following compounds:
a. CsF
c. Ag2S
e. TiO2
b. Li3N
d. MnO2
f. Sr3P2
74. Write the formula for each of the following compounds:
a. zinc chloride
d. aluminum sulfide
b. tin(IV) fluoride
e. mercury(I) selenide
c. calcium nitride
f. silver iodide
75. Name each of the following compounds:
a. BaSO3
c. KMnO4
b. NaNO2
d. K2Cr2O7
76. Write the formula for each of the following compounds:
a. chromium(III) hydroxide
c. lead(IV) carbonate
b. magnesium cyanide
d. ammonium acetate
77. Name each of the following compounds:
a.
c. SO2
O
d. P2S5
N
b.
I
Cl
78. Write the formula for each of the following compounds:
a. diboron trioxide
c. dinitrogen monoxide
b. arsenic pentafluoride
d. sulfur hexachloride
79. Name each of the following compounds:
a. CuI
f. S4N4
b. CuI2
g. SeCl4
c. CoI2
h. NaOCl
d. Na2CO3
i. BaCrO4
e. NaHCO3
j. NH4NO3
80. Name each of the following compounds. Assume the acids are
dissolved in water.
a. HC2H3O2
g. H2SO4
b. NH4NO2
h. Sr3N2
c. Co2S3
i. Al2(SO3)3
d. ICl
j. SnO2
e. Pb3(PO4)2
k. Na2CrO4
f. KClO3
l. HClO
77
81. Elements in the same family often form oxyanions of the same
general formula. The anions are named in a similar fashion.
What are the names of the oxyanions of selenium and tellurium: SeO422, SeO322, TeO422, TeO322?
82. Knowing the names of similar chlorine oxyanions and acids,
deduce the names of the following: IO2, IO22, IO32, IO42,
HIO, HIO2, HIO3, HIO4.
83. Write the formula for each of the following compounds:
a. sulfur difluoride
b. sulfur hexafluoride
c. sodium dihydrogen phosphate
d. lithium nitride
e. chromium(III) carbonate
f. tin(II) fluoride
g. ammonium acetate
h. ammonium hydrogen sulfate
i. cobalt(III) nitrate
j. mercury(I) chloride
k. potassium chlorate
l. sodium hydride
84. Write the formula for each of the following compounds:
a. chromium(VI) oxide
b. disulfur dichloride
c. nickel(II) fluoride
d. potassium hydrogen phosphate
e. aluminum nitride
f. ammonia
g. manganese(IV) sulfide
h. sodium dichromate
i. ammonium sulfite
j. carbon tetraiodide
85. Write the formula for each of the following compounds:
a. sodium oxide
h. copper(I) chloride
b. sodium peroxide
i. gallium arsenide
j. cadmium selenide
c. potassium cyanide
d. copper(II) nitrate
k. zinc sulfide
e. selenium tetrabromide
l. nitrous acid
f. iodous acid
m. diphosphorus pentoxide
g. lead(IV) sulfide
86. Write the formula for each of the following compounds:
a. ammonium hydrogen phosphate
b. mercury(I) sulfide
c. silicon dioxide
d. sodium sulfite
e. aluminum hydrogen sulfate
f. nitrogen trichloride
g. hydrobromic acid
h. bromous acid
i. perbromic acid
j. potassium hydrogen sulfide
k. calcium iodide
l. cesium perchlorate
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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
78
Chapter 2
Atoms, Molecules, and Ions
87. Name the acids illustrated below.
a.
b.
c.
H
N
O
Cl
d.
C
94.
S
P
e.
88. Each of the following compounds is incorrectly named. What
is wrong with each name, and what is the correct name for
each compound?
a. FeCl3, iron chloride
b. NO2, nitrogen(IV) oxide
c. CaO, calcium(II) monoxide
d. Al2S3, dialuminum trisulfide
e. Mg(C2H3O2)2, manganese diacetate
f. FePO4, iron(II) phosphide
g. P2S5, phosphorus sulfide
h. Na2O2, sodium oxide
i. HNO3, nitrate acid
j. H2S, sulfuric acid
Additional Exercises
35
89. Chlorine has two natural isotopes: 37
17Cl and 17Cl. Hydrogen
reacts with chlorine to form the compound HCl. Would a
given amount of hydrogen react with different masses of the
two chlorine isotopes? Does this conflict with the law of definite proportion? Why or why not?
90. What are the symbols for the following nonmetal elements
that are most often present in compounds studied in organic
chemistry: carbon, hydrogen, oxygen, nitrogen, phosphorus,
sulfur? Predict a stable isotope for each of these elements.
91. Four Fe21 ions are key components of hemoglobin, the protein
that transports oxygen in the blood. Assuming that these ions
are 53Fe21, how many protons and neutrons are present in each
­nucleus, and how many electrons are present in each ion?
92. Which of the following statements is(are) true? For the false
statements, correct them.
a. All particles in the nucleus of an atom are charged.
b. The atom is best described as a uniform sphere of matter
in which electrons are embedded.
c. The mass of the nucleus is only a very small fraction of
the mass of the entire atom.
d. The volume of the nucleus is only a very small fraction of
the total volume of the atom.
e. The number of neutrons in a neutral atom must equal the
number of electrons.
93. The isotope of an unknown element, X, has a mass number of
79. The most stable ion of the isotope has 36 electrons and
forms a binary compound with sodium having a formula of
95.
96.
97.
98.
99.
Na2X. Which of the following statements is(are) true? For the
false statements, correct them.
a. The binary compound formed between X and fluorine
will be a covalent compound.
b. The isotope of X contains 38 protons.
c. The isotope of X contains 41 neutrons.
d. The identity of X is strontium, Sr.
For each of the following ions, indicate the total number of
protons and electrons in the ion. For the positive ions in the
list, predict the formula of the simplest compound formed between each positive ion and the oxide ion. Name the compounds. For the negative ions in the list, predict the formula of
the simplest compound formed ­between each negative ion and
the aluminum ion. Name the compounds.
e. S22
a. Fe21
31
b. Fe
f. P32
21
c. Ba
g. Br2
1
d. Cs
h. N32
The formulas and common names for several substances are
given below. Give the systematic names for these substances.
a. sugar of lead
Pb(C2H3O2)2
b. blue vitrol
CuSO4
c. quicklime
CaO
d. Epsom salts
MgSO4
e. milk of magnesia Mg(OH)2
f. gypsum
CaSO4
g. laughing gas
N2O
Identify each of the following elements:
a. a member of the same family as oxygen whose most
stable ion contains 54 electrons
b. a member of the alkali metal family whose most stable
ion contains 36 electrons
c. a noble gas with 18 protons in the nucleus
d. a halogen with 85 protons and 85 electrons
An element’s most stable ion forms an ionic compound with
bromine, having the formula XBr2. If the ion of element X has
a mass number of 230 and has 86 electrons, what is the identity of the element, and how many neutrons does it have?
A certain element has only two naturally occurring isotopes:
one with 18 neutrons and the other with 20 neutrons. The element forms 12 charged ions when in ionic compounds. Predict the identity of the element. What number of electrons
does the 12 charged ion have?
The designations 1A through 8A used for certain families of
the periodic table are helpful for predicting the charges on ions
in binary ionic compounds. In these compounds, the metals
generally take on a positive charge equal to the family number,
while the nonmetals take on a negative charge equal to the
family number minus eight. Thus the compound between sodium and chlorine contains Na1 ions and Cl2 ions and has the
formula NaCl. Predict the formula and the name of the binary
compound formed from the following pairs of elements.
a. Ca and N
e. Ba and I
b. K and O
f. Al and Se
c. Rb and F
g. Cs and P
d. Mg and S
h. In and Br
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
100. By analogy with phosphorus compounds, name the following:
Na3AsO4, H3AsO4, Mg3(SbO4)2.
101. Identify each of the following elements. Give the number of
protons and neutrons in each nucleus.
a. 31
c. 39
19X
15X
127
b. 53 X
d. 173
70 X
102. In a reaction, 34.0 g of chromium(III) oxide reacts with 12.1 g
of aluminum to produce chromium and aluminum oxide. If
23.3 g of chromium is produced, what mass of aluminum oxide is ­produced?
These multiconcept problems (and additional ones) are found interactively online with the same type of assistance a student would get
from an instructor.
103. Complete the following table.
Protons
Neutrons
Electrons
120
50Sn
25
21
12 Mg
56
21
26 Fe
79
34Se
63
29Cu
104. Which of the following is(are) correct?
a. 40Ca21 contains 20 protons and 18 electrons.
b. Rutherford created the cathode-ray tube and was the
founder of the charge-to-mass ratio of an electron.
c. An electron is heavier than a proton.
d. The nucleus contains protons, neutrons, and electrons.
105. What are the formulas of the compounds that correspond to
the names given in the following table?
Formula
Carbon tetrabromide
Cobalt(II) phosphate
Magnesium chloride
Nickel(II) acetate
Calcium nitrate
106. What are the names of the compounds that correspond to the
formulas given in the following table?
Formula
Atom
Gain (G) or Lose (L)
Electrons
Ion Formed
K
Cs
Br
S
108. Which of the following statements is(are) correct?
a. The symbols for the elements magnesium, aluminum, and
xenon are Mn, Al, and Xe, respectively.
b. The elements P, As, and Bi are in the same family on the
periodic table.
c. All of the following elements are expected to gain
electrons to form ions in ionic compounds: Ga, Se, and Br.
d. The elements Co, Ni, and Hg are all transition elements.
e. The correct name for TiO2 is titanium dioxide.
Challenge Problems
35
17Cl
Compound Name
107. Complete the following table to predict whether the given
atom will gain or lose electrons in forming the ion most likely
to form when in ionic compounds.
Se
ChemWork Problems
Atom/Ion
79
Compound Name
Co(NO2)2
AsF5
LiCN
K2SO3
Li3N
PbCrO4
109. The elements in one of the groups in the periodic table are
often called the coinage metals. Identify the elements in this
group based on your own experience.
110. Reaction of 2.0 L of hydrogen gas with 1.0 L of oxygen gas
yields 2.0 L of water vapor. All gases are at the same temperature and pressure. Show how these data support the idea that
oxygen gas is a diatomic molecule. Must we consider hydrogen to be a diatomic molecule to explain these results?
111. A combustion reaction involves the reaction of a substance
with oxygen gas. The complete combustion of any hydrocarbon (binary compound of carbon and hydrogen) produces carbon dioxide and water as the only products. Octane is a hydrocarbon that is found in gasoline. Complete combustion of
octane produces 8 L of carbon dioxide for every 9 L of water
vapor (both measured at the same temperature and pressure).
What is the ratio of carbon atoms to hydrogen atoms in a molecule of octane?
112. A chemistry instructor makes the following claim: “Consider
that if the nucleus were the size of a grape, the electrons would
be about 1 mile away on average.” Is this claim reasonably
accurate? Provide mathematical support.
113. The early alchemists used to do an experiment in which water
was boiled for several days in a sealed glass container. Eventually, some solid residue would appear in the bottom of the
flask, which was interpreted to mean that some of the water in
the flask had been converted into “earth.” When Lavoisier repeated this experiment, he found that the water weighed the
same before and after heating, and the mass of the flask plus
the solid residue equaled the original mass of the flask. Were
the alchemists correct? ­Explain what really happened. (This
experiment is described in the article by A. F. Scott in Scientific American, January 1984.)
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80
Chapter 2
Atoms, Molecules, and Ions
114. Consider the chemical reaction as depicted below. Label as
much as you can using the terms atom, molecule, element,
compound, ionic, gas, and solid.
Cl− Cl− Na+
Na+
Na
Na
+
Cl Cl
115. Each of the following statements is true, but Dalton might
have had trouble explaining some of them with his atomic
theory. Give explanations for the following statements.
a. The space-filling models for ethyl alcohol and dimethyl
ether are shown below.
C
O
H
These two compounds have the same composition by
mass (52% carbon, 13% hydrogen, and 35% oxygen), yet
the two have different melting points, boiling points, and
solubilities in water.
b. Burning wood leaves an ash that is only a small fraction
of the mass of the original wood.
c. Atoms can be broken down into smaller particles.
d. One sample of lithium hydride is 87.4% lithium by mass,
while another sample of lithium hydride is 74.9% lithium
by mass. However, the two samples have the same
chemical properties.
116. You have two distinct gaseous compounds made from element
X and element Y. The mass percents are as follows:
Compound I: 30.43% X, 69.57% Y
Compound II: 63.64% X, 36.36% Y
In their natural standard states, element X and element Y exist
as gases. (Monatomic? Diatomic? Triatomic? That is for you
to determine.) When you react “gas X” with “gas Y” to make
the products, you get the following data (all at the same pressure and temperature):
1 volume “gas X” 1 2 volumes “gas Y” 88n
2 volumes compound I
2 volumes “gas X” 1 1 volume “gas Y” 88n
2 volumes compound II
Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the
relative atomic masses of element X and element Y.
117. A single molecule has a mass of 7.31 3 10223 g. Provide an
example of a real molecule that can have this mass. Assume
the elements that make up the molecule are made of light isotopes where the number of protons equals the number of neutrons in the nucleus of each element.
118. You take three compounds, each consisting of two elements
(X, Y, and/or Z), and decompose them to their respective elements. To determine the relative masses of X, Y, and Z, you
collect and weigh the elements, obtaining the following data:
a.
b.
c.
d.
Elements in Compound
Masses of Elements
1. X and Y
2. Y and Z
3. X and Y
X 5 0.4 g, Y 5 4.2 g
Y 5 1.4 g, Z 5 1.0 g
X 5 2.0 g, Y 5 7.0 g
What are the assumptions needed to solve this problem?
What are the relative masses of X, Y, and Z?
What are the chemical formulas of the three compounds?
If you decompose 21 g of compound XY, how much of
each element is present?
Integrative Problems
These problems require the integration of multiple concepts to find
the solutions.
119. What is the systematic name of Ta2O5? If the charge on the
metal remained constant and then sulfur was substituted for
oxygen, how would the formula change? What is the difference in the total number of protons between Ta2O5 and its sulfur analog?
120. A binary ionic compound is known to contain a cation with
51 protons and 48 electrons. The anion contains one-third the
number of protons as the cation. The number of electrons in the
anion is equal to the number of protons plus 1. What is the formula of this compound? What is the name of this compound?
121. Using the information in Table 2.1, answer the following questions. In an ion with an unknown charge, the total mass of all the
electrons was determined to be 2.55 3 10226 g, while the total
mass of its protons was 5.34 3 10223 g. What is the identity and
charge of this ion? What is the symbol and mass number of a
neutral atom whose total mass of its electrons is 3.92 3 10226 g,
while its neutrons have a mass of 9.35 3 10223 g?
Marathon Problem
This problem is designed to incorporate several concepts and techniques into one situation.
122. You have gone back in time and are working with Dalton on a
table of relative masses. Following are his data.
0.602 g gas A reacts with 0.295 g gas B
0.172 g gas B reacts with 0.401 g gas C
0.320 g gas A reacts with 0.374 g gas C
a. Assuming simplest formulas (AB, BC, and AC), construct
a table of relative masses for Dalton.
b. Knowing some history of chemistry, you tell Dalton that
if he determines the volumes of the gases reacted at
constant temperature and pressure, he need not assume
simplest formulas. You collect the following data:
6 volumes gas A 1 1 volume gas B → 4 volumes product
1 volume gas B 1 4 volumes gas C → 4 volumes product
3 volumes gas A 1 2 volumes gas C → 6 volumes product
rite the simplest balanced equations, and find the actual
W
­relative masses of the elements. Explain your reasoning.
Marathon Problems can be used in class by groups of students to
help facilitate problem-solving skills.
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Chapter 3
Stoichiometry
3.1
Counting by Weighing
3.2
Atomic Masses
3.3
The Mole
3.4
Molar Mass
3.5
Learning to Solve Problems
3.6
ercent Composition of
P
Compounds
3.9
3.7
etermining the Formula of a
D
Compound
3.10 S toichiometric Calculations:
Amounts of Reactants and
Products
3.8
Chemical Equations
3.11 The Concept of Limiting Reactant
Balancing Chemical Equations
Chemical Reactions
The Meaning of a Chemical Equation
Fireworks provide a spectacular example of chemical reactions. (Daff/Dreamstime.com)
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81
C
hemical reactions have a profound effect on our lives. There are many examples: Food is converted to energy in the human body; nitrogen and hydrogen are
combined to form ammonia, which is used as a fertilizer; fuels and plastics are produced from petroleum; the starch in plants is synthesized from carbon dioxide and
water using energy from sunlight; human insulin is produced in laboratories by bacteria; cancer is induced in humans by substances from our environment; and so on, in a
seemingly endless list. The central activity of chemistry is to understand chemical
changes such as these, and the study of reactions occupies a central place in this book.
We will examine why reactions occur, how fast they occur, and the specific pathways
they follow.
In this chapter we will consider the quantities of materials consumed and produced
in chemical reactions. This area of study is called chemical stoichiometry (pronounced stoy?kē ?om9?uh?trē ). To understand chemical stoichiometry, you must first
understand the concept of relative atomic masses.
Experiment 11: Counting
by Weighing
3.1 Counting by Weighing
Suppose you work in a candy store that sells gourmet jelly beans by the bean. People
come in and ask for 50 beans, 100 beans, 1000 beans, and so on, and you have to count
them out—a tedious process at best. As a good problem solver, you try to come up
with a better system. It occurs to you that it might be far more efficient to buy a scale
and count the jelly beans by weighing them. How can you count jelly beans by weighing them? What information about the individual beans do you need to know?
Assume that all of the jelly beans are identical and that each has a mass of
5 g. If a customer asks for 1000 jelly beans, what mass of jelly beans would
be required? Each bean has a mass of 5 g, so you would need 1000 beans
3 5 g/bean, or 5000 g (5 kg). It takes just a few seconds to weigh out
5 kg of jelly beans. It would take much longer to count out 1000 of
them.
In reality, jelly beans are not identical. For example, let’s assume that you weigh 10 beans individually and get the following
results:
Si e d
e Pr
e is/ P
ho to D isc
IBLG: See questions from “Counting by
Weighing and Atomic Masses”
Jelly beans can be counted by weighing.
Bean
Mass
Bean
Mass
1
2
3
4
5
5.1 g
5.2 g
5.0 g
4.8 g
4.9 g
6
7
8
9
10
5.0 g
5.0 g
5.1 g
4.9 g
5.0 g
Can we count these nonidentical beans by weighing? Yes. The key piece of information we need is the average mass of the jelly beans. Let’s compute the average mass
for our 10-bean sample.
Average mass 5
5
82
total mass of beans
number of beans
5.1 g 1 5.2 g 1 5.0 g 1 4.8 g 1 4.9 g 1 5.0 g 1 5.0 g 1 5.1 g 1 4.9 g 1 5.0 g
10
50.0
5
5 5.0 g
10
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3.2
Atomic Masses
83
The average mass of a jelly bean is 5.0 g. Thus, to count out 1000 beans, we need
to weigh out 5000 g of beans. This sample of beans, in which the beans have an
average mass of 5.0 g, can be treated exactly like a sample where all of the beans are
identical. Objects do not need to have identical masses to be counted by weighing.
We simply need to know the average mass of the objects. For purposes of counting,
the objects behave as though they were all identical, as though they each actually
had the average mass.
We count atoms in exactly the same way. Because atoms are so small, we deal with
samples of matter that contain huge numbers of atoms. Even if we could see the atoms,
it would not be possible to count them directly. Thus we determine the number of atoms in a given sample by finding its mass. However, just as with jelly beans, to relate
the mass to a number of atoms, we must know the average mass of the atoms.
3.2 Atomic Masses
As we saw in Chapter 2, the first quantitative information about atomic masses came
from the work of Dalton, Gay-Lussac, Lavoisier, Avogadro, and Berzelius. By observing the proportions in which elements combine to form various compounds, nineteenthcentury chemists calculated relative atomic masses. The modern system of atomic
masses, instituted in 1961, is based on 12C (“carbon twelve”) as the standard. In this
system, 12C is assigned a mass of exactly 12 atomic mass units (u), and the masses of
all other atoms are given relative to this standard.
The most accurate method currently available for comparing the masses of atoms
­involves the use of the mass spectrometer. In this instrument, diagramed in Fig. 3.1,
atoms or molecules are passed into a beam of high-speed electrons, which knock electrons off the atoms or molecules being analyzed and change them into positive ions.
An applied electric field then accelerates these ions into a magnetic field. Because an
accelerating ion produces its own magnetic field, an interaction with the applied magnetic field occurs, which tends to change the path of the ion. The amount of path deflection for each ion depends on its mass—the most massive ions are deflected the
smallest amount—which causes the ions to separate, as shown in Fig. 3.1. A comparison of the positions where the ions hit the detector plate gives very accurate values of
their relative masses. For example, when 12C and 13C are analyzed in a mass spectrometer, the ratio of their masses is found to be
Mass 13C
5 1.0836129
Mass 12C
Detector
plate
Least
massive ions
Ion-accelerating
electric field
Accelerated
ion beam
Geoff Tompkinson/Photo Researchers, Inc.
Positive
ions
Sample
Heating device
to vaporize
sample
Most
massive
ions
Slits
Magnetic field
Electron beam
Figure 3.1 | (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer.
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84
Chapter 3
Stoichiometry
Since the atomic mass unit is defined such that the mass of 12C is exactly 12 atomic
mass units, then on this same scale,
The International Union of Pure and
Applied Chemistry (IUPAC) has declared
that what we refer to as the “average
atomic mass” should be called the
“atomic weight” of an element, which is
dimensionless by custom. However, we
will retain the term “average atomic mass”
because this name accurately describes
what the term represents.
Photo © Cengage Learning. All rights
reserved.
Most elements occur in nature as
mixtures of isotopes; thus atomic masses
are usually average values.
Mass of 13C 5 11.08361292 112 u2 5 13.003355 u
h
Exact number
by definition
The masses of other atoms can be determined in a similar fashion.
The mass for each element is given in the table inside the front cover of this text.
This value, even though it is actually a mass, is sometimes called the atomic weight for
each element.
Look at the value of the atomic mass of carbon given in this table. You might
expect to see 12, since we said the system of atomic masses is based on 12C. However, the number given for carbon is not 12 but 12.01. Why? The reason for this
apparent discrepancy is that the carbon found on earth (natural carbon) is a mixture
of the isotopes 12C, 13C, and 14C. All three isotopes have six protons, but they have
six, seven, and eight neutrons, respectively. Because natural carbon is a mixture of
isotopes, the atomic mass we use for carbon is an average value reflecting the average of the isotopes composing it.
The average atomic mass for carbon is computed as follows: It is known that natural
carbon is composed of 98.89% 12C atoms and 1.11% 13C atoms. The amount of 14C is
negligibly small at this level of precision. Using the masses of 12C (exactly 12 u) and
13
C (13.003355 u), we can calculate the average atomic mass for natural carbon as
follows:
98.89% of 12 u 1 1.11% of 13.0034 u 5
10.98892 112 u2 1 10.01112 113.0034 u2 5 12.01 u
a
100
Relative number of atoms
Ion beam intensity at detector
David Young-Wolff/Alamy
It is much easier to weigh out 600 hex
nuts than to count them one by one.
In this text we will call the average mass for an element the average atomic mass or,
simply, the atomic mass for that element.
Even though natural carbon does not contain a single atom with mass 12.01, for
stoichiometric purposes, we can consider carbon to be composed of only one type of
atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon.
Recall from Section 3.1 that counting by weighing works if you know the average
mass of the units being counted. Counting by weighing works just the same for atoms
as for jelly beans. For natural carbon with an average mass of 12.01 atomic mass units,
to obtain 1000 atoms would require weighing out 12,010 atomic mass units of natural
carbon (a mixture of 12C and 13C).
18
19
20
21
22
23
80
60
40
20
0
24
Mass number
b
91
.3
20
21
9
22
Mass number
c
Figure 3.2 | (a) Neon gas glowing in a discharge tube. The relative intensities of the signals recorded when natural neon is injected
into a mass ­spectrometer, represented in terms of (b) “peaks” and (c) a bar graph. The relative areas of the peaks are 0.9092 ( 20Ne),
0.00257 ( 21Ne), and 0.0882 ( 22Ne); natural neon is therefore 90.92% 20Ne, 0.257% 21Ne, and 8.82% 22Ne.
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3.3 The Mole
85
As in the case of carbon, the mass for each element listed in the table inside the
front cover of the text is an average value based on the isotopic composition of the
naturally occurring element. For instance, the mass listed for hydrogen (1.008) is
the average mass for natural hydrogen, which is a mixture of 1H and 2H (deuterium).
No atom of hydrogen actually has the mass 1.008.
In addition to being useful for determining accurate mass values for individual atoms, the mass spectrometer is used to determine the isotopic composition of a natural
element. For example, when a sample of natural neon is injected into a mass spectrometer, the mass spectrum shown in Fig. 3.2 is obtained. The areas of the “peaks” or the
21
22
heights of the bars indicate the relative abundances of 20
10Ne, 10Ne, and 10Ne atoms.
Example 3.1
The Average Mass of an Element
Arturo Limon/Shutterstock.com
When a sample of natural copper is vaporized and injected into a mass spectrometer, the
results shown in Fig. 3.3 are obtained. Use these data to compute the average mass of
natural copper. (The mass values for 63Cu and 65Cu are 62.93 u and 64.93 u, ­respectively.)
Copper nugget.
Solution
Where are we going?
To calculate the average mass of natural copper
What do we know?
❯ 63Cu mass 5 62.93 u
❯ 65Cu mass 5 64.93 u
Relative number of atoms
How do we get there?
As shown by the graph, of every 100 atoms of natural copper, 69.09 are 63Cu and 30.91
are 65Cu. Thus the mass of 100 atoms of natural copper is
169.09 atoms2 a62.93
100
80
The average mass of a copper atom is
69.09
60
40
6355 u
5 63.55 u/atom
100 atoms
30.91
This mass value is used in doing calculations involving the reactions of copper and is
the value given in the table inside the front cover of this book.
20
0
u
u
b 1 130.91 atoms2 a64.93
b 5 6355 u
atom
atom
63
65
Mass number
Figure 3.3 | Mass spectrum of
natural copper.
Reality Check | When you finish a calculation, you should always check whether
your answer makes sense. In this case our answer of 63.55 u is between the masses of
the atoms that make up natural copper. This makes sense. The answer could not be
smaller than 62.93 u or larger than 64.93 u.
See Exercises 3.37 and 3.38
3.3 The Mole
IBLG: See questions from “The Mole
and Molar Mass”
Because samples of matter typically contain so many atoms, a unit of measure called the
mole has been established for use in counting atoms. For our purposes, it is most convenient to define the mole (abbreviated mol) as the number equal to the number of carbon
atoms in exactly 12 g of pure 12C. Techniques such as mass spectrometry, which count
atoms very precisely, have been used to determine this number as 6.02214 3 1023
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Chapter 3
Stoichiometry
Figure 3.4 | One-mole samples of
Copper
several elements.
Iodine
Aluminum
Sulfur
Iron
Mercury
The SI definition of the mole is the amount
of a substance that contains as many
entities as there are in exactly 12 g of
carbon-12.
Avogadro’s number is 6.022 3 1023. One
mole of anything is 6.022 3 1023 units of
that substance.
Ken O’Donoghue
© Cengage
Learning
86
(6.022 3 1023 will be sufficient for our purposes). This number is called Avogadro’s
number to honor his contributions to chemistry. One mole of something consists of
6.022 3 1023 units of that substance. Just as a dozen eggs is 12 eggs, a mole of eggs is
6.022 31023 eggs.
The magnitude of the number 6.022 3 1023 is very difficult to imagine. To give you
some idea, 1 mole of seconds represents a span of time 4 million times as long as the
earth has already existed, and 1 mole of marbles is enough to cover the entire earth to
a depth of 50 miles! However, since atoms are so tiny, a mole of atoms or molecules is
a perfectly manageable quantity to use in a reaction (Fig. 3.4).
Critical Thinking
What if you were offered $1 million to count from 1 to 6 3 1023 at a rate of one number
each second?
Determine your hourly wage. Would you do it? Could you do it?
How do we use the mole in chemical calculations? Recall that Avogadro’s number
is defined as the number of atoms in exactly 12 g of 12C. This means that 12 g of 12C
contains 6.022 3 1023 atoms. It also means that a 12.01-g sample of natural carbon
contains 6.022 3 1023 atoms (a mixture of 12C, 13C, and 14C atoms, with an average
atomic mass of 12.01). Since the ratio of the masses of the samples (12 gy12.01 g) is
the same as the ratio of the masses of the individual components (12 uy12.01 u), the
two samples contain the same number of atoms (6.022 3 1023).
To be sure this point is clear, think of oranges with an average mass of 0.5 lb each
and grapefruit with an average mass of 1.0 lb each. Any two sacks for which the sack
of grapefruit weighs twice as much as the sack of oranges will contain the same number of pieces of fruit. The same idea extends to atoms. Compare natural carbon (average mass of 12.01) and natural helium (average mass of 4.003). A sample of 12.01 g
of natural carbon contains the same number of atoms as 4.003 g of natural helium.
Both samples contain 1 mole of atoms (6.022 3 1023). Table 3.1 gives more examples
that illustrate this basic idea.
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3.3 The Mole
87
Table 3.1 | Comparison of 1-Mole Samples of Various Elements
Element
Number of Atoms Present
Aluminum
Copper
Iron
Sulfur
Iodine
Mercury
The mass of 1 mole of an element is equal
to its atomic mass in grams.
Mass of Sample (g)
26.98
63.55
55.85
32.07
126.9
200.6
23
6.022 3 10
6.022 3 1023
6.022 3 1023
6.022 3 1023
6.022 3 1023
6.022 3 1023
Thus the mole is defined such that a sample of a natural element with a mass equal
to the element’s atomic mass expressed in grams contains 1 mole of atoms. This definition also fixes the relationship between the atomic mass unit and the gram. Since 6.022
3 1023 atoms of carbon (each with a mass of 12 u) have a mass of 12 g, then
and
16.022 3 1023 atoms2 a
12 u
b 5 12 g
atom
6.022 3 1023 u 5 1 g
h
Exact
number
This relationship can be used to derive the unit factor needed to convert between
atomic mass units and grams.
Critical Thinking
What if you discovered Avogadro’s number was not 6.02 3 1023 but 3.01 3 1023?
Would this affect the relative masses given on the periodic table? If so, how? If not,
why not?
Interactive
Example 3.2
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Determining the Mass of a Sample of Atoms
Americium is an element that does not occur naturally. It can be made in very small
amounts in a device known as a particle accelerator. Compute the mass in grams of a
sample of americium containing six atoms.
Solution
Where are we going?
To calculate the mass of six americium atoms
What do we know?
❯ Mass of 1 atom of Am 5 243 u (from the periodic table inside the front cover)
How do we get there?
The mass of six atoms is
6 atoms 3 243
u
5 1.46 3 103 u
atom
Using the relationship
6.022 3 1023 u 5 1 g
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88
Chapter 3
Stoichiometry
we write the conversion factor for converting atomic mass units to grams:
1g
6.022 3 1023 u
The mass of six americium atoms in grams is
1.46 3 103 u 3
1g
5 2.42 3 10221 g
6.022 3 1023 u
Reality Check Since this sample contains only six atoms, the mass should be very
small as the amount 2.42 3 10221 g indicates.
See Exercise 3.45
To do chemical calculations, you must understand what the mole means and how to
determine the number of moles in a given mass of a substance. These procedures are
­illustrated in Examples 3.3 and 3.4.
Aluminum (Al) is a metal with a high strength-to-mass ratio and a high resistance to
corrosion; thus it is often used for structural purposes. Compute both the number of
moles of atoms and the number of atoms in a 10.0-g sample of aluminum.
(left) Pure aluminum. (right) Aluminum
­alloys are used for many products used
in our kitchens.
Charles D. Winters/Photo Researchers, Inc.
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Determining Moles of Atoms
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Interactive
Example 3.3
Solution
Where are we going?
To calculate the moles and number of atoms in a sample of Al
What do we know?
❯ Sample contains 10.0 g of Al
❯ Mass of 1 mole (6.022 3 1023 atoms) of Al 5 26.93 g
How do we get there?
We can calculate the number of moles of Al in a 10.0-g sample as follows:
10.0 g Al 3
1 mol Al
5 0.371 mol Al atoms
26.98 g Al
The number of atoms in 10.0 g (0.371 mole) of aluminum is
0.371 mol Al 3
6.022 3 1023 atoms
5 2.23 3 1023 atoms
1 mol Al
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3.3 The Mole
89
Reality Check One mole of Al has a mass of 26.98 g and contains 6.022 3 1023 atoms. Our sample is 10.0 g, which is roughly 1y3 of 26.98. Thus the calculated amount
should be on the order of 1y3 of 6 3 1023, which it is.
See Exercise 3.46
Interactive
Example 3.4
Calculating Numbers of Atoms
A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg.
How many silicon (Si) atoms are present in the chip?
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in OWL.
Solution
Where are we going?
To calculate the atoms of Si in the chip
What do we know?
❯ The chip has 5.68 mg of Si
❯ Mass of 1 mole (6.022 3 1023 atoms) of Si 5 28.09 g
How do we get there?
The strategy for doing this problem is to convert from milligrams of silicon to grams
of silicon, then to moles of silicon, and finally to atoms of silicon:
5.68 mg Si 3
Always check to see if your answer is
sensible.
5.68 3 1023 g Si 3
Paying careful attention to units and
making sure the answer is reasonable can
help you detect an inverted ­conversion
factor or a number that was incorrectly
entered in your calculator.
2.02 3 1024 mol Si 3
1 g Si
5 5.68 3 1023 g Si
1000 mg Si
1 mol Si
5 2.02 3 1024 mol Si
28.09 g Si
6.022 3 1023 atoms
5 1.22 3 1020 atoms
1 mol Si
Reality Check Note that 5.68 mg of silicon is clearly much less than 1 mole of silicon (which has a mass of 28.09 g), so the final answer of 1.22 3 1020 atoms (compared
with 6.022 3 1023 atoms) is in the right direction.
See Exercise 3.47
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Interactive
Example 3.5
Russ Lappa/SPL/Photo Researchers, Inc.
Calculating the Number of Moles and Mass
Fragments of cobalt metal.
Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing 5.00 3 1020 atoms
and the mass of the sample.
Solution
Where are we going?
To calculate the number of moles and the mass of a sample of Co
What do we know?
❯ Sample contains 5.00 3 1020 atoms of Co
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90
Chapter 3
Stoichiometry
How do we get there?
Note that the sample of 5.00 3 1020 atoms of cobalt is less than 1 mole (6.022 3 1023
atoms) of cobalt. What fraction of a mole it represents can be determined as follows:
5.00 3 1020 atoms Co 3
1 mol Co
5 8.30 3 1024 mol Co
6.022 3 1023 atoms Co
Since the mass of 1 mole of cobalt atoms is 58.93 g, the mass of 5.00 3 1020 atoms can
be determined as follows:
8.30 3 1024 mol Co 3
58.93 g Co
5 4.89 3 1022 g Co
1 mol Co
Reality Check In this case the sample contains 5 3 1020 atoms, which is approximately 1y1000 of a mole. Thus the sample should have a mass of about (1y1000)(58.93)
> 0.06. Our answer of ,0.05 makes sense.
See Exercise 3.48
3.4 Molar Mass
A chemical compound is, ultimately, a collection of atoms. For example, methane (the
major component of natural gas) consists of molecules that each contain one carbon
and four hydrogen atoms (CH4). How can we calculate the mass of 1 mole of methane;
that is, what is the mass of 6.022 3 1023 CH4 molecules? Since each CH4 molecule
contains one carbon atom and four hydrogen atoms, 1 mole of CH4 molecules contains
1 mole of carbon atoms and 4 moles of hydrogen atoms. The mass of 1 mole of methane can be found by summing the masses of carbon and hydrogen present:
In this case, the term 12.01 limits the
number of significant figures.
A substance’s molar mass is the mass in
grams of 1 mole of the substance.
Interactive
Example 3.6
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in OWL.
Mass of 1 mol C 5 12.01 g
Mass of 4 mol H 5 4 3 1.008 g
Mass of 1 mol CH4 5 16.04 g
Because 16.04 g represents the mass of 1 mole of methane molecules, it makes
sense to call it the molar mass for methane. Thus the molar mass of a substance is the
mass in grams of 1 mole of the compound. Traditionally, the term molecular weight has
been used for this quantity. However, we will use molar mass exclusively in this text.
The molar mass of a known substance is obtained by summing the masses of the component atoms as we did for methane.
Methane is a molecular compound—its components are molecules. Many substances
are ionic—they contain simple ions or polyatomic ions. Examples are NaCl (contains
Na1 and Cl2) and CaCO3 (contains Ca21 and CO322). Because ionic compounds do not
contain molecules, we need a special name for the fundamental unit of these materials. Instead of molecule, we use the term formula unit. Thus CaCO3 is the formula unit
for calcium carbonate, and NaCl is the formula unit for sodium chloride.
Calculating Molar Mass I
Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also
a natural herbicide (weed killer) that kills off competitive plants around the black walnut
tree but does not affect grass and other noncompetitive plants. The formula for ­juglone is
C10H6O3.
a. Calculate the molar mass of juglone.
b. A sample of 1.56 3 1022 g of pure juglone was extracted from black walnut
husks. How many moles of juglone does this sample represent?
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3.4
Molar Mass
91
Solution
a. The molar mass is obtained by summing the masses of the component atoms. In
1 mole of juglone, there are 10 moles of carbon atoms, 6 moles of hydrogen
atoms, and 3 moles of oxygen atoms:
10 C: 10 3 12.01 g 5 120.1 g
6 H: 6 3 1.008 g 5 6.048 g
3 O: 3 3 16.00 g 5 48.00 g
Mass of 1 mol C10H6O3 5 174.1 g
The mass of 1 mole of juglone is 174.1 g, which is the molar mass.
b. The mass of 1 mole of this compound is 174.1 g; thus 1.56 3 1022 g is much less
than a mole. The exact fraction of a mole can be determined as follows:
Juglone
1.56 3 1022 g juglone 3
1 mol juglone
5 8.96 3 1025 mol juglone
174.1 g juglone
See Exercises 3.51 through 3.54
Interactive
Example 3.7
Calculating Molar Mass II
Calcium carbonate (CaCO3), also called calcite, is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals such as clams.
a. Calculate the molar mass of calcium carbonate.
b. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in
grams of this sample? What is the mass of the CO322 ions present?
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Solution
a. Calcium carbonate is an ionic compound composed of Ca21 and CO322 ions. In
1 mole of calcium carbonate, there are 1 mole of Ca21 ions and 1 mole of CO322
ions. The molar mass is calculated by summing the masses of the components:
Charles D. Winters
1 Ca21 : 1 3 40.08 g 5 40.08 g
1 CO322:
1 C:
1 3 12.01 g 5 12.01 g
3 O:
3 3 16.00 g 5 48.00 g
Mass of 1 mol CaCO3 5 100.09 g
A calcite (CaCO3) crystal.
Thus the mass of 1 mole of CaCO3 (1 mole of Ca21 plus 1 mole of CO322) is
100.09 g. This is the molar mass.
b. The mass of 1 mole of CaCO3 is 100.09 g. The sample contains nearly 5 moles,
or close to 500 g. The exact amount is determined as follows:
4.86 mol CaCO3 3
100.09 g CaCO3
5 486 g CaCO3
1 mol CaCO3
To find the mass of carbonate ions (CO322) present in this sample, we must
realize that 4.86 moles of CaCO3 contains 4.86 moles of Ca21 ions and
4.86 moles of CO322 ions. The mass of 1 mole of CO322 ions is
1 C: 1 3 12.01 5 12.01 g
3 O: 3 3 16.00 5 48.00 g
Mass of 1 mol CO322 5 60.01 g
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92
Chapter 3
Stoichiometry
Thus the mass of 4.86 moles of CO322 ions is
4.86 mol CO322 3
60.01 g CO322
5 292 g CO322
1 mol CO322
See Exercises 3.55 through 3.58
Interactive
Example 3.8
Molar Mass and Numbers of Molecules
Isopentyl acetate (C7H14O2) is the compound responsible for the scent of bananas. A
mo­lecular model of isopentyl acetate is shown in the margin below. Interestingly, bees
release about 1 mg (1 3 1026 g) of this compound when they sting. The resulting scent
attracts other bees to join the attack. How many molecules of isopentyl acetate are
released in a typical bee sting? How many atoms of carbon are present?
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Solution
Kenneth Lorenzen
Where are we going?
To calculate the number of molecules of isopentyl acetate and the number of carbon
atoms in a bee sting
Isopentyl acetate is released when a bee
stings.
What do we know?
❯
Mass of isopentyl acetate in a typical bee sting is 1 microgram = 1 3 1026 g
How do we get there?
Since we are given a mass of isopentyl acetate and want to find the number of molecules, we must first compute the molar mass of C7H14O2:
g
5 84.07 g C
mol
g
14 mol H 3 1.008
5 14.11 g H
mol
g
2 mol O 3 16.00
5 32.00 g O
mol
130.18 g
7 mol C 3 12.01
Isopentyl acetate
Carbon
Oxygen
Hydrogen
This means that 1 mole of isopentyl acetate (6.022 3 1023 molecules) has a mass of
130.18 g.
To find the number of molecules released in a sting, we must first determine the
number of moles of isopentyl acetate in 1 3 1026 g:
1 3 1026 g C7H14O2 3
1 mol C7H14O2
5 8 3 1029 mol C7H14O2
130.18 g C7H14O2
Since 1 mole is 6.022 3 1023 units, we can determine the number of molecules:
8 3 1029 mol C7H14O2 3
6.022 3 1023 molecules
5 5 3 1015 molecules
1 mol C7H14O2
To determine the number of carbon atoms present, we must multiply the number of
molecules by 7, since each molecule of isopentyl acetate contains seven carbon atoms:
To show the correct number of significant
figures in each calculation, we round after
each step. In your calculations, always
carry extra significant figures through to
the end, then round.
5 3 1015 molecules 3
7 carbon atoms
5 4 3 1016 carbon atoms
molecule
Note: In keeping with our practice of always showing the correct number of significant figures, we have rounded after each step. However, if extra digits are carried
throughout this problem, the final answer rounds to 3 3 1016.
See Exercises 3.59 through 3.64
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3.5 Learning to Solve Problems
93
John Humble/The Image Bank/Getty Images
3.5 Learning to Solve Problems
Pigeonholes can be used for sorting and
classifying objects like mail.
One of the great rewards of studying chemistry is to become a good problem solver.
Being able to solve complex problems is a talent that will serve you well in all walks
of life. It is our purpose in this text to help you learn to solve problems in a flexible,
creative way based on understanding the fundamental ideas of chemistry. We call this
approach conceptual problem solving.
The ultimate goal is to be able to solve new problems (that is, problems you have
not seen before) on your own. In this text we will provide problems and offer solutions
by explaining how to think about the problems. While the answers to these problems
are ­important, it is perhaps even more important to understand the process—the thinking ­necessary to get the answer. Although at first we will be solving the problem for
you, do not take a passive role. While studying the solution, it is crucial that you interactively think through the problem with us. Do not skip the discussion and jump to the
answer. Usually, the solution will involve asking a series of questions. Make sure that
you understand each step in the process. This active approach should apply to problems outside of chemistry as well. For example, imagine riding with someone in a car
to an unfamiliar destination. If your goal is simply to have the other person get you to
that destination, you will probably not pay much attention to how to get there (passive), and if you have to find this same place in the future on your own, you probably
will not be able to do it. If, however, your goal is to learn how to get there, you would
pay attention to distances, signs, and turns (active). This is how you should read the
solutions in the text (and the text in general).
While actively studying our solutions to problems is helpful, at some point you will
need to know how to think through these problems on your own. If we help you too
much as you solve a problem, you won’t really learn effectively. If we always “drive,”
you won’t interact as meaningfully with the material. Eventually you need to learn to
drive yourself. We will provide more help at the beginning of the text and less as we
proceed to later chapters.
There are two fundamentally different ways you might use to approach a problem.
One way emphasizes memorization. We might call this the “pigeonholing method.” In
this approach, the first step is to label the problem—to decide in which pigeonhole it
fits. The pigeonholing method requires that we provide you with a set of steps that you
memorize and store in the appropriate slot for each different problem you encounter.
The difficulty with this method is that it requires a new pigeonhole each time a problem is changed by even a small amount.
Consider the driving analogy again. Suppose you have memorized how to drive from
your house to the grocery store. Do you know how to drive back from the grocery store
to your house? Not necessarily. If you have only memorized the directions and do not
­understand fundamental principles such as “I traveled north to get to the store, so my
house is south of the store,” you may find yourself stranded. In a more complicated
­example, ­suppose you know how to get from your house to the store (and back) and
from your house to the library (and back). Can you get from the library to the store
without having to go back home? Probably not if you have only memorized directions
and you do not have a “big picture” of where your house, the store, and the library are
relative to one ­another.
The second approach is conceptual problem solving, in which we help you get the
“big picture”—a real understanding of the situation. This approach to problem solving
looks within the problem for a solution. In this method we assume that the problem is
a new one, and we let the problem guide us as we solve it. In this approach we ask a
series of questions as we proceed and use our knowledge of fundamental principles to
answer these questions. Learning this approach requires some patience, but the reward
for learning to solve problems this way is that we become an effective solver of any
new problem that confronts us in daily life or in our work in any field. In summary,
instead of looking outside the problem for a memorized solution, we will look inside
the problem and let the problem help us as we proceed to a solution.
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94
Chapter 3
Stoichiometry
As we have seen in problems we have already considered, there are several organizing principles to help you become a creative problem solver. Although we have
been using these ideas in earlier problems, let’s review and expand on them. Because
as we progress in our study of chemistry the problems become more complicated,
we will need to rely on this approach even more.
1. We need to read the problem and decide on the final goal. Then we sort through
the facts given, focusing on the key words and often drawing a diagram of the
problem. In this part of the analysis we need to state the problem as simply and
as visually as possible. We could summarize this entire process as “Where are
we going?”
2. In order to reach our final goal, we need to decide where to start. For example, in
a stoichiometry problem we always start with the chemical reaction. Then we ask
a series of questions as we proceed, such as, “What are the reactants and products?” “What is the balanced equation?” “What are the amounts of the reactants?”
and so on. Our understanding of the fundamental principles of chemistry will enable us to answer each of these simple questions and eventually will lead us to the
final solution. We might summarize this process as “How do we get there?”
3. Once we get the solution of the problem, then we ask ourselves, “Does it make
sense?” That is, does our answer seem reasonable? We call this the Reality Check.
It always pays to check your answer.
Using a conceptual approach to problem solving will enable you to develop real
confidence as a problem solver. You will no longer panic when you see a problem that
is different in some ways from those you have solved in the past. Although you might
be frustrated at times as you learn this method, we guarantee that it will pay dividends
later and should make your experience with chemistry a positive one that will prepare
you for any career you choose.
To summarize, one of our major goals in this text is to help you become a creative
problem solver. We will do this by, at first, giving you lots of guidance in how to solve
problems. We will “drive,” but we hope you will be paying attention instead of just
“riding along.” As we move forward, we will gradually shift more of the responsibility
to you. As you gain confidence in letting the problem guide you, you will be amazed
at how effective you can be at solving some really complex problems—just like the
ones you will confront in “real life.”
3.6 Percent Composition of Compounds
Experiment 14: Composition 1:
Percentage Composition and Empirical
Formula of ­Magnesium Oxide
There are two common ways of describing the composition of a compound: in terms
of the numbers of its constituent atoms and in terms of the percentages (by mass) of its
elements. We can obtain the mass percents of the elements from the formula of the
compound by comparing the mass of each element present in 1 mole of the compound
to the total mass of 1 mole of the compound.
For example, for ethanol, which has the formula C2H5OH, the mass of each element
present and the molar mass are obtained as follows:
g
mol
g
Mass of H 5 6 mol 3 1.008
mol
g
Mass of O 5 1 mol 3 16.00
mol
Mass of 1 mol C2H5OH
Mass of C 5 2 mol 3 12.01
5 24.02 g
5 6.048 g
5 16.00 g
5 46.07 g
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3.6
Percent Composition of Compounds
95
The mass percent (often called the weight percent) of carbon in ethanol can be computed by comparing the mass of carbon in 1 mole of ethanol to the total mass of 1 mole
of ethanol and multiplying the result by 100:
mass of C in 1 mol C2H5OH
3 100%
mass of 1 mol C2H5OH
24.02 g
5
3 100% 5 52.14%
46.07 g
Mass percent of C 5
The mass percents of hydrogen and oxygen in ethanol are obtained in a similar manner:
mass of H in 1 mol C2H5OH
3 100%
mass of 1 mol C2H5OH
6.048 g
5
3 100% 5 13.13%
46.07 g
mass of O in 1 mol C2H5OH
Mass percent of O 5
3 100%
mass of 1 mol C2H5OH
16.00 g
5
3 100% 5 34.73%
46.07 g
Mass percent of H 5
Reality Check Notice that the percentages add up to 100.00%; this provides a check
that the calculations are correct.
Interactive
Example 3.9
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in OWL.
Calculating Mass Percent
Carvone is a substance that occurs in two forms having different arrangements of the
atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives
caraway seeds their characteristic smell, and the other type is responsible for the smell
of spearmint oil. Compute the mass percent of each element in carvone.
Solution
Where are we going?
To find the mass percent of each element in carvone
What do we know?
❯ Molecular formula is C10H14O
What information do we need to find the mass percent?
❯ Mass of each element (we’ll use 1 mole of carvone)
❯ Molar mass of carvone
How do we get there?
What is the mass of each element in 1 mole of C10H14O?
g
5 120.1 g
mol
g
Mass of H in 1 mol 5 14 mol 3 1.008
5 14.11 g
mol
g
Mass of O in 1 mol 5 1 mol 3 16.00
5 16.00 g
mol
Mass of C in 1 mol 5 10 mol 3 12.01
What is the molar mass of C10H14O?
Carvone
120.1 g 1 14.11 g 1 16.00 g 5 150.2 g
C10 1 H14 1 O
5 C10 H14O
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96
Chapter 3
Stoichiometry
What is the mass percent of each element?
We find the fraction of the total mass contributed by each element and convert it to a
­percentage:
120.1 g C
❯ Mass percent of C 5
3 100% 5 79.96%
150.2 g C10H14O
14.11 g H
❯ Mass percent of H 5
3 100% 5 9.394%
150.2 g C10H14O
16.00 g O
❯ Mass percent of O 5
3 100% 5 10.65%
150.2 g C10H14O
Reality Check Sum the individual mass percent values—they should total to 100%
within round-off errors. In this case, the percentages add up to 100.00%.
See Exercises 3.73 and 3.74
3.7 Determining the Formula
of a Compound
Experiment 14: Composition 2:
Percentage Water in a Hydrate
When a new compound is prepared, one of the first items of interest is the formula of
the compound. This is most often determined by taking a weighed sample of the compound and either decomposing it into its component elements or reacting it with oxygen to produce substances such as CO2, H2O, and N2, which are then collected and
weighed. A device for doing this type of analysis is shown in Fig. 3.5. The results of
such analyses provide the mass of each type of element in the compound, which can be
used to determine the mass percent of each element.
We will see how information of this type can be used to compute the formula of a
compound. Suppose a substance has been prepared that is composed of carbon, hydrogen, and nitrogen. When 0.1156 g of this compound is reacted with oxygen, 0.1638 g
of carbon dioxide (CO2) and 0.1676 g of water (H2O) are collected. Assuming that all
the carbon in the compound is converted to CO2, we can determine the mass of carbon
originally present in the 0.1156-g sample. To do this, we must use the fraction (by
mass) of carbon in CO2. The molar mass of CO2 is
g
5 12.01 g
mol
g
O: 2 mol 3 16.00
5 32.00 g
mol
Molar mass of CO2 5 44.01 g/mol
C: 1 mol 3 12.01
CO2
Furnace
CO2, H2O, O2, and other gases
O2 and
other gases
Sample
O2
H2O absorber
such as Mg(ClO4)2
CO2 absorber
such as NaOH
Figure 3.5 | A schematic diagram of the combustion device used to analyze substances
for carbon and hydrogen. The sample is burned in the presence of excess oxygen, which
converts all its carbon to carbon dioxide and all its hydrogen to water. These products are
collected by absorption using appropriate materials, and their amounts are determined
by measuring the increase in masses of the absorbents.
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3.7
Determining the Formula of a Compound
97
The fraction of carbon present by mass is
Mass of C
12.01 g C
5
Total mass of CO2
44.01 g CO2
This factor can now be used to determine the mass of carbon in 0.1638 g of CO2:
0.1638 g CO2 3
12.01 g C
5 0.04470 g C
44.01 g CO2
Remember that this carbon originally came from the 0.1156-g sample of unknown
compound. Thus the mass percent of carbon in this compound is
0.04470 g C
3 100% 5 38.67% C
0.1156 g compound
H2O
The same procedure can be used to find the mass percent of hydrogen in the unknown
compound. We assume that all the hydrogen present in the original 0.1156 g of compound
was converted to H2O. The molar mass of H2O is 18.02 g, and the fraction of hydrogen by
mass in H2O is
Mass of H
2.016 g H
5
Mass of H2O
18.02 g H2O
Therefore, the mass of hydrogen in 0.1676 g of H2O is
0.1676 g H2O 3
2.016 g H
5 0.01875 g H
18.02 g H2O
The mass percent of hydrogen in the compound is
0.01875 g H
3 100% 5 16.22% H
0.1156 g compound
PowerLecture: Oxidation of Zinc
with Iodine
The unknown compound contains only carbon, hydrogen, and nitrogen. So far we have
determined that it is 38.67% carbon and 16.22% hydrogen. The remainder must be
nitrogen:
100.00% 2 138.67% 1 16.22%2 5 45.11% N
h
%C
h
%H
We have determined that the compound contains 38.67% carbon, 16.22% hydrogen,
and 45.11% nitrogen. Next we use these data to obtain the formula.
Since the formula of a compound indicates the numbers of atoms in the compound,
we must convert the masses of the elements to numbers of atoms. The easiest way to
do this is to work with 100.00 g of the compound. In the present case, 38.67% carbon
by mass means 38.67 g of carbon per 100.00 g of compound; 16.22% hydrogen means
16.22 g of hydrogen per 100.00 g of compound; and so on. To determine the formula,
we must calculate the number of carbon atoms in 38.67 g of carbon, the number of
hydrogen atoms in 16.22 g of hydrogen, and the number of nitrogen atoms in 45.11 g
of nitrogen. We can do this as follows:
1 mol C
5 3.220 mol C
12.01 g C
1 mol H
16.22 g H 3
5 16.09 mol H
1.008 g H
1 mol N
45.11 g N 3
5 3.220 mol N
14.01 g N
38.67 g C 3
Thus 100.00 g of this compound contains 3.220 moles of carbon atoms, 16.09 moles
of hydrogen atoms, and 3.220 moles of nitrogen atoms.
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98
Chapter 3
Stoichiometry
We can find the smallest whole-number ratio of atoms in this compound by dividing each of the mole values above by the smallest of the three:
3.220
5 1.000 5 1
3.220
16.09
5 4.997 5 5
H:
3.220
C:
N:
Molecular formula 5 (empirical formula)n,
where n is an integer.
3.220
5 1.000 5 1
3.220
Thus the formula might well be CH5N. However, it also could be C2H10N2 or
C3H15N3, and so on—that is, some multiple of the smallest whole-number ratio. Each
of these alternatives also has the correct relative numbers of atoms. That is, any molecule that can be represented as (CH5N)n, where n is an integer, has the empirical
formula CH5N. To be able to specify the exact formula of the molecule involved, the
molecular formula, we must know the molar mass.
Suppose we know that this compound with empirical formula CH5N has a molar
mass of 31.06 g/mol. How do we determine which of the possible choices represents
the mo­lecular formula? Since the molecular formula is always a whole-number multiple of the empirical formula, we must first find the empirical formula mass for CH5N:
1 C: 1 3 12.01 g 5 12.01 g
5 H: 5 3 1.008 g 5 5.040 g
1 N: 1 3 14.01 g 5 14.01 g
Formula mass of CH5N 5 31.06 g/mol
This is the same as the known molar mass of the compound. Thus in this case the empirical formula and the molecular formula are the same; this substance consists of
molecules with the formula CH5N. It is quite common for the empirical and molecular
formulas to be different; some examples where this is the case are shown in Fig. 3.6.
Problem-Solving Strategy
Empirical Formula Determination
❯
❯
Numbers very close to whole numbers,
such as 9.92 and 1.08, should be rounded
to whole numbers. Numbers such as 2.25,
4.33, and 2.72 should not be rounded to
whole numbers.
❯
❯
Since mass percentage gives the number of grams of a particular element per
100 g of compound, base the calculation on 100 g of compound. Each
percent will then represent the mass in grams of that element.
Determine the number of moles of each element present in 100 g of compound using the atomic masses of the elements present.
Divide each value of the number of moles by the smallest of the values. If
each resulting number is a whole number (after appropriate rounding), these
numbers represent the subscripts of the elements in the empirical formula.
If the numbers obtained in the previous step are not whole numbers, multiply
each number by an integer so that the results are all whole numbers.
Critical Thinking
One part of the problem-solving strategy for empirical formula determination is to
base the calculation on 100 g of compound. What if you chose a mass other than 100 g? Would this work? What if you chose to base the calculation on 100 moles of
compound? Would this work?
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3.7
Determining the Formula of a Compound
99
Figure 3.6 | Examples of substances
whose empirical and molecular
formulas differ. Notice that molecular
formula 5 (empirical formula)n, where
n is an integer.
C6H6 = (CH)6
S8 = (S)8
C6H12O6 = (CH2O)6
Problem-Solving Strategy
Determining Molecular Formula from Empirical Formula
❯
❯
❯
Obtain the empirical formula.
Compute the mass corresponding to the empirical formula.
Calculate the ratio:
Molar mass
Empirical formula mass
❯
The integer from the previous step represents the number of empirical formula
units in one molecule. When the empirical formula subscripts are multiplied by
this integer, the molecular formula results. This procedure is summarized by the
equation:
Molecular formula 5 empirical formula 3
Interactive
Example 3.10
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.com to try this Interactive Example
in OWL.
molar mass
empirical formula mass
Determining Empirical and Molecular Formulas I
Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percents):
71.65% Cl 24.27% C 4.07% H
The molar mass is known to be 98.96 g/mol.
Solution
Where are we going?
To find the empirical and molecular formulas for the given compound
What do we know?
❯ Percent of each element
❯ Molar mass of the compound is 98.96 g/mol
What information do we need to find the empirical formula?
❯ Mass of each element in 100.00 g of compound
❯ Moles of each element
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100
Chapter 3
Stoichiometry
How do we get there?
What is the mass of each element in 100.00 g of compound?
Cl 71.65 g C 24.27 g H 4.07 g
What are the moles of each element in 100.00 g of compound?
1 mol Cl
5 2.021 mol Cl
35.45 g Cl
1 mol C
24.27 g C 3
5 2.021 mol C
12.01 g C
1 mol H
4.07 g H 3
5 4.04 mol H
1.008 g H
71.65 g Cl 3
What is the empirical formula for the compound?
Dividing each mole value by 2.021 (the smallest number of moles present), we find the
empirical formula ClCH2.
What is the molecular formula for the compound?
Compare the empirical formula mass to the molar mass.
Empirical formula mass 5 49.48 g/mol (Confirm this!)
Molar mass is given 5 98.96 g/mol
❯
Figure 3.7 | The two forms of
dichloroethane.
Molar mass
98.96 g/mol
5
52
Empirical formula mass
49.48 g/mol
Molecular formula 5 1ClCH22 2 5 Cl2C2H4
This substance is composed of molecules with the formula Cl2C2H4.
Note: The method we use here allows us to determine the molecular formula of a
compound but not its structural formula. The compound Cl2C2H4 is called dichloro­
ethane. There are two forms of this compound, shown in Fig. 3.7. The form at the
bottom was formerly used as an additive in leaded gasoline.
See Exercises 3.87 and 3.88
Interactive
Example 3.11
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in OWL.
Determining Empirical and Molecular Formulas II
A white powder is analyzed and found to contain 43.64% phosphorus and 56.36%
oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?
Solution
Where are we going?
To find the empirical and molecular formulas for the given compound
What do we know?
❯ Percent of each element
❯ Molar mass of the compound is 283.88 g/mol
What information do we need to find the empirical formula?
❯ Mass of each element in 100.00 g of compound
❯ Moles of each element
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3.7
Determining the Formula of a Compound
101
How do we get there?
What is the mass of each element in 100.00 g of compound?
P 43.64 g O 56.36 g
What are the moles of each element in 100.00 g of compound?
1 mol P
5 1.409 mol P
30.97 g P
1 mol O
56.36 g O 3
5 3.523 mol O
16.00 g O
43.64 g P 3
What is the empirical formula for the compound?
Dividing each mole value by the smaller one gives
1.409
5 1 P and
1.409
3.523
5 2.5 O
1.409
This yields the formula PO2.5. Since compounds must contain whole numbers of atoms,
the empirical formula should contain only whole numbers. To obtain the simplest set of
whole numbers, we multiply both numbers by 2 to give the empirical formula P2O5.
What is the molecular formula for the compound?
Compare the empirical formula mass to the molar mass.
Figure 3.8 | The structure of P4O10.
Note that some of the oxygen atoms
act as “bridges” between the phosphorus atoms. This compound has a great
affinity for water and is often used as
a desiccant, or ­drying agent.
Empirical formula mass 5 141.94 g/mol (Confirm this!)
Molar mass is given 5 283.88 g/mol
Molar mass
283.88
5
52
Empirical formula mass
141.94
❯
The molecular formula is (P2O5)2, or P4O10.
Note: The structural formula for this interesting compound is given in Fig. 3.8.
See Exercise 3.89
In Examples 3.10 and 3.11 we found the molecular formula by comparing the empirical formula mass with the molar mass. There is an alternate way to obtain the
mo­lecular formula. For example, in Example 3.10 we know the molar mass of the
compound is 98.96 g/mol. This means that 1 mole of the compound weighs 98.96 g.
Since we also know the mass percentages of each element, we can compute the mass
of each element present in 1 mole of compound:
71.65 g Cl
98.96 g
70.90 g Cl
3
5
100.0 g compound
mol
mol compound
24.27 g C
98.96 g
24.02 g C
Carbon:
3
5
100.0 g compound
mol
mol compound
4.07 g H
98.96 g
4.03 g H
Hydrogen:
3
5
100.0 g compound
mol
mol compound
Chlorine:
Now we can compute moles of atoms present per mole of compound:
70.90 g Cl
1 mol Cl
2.000 mol Cl
3
5
mol compound
35.45 g Cl
mol compound
24.02 g C
1 mol C
2.000 mol C
Carbon:
3
5
mol compound
12.01 g C
mol compound
4.03 g H
1 mol H
4.00 mol H
Hydrogen:
3
5
mol compound
1.008 g H
mol compound
Chlorine:
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102
Chapter 3
Stoichiometry
Thus 1 mole of the compound contains 2 moles of Cl atoms, 2 moles of C atoms,
and 4 moles of H atoms, and the molecular formula is Cl2C2H4, as obtained in Example 3.10.
Problem-Solving Strategy
Determining Molecular Formula from Mass Percent and Molar Mass
❯
❯
❯
Interactive
Example 3.12
Using the mass percentages and the molar mass, determine the mass of each
element present in 1 mole of compound.
Determine the number of moles of each element present in 1 mole of
compound.
The integers from the previous step represent the subscripts in the molecular
formula.
Determining a Molecular Formula
Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon,
5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass
of 194.2 g/mol. Determine the molecular formula of caffeine.
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.com to try this Interactive Example
in OWL.
Solution
Ken O’Donoghue © Cengage Learning
Where are we going?
To find the molecular formula for caffeine
Computer-generated molecule of caffeine.
What do we know?
❯ Percent of each element
❯ 49.48% C
❯  5.15% H
❯ Molar mass of caffeine is 194.2 g/mol
❯
❯
28.87% N
16.49% O
What information do we need to find the molecular formula?
❯ Mass of each element (in 1 mole of caffeine)
❯ Mole of each element (in 1 mole of caffeine)
How do we get there?
What is the mass of each element in 1 mole (194.2 g) of caffeine?
49.48 g C
100.0 g caffeine
5.15 g H
100.0 g caffeine
28.87 g N
100.0 g caffeine
16.49 g O
100.0 g caffeine
194.2 g
mol
194.2 g
3
mol
194.2 g
3
mol
194.2 g
3
mol
3
96.09 g C
mol caffeine
10.0 g H
5
mol caffeine
56.07 g N
5
mol caffeine
32.02 g O
5
mol caffeine
5
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3.8
Chemical Equations
103
What are the moles of each element in 1 mole of caffeine?
Carbon:
Hydrogen:
Nitrogen:
Oxygen:
96.09 g C
1 mol C
3
mol caffeine
12.01 g C
10.0 g H
1 mol H
3
mol caffeine
1.008 g H
56.07 g N
1 mol N
3
mol caffeine
14.01 g N
32.02 g O
1 mol O
3
mol caffeine
16.00 g O
8.001 mol C
mol caffeine
9.92 mol H
5
mol caffeine
4.002 mol N
5
mol caffeine
2.001 mol O
5
mol caffeine
5
Rounding the numbers to integers gives the molecular formula for caffeine: C8H10N4O2.
See Exercise 3.90
3.8 Chemical Equations
Chemical Reactions
IBLG: See questions from “Balancing
Chemical Equations”
A chemical change involves a reorganization of the atoms in one or more substances.
For example, when the methane (CH4) in natural gas combines with oxygen (O2) in the
air and burns, carbon dioxide (CO2) and water (H2O) are formed. This process is represented by a chemical equation with the reactants (here methane and oxygen) on the
left side of an arrow and the products (carbon dioxide and water) on the right side:
CH4 1 O2 h CO2 1 H2O
Reactants
Products
Notice that the atoms have been reorganized. Bonds have been broken, and new ones
have been formed. It is important to recognize that in a chemical reaction, atoms are
neither created nor destroyed. All atoms present in the reactants must be accounted for
among the products. In other words, there must be the same number of each type of
atom on the products side and on the reactants side of the arrow. Making sure that this
rule is obeyed is called balancing a chemical equation for a reaction.
The equation (shown above) for the reaction between CH4 and O2 is not balanced.
We can see this from the following representation of the reaction:
+
+
Notice that the number of oxygen atoms (in O2) on the left of the arrow is two, while on
the right there are three O atoms (in CO2 and H2O). Also, there are four hydrogen atoms
(in CH4) on the left and only two (in H2O) on the right. Remember that a chemical reaction is simply a rearrangement of the atoms (a change in the way they are organized).
Atoms are neither created nor destroyed in a chemical reaction. Thus the reactants and
products must occur in numbers that give the same number of each type of atom among
both the reactants and products. Simple trial and error will allow us to figure this out for
the reaction of methane with oxygen. The needed numbers of molecules are
+
+
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104
Chapter 3
Stoichiometry
Notice that now we have the same number of each type of atom represented among the
reactants and the products.
We can represent the preceding situation in a shorthand manner by the following
chemical equation:
CH4 1 2O2 h CO2 1 2H2O
We can check that the equation is balanced by comparing the number of each type of
atom on both sides:
CH4 1 2O2 h CO2 1 2H2O
p h
1C 4H
6
h
6
6
h h
h h
1C
4H
4 O 2 O 2 O
To summarize, we have
Reactants
Products
1C
4H
4O
1C
4H
4O
The Meaning of a Chemical Equation
The chemical equation for a reaction gives two important types of information: the
nature of the reactants and products and the relative numbers of each.
The reactants and products in a specific reaction must be identified by experiment.
Besides specifying the compounds involved in the reaction, the equation often gives
the physical states of the reactants and products:
State
Symbol
Solid
Liquid
Gas
Dissolved in water (in aqueous solution)
(s)
(l)
(g)
(aq)
For example, when hydrochloric acid in aqueous solution is added to solid sodium
hydrogen carbonate, the products carbon dioxide gas, liquid water, and sodium chloride (which dissolves in the water) are formed:
HCl 1aq2 1 NaHCO3 1s2 h CO2 1g2 1 H2O 1l2 1 NaCl 1aq2
The relative numbers of reactants and products in a reaction are indicated by the
coefficients in the balanced equation. (The coefficients can be determined because we
know that the same number of each type of atom must occur on both sides of the equation.) For example, the balanced equation
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1g2
can be interpreted in several equivalent ways, as shown in Table 3.2. Note that the total
mass is 80 g for both reactants and products. We expect the mass to remain constant,
since chemical reactions involve only a rearrangement of atoms. Atoms, and therefore
mass, are conserved in a chemical reaction.
From this discussion you can see that a balanced chemical equation gives you a
great deal of information.
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3.9
Balancing Chemical Equations
105
Table 3.2 | Information Conveyed by the Balanced Equation for the Combustion of Methane
Reactants
Products
CH4 1g2 1 2O2 1g2
CO2 1g2 1 2H2O 1g2
88n
88n
1 molecule 1 2 molecules
1 mole 1 2 moles
6.022 3 1023 molecules 1 2 16.022 3 1023 molecules2
16 g 1 2 132 g2
80 g reactants
88n
88n
88n
1 molecule 1 2 molecules
1 mole 1 2 moles
6.022 3 1023 molecules 1 2 16.022 3 1023 molecules2
44 g 1 2 118 g2
80 g products
3.9 Balancing Chemical Equations
PowerLecture: Conservation of Mass
and Balancing Equations
An unbalanced chemical equation is of limited use. Whenever you see an equation,
you should ask yourself whether it is balanced. The principle that lies at the heart of
the balancing process is that atoms are conserved in a chemical reaction. The same
number of each type of atom must be found among the reactants and products. It is
also important to recognize that the identities of the reactants and products of a reaction are determined by experimental observation. For example, when liquid ethanol is
burned in the presence of sufficient oxygen gas, the products are always carbon dioxide and water. When the equation for this reaction is balanced, the identities of the
reactants and products must not be changed. The formulas of the compounds must
never be changed in balancing a chemical equation. That is, the subscripts in a formula
cannot be changed, nor can atoms be added or subtracted from a formula.
Critical Thinking
What if a friend was balancing chemical equations by changing the values of the subscripts instead of using the coefficients? How would you explain to your friend that
this was the wrong thing to do?
In balancing equations, start with the
most complicated molecule.
Most chemical equations can be balanced by inspection, that is, by trial and error.
It is always best to start with the most complicated molecules (those containing the
greatest number of atoms). For example, consider the reaction of ethanol with oxygen,
given by the unbalanced equation
C2H5OH 1l2 1 O2 1g2 h CO2 1g2 1 H2O 1g2
which can be represented by the following molecular models:
+
+
Notice that the carbon and hydrogen atoms are not balanced. There are two carbon
atoms on the left and one on the right, and there are six hydrogens on the left and two on
the right. We need to find the correct numbers of reactants and products so that we
have the same number of all types of atoms among the reactants and products. We
will balance the equation “by inspection” (a systematic trial-and-error procedure).
The most complicated molecule here is C2H5OH. We will begin by balancing the
products that contain the atoms in C2H5OH. Since C2H5OH contains two carbon atoms, we place the coefficient 2 before the CO2 to balance the carbon atoms:
C2H5OH(l) + O2(g)
2 C atoms
2CO2(g) + H2O(g)
2 C atoms
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106
Chapter 3
Stoichiometry
Since C2H5OH contains six hydrogen atoms, the hydrogen atoms can be balanced by
­placing a 3 before the H2O:
2CO2(g) + 3H2O(g)
C2H5OH(l ) + O2(g)
(5 + 1) H
(3 × 2) H
Last, we balance the oxygen atoms. Note that the right side of the preceding equation
contains seven oxygen atoms, whereas the left side has only three. We can correct this
by putting a 3 before the O2 to produce the balanced equation:
2CO2(g) + 3H2O(g)
C2H5OH(l) + 3O2(g)
1O
(2 × 2) O
6O
3O
7O
7O
Now we check:
2CO2(g) + 3H2O(g)
C2H5OH(l) + 3O2(g)
2 C atoms
6 H atoms
7 O atoms
2 C atoms
6 H atoms
7 O atoms
The equation is balanced.
The balanced equation can be represented as follows:
+
+
You can see that all the elements balance.
Problem-Solving Strategy
Writing and Balancing the Equation for a Chemical Reaction
1. Determine what reaction is occurring. What are the reactants, the products,
and the physical states involved?
2. Write the unbalanced equation that summarizes the reaction described in
Step 1.
3. Balance the equation by inspection, starting with the most complicated
molecule(s). Determine what coefficients are necessary so that the same
number of each type of atom appears on both reactant and product sides.
Do not change the identities (formulas) of any of the reactants or products.
Critical Thinking
One part of the problem-solving strategy for balancing chemical equations is “starting
with the most complicated molecule.” What if you started with a different molecule?
Could you still eventually balance the chemical equation? How would this approach
be different from the suggested technique?
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3.9
Interactive
Example 3.13
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in OWL.
Chromate and dichromate compounds are
carcinogens (cancer-inducing agents) and
should be handled very carefully.
Balancing Chemical Equations
107
Balancing a Chemical Equation I
Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH4)2Cr2O7, a vivid orange compound, is ignited, a spectacular reaction
occurs, as shown in the two photographs. Although the reaction is actually somewhat
more complex, let’s assume here that the products are solid chromium(III) oxide, nitrogen gas (consisting of N2 molecules), and water vapor. Balance the equation for this
reaction.
Solution
1. From the description given, the reactant is solid ammonium dichromate,
(NH4)2Cr2O7(s), and the products are nitrogen gas, N2(g), water vapor, H2O(g),
and solid chromium(III) oxide, Cr2O3(s). The formula for chromium(III) oxide
can be ­determined by recognizing that the Roman numeral III means that
Cr31 ions are present. For a neutral compound, the formula must then be
Cr2O3, since each oxide ion is O22.
2. The unbalanced equation is
1NH42 2Cr2O7 1s2 S Cr2O3 1s2 1 N2 1g2 1 H2O 1g2
3. Note that nitrogen and chromium are balanced (two nitrogen atoms and two
chromium atoms on each side), but hydrogen and oxygen are not. A coefficient of 4 for H2O balances the hydrogen atoms:
1NH42 2Cr2O7 1s2 S Cr2O3 1s2 1 N2 1g2 1 4H2O 1g2
14 3 22 H
14 3 22 H
Note that in balancing the hydrogen we also have balanced the oxygen, since there are
seven oxygen atoms in the reactants and in the products.
PowerLecture: Ammonium Dichromate
Volcano
Reality Check
2 N, 8 H, 2 Cr, 7 O S 2 N, 8 H, 2 Cr, 7 O
Reactant
atoms
Product
atoms
Photos: Ken O’Donoghue © Cengage Learning
The equation is balanced.
Decomposition of ammonium dichromate.
See Exercises 3.95 through 3.98
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108
Chapter 3
Stoichiometry
Interactive
Example 3.14
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in OWL.
Balancing a Chemical Equation II
At 10008C, ammonia gas, NH3(g), reacts with oxygen gas to form gaseous nitric oxide,
NO(g), and water vapor. This reaction is the first step in the commercial production of
nitric acid by the Ostwald process. Balance the equation for this reaction.
Solution
1, 2. The unbalanced equation for the reaction is
NH3 1g2 1 O2 1g2 S NO 1g2 1 H2O 1g2
3. Because all the molecules in this equation are of about equal complexity,
where we start in balancing it is rather arbitrary. Let’s begin by balancing the
hydrogen. A coefficient of 2 for NH3 and a coefficient of 3 for H2O give six
atoms of hydrogen on both sides:
2NH3 1g2 1 O2 1g2 S NO 1g2 1 3H2O 1g2
The nitrogen can be balanced with a coefficient of 2 for NO:
2NH3 1g2 1 O2 1g2 S 2NO 1g2 1 3H2O 1g2
Finally, note that there are two atoms of oxygen on the left and five on the right. The
oxygen can be balanced with a coefficient of 52 for O2:
5
O 1g2 S 2NO 1g2 1 3H2O 1g2
2 2
However, the usual custom is to have whole-number coefficients. We simply multiply
the entire equation by 2.
2NH3 1g2 1
4NH3 1g2 1 5O2 1g2 S 4NO 1g2 1 6H2O 1g2
Reality Check There are 4 N, 12 H, and 10 O on both sides, so the equation is
balanced.
We can represent this balanced equation visually as
+
+
See Exercises 3.99 through 3.104
3.10Stoichiometric Calculations: Amounts
of Reactants and Products
IBLG: See questions from “Reaction
Stoichiometry”
As we have seen in previous sections of this chapter, the coefficients in chemical equations represent numbers of molecules, not masses of molecules. However, when a reaction is to be run in a laboratory or chemical plant, the amounts of substances needed
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3.10
Stoichiometric Calculations: Amounts of Reactants and Products
109
Chemical connections
High Mountains—Low Octane
available, then less fuel is required to
achieve this optimal ratio. In turn, the
lower volumes of oxygen and fuel
result in a lower pressure in the
cylinder. Because high pressure tends
to promote knocking, the lower
pressure within engine cylinders at
higher elevations promotes a more
controlled ­combustion of the air–fuel
mixture, and therefore octane
­requirements are lower. While
consumers in the Rocky Mountain
states can purchase three grades of
gasoline, the octane ratings of these
fuel blends are different from those in
the rest of the United States. In
Denver, Colorado, regular gasoline is
85 octane, midgrade is 87 octane, and
premium is 91 octane—2 points lower
than gasoline sold in most of the rest
of the country.
The next time you visit a gas station,
take a moment to note the octane
rating that accompanies the grade of
gasoline that you are purchasing. The
gasoline is priced according to its
octane rating—a measure of the fuel’s
antiknock properties. In a conventional
internal combustion engine, gasoline
vapors and air are drawn into the
combustion cylinder on the downward
stroke of the piston. This air–fuel
mixture is compressed on the upward
piston stroke (compression stroke),
and a spark from the sparkplug ignites
the mix. The rhythmic combustion of
the air–fuel mix occurring sequentially
in several cylinders furnishes the power
to propel the vehicle down the road.
Excessive heat and pressure (or
poor-quality fuel) within the cylinder
may cause the premature combustion
of the mixture—commonly known as
engine “knock” or “ping.” Over time,
this engine knock can damage the
engine, resulting in inefficient
performance and costly repairs.
A consumer typically is faced with
three choices of gasoline, with octane
ratings of 87 (regular), 89 (midgrade),
and 93 (premium). But if you happen
to travel or live in the higher elevations of the Rocky Mountain states,
you might be surprised to find
different octane ratings at the
gasoline pumps. The reason for this
provides a lesson in stoichiometry. At higher elevations the air is less
dense—the volume of oxygen per unit
volume of air is smaller. Most engines
are designed to achieve a 14;1
oxygen-to-fuel ratio in the cylinder
prior to combustion. If less oxygen is
PowerLecture: Oxygen, Hydrogen,
Soap Bubbles, and Balloons
cannot be determined by counting molecules directly. Counting is always done by weighing. In this section we will see how chemical equations can be used to determine the
masses of reacting chemicals.
To develop the principles for dealing with the stoichiometry of reactions, we will
consider the reaction of propane with oxygen to produce carbon dioxide and water. We
will consider the question: “What mass of oxygen will react with 96.1 g of propane?”
In doing stoichiometry, the first thing we must do is write the balanced chemical equation for the reaction. In this case the balanced equation is
Before doing any calculations involving a
chemical reaction, be sure the equation
for the reaction is balanced.
C3H8 1g2 1 5O2 1g2 h 3CO2 1g2 1 4H2O 1g2
which can be visualized as
+
+
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110
Chapter 3
Stoichiometry
This equation means that 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles
of CO2 and 4 moles of H2O. To use this equation to find the masses of reactants and
products, we must be able to convert between masses and moles of substances. Thus
we must first ask: “How many moles of propane are present in 96.1 g of propane?” The
molar mass of propane to three significant figures is 44.1 (that is, 3 3 12.01 1 8 3
1.008). The moles of propane can be calculated as follows:
96.1 g C3H8 3
1 mol C3H8
5 2.18 mol C3H8
44.1 g C3H8
Next we must take into account the fact that each mole of propane reacts with
5 moles of oxygen. The best way to do this is to use the balanced equation to construct
a mole ratio. In this case we want to convert from moles of propane to moles of oxygen. From the balanced equation, we see that 5 moles of O2 are required for each mole
of C3H8, so the appropriate ratio is
5 mol O2
1 mol C3H8
Multiplying the number of moles of C3H8 by this factor gives the number of moles
of O2 required:
5 mol O2
5 10.9 mol O2
1 mol C3H8
2.18 mol C3H8 3
Notice that the mole ratio is set up so that the moles of C3H8 cancel out, and the units
that result are moles of O2.
Since the original question asked for the mass of oxygen needed to react with
96.1 g of propane, the 10.9 moles of O2 must be converted to grams. Since the molar
mass of O2 is 32.0 g/mol,
10.9 mol O2 3
32.0 g O2
5 349 g O2
1 mol O2
Therefore, 349 g of oxygen are required to burn 96.1 g of propane.
This example can be extended by asking: “What mass of carbon dioxide is produced when 96.1 g of propane are combusted with oxygen?” In this case we must
convert between moles of propane and moles of carbon dioxide. This can be accomplished by looking at the balanced equation, which shows that 3 moles of CO2 are
produced for each mole of C3H8 reacted. The mole ratio needed to convert from moles
of propane to moles of carbon dioxide is
3 mol CO2
1 mol C3H8
The conversion is
2.18 mol C3H8 3
3 mol CO2
5 6.54 mol CO2
1 mol C3H8
Then, using the molar mass of CO2 (44.0 g/mol), we calculate the mass of CO2
­produced:
6.54 mol CO2 3
44.0 g CO2
5 288 g CO2
1 mol CO2
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3.10
Experiment 13: Stoichiometry 2:
Spectrophotometric Determination
of the Stoichiometry of an Iron(III)
Phenol Reaction
Stoichiometric Calculations: Amounts of Reactants and Products
111
We will now summarize the sequence of steps needed to carry out stoichiometric
­calculations.
96.1 g C3H8
1 mol C3H8
44.1 g C3H8
2.18 mol C3H8
44.0 g CO2
1 mol CO2
3 mol CO2
1 mol C3H8
6.54 mol CO2
288 g CO2
Critical Thinking
Your lab partner has made the observation that you always take the mass of chemicals in lab, but then you use mole ratios to balance the equation. “Why not use the
masses in the equation?” your partner asks. What if your lab partner decided to
balance equations by using masses as coefficients? Is this even possible? Why or
why not?
Problem-Solving Strategy
Calculating Masses of Reactants and Products in Chemical Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product to moles of that
substance.
3. Use the balanced equation to set up the appropriate mole ratios.
4. Use the appropriate mole ratios to calculate the number of moles of the
desired reactant or product.
5. Convert from moles back to grams if required by the problem.
Balanced chemical
equation
Find appropriate
mole ratio
Moles desired substance
Moles known substance
Mass of
known
substance
Convert
to moles
Moles of
known
substance
Use mole ratio
to convert
Moles of
desired
substance
Convert
to grams
Mass of
desired
substance
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112
Chapter 3
Stoichiometry
Interactive
Example 3.15
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in OWL.
Chemical Stoichiometry I
Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide
from the living environment by forming solid lithium carbonate and liquid water.
What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium
hydroxide?
Solution
Where are we going?
To find the mass of CO2 absorbed by 1.00 kg LiOH
What do we know?
❯
❯
Chemical reaction
LiOH 1s2 1 CO2 1g2 h Li2CO3 1s2 1 H2O 1l2
1.00 kg LiOH
What information do we need to find the mass of CO2?
❯
Balanced equation for the reaction
How do we get there?
1. What is the balanced equation?
2LiOH 1s2 1 CO2 1g2 S Li2CO3 1s2 1 H2O 1l2
2. What are the moles of LiOH?
To find the moles of LiOH, we need to know the molar mass.
What is the molar mass for LiOH?
6.941 1 16.00 1 1.008 5 23.95 g /mol
Now we use the molar mass to find the moles of LiOH:
1.00 kg LiOH 3
1000 g LiOH
1 mol LiOH
3
5 41.8 mol LiOH
1 kg LiOH
23.95 g LiOH
3. What is the mole ratio between CO2 and LiOH in the balanced equation?
1 mol CO2
2 mol LiOH
4. What are the moles of CO2?
41.8 mol LiOH 3
1 mol CO2
5 20.9 mol CO2
2 mol LiOH
5. What is the mass of CO2 formed from 1.00 kg LiOH?
20.9 mol CO2 3
❯
44.0 g CO2
5 9.20 3 102 g CO2
1 mol CO2
Thus 920. g of CO2(g) will be absorbed by 1.00 kg of LiOH(s).
See Exercises 3.105 and 3.106
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3.10
Interactive
Example 3.16
Stoichiometric Calculations: Amounts of Reactants and Products
113
Chemical Stoichiometry II
Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric
acid secreted by the stomach:
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NaHCO3 1s2 1 HCl 1aq2 h NaCl 1aq2 1 H2O 1l2 1 CO2 1aq2
Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also
used as an antacid:
Mg 1OH2 2 1s2 1 2HCl 1aq2 h 2H2O 1l2 1 MgCl2 1aq2
Which is the more effective antacid per gram, NaHCO3 or Mg(OH)2?
Solution
Where are we going?
To compare the acid neutralizing power of NaHCO3 and Mg(OH)2 per gram
What do we know?
❯ Balanced equations for the reactions
❯
❯
1.00 g NaHCO3
1.00 g Mg(OH)2
How do we get there?
For NaHCO3
1. What is the balanced equation?
NaHCO3 1s2 1 HCl 1aq2 h NaCl 1aq2 1 H2O 1l2 1 CO2 1aq2
2. What are the moles of NaHCO3 in 1.00 g?
To find the moles of NaHCO3, we need to know the molar mass (84.01 g/mol).
1.00 g NaHCO3 3
1 mol NaHCO3
5 1.19 3 1022 mol NaHCO3
84.01 g NaHCO3
3. What is the mole ratio between HCl and NaHCO3 in the balanced equation?
1 mol HCl
1 mol NaHCO3
4. What are the moles of HCl?
1.19 3 1022 mol NaHCO3 3
1 mol HCl
5 1.19 3 1022 mol HCl
1 mol NaHCO3
Thus 1.00 g of NaHCO3 will neutralize 1.19 3 1022 mole of HCl.
For Mg(OH)2
Charles D. Winters
1. What is the balanced equation?
Mg 1OH2 2 1s2 1 2HCl 1aq2 h 2H2O 1l2 1 MgCl2 1aq2
2. What are the moles of Mg(OH)2 in 1.00 g?
To find the moles of Mg(OH)2, we need to know the molar mass (58.32 g/mol).
Milk of magnesia contains a suspension
of Mg(OH)2(s).
1.00 g Mg 1OH2 2 3
1 mol Mg 1OH2 2
5 1.71 3 1022 mol Mg 1OH2 2
58.32 g Mg 1OH2 2
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114
Chapter 3
Stoichiometry
3. What is the mole ratio between HCl and Mg(OH)2 in the balanced equation?
4. What are the moles of HCl?
2 mol HCl
1 mol Mg 1OH2 2
1.71 3 1022 mol Mg 1OH2 2 3
❯
2 mol HCl
5 3.42 3 1022 mol HCl
1 mol Mg 1OH2 2
Thus 1.00 g of Mg(OH)2 will neutralize 3.42 3 1022 mole of HCl.
Since 1.00 g NaHCO3 neutralizes 1.19 3 1022 mole of HCl and 1.00 g
Mg(OH)2 neutralizes 3.42 3 1022 mole of HCl, Mg(OH)2 is the more effective
antacid.
See Exercises 3.107 and 3.108
3.11 The Concept of Limiting Reactant
IBLG: See questions from “Calculations
Involving a Limited Reactant”
Suppose you have a part-time job in a sandwich shop. One very popular sandwich is
always made as follows:
2 slices bread 1 3 slices meat 1 1 slice cheese h sandwich
Assume that you come to work one day and find the following quantities of ingredients:
8 slices bread
9 slices meat
5 slices cheese
How many sandwiches can you make? What will be left over?
To solve this problem, let’s see how many sandwiches we can make with each
­component:
Bread:
8 slices bread 3
1 sandwich
5 4 sandwiches
2 slices bread
Meat:
9 slices meat 3
1 sandwich
5 3 sandwiches
3 slices meat
Cheese:
5 slices cheese 3
1 sandwich
5 5 sandwiches
1 slice cheese
How many sandwiches can you make? The answer is three. When you run out of meat,
you must stop making sandwiches. The meat is the limiting ingredient (Fig. 3.9).
What do you have left over? Making three sandwiches requires six pieces of
bread. You started with eight slices, so you have two slices of bread left. You also
used three pieces of cheese for the three sandwiches, so you have two pieces of
cheese left.
In this example, the ingredient present in the largest number (the meat) was actually
the component that limited the number of sandwiches you could make. This situation
arose because each sandwich required three slices of meat—more than the quantity
required of any other ingredient.
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3.11
Figure 3.9 | Making sandwiches.
Bread
Meat
The Concept of Limiting Reactant
Cheese
115
Sandwich
When molecules react with each other to form products, considerations very similar
to those involved in making sandwiches arise. We can illustrate these ideas with the
reaction of N2(g) and H2(g) to form NH3(g):
N2 1g2 1 3H2 1g2 h 2NH3 1g2
Consider the following container of N2(g) and H2(g):
H2
N2
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116
Chapter 3
Stoichiometry
What will this container look like if the reaction between N2 and H2 proceeds to
completion? To answer this question, you need to remember that each N2 requires
3 H2 molecules to form 2 NH3. To make things clear, we will circle groups of
reactants:
H2
N2
NH3
Before the reaction
After the reaction
In this case, the mixture of N2 and H2 contained just the number of molecules
needed to form NH3 with nothing left over. That is, the ratio of the number of H2 molecules to N2 molecules was
15H2
3H2
5
5N2
1N2
This ratio exactly matches the numbers in the balanced equation
3H2 1g2 1 N2 1g2 h 2NH3 1g2
This type of mixture is called a stoichiometric mixture—one that contains the
relative amounts of reactants that match the numbers in the balanced equation. In this
case all reactants will be consumed to form products.
Now consider another container of N2(g) and H2(g):
H2
N2
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3.11
The Concept of Limiting Reactant
117
What will the container look like if the reaction between N2(g) and H2(g) proceeds
to completion? Remember that each N2 requires 3 H2. Circling groups of reactants,
we have
H2
N2
NH3
Before the reaction
After the reaction
In this case, the hydrogen (H2) is limiting. That is, the H2 molecules are used up before
all the N2 molecules are consumed. In this situation the amount of hydrogen limits the
amount of product (ammonia) that can form—hydrogen is the limiting reactant. Some
N2 molecules are left over in this case because the reaction runs out of H2 molecules
first. To determine how much product can be formed from a given mixture of reactants,
we have to look for the reactant that is limiting—the one that runs out first and thus
limits the amount of product that can form. In some cases, the mixture of reactants
might be stoichiometric—that is, all reactants run out at the same time. In general,
however, you cannot assume that a given mixture of reactants is a stoichiometric mixture, so you must determine whether one of the reactants is limiting. The reactant that
runs out first and thus limits the amounts of products that can form is called the limiting reactant.
To this point we have considered examples where the numbers of reactant molecules could be counted. In “real life” you can’t count the molecules directly—you
can’t see them, and even if you could, there would be far too many to count. Instead,
you must count by weighing. We must therefore explore how to find the limiting reactant, given the masses of the reactants.
A. Determination of Limiting Reactant Using
Reactant Quantities
There are two ways to determine the limiting reactant in a chemical reaction. One involves comparing the moles of reactants to see which runs out first. We will consider
this approach here.
In the laboratory or chemical plant, we work with much larger quantities than the
few molecules of the preceding example. Therefore, we must learn to deal with limiting reactants using moles. The ideas are exactly the same, except that we are using
moles of molecules instead of individual molecules. For example, suppose 25.0 kg of
nitrogen and 5.00 kg of hydrogen are mixed and reacted to form ammonia. How do we
calculate the mass of ammonia produced when this reaction is run to completion (until
one of the reactants is completely consumed)?
As in the preceding example, we must use the balanced equation
N2 1g2 1 3H2 1g2 h 2NH3 1g2
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118
Chapter 3
Stoichiometry
to determine whether nitrogen or hydrogen is the limiting reactant and then to determine the amount of ammonia that is formed. We first calculate the moles of reactants
present:
25.0 kg N2 3
1000 g N2
1 mol N2
3
5 8.93 3 102 mol N2
1 kg N2
28.0 g N2
5.00 kg H2 3
1000 g H2
1 mol H2
3
5 2.48 3 103 mol H2
1 kg H2
2.016 g H2
Since 1 mole of N2 reacts with 3 moles of H2, the number of moles of H2 that will react
exactly with 8.93 3 102 moles of N2 is
8.93 3 102 mol N2 3
Always determine which reactant is
limiting.
3 mol H2
5 2.68 3 103 mol H2
1 mol N2
Thus 8.93 3 102 moles of N2 requires 2.68 3 103 moles of H2 to react completely.
However, in this case, only 2.48 3 103 moles of H2 are present. This means that the
hydrogen will be consumed before the nitrogen. Thus hydrogen is the limiting reactant
in this particular situation, and we must use the amount of hydrogen to compute the
quantity of ammonia formed:
2.48 3 103 mol H2 3
2 mol NH3
5 1.65 3 103 mol NH3
3 mol H2
Converting moles to kilograms gives
1.65 3 103 mol NH3 3
17.0 g NH3
5 2.80 3 104 g NH3 5 28.0 kg NH3
1 mol NH3
Note that to determine the limiting reactant, we could have started instead with the
given amount of hydrogen and calculated the moles of nitrogen required:
2.48 3 103 mol H2 3
1 mol N2
5 8.27 3 102 mol N2
3 mol H2
Thus 2.48 3 103 moles of H2 requires 8.27 3 102 moles of N2. Since 8.93 3 102 moles
of N2 are actually present, the nitrogen is in excess. The hydrogen will run out first, and
thus again we find that hydrogen limits the amount of ammonia formed.
A related but simpler way to determine which reactant is limiting is to compare the
mole ratio of the substances required by the balanced equation with the mole ratio of
reactants actually present. For example, in this case the mole ratio of H2 to N2 required
by the balanced equation is
3 mol H2
1 mol N2
That is,
mol H2
3
1required2 5 5 3
mol N2
1
In this experiment we have 2.48 3 103 moles of H2 and 8.93 3 102 moles of N2. Thus
the ratio
mol H2
2.48 3 103
1actual2 5
5 2.78
mol N2
8.93 3 102
Since 2.78 is less than 3, the actual mole ratio of H2 to N2 is too small, and H2 must be
limiting. If the actual H2 to N2 mole ratio had been greater than 3, then the H2 would
have been in excess and the N2 would be limiting.
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3.11
The Concept of Limiting Reactant
119
B. Determination of Limiting Reactant Using
Quantities of Products Formed
A second method for determining which reactant in a chemical reaction is limiting is
to consider the amounts of products that can be formed by completely consuming each
reactant. The reactant that produces the smallest amount of product must run out first
and thus be limiting. To see how this works, consider again the reaction of 25.0 kg
(8.93 3 102 moles) of nitrogen with 5.00 kg (2.48 3 103 moles) of hydrogen.
We will now use these amounts of reactants to determine how much NH3 would
form. Since 1 mole of N2 forms 2 moles of NH3, the amount of NH3 that would be
formed if all of the N2 was used up is calculated as follows:
8.93 3 102 mol N2 3
2 mol NH3
5 1.79 3 103 mol NH3
1 mol N2
Next we will calculate how much NH3 would be formed if the H2 was completely
used up:
2.48 3 103 mol H2 3
2 mol NH3
5 1.65 3 103 mol NH3
3 mol H2
Because a smaller amount of NH3 is produced from the H2 than from the N2, the
amount of H2 must be limiting.
Thus because the H2 is the limiting reactant, the amount of NH3 that can form is
1.65 3 103 moles. Converting moles to kilograms gives:
1.65 3 103 mol NH3 3
Interactive
Example 3.17
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.com to try this Interactive Example
in OWL.
Experiment 12: Stoichiometry 1:
Limiting ­Reactant
17 g NH3
5 2.80 3 104 g NH3 5 28.0 kg NH3
1 mol NH3
Stoichiometry: Limiting Reactant
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide
at high temperatures. The other products of the reaction are solid copper and water
vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the
limiting reactant? How many grams of N2 will be formed?
Solution
Where are we going?
To find the limiting reactant
To find the mass of N2 produced
What do we know?
❯ The chemical reaction
❯
❯
NH3 1g2 1 CuO 1s2 h N2 1g2 1 Cu 1s2 1 H2O 1g2
18.1 g NH3
90.4 g CuO
What information do we need?
❯ Balanced equation for the reaction
❯ Moles of NH3
❯ Moles of CuO
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120
Chapter 3
Stoichiometry
How do we get there?
To find the limiting reactant
What is the balanced equation?
2NH3 1g2 1 3CuO 1s2 h N2 1g2 1 3Cu 1s2 1 3H2O 1g2
What are the moles of NH3 and CuO?
To find the moles, we need to know the molar masses.
NH3 17.03 g/mol
CuO 79.55 g/mol
1 mol NH3
5 1.06 mol NH3
17.03 g NH3
1 mol CuO
90.4 g CuO 3
5 1.14 mol CuO
79.55 g CuO
18.1 g NH3 3
A. First we will determine the limiting reactant by comparing the moles of reactants
to see which one is consumed first.
What is the mole ratio between NH3 and CuO in the balanced equation?
3 mol CuO
2 mol NH3
How many moles of CuO are required to react with 1.06 moles of NH3?
1.06 mol NH3 3
❯
3 mol CuO
5 1.59 mol CuO
2 mol NH3
Thus 1.59 moles of CuO are required to react with 1.06 moles of NH3. Since
only 1.14 moles of CuO are actually present, the amount of CuO is limiting;
CuO will run out before NH3 does. We can verify this conclusion by comparing
the mole ratio of CuO and NH3 required by the balanced equation:
mol CuO
3
1required2 5 5 1.5
mol NH3
2
with the mole ratio actually present:
❯
mol CuO
1.14
1actual2 5
5 1.08
mol NH3
1.06
Since the actual ratio is too small (less than 1.5), CuO is the limiting reactant.
B. Alternatively we can determine the limiting reactant by computing the moles of
N2 that would be formed by complete consumption of NH3 and CuO:
1 mol N2
5 0.530 mol N2
2 mol NH3
1 mol N2
1.14 mol CuO 3
5 0.380 mol N2
3 mol CuO
1.06 mol NH3 3
As before, we see that the CuO is limiting since it produces the smaller amount of N2.
To find the mass of N2 produced
What are the moles of N2 formed?
Because CuO is the limiting reactant, we must use the amount of CuO to calculate the
amount of N2 formed.
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3.11
The Concept of Limiting Reactant
121
What is the mole ratio between N2 and CuO in the balanced equation?
1 mol N2
3 mol CuO
What are the moles of N2?
1.14 mol CuO 3
1 mol N2
5 0.380 mol N2
3 mol CuO
What mass of N2 is produced?
Using the molar mass of N2 (28.02 g/mol), we can calculate the mass of N2 produced:
j
0.380 mol N2 3
28.02 g N2
5 10.6 g N2
1 mol N2
See Exercises 3.117 through 3.122
The amount of a product formed when the limiting reactant is completely consumed is called the theoretical yield of that product. In Example 3.17, 10.6 g of nitrogen represent the theoretical yield. This is the maximum amount of nitrogen that can
be produced from the quantities of reactants used. Actually, the amount of product
predicted by the theoretical yield is seldom obtained because of side reactions (other
reactions that involve one or more of the reactants or products) and other complications. The actual yield of product is often given as a percentage of the theoretical yield.
This is called the percent yield:
Percent yield is important as an indicator
of the efficiency of a particular laboratory
or industrial reaction.
Actual yield
3 100% 5 percent yield
Theoretical yield
For example, if the reaction considered in Example 3.17 actually gave 6.63 g of
n­ itrogen instead of the predicted 10.6 g, the percent yield of nitrogen would be
6.63 g N2
3 100% 5 62.5%
10.6 g N2
Interactive
Example 3.18
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.com to try this Interactive Example
in OWL.
Calculating Percent Yield
Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a
fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combining gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is
­reacted with 8.60 kg H2(g). Calculate the theoretical yield of methanol. If 3.57 3 104 g
CH3OH is actually produced, what is the percent yield of methanol?
Solution
Methanol
Where are we going?
To calculate the theoretical yield of methanol
To calculate the percent yield of methanol
What do we know?
❯ The chemical reaction
❯
❯
❯
H2 1g2 1 CO 1g2 h CH3OH 1l2
68.5 kg CO(g)
8.60 kg H2(g)
3.57 3 104 g CH3OH is produced
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122
Chapter 3
Stoichiometry
What information do we need?
❯ Balanced equation for the reaction
❯ Moles of H2
❯ Moles of CO
❯ Which reactant is limiting
❯ Amount of CH3OH produced
How do we get there?
To find the limiting reactant
What is the balanced equation?
2H2 1g2 1 CO 1g2 h CH3OH 1l2
What are the moles of H2 and CO?
To find the moles, we need to know the molar masses.
H2 2.016 g/mol
CO 28.02 g/mol
1000 g CO
1 mol CO
3
5 2.44 3 103 mol CO
1 kg CO
28.02 g CO
1000 g H2
1 mol H2
8.60 kg H2 3
3
5 4.27 3 103 mol H2
1 kg H2
2.016 g H2
68.5 kg CO 3
A. Determination of Limiting Reactant Using Reactant Quantities
What is the mole ratio between H2 and CO in the balanced equation?
2 mol H2
1 mol CO
How does the actual mole ratio compare to the stoichiometric ratio?
To determine which reactant is limiting, we compare the mole ratio of H2 and CO
required by the balanced equation
mol H2
2
1required2 5 5 2
mol CO
1
with the actual mole ratio
❯
mol H2
4.27 3 103
1actual2 5
5 1.75
mol CO
2.44 3 103
Since the actual mole ratio of H2 to CO is smaller than the required ratio, H2 is
limiting.
B. Determination of Limiting Reactant Using Quantities of Products Formed
We can also determine the limiting reactant by calculating the amounts of CH3OH
formed by complete consumption of CO(g) and H2(g):
1 mol CH3OH
5 2.44 3 103 mol CH3OH
1 mol CO
1 mol CH3OH
4.27 3 103 mol H2 3
5 2.14 3 103 mol CH3OH
2 mol H2
2.44 3 103 mol CO 3
Since complete consumption of the H2 produces the smaller amount of CH3OH, the H2
is the limiting reactant as we determined above.
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3.11
The Concept of Limiting Reactant
123
To calculate the theoretical yield of methanol
What are the moles of CH3OH formed?
We must use the amount of H2 and the mole ratio between H2 and CH3OH to determine
the maximum amount of methanol that can be produced:
4.27 3 103 mol H2 3
1 mol CH3OH
5 2.14 3 103 mol CH3OH
2 mol H2
What is the theoretical yield of CH3OH in grams?
2.14 3 103 mol CH3OH 3
32.04 g CH3OH
5 6.86 3 104 g CH3OH
1 mol CH3OH
Thus, from the amount of reactants given, the maximum amount of CH3OH
that can be formed is 6.86 3 104 g. This is the theoretical yield.
❯
What is the percent yield of CH3OH?
Actual yield 1grams2
3.57 3 104 g CH3OH
3 100 5
3 100% 5 52.0%
Theoretical yield 1grams2
6.86 3 104 g CH3OH
❯
See Exercises 3.123 and 3.124
Problem-Solving Strategy
Solving a Stoichiometry Problem Involving Masses of Reactants and Products
1.
2.
3.
4.
Write and balance the equation for the reaction.
Convert the known masses of substances to moles.
Determine which reactant is limiting.
Using the amount of the limiting reactant and the appropriate mole ratios,
compute the number of moles of the desired product.
5. Convert from moles to grams, using the molar mass.
This process is summarized in the diagram below:
Balanced chemical
equation
Find appropriate
mole ratio
Masses of
known
substances
Convert
to moles
Moles of
known
substances
Moles desired substance
Moles limiting reactant
Find
limiting
reactant
Moles
limiting
reactant
Use mole ratio
to convert
Moles of
desired
product
Convert
to grams
Mass of
desired
product
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124
Chapter 3
Stoichiometry
For review
Key terms
Stoichiometry
chemical stoichiometry
❯Deals
Section 3.2
❯
mass spectrometer
average atomic mass
❯
Section 3.3
Mole
mole
Avogadro’s number
❯
Section 3.4
❯
molar mass
Section 3.5
conceptual problem solving
Section 3.6
❯
with the amounts of substances consumed and/or produced in a chemical ­reaction.
We count atoms by measuring the mass of the sample.
To relate mass and the number of atoms, the average atomic mass is required.
The amount of carbon atoms in exactly 12 g of pure 12C
6.022 3 1023 units of a substance
The mass of 1 mole of an element 5 the atomic mass in grams
Molar mass
❯
❯
mass percent
Mass (g) of 1 mole of a compound or element
Obtained for a compound by finding the sum of the average masses of its constituent atoms
Section 3.7
Percent composition
empirical formula
molecular formula
❯
Section 3.8
❯
chemical equation
reactants
products
balancing a chemical equation
Empirical formula
Section 3.10
❯
mole ratio
Section 3.11
stoichiometric mixture
limiting reactant
theoretical yield
percent yield
❯
The mass percent of each element in a compound
mass of element in 1 mole of substance
3 100%
Mass percent 5
mass of 1 mole of substance
The simplest whole-number ratio of the various types of atoms in a compound
Can be obtained from the mass percent of elements in a compound
Molecular formula
❯
❯
For molecular substances:
❯ The formula of the constituent molecules
❯ Always an integer multiple of the empirical formula
For ionic substances:
❯ The same as the empirical formula
Chemical reactions
❯
❯
❯
Reactants are turned into products.
Atoms are neither created nor destroyed.
All of the atoms present in the reactants must also be present in the products.
Characteristics of a chemical equation
❯
❯
❯
Represents a chemical reaction
Reactants on the left side of the arrow, products on the right side
When balanced, gives the relative numbers of reactant and product molecules or ions
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For Review
125
Stoichiometry calculations
❯
❯
Amounts of reactants consumed and products formed can be determined from the balanced
chemical equation.
The limiting reactant is the one consumed first, thus limiting the amount of product that can
form.
Yield
❯
❯
❯
Review questions
The theoretical yield is the maximum amount that can be produced from a given amount of
the limiting reactant.
The actual yield, the amount of product actually obtained, is always less than the theoretical
yield.
actual yield 1g2
Percent yield 5
3 100%
theoretical yield 1g2
Answers to the Review Questions can be found on the Student website (accessible from www.cengagebrain.com).
1. Explain the concept of “counting by weighing” using
marbles as your example.
2. Atomic masses are relative masses. What does this
mean?
3. The atomic mass of boron (B) is given in the periodic
table as 10.81, yet no single atom of boron has a mass
of 10.81 u. Explain.
4. What three conversion factors and in what order would
you use them to convert the mass of a compound into
atoms of a particular element in that compound—for
example, from 1.00 g aspirin (C9H8O4) to number of
hydrogen atoms in the 1.00-g sample?
5. Fig. 3.5 illustrates a schematic diagram of a combustion
device used to analyze organic compounds. Given that a
certain amount of a compound containing carbon,
hydrogen, and oxygen is combusted in this device,
explain how the data relating to the mass of CO2
produced and the mass of H2O produced can be
manipulated to determine the empirical formula.
6. What is the difference between the empirical and
molecular formulas of a compound? Can they ever be
the same? Explain.
7. Consider the hypothetical reaction between A2 and AB
pictured below.
What is the balanced equation? If 2.50 moles of A2 are
reacted with excess AB, what amount (moles) of product
will form? If the mass of AB is 30.0 u and the mass of
A2 are 40.0 u, what is the mass of the product? If 15.0 g
of AB is reacted, what mass of A2 is required to react
with all of the AB, and what mass of product is formed?
8. What is a limiting reactant problem? Explain the
method you are going to use to solve limiting reactant
problems.
9. Consider the following mixture of SO2(g) and O2(g).
O2
SO2
?
If SO2(g) and O2(g) react to form SO3(g), draw a
representation of the product mixture assuming the
reaction goes to completion. What is the limiting
reactant in the reaction? If 96.0 g of SO2 react with
32.0 g O2, what mass of product will form?
10. Why is the actual yield of a reaction often less than the
theoretical yield?
A2
AB
A2B
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126
Chapter 3
Stoichiometry
A discussion of the Active Learning ­Questions can be found online in the ­Instructor’s Resource
Guide and on PowerLecture. The questions allow students to explore their understanding of
concepts through discussion and peer teaching. The real value of these questions is the learning
that occurs while students talk to each other about chemical concepts.
Active Learning Questions
These questions are designed to be used by groups of students in
class.
1. The following are actual student responses to the question:
Why is it necessary to balance chemical equations?
a. The chemicals will not react until you have added the correct mole ratios.
b. The correct products will not be formed unless the right
amount of reactants have been added.
c. A certain number of products cannot be formed without a
certain number of reactants.
d. The balanced equation tells you how much reactant you
need and allows you to predict how much product you’ll
make.
e. A mole-to-mole ratio must be established for the reaction
to occur as written.
Justify the best choice, and for choices you did not pick, explain what is wrong with them.
2. What information do we get from a chemical formula? From
a chemical equation?
3. You are making cookies and are missing a key ingredient—
eggs. You have most of the other ingredients needed to make
the cookies, except you have only 1.33 cups of butter and no
eggs. You note that the recipe calls for two cups of butter and
three eggs (plus the other ingredients) to make six dozen cookies. You call a friend and have him bring you some eggs.
a. What number of eggs do you need?
b. If you use all the butter (and get enough eggs), what number of cookies will you make?
Unfortunately, your friend hangs up before you tell him how
many eggs you need. When he arrives, he has a surprise for
you—to save time, he has broken them all in a bowl for you.
You ask him how many he brought, and he replies, “I can’t remember.” You weigh the eggs and find that they weigh 62.1 g.
Assuming that an average egg weighs 34.21 g,
a. What quantity of butter is needed to react with all the eggs?
b. What number of cookies can you make?
c. Which will you have left over, eggs or butter?
d. What quantity is left over?
4. Nitrogen gas (N2) and hydrogen gas (H2) react to form ammonia gas (NH3).
Consider the mixture of N2 (
) and H2 (
) in a
closed container as illustrated below:
Assuming the reaction goes to completion, draw a representation of the product mixture. Explain how you arrived at this
representation.
5. For the preceding question, which of the following equations
best represents the reaction?
a. 6N2 1 6H2 h 4NH3 1 4N2
b. N2 1 H2 h NH3
c. N 1 3H h NH3
d. N2 1 3H2 h 2NH3
e. 2N2 1 6H2 h 4NH3
Justify your choice, and for choices you did not pick, explain
what is wrong with them.
6. You know that chemical A reacts with chemical B. You react
10.0 g A with 10.0 g B. What information do you need to determine the amount of product that will be produced? Explain.
7. A new grill has a mass of 30.0 kg. You put 3.0 kg of charcoal
in the grill. You burn all the charcoal and the grill has a mass
of 30.0 kg. What is the mass of the gases given off? Explain.
8. Consider an iron bar on a balance as shown.
75.0 g
As the iron bar rusts, which of the following is true? Explain
your answer.
a. The balance will read less than 75.0 g.
b. The balance will read 75.0 g.
c. The balance will read greater than 75.0 g.
d. The balance will read greater than 75.0 g, but if the bar is
removed, the rust is scraped off, and the bar replaced, the
balance will read 75.0 g.
9. You may have noticed that water sometimes drips from the
exhaust of a car as it is running. Is this evidence that there is at
least a small amount of water originally present in the gasoline? Explain.
Questions 10 and 11 deal with the following situation: You react
chemical A with chemical B to make one product. It takes 100 g of
A to react completely with 20 g of B.
10. What is the mass of the product?
a. less than 10 g
b. between 20 and 100 g
c. between 100 and 120 g
d. exactly 120 g
e. more than 120 g
11. What is true about the chemical properties of the product?
a. The properties are more like chemical A.
b. The properties are more like chemical B.
c. The properties are an average of those of chemical A and
chemical B.
d. The properties are not necessarily like either chemical A
or chemical B.
e. The properties are more like chemical A or more like
chemical B, but more information is needed.
Justify your choice, and for choices you did not pick, explain
what is wrong with them.
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For Review
12. Is there a difference between a homogeneous mixture of hydrogen and oxygen in a 2:1 mole ratio and a sample of water
vapor? Explain.
13. Chlorine exists mainly as two isotopes, 37Cl and 35Cl. Which is
more abundant? How do you know?
14. The average mass of a carbon atom is 12.011. Assuming you
could pick up one carbon atom, estimate the chance that you
would randomly get one with a mass of 12.011. Support your
answer.
15. Can the subscripts in a chemical formula be fractions? Explain.
Can the coefficients in a balanced chemical equation be fractions? Explain. Changing the subscripts of chemicals can balance the equations mathematically. Why is this unacceptable?
16. Consider the equation 2A 1 B h A2B. If you mix 1.0 mole
of A with 1.0 mole of B, what amount (moles) of A2B can be
­produced?
17. According to the law of conservation of mass, mass cannot be
gained or destroyed in a chemical reaction. Why can’t you
simply add the masses of two reactants to determine the total
mass of product?
18. Which of the following pairs of compounds have the same
­empirical formula?
a. acetylene, C2H2, and benzene, C6H6
b. ethane, C2H6, and butane, C4H10
c. nitrogen dioxide, NO2, and dinitrogen tetroxide, N2O4
d. diphenyl ether, C12H10O, and phenol, C6H5OH
19. Atoms of three different elements are represented by O, h,
and D. Which compound is left over when three molecules of
OD and three molecules of hhD react to form OhD and
ODD?
20. In chemistry, what is meant by the term “mole”? What is the
importance of the mole concept?
21. Which (if any) of the following is (are) true regarding the limiting ­reactant in a chemical reaction?
a. The limiting reactant has the lowest coefficient in a balanced equation.
b. The limiting reactant is the reactant for which you have
the fewest number of moles.
c. The limiting reactant has the lowest ratio of moles
available/coefficient in the balanced equation.
d. The limiting reactant has the lowest ratio of coefficient in
the balanced equation/moles available.
Justify your choice. For those you did not choose, explain why
they are incorrect.
22. Consider the equation 3A 1 B S C 1 D. You react 4 moles
of A with 2 moles of B. Which of the following is true?
a. The limiting reactant is the one with the higher molar mass.
b. A is the limiting reactant because you need 6 moles of A
and have 4 moles.
c. B is the limiting reactant because you have fewer moles
of B than A.
d. B is the limiting reactant because three A molecules react
with each B molecule.
e. Neither reactant is limiting.
Justify your choice. For those you did not choose, explain why
they are incorrect.
127
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of the book and a
solution appears in the Solutions Guide, as found on PowerLecture.
Questions
23. Reference Section 3.2 to find the atomic masses of 12C and
13
C, the relative abundance of 12C and 13C in natural carbon,
and the average mass (in u) of a carbon atom. If you had a
sample of natural carbon containing exactly 10,000 atoms, determine the number of 12C and 13C atoms present. What would
be the average mass (in u) and the total mass (in u) of the carbon atoms in this 10,000-atom sample? If you had a sample of
natural carbon containing 6.0221 3 1023 atoms, determine the
number of 12C and 13C atoms present. What would be the average mass (in u) and the total mass (in u) of this 6.0221 3 1023
atom sample? Given that 1 g 5 6.0221 3 1023 u, what is the
total mass of 1 mole of natural carbon in units of grams?
24. Avogadro’s number, molar mass, and the chemical formula of
a compound are three useful conversion factors. What unit conversions can be accomplished using these conversion factors?
25. If you had a mole of U.S. dollar bills and equally distributed
the money to all of the people of the world, how rich would
every person be? Assume a world population of 7 billion.
26. Describe 1 mole of CO2 in as many ways as you can.
27. Which of the following compounds have the same empirical
formulas?
a.
b.
c.
d.
28. What is the difference between the molar mass and the empirical formula mass of a compound? When are these masses the
same, and when are they different? When different, how is the
molar mass related to the empirical formula mass?
29. How is the mass percent of elements in a compound different
for a 1.0-g sample versus a 100.-g sample versus a 1-mole
sample of the compound?
30. A balanced chemical equation contains a large amount of information. What information is given in a balanced equation?
31. The reaction of an element X with element Y is represented in
the following diagram. Which of the equations best describes
this reaction?
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X
Y
Chapter 3
Stoichiometry
a. 3X 1 8Y S X3Y8
b. 3X 1 6Y S X3Y6
c. X 1 2Y S XY2
d. 3X 1 8Y S 3XY2 1 2Y
32. Hydrogen gas and oxygen gas react to form water, and this
reaction can be depicted as follows:
Isotope
+
28
Explain why this equation is not balanced, and draw a picture
of the balanced equation.
33. What is the theoretical yield for a reaction, and how does this
quantity depend on the limiting reactant?
34. What does it mean to say a reactant is present “in excess” in a
process? Can the limiting reactant be present in excess? Does
the presence of an excess of a reactant affect the mass of products expected for a reaction?
35. Consider the following generic reaction:
A2B2 1 2C h 2CB 1 2A
What steps and information are necessary to perform the following determinations assuming that 1.00 3 104 molecules of
A2B2 are reacted with excess C?
a. mass of CB produced
b. atoms of A produced
c. moles of C reacted
d. percent yield of CB
36. Consider the following generic reaction:
Y2 1 2XY h 2XY2
In a limiting reactant problem, a certain quantity of each reactant is given and you are usually asked to calculate the mass of
product formed. If 10.0 g of Y2 is reacted with 10.0 g of XY,
outline two methods you could use to determine which reactant is limiting (runs out first) and thus determines the mass of
product formed.
Exercises
In this section similar exercises are paired.
Atomic Masses and the Mass Spectrometer
37. An element consists of 1.40% of an isotope with mass 203.973
u, 24.10% of an isotope with mass 205.9745 u, 22.10% of an
isotope with mass 206.9759 u, and 52.40% of an isotope with
mass 207.9766 u. Calculate the average atomic mass, and
identify the element.
38. An element “X” has five major isotopes, which are listed below along with their abundances. What is the element?
Isotope
46
X
X
48
X
49
X
50
X
47
39. The element rhenium (Re) has two naturally occurring isotopes, 185Re and 187Re, with an average atomic mass of
186.207 u. Rhenium is 62.60% 187Re, and the atomic mass of
187
Re is 186.956 u. Calculate the mass of 185Re.
40. Assume silicon has three major isotopes in nature as shown in
the table below. Fill in the missing information.
Percent Natural Abundance
Mass (u)
8.00%
7.30%
73.80%
5.50%
5.40%
45.95232
46.951764
47.947947
48.947841
49.944792
Si
Si
30
Si
29
Mass (u)
Abundance
27.98
_________
29.97
_________
4.70%
3.09%
41. The element europium exists in nature as two isotopes: 151Eu
has a mass of 150.9196 u and 153Eu has a mass of 152.9209 u.
The average atomic mass of europium is 151.96 u. Calculate
the relative abundance of the two europium isotopes.
42. The element silver (Ag) has two naturally occurring isotopes:
109
Ag and 107Ag with a mass of 106.905 u. Silver consists of
51.82% 107Ag and has an average atomic mass of 107.868 u.
Calculate the mass of 109Ag.
43. The mass spectrum of bromine (Br2) consists of three peaks
with the following characteristics:
Mass (u)
Relative Size
157.84
159.84
161.84
0.2534
0.5000
0.2466
How do you interpret these data?
44. The stable isotopes of iron are 54Fe, 56Fe, 57Fe, and 58Fe. The
mass spectrum of iron looks like the following:
Relative number of atoms
128
100
91.75
80
60
40
20
0
5.85
54
2.12 0.28
56 57 58
Mass number
Use the data on the mass spectrum to estimate the average
atomic mass of iron, and compare it to the value given in the
table inside the front cover of this book.
Moles and Molar Masses
45. Calculate the mass of 500. atoms of iron (Fe).
46. What number of Fe atoms and what amount (moles) of Fe
atoms are in 500.0 g of iron?
47. Diamond is a natural form of pure carbon. What number of
atoms of carbon are in a 1.00-carat diamond (1.00 carat 5
0.200 g)?
48. A diamond contains 5.0 3 1021 atoms of carbon. What amount
(moles) of carbon and what mass (grams) of carbon are in this
diamond?
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For Review
49. Aluminum metal is produced by passing an electric current
through a solution of aluminum oxide (Al2O3) dissolved in
molten cryolite (Na3AlF6). Calculate the molar masses of
Al2O3 and Na3AlF6.
50. The Freons are a class of compounds containing carbon, chlorine, and fluorine. While they have many valuable uses, they
have been shown to be responsible for depletion of the ozone in
the upper atmosphere. In 1991, two replacement compounds
for Freons went into production: HFC-134a (CH2FCF3) and
HCFC-124 (CHClFCF3). Calculate the molar masses of these
two compounds.
51. Calculate the molar mass of the following substances.
a.
H b.
N
H
N
c. (NH4)2Cr2O7
52. Calculate the molar mass of the following substances.
a.
O
P
b. Ca3(PO4)2 c. Na2HPO4
53. What amount (moles) of compound is present in 1.00 g of
each of the compounds in Exercise 51?
54. What amount (moles) of compound is present in 1.00 g of
each of the compounds in Exercise 52?
55. What mass of compound is present in 5.00 moles of each of
the compounds in Exercise 51?
56. What mass of compound is present in 5.00 moles of each of
the compounds in Exercise 52?
57. What mass of nitrogen is present in 5.00 moles of each of the
compounds in Exercise 51?
58. What mass of phosphorus is present in 5.00 moles of each of
the compounds in Exercise 52?
59. What number of molecules (or formula units) are present in
1.00 g of each of the compounds in Exercise 51?
60. What number of molecules (or formula units) are present in
1.00 g of each of the compounds in Exercise 52?
61. What number of atoms of nitrogen are present in 1.00 g of
each of the compounds in Exercise 51?
62. What number of atoms of phosphorus are present in 1.00 g of
each of the compounds in Exercise 52?
63. Freon-12 (CCl2F2) is used as a refrigerant in air conditioners
and as a propellant in aerosol cans. Calculate the number of
molecules of Freon-12 in 5.56 mg of Freon-12. What is the
mass of chlorine in 5.56 mg of Freon-12?
64. Bauxite, the principal ore used in the production of aluminum,
has a molecular formula of Al2O3 ? 2H2O. The ?H2O in the
formula are called waters of hydration. Each formula unit of
the compound contains two water molecules.
129
a. What is the molar mass of bauxite?
b. What is the mass of aluminum in 0.58 mole of bauxite?
c. How many atoms of aluminum are in 0.58 mole of
bauxite?
d. What is the mass of 2.1 3 1024 formula units of bauxite?
65. What amount (moles) is represented by each of these samples?
a. 150.0 g Fe2O3
b. 10.0 mg NO2
c. 1.5 3 1016 molecules of BF3
66. What amount (moles) is represented by each of these samples?
a. 20.0 mg caffeine, C8H10N4O2
b. 2.72 3 1021 molecules of ethanol, C2H5OH
c. 1.50 g of dry ice, CO2
67. What number of atoms of nitrogen are present in 5.00 g of
each of the following?
a. glycine, C2H5O2N
b. magnesium nitride
c. calcium nitrate
d. dinitrogen tetroxide
68. Complete the following table.
Mass of
Sample
Moles of
Sample
Molecules
in Sample
Total Atoms
in Sample
4.24 g C6H6
_________
_________
_________
0.224 mol H2O
_________
_________
_________
_________
_________
_________
_________
_________
2.71 3 1022
molecules CO2
_________
3.35 3 1022
total atoms
in CH3OH
sample
69. Ascorbic acid, or vitamin C (C6H8O6), is an essential vitamin.
It cannot be stored by the body and must be present in the diet.
What is the molar mass of ascorbic acid? Vitamin C tablets are
taken as a dietary supplement. If a typical tablet contains
500.0 mg vitamin C, what amount (moles) and what number
of molecules of vitamin C does it contain?
70. The molecular formula of acetylsalicylic acid (aspirin), one of
the most commonly used pain relievers, is C9H8O4.
a. Calculate the molar mass of aspirin.
b. A typical aspirin tablet contains 500. mg C9H8O4. What
amount (moles) of C9H8O4 molecules and what number of
molecules of acetylsalicylic acid are in a 500.-mg tablet?
71. Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a sedative and hypnotic. It is the compound used to make “Mickey
Finns” in detective stories.
a. Calculate the molar mass of chloral hydrate.
b. What amount (moles) of C2H3Cl3O2 molecules are in
500.0 g chloral hydrate?
c. What is the mass in grams of 2.0 3 1022 mole of chloral
hydrate?
d. What number of chlorine atoms are in 5.0 g chloral
hydrate?
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130
Chapter 3
Stoichiometry
e. What mass of chloral hydrate would contain 1.0 g Cl?
f. What is the mass of exactly 500 molecules of chloral
hydrate?
72. Dimethylnitrosamine, (CH3)2N2O, is a carcinogenic (cancercausing) substance that may be formed in foods, beverages, or
gastric juices from the reaction of nitrite ion (used as a food
preservative) with other substances.
a. What is the molar mass of dimethylnitrosamine?
b. How many moles of (CH3)2N2O molecules are present in
250 mg dimethylnitrosamine?
c. What is the mass of 0.050 mole of dimethylnitrosamine?
d. How many atoms of hydrogen are in 1.0 mole of
dimethylnitrosamine?
e. What is the mass of 1.0 3 106 molecules of
dimethylnitrosamine?
f. What is the mass in grams of one molecule of
dimethylnitrosamine?
80. Considering your answer to Exercise 79, which type of formula, empirical or molecular, can be obtained from elemental
analysis that gives percent composition?
81. Give the empirical formula for each of the compounds represented below.
a.
b.
c.
H
O
Percent Composition
73. Calculate the percent composition by mass of the following
compounds that are important starting materials for synthetic
polymers:
a. C3H4O2 (acrylic acid, from which acrylic plastics are
made)
b. C4H6O2 (methyl acrylate, from which Plexiglas is made)
c. C3H3N (acrylonitrile, from which Orlon is made)
74. In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77 K) was discovered.
The approximate formula of this substance is YBa2Cu3O7.
Calculate the percent composition by mass of this material.
75. The percent by mass of nitrogen for a compound is found to be
46.7%. Which of the following could be this species?
N
O
76. Arrange the following substances in order of increasing mass
percent of carbon.
a. caffeine, C8H10N4O2
b. sucrose, C12H22O11
c. ethanol, C2H5OH
77. Fungal laccase, a blue protein found in wood-rotting fungi, is
0.390% Cu by mass. If a fungal laccase molecule contains
four copper atoms, what is the molar mass of fungal laccase?
78. Hemoglobin is the protein that transports oxygen in mammals.
Hemoglobin is 0.347% Fe by mass, and each hemoglobin
molecule contains four iron atoms. Calculate the molar mass
of hemoglobin.
Empirical and Molecular Formulas
79. Express the composition of each of the following compounds
as the mass percents of its elements.
a. formaldehyde, CH2O
b. glucose, C6H12O6
c. acetic acid, HC2H3O2
N
C
P
d.
82. Determine the molecular formulas to which the following empirical formulas and molar masses pertain.
a. SNH (188.35 g/mol)
c. CoC4O4 (341.94 g/mol)
b. NPCl2 (347.64 g/mol)
d. SN (184.32 g/mol)
83. A compound that contains only carbon, hydrogen, and oxygen
is 48.64% C and 8.16% H by mass. What is the empirical formula of this substance?
84. The most common form of nylon (nylon-6) is 63.68% carbon,
12.38% nitrogen, 9.80% hydrogen, and 14.14% oxygen. Calculate the empirical formula for nylon-6.
85. There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound,
with oxygen gas escaping into the atmosphere while leaving a
residue of pure mercury. Heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. Heating 0.4172 g of the
other compound results in a mass loss of 0.016 g. Determine the
empirical formula of each compound.
86. A sample of urea contains 1.121 g N, 0.161 g H, 0.480 g C,
and 0.640 g O. What is the empirical formula of urea?
87. A compound containing only sulfur and nitrogen is 69.6% S
by mass; the molar mass is 184 g/mol. What are the empirical
and molecular formulas of the compound?
88. Determine the molecular formula of a compound that contains
26.7% P, 12.1% N, and 61.2% Cl, and has a molar mass of
580 g/mol.
89. A compound contains 47.08% carbon, 6.59% hydrogen, and
46.33% chlorine by mass; the molar mass of the compound is
153 g/mol. What are the empirical and molecular formulas of
the compound?
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For Review
90. Maleic acid is an organic compound composed of 41.39% C,
3.47% H, and the rest oxygen. If 0.129 mole of maleic acid
has a mass of 15.0 g, what are the empirical and molecular
formulas of maleic acid?
91. One of the components that make up common table sugar is
fructose, a compound that contains only carbon, hydrogen,
and oxygen. Complete combustion of 1.50 g of fructose produced 2.20 g of carbon dioxide and 0.900 g of water. What is
the empirical formula of fructose?
92. A compound contains only C, H, and N. Combustion of
35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg
H2O. What is the empirical formula of the compound?
93. Cumene is a compound containing only carbon and hydrogen
that is used in the production of acetone and phenol in the
chemical industry. Combustion of 47.6 mg cumene produces
some CO2 and 42.8 mg water. The molar mass of cumene is
between 115 and 125 g/mol. Determine the empirical and molecular ­formulas.
94. A compound contains only carbon, hydrogen, and oxygen.
­Combustion of 10.68 mg of the compound yields 16.01 mg
CO2 and 4.37 mg H2O. The molar mass of the compound is
176.1 g/mol. What are the empirical and molecular formulas
of the compound?
Balancing Chemical Equations
95. Give the balanced equation for each of the following chemical
reactions:
a. Glucose (C6H12O6) reacts with oxygen gas to produce
gaseous carbon dioxide and water vapor.
b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide
gas.
c. Carbon disulfide liquid reacts with ammonia gas to produce
hydrogen sulfide gas and solid ammonium thiocyanate
(NH4SCN).
96. Give the balanced equation for each of the following.
a. The combustion of ethanol (C2H5OH) forms carbon dioxide and water vapor. A combustion reaction refers to a
reaction of a substance with oxygen gas.
b. Aqueous solutions of lead(II) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of
lead(II) phosphate with aqueous sodium nitrate as the
other product.
c. Solid zinc reacts with aqueous HCl to form aqueous zinc
chloride and hydrogen gas.
d. Aqueous strontium hydroxide reacts with aqueous hydro­
bromic acid to produce water and aqueous strontium
bromide.
97. A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated
hydrogen peroxide (H2O2 ) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV)
oxide is a catalyst for the decomposition of hydrogen peroxide
and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.
131
98. Iron oxide ores, commonly a mixture of FeO and Fe2O3, are
given the general formula Fe3O4. They yield elemental iron
when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations
for these processes:
Fe3O4 1s2 1 H2 1g2 h Fe 1s2 1 H2O 1g2
Fe3O4 1s2 1 CO 1g2 h Fe 1s2 1 CO2 1g2
99. Balance the following equations:
a. Ca 1OH2 2 1aq2 1 H3PO4 1aq2 S H2O 1l2 1 Ca3 1PO42 2 1s2
b. Al 1OH2 3 1s2 1 HCl 1aq2 S AlCl3 1aq2 1 H2O 1l2
c. AgNO3 1aq2 1 H2SO4 1aq2 S Ag2SO4 1s2 1 HNO3 1aq2
100. Balance each of the following chemical equations.
a. KO2 1s2 1 H2O 1l2 S KOH 1aq2 1 O2 1g2 1 H2O2 1aq2
b. Fe2O3 1s2 1 HNO3 1aq2 S Fe 1NO32 3 1aq2 1 H2O 1l2
c. NH3 1g2 1 O2 1g2 S NO 1g2 1 H2O 1g2
d. PCl5 1l2 1 H2O 1l2 S H3PO4 1aq2 1 HCl 1g2
e. CaO 1s2 1 C 1s2 S CaC2 1s2 1 CO2 1g2
f. MoS2 1s2 1 O2 1g2 S MoO3 1s2 1 SO2 1g2
g. FeCO3 1s2 1 H2CO3 1aq2 S Fe 1HCO32 2 1aq2
101. Balance the following equations representing combustion
reactions:
a.
(l)
+
(g)
H
(g)
C
+
(g)
O
b.
(g)
+
(g)
(g)
+
c. C12H22O11 1s2 1 O2 1g2 S CO2 1g2 1 H2O 1g2
d. Fe 1s2 1 O2 1g2 S Fe2O3 1s2
e. FeO 1s2 1 O2 1g2 S Fe2O3 1s2
102. Balance the following equations:
a. Cr 1s2 1 S8 1s2 S Cr2S3 1s2
Heat
b. NaHCO3 1s2 h Na2CO3 1s2 1 CO2 1g2 1 H2O 1g2
Heat
c. KClO3 1s2 h KCl 1s2 1 O2 1g2
d. Eu 1s2 1 HF 1g2 S EuF3 1s2 1 H2 1g2
103. Silicon is produced for the chemical and electronics industries
by the following reactions. Give the balanced equation for
each reaction.
Electric
a. SiO2 1s2 1 C 1s2 88888n
Si 1s2 1 CO 1g2
arc furnace
b. Liquid silicon tetrachloride is reacted with very pure solid
magnesium, producing solid silicon and solid magnesium
chloride.
c. Na2SiF6 1s2 1 Na 1s2 S Si 1s2 1 NaF 1s2
104. Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, CaSiO3. Glass can be
etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble
products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is
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(g)
132
Chapter 3
Stoichiometry
often graduated by using this process. Balance the following
equation for the reaction of hydrofluoric acid with calcium
silicate.
CaSiO3 1s2 1 HF 1aq2 h CaF2 1aq2 1 SiF4 1g2 1 H2O 1l2
Reaction Stoichiometry
105. Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel
rocket motors. The reaction is
Fe2O3 1s2 1 2Al 1s2 h 2Fe 1l2 1 Al2O3 1s2
What masses of iron(III) oxide and aluminum must be used to
produce 15.0 g iron? What is the maximum mass of aluminum
oxide that could be produced?
106. The reaction between potassium chlorate and red phosphorus
takes place when you strike a match on a matchbox. If you
were to react 52.9 g of potassium chlorate (KClO3) with excess red phosphorus, what mass of tetraphosphorus decaoxide
(P4O10) could be produced?
KClO3 1s2 1 P4 1s2 h P4O10 1s2 1 KCl 1s2
1unbalanced2
107. The reusable booster rockets of the U.S. space shuttle employ
a mixture of aluminum and ammonium perchlorate for fuel. A
possible equation for this reaction is
3Al 1s2 1 3NH4ClO4 1s2 h
Al2O3 1s2 1 AlCl3 1s2 1 3NO 1g2 1 6H2O 1g2
What mass of NH4ClO4 should be used in the fuel mixture for
every kilogram of Al?
108. One of relatively few reactions that takes place directly between two solids at room temperature is
Ba 1OH2 2 # 8H2O 1s2 1 NH4SCN 1s2 h
Ba 1SCN2 2 1s2 1 H2O 1l2 1 NH3 1g2
In this equation, the ? 8H2O in Ba(OH)2 ? 8H2O indicates the
presence of eight water molecules. This compound is called
barium hydroxide octahydrate.
a. Balance the equation.
b. What mass of ammonium thiocyanate (NH4SCN) must be
used if it is to react completely with 6.5 g barium hydroxide octahydrate?
111. Bacterial digestion is an economical method of sewage treatment. The reaction
bacteria
5CO2 1g2 1 55NH41 1aq2 1 76O2 1g2 88888n
2
1
2
1
2
C5H7O2N s 1 54NO2 aq 1 52H2O 1l2 1 109H 1 1aq2
bacterial tissue
is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. What mass of bacterial tissue is produced in a treatment plant for every 1.0 3 104 kg of
wastewater containing 3.0% NH41 ions by mass? Assume that
95% of the ammonium ions are consumed by the ­bacteria.
112. Phosphorus can be prepared from calcium phosphate by the
following reaction:
2Ca3 1PO42 2 1s2 1 6SiO2 1s2 1 10C 1s2 h
6CaSiO3 1s2 1 P4 1s2 1 10CO 1g2
Phosphorite is a mineral that contains Ca3(PO4)2 plus other
non-phosphorus-containing compounds. What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phorphorite sample is 75% Ca3(PO4)2 by mass?
Assume an excess of the other reactants.
113. Coke is an impure form of carbon that is often used in the industrial production of metals from their oxides. If a sample of
coke is 95% carbon by mass, determine the mass of coke
needed to react completely with 1.0 ton of copper(II) oxide.
2CuO 1s2 1 C 1s2 h 2Cu 1s2 1 CO2 1g2
114. The space shuttle environmental control system handles excess CO2 (which the astronauts breathe out; it is 4.0% by mass
of exhaled air) by reacting it with lithium hydroxide, LiOH,
pellets to form lithium carbonate, Li2CO3, and water. If there
are seven astronauts on board the shuttle, and each exhales
20. L of air per minute, how long could clean air be generated
if there were 25,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL.
Limiting Reactants and Percent Yield
115. Consider the reaction between NO(g) and O2(g) represented
­below.
O2
NO
NO2
109. Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz:
3NaHCO3 1aq2 1 C6H8O7 1aq2 h
3CO2 1g2 1 3H2O 1l2 1 Na3C6H5O7 1aq2
a. What mass of C6H8O7 should be used for every 1.0 3 102
mg NaHCO3?
b. What mass of CO2(g) could be produced from such a
­mixture?
110. Aspirin (C9H8O4) is synthesized by reacting salicylic acid
(C7H6O3) with acetic anhydride (C4H6O3). The balanced equation is
C7H6O3 1 C4H6O3 h C9H8O4 1 HC2H3O2
a. What mass of acetic anhydride is needed to completely
consume 1.00 3 102 g salicylic acid?
b. What is the maximum mass of aspirin (the theoretical
yield) that could be produced in this reaction?
What is the balanced equation for this reaction, and what is the
limiting reactant?
116. Consider the following reaction:
4NH3 1g2 1 5O2 1g2 h 4NO 1g2 1 6H2O 1g2
If a container were to have 10 molecules of O2 and 10 molecules of NH3 initially, how many total molecules (reactants
plus products) would be present in the container after this reaction goes to completion?
117. Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
N2 1g2 1 3H2 1g2 h 2NH3 1g2
a. What is the maximum mass of ammonia that can be
produced from a mixture of 1.00 3 103 g N2 and
5.00 3 102 g H2?
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For Review
b. What mass of which starting material would remain
unreacted?
118. Consider the following unbalanced equation:
Ca3 1PO42 2 1s2 1 H2SO4 1aq2 h CaSO4 1s2 1 H3PO4 1aq2
What masses of calcium sulfate and phosphoric acid can be
produced from the reaction of 1.0 kg calcium phosphate with
1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?
119. Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen
gas (which inhibits the growth of anaerobic microorganisms);
and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline
earth metal peroxide, such as barium peroxide:
BaO2 1s2 1 2HCl 1aq2 h H2O2 1aq2 1 BaCl2 1aq2
What mass of hydrogen peroxide should result when 1.50 g
barium peroxide is treated with 25.0 mL hydrochloric acid solution containing 0.0272 g HCl per mL? What mass of which
reagent is left unreacted?
120. Silver sulfadiazine burn-treating cream creates a barrier against
bacterial invasion and releases antimicrobial agents directly into
the wound. If 25.0 g Ag2O is reacted with 50.0 g C10H10N4SO2,
what mass of silver sulfadiazine, AgC10H9N4SO2, can be produced, assuming 100% yield?
Ag2O 1s2 1 2C10H10N4SO2 1s2 h 2AgC10H9N4SO2 1s2 1 H2O 1l2
121. Hydrogen cyanide is produced industrially from the reaction
of gaseous ammonia, oxygen, and methane:
2NH3 1g2 1 3O2 1g2 1 2CH4 1g2 h 2HCN 1g2 1 6H2O 1g2
If 5.00 3 103 kg each of NH3, O2, and CH4 are reacted, what
mass of HCN and of H2O will be produced, assuming 100%
yield?
122. Acrylonitrile (C3H3N) is the starting material for many synthetic
carpets and fabrics. It is produced by the following reaction.
2C3H6 1g2 1 2NH3 1g2 1 3O2 1g2 h 2C3H3N 1g2 1 6H2O 1g2
If 15.0 g C3H6, 10.0 g O2, and 5.00 g NH3 are reacted, what
mass of acrylonitrile can be produced, assuming 100% yield?
123. The reaction of ethane gas (C2H6) with chlorine gas produces
C2H5Cl as its main product (along with HCl). In addition, the
reaction invariably produces a variety of other minor products,
including C2H4Cl2, C2H3Cl3, and others. Naturally, the production of these minor products reduces the yield of the main
product. Calculate the percent yield of C2H5Cl if the reaction
of 300. g of ethane with 650. g of chlorine produced 490. g of
C2H5Cl.
124. DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction:
2C6H5Cl 1 C2HOCl3 h C14H9Cl5 1 H2O
chlorobenzene
chloral
DDT
In a government lab, 1142 g of chlorobenzene is reacted with
485 g of chloral.
a. What mass of DDT is formed, assuming 100% yield?
b. Which reactant is limiting? Which is in excess?
133
c. What mass of the excess reactant is left over?
d. If the actual yield of DDT is 200.0 g, what is the percent
yield?
125. Bornite (Cu3FeS3) is a copper ore used in the production of
copper. When heated, the following reaction occurs:
2Cu3FeS3 1s2 1 7O2 1g2 h 6Cu 1s2 1 2FeO 1s2 1 6SO2 1g2
If 2.50 metric tons of bornite is reacted with excess O2 and the
process has an 86.3% yield of copper, what mass of copper is
produced?
126. Consider the following unbalanced reaction:
P4 1s2 1 F2 1g2 h PF3 1g2
What mass of F2 is needed to produce 120. g of PF3 if the reaction has a 78.1% yield?
Additional Exercises
127. In using a mass spectrometer, a chemist sees a peak at a mass
of 30.0106. Of the choices 12C21H6, 12C1H216O, and 14N16O,
which is responsible for this peak? Pertinent masses are 1H,
1.007825; 16O, 15.994915; and 14N, 14.003074.
128. Boron consists of two isotopes, 10B and 11B. Chlorine also has
two isotopes, 35Cl and 37Cl. Consider the mass spectrum of BCl3.
How many peaks would be present, and what approximate mass
would each peak correspond to in the BCl3 mass spectrum?
129. A given sample of a xenon fluoride compound contains molecules of the type XeFn, where n is some whole number. Given
that 9.03 3 1020 molecules of XeFn weigh 0.368 g, determine
the value for n in the formula.
130. Aspartame is an artificial sweetener that is 160 times sweeter
than sucrose (table sugar) when dissolved in water. It is marketed as NutraSweet. The molecular formula of aspartame is
C14H18N2O5.
a. Calculate the molar mass of aspartame.
b. What amount (moles) of molecules are present in 10.0 g
­aspartame?
c. Calculate the mass in grams of 1.56 mole of aspartame.
d. What number of molecules are in 5.0 mg aspartame?
e. What number of atoms of nitrogen are in 1.2 g aspartame?
f. What is the mass in grams of 1.0 3 109 molecules of
­aspartame?
g. What is the mass in grams of one molecule of aspartame?
131. Anabolic steroids are performance enhancement drugs whose
use has been banned from most major sporting activities. One
anabolic steroid is fluoxymesterone (C20H29FO3). Calculate
the percent composition by mass of fluoxymesterone.
132. Many cereals are made with high moisture content so that the
cereal can be formed into various shapes before it is dried. A
cereal product containing 58% H2O by mass is produced at the
rate of 1000. kg/h. What mass of water must be evaporated per
hour if the final product contains only 20.% water?
133. The compound adrenaline contains 56.79% C, 6.56% H,
28.37% O, and 8.28% N by mass. What is the empirical formula for adrenaline?
134. Adipic acid is an organic compound composed of 49.31% C,
43.79% O, and the rest hydrogen. If the molar mass of adipic
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135.
136.
137.
138.
139.
140.
141.
Chapter 3
Stoichiometry
acid is 146.1 g/mol, what are the empirical and molecular formulas for adipic acid?
Vitamin B12, cyanocobalamin, is essential for human nutrition.
It is concentrated in animal tissue but not in higher plants. Although nutritional requirements for the vitamin are quite low,
people who abstain completely from animal products may develop a deficiency anemia. Cyanocobalamin is the form used in
vitamin supplements. It contains 4.34% cobalt by mass. Calculate the molar mass of cyanocobalamin, assuming that there is
one atom of cobalt in every molecule of cyanocobalamin.
Some bismuth tablets, a medication used to treat upset stomachs, contain 262 mg of bismuth subsalicylate, C7H5BiO4, per
tablet. Assuming two tablets are digested, calculate the mass
of bismuth consumed.
The empirical formula of styrene is CH; the molar mass of
styrene is 104.14 g/mol. What number of H atoms are present
in a 2.00-g sample of styrene?
Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H,
and O. Combustion of 19.81 mg terephthalic acid produces
41.98 mg CO2 and 6.45 mg H2O. If 0.250 mole of terephthalic
acid has a mass of 41.5 g, determine the molecular formula for
terephthalic acid.
A sample of a hydrocarbon (a compound consisting of only
carbon and hydrogen) contains 2.59 3 1023 atoms of hydrogen
and is 17.3% hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles)
of compound is present, and what is the mass of the sample?
A binary compound between an unknown element E and hydrogen contains 91.27% E and 8.73% H by mass. If the formula of the compound is E3H8, calculate the atomic mass of E.
A 0.755-g sample of hydrated copper(II) sulfate
CuSO4 # xH2O
was heated carefully until it had changed completely to anhydrous copper(II) sulfate (CuSO4) with a mass of 0.483 g. Determine the value of x. [This number is called the number of waters
of hydration of copper(II) sulfate. It specifies the number of water
molecules per formula unit of CuSO4 in the hydrated crystal.]
142. ABS plastic is a tough, hard plastic used in applications
requiring shock resistance. The polymer consists of three
monomer units: acrylonitrile (C3H3N), butadiene (C4H6), and
styrene (C8H8).
a. A sample of ABS plastic contains 8.80% N by mass. It
took 0.605 g of Br2 to react completely with a 1.20-g
sample of ABS plastic. Bromine reacts 1:1 (by moles)
with the butadiene molecules in the polymer and nothing
else. What is the percent by mass of acrylonitrile and
butadiene in this polymer?
b. What are the relative numbers of each of the monomer
units in this polymer?
143. A sample of LSD (d-lysergic acid diethylamide, C24H30N3O)
is added to some table salt (sodium chloride) to form a mixture. Given that a 1.00-g sample of the mixture undergoes
combustion to produce 1.20 g of CO2, what is the mass percent of LSD in the mixture?
144. Methane (CH4) is the main component of marsh gas. Heating
methane in the presence of sulfur produces carbon disulfide
and hydrogen sulfide as the only products.
a. Write the balanced chemical equation for the reaction of
methane and sulfur.
b. Calculate the theoretical yield of carbon disulfide when
120. g of methane is reacted with an equal mass of sulfur.
145. A potential fuel for rockets is a combination of B5H9 and O2.
The two react according to the following balanced equation:
2B5H9 1l2 1 12O2 1g2 h 5B2O3 1s2 1 9H2O 1g2
If one tank in a rocket holds 126 g B5H9 and another tank
holds 192 g O2, what mass of water can be produced when the
entire contents of each tank react together?
146. A 0.4230-g sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and
oxygen gas. Determine the percent of sodium nitrate in the
original sample.
147. An iron ore sample contains Fe2O3 plus other impurities. A
752-g sample of impure iron ore is heated with excess carbon,
producing 453 g of pure iron by the following reaction:
Fe2O3 1s2 1 3C 1s2 h 2Fe 1s2 1 3CO 1g2
What is the mass percent of Fe2O3 in the impure iron ore sample? Assume that Fe2O3 is the only source of iron and that the
reaction is 100% efficient.
148. Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows:
Zn 1s2 1 2HCl 1aq2 h ZnCl2 1aq2 1 H2 1g2
(Cu does not react with HCl.) When 0.5065 g of a certain brass
alloy is reacted with excess HCl, 0.0985 g ZnCl2 is eventually
isolated.
a. What is the composition of the brass by mass?
b. How could this result be checked without changing the
above procedure?
149. Vitamin A has a molar mass of 286.4 g/mol and a general
­molecular formula of CxHyE, where E is an unknown element.
If ­vitamin A is 83.86% C and 10.56% H by mass, what is the
molecular formula of vitamin A?
150. You have seven closed containers, each with equal masses of
chlorine gas (Cl2). You add 10.0 g of sodium to the first sample, 20.0 g of sodium to the second sample, and so on (adding
70.0 g of sodium to the seventh sample). Sodium and chlorine
react to form sodium chloride according to the equation
2Na 1s2 1 Cl2 1g2 h 2NaCl 1s2
After each reaction is complete, you collect and measure the
amount of sodium chloride formed. A graph of your results is
shown below.
Mass of NaCl (g)
134
0
20
40
60
80
Mass of Sodium (g)
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For Review
Answer the following questions:
a. Explain the shape of the graph.
b. Calculate the mass of NaCl formed when 20.0 g of
sodium is used.
c. Calculate the mass of Cl2 in each container.
d. Calculate the mass of NaCl formed when 50.0 g of
sodium is used.
e. Identify the leftover reactant, and determine its mass for
parts b and d above.
151. A substance X2Z has the composition (by mass) of 40.0% X
and 60.0% Z. What is the composition (by mass) of the compound XZ 2?
ChemWork Problems
These multiconcept problems (and additional ones) are found inter­
actively online with the same type of assistance a student would
get from an instructor.
152. Consider samples of phosphine (PH3), water (H2O), hydrogen
sulfide (H2S), and hydrogen fluoride (HF), each with a mass of
119 g. Rank the compounds from the least to the greatest number of hydrogen atoms contained in the samples.
153. Calculate the number of moles for each compound in the following table.
Compound
Mass
Moles
Magnesium phosphate
Calcium nitrate
Potassium chromate
Dinitrogen pentoxide
326.4 g
303.0 g
141.6 g
406.3 g
_________
_________
_________
_________
154. Arrange the following substances in order of increasing mass
percent of nitrogen.
a. NO
c. NH3
b. N2O
d. SNH
155. Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the
elements carbon, hydrogen, and oxygen. Complete combustion of a 0.345-g sample of p-cresol produced 0.983 g carbon
dioxide and 0.230 g water. Determine the empirical formula
for p-cresol.
156. A compound with molar mass 180.1 g/mol has the following
composition by mass:
C
H
O
40.0%
6.70%
53.3%
Determine the empirical and molecular formulas of the
compound.
157. Which of the following statements about chemical equations
is(are) true?
a. When balancing a chemical equation, you can never
change the coefficient in front of any chemical formula.
b. The coefficients in a balanced chemical equation refer to
the number of grams of reactants and products.
c. In a chemical equation, the reactants are on the right and
the products are on the left.
135
d. When balancing a chemical equation, you can never
change the subscripts of any chemical formula.
e. In chemical reactions, matter is neither created nor
destroyed so a chemical equation must have the same number of atoms on both sides of the equation.
158. Consider the following unbalanced chemical equation for the
combustion of pentane (C5H12):
C5H12 1l2 1 O2 1g2 h CO2 1g2 1 H2O 1l2
If 20.4 g of pentane are burned in excess oxygen, what mass of
water can be produced, assuming 100% yield?
159. Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. The unbalanced chemical equation for
this reaction is given below:
SO2 1g2 1 NaOH 1s2 h Na2SO3 1s2 1 H2O 1l2
Assuming you react 38.3 g sulfur dioxide with 32.8 g sodium
hydroxide and assuming that the reaction goes to completion,
calculate the mass of each product formed.
Challenge Problems
160. Gallium arsenide, GaAs, has gained widespread use in semiconductor devices that convert light and electrical signals in
fiber-optic communications systems. Gallium consists of
60.% 69Ga and 40.% 71Ga. Arsenic has only one naturally occurring isotope, 75As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas
GaAs and Ga2As2. What would the distribution of peaks look
like for these two fragments?
161. Consider the following data for three binary compounds of
hydrogen and nitrogen:
I
II
III
% H (by Mass)
% N (by Mass)
17.75
12.58
2.34
82.25
87.42
97.66
When 1.00 L of each gaseous compound is decomposed to its
elements, the following volumes of H2(g) and N2(g) are
obtained:
I
II
III
H2 (L)
N2 (L)
1.50
2.00
0.50
0.50
1.00
1.50
Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the
atomic masses of hydrogen and nitrogen.
162. Natural rubidium has the average mass of 85.4678 u and is
composed of isotopes 85Rb (mass 5 84.9117 u) and 87Rb. The
ratio of atoms 85Rby87Rb in natural rubidium is 2.591. Calculate the mass of 87Rb.
163. A compound contains only carbon, hydrogen, nitrogen, and
oxygen. Combustion of 0.157 g of the compound produced
0.213 g CO2 and 0.0310 g H2O. In another experiment, it is
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136
Chapter 3
Stoichiometry
found that 0.103 g of the compound produces 0.0230 g NH3.
What is the empirical formula of the compound? Hint: Combustion involves reacting with excess O2. Assume that all the
carbon ends up in CO2 and all the hydrogen ends up in H2O.
Also assume that all the nitrogen ends up in the NH3 in the
second experiment.
164. Nitric acid is produced commercially by the Ostwald process,
represented by the following equations:
4NH3 1g2 1 5O2 1g2 h 4NO 1g2 1 6H2O 1g2
2NO 1g2 1 O2 1g2 h 2NO2 1g2
3NO2 1g2 1 H2O 1l2 h 2HNO3 1aq2 1 NO 1g2
165.
166.
167.
168.
169.
170.
171.
What mass of NH3 must be used to produce 1.0 3 106 kg
HNO3 by the Ostwald process? Assume 100% yield in each
reaction, and assume that the NO produced in the third step is
not recycled.
When the supply of oxygen is limited, iron metal reacts with
oxygen to produce a mixture of FeO and Fe2O3. In a certain
experiment, 20.00 g iron metal was reacted with 11.20 g oxygen gas. After the experiment, the iron was totally consumed,
and 3.24 g oxygen gas remained. Calculate the amounts of
FeO and Fe2O3 formed in this experiment.
A 9.780-g gaseous mixture contains ethane (C2H6) and propane (C3H8). Complete combustion to form carbon dioxide
and water requires 1.120 mole of oxygen gas. Calculate the
mass percent of ethane in the original mixture.
Zinc and magnesium metal each reacts with hydrochloric acid to
make chloride salts of the respective metals, and hydrogen gas.
A 10.00-g mixture of zinc and magnesium produces 0.5171 g of
hydrogen gas upon being mixed with an excess of hydrochloric
acid. Determine the percent magnesium by mass in the original
mixture.
A gas contains a mixture of NH3(g) and N2H4(g), both of which
react with O2(g) to form NO2(g) and H2O(g). The gaseous
mixture (with an initial mass of 61.00 g) is reacted with
10.00 moles O2, and after the reaction is complete, 4.062 moles
of O2 remains. Calculate the mass percent of N2H4(g) in the
original gaseous mixture.
Consider a gaseous binary compound with a molar mass of
62.09 g/mol. When 1.39 g of this compound is completely
burned in excess oxygen, 1.21 g of water is formed. Determine
the formula of the compound. Assume water is the only product that contains hydrogen.
A 2.25-g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the
formula of the scandium chloride produced in the reaction?
In the production of printed circuit boards for the electronics
industry, a 0.60-mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a
chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching
reaction is
Cu 1NH32 4Cl2 1aq2 1 4NH3 1aq2 1 Cu 1s2 h 2Cu 1NH32 4Cl 1aq2
A plant needs to manufacture 10,000 printed circuit boards,
each 8.0 3 16.0 cm in area. An average of 80.% of the copper
is removed from each board (density of copper 5 8.96 g/cm3).
What masses of Cu(NH3)4Cl2 and NH3 are needed to do this?
Assume 100% yield.
172. The aspirin substitute, acetaminophen (C8H9O2N), is produced by the following three-step synthesis:
I. C6H5O3N 1s2 1 3H2 1g2 1 HCl 1aq2 h
C6H8ONCl 1s2 1 2H2O 1l2
1
1
2
2
II. C6H8ONCl s 1 NaOH aq h
C6H7ON 1s2 1 H2O 1l2 1 NaCl 1aq2
1
2
III. C6H7ON s 1 C4H6O3 1l2 h
C8H9O2N 1s2 1 HC2H3O2 1l2
173.
174.
175.
176.
177.
The first two reactions have percent yields of 87% and 98%
by mass, respectively. The overall reaction yields 3 moles of
acetaminophen product for every 4 moles of C6H5O3N reacted.
a. What is the percent yield by mass for the overall process?
b. What is the percent yield by mass of Step III?
An element X forms both a dichloride (XCl2) and a tetrachloride
(XCl4). Treatment of 10.00 g XCl2 with excess chlorine forms
12.55 g XCl4. Calculate the atomic mass of X, and identify X.
When M2S3(s) is heated in air, it is converted to MO2(s). A
4.000‑g sample of M2S3(s) shows a decrease in mass of 0.277 g
when it is heated in air. What is the average atomic mass of M?
When aluminum metal is heated with an element from Group
6A of the periodic table, an ionic compound forms. When the
experiment is performed with an unknown Group 6A element,
the product is 18.56% Al by mass. What is the formula of the
compound?
Consider a mixture of potassium chloride and potassium nitrate that is 43.2% potassium by mass. What is the percent
KCl by mass of the original mixture?
Ammonia reacts with O2 to form either NO(g) or NO2(g) according to these unbalanced equations:
NH3 1g2 1 O2 1g2 h NO 1g2 1 H2O 1g2
NH3 1g2 1 O2 1g2 h NO2 1g2 1 H2O 1g2
In a certain experiment 2.00 moles of NH3(g) and 10.00 moles
of O2(g) are contained in a closed flask. After the reaction is
­complete, 6.75 moles of O2(g) remains. Calculate the number
of moles of NO(g) in the product mixture: (Hint: You cannot
do this problem by adding the balanced equations because you
cannot assume that the two reactions will occur with equal
probability.)
178. You take 1.00 g of an aspirin tablet (a compound consisting
solely of carbon, hydrogen, and oxygen), burn it in air, and collect 2.20 g CO2 and 0.400 g H2O. You know that the molar
mass of aspirin is between 170 and 190 g/mol. Reacting 1 mole
of salicylic acid with 1 mole of acetic anhydride (C4H6O3)
gives you 1 mole of aspirin and 1 mole of acetic acid (C2H4O2).
Use this information to determine the molecular formula of
salicylic acid.
Integrative Problems
These problems require the integration of multiple concepts to find
the solutions.
179. With the advent of techniques such as scanning tunneling microscopy, it is now possible to “write” with individual atoms
by manipulating and arranging atoms on an atomic surface.
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For Review
a. If an image is prepared by manipulating iron atoms and
their total mass is 1.05 3 10220 g, what number of iron
atoms were used?
b. If the image is prepared on a platinum surface that is
exactly 20 platinum atoms high and 14 platinum atoms
wide, what is the mass (grams) of the atomic surface?
c. If the atomic surface were changed to ruthenium atoms
and the same surface mass as determined in part b is used,
what number of ruthenium atoms is needed to construct
the ­surface?
180. Tetrodotoxin is a toxic chemical found in fugu pufferfish, a
popular but rare delicacy in Japan. This compound has an LD50
(the amount of substance that is lethal to 50.% of a population
sample) of 10. mg per kg of body mass. Tetrodotoxin is 41.38%
carbon by mass, 13.16% nitrogen by mass, and 5.37% hydrogen by mass, with the remaining amount consisting of oxygen.
What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin have a mass of 1.59 3 10221 g, what is the
molecular formula of tetrodotoxin? What number of molecules
of tetrodotoxin would be the LD50 dosage for a person weighing 165 lb?
181. An ionic compound MX3 is prepared according to the following unbalanced chemical equation.
M 1 X2 h MX3
A 0.105-g sample of X2 contains 8.92 3 1020 molecules. The
compound MX3 consists of 54.47% X by mass. What are the
identities of M and X, and what is the correct name for MX3?
Starting with 1.00 g each of M and X2, what mass of MX3 can
be prepared?
182. The compound As2I4 is synthesized by reaction of arsenic
metal with arsenic triiodide. If a solid cubic block of arsenic
137
(d 5 5.72 g/cm3) that is 3.00 cm on edge is allowed to react
with 1.01 3 1024 molecules of arsenic triiodide, what mass of
As2I4 can be prepared? If the percent yield of As2I4 was 75.6%,
what mass of As2I4 was actually isolated?
Marathon Problems
These problems are designed to incorporate several concepts and
techniques into one situation.
183. A 2.077-g sample of an element, which has an atomic mass between 40 and 55, reacts with oxygen to form 3.708 g of an oxide.
Determine the formula of the oxide (and identify the element).
184. Consider the following balanced chemical equation:
A 1 5B h 3C 1 4D
a. Equal masses of A and B are reacted. Complete each of
the following with either “A is the limiting reactant
because ________”; “B is the limiting reactant because
________”; or “we cannot determine the limiting reactant
because ________.”
i. If the molar mass of A is greater than the molar mass
of B, then
ii.If the molar mass of B is greater than the molar mass
of A, then
b. The products of the reaction are carbon dioxide (C) and
water (D). Compound A has a similar molar mass to carbon dioxide. Compound B is a diatomic molecule. Identify compound B, and support your answer.
c. Compound A is a hydrocarbon that is 81.71% carbon by
mass. Determine its empirical and molecular formulas.
Marathon Problems can be used in class by groups of students to
help facilitate problem-solving skills.
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Chapter 4
Types of Chemical Reactions
and Solution Stoichiometry
4.1
Water, the Common Solvent
4.4
Types of Chemical Reactions
4.2
The Nature of Aqueous Solutions:
Strong and Weak Electrolytes
4.5
Precipitation Reactions
Oxidation States
Strong Electrolytes
4.6
Describing Reactions in Solution
4.7
Stoichiometry of Precipitation
Reactions
The Characteristics of Oxidation–
Reduction Reactions
4.8
Acid–Base Reactions
Weak Electrolytes
Nonelectrolytes
4.3
The Composition of Solutions
Dilution
Acid–Base Titrations
4.9
Oxidation–Reduction Reactions
4.10 Balancing Oxidation–Reduction
Equations
Oxidation States Method of Balancing
Oxidation–Reduction Reactions
Sodium reacts violently when water is dripped on it. (Charles D.Winters)
138
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M
uch of the chemistry that affects each of us occurs among substances dissolved in water. For example, virtually all the chemistry that makes life possible occurs in an aqueous environment. Also, various medical tests involve aqueous
reactions, depending heavily on analyses of blood and other body fluids. In addition to
the common tests for sugar, cholesterol, and iron, analyses for specific chemical markers allow detection of many diseases before obvious symptoms occur.
Aqueous chemistry is also important in our environment. In recent years, contamination of the groundwater by substances such as chloroform and nitrates has been
widely publicized. Water is essential for life, and the maintenance of an ample supply
of clean water is crucial to all civilization.
To understand the chemistry that occurs in such diverse places as the human body,
the atmosphere, the groundwater, the oceans, the local water treatment plant, your hair
as you shampoo it, and so on, we must understand how substances dissolved in water
react with each other.
However, before we can understand solution reactions, we need to discuss the nature of solutions in which water is the dissolving medium, or solvent. These solutions
are called aqueous solutions. In this chapter we will study the nature of materials after
they are dissolved in water and various types of reactions that occur among these substances. You will see that the procedures developed in Chapter 3 to deal with chemical
reactions work very well for reactions that take place in aqueous solutions. To understand the types of reactions that occur in aqueous solutions, we must first explore the
types of species present. This requires an understanding of the nature of water.
4.1 Water, the Common Solvent
Water is one of the most important substances on the earth. It is essential for sustaining
the reactions that keep us alive, but it also affects our lives in many indirect ways.
Water helps moderate the earth’s temperature; it cools automobile engines, nuclear
power plants, and many industrial processes; it provides a means of transportation on
the earth’s surface and a medium for the growth of a myriad of creatures we use as
food; and much more.
One of the most valuable properties of water is its ability to dissolve many different
substances. For example, salt “disappears” when you sprinkle it into the water used to
cook vegetables, as does sugar when you add it to your iced tea. In each case the “disappearing” substance is obviously still present—you can taste it. What happens when
a solid dissolves? To understand this process, we need to consider the nature of water.
Liquid water consists of a collection of H2O molecules. An individual H2O molecule
is “bent” or V-shaped, with an HOOOH angle of approximately 105 degrees:
H
105°
O
H
The OOH bonds in the water molecule are covalent bonds formed by electron sharing between the oxygen and hydrogen atoms. However, the electrons of the bond are
not shared equally between these atoms. For reasons we will discuss in later chapters,
oxygen has a greater attraction for electrons than does hydrogen. If the electrons were
shared equally between the two atoms, both would be electrically neutral because, on
average, the number of electrons around each would equal the number of protons in
that nucleus. However, because the oxygen atom has a greater attraction for electrons,
the shared electrons tend to spend more time close to the oxygen than to either of the
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139
140
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
δ+
hydrogens. Thus the oxygen atom gains a slight excess of negative charge, and the hydrogen atoms become slightly positive. This is shown in Fig. 4.1, where d (delta) indicates a partial charge (less than one unit of charge). Because of this unequal charge
distribution, water is said to be a polar molecule. It is this polarity that gives water its
great ability to dissolve compounds.
A schematic of an ionic solid dissolving in water is shown in Fig. 4.2. Note that the
“positive ends” of the water molecules are attracted to the negatively charged anions
and that the “negative ends” are attracted to the positively charged cations. This process is called hydration. The hydration of its ions tends to cause a salt to “fall apart”
in the water, or to dissolve. The strong forces present among the positive and negative
ions of the solid are replaced by strong water–ion interactions.
It is very important to recognize that when ionic substances (salts) dissolve in water, they break up into the individual cations and anions. For instance, when ammonium nitrate (NH4NO3) dissolves in water, the resulting solution contains NH41 and
NO32 ions moving around independently. This process can be represented as
H
2δ−
O
105°
H
δ+
Figure 4.1 | (top) The water
molecule is polar. ­(bottom) A
space-filling model of the ­water
molecule.
H O(l)
NH4NO3 1s2 888n NH41 1aq2 1 NO32 1aq2
2
where (aq) designates that the ions are hydrated by unspecified numbers of water
molecules.
The solubility of ionic substances in water varies greatly. For example, sodium
chloride is quite soluble in water, whereas silver chloride (contains Ag1 and Cl2 ions)
is only very slightly soluble. The differences in the solubilities of ionic compounds in
water ­typically depend on the relative attractions of the ions for each other (these
forces hold the solid together) and the attractions of the ions for water molecules
(which cause the solid to disperse [dissolve] in water). Solubility is a complex topic
that we will explore in much more detail in Chapter 11. However, the most important
thing to remember at this point is that when an ionic solid does dissolve in water, the
ions become hydrated and are dispersed (move around independently).
Water also dissolves many nonionic substances. Ethanol (C2H5OH), for example, is
very soluble in water. Wine, beer, and mixed drinks are aqueous solutions of ethanol
and other substances. Why is ethanol so soluble in water? The answer lies in the
PowerLecture: Dissolution of a Solid
in a Liquid
+
Anion
–
–
+
–
+
–
+
–
+
–
+
+
δ+
–
+
+
–
–
–
+
+
+
2δ−
δ+
+
–
–
–
+
2δ−
δ+
+
Cation
δ+
–
Figure 4.2 | Polar water molecules interact with the positive and negative ions of a salt, assisting in the dissolving process.
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4.2
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
Figure 4.3 | (a) The ethanol
molecule contains a polar OOH bond
similar to those in the water molecule.
(b) The polar water molecule interacts
strongly with the polar OOH bond in
ethanol. This is a case of “like
dissolving like.”
δ−
H
O
H C
H
δ−
H
H
O
C H
H C
C H
H
H
H
δ+
Polar bond
a
δ−
H
δ+
141
δ+
H
O
H
δ+
b
structure of the alcohol molecules, which is shown in Fig. 4.3(a). The molecule contains a polar OOH bond like those in water, which makes it very compatible with
water. The interaction of water with ethanol is represented in Fig. 4.3(b).
Many substances do not dissolve in water. Pure water will not, for example, dissolve animal fat, because fat molecules are nonpolar and do not interact effectively
with polar water molecules. In general, polar and ionic substances are expected to be
more soluble in water than nonpolar substances. “Like dissolves like” is a useful rule
for predicting solubility. We will explore the basis for this generalization when we
discuss the details of solution formation in Chapter 11.
Critical Thinking
What if no ionic solids were soluble in water? How would this affect the way reactions
occur in aqueous solutions?
4.2 The Nature of Aqueous Solutions:
Strong and Weak Electrolytes
PowerLecture:
Electrolytes
Electrolyte Behavior
IBLG: See questions from “Aqueous
Solutions”
An electrolyte is a substance that when
dissolved in water produces a solution
that can conduct electricity.
As we discussed in Chapter 1, a solution is a homogeneous mixture. It is the same
throughout (the first sip of a cup of coffee is the same as the last), but its composition can
be varied by changing the amount of dissolved substances (you can make weak or
strong coffee). In this section we will consider what happens when a substance, the
solute, is dissolved in liquid water, the solvent.
One useful property for characterizing a solution is its electrical conductivity, its
ability to conduct an electric current. This characteristic can be checked conveniently
by using an apparatus like the ones shown in Fig. 4.4. If the solution in the container
conducts electricity, the bulb lights. Pure water is not an electrical conductor. However,
some aqueous solutions conduct current very efficiently, and the bulb shines very
brightly; these solutions contain strong electrolytes. Other solutions conduct only a
small current, and the bulb glows dimly; these solutions contain weak electrolytes.
Some solutions permit no current to flow, and the bulb remains unlit; these solutions
contain nonelectrolytes.
The basis for the conductivity properties of solutions was first correctly identified
by Svante Arrhenius (1859–1927), then a Swedish graduate student in physics, who
carried out research on the nature of solutions at the University of Uppsala in the early
1880s. Arrhenius came to believe that the conductivity of solutions arose from the
presence of ions, an idea that was at first scorned by the majority of the scientific establishment. However, in the late 1890s when atoms were found to contain charged
particles, the ionic theory suddenly made sense and became widely accepted.
As Arrhenius postulated, the extent to which a solution can conduct an electric
­current depends directly on the number of ions present. Some materials, such as
sodium chloride, readily produce ions in aqueous solution and thus are strong
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142
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Figure 4.4 | Electrical conductivity
Photos © Ken O’Donoghue © Cengage Learning
of aqueous solutions. The circuit will
be completed and will allow current
to flow only when there are charge
carriers (ions) in the solution. Note:
Water molecules are present but
not shown in these pictures. (a) A
hydrochloric acid solution, which is a
strong electrolyte, contains ions that
readily ­conduct the current and give a
brightly lit bulb. (b) An acetic acid
solution, which is a weak electrolyte,
contains only a few ions and does not
conduct as much ­current as a strong
electrolyte. The bulb is only dimly lit.
(c) A sucrose solution, which is a
nonelectrolyte, contains no ions and
does not conduct a current. The bulb
remains unlit.
+
−
+
−
PowerLecture: Conductiveness of
Aqueous Solutions
+
+
−
−
−
+
−
+
a
Many ions
b
c
Few ions
No ions
electrolytes. Other substances, such as acetic acid, produce relatively few ions when
dissolved in water and are weak electrolytes. A third class of materials, such as sugar,
form virtually no ions when dissolved in water and are nonelectrolytes.
Strong Electrolytes
Strong electrolytes are substances that are completely ionized when they are dissolved
in water, as represented in Fig. 4.4(a). We will consider several classes of strong electrolytes: (1) soluble salts, (2) strong acids, and (3) strong bases.
As shown in Fig. 4.2, a salt consists of an array of cations and anions that separate
and become hydrated when the salt dissolves. For example, when NaCl dissolves in
water, it produces hydrated Na1 and Cl2 ions in the solution (Fig. 4.5). Virtually no
NaCl(s)
dissolves
Figure 4.5 | When solid NaCl
Na+
Cl−
dissolves, the Na and Cl ions are
randomly dispersed in the water.
1
2
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4.2
+
–
H+
Cl−
– +
+ +
–
–
+ –
–
–
+
–
+
+
–
+ –
– +
+
Figure 4.6 | HCl(aq) is completely
ionized.
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
143
NaCl units are present. Thus NaCl is a strong electrolyte. It is important to recognize
that these aqueous solutions contain millions of water molecules that we will not include in our molecular-level drawings.
One of Arrhenius’s most important discoveries concerned the nature of acids. Acidity was first associated with the sour taste of citrus fruits. In fact, the word acid comes
directly from the Latin word acidus, meaning “sour.” The mineral acids sulfuric acid
(H2SO4) and nitric acid (HNO3), so named because they were originally obtained by
the treatment of minerals, were discovered around 1300.
Although acids were known for hundreds of years before the time of Arrhenius, no
one had recognized their essential nature. In his studies of solutions, Arrhenius found
that when the substances HCl, HNO3, and H2SO4 were dissolved in water, they behaved as strong electrolytes. He postulated that this was the result of ionization reactions in water, for example:
HO
HCl 888n
H1 1aq2 1 Cl2 1aq2
2
H2
O
HNO3 888n
H1 1aq2 1 NO32 1aq2
H2
The Arrhenius definition of an acid is
a substance that produces H1 ions in
­solution.
O
H2SO4 888n
H1 1aq2 1 HSO42 1aq2
Thus Arrhenius proposed that an acid is a substance that produces H1 ions (protons)
when it is dissolved in water.
We now understand that the polar nature of water plays a very important role in
causing acids to produce H1 in solution. In fact, it is most appropriate to represent the
“ionization” of an acid as follows:
HA 1aq2 1 H2O 1l2 h H3O1 1aq2 1 A2 1aq2
Strong electrolytes dissociate (ionize)
completely in aqueous solution.
Perchloric acid, HClO4(aq), is another
strong acid.
which emphasizes the important role of water in this process. We will have much more
to say about this process in Chapter 14.
Studies of conductivity show that when HCl, HNO3, and H2SO4 are placed in ­water,
virtually every molecule ionizes. These substances are strong electrolytes and are thus
called strong acids. All three are very important chemicals, and much more will be
said about them as we proceed. However, at this point the following facts are
important:
Sulfuric acid, nitric acid, and hydrochloric acid are aqueous solutions and should
be written in chemical equations as H2SO4(aq), HNO3(aq), and HCl(aq), respectively, although they often appear without the (aq) symbol.
A strong acid is one that completely dissociates into its ions. Thus, if 100 molecules
of HCl are dissolved in water, 100 H1 ions and 100 Cl2 ions are produced. Virtually no HCl molecules exist in aqueous solutions (Fig. 4.6).
–
OH−
+
Na+
Sulfuric acid is a special case. The formula H2SO4 indicates that this acid can
produce two H1 ions per molecule when dissolved in water. However, only the
first H1 ion is completely dissociated. The second H1 ion can be pulled off under
certain conditions, which we will discuss later. Thus an aqueous solution of H2SO4
contains mostly H1 ions and HSO42 ions.
– +
–
+
–
+
+ –
–
–
+
+
+
– + – +
+
–
–
Another important class of strong electrolytes consists of the strong bases,
soluble ionic compounds containing the hydroxide ion (OH2). When these compounds are dissolved in water, the cations and OH2 ions separate and move independently. Solutions containing bases have a bitter taste and a slippery feel. The
most common basic solutions are those produced when solid sodium hydroxide
(NaOH) or potassium hydroxide (KOH) is dissolved in water to produce ions, as
follows (Fig. 4.7):
Figure 4.7 | An aqueous solution of
sodium ­hydroxide.
Unless otherwise noted, all art on this page is © Cengage Learning 2014.
HO
NaOH 1s2 888n
Na1 1aq2 1 OH2 1aq2
2
H2
O
KOH 1s2 888n
K1 1aq2 1 OH2 1aq2
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144
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Chemical connections
Science is a human endeavor, subject
to human frailties and governed by
personalities, politics, and prejudices.
One of the best illustrations of the
often bumpy path of the advancement
of scientific knowledge is the story of
Swedish chemist Svante Arrhenius.
When Arrhenius began studies
toward his doctorate at the University
of Uppsala around 1880, he chose to
investigate the passage of electricity
through solutions, a mystery that had
baffled scientists for a century. The
first experiments had been done in the
1770s by Cavendish, who compared the
conductivity of salt solution with that
of rain water using his own physiologic
reaction to the electric shocks he
received! Arrhenius had an array of
instruments to measure electric
current, but the process of carefully
weighing, measuring, and recording
data from a multitude of experiments
was a tedious one.
After his long series of experiments
was performed, ­Arrhenius quit his
laboratory bench and returned to his
country home to try to formulate a
model that could account for his data.
He wrote, “I got the idea in the night of
the 17th of May in the year 1883, and I
could not sleep that night until I had
worked through the whole problem.”
His idea was that ions were responsible for conducting electricity
through a solution.
Back at Uppsala, Arrhenius took his
doctoral dissertation containing the
new theory to his advisor, Professor
Cleve, an eminent chemist and the
discoverer of the elements holmium
and thulium. Cleve’s uninterested
response was what Arrhenius had
expected. It was in keeping with
Cleve’s resistance to new ideas—he
had not even accepted Mendeleev’s
periodic table, introduced 10 years
earlier.
It is a long-standing custom that
before a doctoral degree is granted,
the dissertation must be defended
before a panel of professors. Although
this procedure is still followed at most
universities today, the problems are
usually worked out in private with the
evaluating professors before the actual
defense. However, when Arrhenius did
it, the dissertation defense was an
open debate, which could be rancorous and humiliating. Knowing that
it would be unwise to antagonize his
professors, Arrhenius downplayed his
convictions about his new theory as he
defended his dissertation. His
diplomacy paid off: He was awarded
his degree, albeit reluctantly, because
the professors still did not believe his
model and considered him to be a
marginal scientist, at best.
Such a setback could have ended
his scientific career, but Arrhenius was
a crusader; he was determined to see
his theory triumph. He promptly
embarked on a political campaign,
enlisting the aid of several prominent
scientists, to get his theory accepted.
Royal Swedish Academy of Sciences
Arrhenius: A Man with Solutions
Svante August Arrhenius.
Ultimately, the ionic theory
triumphed. Arrhenius’s fame spread,
and honors were heaped on him,
culminating in the Nobel Prize in
chemistry in 1903. Not one to rest on
his laurels, Arrhenius turned to new
fields, including astronomy; he
formulated a new theory that the solar
system may have come into being
through the collision of stars. His
exceptional versatility led him to study
the use of serums to fight disease,
energy resources and conservation,
and the origin of life.
Additional insight on Arrhenius and his
scientific career can be ­obtained from
his address on receiving the Willard
Gibbs Award. See Journal of the American
Chemical Society 36 (1912): 353.
Weak Electrolytes
Weak electrolytes dissociate (ionize) only
to a small extent in aqueous solution.
Weak electrolytes are substances that exhibit a small degree of ionization in water.
That is, they produce relatively few ions when dissolved in water, as shown in
Fig. 4.4(b). The most common weak electrolytes are weak acids and weak bases.
The main acidic component of vinegar is acetic acid (HC2H3O2). The formula is
­written to indicate that acetic acid has two chemically distinct types of hydrogen
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4.3
The Composition of Solutions
145
atoms. Formulas for acids are often written with the acidic hydrogen atom or atoms
(any that will produce H1 ions in solution) listed first. If any nonacidic hydrogens are
present, they are written later in the formula. Thus the formula HC2H3O2 indicates one
acidic and three nonacidic hydrogen atoms. The dissociation reaction for acetic acid in
water can be written as follows:
Hydrogen
Oxygen
Carbon
–
HC2H3O2 1aq2 1 H2O 1l2 m H3O1 1aq2 1 C2H3O22 1aq2
+
Figure 4.8 | Acetic acid (HC2H3O2)
exists in water mostly as undissociated
molecules. Only a small percentage of
the molecules are ionized.
Acetic acid is very different from the strong acids because only about 1% of its molecules dissociate in aqueous solutions at typical concentrations. For example, in a solution containing 0.1 mole of HC2H3O2 per liter, for every 100 molecules of HC2H3O2
originally dissolved in water, approximately 99 molecules of HC2H3O2 remain intact
(Fig. 4.8). That is, only one molecule out of every 100 dissociates (to produce one H1
ion and one C2H3O22 ion). The double arrow indicates the reaction can occur in ­either
direction.
Because acetic acid is a weak electrolyte, it is called a weak acid. Any acid, such
as acetic acid, that dissociates (ionizes) only to a slight extent in aqueous solutions is
called a weak acid. We will explore the subject of weak acids in detail in Chapter 14.
The most common weak base is ammonia (NH3). When ammonia is dissolved in
­water, it reacts as follows:
NH3 1aq2 1 H2O 1l2 m NH41 1aq2 1 OH2 1aq2
The solution is basic because OH2 ions are produced. Ammonia is called a weak base
because the resulting solution is a weak electrolyte; that is, very few ions are formed.
In fact, in a solution containing 0.1 mole of NH3 per liter, for every 100 molecules of
NH3 originally dissolved, only one NH41 ion and one OH2 ion are produced; 99 molecules of NH3 remain unreacted (Fig. 4.9). The double arrow indicates the reaction can
occur in either direction.
Hydrogen
Oxygen
Nonelectrolytes
Nitrogen
+
–
Figure 4.9 | The reaction of NH3 in
water.
Nonelectrolytes are substances that dissolve in water but do not produce any ions, as
shown in Fig. 4.4(c). An example of a nonelectrolyte is ethanol (see Fig. 4.3 for the
structural formula). When ethanol dissolves, entire C2H5OH molecules are dispersed
in the water. Since the molecules do not break up into ions, the resulting solution does
not conduct an electric current. Another common nonelectrolyte is table sugar (sucrose, C12H22O11), which is very soluble in water but which produces no ions when it
dissolves. The sucrose molecules remain intact.
4.3 The Composition of Solutions
Chemical reactions often take place when two solutions are mixed. To perform stoichiometric calculations in such cases, we must know two things: (1) the nature of
the reaction, which depends on the exact forms the chemicals take when dissolved,
and (2) the amounts of chemicals present in the solutions, usually expressed as
concentrations.
The concentration of a solution can be described in many different ways, as we will
see in Chapter 11. At this point we will consider only the most commonly used expression of concentration, molarity (M), which is defined as moles of solute per volume of
solution in liters:
M 5 molarity 5
moles of solute
liters of solution
A solution that is 1.0 molar (written as 1.0 M) contains 1.0 mole of solute per liter of
solution.
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146
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Interactive
Example 4.1
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Calculation of Molarity I
Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in
enough water to make 1.50 L of solution.
Solution
Where are we going?
To find the molarity of NaOH solution
What do we know?
❯ 11.5 g NaOH
❯ 1.50 L solution
What information do we need to find molarity?
❯ Moles solute
mol solute
❯ Molarity 5
L solution
How do we get there?
What are the moles of NaOH (40.00 g/mol)?
11.5 g NaOH 3
1 mol NaOH
5 0.288 mol NaOH
40.00 g NaOH
What is the molarity of the solution?
j
Molarity 5
mol solute
0.288 mol NaOH
5
5 0.192 M NaOH
L solution
1.50 L solution
Reality Check | The units are correct for molarity.
See Exercises 4.27 and 4.28
Interactive
Example 4.2
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Calculation of Molarity II
Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in
enough water to make 26.8 mL of solution.
Solution
Where are we going?
To find the molarity of HCl solution
What do we know?
❯ 1.56 g HCl
❯ 26.8 mL solution
What information do we need to find molarity?
❯ Moles solute
mol solute
❯ Molarity 5
L solution
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4.3
The Composition of Solutions
147
How do we get there?
What are the moles of HCl (36.46 g/mol)?
1.56 g HCl 3
1 mol HCl
5 4.28 3 1022 mol HCl
36.46 g HCl
What is the volume of solution (in liters)?
26.8 mL 3
1L
5 2.68 3 1022 L
1000 mL
What is the molarity of the solution?
j
Molarity 5
4.28 3 1022 mol HCl
5 1.60 M HCl
2.68 3 1022 L solution
Reality Check | The units are correct for molarity.
See Exercises 4.27 and 4.28
Interactive
Example 4.3
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Concentration of Ions I
Give the concentration of each type of ion in the following solutions:
a. 0.50 M Co(NO3)2
b. 1 M Fe(ClO4)3
Solution
Where are we going?
To find the molarity of each ion in the solution
What do we know?
❯ 0.50 M Co(NO3)2
❯ 1 M Fe(ClO4)3
What information do we need to find the molarity of each ion?
❯ Moles of each ion
How do we get there?
For Co(NO3)2
What is the balanced equation for dissolving the ions?
HO
Co 1NO32 2 1s2 888n
Co21 1aq2 1 2NO32 1aq2
2
What is the molarity for each ion?
j
Photo © Cengage Learning. All rights reserved.
j
Co21 1 3 0.50 M 5 0.50 M Co21
NO32 2 3 0.50 M 5 1.0 M NO32
For Fe(ClO4)3
What is the balanced equation for dissolving the ions?
HO
Fe 1ClO42 3 1s2 888n
Fe31 1aq2 1 3ClO42 1aq2
2
What is the molarity for each ion?
An aqueous solution of Co(NO3)2.
j
j
Fe31
ClO42
1 3 1 M 5 1 M Fe31
3 3 1 M 5 3 M ClO42
See Exercises 4.29 and 4.30
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148
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Often chemists need to determine the number of moles of solute present in a given
volume of a solution of known molarity. The procedure for doing this is easily derived
from the definition of molarity. If we multiply the molarity of a solution by the volume (in liters) of a particular sample of the solution, we get the moles of solute present in that sample:
M5
moles of solute
liters of solution
Liters of solution 3 molarity 5 liters of solution 3
moles of solute
5 moles of solute
liters of solution
This procedure is demonstrated in Examples 4.4 and 4.5.
Interactive
Example 4.4
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in OWL.
Concentration of Ions II
Calculate the number of moles of Cl2 ions in 1.75 L of 1.0 3 1023 M ZnCl2.
Solution
Where are we going?
To find the moles of Cl2 ion in the solution
What do we know?
❯ 1.0 3 1023 M ZnCl2
❯ 1.75 L
What information do we need to find moles of Cl2?
❯ Balanced equation for dissolving ZnCl2
How do we get there?
What is the balanced equation for dissolving the ions?
HO
ZnCl2 1s2 888n
Zn21 1aq2 1 2Cl2 1aq2
2
What is the molarity of Cl2 ion in the solution?
2 3 11.0 3 1023 M2 5 2.0 3 1023 M Cl2
How many moles of Cl2?
j
1.75 L solution 3 2.0 3 1023 M Cl2 5 1.75 L solution 3
2.0 3 1023 mol Cl2
L solution
5 3.5 3 1023 mol Cl2
See Exercise 4.31
Interactive
Example 4.5
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in OWL.
Concentration and Volume
Typical blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg
of NaCl?
Solution
Where are we going?
To find the volume of blood containing 1.0 mg of NaCl
What do we know?
❯ 0.14 M NaCl
❯ 1.0 mg NaCl
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4.3
The Composition of Solutions
149
What information do we need to find volume of blood containing 1.0 mg of NaCl?
❯ Moles of NaCl (in 1.0 mg)
How do we get there?
What are the moles of NaCl (58.44 g/mol)?
1.0 mg NaCl 3
1 g NaCl
1 mol NaCl
3
5 1.7 3 1025 mol NaCl
1000 mg NaCl
58.44 g NaCl
hat volume of 0.14 M NaCl contains 1.0 mg (1.7 3 1025 mole) of NaCl?
W
There is some volume, call it V, that when multiplied by the molarity of this solution
will yield 1.7 3 1025 mole of NaCl. That is,
V3
0.14 mol NaCl
5 1.7 3 1025 mol NaCl
L solution
We want to solve for the volume:
V5
j
1.7 3 1025 mol NaCl
5 1.2 3 1024 L solution
0.14 mol NaCl
L solution
Thus 0.12 mL of blood contains 1.7 3 1025 mole of NaCl or 1.0 mg of NaCl.
See Exercises 4.33 and 4.34
A standard solution is a solution whose concentration is accurately known. Standard solutions, often used in chemical analysis, can be prepared as shown in Fig. 4.10
and in Example 4.6.
Interactive
Example 4.6
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in OWL.
Solutions of Known Concentration
To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of an aqueous
0.200-M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 must be
weighed out to make this solution?
Solution
Where are we going?
To find the mass of K2Cr2O7 required for the solution
Wash
bottle
Figure 4.10 | Steps involved in the
preparation of a standard aqueous
solution. (a) Put a weighed amount
of a substance (the solute) into the
volumetric flask, and add a small
quantity of water. (b) Dissolve the
solid in the water by gently swirling
the flask (with the stopper in place).
(c) Add more water (with gentle
swirling) until the level of the solution
just reaches the mark etched on the
neck of the flask. Then mix the
solution thoroughly by inverting the
flask several times.
Volume marker
(calibration mark)
Weighed
amount
of solute
a
b
c
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150
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
What do we know?
❯ 1.00 L of 0.200 M K2Cr2O7 is required
What information do we need to find the mass of K2Cr2O7?
❯ Moles of K2Cr2O7 in the required solution
How do we get there?
What are the moles of K2Cr2O7 required?
M 3 V 5 mol
1.00 L solution 3
0.200 mol K2Cr2O7
5 0.200 mol K2Cr2O7
L solution
What mass of K2Cr2O7 is required for the solution?
0.200 mol K2Cr2O7 3
j
294.20 g K2Cr2O7
5 58.8 g K2Cr2O7
mol K2Cr2O7
To make 1.00 L of 0.200 M K2Cr2O7, the chemist must weigh out 58.8 g K2Cr2O7,
transfer it to a 1.00-L volumetric flask, and add distilled water to the mark on the
flask.
See Exercises 4.35a,c and 4.36c,e
PowerLecture: Dilution
Dilution with water does not alter the
numbers of moles of solute present.
Dilution
To save time and space in the laboratory, routinely used solutions are often purchased
or prepared in concentrated form (called stock solutions). Water is then added to
achieve the molarity desired for a particular solution. This process is called dilution.
For example, the common acids are purchased as concentrated solutions and diluted as
needed. A typical dilution calculation involves determining how much water must be
added to an amount of stock solution to achieve a solution of the desired concentration.
The key to doing these calculations is to remember that
Moles of solute after dilution 5 moles of solute before dilution
because only water (no solute) is added to accomplish the dilution.
For example, suppose we need to prepare 500. mL of 1.00 M acetic acid (HC2H3O2)
from a 17.4-M stock solution of acetic acid. What volume of the stock solution is required? The first step is to determine the number of moles of acetic acid in the final
solution by multiplying the volume by the molarity (remembering that the volume
must be changed to liters):
500. mL solution 3
1 L solution
1.00 mol HC2H3O2
3
5 0.500 mol HC2H3O2
1000 mL solution
L solution
Thus we need to use a volume of 17.4 M acetic acid that contains 0.500 mole of
HC2H3O2. That is,
V3
17.4 mol HC2H3O2
5 0.500 mol HC2H3O2
L solution
Solving for V gives
V5
0.500 mol HC2H3O2
5 0.0287 L or 28.7 mL solution
17.4 mol HC2H3O2
L solution
Thus to make 500 mL of a 1.00-M acetic acid solution, we can take 28.7 mL of
17.4 M acetic acid and dilute it to a total volume of 500 mL with distilled water.
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4.3
The Composition of Solutions
151
A dilution procedure typically involves two types of glassware: a pipet and a
v­ olumetric flask. A pipet is a device for accurately measuring and transferring a
given volume of solution. There are two common types of pipets: volumetric (or
transfer) pipets and measuring pipets (Fig. 4.11). Volumetric pipets come in specific
sizes, such as 5 mL, 10 mL, 25 mL, and so on. Measuring pipets are used to measure
­volumes for which a ­volumetric pipet is not available. For example, we would use a
measuring pipet as shown in Fig. 4.12 on page 153 to deliver 28.7 mL of 17.4 M
acetic acid into a 500-mL volumetric flask and then add water to the mark to perform
the dilution described above.
Interactive
Example 4.7
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in OWL.
Calibration
mark
Concentration and Volume
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10-M H2SO4
­solution?
Solution
Where are we going?
To find the volume of H2SO4 required to prepare the solution
What do we know?
❯ 1.5 L of 0.10 M H2SO4 is required
❯ We have 16 M H2SO4
What information do we need to find the volume of H2SO4?
❯ Moles of H2SO4 in the required solution
How do we get there?
What are the moles of H2SO4 required?
M 3 V 5 mol
1.5 L solution 3
a
b
What volume of 16 M H2SO4 contains 0.15 mole of H2SO4?
V3
Figure 4.11 | (a) A measuring pipet
is graduated and can be used to
measure various volumes of liquid
accurately. (b) A volumetric (transfer)
pipet is designed to measure one
volume accurately. When filled to the
mark, it delivers the volume indicated
on the pipet.
0.10 mol H2SO4
5 0.15 mol H2SO4
L solution
16 mol H2SO4
5 0.15 mol H2SO4
L solution
Solving for V gives
V5
0.15 mol H2SO4
5 9.4 3 1023 L or 9.4 mL solution
16 mol H2SO4
1 L solution
To make 1.5 L of 0.10 M H2SO4 using 16 M H2SO4, we must take 9.4 mL of the
concentrated acid and dilute it with water to 1.5 L. The correct way to do this is to
add the 9.4 mL of acid to about 1 L of distilled water and then dilute to 1.5 L by
adding more water.
j
When diluting an acid, “Do what you
oughta, always add acid to water.”
See Exercises 4.35b,d, and 4.36a,b,d
As noted earlier, the central idea in performing the calculations associated with dilutions is to recognize that the moles of solute are not changed by the dilution. Another
way to express this condition is by the following equation:
M1V1 5 M2V2
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152
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Chemical connections
One of the major impacts of modern
technology is to make things smaller.
The best example is the computer.
Calculations that 30 years ago required
a machine the size of a large room now
can be carried out on a hand-held
calculator. This tendency toward
miniaturization is also having a major
impact on the science of chemical
analysis. Using the techniques of
computer chip makers, researchers are
now constructing minuscule laboratories on the surface of a tiny chip made
of silicon, glass, or plastic (see photo).
Instead of electrons, 1026 to 1029 L of
liquids moves between reaction
chambers on the chip through tiny
capillaries. The chips typically contain
no moving parts. Instead of conventional pumps, the chip-based laboratories use voltage differences to move
liquids that contain ions from one
reaction chamber to another.
Microchip laboratories have many
advantages. They require only tiny
amounts of sample. This is especially
advantageous for expensive, difficultto-prepare materials or in cases such
as criminal investigations, where only
small amounts of evidence may exist.
The chip laboratories also minimize
contamination because they represent
a “closed system” once the material
has been introduced to the chip. In
addition, the chips can be made to
be disposable to prevent crosscontamination of different samples.
The chip laboratories present some
difficulties not found in macroscopic
laboratories. The main problem
concerns the large surface area of the
capillaries and reaction chambers
relative to the sample volume.
Molecules or biological cells in the
sample solution encounter so much
“wall” that they may undergo
unwanted reactions with the wall
materials. Glass seems to present the
least of these problems, and the walls
of silicon chip laboratories can be
protected by formation of relatively
inert silicon dioxide. Because plastic is
inexpensive, it seems a good choice for
disposable chips, but plastic also is the
most reactive with the samples and the
least durable of the available materials.
PerkinElmer, Inc. is working toward
creating a miniature chemistry
laboratory about the size of a toaster
that can be used with “plug-in”
chip-based laboratories. Various chips
would be furnished with the unit that
would be appropriate for different
types of analyses. The entire unit
would be connected to a computer to
collect and analyze the data. There is
even the possibility that these
“laboratories” could be used in the
© 2012 PerkinElmer, Inc. All rights reserved. Printed with permission.
Tiny Laboratories
Plastic chips such as this one made by
PerkinElmer, Inc. are being used to
perform laboratory procedures
traditionally done with test tubes.
home to perform analyses such as
blood sugar and blood cholesterol and
to check for the presence of bacteria
such as E. coli and many others. This
would revolutionize the healthcare
industry.
Adapted from “The Incredible Shrinking
Laboratory,” by Corinna Wu, as appeared
in Science News, Vol. 154, August 15, 1998,
p. 104.
where M1 and V1 represent the molarity and volume of the original solution (before
dilution) and M2 and V2 represent the molarity and volume of the diluted solution. This
equation makes sense because
M1 3 V1 5 mol solute before dilution
5 mol solute after dilution 5 M2 3 V2
Repeat Example 4.7 using the equation M1V1 5 M2V2. Note that in doing so,
M1 5 16 M
M2 5 0.10 M
V2 5 1.5 L
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4.5
Precipitation Reactions
153
Figure 4.12 | (a) A measuring pipet
is used to transfer 28.7 mL of 17.4 M
acetic acid solution to a volumetric
flask. (b) Water is added to the flask to
the calibration mark. (c) The resulting
solution is 1.00 M acetic acid.
500 mL
a
b
c
and V1 is the unknown quantity sought. The equation M1V1 5 M2V2 always holds for a
dilu­tion. This equation will be easy for you to remember if you understand where it
comes from.
4.4 Types of Chemical Reactions
Experiment 26: Classification of
Chemical Reactions
Although we have considered many reactions so far in this text, we have examined
only a tiny fraction of the millions of possible chemical reactions. To make sense of all
these reactions, we need some system for grouping reactions into classes. Although
there are many different ways to do this, we will use the system most commonly used
by practicing chemists:
Types of Solution Reactions
❯ Precipitation reactions
❯ Acid–base reactions
❯ Oxidation–reduction reactions
Virtually all reactions can be put into one of these classes. We will define and illustrate
each type in the following sections.
4.5 Precipitation Reactions
IBLG: See questions from
“Precipitation Reactions”
When two solutions are mixed, an insoluble substance sometimes forms; that is, a
solid forms and separates from the solution. Such a reaction is called a precipitation
reaction, and the solid that forms is called a precipitate. For example, a precipitation
reaction occurs when an aqueous solution of potassium chromate, K2CrO4(aq), which
is yellow, is added to a colorless aqueous solution containing barium nitrate,
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154
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Ba(NO3)2(aq). As shown in Fig. 4.13, when these solutions are mixed, a yellow solid
forms. What is the equation that describes this chemical change? To write the equation,
we must know the identities of the reactants and products. The reactants have already
been described: K2CrO4(aq) and Ba(NO3)2(aq). Is there some way we can predict the
identities of the products? In particular, what is the yellow solid?
The best way to predict the identity of this solid is to think carefully about what
products are possible. To do this, we need to know what species are present in the solution ­after the two reactant solutions are mixed. First, let’s think about the nature of each
reactant solution. The designation Ba(NO3)2(aq) means that barium nitrate (a white
solid) has been dissolved in water. Notice that barium nitrate contains the Ba21 and
NO32 ions. Remember: In virtually every case, when a solid containing ions dissolves
in water, the ions separate and move around independently. That is, Ba(NO3)2(aq) does
not contain Ba(NO3)2 units; it contains separated Ba21 and NO32 ions [Fig. 4.14(a)].
Similarly, since solid potassium chromate contains the K1 and CrO422 ions, an
aqueous solution of potassium chromate (which is prepared by dissolving solid K2CrO4
in ­water) contains these separated ions [Fig. 4.14(b)].
We can represent the mixing of K2CrO4(aq) and Ba(NO3)2(aq) in two ways. First,
we can write
A precipitation reaction also can be called
a double displacement reaction.
PowerLecture: Precipitation Reactions
The quantitative aspects of precipitation
reactions are covered in Chapter 15.
When ionic compounds dissolve in
water, the resulting solution contains the
separated ions.
K2CrO4 1aq2 1 Ba 1NO32 2 1aq2 h products













2K1 1aq2 1 CrO422 1aq2 1 Ba21 1aq2 1 2NO32 1aq2 h products













Richard Megna/Fundamental Photographs © Cengage Learning
However, a much more accurate representation is
The ions in
K2CrO4(aq)
The ions in
Ba(NO3)2(aq)
Thus the mixed solution contains the ions:
K1
CrO422
Ba21
NO32
as illustrated in Fig. 4.15(a).
How can some or all of these ions combine to form a yellow solid? This is not an
easy question to answer. In fact, predicting the products of a chemical reaction is one
of the hardest things a beginning chemistry student is asked to do. Even an experienced chemist, when confronted with a new reaction, is often not sure what will happen. The chemist tries to think of the various possibilities, considers the likelihood of
Photos © Cengage Learning. All rights reserved.
Figure 4.13 | When yellow aqueous
potassium ­chromate is added to a
colorless barium nitrate solution,
yellow barium chromate precipitates.
K+
Ba2+
NO3−
Figure 4.14 | Reactant solutions:
(a) Ba(NO3)2(aq) and (b) K2CrO4(aq).
a
b
CrO42−
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4.5
155
K+
Ba2+
NO3−
© Cengage Learning
Figure 4.15 | The reaction of
K2CrO4(aq) and Ba(NO3)2(aq). (a) The
molecular-level “picture” of the mixed
solution before any ­reaction has
occurred. (b) The molecular-level
“picture” of the solution after the
reaction has occurred to form
BaCrO4(s). Note: BaCrO4(s) is not
molecular. It actually contains Ba21
and CrO422 ions packed together in a
lattice. (c) A photo of the solution
after the reaction has occurred,
showing the solid BaCrO4 on the
bottom.
Precipitation Reactions
CrO42−
a
b
c
each possibility, and then makes a prediction (an educated guess). Only after identifying
each product experimentally is the chemist sure what reaction has taken place. However, an educated guess is very useful because it provides a place to start. It tells us
what kinds of products we are most likely to find. We already know some things that
will help us predict the products of the above reaction.
1. When ions form a solid compound, the compound must have a zero net charge.
Thus the products of this reaction must contain both anions and cations. For example, K1 and Ba21 could not combine to form the solid, nor could CrO422 and
NO32.
2. Most ionic materials contain only two types of ions: one type of cation and one
type of anion (for example, NaCl, KOH, Na2SO4, K2CrO4, Co(NO3)2, NH4Cl,
Na2CO3).
The possible combinations of a given cation and a given anion from the list of ions
K1, CrO422, Ba21, and NO32 are
K2CrO4
KNO3
BaCrO4
Ba 1NO32 2
Which of these possibilities is most likely to represent the yellow solid? We know it’s
not K2CrO4 or Ba(NO3)2. They are the reactants. They were present (dissolved) in the
separate solutions that were mixed. The only real possibilities for the solid that
formed are
KNO3 and BaCrO4
To decide which of these most likely represents the yellow solid, we need more facts.
An experienced chemist knows that the K1 ion and the NO32 ion are both colorless.
Thus, if the solid is KNO3, it should be white, not yellow. On the other hand, the
CrO422 ion is yellow (note in Fig. 4.14 that K2CrO4(aq) is yellow). Thus the yellow
solid is almost certainly BaCrO4. Further tests show that this is the case.
So far we have determined that one product of the reaction between K2CrO4(aq) and
Ba(NO3)2(aq) is BaCrO4(s), but what happened to the K1 and NO32 ions? The answer
is that these ions are left dissolved in the solution; KNO3 does not form a solid when the
K1 and NO32 ions are present in this much water. In other words, if we took solid
KNO3 and put it in the same quantity of water as is present in the mixed solution, it
would dissolve. Thus, when we mix K2CrO4(aq) and Ba(NO3)2(aq), BaCrO4(s) forms,
but KNO3 is left behind in solution (we write it as KNO3(aq)). Thus the overall equation
for this precipitation reaction using the formulas of the reactants and products is
K2CrO4 1aq2 1 Ba 1NO32 2 1aq2 h BaCrO4 1s2 1 2KNO3 1aq2
As long as water is present, the KNO3 remains dissolved as separated ions. (See Fig.
4.15 to help visualize what is happening in this reaction. Note the solid BaCrO4 on the
bottom of the container, while the K1 and NO32 ions remain dispersed in the solution.)
If we ­removed the solid BaCrO4 and then evaporated the water, white solid KNO3
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156
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Figure 4.16 | Precipitation of silver
chloride by mixing solutions of silver
nitrate and potassium chloride. The
K1 and NO32 ions remain in solution.
AgNO3 1aq2 1 KCl 1aq2 h unknown white solid
Remembering that when ionic substances dissolve in water, the ions separate, we can
write
In silver
nitrate
solution
In potassium
chloride
solution














Ag1, NO32 1 K1, Cl2 h Ag1, NO32, K1, Cl2 h white solid





Photo © Cengage Learning. All rights reserved.
would be ­obtained; the K1 and NO32 ions would assemble themselves into solid KNO3
when the water is removed.
Now let’s consider another example. When an aqueous solution of silver nitrate
is added to an aqueous solution of potassium chloride, a white precipitate forms
(Fig. 4.16). We can represent what we know so far as
Combined solution,
before reaction
Since we know the white solid must contain both positive and negative ions, the possible compounds that can be assembled from this collection of ions are
AgNO3
KCl
AgCl
KNO3
Since AgNO3 and KCl are the substances dissolved in the two reactant solutions, we
know that they do not represent the white solid product. Therefore, the only real possibilities are
AgCl and KNO3
From the first example considered, we know that KNO3 is quite soluble in water. Thus
solid KNO3 will not form when the reactant solids are mixed. The product must be
AgCl(s) (which can be proved by experiment to be true). The overall equation for the
reaction now can be written
AgNO3 1aq2 1 KCl 1aq2 h AgCl 1s2 1 KNO3 1aq2
Figure 4.17 shows the result of mixing aqueous solutions of AgNO3 and KCl, including a microscopic visualization of the reaction.
Notice that in these two examples we had to apply both concepts (solids must have
a zero net charge) and facts (KNO3 is very soluble in water, CrO422 is yellow, and so
on). Doing chemistry requires both understanding ideas and remembering key information. Predicting the identity of the solid product in a precipitation reaction requires
knowledge of the solubilities of common ionic substances. As an aid in predicting the
products of precipitation reactions, some simple solubility rules are given in Table 4.1.
You should memorize these rules.
Table 4.1 | Simple Rules for the Solubility of Salts in Water
1. Most nitrate (NO32) salts are soluble.
2. Most salts containing the alkali metal ions (Li1, Na1, K1, Cs1, Rb1) and the ammonium ion
(NH41) are soluble.
3. Most chloride, bromide, and iodide salts are soluble. Notable exceptions are salts containing the ions Ag1, Pb21, and Hg221.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, Hg2SO4, and
CaSO4.
5. Most hydroxides are only slightly soluble. The important soluble hydroxides are NaOH and
KOH. The compounds Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble.
6. Most sulfide (S22), carbonate (CO322), chromate (CrO422), and phosphate (PO432) salts are
only slightly soluble, except for those containing the cations in Rule 2.
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Precipitation Reactions
Photo
FPO
Solutions are mixed
Cl–
K+
Ag+
157
Photos © Cengage Learning. All rights reserved.
4.5
NO3–
Ag+
Figure 4.17 | Photos and accompanying molecular-level representations illustrating the reaction of KCl(aq) with AgNO3(aq) to
form AgCl(s). Note that it is not possible to have a photo of the mixed solution before the reaction occurs, because it is an imaginary step that we use to help visualize the reaction. Actually, the reaction occurs immediately when the two solutions are mixed.
PowerLecture: Reactions of Silver(I)
The phrase slightly soluble used in the solubility rules in Table 4.1 means that the
tiny amount of solid that dissolves is not noticeable. The solid appears to be insoluble
to the naked eye. Thus the terms insoluble and slightly soluble are often used
interchangeably.
Note that the information in Table 4.1 allows us to predict that AgCl is the white
solid formed when solutions of AgNO3 and KCl are mixed. Rules 1 and 2 indicate that
KNO3 is soluble, and Rule 3 states that AgCl is insoluble.
When solutions containing ionic substances are mixed, it will be helpful in determining the products if you think in terms of ion interchange. For example, in the preceding discussion we considered the results of mixing AgNO3(aq) and KCl(aq). In determining the products, we took the cation from one reactant and combined it with the
anion of the other reactant:
Ag1
1
NO32
1
K1
1
Cl2
h
r
p
Possible
solid
products
To begin, focus on the ions in solution
before any reaction occurs.
The solubility rules in Table 4.1 allow us to predict whether either product forms as a
solid.
The key to dealing with the chemistry of an aqueous solution is first to focus on the
actual components of the solution before any reaction occurs and then to figure out
how these components will react with each other. Example 4.8 illustrates this process
for three different reactions.
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158
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Interactive
Example 4.8
Predicting Reaction Products
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in OWL.
Using the solubility rules in Table 4.1, predict what will happen when the following
pairs of solutions are mixed.
a. KNO3(aq) and BaCl2(aq)
b. Na2SO4(aq) and Pb(NO3)2(aq)
c. KOH(aq) and Fe(NO3)3(aq)
Solution
a. The formula KNO3(aq) represents an aqueous solution obtained by dissolving
solid KNO3 in water to form a solution containing the hydrated ions K1(aq) and
NO32(aq). Likewise, BaCl2(aq) represents a solution formed by dissolving solid
BaCl2 in water to produce Ba21(aq) and Cl2(aq). When these two solutions are
mixed, the resulting solution contains the ions K1, NO32, Ba21, and Cl2. All ions
are hydrated, but the (aq) is omitted for simplicity. To look for possible solid
products, combine the cation from one reactant with the anion from the other:
K1
1
NO32
1
Ba21
1
Cl2
h
Photo © Cengage Learning. All rights reserved.
r
p
Possible
solid
products
Solid Fe(OH)3 forms when aqueous KOH
and Fe(NO3)3 are mixed.
Note from Table 4.1 that the rules predict that both KCl and Ba(NO3)2 are soluble in
water. Thus no precipitate forms when KNO3(aq) and BaCl2(aq) are mixed. All the
ions remain dissolved in solution. No chemical reaction occurs.
b. Using the same procedures as in part a, we find that the ions present in the
combined solution before any reaction occurs are Na1, SO422, Pb21, and NO32.
The possible salts that could form precipitates are
Na1
1
SO422
1
Pb21
1
NO32
h
The compound NaNO3 is soluble, but PbSO4 is insoluble (see Rule 4 in Table
4.1). When these solutions are mixed, PbSO4 will precipitate from the solution.
The balanced equation is
Na2SO4 1aq2 1 Pb 1NO32 2 1aq2 h PbSO4 1s2 1 2NaNO3 1aq2
c. The combined solution (before any reaction occurs) contains the ions K1,
OH2, Fe31, and NO32. The salts that might precipitate are KNO3 and Fe(OH)3.
The solubility rules in Table 4.1 indicate that both K1 and NO32 salts are
soluble. However, Fe(OH)3 is only slightly soluble (Rule 5) and hence will
precipitate. The balanced equation is
3KOH 1aq2 1 Fe 1NO32 3 1aq2 h Fe 1OH2 3 1s2 1 3KNO3 1aq2
See Exercises 4.45 and 4.46
Experiment 26: Classification
of Chemical Reactions
4.6 Describing Reactions in Solution
In this section we will consider the types of equations used to represent reactions in
solution. For example, when we mix aqueous potassium chromate with aqueous barium nitrate, a reaction occurs to form a precipitate (BaCrO4) and dissolved potassium
nitrate. So far we have written the overall or formula equation for this reaction:
K2CrO4 1aq2 1 Ba 1NO32 2 1aq2 h BaCrO4 1s2 1 2KNO3 1aq2
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4.6
Describing Reactions in Solution
159
Although the formula equation shows the reactants and products of the reaction, it
does not give a correct picture of what actually occurs in solution. As we have seen,
aqueous solutions of potassium chromate, barium nitrate, and potassium nitrate contain individual ions, not collections of ions, as implied by the formula equation. Thus
the complete ionic equation
A strong electrolyte is a substance that
completely breaks apart into ions when
dissolved in water.
Net ionic equations include only those
components that undergo changes in the
reaction.
2K1 1aq2 1 CrO422 1aq2 1 Ba21 1aq2 1 2NO32 1aq2 h
BaCrO4 1s2 1 2K1 1aq2 1 2NO32 1aq2
better represents the actual forms of the reactants and products in solution. In a complete ionic equation, all substances that are strong electrolytes are represented as ions.
The complete ionic equation reveals that only some of the ions participate in the
reaction. The K1 and NO32 ions are present in solution both before and after the reaction. The ions that do not participate directly in the reaction are called spectator ions.
The ions that participate in this reaction are the Ba21 and CrO422 ions, which combine
to form solid BaCrO4:
Ba21 1aq2 1 CrO422 1aq2 h BaCrO4 1s2
This equation, called the net ionic equation, includes only those solution components
directly involved in the reaction. Chemists usually write the net ionic equation for a
reaction in solution because it gives the actual forms of the reactants and products and
includes only the species that undergo a change.
Three Types of Equations Are Used to Describe
Reactions in Solution
❯ The formula equation gives the overall reaction stoichiometry but not necessarily
the actual forms of the reactants and products in solution.
❯ The complete ionic equation represents as ions all reactants and products that are
strong electrolytes.
❯ The net ionic equation includes only those solution components undergoing a
change. Spectator ions are not included.
Interactive
Example 4.9
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in OWL.
Writing Equations for Reactions
For each of the following reactions, write the formula equation, the complete ionic
equation, and the net ionic equation.
a. Aqueous potassium chloride is added to aqueous silver nitrate to form a silver
chloride precipitate plus aqueous potassium nitrate.
b. Aqueous potassium hydroxide is mixed with aqueous iron(III) nitrate to form a
precipitate of iron(III) hydroxide and aqueous potassium nitrate.
Solution
a. Formula Equation
KCl 1aq2 1 AgNO3 1aq2 h AgCl 1s2 1 KNO3 1aq2
Complete Ionic Equation
(Remember: Any ionic compound dissolved in water will be present as the
separated ions.)
K1 1aq2 1 Cl2 1aq2 1 Ag1 1aq2 1 NO32 1aq2 h AgCl 1s2 1 K1 1aq2 1 NO32 1aq2
h
h
Spectator
Spectator
ion
ion
h
h
Solid,
Spectator
not written
ion
as separate ions
h
Spectator
ion
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Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Canceling the spectator ions
K 1aq2 1 Cl2 1aq2 1 Ag1 1aq2 1 NO32 1aq2 h AgCl 1s2 1 K1 1aq2 1 NO32 1aq2
1
gives the following net ionic equation.
Net Ionic Equation
b. Formula Equation
Cl2 1aq2 1 Ag1 1aq2 h AgCl 1s2
3KOH 1aq2 1 Fe 1NO32 3 1aq2 h Fe 1OH2 3 1s2 1 3KNO3 1aq2
Complete Ionic Equation
3K1 1aq2 1 3OH2 1aq2 1 Fe31 1aq2 1 3NO32 1aq2 h
Fe 1OH2 3 1s2 1 3K1 1aq2 1 3NO32 1aq2
Net Ionic Equation
3OH2 1aq2 1 Fe31 1aq2 h Fe 1OH2 3 1s2
See Exercises 4.47 through 4.52
4.7 Stoichiometry of Precipitation
Reactions
Experiment 26: Classification of
Chemical Reactions
In Chapter 3 we covered the principles of chemical stoichiometry: the procedures for
calculating quantities of reactants and products involved in a chemical reaction. Recall that in performing these calculations we first convert all quantities to moles and
then use the coefficients of the balanced equation to assemble the appropriate mole
ratios. In cases where reactants are mixed, we must determine which reactant is limiting, since the reactant that is consumed first will limit the amounts of products formed.
These same principles apply to reactions that take place in solutions. However, two
points about solution reactions need special emphasis. The first is that it is sometimes
difficult to tell immediately what reaction will occur when two solutions are mixed.
Usually we must do some thinking about the various possibilities and then decide
what probably will happen. The first step in this process always should be to write
down the species that are actually present in the solution, as we did in Section 4.5.
The second special point about solution reactions is that to obtain the moles of reactants we must use the volume of the solution and its molarity. This procedure was
covered in Section 4.3.
We will introduce stoichiometric calculations for reactions in solution in Example 4.10.
Critical Thinking
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Interactive
Example 4.10
What if all ionic solids were soluble in water? How would this affect stoichiometry
calculations for reactions in aqueous solution?
Determining the Mass of Product Formed I
Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100-M AgNO3
solution to precipitate all the Ag1 ions in the form of AgCl.
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4.7
Stoichiometry of Precipitation Reactions
161
Solution
Where are we going?
To find the mass of solid NaCl required to precipitate the Ag1
What do we know?
❯ 1.50 L of 0.100 M AgNO3
What information do we need to find the mass of NaCl?
❯ Moles of Ag1 in the solution
How do we get there?
What are the ions present in the combined solution?
Ag1
Species present
Write the
reaction
Balanced net
ionic equation
Determine moles
of products
Check units
of products
Na1
Cl2
What is the balanced net ionic equation for the reaction?
Note from Table 4.1 that NaNO3 is soluble and that AgCl is insoluble. Therefore, solid
AgCl forms according to the ­following net ionic equation:
Ag1 1aq2 1 Cl2 1aq2 h AgCl 1s2
What are the moles of Ag1 ions present in the solution?
Determine moles
of reactants
Identify limiting
reactant
NO32
1.50 L 3
0.100 mol Ag1
5 0.150 mol Ag1
L
How many moles of Cl 2 are required to react with all the Ag1?
Because Ag1 and Cl2 react in a 1:1 ratio, 0.150 mole of Cl2 and thus 0.150 mole of
NaCl are required.
What mass of NaCl is required?
j
0.150 mol NaCl 3
58.44 g NaCl
5 8.77 g NaCl
mol NaCl
See Exercise 4.55
Notice from Example 4.10 that the procedures for doing stoichiometric calculations
for solution reactions are very similar to those for other types of reactions. It is useful
to think in terms of the following steps for reactions in solution.
Problem-Solving Strategy
Solving Stoichiometry Problems for Reactions in Solution
1. Identify the species present in the combined solution, and determine what
reaction occurs.
2. Write the balanced net ionic equation for the reaction.
3. Calculate the moles of reactants.
4. Determine which reactant is limiting.
5. Calculate the moles of product or products, as required.
6. Convert to grams or other units, as required.
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162
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Interactive
Example 4.11
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in OWL.
Determining the Mass of Product Formed II
When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of
0.0250 M Na2SO4 are mixed.
Solution
Where are we going?
To find the mass of solid PbSO4 formed
What do we know?
❯ 1.25 L of 0.0500 M Pb(NO3)2
❯ 2.00 L of 0.0250 M Na2SO4
❯ Chemical reaction
Pb21 1aq2 1 SO422 1aq2 h PbSO4 1s2
What information do we need?
❯
The limiting reactant
How do we get there?
1. What are the ions present in the combined solution?
Na+ SO42–
Pb2+ NO3–
Na1
PbSO4(s)
Pb21 1aq2 1 SO422 1aq2 h PbSO4 1s2
3. What are the moles of reactants present in the solution?
1.25 L 3
Determine moles
of products
2.00 L 3
Grams needed
Convert
to grams
15.2 g PbSO4
NO2
3
2. What is the balanced net ionic equation for the reaction?
Determine moles
of reactants
SO42– is limiting
Pb21
What is the reaction?
Since NaNO3 is soluble and PbSO4 is insoluble, solid PbSO4 will form.
Write the reaction
Pb2+(aq) + SO42–(aq)
SO422
0.0500 mol Pb21
5 0.0625 mol Pb21
L
0.0250 mol SO422
5 0.0500 mol SO422
L
4. Which reactant is limiting?
Because Pb21 and SO422 react in a 1:1 ratio, the amount of SO422 will be limiting
(0.0500 mol SO422 is less than 0.0625 mole of Pb21).
5. What number of moles of PbSO4 will be formed?
Since SO422 is limiting, only 0.0500 mole of solid PbSO4 will be formed.
6. What mass of PbSO4 will be formed?
j
0.0500 mol PbSO4 3
303.3 g PbSO4
5 15.2 g PbSO4
1 mol PbSO4
See Exercises 4.57 and 4.58
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4.8
Acid–Base Reactions
163
4.8 Acid–Base Reactions
Experiment 26: Classification of
Chemical Reactions
PowerLecture: Proton Transfer
The Brønsted–Lowry concept of acids
and bases will be discussed in detail in
Chapter 14.
Earlier in this chapter we considered Arrhenius’s concept of acids and bases: An acid
is a substance that produces H1 ions when dissolved in water, and a base is a substance that produces OH2 ions. Although these ideas are fundamentally correct, it is
convenient to have a more general definition of a base, which includes substances that
do not contain OH2 ions. Such a definition was provided by Johannes N. Brønsted
(1879–1947) and Thomas M. Lowry (1874–1936), who defined acids and bases as
follows:
An acid is a proton donor.
A base is a proton acceptor.
How do we know when to expect an acid–base reaction? One of the most difficult
tasks for someone inexperienced in chemistry is to predict what reaction might occur
when two solutions are mixed. With precipitation reactions, we found that the best way
to deal with this problem is to focus on the species actually present in the mixed solution. This idea also applies to acid–base reactions. For example, when an aqueous solution of hydrogen chloride (HCl) is mixed with an aqueous solution of sodium hydroxide (NaOH), the combined solution contains the ions H1, Cl2, Na1, and OH2.
The separated ions are present because HCl is a strong acid and NaOH is a strong base.
How can we predict what reaction occurs, if any? First, will NaCl precipitate? From
Table 4.1 we can see that NaCl is soluble in water and thus will not precipitate. Therefore, the Na1 and Cl2 ions are spectator ions. On the other hand, because water is a
nonelectrolyte, large quantities of H1 and OH2 ions cannot coexist in solution. They
react to form H2O molecules:
H1 1aq2 1 OH2 1aq2 h H2O 1l2
This is the net ionic equation for the reaction that occurs when aqueous solutions of
HCl and NaOH are mixed.
Next, consider mixing an aqueous solution of acetic acid (HC2H3O2) with an
aqueous solution of potassium hydroxide (KOH). In our earlier discussion of conductivity we said that an aqueous solution of acetic acid is a weak electrolyte. This
tells us that acetic acid does not dissociate into ions to any great extent. In fact, in
0.1 M HC2H3O2 approximately 99% of the HC2H3O2 molecules remain undissociated. However, when solid KOH is dissolved in water, it dissociates completely to
produce K1 and OH2 ions. Therefore, in the solution formed by mixing aqueous
solutions of HC2H3O2 and KOH, before any reaction occurs, the principal species
are HC2H3O2, K1, and OH2. What reaction will occur? A possible precipitation reaction could occur between K1 and OH2. However, we know that KOH is soluble,
so precipitation does not occur. Another possibility is a reaction involving the hydroxide ion (a proton acceptor) and some proton donor. Is there a source of protons
in the solution? The answer is yes—the HC2H3O2 molecules. The OH2 ion has such
a strong affinity for protons that it can strip them from the HC2H3O2 molecules. The
net ionic equation for this reaction is
OH2 1aq2 1 HC2H3O2 1aq2 h H2O 1l2 1 C2H3O22 1aq2
This reaction illustrates a very important general principle: The hydroxide ion is
such a strong base that for purposes of stoichiometric calculations it can be assumed
to react completely with any weak acid that we will encounter. Of course, OH2 ions
also react completely with the H1 ions in solutions of strong acids.
We will now deal with the stoichiometry of acid–base reactions in aqueous solutions. The procedure is fundamentally the same as that used previously for precipitation
reactions.
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164
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Species present
Write the
reaction
Balanced net
ionic equation
Determine moles
of reactants
Identify limiting
reactant
Determine moles
of products
Check units
of products
Interactive
Example 4.12
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in OWL.
Problem-Solving Strategy
Performing Calculations for Acid–Base Reactions
1. List the species present in the combined solution before any reaction
occurs, and decide what reaction will occur.
2. Write the balanced net ionic equation for this reaction.
3. Calculate the moles of reactants. For reactions in solution, use the volumes
of the original solutions and their molarities.
4. Determine the limiting reactant where appropriate.
5. Calculate the moles of the required reactant or product.
6. Convert to grams or volume (of solution), as required.
An acid–base reaction is often called a neutralization reaction. When just enough
base is added to react exactly with the acid in a solution, we say the acid has been
neutralized.
Neutralization Reactions I
What volume of a 0.100-M HCl solution is needed to neutralize 25.0 mL of 0.350 M
NaOH?
Solution
Where are we going?
To find the volume of 0.100-M HCl required for neutralization
What do we know?
❯ 25 mL of 0.350 M NaOH
❯ 0.100 M HCl
❯ The chemical reaction
H1 1aq2 1 OH2 1aq2 h H2O 1l2
How do we get there?
Use the Problem-Solving Strategy for Performing Calculations for Acid–Base
Reactions.
1. What are the ions present in the combined solution?
H1
Cl2
Na1
OH2
What is the reaction?
The two possibilities are
Na1 1aq2 1 Cl2 1aq2 h NaCl 1s2
H1 1aq2 1 OH2 1aq2 h H2O 1l2
Since we know that NaCl is soluble, the first reaction does not take place (Na1 and Cl2
are spectator ions). However, as we have seen before, the reaction of the H1 and OH2
ions to form H2O does occur.
2. What is the balanced net ionic equation for the reaction?
H1 1aq2 1 OH2 1aq2 h H2O 1l2
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4.8
25.0 mL NaOH 3
Write the reaction
H+(aq) + OH–(aq)
H2O(l )
8.75 × 10–3
No limiting
reactant
Moles H+
165
3. What are the moles of reactant present in the solution?
H+ Cl –
Na+ OH –
Moles OH–
Acid–Base Reactions
8.75 × 10–3
4. Which reactant is limiting?
This problem requires the addition of just enough H1 to react exactly with the
OH2 ions present. We do not need to be concerned with limiting reactant here.
5. What moles of H1 are needed?
Since H1 and OH2 ions react in a 1:1 ratio, 8.75 3 1023 mole of H1 is required
to neutralize the OH2 ions present.
6. What volume of HCl is required?
0.100 mol H1
5 8.75 3 1023 mol H1
L
V3
Volume needed
Convert to volume
1L
0.350 mol OH2
3
5 8.75 3 1023 mol OH2
1000 mL
L NaOH
Solving for V gives
87.5 mL of 0.100 M HCl needed
j
V5
8.75 3 1023 mol H1
5 8.75 3 1022 L
0.100 mol H1
L
See Exercises 4.69 and 4.70
Interactive
Example 4.13
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in OWL.
Neutralization Reactions II
In a certain experiment, 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320 M KOH are
mixed. What is the concentration of H1 or OH2 ions in excess after the reaction goes
to completion?
Solution
Where are we going?
To find the concentration of H1 or OH2 in excess after the reaction is complete
What do we know?
❯ 28.0 mL of 0.250 M HNO3
❯ 53.0 mL of 0.320 M KOH
❯ The chemical reaction
H1 1aq2 1 OH2 1aq2 h H2O 1l2
How do we get there?
Use the Problem-Solving Strategy for Performing Calculations for Acid–Base
Reactions.
1. What are the ions present in the combined solution?
H1
NO32
K1
OH2
2. What is the balanced net ionic equation for the reaction?
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H1 1aq2 1 OH2 1aq2 h H2O 1l2
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166
Chapter 4
H+
K
+
Types of Chemical Reactions and Solution Stoichiometry
3. What are the moles of reactant present in the solution?
NO3–
OH –
28.0 mL HNO3 3
Write the reaction
53.0 mL KOH 3
H+(aq) + OH–(aq)
H2O(l )
Find moles H+, OH–
Limiting
reactant is H+
Find moles OH– that react
Concentration
of OH– needed
Find excess OH– concentration
0.123 M OH–
1L
0.250 mol H1
3
5 7.00 3 1023 mol H1
1000 mL
L HNO3
1L
0.320 mol OH2
3
5 1.70 3 1022 mol OH2
1000 mL
L KOH
4. Which reactant is limiting?
Since H1 and OH2 ions react in a 1:1 ratio, the limiting reactant is H1.
5. What amount of OH2 will react?
7.00 3 1023 mole of OH2 is required to neutralize the H1 ions present.
What amount of OH2 ions are in excess?
The amount of OH2 ions in excess is obtained from the following difference:
Original amount 2 amount consumed 5 amount in excess
22
1.70 3 10
mol OH2 2 7.00 3 1023 mol OH2 5 1.00 3 1022 mol OH2
What is the volume of the combined solution?
The volume of the combined solution is the sum of the individual volumes:
Original volume of HNO3 1 original volume of KOH 5 total volume
28.0 mL 1 53.0 mL 5 81.0 mL 5 8.10 3 1022 L
6. What is the molarity of the OH2 ions in excess?
j
mol OH2
1.00 3 1022 mol OH2
5
5 0.123 M OH2
L solution
8.10 3 1022 L
Reality Check | This calculated molarity is less than the initial molarity, as it
should be.
See Exercises 4.71 and 4.72
Acid–Base Titrations
Experiment 37: Acid–Base Titrations
2: ­Evaluation of Commercial Antacid
Tablets
Ideally, the endpoint and stoichiometric
point should coincide.
PowerLecture: Acid–Base Titration
Volumetric analysis is a technique for determining the amount of a certain substance
by doing a titration. A titration involves delivery (from a buret) of a measured volume
of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). The titrant contains a substance that reacts in a
known manner with the analyte. The point in the titration where enough titrant has been
added to react exactly with the analyte is called the equivalence point or the stoichiometric point. This point is often marked by an indicator, a substance added at the beginning of the titration that changes color at (or very near) the equivalence point. The
point where the indicator actually changes color is called the endpoint of the titration.
The goal is to choose an indicator such that the endpoint (where the indicator changes
color) occurs exactly at the equivalence point (where just enough titrant has been added
to react with all the analyte).
Requirements for a Successful Titration
❯ The exact reaction between titrant and analyte must be known (and rapid).
❯ The stoichiometric (equivalence) point must be marked accurately.
❯ T he volume of titrant required to reach the stoichiometric point must be known
accurately.
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Acid–Base Reactions
167
Photos © Cengage Learning. All rights reserved.
4.8
a
b
c
Figure 4.18 | The titration of an acid with a base. (a) The titrant (the base) is in the buret, and the flask contains the acid solution
along with a small amount of indicator. (b) As base is added drop by drop to the acid solution in the flask during the titration, the
indicator changes color, but the color disappears on mixing. (c) The stoichiometric (equivalence) point is marked by a permanent
indicator color change. The volume of base added is the difference between the final and initial buret readings.
When the analyte is a base or an acid, the required titrant is a strong acid or strong
base, respectively. This procedure is called an acid–base titration. An indicator very
commonly used for acid–base titrations is phenolphthalein, which is colorless in an
acidic solution and pink in a basic solution. Thus, when an acid is titrated with a base,
the phenolphthalein remains colorless until after the acid is consumed and the first
drop of excess base is added. In this case, the endpoint (the solution changes from
colorless to pink) occurs approximately one drop of base beyond the stoichiometric
point. This type of titration is illustrated in Fig. 4.18.
We will deal with the acid–base titrations only briefly here but will return to the
topic of titrations and indicators in more detail in Chapter 15. The titration of an acid
with a standard solution containing hydroxide ions is described in Example 4.15. In
Example 4.14 we show how to determine accurately the concentration of a sodium
hydroxide solution. This procedure is called standardizing the solution.
Interactive
Example 4.14
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.com to try this Interactive Example
in OWL.
Neutralization Titration
A student carries out an experiment to standardize (determine the exact concentration
of) a sodium hydroxide solution. To do this, the student weighs out a 1.3009-g sample
of potassium hydrogen phthalate (KHC8H4O4, often abbreviated KHP). KHP (molar
mass 204.22 g/mol) has one acidic hydrogen. The student dissolves the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution
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168
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
with the sodium hydroxide solution to the phenolphthalein endpoint. The difference
between the final and initial buret readings indicates that 41.20 mL of the sodium hydroxide solution is required to react exactly with the 1.3009 g KHP. Calculate the
concentration of the sodium ­hydroxide solution.
Solution
Where are we going?
To find the concentration of NaOH solution
What do we know?
❯ 1.3009 g KHC8H4O4 (KHP), molar mass (204.22 g/mol)
❯ 41.20 mL NaOH solution to neutralize KHP
❯ The chemical reaction
HC8H4O42 1aq2 1 OH2 1aq2 h H2O 1l2 1 C8H4O422 1aq2
How do we get there?
Use the Problem-Solving Strategy for Performing Calculations for Acid–Base
Reactions.
1. What are the ions present in the combined solution?
K1
–
HC8H4O42
Na1
OH2
2. What is the balanced net ionic equation for the reaction?
HC8H4O4−
Hydrogen phthalate ion
HC8H4O42 1aq2 1 OH2 1aq2 h H2O 1l2 1 C8H4O422 1aq2
3. What are the moles of KHP?
1.3009 g KHC8H4O4 3
1 mol KHC8H4O4
5 6.3701 3 1023 mol KHC8H4O4
204.22 g KHC8H4O4
4. Which reactant is limiting?
This problem requires the addition of just enough OH2 ions to react exactly with
the KHP present. We do not need to be concerned with ­limiting reactant here.
5. What moles of OH2 are required?
6.3701 3 1023 mole of OH2 is required to neutralize the KHP present.
6. What is the molarity of the NaOH solution?
j
mol NaOH
6.3701 3 1023 mol NaOH
5
L solution
4.120 3 1022 L
5 0.1546 M
Molarity of NaOH 5
This standard sodium hydroxide solution can now be used in other experiments (see
­Example 4.15).
See Exercises 4.73 and 4.78
Critical Thinking
In Example 4.14 you determined the concentration of an aqueous solution of NaOH
using phenolphthalein as an indicator. What if you used an indicator for which the
endpoint of the titration occurs after the equivalence point? How would this affect
your calculated concentration of NaOH?
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4.8
Interactive
Example 4.15
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in OWL.
Acid–Base Reactions
169
Neutralization Analysis
An environmental chemist analyzed the effluent (the released waste material) from an
industrial process known to produce the compounds carbon tetrachloride (CCl4) and
benzoic acid (HC7H5O2), a weak acid that has one acidic hydrogen atom per molecule.
A sample of this effluent weighing 0.3518 g was shaken with water, and the resulting
aqueous solution required 10.59 mL of 0.1546 M NaOH for neutralization. Calculate
the mass percent of HC7H5O2 in the original sample.
Solution
Where are we going?
To find the mass percent of HC7H5O2 in the original sample
What do we know?
❯ 0.3518 g effluent (original sample)
❯ 10.59 mL 0.1546 M NaOH for neutralization of HC7H5O2
❯ The chemical reaction
HC7H5O2 1aq2 1 OH2 1aq2 h H2O 1l2 1 C7H5O22 1aq2
How do we get there?
Use the Problem-Solving Strategy for Performing Calculations for Acid–Base
Reactions.
1. What are the species present in the combined solution?
Na1
HC7H5O2
OH2
2. What is the balanced net ionic equation for the reaction?
HC7H5O2 1aq2 1 OH2 1aq2 h H2O 1l2 1 C7H5O22 1aq2
3. What are the moles of OH2 required?
10.59 mL NaOH 3
1L
0.1546 mol OH2
3
5 1.637 3 1023 mol OH2
1000 mL
L NaOH
4. Which reactant is limiting?
This problem requires the addition of just enough OH2 ions to react exactly with the
HC7H5O2 present. We do not need to be concerned with limiting reactant here.
5. What mass of HC7H5O2 is present?
1.637 3 1023 mol HC7H5O2 3
122.12 g HC7H5O2
5 0.1999 g HC7H5O2
1 mol HC7H5O2
6. What is the mass percent of the HC7H5O2 in the effluent?
j
0.1999 g
3 100% 5 56.82%
0.3518 g
Reality Check | The calculated percent of HC7H5O2 is less than 100%, as it should be.
See Exercise 4.77
The first step in the analysis of a complex
solution is to write down the components
and focus on the chemistry of each one.
When a strong electrolyte is present,
write it as separated ions.
In doing problems involving titrations, you must first decide what reaction is
o­ ccurring. Sometimes this seems difficult because the titration solution contains several components. The key to success in doing solution reactions is to first write down
all the components in the solution and focus on the chemistry of each one. We have
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170
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
been emphasizing this approach in dealing with the reactions between ions in solution.
Make it a habit to write down the components of solutions before trying to decide what
­reaction(s) might take place as you attempt the end-of-chapter problems ­involving
­titrations.
4.9 Oxidation–Reduction Reactions
We have seen that many important substances are ionic. Sodium chloride, for example,
can be formed by the reaction of elemental sodium and chlorine:
2Na 1s2 1 Cl2 1g2 h 2NaCl 1s2
Experiment 26: Classification of
Chemical Reactions
Photos © Cengage Learning. All rights reserved.
IBLG: See questions from ­“Oxidation
Reduction”
In this reaction, solid sodium, which contains neutral sodium atoms, reacts with chlorine gas, which contains diatomic Cl2 molecules, to form the ionic solid NaCl, which
contains Na1 and Cl2 ions. This process is represented in Fig. 4.19. Reactions like this
one, in which one or more electrons are transferred, are called oxidation–reduction
reactions or redox reactions.
Many important chemical reactions involve oxidation and reduction. Photosynthesis, which stores energy from the sun in plants by converting carbon dioxide and water
to sugar, is a very important oxidation–reduction reaction. In fact, most reactions used
for energy production are redox reactions. In humans, the oxidation of sugars, fats, and
proteins provides the energy necessary for life. Combustion reactions, which provide
Cl−
Na
Na+
Cl−
Na+
Na
CCl
l
Cl Cl
2Na(s)
Sodium
+
Cl2(g)
Chlorine
2NaCl(s)
Sodium chloride
Figure 4.19 | The reaction of solid sodium and gaseous chlorine to form solid sodium chloride.
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4.9
PowerLecture:
Oxidation of Zinc with Iodine
Spontaneous Reaction of Phosphorus
(Barking Dogs)
Reaction of Magnesium and Carbon
­Dioxide
Combustion Reaction: Sugar and
Potassium Chlorate
Oxidation–Reduction Reactions
171
most of the energy to power our civilization, also involve oxidation and reduction. An
example is the reaction of methane with oxygen:
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1g2 1 energy
Even though none of the reactants or products in this reaction is ionic, the reaction is
still assumed to involve a transfer of electrons from carbon to oxygen. To explain this,
we must introduce the concept of oxidation states.
Oxidation States
The concept of oxidation states (also called oxidation numbers) provides a way to
keep track of electrons in oxidation–reduction reactions, particularly redox reactions
­involving covalent substances. Recall that electrons are shared by atoms in covalent
bonds. The oxidation states of atoms in covalent compounds are obtained by arbitrarily assigning the electrons (which are actually shared) to particular atoms. We do
this as follows: For a covalent bond between two identical atoms, the electrons are split
equally between the two. In cases where two different atoms are involved (and the
electrons are thus shared unequally), the shared electrons are assigned completely to
the atom that has the stronger attraction for electrons. For example, recall from the
discussion of the water molecule in Section 4.1 that oxygen has a greater attraction for
electrons than does hydrogen. Therefore, in assigning the oxidation state of oxygen
and hydrogen in H2O, we assume that the oxygen atom actually possesses all the electrons. Recall that a hydrogen atom has one electron. Thus, in water, oxygen has formally “taken” the electrons from two hydrogen atoms. This gives the oxygen an excess
of two electrons (its oxidation state is 22) and leaves each hydrogen with no electrons
(the oxidation state of each hydrogen is thus 11).
We define the oxidation states (or oxidation numbers) of the atoms in a covalent
compound as the imaginary charges the atoms would have if the shared electrons were
divided equally between identical atoms bonded to each other or, for different atoms,
were all assigned to the atom in each bond that has the greater attraction for electrons.
Of course, for ionic compounds containing monatomic ions, the oxidation states of the
ions are equal to the ion charges.
These considerations lead to a series of rules for assigning oxidation states that are
summarized in Table 4.2. Application of these simple rules allows the assignment of
oxidation states in most compounds. To apply these rules, recognize that the sum of the
Table 4.2 | Rules for Assigning Oxidation States
The Oxidation State of . . .
Summary
Examples
• An atom in an element
is zero
Element: 0
Na(s), O2(g), O3(g), Hg(l)
• A monatomic ion is the
same as its charge
Monatomic ion:
charge of ion
Na1, Cl2
• Fluorine is 21 in its
compounds
Fluorine: 21
HF, PF3
• Oxygen is usually 22 in
its compounds
Exception:
peroxides (containing O222),
in which oxygen is 21
Oxygen: 22
H2O, CO2
• Hydrogen is 11 in its
covalent compounds
Hydrogen: 11
H2O, HCl, NH3
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172
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
oxidation states must be zero for an electrically neutral compound. For an ion, the sum
of the oxidation states must equal the charge of the ion. The principles are illustrated
by ­Example 4.16.
It is worthwhile to note at this point that the convention is to write actual charges on
ions as n1 or n2, the number being written before the plus or minus sign. On the other
hand, oxidation states (not actual charges) are written 1n or 2n, the number being
written after the plus or minus sign.
Critical Thinking
What if the oxidation state for oxygen was defined as 21 instead of 22? What effect,
if any, would it have on the oxidation state of hydrogen?
Interactive
Example 4.16
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.com to try this Interactive Example
in OWL.
Assigning Oxidation States
Assign oxidation states to all atoms in the following.
a. CO2
b. SF6
c. NO32
Solution
a. Since we have a specific rule for the oxidation state of oxygen, we will assign its
value first. The oxidation state of oxygen is 22. The oxidation state of the carbon
atom can be determined by recognizing that since CO2 has no charge, the sum of
the oxidation states for oxygen and carbon must be zero. Since each oxygen is
22 and there are two oxygen atoms, the carbon atom must be assigned an
oxidation state of 14:
CO2
p r
14 22 for each oxygen
Experiment 39: Determination of Iron
by ­Redox Titration
Reality Check | We can check the assigned oxidation states by noting that when the
number of atoms is taken into account, the sum is zero as required:
1 11 42 1 2 1222 5 0
p
No. of C
atoms
h
No. of O
atoms
b. Since we have no rule for sulfur, we first assign the oxidation state of each
fluorine as 21. The sulfur must then be assigned an oxidation state of 16 to
balance the total of 26 from the fluorine atoms:
SF6
p r
16 21 for each fluorine
Reality Check | 16 1 6(21) 5 0
c. Oxygen has an oxidation state of 22. Because the sum of the oxidation states of
the three oxygens is 26 and the net charge on the NO32 ion is 12, the nitrogen
must have an oxidation state of 15:
NO32
p r
15 22 for each oxygen
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4.9
Oxidation–Reduction Reactions
173
Reality Check | 15 1 3(22) 5 21
Note that in this case the sum must be 21 (the overall charge on the ion).
Photo © Cengage Learning. All rights reserved.
See Exercises 4.79 through 4.82
Magnetite is a magnetic ore containing
Fe3O4. Note that the compass needle
points toward the ore.
We need to make one more point about oxidation states, and this can be illustrated
by the compound Fe3O4, which is the main component in magnetite, an iron ore that
accounts for the reddish color of many types of rocks and soils. To determine the oxidation states in Fe3O4, we first assign each oxygen atom its usual oxidation state of
22. The three iron atoms must yield a total of 18 to balance the total of 28 from the
four oxygens. This means that each iron atom has an oxidation state of 183 . A noninteger value for the oxidation state may seem strange because charge is expressed in
whole numbers. However, although they are rare, noninteger oxidation states do occur because of the rather arbitrary way that electrons are divided up by the rules in
Table 4.2. For Fe3O4, for example, the rules assume that all the iron atoms are equal,
when in fact this compound can best be viewed as containing four O22 ions, two Fe31
ions, and one Fe21 ion per formula unit. (Note that the “average” charge on iron
works out to be 831, which is equal to the oxidation state we determined above.) Noninteger oxidation states should not intimidate you. They are used in the same way as
integer oxidation states—for keeping track of electrons.
The Characteristics of Oxidation–Reduction
Reactions
Oxidation–reduction reactions are characterized by a transfer of electrons. In some
cases, the transfer occurs in a literal sense to form ions, such as in the reaction
2Na 1s2 1 Cl2 1g2 h 2NaCl 1s2
However, sometimes the transfer is less obvious. For example, consider the combustion of methane (the oxidation state for each atom is given):
8n
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1g2
Oxidation h
h
h
state
24 11
0
14 22
11
22
(each H) (each O) (each H)
88n
88n
8n
A helpful mnemonic device is OIL RIG
(Oxidation Involves Loss; Reduction
Involves Gain). Another common
mnemonic is LEO says GER. (Loss of
Electrons, Oxidation; Gain of Electrons,
Reduction).
Note that the oxidation state for oxygen in O2 is 0 because it is in elemental form. In
this reaction there are no ionic compounds, but we can still describe the process in
terms of a transfer of electrons. Note that carbon undergoes a change in oxidation state
from 24 in CH4 to 14 in CO2. Such a change can be accounted for by a loss of eight
electrons (the symbol e2 stands for an electron):
CH4 h CO2 1 8e2
h
24
h
14
On the other hand, each oxygen changes from an oxidation state of 0 in O2 to 22 in
H2O and CO2, signifying a gain of two electrons per atom. Since four oxygen atoms
are involved, this is a gain of eight electrons:
h
0
8n
2O2 1 8e2 h CO2 1 2H2O
88n
4(22) 5 28
No change occurs in the oxidation state of hydrogen, and it is not formally involved in
the electron-transfer process.
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X
2Na 1s2 1 Cl2 1g2 h 2NaCl 1s2
h
0
on
tr er
ec
el nsf
a
tr
M+
With this background, we can now define some important terms. Oxidation is an
increase in oxidation state (a loss of electrons). Reduction is a decrease in oxidation
state (a gain of electrons). Thus in the reaction
X−
h
0
Reduced
loses electrons
gains electrons
oxidation state
increases
oxidation state
decreases
reducing agent
oxidizing agent
Figure 4.20 | A summary of an
oxidation–reduction process, in which
M is oxidized and X is reduced.
21
sodium is oxidized and chlorine is reduced. In addition, Cl2 is called the oxidizing
agent (electron acceptor), and Na is called the reducing agent (electron donor).
These terms are summarized in Fig. 4.20.
Concerning the reaction
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1g2
h
24 11
h
0
8n
Oxidized
h
11
h
14 22
h
11
8n
M
Types of Chemical Reactions and Solution Stoichiometry
8n
Chapter 4
8n
174
22
we can say the following:
Methane is oxidized because there has been an increase in carbon’s oxidation state
(the carbon atom has formally lost electrons).
Oxygen is reduced because there has been a decrease in its oxidation state (oxygen
has formally gained electrons).
CH4 is the reducing agent.
O2 is the oxidizing agent.
Note that when the oxidizing or reducing agent is named, the whole compound is
specified, not just the element that undergoes the change in oxidation state.
Interactive
Example 4.17
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in OWL.
Oxidation–Reduction Reactions
Metallurgy, the process of producing a metal from its ore, always involves oxidation–
­reduction reactions. In the metallurgy of galena (PbS), the principal lead-containing
ore, the first step is the conversion of lead sulfide to its oxide (a process called
roasting):
2PbS 1s2 1 3O2 1g2 h 2PbO 1s2 1 2SO2 1g2
The oxide is then treated with carbon monoxide to produce the free metal:
PbO 1s2 1 CO 1g2 h Pb 1s2 1 CO2 1g2
For each reaction, identify the atoms that are oxidized and reduced, and specify the
oxidizing and reducing agents.
Solution
For the first reaction, we can assign the following oxidation states:
Oxidation is an increase in oxidation state.
Reduction is a decrease in ­oxidation state.
h
12 22
8n
8n
h
0
h
14
8n
2PbS 1s2 1 3O2 1g2 h 2PbO 1s2 1 2SO2 1g2
h
12 22
22 (each O)
The oxidation state for the sulfur atom increases from 22 to 14. Thus sulfur is oxidized. The oxidation state for each oxygen atom decreases from 0 to 22. Oxygen is
reduced. The oxidizing agent (that accepts the electrons) is O2, and the reducing agent
(that donates electrons) is PbS.
For the second reaction we have
h
0
8n
h
12 22
8n
8n
PbO 1s2 1 CO 1g2 h Pb 1s2 1 CO2 1g2
h
12 22
h
14 22 (each O)
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4.10
An oxidizing agent is reduced and a
­reducing agent is oxidized in a redox
­reaction.
175
Balancing Oxidation–Reduction Equations
Lead is reduced (its oxidation state decreases from 12 to 0), and carbon is oxidized
(its oxidation state increases from 12 to 14). PbO is the oxidizing agent, and CO is
the ­reducing agent.
See Exercises 4.83 and 4.84
Critical Thinking
Dalton believed that atoms were indivisible. Thomson and Rutherford helped to show
that this was not true. What if atoms were indivisible? How would this affect the types
of reactions you have learned about in this chapter?
4.10 Balancing Oxidation–Reduction
Equations
It is important to be able to balance oxidation–reduction reactions. One method involves the use of oxidation states (discussed in this section), and the other method
(normally used for more complex reactions) involves separating the reaction into two
half-reactions. We’ll discuss the second method for balancing oxidation–reduction reactions in Chapter 18.
Oxidation States Method of Balancing
Oxidation–Reduction Reactions
Consider the reaction between solid copper and silver ions in aqueous solution:
Cu 1s2 1 Ag1 1aq2 h Ag 1s2 1 Cu21 1aq2
We can tell this is a redox reaction by assigning oxidation states as follows:
1 e2 gained
Cu
Ag1
1
0
h
11
Ag
1
0
Cu21
12
2 e lost
2
We know that in an oxidation–reduction reaction we must ultimately have equal numbers of electrons gained and lost, and we can use this principle to balance redox equations. For example, in this case, 2 Ag1 ions must be reduced for every Cu atom
oxidized:
Cu 1s2 1 2Ag1 1aq2 h 2Ag 1s2 1 Cu21 1aq2
This gives us the balanced equation.
Now consider a more complex reaction:
H1 1aq2 1 Cl2 1aq2 1 Sn 1s2 1 NO32 1aq2 h SnCl622 1aq2 1 NO2 1g2 1 H2O 1l2
To balance this equation by oxidation states, we first need to assign the oxidation states
to all the atoms in the reactants and products.
H1
11
1
Cl2
21
0
1
Sn
15
1
NO32
22
14
h
14
SnCl622
21
1
NO2
11
1
22
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H2 O
22
176
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Note that hydrogen, chlorine, and oxygen do not change oxidation states and are not
involved in electron exchange. Thus we focus our attention on Sn and N:
4 e2 lost
0
14
Sn
1
NO32
h
SnCl622
1
15
NO2
14
1 e2 gained
This means we need a coefficient of 4 for the N-containing species.
H1 1 Cl2 1 Sn 1 4NO32 h SnCl622 1 4NO2 1 H2O
Now we balance the rest of the equation by inspection.
Balance Cl2:
H1 1 6Cl2 1 Sn 1 4NO32 h SnCl622 1 4NO2 1 H2O
Balance O:
H1 1 6Cl2 1 Sn 1 4NO32 h SnCl622 1 4NO2 1 4H2O
Balance H:
8H1 1 6Cl2 1 Sn 1 4NO32 h SnCl622 1 4NO2 1 4H2O
This gives the final balanced equation. Now we will write the equation with the
states included:
8H1 1aq2 1 6Cl2 1aq2 1 Sn 1s2 1 4NO32 1aq2 h SnCl622 1aq2 1 4NO2 1g2 1 4H2O 1l2
Problem-Solving Strategy
Balancing Oxidation–Reduction Reactions by Oxidation States
1.
2.
3.
4.
5.
6.
Example 4.18
Write the unbalanced equation.
Determine the oxidation states of all atoms in the reactants and products.
Show electrons gained and lost using “tie lines.”
Use coefficients to equalize the electrons gained and lost.
Balance the rest of the equation by inspection.
Add appropriate states.
Balancing Oxidation–Reduction Reactions
Balance the reaction between solid lead(II) oxide and ammonia gas to produce nitrogen gas, liquid water, and solid lead.
Solution
We’ll use the Problem-Solving Strategy for Balancing Oxidation–Reduction Reactions by Oxidation States.
1. What is the unbalanced equation?
PbO 1s2 1 NH3 1g2 h N2 1g2 1 H2O 1l2 1 Pb 1s2
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For Review
177
2. What are the oxidation states for each atom?
22
PbO
0
23
1
12
NH3
h
N2
22
H2O
1
11
1
Pb
0
11
3. How are electrons gained and lost?
3 e2 lost (each atom)
PbO
0
23
1
NH3
h
N2
H2O
1
1
Pb
0
12
2 e gained
2
The oxidation states of all other atoms are unchanged.
4. What coefficients are needed to equalize the electrons gained and lost?
3 e2 lost (each atom) multiply by 2
PbO
0
23
1
NH3
h
N2
H2O
1
1
12
Pb
0
2 e2 gained multiply by 3
3PbO 1 2NH3 h N2 1 H2O 1 3Pb
5. What coefficients are needed to balance the remaining elements?
Balance O:
3PbO 1 2NH3 h N2 1 3H2O 1 3Pb
All the elements are now balanced. The balanced equation with states is:
j
3PbO 1s2 1 2NH3 1g2 h N2 1g2 1 3H2O 1l2 1 3Pb 1s2
See Exercises 4.87 and 4.88
For review
Key terms
Chemical reactions in solution are very important in everyday life.
aqueous solution
Water is a polar solvent that dissolves many ionic and polar substances.
Section 4.1
polar molecule
hydration
solubility
Section 4.2
solute
solvent
electrical conductivity
strong electrolyte
weak electrolyte
nonelectrolyte
acid
strong acid
strong base
Electrolytes
❯
❯
❯
Strong electrolyte: 100% dissociated to produce separate ions; strongly conducts an electric
current
Weak electrolyte: Only a small percentage of dissolved molecules produce ions; weakly
conducts an electric current
Nonelectrolyte: Dissolved substance produces no ions; does not conduct an electric current
Acids and bases
❯
Arrhenius model
Acid: produces H1
❯ Base: produces OH2
❯
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178
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
Key terms
Acids and bases
weak acid
weak base
❯
Brønsted–Lowry model
Acid: proton donor
❯ Base: proton acceptor
Strong acid: completely dissociates into separated H1 and anions
Weak acid: dissociates to a slight extent
❯
Section 4.3
molarity
standard solution
dilution
❯
Section 4.5
Molarity
precipitation reaction
precipitate
❯
❯
One way to describe solution composition
Section 4.6
formula equation
complete ionic equation
spectator ions
net ionic equation
Section 4.8
acid
base
neutralization reaction
volumetric analysis
titration
stoichiometric (equivalence)
point
indicator
endpoint
❯
❯
moles of solute
volume of solution 1L2
Moles solute 5 volume of solution (L) 3 molarity
Standard solution: molarity is accurately known
Dilution
❯
❯
Solvent is added to reduce the molarity
Moles of solute after dilution 5 moles of solute before dilution
M1V1 5 M2V2
Types of equations that describe solution reactions
❯
❯
Section 4.9
oxidation–reduction (redox)
reaction
oxidation state
oxidation
reduction
oxidizing agent (electron
acceptor)
reducing agent (electron donor)
Molarity 1M2 5
❯
Formula equation: All reactants and products are written as complete formulas
Complete ionic equation: All reactants and products that are strong electrolytes are written
as separated ions
Net ionic equation: Only those compounds that undergo a change are written; spectator ions
are not included
Solubility rules
❯
❯
Based on experiment observation
Help predict the outcomes of precipitation reactions
Important types of solution reactions
❯
❯
❯
Acid–base reactions: involve a transfer of H1 ions
Precipitation reactions: formation of a solid occurs
Oxidation–reduction reactions: involve electron transfer
Titrations
❯
❯
❯
Measures the volume of a standard solution (titrant) needed to react with a substance in
solution
Stoichiometric (equivalence) point: the point at which the required amount of titrant has
been added to exactly react with the substance being analyzed
Endpoint: the point at which a chemical indicator changes color
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For Review
179
Oxidation–reduction reactions
❯
❯
❯
❯
❯
❯
Review questions
Oxidation states are assigned using a set of rules to keep track of electron flow.
Oxidation: increase in oxidation state (a loss of electrons)
Reduction: decrease in oxidation state (a gain of electrons)
Oxidizing agent: gains electrons (is reduced)
Reducing agent: loses electrons (is oxidized)
Equations for oxidation–reduction reactions can be balanced by the oxidation states method.
Answers to the Review Questions can be found on the Student website (accessible from www.cengagebrain.com).
1. The (aq) designation listed after a solute indicates the
process of hydration. Using KBr(aq) and C2H5OH(aq)
as your examples, explain the process of hydration for
soluble ionic compounds and for soluble covalent
compounds.
2. Characterize strong electrolytes versus weak electrolytes versus nonelectrolytes. Give examples of each.
How do you experimentally determine whether a
soluble substance is a strong electrolyte, weak electrolyte, or nonelectrolyte?
3. Distinguish between the terms slightly soluble and weak
electrolyte.
4. Molarity is a conversion factor relating moles of solute
in solution to the volume of the solution. How does one
use molarity as a conversion factor to convert from
moles of solute to volume of solution, and from volume
of solution to moles of solute present?
5. What is a dilution? What stays constant in a dilution?
Explain why the equation M1V1 5 M2V2 works for
dilution problems.
6. When the following beakers are mixed, draw a
molecular-level representation of the product mixture
(see Fig. 4.17).
Na+
+
Br–
Pb2+
NO3–
Al3+
+
Cl–
K+
OH –
7. Differentiate between the formula equation, the complete
ionic equation, and the net ionic equation. For each
reaction in Question 6, write all three balanced equations.
8. What is an acid–base reaction? Strong bases are soluble
ionic compounds that contain the hydroxide ion. List
the strong bases. When a strong base reacts with an
acid, what is always produced? Explain the terms
titration, stoichiometric point, neutralization, and
standardization.
9. Define the terms oxidation, reduction, oxidizing agent,
and reducing agent. Given a chemical reaction, how can
you tell if it is a redox reaction?
10. Consider the steps involved in balancing oxidation–
reduction reactions by using oxidation states. The key
to the oxidation states method is to equalize the
electrons lost by the species oxidized with the electrons
gained by the species reduced. First of all, how do you
recognize what is oxidized and what is reduced?
Second, how do you balance the electrons lost with the
electrons gained? Once the electrons are balanced, what
else is needed to balance the oxidation–reduction
reaction?
A discussion of the Active Learning ­Questions can be found online in the ­Instructor’s Resource Guide and on PowerLecture. The questions
allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the
learning that occurs while students talk to each other about chemical concepts.
Active Learning Questions
These questions are designed to be used by groups of students in
class.
1. Assume you have a highly magnified view of a solution of
HCl that allows you to “see” the HCl. Draw this magnified
view. If you dropped in a piece of magnesium, the magnesium
would disappear and hydrogen gas would be released. Represent this change using symbols for the elements, and write out
the balanced equation.
2. You have a solution of table salt in water. What happens to the
salt concentration (increases, decreases, or stays the same) as
the solution boils? Draw pictures to explain your answer.
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180
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
3. You have a sugar solution (solution A) with concentration x.
You pour one-fourth of this solution into a beaker, and add an
equivalent volume of water (solution B).
a. What is the ratio of sugar in solutions A and B?
b. Compare the volumes of solutions A and B.
c. What is the ratio of the concentrations of sugar in solutions A and B?
4. You add an aqueous solution of lead nitrate to an aqueous
solution of potassium iodide. Draw highly magnified views
of each solution individually, and the mixed solution, including any product that forms. Write the balanced equation
for the reaction.
5. Order the following molecules from lowest to highest oxidation state of the nitrogen atom: HNO3, NH4Cl, N2O, NO2,
NaNO2.
6. Why is it that when something gains electrons, it is said to be
reduced? What is being reduced?
7. Consider separate aqueous solutions of HCl and H2SO4 with
the same molar concentrations. You wish to neutralize an
aqueous solution of NaOH. For which acid solution would
you need to add more volume (in milliliters) to neutralize the
base?
a. the HCl solution
b. the H2SO4 solution
c. You need to know the acid concentrations to answer this
­question.
d. You need to know the volume and concentration of the
NaOH solution to answer this question.
e. c and d
Explain.
8. Draw molecular-level pictures to differentiate between concentrated and dilute solutions.
9. You need to make 150.0 mL of a 0.10-M NaCl solution. You
have solid NaCl, and your lab partner has a 2.5-M NaCl solution. Explain how you each make the 0.10-M NaCl solution.
10. The exposed electrodes of a light bulb are placed in a solution
of H2SO4 in an electrical circuit such that the light bulb is
glowing. You add a dilute salt solution, and the bulb dims.
Which of the following could be the salt in the solution?
a. Ba(NO3)2
c. K2SO4
b. NaNO3
d. Ca(NO3)2
Justify your choices. For those you did not choose, explain
why they are incorrect.
11. You have two solutions of chemical A. To determine which has
the highest concentration of A (molarity), which of the following must you know (there may be more than one answer)?
a. the mass in grams of A in each solution
b. the molar mass of A
c. the volume of water added to each solution
d. the total volume of the solution
Explain.
12. Which of the following must be known to calculate the molarity of a salt solution (there may be more than one answer)?
a. the mass of salt added
b. the molar mass of the salt
c. the volume of water added
d. the total volume of the solution
Explain.
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of this book and a
solution appears in the Solutions Guide, as found on PowerLecture.
Questions
13. Differentiate between what happens when the following are
added to water.
a. polar solute versus nonpolar solute
b. KF versus C6H12O6
c. RbCl versus AgCl
d. HNO3 versus CO
14. A typical solution used in general chemistry laboratories is
3.0 M HCl. Describe, in detail, the composition of 2.0 L of a
3.0-M HCl solution. How would 2.0 L of a 3.0-M HC2H3O2
solution differ from the same quantity of the HCl solution?
15. Which of the following statements is(are) true? For the false
statements, correct them.
a. A concentrated solution in water will always contain a
strong or weak electrolyte.
b. A strong electrolyte will break up into ions when dissolved in water.
c. An acid is a strong electrolyte.
d. All ionic compounds are strong electrolytes in water.
16. A student wants to prepare 1.00 L of a 1.00-M solution of
NaOH (molar mass 5 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 M
NaOH is available, how would the student prepare the solution? To help ensure three significant figures in the NaOH molarity, to how many significant figures should the volumes and
mass be determined?
17. List the formulas of three soluble bromide salts and three insoluble bromide salts. Do the same exercise for sulfate salts,
hydroxide salts, and phosphate salts (list three soluble salts
and three insoluble salts). List the formulas for six insoluble
Pb21 salts and one soluble Pb21 salt.
18. When 1.0 mole of solid lead nitrate is added to 2.0 moles of
aqueous potassium iodide, a yellow precipitate forms. After
the precipitate settles to the bottom, does the solution above
the precipitate conduct electricity? Explain. Write the complete ionic equation to help you answer this question.
19. What is an acid and what is a base? An acid–base reaction is
sometimes called a proton-transfer reaction. Explain.
20. A student had 1.00 L of a 1.00-M acid solution. Much to the
surprise of the student, it took 2.00 L of 1.00 M NaOH solution to react completely with the acid. Explain why it took
twice as much NaOH to react with all of the acid.
In a different experiment, a student had 10.0 mL of 0.020 M
HCl. Again, much to the surprise of the student, it took only
5.00 mL of 0.020 M strong base to react completely with the
HCl. Explain why it took only half as much strong base to react with all of the HCl.
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For Review
21. Differentiate between the following terms.
a. species reduced versus the reducing agent
b. species oxidized versus the oxidizing agent
c. oxidation state versus actual charge
22. How does one balance redox reactions by the oxidation states
method?
Exercises
In this section similar exercises are paired.
Aqueous Solutions: Strong and Weak Electrolytes
23. Show how each of the following strong electrolytes “breaks
up” into its component ions upon dissolving in water by drawing molecular-level pictures.
a. NaBr
f. FeSO4
b. MgCl2
g. KMnO4
c. Al(NO3)3
h. HClO4
d. (NH4)2SO4
i. NH4C2H3O2 (ammonium acetate)
e. NaOH
24. Match each name below with the following microscopic pictures of that compound in aqueous solution.
2+
2−
i.
−
+
2−
2+
+
ii.
+
2− +
−
+
−
+
+
2+
2−
iii.
−
−
−
2+
−
iv.
a. barium nitrate
c. potassium carbonate
b. sodium chloride
d. magnesium sulfate
Which picture best represents HNO3(aq)? Why aren’t any of
the pictures a good representation of HC2H3O2(aq)?
25. Calcium chloride is a strong electrolyte and is used to “salt”
streets in the winter to melt ice and snow. Write a reaction to
show how this substance breaks apart when it dissolves in
water.
26. Commercial cold packs and hot packs are available for treating athletic injuries. Both types contain a pouch of water and
a dry chemical. When the pack is struck, the pouch of water
breaks, dissolving the chemical, and the solution becomes
either hot or cold. Many hot packs use magnesium sulfate,
and many cold packs use ammonium nitrate. Write reactions
to show how these strong electrolytes break apart when they
dissolve in water.
Solution Concentration: Molarity
27. Calculate the molarity of each of these solutions.
a. A 5.623-g sample of NaHCO3 is dissolved in enough
water to make 250.0 mL of solution.
b. A 184.6-mg sample of K2Cr2O7 is dissolved in enough
water to make 500.0 mL of solution.
c. A 0.1025-g sample of copper metal is dissolved in 35 mL
of concentrated HNO3 to form Cu21 ions and then water
181
is added to make a total volume of 200.0 mL. (Calculate
the molarity of Cu21.)
28. A solution of ethanol (C2H5OH) in water is prepared by dissolving 75.0 mL of ethanol (density 5 0.79 g/cm3) in enough
water to make 250.0 mL of solution. What is the molarity of
the ethanol in this solution?
29. Calculate the concentration of all ions present in each of the
following solutions of strong electrolytes.
a. 0.100 mole of Ca(NO3)2 in 100.0 mL of solution
b. 2.5 moles of Na2SO4 in 1.25 L of solution
c. 5.00 g of NH4Cl in 500.0 mL of solution
d. 1.00 g K3PO4 in 250.0 mL of solution
30. Calculate the concentration of all ions present in each of the
following solutions of strong electrolytes.
a. 0.0200 mole of sodium phosphate in 10.0 mL of solution
b. 0.300 mole of barium nitrate in 600.0 mL of solution
c. 1.00 g of potassium chloride in 0.500 L of solution
d. 132 g of ammonium sulfate in 1.50 L of solution
31. Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: 100.0 mL
of 0.30 M AlCl3, 50.0 mL of 0.60 M MgCl2, or 200.0 mL of
0.40 M NaCl?
32. Which of the following solutions of strong electrolytes contains the largest number of ions: 100.0 mL of 0.100 M NaOH,
50.0 mL of 0.200 M BaCl2, or 75.0 mL of 0.150 M Na3PO4?
33. What mass of NaOH is contained in 250.0 mL of a 0.400 M
sodium hydroxide solution?
34. If 10. g of AgNO3 is available, what volume of 0.25 M AgNO3
solution can be prepared?
35. Describe how you would prepare 2.00 L of each of the following solutions.
a. 0.250 M NaOH from solid NaOH
b. 0.250 M NaOH from 1.00 M NaOH stock solution
c. 0.100 M K2CrO4 from solid K2CrO4
d. 0.100 M K2CrO4 from 1.75 M K2CrO4 stock solution
36. How would you prepare 1.00 L of a 0.50-M solution of each of
the following?
a. H2SO4 from “concentrated” (18 M) sulfuric acid
b. HCl from “concentrated” (12 M) reagent
c. NiCl2 from the salt NiCl2 ? 6H2O
d. HNO3 from “concentrated” (16 M) reagent
e. Sodium carbonate from the pure solid
37. A solution is prepared by dissolving 10.8 g ammonium sulfate
in enough water to make 100.0 mL of stock solution. A
10.00-mL sample of this stock solution is added to 50.00 mL
of water. Calculate the concentration of ammonium ions and
sulfate ions in the final solution.
38. A solution was prepared by mixing 50.00 mL of 0.100 M
HNO3 and 100.00 mL of 0.200 M HNO3. Calculate the molarity of the final solution of nitric acid.
39. Calculate the sodium ion concentration when 70.0 mL of
3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium
bicarbonate.
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182
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
40. Suppose 50.0 mL of 0.250 M CoCl2 solution is added to
25.0 mL of 0.350 M NiCl2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing.
Assume that the volumes are additive.
41. A standard solution is prepared for the analysis of fluoxymesterone (C20H29FO3), an anabolic steroid. A stock solution is
first prepared by dissolving 10.0 mg of fluoxymesterone in
enough water to give a total volume of 500.0 mL. A 100.0-mL
aliquot (portion) of this solution is diluted to a final volume of
100.0 mL. Calculate the concentration of the final solution in
terms of molarity.
42. A stock solution containing Mn21 ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were
then prepared by dilution:
c. K2CO3(aq) 1 MgI2(aq)
d. Na2CrO4(aq) 1 AlBr3(aq)
47. For the reactions in Exercise 45, write the balanced formula
equation, complete ionic equation, and net ionic equation. If
no precipitate forms, write “No reaction.”
48. For the reactions in Exercise 46, write the balanced formula
equation, complete ionic equation, and net ionic equation. If
no precipitate forms, write “No reaction.”
49. Write the balanced formula and net ionic equation for the
reaction that occurs when the contents of the two beakers
are added together. What colors represent the spectator ions
in each reaction?
Cu2+
For solution A, 50.00 mL of stock solution was diluted to
1000.0 mL.
For solution B, 10.00 mL of solution A was diluted to 250.0 mL.
For solution C, 10.00 mL of solution B was diluted to 500.0 mL.
Calculate the concentrations of the stock solution and solutions A, B, and C.
+
45. When the following solutions are mixed together, what precipitate (if any) will form?
a. FeSO4(aq) 1 KCl(aq)
b. Al(NO3)3(aq) 1 Ba(OH)2(aq)
c. CaCl2(aq) 1 Na2SO4(aq)
d. K2S(aq) 1 Ni(NO3)2(aq)
46. When the following solutions are mixed together, what precipitate (if any) will form?
a. Hg2(NO3)2(aq) 1 CuSO4(aq)
b. Ni(NO3)2(aq) 1 CaCl2(aq)
Na+
S2–
a.
Precipitation Reactions
43. On the basis of the general solubility rules given in Table 4.1,
predict which of the following substances are likely to be soluble in water.
a. aluminum nitrate
b. magnesium chloride
c. rubidium sulfate
d. nickel(II) hydroxide
e. lead(II) sulfide
f. magnesium hydroxide
g. iron(III) phosphate
44. On the basis of the general solubility rules given in Table 4.1,
predict which of the following substances are likely to be soluble in water.
a. zinc chloride
b. lead(II) nitrate
c. lead(II) sulfate
d. sodium iodide
e. cobalt(III) sulfide
f. chromium(III) hydroxide
g. magnesium carbonate
h. ammonium carbonate
SO42–
Co2+
+
Cl−
Na+
OH −
b.
Ag+
+
NO3−
K+
I−
c.
50. Give an example how each of the following insoluble ionic
compounds could be produced using a precipitation reaction.
Write the balanced formula equation for each reaction.
a. Fe(OH)3(s)
c. PbSO4(s)
b. Hg2Cl2(s)
d. BaCrO4(s)
51. Write net ionic equations for the reaction, if any, that occurs
when aqueous solutions of the following are mixed.
a. ammonium sulfate and barium nitrate
b. lead(II) nitrate and sodium chloride
c. sodium phosphate and potassium nitrate
d. sodium bromide and rubidium chloride
e. copper(II) chloride and sodium hydroxide
52. Write net ionic equations for the reaction, if any, that occurs
when aqueous solutions of the following are mixed.
a. chromium(III) chloride and sodium hydroxide
b. silver nitrate and ammonium carbonate
c. copper(II) sulfate and mercury(I) nitrate
d. strontium nitrate and potassium iodide
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
For Review
53. Separate samples of a solution of an unknown soluble ionic
compound are treated with KCl, Na2SO4, and NaOH. A precipitate forms only when Na2SO4 is added. Which cations
could be present in the unknown soluble ionic compound?
54. A sample may contain any or all of the following ions: Hg221,
Ba21, and Mn21.
a. No precipitate formed when an aqueous solution of NaCl
was added to the sample solution.
b. No precipitate formed when an aqueous solution of
Na2SO4 was added to the sample solution.
c. A precipitate formed when the sample solution was made
­basic with NaOH.
Which ion or ions are present in the sample solution?
55. What mass of Na2CrO4 is required to precipitate all of the silver ions from 75.0 mL of a 0.100-M solution of AgNO3?
56. What volume of 0.100 M Na3PO4 is required to precipitate all
the lead(II) ions from 150.0 mL of 0.250 M Pb(NO3)2?
57. What mass of solid aluminum hydroxide can be produced
when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of
0.100 M KOH?
58. What mass of barium sulfate can be produced when 100.0 mL
of a 0.100-M solution of barium chloride is mixed with
100.0 mL of a 0.100-M solution of iron(III) sulfate?
59. What mass of solid AgBr is produced when 100.0 mL of
0.150 M AgNO3 is added to 20.0 mL of 1.00 M NaBr?
60. What mass of silver chloride can be prepared by the reaction
of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M
calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.
61. A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is
mixed with 100.0 mL of 0.200 M aqueous magnesium nitrate.
a. Write a balanced chemical equation for any reaction that
­occurs.
b. What precipitate forms?
c. What mass of precipitate is produced?
d. Calculate the concentration of each ion remaining in solution after precipitation is complete.
62. The drawings below represent aqueous solutions. Solution A
is 2.00 L of a 2.00-M aqueous solution of copper(II) nitrate.
Solution B is 2.00 L of a 3.00-M aqueous solution of potassium hydroxide.
K+
OH–
Cu2+
NO3–
A
B
a. Draw a picture of the solution made by mixing solutions
A and B together after the precipitation reaction takes
place. Make sure this picture shows the correct relative
volume compared to solutions A and B, and the correct
relative number of ions, along with the correct relative
amount of solid formed.
183
b. Determine the concentrations (in M) of all ions left in
solution (from part a) and the mass of solid formed.
63. A 1.42-g sample of a pure compound, with formula M2SO4,
was dissolved in water and treated with an excess of aqueous
calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected,
dried, and found to weigh 1.36 g. Determine the atomic mass
of M, and identify M.
64. You are given a 1.50-g mixture of sodium nitrate and sodium
chloride. You dissolve this mixture into 100 mL of water and
then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure.
The white solid has a mass of 0.641 g.
a. If you had an extremely magnified view of the solution (to
the atomic-molecular level), list the species you would
see (include charges, if any).
b. Write the balanced net ionic equation for the reaction that
produces the solid. Include phases and charges.
c. Calculate the percent sodium chloride in the original
unknown mixture.
Acid–Base Reactions
65. Write the balanced formula, complete ionic, and net ionic
equations for each of the following acid–base reactions.
a. HClO4 1aq2 1 Mg 1OH2 2 1s2 S
b. HCN 1aq2 1 NaOH 1aq2 S
c. HCl 1aq2 1 NaOH 1aq2 S
66. Write the balanced formula, complete ionic, and net ionic
equations for each of the following acid–base reactions.
a. HNO3 1aq2 1 Al 1OH2 3 1s2 S
b. HC2H3O2 1aq2 1 KOH 1aq2 S
c. Ca 1OH2 2 1aq2 1 HCl 1aq2 S
67. Write the balanced formula equation for the acid–base reactions that occur when the following are mixed.
a. potassium hydroxide (aqueous) and nitric acid
b. barium hydroxide (aqueous) and hydrochloric acid
c. perchloric acid [HClO4(aq)] and solid iron(III) hydroxide
d. solid silver hydroxide and hydrobromic acid
e. aqueous strontium hydroxide and hydroiodic acid
68. What acid and what base would react in aqueous solution so
that the following salts appear as products in the formula equation? Write the balanced formula equation for each reaction.
a. potassium perchlorate
b. cesium nitrate
c. calcium iodide
69. What volume of each of the following acids will react completely with 50.00 mL of 0.200 M NaOH?
a. 0.100 M HCl
b. 0.150 M HNO3
c. 0.200 M HC2H3O2 (1 acidic hydrogen)
70. What volume of each of the following bases will react completely with 25.00 mL of 0.200 M HCl?
a. 0.100 M NaOH
c. 0.250 M KOH
b. 0.0500 M Sr(OH)2
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184
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
71. Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL
of 0.0550 M Ba(OH)2 solution. What is the concentration of
the excess H1 or OH2 ions left in this solution?
72. A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
50.0 mL of 0.100 M hydrochloric acid, 100.0 mL of 0.200 M of
nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and
200.0 mL of 0.100 M rubidium hydroxide. Did the acids and
bases exactly neutralize each other? If not, calculate the concentration of excess H1 or OH2 ions left in solution.
73. A 25.00-mL sample of hydrochloric acid solution requires
24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric
acid ­solution?
74. A 10.00-mL sample of vinegar, an aqueous solution of acetic
acid (HC2H3O2), is titrated with 0.5062 M NaOH, and
16.58 mL is required to reach the equivalence point.
a. What is the molarity of the acetic acid?
b. If the density of the vinegar is 1.006 g/cm3, what is the
mass ­percent of acetic acid in the vinegar?
75. What volume of 0.0200 M calcium hydroxide is required to
neutralize 35.00 mL of 0.0500 M nitric acid?
76. A 30.0-mL sample of an unknown strong base is neutralized
after the addition of 12.0 mL of a 0.150 M HNO3 solution. If
the unknown base concentration is 0.0300 M, give some possible identities for the unknown base.
77. A student titrates an unknown amount of potassium hydrogen
phthalate (KHC8H4O4, often abbreviated KHP) with 20.46 mL
of a 0.1000-M NaOH solution. KHP (molar mass 5 204.22 g/
mol) has one acidic hydrogen. What mass of KHP was titrated
(reacted completely) by the sodium hydroxide solution?
78. The concentration of a certain sodium hydroxide solution was
determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid
with one acidic hydrogen and a molar mass of 204.22 g/mol. In
the titration, 34.67 mL of the sodium hydroxide solution was
required to react with 0.1082 g KHP. Calculate the molarity of
the sodium hydroxide.
Oxidation–Reduction Reactions
79. Assign oxidation states for all atoms in each of the following
compounds.
a. KMnO4
f. Fe3O4
b. NiO2
g. XeOF4
c. Na4Fe(OH)6
h. SF4
d. (NH4)2HPO4
i. CO
e. P4O6
j. C6H12O6
80. Assign oxidation states for all atoms in each of the following
compounds.
a. UO221
f. Mg2P2O7
b. As2O3
g. Na2S2O3
c. NaBiO3
h. Hg2Cl2
d. As4
i. Ca(NO3)2
e. HAsO2
81. Assign the oxidation state for nitrogen in each of the following.
a. Li3N
f. NO2
b. NH3
g. NO22
c. N2H4
h. NO32
d. NO
i. N2
e. N2O
82. Assign oxidation numbers to all the atoms in each of the
­following.
a. SrCr2O7
g. PbSO3
b. CuCl2
h. PbO2
c. O2
i. Na2C2O4
d. H2O2
j. CO2
e. MgCO3
k. (NH4)2Ce(SO4)3
f. Ag
l. Cr2O3
83. Specify which of the following are oxidation–reduction reactions, and identify the oxidizing agent, the reducing agent, the
substance being oxidized, and the substance being reduced.
a. Cu 1s2 1 2Ag1 1aq2 S 2Ag 1s2 1 Cu21 1aq2
b. HCl 1g2 1 NH3 1g2 S NH4Cl 1s2
c. SiCl4 1l2 1 2H2O 1l2 S 4HCl 1aq2 1 SiO2 1s2
d. SiCl4 1l2 1 2Mg 1s2 S 2MgCl2 1s2 1 Si 1s2
2
e. Al 1OH2 2
4 1aq2 S AlO2 1aq2 1 2H2O 1l2
84. Specify which of the following equations represent oxidation–
reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced.
a. CH4 1g2 1 H2O 1g2 S CO 1g2 1 3H2 1g2
b. 2AgNO3 1aq2 1 Cu 1s2 S Cu 1NO32 2 1aq2 1 2Ag 1s2
c. Zn 1s2 1 2HCl 1aq2 S ZnCl2 1aq2 1 H2 1g2
d. 2H1 1aq2 1 2CrO422 1aq2 S Cr2O722 1aq2 1 H2O 1l2
85. Consider the reaction between sodium metal and fluorine (F2)
gas to form sodium fluoride. Using oxidation states, how
many electrons would each sodium atom lose, and how many
electrons would each fluorine atom gain? How many sodium
atoms are needed to react with one fluorine molecule? Write a
balanced equation for this reaction.
86. Consider the reaction between oxygen (O2) gas and magnesium metal to form magnesium oxide. Using oxidation states,
how many electrons would each oxygen atom gain, and how
many electrons would each magnesium atom lose? How many
magnesium atoms are needed to react with one oxygen molecule? Write a balanced equation for this reaction.
87. Balance each of the following oxidation–reduction reactions
by using the oxidation states method.
a. C2H6 1g2 1 O2 1g2 S CO2 1g2 1 H2O 1g2
b. Mg 1s2 1 HCl 1aq2 S Mg21 1aq2 1 Cl2 1aq2 1 H2 1g2
c. Co31 1aq2 1 Ni 1s2 S Co21 1aq2 1 Ni21 1aq2
d. Zn 1s2 1 H2SO4 1aq2 S ZnSO4 1aq2 1 H2 1g2
88. Balance each of the following oxidation–reduction reactions
by using the oxidation states method.
a. Cl2 1g2 1 Al 1s2 S Al31 1aq2 1 Cl2 1aq2
b. O2 1g2 1 H2O 1l2 1 Pb 1s2 S Pb 1OH2 2 1s2
21
c. H1 1aq2 1 MnO2
4 1aq2 1 Fe 1aq2 S
Mn21 1aq2 1 Fe31 1aq2 1 H2O 1l2
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For Review
Additional Exercises
89. You wish to prepare 1 L of a 0.02-M potassium iodate solution. You require that the final concentration be within 1% of
0.02 M and that the concentration must be known accurately to
the fourth decimal place. How would you prepare this solution? Specify the glassware you would use, the accuracy
needed for the balance, and the ranges of acceptable masses of
KIO3 that can be used.
90. The figures below are molecular-level representations of four
aqueous solutions of the same solute. Arrange the solutions
from most to least concentrated.
Solution A (1.0 L)
Solution B (4.0 L)
Solution C (2.0 L)
Solution D (2.0 L)
91. An average human being has about 5.0 L of blood in his or her
body. If an average person were to eat 32.0 g of sugar (sucrose,
C12H22O11, 342.30 g/mol), and all that sugar were dissolved
into the bloodstream, how would the molarity of the blood
sugar change?
92. A 230.-mL sample of a 0.275-M CaCl2 solution is left on a hot
plate overnight; the following morning, the solution is 1.10 M.
What volume of water evaporated from the 0.275 M CaCl2
­solution?
93. Using the general solubility rules given in Table 4.1, name
three reagents that would form precipitates with each of the
following ions in aqueous solution. Write the net ionic equation for each of your suggestions.
a. chloride ion
d. sulfate ion
e. mercury(I) ion, Hg221
b. calcium ion
c. iron(III) ion
f. silver ion
94. Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500 M
silver nitrate is added dropwise until precipitate formation is
complete. The mass of the white precipitate formed is 0.641 g.
a. Calculate the mass percent of magnesium chloride in the
mixture.
b. Determine the minimum volume of silver nitrate that must
have been added to ensure complete formation of the
precipitate.
95. A 1.00-g sample of an alkaline earth metal chloride is treated
with excess silver nitrate. All of the chloride is recovered as
1.38 g of silver chloride. Identify the metal.
185
96. A mixture contains only NaCl and Al2(SO4)3. A 1.45-g sample
of the mixture is dissolved in water and an excess of NaOH is
added, producing a precipitate of Al(OH)3. The precipitate is
filtered, dried, and weighed. The mass of the precipitate is
0.107 g. What is the mass percent of Al2(SO4)3 in the sample?
97. The thallium (present as Tl2SO4) in a 9.486-g pesticide sample
was precipitated as thallium(I) iodide. Calculate the mass percent of Tl2SO4 in the sample if 0.1824 g of TlI was recovered.
98. A mixture contains only NaCl and Fe(NO3)3. A 0.456-g sample
of the mixture is dissolved in water, and an excess of NaOH is
added, producing a precipitate of Fe(OH)3. The precipitate is
filtered, dried, and weighed. Its mass is 0.107 g. Calculate the
following.
a. the mass of iron in the sample
b. the mass of Fe(NO3)3 in the sample
c. the mass percent of Fe(NO3)3 in the sample
99. A student added 50.0 mL of an NaOH solution to 100.0 mL
of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation
of 2.06 g of precipitate. Determine the concentration of the
NaOH solution.
100. Some of the substances commonly used in stomach antacids
are MgO, Mg(OH)2, and Al(OH)3.
a. Write a balanced equation for the neutralization of hydrochloric acid by each of these substances.
b. Which of these substances will neutralize the greatest
amount of 0.10 M HCl per gram?
101. Acetylsalicylic acid is the active ingredient in aspirin. It
took 35.17 mL of 0.5065 M sodium hydroxide to react completely with 3.210 g of acetylsalicylic acid. Acetylsalicylic
acid has one acidic hydrogen. What is the molar mass of
acetylsalicylic acid?
102. When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What
volume of 5.0 M HCl is required to react completely with
3.00 g of magnesium?
103. A 2.20-g sample of an unknown acid (empirical formula 5
C3H4O3) is dissolved in 1.0 L of water. A titration required
25.0 mL of 0.500 M NaOH to react completely with all the
acid present. Assuming the unknown acid has one acidic
­proton per molecule, what is the molecular formula of the unknown acid?
104. Carminic acid, a naturally occurring red pigment extracted
from the cochineal insect, contains only carbon, hydrogen,
and oxygen. It was commonly used as a dye in the first half
of the nineteenth century. It is 53.66% C and 4.09% H by
mass. A titration required 18.02 mL of 0.0406 M NaOH to
neutralize 0.3602 g carminic acid. Assuming that there is
only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?
105. Chlorisondamine chloride (C14H20Cl6N2) is a drug used in the
treatment of hypertension. A 1.28-g sample of a medication
containing the drug was treated to destroy the organic material
and to release all the chlorine as chloride ion. When the filtered ­solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered.
Calculate the mass percent of chlorisondamine chloride in the
medication, assuming the drug is the only source of chloride.
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186
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
106. Saccharin (C7H5NO3S) is sometimes dispensed in tablet form.
Ten tablets with a total mass of 0.5894 g were dissolved in
water. The saccharin was oxidized to convert all the sulfur to
sulfate ion, which was precipitated by adding an excess of
barium chloride solution. The mass of BaSO4 obtained was
0.5032 g. What is the average mass of saccharin per tablet?
What is the average mass percent of saccharin in the tablets?
107. Douglasite is a mineral with the formula 2KCl # FeCl2 # 2H2O.
Calculate the mass percent of douglasite in a 455.0-mg sample
if it took 37.20 mL of a 0.1000-M AgNO3 solution to precipitate all the Cl2 as AgCl. Assume the douglasite is the only
source of chloride ion.
108. Many oxidation–reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection.
In each reaction, identify the substance reduced and the substance ­oxidized.
a. Al 1s2 1 HCl 1aq2 S AlCl3 1aq2 1 H2 1g2
b. CH4 1g2 1 S 1s2 S CS2 1l2 1 H2S 1g2
c. C3H8 1g2 1 O2 1g2 S CO2 1g2 1 H2O 1l2
d. Cu 1s2 1 Ag1 1aq2 S Ag 1s2 1 Cu21 1aq2
109. The blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of Cr31(aq) and
carbon dioxide. The reaction can be monitored because the
dichromate ion (Cr2O722) is orange in solution, and the Cr31
ion is green. The balanced equation is
16H1 1aq2 1 2Cr2O722 1aq2 1 C2H5OH 1aq2 h
4Cr31 1aq2 1 2CO2 1g2 1 11H2O 1l2
This reaction is an oxidation–reduction reaction. What species
is reduced, and what species is oxidized? How many electrons
are transferred in the balanced equation above?
ChemWork Problems
These multiconcept problems (and additional ones) are found inter­
actively online with the same type of assistance a student would
get from an instructor.
110. Calculate the concentration of all ions present when 0.160 g
of MgCl2 is dissolved in 100.0 mL of solution.
111. A solution is prepared by dissolving 0.6706 g oxalic acid
(H2C2O4) in enough water to make 100.0 mL of solution. A
10.00-mL aliquot (portion) of this solution is then diluted to a
final volume of 250.0 mL. What is the final molarity of the
oxalic acid solution?
112. For the following chemical reactions, determine the precipitate produced when the two reactants listed below are mixed
together. Indicate “none” if no precipitate will form.
Formula of Precipitate
Sr(NO3)2(aq) 1 K3PO4(aq) 88n
K2CO3(aq) 1 AgNO3(aq) 88n
NaCl(aq) 1 KNO3(aq) 88n
KCl(aq) 1 AgNO3(aq) 88n
FeCl3(aq) 1 Pb(NO3)2(aq) 88n
_______________ (s)
_______________ (s)
_______________ (s)
_______________ (s)
_______________ (s)
113. What volume of 0.100 M NaOH is required to precipitate all
of the nickel(II) ions from 150.0 mL of a 0.249-M solution of
Ni(NO3)2?
114. A 500.0-mL sample of 0.200 M sodium phosphate is mixed
with 400.0 mL of 0.289 M barium chloride. What is the mass
of the solid produced?
115. A 450.0-mL sample of a 0.257-M solution of silver nitrate is
mixed with 400.0 mL of 0.200 M calcium chloride. What is the
concentration of Cl2 in solution after the reaction is complete?
116. The zinc in a 1.343-g sample of a foot powder was precipitated as ZnNH4PO4. Strong heating of the precipitate yielded
0.4089 g Zn2P2O7. Calculate the mass percent of zinc in the
sample of foot powder.
117. A 50.00-mL sample of aqueous Ca(OH)2 requires 34.66 mL of a
0.944-M nitric acid for neutralization. Calculate the concentration (molarity) of the original solution of calcium hydroxide.
118. When organic compounds containing sulfur are burned, sulfur
dioxide is produced. The amount of SO2 formed can be determined by the reaction with hydrogen peroxide:
H2O2 1aq2 1 SO2 1g2 h H2SO4 1aq2
The resulting sulfuric acid is then titrated with a standard
NaOH solution. A 1.302-g sample of coal is burned and the
SO2 is collected in a solution of hydrogen peroxide. It took
28.44 mL of a 0.1000-M NaOH solution to titrate the resulting
sulfuric acid. Calculate the mass percent of sulfur in the coal
sample. Sulfuric acid has two acidic hydrogens.
119. Assign the oxidation state for the element listed in each of the
following compounds:
Oxidation State
S in MgSO4
Pb in PbSO4
O in O2
Ag in Ag
Cu in CuCl2
_______________
_______________
_______________
_______________
_______________
Challenge Problems
120. A 10.00-g sample consisting of a mixture of sodium chloride
and potassium sulfate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead(II) nitrate to form
21.75 g of solid. Determine the mass percent of sodium chloride in the original mixture.
121. The units of parts per million (ppm) and parts per billion (ppb)
are commonly used by environmental chemists. In general,
1 ppm means 1 part of solute for every 106 parts of solution.
Mathematically, by mass:
ppm 5
mg solute
mg solute
5
g solution
kg solution
In the case of very dilute aqueous solutions, a concentration of
1.0 ppm is equal to 1.0 mg of solute per 1.0 mL, which equals
1.0 g solution. Parts per billion is defined in a similar fashion.
Calculate the molarity of each of the following aqueous
solutions.
a. 5.0 ppb Hg in H2O
b. 1.0 ppb CHCl3 in H2O
c. 10.0 ppm As in H2O
d. 0.10 ppm DDT (C14H9Cl5) in H2O
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For Review
122. In the spectroscopic analysis of many substances, a series of
standard solutions of known concentration are measured to
generate a calibration curve. How would you prepare standard
solutions containing 10.0, 25.0, 50.0, 75.0, and 100. ppm of
copper from a commercially produced 1000.0-ppm solution?
Assume each ­solution has a final volume of 100.0 mL. (See
Exercise 121 for definitions.)
123. In most of its ionic compounds, cobalt is either Co(II) or Co(III).
One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained.
A 0.256-g sample of the compound was dissolved in water, and
excess silver nitrate was added. The silver chloride was filtered,
dried, and weighed, and it had a mass of 0.308 g. A second
sample of 0.416 g of the compound was dissolved in water, and
an excess of sodium hydroxide was added. The hydroxide salt
was filtered and heated in a flame, forming cobalt(III) oxide.
The mass of cobalt(III) oxide formed was 0.145 g.
a. What is the percent composition, by mass, of the
compound?
b. Assuming the compound contains one cobalt ion per formula unit, what is the formula?
c. Write balanced equations for the three reactions
described.
124. Polychlorinated biphenyls (PCBs) have been used extensively
as dielectric materials in electrical transformers. Because PCBs
have been shown to be potentially harmful, analysis for their
presence in the environment has become very important. PCBs
are manufactured according to the following generic reaction:
C12H10 1 nCl2 S C12H102nCln 1 nHCl
125.
126.
127.
128.
This reaction results in a mixture of PCB products. The mixture is analyzed by decomposing the PCBs and then precipitating the resulting Cl2 as AgCl.
a. Develop a general equation that relates the average value
of n to the mass of a given mixture of PCBs and the mass
of AgCl produced.
b. A 0.1947-g sample of a commercial PCB yielded 0.4791 g
of AgCl. What is the average value of n for this sample?
Consider the reaction of 19.0 g of zinc with excess silver nitrite
to produce silver metal and zinc nitrite. The reaction is stopped
before all the zinc metal has reacted and 29.0 g of solid metal is
present. Calculate the mass of each metal in the 29.0-g mixture.
A mixture contains only sodium chloride and potassium chloride. A 0.1586-g sample of the mixture was dissolved in water.
It took 22.90 mL of 0.1000 M AgNO3 to completely precipitate all the chloride present. What is the composition (by mass
percent) of the mixture?
You are given a solid that is a mixture of Na2SO4 and K2SO4.
A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of BaCl2 is added. The BaSO4 that
is formed is filtered, dried, and weighed. Its mass is 0.298 g.
What mass of SO422 ion is in the sample? What is the mass
percent of SO422 ion in the sample? What are the percent compositions by mass of Na2SO4 and K2SO4 in the sample?
Zinc and magnesium metal each react with hydrochloric acid
according to the following equations:
Zn 1s2 1 2HCl 1aq2 h ZnCl2 1aq2 1 H2 1g2
Mg 1s2 1 2HCl 1aq2 h MgCl2 1aq2 1 H2 1g2
187
A 10.00-g mixture of zinc and magnesium is reacted with the
stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156 mL of 3.00 M silver nitrate to
produce the maximum possible amount of silver chloride.
a. Determine the percent magnesium by mass in the original
mixture.
b. If 78.0 mL of HCl was added, what was the concentration
of the HCl?
129. You made 100.0 mL of a lead(II) nitrate solution for lab but
forgot to cap it. The next lab session you noticed that there was
only 80.0 mL left (the rest had evaporated). In addition, you
forgot the initial concentration of the solution. You decide to
take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass
of 3.407 g. What was the concentration of the original lead(II)
nitrate solution?
130. Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations.
copper 1II2 sulfate 1aq2 1 iron 1s2 h
copper 1s2 1 iron 1II2 sulfate 1aq2
copper 1II2 sulfate 1aq2 1 iron 1s2 h
copper 1s2 1 iron 1III2 sulfate 1aq2
131.
132.
133.
134.
135.
You place 87.7 mL of a 0.500-M solution of copper(II) sulfate
in a beaker. You then add 2.00 g of iron filings to the copper(II)
sulfate solution. After one of the above reactions occurs, you
isolate 2.27 g of copper. Which equation above describes the
reaction that occurred? Support your answer.
Consider an experiment in which two burets, Y and Z, are simultaneously draining into a beaker that initially contained
275.0 mL of 0.300 M HCl. Buret Y contains 0.150 M NaOH
and buret Z contains 0.250 M KOH. The stoichiometric point
in the titration is reached 60.65 minutes after Y and Z were
started simultaneously. The total volume in the beaker at the
stoichiometric point is 655 mL. Calculate the flow rates of
burets Y and Z. Assume the flow rates remain constant during
the experiment.
Complete and balance each acid–base reaction.
a. H3PO4 1aq2 1 NaOH 1aq2 S
Contains three acidic hydrogens
b. H2SO4 1aq2 1 Al 1OH2 3 1s2 S
Contains two acidic hydrogens
c. H2Se 1aq2 1 Ba 1OH2 2 1aq2 S
Contains two acidic hydrogens
d. H2C2O4 1aq2 1 NaOH 1aq2 S
Contains two acidic hydrogens
What volume of 0.0521 M Ba(OH)2 is required to neutralize
exactly 14.20 mL of 0.141 M H3PO4? Phosphoric acid contains three acidic hydrogens.
A 10.00-mL sample of sulfuric acid from an automobile battery requires 35.08 mL of 2.12 M sodium hydroxide solution
for complete neutralization. What is the molarity of the sulfuric acid? Sulfuric acid contains two acidic hydrogens.
A 0.500-L sample of H2SO4 solution was analyzed by taking a
100.0-mL aliquot and adding 50.0 mL of 0.213 M NaOH. After the reaction occurred, an excess of OH2 ions remained in
the solution. The excess base required 13.21 mL of 0.103 M
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188
Chapter 4
Types of Chemical Reactions and Solution Stoichiometry
HCl for neutralization. Calculate the molarity of the original
sample of H2SO4. Sulfuric acid has two acidic hydrogens.
136. A 6.50-g sample of a diprotic acid requires 137.5 mL of a
0.750 M NaOH solution for complete neutralization. Determine the molar mass of the acid.
137. Citric acid, which can be obtained from lemon juice, has the
mo­lecular formula C6H8O7. A 0.250-g sample of citric acid
dissolved in 25.0 mL of water requires 37.2 mL of 0.105 M
NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?
138. A stream flows at a rate of 5.00 3 104 liters per second (L/s)
upstream of a manufacturing plant. The plant discharges
3.50 3 103 L/s of water that contains 65.0 ppm HCl into the
stream. (See Exercise 121 for definitions.)
a. Calculate the stream’s total flow rate downstream from
this plant.
b. Calculate the concentration of HCl in ppm downstream
from this plant.
c. Further downstream, another manufacturing plant diverts
1.80 3 104 L/s of water from the stream for its own use.
This plant must first neutralize the acid and does so by
adding lime:
here. If the percent yield of the reaction was 88.0%, what mass
of chromium(III) chromate was isolated?
142. The vanadium in a sample of ore is converted to VO21. The
VO21 ion is subsequently titrated with MnO42 in acidic solution to form V(OH)41 and manganese(II) ion. The unbalanced
titration reaction is
CaO 1s2 1 2H1 1aq2 h Ca21 1aq2 1 H2O 1l2
What mass of CaO is consumed in an 8.00-h work day by
this plant?
d. The original stream water contained 10.2 ppm Ca21.
Although no calcium was in the waste water from the first
plant, the waste water of the second plant contains Ca21
from the neutralization process. If 90.0% of the water
used by the second plant is returned to the stream, calculate the concentration of Ca21 in ppm downstream of the
second plant.
139. It took 25.06 60.05 mL of a sodium hydroxide solution to titrate a 0.4016-g sample of KHP (see Exercise 77). Calculate
the concentration and uncertainty in the concentration of the
sodium hydroxide solution. (See Appendix 1.5.) Neglect any
uncertainty in the mass.
Marathon Problems
Integrative Problems
These problems require the integration of multiple concepts to find
the solutions.
140. Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of
ethylene or propylene in the presence of a catalytic transition
metal compound. It is composed solely of C, F, and B; it is
42.23% C and 55.66% F by mass.
a. What is the empirical formula of BARF?
b. A 2.251-g sample of BARF dissolved in 347.0 mL of
solution produces a 0.01267-M solution. What is the
molecular formula of BARF?
141. In a 1-L beaker, 203 mL of 0.307 M ammonium chromate was
mixed with 137 mL of 0.269 M chromium(III) nitrite to produce ammonium nitrite and chromium(III) chromate. Write
the balanced chemical equation for the reaction occurring
MnO42 1aq2 1 VO21 1aq2 1 H2O 1l2 h
V 1OH2 41 1aq2 1 Mn21 1aq2 1 H1 1aq2
To titrate the solution, 26.45 mL of 0.02250 M MnO42 was
required. If the mass percent of vanadium in the ore was
58.1%, what was the mass of the ore sample? Hint: Balance
the titration reaction by the oxidation states method.
143. The unknown acid H2X can be neutralized completely by
OH2 according to the following (unbalanced) equation:
H2X 1aq2 1 OH2 1aq2 h X22 1aq2 1 H2O 1l2
The ion formed as a product, X22, was shown to have 36 total
electrons. What is element X? Propose a name for H2X. To
completely neutralize a sample of H2X, 35.6 mL of 0.175 M
OH2 solution was required. What was the mass of the H2X
sample used?
These problems are designed to incorporate several concepts and
techniques into one situation.
144. Three students were asked to find the identity of the metal in a
particular sulfate salt. They dissolved a 0.1472-g sample of the
salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate
had been filtered and dried, it weighed 0.2327 g.
Each student analyzed the data independently and came to
different conclusions. Pat decided that the metal was titanium.
Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt?
Look for information on the sulfates of gallium, sodium, and
titanium in this text and reference books such as the CRC
Handbook of Chemistry and Physics. What further tests would
you suggest to determine which student is most likely correct?
145. You have two 500.0-mL aqueous solutions. Solution A is a
solution of a metal nitrate that is 8.246% nitrogen by mass.
The ionic compound in solution B consists of potassium, chromium, and oxygen; chromium has an oxidation state of 16
and there are 2 potassiums and 1 chromium in the formula.
The masses of the solutes in each of the solutions are the same.
When the solutions are added together, a blood-red precipitate
forms. After the reaction has gone to completion, you dry the
solid and find that it has a mass of 331.8 g.
a. Identify the ionic compounds in solution A and solution B.
b. Identify the blood-red precipitate.
c. Calculate the concentration (molarity) of all ions in the
orig­inal solutions.
d. Calculate the concentration (molarity) of all ions in the
final solution.
Marathon Problems can be used in class by groups of students to
help facilitate problem-solving skills.
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Chapter 5
Gases
5.1
Pressure
5.5
Units of Pressure
5.2
5.3
5.4
The Gas Laws of Boyle, Charles,
and Avogadro
5.6
Dalton’s Law of Partial Pressures
Deriving the Ideal Gas Law
Collecting a Gas over Water
The Meaning of Temperature
The Kinetic Molecular Theory of
Gases
Root Mean Square Velocity
5.7
Effusion and Diffusion
Boyle’s Law
Pressure and Volume (Boyle’s Law)
Effusion
Charles’s Law
Pressure and Temperature
Diffusion
Avogadro’s Law
Volume and Temperature (Charles’s Law)
The Ideal Gas Law
Volume and Number of Moles
(Avogadro’s Law)
Gas Stoichiometry
Molar Mass of a Gas
Mixture of Gases (Dalton’s Law)
5.8
5.9
Real Gases
Characteristics of Several Real
Gases
5.10 Chemistry in the Atmosphere
The artistry of this sunset over Lamarck Col in California’s Sierra Nevada mountains results from reflections
on the clouds in our gaseous atmosphere. (Jerry Dodrill/Aurora Photos/Getty Images)
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189
M
atter exists in three distinct physical states: gas, liquid, and solid. Although
relatively few substances exist in the gaseous state under typical conditions,
gases are very important. For example, we live immersed in a gaseous solution. The
earth’s atmosphere is a mixture of gases that consists mainly of elemental nitrogen (N2)
and oxygen (O2). The atmosphere both supports life and acts as a waste receptacle for
the exhaust gases that accompany many industrial processes. The chemical reactions of
these waste gases in the atmosphere lead to various types of pollution, including smog
and acid rain. The gases in the atmosphere also shield us from harmful radiation from
the sun and keep the earth warm by reflecting heat radiation back toward the earth. In
fact, there is now great concern that an increase in atmospheric carbon dioxide, a product of the combustion of fossil fuels, is causing a dangerous warming of the earth.
In this chapter we will look carefully at the properties of gases. First we will see
how measurements of gas properties lead to various types of laws—statements that
show how the properties are related to each other. Then we will construct a model to
explain why gases behave as they do. This model will show how the behavior of the
individual particles of a gas leads to the observed properties of the gas itself (a collection of many, many particles).
The study of gases provides an excellent example of the scientific method in action. It illustrates how observations lead to natural laws, which in turn can be accounted for by models.
5.1 Pressure
As a gas, water occupies 1200 times as
much space as it does as a liquid at 258C
and atmospheric pressure.
A gas uniformly fills any container, is easily compressed, and mixes completely with
any other gas. One of the most obvious properties of a gas is that it exerts pressure on
its surroundings. For example, when you blow up a balloon, the air inside pushes
against the elastic sides of the balloon and keeps it firm.
As mentioned earlier, the gases most familiar to us form the earth’s atmosphere.
The pressure exerted by this gaseous mixture that we call air can be dramatically demonstrated by the experiment shown in Fig. 5.1. A small volume of water is placed in a
190
a
Charles D. Winters
by the gases in the atmosphere can
be demonstrated by boiling water
in a large metal can (a) and then
turning off the heat and sealing the
can. As the can cools, the water vapor
condenses, lowering the gas pressure
inside the can. This causes the can to
crumple (b).
Charles D. Winters
Figure 5.1 | The pressure exerted
b
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5.1 Pressure
Vacuum
h = 760 mm Hg
for standard
atmosphere
Figure 5.2 | A torricellian barometer. The tube, completely filled with
mercury, is inverted in a dish of
mercury. Mercury flows out of the
tube until the pressure of the column
of mercury (shown by the black arrow)
“standing on the surface” of the
mercury in the dish is equal to the
pressure of the air (shown by the
purple arrows) on the rest of the
surface of the mercury in the dish.
Soon after Torricelli died, a German
physicist named Otto von Guericke
invented an air pump. In a famous
demonstration for the King of Prussia in
1663, Guericke placed two hemispheres
together, pumped the air out of the
resulting sphere through a valve, and
showed that teams of horses could not
pull the hemispheres apart. Then, after
secretly opening the air valve, Guericke
easily separated the hemispheres by hand.
The King of Prussia was so impressed that
he awarded Guericke a lifetime pension!
metal can, and the water is boiled, which fills the can with steam. The can is then
sealed and allowed to cool. Why does the can collapse as it cools? It is the atmospheric
pressure that crumples the can. When the can is cooled after being sealed so that no air
can flow in, the water vapor (steam) condenses to a very small volume of liquid water.
As a gas, the water filled the can, but when it is condensed to a liquid, the liquid does
not come close to filling the can. The H2O molecules formerly present as a gas are now
collected in a very small volume of liquid, and there are very few molecules of gas left
to exert pressure outward and counteract the air pressure. As a result, the pressure exerted by the gas molecules in the atmosphere smashes the can.
A device to measure atmospheric pressure, the barometer, was invented in 1643 by
an Italian scientist named Evangelista Torricelli (1608–1647), who had been a student
of Galileo. Torricelli’s barometer is constructed by filling a glass tube with liquid mercury and inverting it in a dish of mercury (Fig. 5.2). Notice that a large quantity of
mercury stays in the tube. In fact, at sea level the height of this column of mercury
averages 760 mm. Why does this mercury stay in the tube, seemingly in defiance of
gravity? Figure 5.2 illustrates how the pressure exerted by the atmospheric gases on
the surface of mercury in the dish keeps the mercury in the tube.
Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity—in other words, it results from the weight of the air. Changing weather conditions cause the atmospheric pressure to vary, so the height of the
column of Hg supported by the atmosphere at sea level varies; it is not always 760 mm.
The meteorologist who says a “low” is approaching means that the atmospheric pressure is going to decrease. This condition often occurs in conjunction with a storm.
Atmospheric pressure also varies with altitude. For example, when Torricelli’s experiment is done in Breckenridge, Colorado (elevation 9600 feet), the atmosphere supports a column of mercury only about 520 mm high because the air is “thinner.” That is,
there is less air pushing down on the earth’s surface at Breckenridge than at sea level.
Units of Pressure
Because instruments used for measuring pressure, such as the manometer (Fig. 5.3),
often contain mercury, the most commonly used units for pressure are based on the
height of the mercury column (in millimeters) that the gas pressure can support. The
Atmospheric
pressure (Patm )
Atmospheric
pressure (Patm )
h
Figure 5.3 | A simple manometer, a
device for measuring the pressure of a
gas in a container. The pressure of the
gas is given by h (the difference in
mercury levels) in units of torr
(equivalent to mm Hg). (a) Gas
pressure 5 atmospheric pressure 2 h.
(b) Gas pressure 5 atmospheric
pressure 1 h.
191
h
Pgas < Patm
Pgas > Patm
Pgas = Patm – h
a
Pgas = Patm + h
b
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192
Chapter 5
Gases
unit mm Hg (millimeter of mercury) is often called the torr in honor of Torricelli. The
terms torr and mm Hg are used interchangeably by chemists. A related unit for pressure is the standard atmosphere (abbreviated atm):
1 standard atmosphere 5 1 atm 5 760 mm Hg 5 760 torr
However, since pressure is defined as force per unit area,
Vanessa Vick/Photo Researchers, Inc.
Pressure 5
Checking tire pressure.
Interactive
Example 5.1
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
1 atm 5 760 mm Hg
5 760 torr
5 101,325 Pa
5 29.92 in Hg
5 14.7 lb/in2
force
area
the fundamental units of pressure involve units of force divided by units of area. In the
SI system, the unit of force is the newton (N) and the unit of area is meters squared
(m2). (For a review of the SI system, see Chapter 1.) Thus the unit of pressure in the SI
system is newtons per meter squared (N/m2) and is called the pascal (Pa). In terms of
pascals, the standard atmosphere is
1 standard atmosphere 5 101,325 Pa
Thus 1 atmosphere is about 105 pascals. Since the pascal is so small, and since it is
not commonly used in the United States, we will use it sparingly in this book. However, converting from torrs or atmospheres to pascals is straightforward, as shown in
Example 5.1.
Pressure Conversions
The pressure of a gas is measured as 49 torr. Represent this pressure in both atmospheres and pascals.
Solution
1 atm
5 6.4 3 1022 atm
760 torr
101,325 Pa
6.4 3 1022 atm 3
5 6.5 3 103 Pa
1 atm
49 torr 3
See Exercises 5.37 and 5.38
5.2 The Gas Laws of Boyle, Charles,
PowerLecture: Boyle’s Law: A Graphical
and Avogadro
View
IBLG: See questions from
“Gas Laws”
In this section we will consider several mathematical laws that relate the properties of
gases. These laws derive from experiments involving careful measurements of the relevant gas properties. From these experimental results, the mathematical relationships
among the properties can be discovered. These relationships are often represented pictorially by means of graphs (plots).
We will take a historical approach to these laws to give you some perspective on the
scientific method in action.
Boyle’s Law
Boyle’s law: V ~ 1yP at constant
temperature
Graphing is reviewed in Appendix 1.3.
The first quantitative experiments on gases were performed by an Irish chemist, Robert
Boyle (1627–1691). Using a J-shaped tube closed at one end (Fig. 5.4), which he reportedly set up in the multistory entryway of his house, Boyle studied the relationship between the pressure of the trapped gas and its volume. Representative values from
Boyle’s experiments are given in Table 5.1. These data show that the product of the
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5.2
Mercury
added
Gas
Gas
The Gas Laws of Boyle, Charles, and Avogadro
193
Table 5.1 | Actual Data from Boyle’s Experiment
Pressure (in Hg)
Volume (in3)
Pressure 3 Volume
(in Hg 3 in3)
117.5
87.2
70.7
58.8
44.2
35.3
29.1
12.0
16.0
20.0
24.0
32.0
40.0
48.0
14.1 3 102
14.0 3 102
14.1 3 102
14.1 3 102
14.1 3 102
14.1 3 102
14.0 3 102
h
h
pressure and volume for the trapped air sample is constant within the accuracies of
Boyle’s measurements (note the third column in Table 5.1). This behavior can be represented by the equation
PV 5 k
Mercury
which is called Boyle’s law, where k is a constant for a given sample of air at a specific
temperature.
It is convenient to represent the data in Table 5.1 by using two different plots. The first
type of plot, P versus V, forms a curve called a hyperbola [Fig. 5.5(a)]. Looking at this
plot, note that as the pressure drops by about half (from 58.8 to 29.1), the volume doubles
(from 24.0 to 48.0). In other words, there is an inverse relationship between pressure and
volume. The second type of plot can be obtained by rearranging Boyle’s law to give
Figure 5.4 | A J-tube similar to the
one used by Boyle. When mercury is
added to the tube, pressure on the
trapped gas is increased, ­resulting in a
decreased volume.
V5
k
1
5k
P
P
which is the equation for a straight line of the type
y 5 mx 1 b
P (in Hg)
100
50
P
P
2
0
20
40
V
2V
V (in3)
V (in3)
a
40
slope = k
20
0
0
0.01
0.02
1/P (in Hg)
b
0.03
60
where m represents the slope and b is the intercept of the straight line. In this case, y 5
V, x 5 1yP, m 5 k, and b 5 0. Thus a plot of V versus 1yP using Boyle’s data gives a
straight line with an intercept of zero [Fig. 5.5(b)].
In the three centuries since Boyle carried out his studies, the sophistication of measuring techniques has increased tremendously. The results of highly accurate measurements show that Boyle’s law holds precisely only at very low pressures. Measurements
at higher pressures reveal that PV is not constant but varies as the pressure is varied.
Results for several gases at pressures below 1 atm are shown in Fig. 5.6. Note the very
small changes that occur in the product PV as the pressure is changed at these low
pressures. Such changes become more significant at much higher pressures, where the
complex nature of the dependence of PV on pressure becomes more obvious. We will
discuss these deviations and the reasons for them in detail in Section 5.8. A gas that
strictly obeys Boyle’s law is called an ideal gas. We will describe the characteristics of
an ideal gas more completely in Section 5.3.
One common use of Boyle’s law is to predict the new volume of a gas when the
pressure is changed (at constant temperature), or vice versa. Because deviations from
Boyle’s law are so slight at pressures close to 1 atm, in our calculations we will assume
that gases obey Boyle’s law (unless stated otherwise).
Figure 5.5 | Plotting Boyle’s data from Table 5.1. (a) A plot of P versus V shows that the
volume doubles as the pressure is halved. (b) A plot of V versus 1yP gives a straight line. The
slope of this line equals the value of the constant k.
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194
Chapter 5
Gases
Figure 5.6 | A plot of PV versus P for
Interactive
Example 5.2
Ideal
22.45
Ne
22.40
PV (L • atm)
several gases at pressures below 1 atm.
An ideal gas is expected to have a
constant value of PV, as shown by the
dotted line. Carbon dioxide shows the
largest change in PV, and this change
is actually quite small: PV changes
from about 22.39 L ? atm at 0.25 atm
to 22.26 L ? atm at 1.00 atm. Thus
Boyle’s law is a good approximation at
these relatively low pressures.
O2
22.35
CO2
22.30
22.25
0
0.25
0.50 0.75
P (atm)
1.00
Boyle’s Law I
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is
found in the exhaust of automobiles and power plants. Consider a 1.53-L sample of
gaseous SO2 at a pressure of 5.6 3 103 Pa. If the pressure is changed to 1.5 3 104 Pa
at a constant temperature, what will be the new volume of the gas?
Solution
Where are we going?
To calculate the new volume of gas
What do we know?
❯
❯
P1 5 5.6 3 103 Pa ❯ P2 5 1.5 3 104 Pa
V1 5 1.53 L
❯ V2 5 ?
What information do we need?
❯
Boyle’s law also can be written as
Boyle’s law
P1V1 5 P2V2
3
5.6 × 10 Pa
PV 5 k
4
1.5 × 10 Pa
How do we get there?
What is Boyle’s law (in a form useful with our knowns)?
P1V1 5 P2V2
What is V2?
V2 5
V = 1.53 L
V=?
As pressure increases, the volume of SO2
decreases.
❯
P1V1
5.6 3 103 Pa 3 1.53 L
5
5 0.57 L
P2
1.5 3 104 Pa
The new volume will be 0.57 L.
Reality Check | The new volume (0.57 L) is smaller than the original volume. As
pressure increases, the volume should decrease, so our answer is reasonable.
See Exercise 5.43
The fact that the volume decreases in Example 5.2 makes sense because the pressure was increased. To help eliminate errors, make it a habit to check whether an answer to a problem makes physical (common!) sense.
We mentioned before that Boyle’s law is only approximately true for real gases. To
determine the significance of the deviations, studies of the effect of changing pressure
on the volume of a gas are often done, as shown in Example 5.3.
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5.2
Example 5.3
The Gas Laws of Boyle, Charles, and Avogadro
195
Boyle’s Law II
In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume
measurements were made at various pressures, using 1.0 mole of NH3 gas at a temperature of 08C. Using the results listed below, calculate the Boyle’s law constant for
NH3 at the various pressures.
Experiment
Pressure (atm)
Volume (L)
1
2
3
4
5
6
0.1300
0.2500
0.3000
0.5000
0.7500
1.000
172.1
89.28
74.35
44.49
29.55
22.08
22.6
PV (L • atm)
22.5
22.4
Solution
To determine how closely NH3 gas follows Boyle’s law under these conditions, we
calculate the value of k (in L ? atm) for each set of values:
22.3
22.2
Experiment
k 5 PV
22.1
0
0.20 0.40 0.60 0.80 1.00
P (atm)
Figure 5.7 | A plot of PV versus P
for 1 mole of ammonia. The dashed line
shows the extrapolation of the data to
zero pressure to give the “ideal” value
of PV of 22.41 L ? atm.
1
22.37
2
22.32
3
22.31
4
22.25
5
22.16
6
22.08
Although the deviations from true Boyle’s law behavior are quite small at these
low pressures, note that the value of k changes regularly in one direction as the pressure is increased. Thus, to calculate the “ideal” value of k for NH3, we can plot PV
versus P (Fig. 5.7), and extrapolate (extend the line beyond the experimental points)
back to zero pressure, where, for reasons we will discuss later, a gas behaves most ideally. The value of k obtained by this extrapolation is 22.41 L ? atm. Notice that this is
the same value obtained from similar plots for the gases CO2, O2, and Ne at 08C (see
Fig. 5.6).
See Exercise 5.125
Charles’s Law
PowerLecture: Liquid Nitrogen and
Balloons
6
5
V (L)
4
He
CH4
3
H2O
2
H2
1
N2O
–300 –200 –100 0 100 200 300
–273°C
T (°C)
In the century following Boyle’s findings, scientists continued to study the properties
of gases. One of these scientists was a French physicist, Jacques Charles (1746–1823),
who was the first person to fill a balloon with hydrogen gas and who made the first solo
balloon flight. Charles found in 1787 that the volume of a gas at constant pressure increases linearly with the temperature of the gas. That is, a plot of the volume of a gas
(at constant pressure) versus its temperature (8C) gives a straight line. This behavior is
shown for samples of several gases in Fig. 5.8. The slopes of the lines in this graph are
different ­because the samples contain different numbers of moles of gas. A very interesting feature of these plots is that the volumes of all the gases extrapolate to zero at
the same temperature, 22738C. On the Kelvin temperature scale, this point is defined
as 0 K, which leads to the following relationship between the Kelvin and Celsius
scales:
K 5 °C 1 273
Figure 5.8 | Plots of V versus T (8C) for several gases. The solid lines represent experimental measurements of gases. The dashed lines represent extrapolation of the data into
regions where these gases would become liquids or solids. Note that the samples of the
various gases contain different numbers of moles.
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196
Chapter 5
Gases
Figure 5.9 | Plots of V versus T as in
He
6
Fig. 5.8, except here the Kelvin scale is
used for temperature.
5
CH4
V (L)
4
3
H2O
2
H2
1
N2O
73 173 273 373 473 573
0
T (K)
When the volumes of the gases shown in Fig. 5.8 are plotted versus temperature on the
Kelvin scale, the plots in Fig. 5.9 result. In this case, the volume of each gas is directly
proportional to temperature and extrapolates to zero when the temperature is 0 K. This
behavior is represented by the equation known as Charles’s law,
V 5 bT
PowerLecture: Charles’s Law: A Graphical
View
Charles’s law: V ~ T (expressed in K) at
constant pressure.
where T is in kelvins and b is a proportionality constant.
Before we illustrate the uses of Charles’s law, let us consider the importance of 0 K.
At temperatures below this point, the extrapolated volumes would become negative.
The fact that a gas cannot have a negative volume suggests that 0 K has a special significance. In fact, 0 K is called absolute zero, and there is much evidence to suggest
that this ­temperature cannot be attained. Temperatures of approximately 0.000000001 K
have been produced in laboratories, but 0 K has never been reached.
Critical Thinking
According to Charles’s law, the volume of a gas is directly related to its temperature in
kelvins at constant pressure and number of moles. What if the volume of a gas was directly related to its temperature in degrees Celsius at constant pressure and number
of moles? What differences would you notice?
Interactive
Example 5.4
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.com to try this Interactive Example
in OWL.
Experiment 17: Gas Laws 1: Charles’s
Law and Absolute Zero
Charles’s Law
A sample of gas at 158C and 1 atm has a volume of 2.58 L. What volume will this gas
occupy at 388C and 1 atm?
Solution
Where are we going?
To calculate the new volume of gas
What do we know?
❯ T1 5 15°C 1 273 5 288 K ❯ T2 5 38°C 1 273 5 311 K
❯ V1 5 2.58 L
❯ V2 5 ?
What information do we need?
Charles’s law also can be written as
V1
V2
5
T1
T2
❯
Charles’s law
V
5b
T
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5.2
N2
H2
The Gas Laws of Boyle, Charles, and Avogadro
197
How do we get there?
What is Charles’s law (in a form useful with our knowns)?
V1
V
5 2
T1
T2
What is V2?
j
Ar
CH4
V2 5 a
T2
311 K
b V1 5 a
b 2.58 L 5 2.79 L
T1
288 K
Reality Check | The new volume is greater than the original volume, which makes
physical sense because the gas will expand as it is heated.
See Exercise 5.44
Avogadro’s Law
In Chapter 2 we noted that in 1811 the Italian chemist Avogadro postulated that equal
volumes of gases at the same temperature and pressure contain the same number of
“particles.” This observation is called Avogadro’s law, which is illustrated by Fig.
5.10. Stated mathematically, Avogadro’s law is
Figure 5.10 | These balloons each
hold 1.0 L gas at 258C and 1 atm. Each
balloon contains 0.041 mole of gas, or
2.5 3 1022 molecules.
Interactive
Example 5.5
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in OWL.
V 5 an
where V is the volume of the gas, n is the number of moles of gas particles, and a is a
proportionality constant. This equation states that for a gas at constant temperature
and pressure, the volume is directly proportional to the number of moles of gas. This
relationship is obeyed closely by gases at low pressures.
Avogadro’s Law
Suppose we have a 12.2-L sample containing 0.50 mole of oxygen gas (O2) at a pressure of 1 atm and a temperature of 258C. If all this O2 were converted to ozone (O3) at
the same temperature and pressure, what would be the volume of the ozone?
Solution
Where are we going?
To calculate the volume of the ozone produced by 0.50 mole of oxygen
What do we know?
❯
❯
n1 5 0.50 mol O2 ❯ n2 5 ? mol O3
V1 5 12.2 L O2 ❯ V2 5 ? L O3
What information do we need?
❯
❯
Avogadro’s law also can be written as
V1
V2
5
n1
n2
❯
Balanced equation
Moles of O3
Avogadro’s law
V 5 an
How do we get there?
How many moles of O3 are produced by 0.50 mole of O2?
What is the balanced equation?
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3O2 1g2 h 2O3 1g2
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198
Chapter 5
Gases
What is the mole ratio between O3 and O2?
2 mol O3
3 mol O2
Now we can calculate the moles of O3 formed.
0.50 mol O2 3
2 mol O3
5 0.33 mol O3
3 mol O2
What is the volume of O3 produced?
Avogadro’s law states that V 5 an, which can be rearranged to give
V
5a
n
Since a is a constant, an alternative representation is
V1
V
5a5 2
n1
n2
where V1 is the volume of n1 moles of O2 gas and V2 is the volume of n2 moles of O3
gas. In this case we have
❯
❯
j
n1 5 0.50 mol ❯ n2 5 0.33 mol
V1 5 12.2 L ❯ V2 5 ?
Solving for V2 gives
V2 5 a
n2
0.33 mol
b V1 5 a
b12.2 L 5 8.1 L
n1
0.50 mol
Reality Check | Note that the volume decreases, as it should, since fewer moles of gas
molecules will be present after O2 is converted to O3.
See Exercises 5.45 and 5.46
5.3 The Ideal Gas Law
PowerLecture: The Ideal Gas Law
Experiment 19: Gas Laws 3: Molar Mass
of a Volatile Liquid
We have considered three laws that describe the behavior of gases as revealed by experimental observations:
Boyle’s law:
V5
k
P
Charles’s law:
V 5 bT
Avogadro’s law:
V 5 an
1at constant T and n2
1at constant P and n2
1at constant T and P2
These relationships, which show how the volume of a gas depends on pressure, temperature, and number of moles of gas present, can be combined as follows:
V 5 Ra
R 5 0.08206
L # atm
K # mol
Tn
b
P
where R is the combined proportionality constant called the universal gas constant.
When the pressure is expressed in atmospheres and the volume in liters, R has the
value 0.08206 L ? atm/K ? mol. The preceding equation can be rearranged to the more
familiar form of the ideal gas law:
PV 5 nRT
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5.3
The Ideal Gas Law
199
The ideal gas law is an equation of state for a gas, where the state of the gas is its
condition at a given time. A particular state of a gas is described by its pressure, volume, temperature, and number of moles. Knowledge of any three of these properties is
enough to completely define the state of a gas, since the fourth property can then be
determined from the equation for the ideal gas law.
It is important to recognize that the ideal gas law is an empirical equation—it is
based on experimental measurements of the properties of gases. A gas that obeys this
equation is said to behave ideally. The ideal gas equation is best regarded as a limiting
law—it expresses behavior that real gases approach at low pressures and high temperatures. Therefore, an ideal gas is a hypothetical substance. However, most gases
obey the ideal gas equation closely enough at pressures below 1 atm that only minimal
errors result from ­assuming ideal behavior. Unless you are given information to the
contrary, you should ­assume ideal gas behavior when solving problems involving
gases in this text.
The ideal gas law can be used to solve a variety of problems. Example 5.6 demonstrates one type, where you are asked to find one property characterizing the state of a
gas, given the other three.
PowerLecture: The Ideal Gas Law,
PV 5 nRT
The ideal gas law applies best at pressures
smaller than 1 atm.
Interactive
Example 5.6
Ideal Gas Law I
A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 08C and a
pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.
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Solution
Where are we going?
To calculate the moles of H2
What do we know?
❯
❯
❯
Ken O’Donoghue © Cengage Learning
❯
The reaction of zinc with hydrochloric acid
to produce bubbles of hydrogen gas.
n 5 ? mol H2
V 5 8.56 L
P 5 1.5 atm
T 5 0°C 1 273 5 273 K
What information do we need?
❯ Ideal gas law
PV 5 nRT
❯
R 5 0.08206 L ? atm/K ? mol
How do we get there?
How many moles of H2 are present in the sample?
❯
Solve the ideal gas equation for n:
n5
11.5 atm2 18.56 L2
5 0.57 mol
L # atm
a0.08206 #
b 1273 K2
K mol
See Exercises 5.47 through 5.54
Gas law problems can be solved in a variety of ways. They can be classified as a
Boyle’s law, Charles’s law, or Avogadro’s law problem and solved, but now we need to
remember the specific law and when it applies. The real advantage of using the ideal gas
law is that it applies to virtually any problem dealing with gases and is easy to remember.
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200
Chapter 5
Gases
The basic assumption we make when using the ideal gas law to describe a change
in state for a gas is that the equation applies equally well to both the initial and final
states. In dealing with changes in state, we always place the variables that change on
one side of the equal sign and the constants on the other. Let’s see how this might work
in several examples.
Interactive
Example 5.7
Ideal Gas Law II
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in OWL.
Suppose we have a sample of ammonia gas with a volume of 7.0 mL at a pressure of
1.68 atm. The gas is compressed to a volume of 2.7 mL at a constant temperature. Use
the ideal gas law to calculate the final pressure.
Solution
Where are we going?
To use the ideal gas equation to determine the final pressure
What do we know?
❯
❯
10
10
9
9
8
8
7.0 mL
7
6
5
5
4
4
3
3
2
2
1
What information do we need?
❯ Ideal gas law
7
6
PV 5 nRT
❯
2.7 mL
R 5 0.08206 L ? atm/K ? mol
How do we get there?
What are the variables that change?
1
mL
P1 5 1.68 atm ❯ P2 5 ?
V1 5 7.0 mL ❯ V2 5 2.7 mL
mL
P, V
1.68
3
2
1
atm
0
4
5
3
2
1
atm
0
4
5
As pressure increases, the volume
­decreases.
4.4
What are the variables that remain constant?
n, R, T
Write the ideal gas law, collecting the change variables on one side of the equal sign
and the variables that do not change on the other.
PV 5 nRT
p
r
Change Remain constant
Since n and T remain the same in this case, we can write P1V1 5 nRT and P2V2 5
nRT. Combining these gives
P1V1 5 nRT 5 P2V2
or
P1V1 5 P2V2
We are given P1 5 1.68 atm, V1 5 7.0 mL, and V2 5 2.7 mL. Solving for P2 thus gives
j
PowerLecture: Collapsing Can
P2 5 a
V1
7.0 mL
bP1 5 a
b 1.68 atm 5 4.4 atm
V2
2.7 mL
Reality Check | Does this answer make sense? The volume decreased (at constant
temperature), so the pressure should increase, as the result of the calculation indicates.
Note that the calculated final pressure is 4.4 atm. Most gases do not behave ideally
above 1 atm. Therefore, we might find that if we measured the pressure of this gas
sample, the observed pressure would differ slightly from 4.4 atm.
See Exercises 5.55 and 5.56
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5.3
Interactive
Example 5.8
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in OWL.
The Ideal Gas Law
201
Ideal Gas Law III
A sample of methane gas that has a volume of 3.8 L at 58C is heated to 868C at constant
pressure. Calculate its new volume.
Solution
Where are we going?
To use the ideal gas equation to determine the final volume
What do we know?
❯ T1 5 5°C 1 273 5 278 K ❯ T2 5 86°C 1 273 5 359 K
❯ V1 5 3.8 L
❯ V2 5 ?
What information do we need?
❯ Ideal gas law
PV 5 nRT
❯
R 5 0.08206 L ? atm/K ? mol
How do we get there?
What are the variables that change?
V, T
What are the variables that remain constant?
n, R, P
rite the ideal gas law, collecting the change variables on one side of the equal sign
W
and the variables that do not change on the other.
V
nR
5
T
P
which leads to
V1
nR
5
T1
P
V2
nR
5
T2
P
and
Combining these gives
V1
nR
V
5
5 2
T1
P
T2
❯
or
V1
V
5 2
T1
T2
Solving for V2:
V2 5
1359 K2 13.8 L2
T2V1
5
5 4.9 L
T1
278 K
Reality Check | Is the answer sensible? In this case the temperature increased (at constant pressure), so the volume should increase. Thus the answer makes sense.
See Exercises 5.57 through 5.59
The problem in Example 5.8 could be described as a Charles’s law problem,
whereas the problem in Example 5.7 might be called a Boyle’s law problem. In both
cases, however, we started with the ideal gas law. The real advantage of using the ideal
gas law is that it applies to virtually any problem dealing with gases and is easy to
­remember.
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202
Chapter 5
Gases
Interactive
Example 5.9
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in OWL.
Ideal Gas Law IV
A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to
air, has a pressure of 345 torr at a temperature of 2158C and a volume of 3.48 L. If
conditions are changed so that the temperature is 368C and the pressure is 468 torr,
what will be the volume of the sample?
Solution
Where are we going?
To use the ideal gas equation to determine the final volume
What do we know?
❯ T1 5 15°C 1 273 5 258 K ❯ T2 5 36°C 1 273 5 309 K
❯ V1 5 3.48 L
❯ V2 5 ?
❯ P1 5 345 torr
❯ P2 5 468 torr
What information do we need?
❯ Ideal gas law
PV 5 nRT
❯
R 5 0.08206 L ? atm/K ? mol
How do we get there?
What are the variables that change?
P, V, T
What are the variables that remain constant?
n, R
Write the ideal gas law, collecting the change variables on one side of the equal sign
and the variables that do not change on the other.
PV
5 nR
T
which leads to
P1V1
PV
5 nR 5 2 2
T1
T2
❯
or
P1V1
PV
5 2 2
T1
T2
Solving for V2:
V2 5
1309 K2 1345 torr2 13.48 L2
T2P1V1
5
5 3.07 L
1258 K2 1468 torr2
T1P2
See Exercises 5.61 and 5.62
Always convert the temperature to the
Kelvin scale when applying the ideal
gas law.
Since the equation used in Example 5.9 involves a ratio of pressures, it was unnecessary to convert pressures to units of atmospheres. The units of torrs cancel. (You
345
468
will obtain the same answer by inserting P1 5
and P2 5
into the equation.)
760
760
However, temperature must always be converted to the Kelvin scale; since this conversion involves addition of 273, the conversion factor does not cancel. Be careful.
One of the many other types of problems dealing with gases that can be solved ­­using
the ideal gas law is illustrated in Example 5.10.
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5.4
Interactive
Example 5.10
Gas Stoichiometry
203
Ideal Gas Law V
A sample containing 0.35 mole of argon gas at a temperature of 138C and a pressure of
568 torr is heated to 568C and a pressure of 897 torr. Calculate the change in volume that
occurs.
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in OWL.
Solution
Where are we going?
To use the ideal gas equation to determine the final volume
Roger Ressmeyer/Corbis
What do we know?
A Twyman–Green interferometer emits
green argon laser light. Interferometers
can measure extremely small distances,
useful in configuring telescope mirrors.
State 1
State 2
n1 5 0.35 mol
n2 5 0.35 mol
1 atm
P1 5 568 torr 3
5 0.747 atm
760 torr
P2 5 897 torr 3
T1 5 138C 1 273 5 286 K
T2 5 568C 1 273 5 329 K
1 atm
5 1.18 atm
760 torr
What information do we need?
❯ Ideal gas law
PV 5 nRT
❯
❯
R 5 0.08206 L ? atm/K ? mol
V1 and V2
How do we get there?
What is V1?
V1 5
What is V2?
V2 5
10.35 mol2 10.08206 L # atm/K # mol2 1286 K2
n1RT1
5
5 11 L
10.747 atm2
P1
10.35 mol2 10.08206 L # atm/K # mol2 1329 K2
n2RT2
5
5 8.0 L
11.18 atm2
P2
What is the change in volume DV?
j
DV 5 V2 2 V1 5 8.0 L 2 11 L 5 23 L
The change in volume is negative because the volume decreases.
Note: For this problem (unlike Example 5.9), the pressures must be converted from
torrs to atmospheres as required by the atmospheres part of the units for R since each
volume was found separately, and the conversion factor does not cancel.
See Exercise 5.63
When 273.15 K is used in this
calculation, the molar volume
obtained in Example 5.3 is the
same value as 22.41 L.
5.4 Gas Stoichiometry
Suppose we have 1 mole of an ideal gas at 08C (273.2 K) and 1 atm. From the ideal gas
law, the volume of the gas is given by
IBLG: See questions from “Gas
Stoichiometry”
V5
11.000 mol2 10.08206 L # atm/K # mol2 1273.2 K2
nRT
5
5 22.42 L
P
1.000 atm
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204
Chapter 5
Gases
Table 5.2 | Molar Volumes for Various Gases
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All rights reserved
at 08C and 1 atm
Figure 5.11 | 22.42 L of a gas would
just fit into this beach ball.
STP: 08C and 1 atm
Gas
Molar Volume (L)
Oxygen (O2)
Nitrogen (N2)
Hydrogen (H2)
Helium (He)
Argon (Ar)
Carbon dioxide (CO2)
Ammonia (NH3)
22.397
22.402
22.433
22.434
22.397
22.260
22.079
This volume of 22.42 L is the molar volume of an ideal gas (at 08C and 1 atm). The
measured molar volumes of several gases are listed in Table 5.2. Note that the molar
­volumes of some of the gases are very close to the ideal value, while others deviate significantly. Later in this chapter, we will discuss some of the reasons for the deviations.
The conditions 08C and 1 atm, called standard temperature and pressure (abbreviated STP), are common reference conditions for the properties of gases. For example, the molar volume of an ideal gas is 22.42 L at STP (Fig. 5.11).
Critical Thinking
What if STP was defined as normal room temperature (22°C) and 1 atm? How would
this affect the molar volume of an ideal gas? Include an explanation and a number.
Interactive
Example 5.11
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in OWL.
Gas Stoichiometry I
A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are
present?
Solution
We could solve this problem by using the ideal gas equation, but we can take a shortcut
by using the molar volume of an ideal gas at STP. Since 1 mole of an ideal gas at STP
has a volume of 22.42 L, 1.75 L N2 at STP will contain less than 1 mole. We can find
how many moles using the ratio of 1.75 L to 22.42 L:
1.75 L N2 3
1 mol N2
5 7.81 3 1022 mol N2
22.42 L N2
See Exercises 5.65 and 5.66
Many chemical reactions involve gases. By assuming ideal behavior for these
gases, we can carry out stoichiometric calculations if the pressure, volume, and temperature of the gases are known.
Interactive
Example 5.12
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in OWL.
Gas Stoichiometry II
Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate
(CaCO3). Calculate the volume of CO2 at STP produced from the decomposition of
152 g CaCO3 by the reaction
CaCO3 1s2 h CaO 1s2 1 CO2 1g2
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5.4
Gas Stoichiometry
205
Solution
Where are we going?
To use stoichiometry to determine the volume of CO2 produced
What do we know?
CaCO3 1s2 h CaO 1s2 1 CO2 1g2
❯
What information do we need?
❯ Molar volume of a gas at STP is 22.42 L
How do we get there?
We need to use the strategy for solving stoichiometry problems that we learned in
Chapter 3.
1. What is the balanced equation?
CaCO3 1s2 h CaO 1s2 1 CO2 1g2
2. What are the moles of CaCO3 (100.09 g/mol)?
152 g CaCO3 3
1 mol CaCO3
5 1.52 mol CaCO3
100.09 g CaCO3
3. What is the mole ratio between CO2 and CaCO3 in the balanced equation?
1 mol CO2
1 mol CaCO3
4. What are the moles of CO2?
1.52 moles of CO2, which is the same as the moles of CaCO3 because the mole
ratio is 1.
5. What is the volume of CO2 produced?
We can compute this by using the molar volume since the sample is at STP:
1.52 mol CO2 3
❯
22.42 L CO2
5 34.1 L CO2
1 mol CO2
Thus the decomposition of 152 g CaCO3 produces 34.1 L CO2 at STP.
See Exercises 5.67 through 5.70
Remember that the molar volume of
an ideal gas is 22.42 L when measured
at STP.
Interactive
Example 5.13
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in OWL.
Note that in Example 5.12 the final step involved calculation of the volume of gas
from the number of moles. Since the conditions were specified as STP, we were able
to use the molar volume of a gas at STP. If the conditions of a problem are different
from STP, the ideal gas law must be used to compute the volume.
Gas Stoichiometry III
A sample of methane gas having a volume of 2.80 L at 258C and 1.65 atm was mixed
with a sample of oxygen gas having a volume of 35.0 L at 318C and 1.25 atm. The
mixture was then ignited to form carbon dioxide and water. Calculate the volume of
CO2 formed at a pressure of 2.50 atm and a temperature of 1258C.
Solution
Where are we going?
To determine the volume of CO2 produced
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206
Chapter 5
Gases
What do we know?
P
V
T
CH4
O2
CO2
1.65 atm
2.80 L
258C 1 273 5 298 K
1.25 atm
35.0 L
318C 1 273 5 304 K
2.50 atm
?
1258C 1 273 5 398 K
What information do we need?
❯ Balanced chemical equation for the reaction
❯ Ideal gas law
PV 5 nRT
❯
R 5 0.08206 L ? atm/K ? mol
How do we get there?
We need to use the strategy for solving stoichiometry problems that we learned in
­Chapter 3.
1. What is the balanced equation?
From the description of the reaction, the unbalanced equation is
CH4 1g2 1 O2 1g2 h CO2 1g2 1 H2O 1g2
which can be balanced to give
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1g2
2. What is the limiting reactant?
We can determine this by using the ideal gas law to determine the moles for each
­reactant:
11.65 atm2 12.80 L2
PV
5
5 0.189 mol
10.08206 L # atm/K # mol2 1298 K2
RT
11.25 atm2 135.0 L2
PV
5
5 1.75 mol
nO2 5
10.08206 L # atm/K # mol2 1304 K2
RT
nCH4 5
In the balanced equation for the combustion reaction, 1 mole of CH4 requires
2 moles of O2. Thus the moles of O2 required by 0.189 mole of CH4 can be
calculated as follows:
0.189 mol CH4 3
2 mol O2
5 0.378 mol O2
1 mol CH4
The limiting reactant is CH4.
3. What are the moles of CO2?
Since CH4 is limiting, we use the moles of CH4 to determine the moles of CO2
­produced:
0.189 mol CH4 3
1 mol CO2
5 0.189 mol CO2
1 mol CH4
4. What is the volume of CO2 produced?
Since the conditions stated are not STP, we must use the ideal gas law to calculate the volume:
V5
nRT
P
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5.4
Gas Stoichiometry
207
In this case n 5 0.189 mol, T 5 1258C 1 273 5 398 K, P 5 2.50 atm, and
R 5 0.08206 L ? atm/K ? mol. Thus
j
V5
10.189 mol2 10.08206 L # atm/K # mol2 1398 K2
5 2.47 L
2.50 atm
This represents the volume of CO2 produced under these conditions.
See Exercises 5.71 and 5.74
Molar Mass of a Gas
One very important use of the ideal gas law is in the calculation of the molar mass
(mo­lecular weight) of a gas from its measured density. To see the relationship between
gas density and molar mass, consider that the number of moles of gas, n, can be expressed as
n5
grams of gas
mass
m
5
5
molar mass
molar mass
molar mass
Substitution into the ideal gas equation gives
P5
Density 5
mass
volume
1m /molar mass2 RT
nRT
m 1RT 2
5
5
1
V
V
V molar mass2
However, myV is the gas density, d, in units of grams per liter. Thus
P5
dRT
molar mass
or
Molar mass 5
dRT
P
(5.1)
Thus, if the density of a gas at a given temperature and pressure is known, its molar
mass can be calculated.
Interactive
Example 5.14
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.com to try this Interactive Example
in OWL.
Gas Density/Molar Mass
The density of a gas was measured at 1.50 atm and 278C and found to be 1.95 g/L.
Calculate the molar mass of the gas.
Solution
Where are we going?
To determine the molar mass of the gas
What do we know?
❯
PowerLecture: Changes in Gas Volume,
­Pressure, and Concentration
❯
❯
P 5 1.50 atm
T 5 278C 1 273 5 300. K
d 5 1.95 g/L
What information do we need?
dRT
❯ Molar mass 5
P
❯ R 5 0.08206 L ? atm/K ? mol
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208
Chapter 5
Gases
How do we get there?
j
g
L # atm
a1.95 b a0.08206 #
b 1300. K2
dRT
L
K mol
Molar mass 5
5
5 32.0 g/mol
P
1.50 atm
Reality Check | These are the units expected for molar mass.
See Exercises 5.77 through 5.80
You could memorize the equation involving gas density and molar mass, but it is
better simply to remember the ideal gas equation, the definition of density, and the
relationship between number of moles and molar mass. You can then derive the appropriate equation when you need it. This approach ensures that you understand the
concepts and means one less equation to memorize.
5.5 Dalton’s Law of Partial Pressures
Among the experiments that led John Dalton to propose the atomic theory were his
studies of mixtures of gases. In 1803 Dalton summarized his observations as follows:
For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. This statement, known as Dalton’s
law of partial pressures, can be expressed as follows:
PTOTAL 5 P1 1 P2 1 P3 1 c
where the subscripts refer to the individual gases (gas 1, gas 2, and so on). The symbols P1, P2, P3, and so on represent each partial pressure, the pressure that a particular
gas would exert if it were alone in the container.
Assuming that each gas behaves ideally, the partial pressure of each gas can be
calculated from the ideal gas law:
P1 5
n1RT
,
V
P2 5
n2RT
,
V
P3 5
n3RT
,
V
c
The total pressure of the mixture PTOTAL can be represented as
PTOTAL 5 P1 1 P2 1 P3 1 c5
n1RT
n RT
n RT c
1 2
1 3
1
V
V
V
5 1n1 1 n2 1 n3 1 c2 a
5 nTOTAL a
RT
b
V
RT
b
V
where nTOTAL is the sum of the numbers of moles of the various gases. Thus, for a mixture of ideal gases, it is the total number of moles of particles that is important, not the
identity or composition of the involved gas particles. This idea is illustrated in Fig. 5.12.
Figure 5.12 | The partial pressure
of each gas in a mixture of gases in a
container depends on the number of
moles of that gas. The total pressure is
the sum of the partial pressures and
depends on the total moles of gas
particles present, no matter what they
are. Note that the volume remains
constant.
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5.5
Dalton’s Law of Partial Pressures
209
This important observation indicates some fundamental characteristics of an ideal
gas. The fact that the pressure exerted by an ideal gas is not affected by the identity (composition) of the gas particles reveals two things about ideal gases: (1) the volume of the
individual gas particle must not be important, and (2) the forces among the particles must
not be important. If these factors were important, the pressure exerted by the gas would
depend on the nature of the individual particles. These observations will strongly influence the model that we will eventually construct to explain ideal gas behavior.
Interactive
Example 5.15
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in OWL.
Dalton’s Law I
Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the
bends.” For a particular dive, 46 L He at 258C and 1.0 atm and 12 L O2 at 258C and
1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure
of each gas and the total pressure in the tank at 258C.
Solution
Where are we going?
To determine the partial pressure of each gas
To determine the total pressure in the tank at 258C
What do we know?
P
V
T
He
O2
Tank
1.00 atm
46 L
258C 1 273 5 298 K
1.00 atm
12 L
258C 1 273 5 298 K
? atm
5.0 L
258C 1 273 5 298 K
What information do we need?
❯ Ideal gas law
PV 5 nRT
❯
R 5 0.08206 L ? atm/K ? mol
How do we get there?
How many moles are present for each gas?
n5
PV
RT
11.0 atm2 146 L2
5 1.9 mol
10.08206 L # atm/K # mol2 1298 K2
11.0 atm2 112 L2
5 0.49 mol
nO2 5
10.08206 L # atm/K # mol2 1298 K2
nHe 5
What is the partial pressure for each gas in the tank?
The tank containing the mixture has a volume of 5.0 L, and the temperature is 258C.
We can use these data and the ideal gas law to calculate the partial pressure of
each gas:
j
j
nRT
V
11.9 mol2 10.08206 L # atm/K # mol2 1298 K2
PHe 5
5 9.3 atm
5.0 L
10.49 mol2 10.08206 L # atm/K # mol2 1298 K2
PO2 5
5 2.4 atm
5.0 L
P5
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210
Chapter 5
Gases
Chemical connections
Separating Gases
Assume you work for an oil company
that owns a huge natural gas reservoir
containing a mixture of methane and
nitrogen gases. In fact, the gas mixture
contains so much nitrogen that it is
unusable as a fuel. Your job is to
separate the nitrogen (N2) from the
methane (CH4). How might you
accomplish this task? You clearly need
some sort of “molecular filter” that will
stop the slightly larger methane
molecules (size  430 pm) and allow
the nitrogen molecules (size  410 pm)
to pass through. To accomplish the
separation of molecules so similar in
size will require a very precise “filter.”
The good news is that such a filter
exists. Recent work by Steven Kuznicki
and Valerie Bell at Engelhard Corporation in New Jersey and Michael
Tsapatsis at the University of
Massachusetts has produced a
“molecular sieve” in which the pore
(passage) sizes can be adjusted
precisely enough to separate N2
molecules from CH4 molecules. The
material involved is a special hydrated
titanosilicate (contains H2O, Ti, Si, O,
and Sr) compound patented by
Engelhard known as ETS-4 (Engelhard
TitanoSilicate-4). When sodium ions
are substituted for the strontium ions
in ETS‑4 and the new material is
carefully dehydrated, a uniform and
controllable pore-size reduction
occurs (see figure). The researchers
have shown that the material can be
used to separate N2 ( 410 pm) from
O2 ( 390 pm). They have also shown
that it is possible to reduce the
nitrogen content of natural gas from
18% to less than 5% with a 90%
recovery of methane.
Dehydration
d
d
Molecular sieve framework of titanium (blue), silicon (green), and oxygen (red) atoms
contracts on heating—at room temperature (left), d 5 4.27 Å; at 2508C (right),
d 5 3.94 Å.
What is the total pressure of the mixture of gases in the tank?
The total pressure is the sum of the partial pressures:
j
PTOTAL 5 PHe 1 PO2 5 9.3 atm 1 2.4 atm 5 11.7 atm
See Exercises 5.83 and 5.84
At this point we need to define the mole fraction: the ratio of the number of moles
of a given component in a mixture to the total number of moles in the mixture. The
Greek lowercase letter chi ( x ) is used to symbolize the mole fraction. For example, for
a given component in a mixture, the mole fraction x1 is
x1 5
n1
nTOTAL
5
n1
n1 1 n2 1 n3 1 c
From the ideal gas equation, we know that the number of moles of a gas is directly
proportional to the pressure of the gas, since
n 5 Pa
V
b
RT
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5.5
Dalton’s Law of Partial Pressures
211
Chemical connections
The Chemistry of Air Bags
N2 gas. Originally, sodium azide, which
decomposes to produce N2,
2NaN3 1s2 S 2Na 1s2 1 3N2 1g2
was used, but it has now been replaced
by less toxic materials.
The sensing devices that trigger the
air bags must react very rapidly. For
example, consider a car hitting a
concrete bridge abutment. When this
happens, an internal accelerometer
sends a message to the control
module that a collision possibly is
occurring. The microprocessor then
analyzes the measured deceleration
from several accelerometers and door
pressure sensors and decides whether
air bag deployment is appropriate. All
this happens within 8 to 40 ms of the
initial impact.
Because an air bag must provide
the appropriate cushioning effect, the
bag begins to vent even as it is being
filled. In fact, the maximum pressure in
the bag is 5 pounds per square inch
(psi), even in the middle of a collision
event. Air bags represent a case where
an explosive chemical reaction saves
lives rather than the reverse.
Courtesy, Chrysler
The inclusion of air bags in modern
automobiles has led to a significant
reduction in the number of injuries
as a result of car crashes. Air bags
are stored in the steering wheel and
dashboard of all cars, and many
autos now have additional air bags
that protect the occupant’s knees,
head, and shoulders. In fact, some
auto manufacturers now include air
bags in the seat belts. Also, because
deployment of an air bag can
severely injure a child, all cars now
have “smart” air bags that deploy
with an inflation force that is
proportional to the seat occupant’s
weight.
The term “air bag” is really a
misnomer because air is not involved
in the inflation process. Rather, an air
bag inflates rapidly (in about 30 ms)
due to the explosive production of
Inflated air bags.
That is, for each component in the mixture,
n1 5 P1 a
V
b,
RT
n2 5 P2 a
V
b,
RT
c
Therefore, we can represent the mole fraction in terms of pressures:
P1 1V/RT 2
1
2
1
P1 V/RT 1 P2 V/RT 2 1 P3 1V/RT 2 1 c
n1
n2





nTOTAL
5





n1





x1 5





n1
n3
1V/RT 2 P1
5
1V/RT 2 1P1 1 P2 1 P3 1 c2
P1
P1
5
5
P1 1 P2 1 P3 1 c PTOTAL
In fact, the mole fraction of each component in a mixture of ideal gases is directly related to its partial pressure:
n2
P2
x2 5
5
nTOTAL
PTOTAL
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212
Chapter 5
Gases
Interactive
Example 5.16
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in OWL.
Dalton’s Law II
The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.
Solution
Where are we going?
To determine the mole fraction of O2
What do we know?
❯
❯
PO2
5 156 torr
PTOTAL 5 743 torr
How do we get there?
The mole fraction of O2 can be calculated from the equation
j
xO2 5
PO2
PTOTAL
5
156 torr
5 0.210
743 torr
Note that the mole fraction has no units.
See Exercise 5.89
The expression for the mole fraction,
x1 5
P1
PTOTAL
can be rearranged to give
P1 5 x1 3 PTOTAL
That is, the partial pressure of a particular component of a gaseous mixture is the mole
fraction of that component times the total pressure.
Interactive
Example 5.17
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in OWL.
Dalton’s Law III
The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2
in air when the atmospheric pressure is 760. torr.
Solution
The partial pressure of N2 can be calculated as follows:
PN2 5 xN2 3 PTOTAL 5 0.7808 3 760. torr 5 593 torr
See Exercise 5.90
Collecting a Gas over Water
A mixture of gases results whenever a gas is collected by displacement of water. For
example, Fig. 5.13 shows the collection of oxygen gas produced by the decomposition
of solid potassium chlorate. In this situation, the gas in the bottle is a mixture of water
vapor and the oxygen being collected. Water vapor is present because molecules of
water escape from the surface of the liquid and collect in the space above the liquid.
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5.5
Figure 5.13 | The production of
oxygen by thermal ­decomposition of
KClO3. The MnO2 is mixed with the
KClO3 to make the ­reaction faster.
Vapor pressure will be discussed in detail
in Chapter 10. A table of water vapor
pressure values is given in Section 10.8.
Interactive
Example 5.18
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in OWL.
Experiment 16: Preparation and
Properties of Hydrogen and Oxygen
Gases
Dalton’s Law of Partial Pressures
213
Oxygen plus
water vapor
KClO3 and MnO3
Molecules of water also return to the liquid. When the rate of escape equals the rate of
return, the number of water molecules in the vapor state remains constant, and thus the
pressure of water vapor remains constant. This pressure, which depends on temperature, is called the vapor pressure of water.
Gas Collection over Water
A sample of solid potassium chlorate (KClO3) was heated in a test tube (see Fig. 5.13)
and decomposed by the following reaction:
2KClO3 1s2 h 2KCl 1s2 1 3O2 1g2
The oxygen produced was collected by displacement of water at 228C at a total pressure of 754 torr. The volume of the gas collected was 0.650 L, and the vapor pressure
of water at 228C is 21 torr. Calculate the partial pressure of O2 in the gas collected and
the mass of KClO3 in the sample that was decomposed.
Solution
Where are we going?
To determine the partial pressure of O2 in the gas collected
Calculate the mass of KClO3 in the original sample
What do we know?
P
V
T
Gas Collected
Water Vapor
754 torr
0.650 L
228C 1 273 5 295 K
21 torr
228C 1 273 5 295 K
How do we get there?
What is the partial pressure of O2?
PTOTAL 5 PO2 1 PH2O 5 PO2 1 21 torr 5 754 torr
j
PO2 5 754 torr 2 21 torr 5 733 torr
What is the number of moles of O2?
Now we use the ideal gas law to find the number of moles of O2:
nO2 5
PO2V
RT
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214
Chapter 5
Gases
In this case, the partial pressure of the O2 is
PO2 5 733 torr 5
733 torr
5 0.964 atm
760 torr /atm
To find the moles of O2 produced, we use
V 5 0.650 L
T 5 22°C 1 273 5 295 K
R 5 0.08206 L # atm/K # mol
nO2 5
10.964 atm2 10.650 L2
5 2.59 3 1022 mol
10.08206 L # atm/K # mol2 1295 K2
How many moles of KClO3 are required to produce this amount of O2?
Use the stoichiometry problem-solving strategy:
1. What is the balanced equation?
2KClO3 1s2 h 2KCl 1s2 1 3O2 1g2
2. What is the mole ratio between KClO3 and O2 in the balanced equation?
2 mol KClO3
3 mol O2
3. What are the moles of KClO3?
2.59 3 1022 mol O2 3
2 mol KClO3
5 1.73 3 1022 mol KClO3
3 mol O2
4. What is the mass of KClO3 (molar mass 122.6 g/mol) in the original sample?
1.73 3 1022 mol KClO3 3
❯
122.6 g KClO3
5 2.12 g KClO3
1 mol KClO3
Thus the original sample contained 2.12 g KClO3.
See Exercises 5.91 through 5.93
5.6 The Kinetic Molecular Theory of Gases
IBLG: See questions from “Partial
Pressures”
IBLG: See questions from “The Kinetic
Molecular Theory of Gases and Real
Gases”
We have so far considered the behavior of gases from an experimental point of view.
Based on observations from different types of experiments, we know that at pressures
of less than 1 atm most gases closely approach the behavior described by the ideal gas
law. Now we want to construct a model to explain this behavior.
Before we do this, let’s briefly review the scientific method. Recall that a law is a
way of generalizing behavior that has been observed in many experiments. Laws are
very useful, since they allow us to predict the behavior of similar systems. For example, if a chemist prepares a new gaseous compound, a measurement of the gas density
at known pressure and temperature can provide a reliable value for the compound’s
molar mass.
However, although laws summarize observed behavior, they do not tell us why nature behaves in the observed fashion. This is the central question for scientists. To try
to answer this question, we construct theories (build models). The models in chemistry
consist of speculations about what the individual atoms or molecules (microscopic
particles) might be doing to cause the observed behavior of the macroscopic systems
(collections of very large numbers of atoms and molecules).
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The Kinetic Molecular Theory of Gases
215
Photos © Cengage Learning. All rights reserved.
5.6
a
b
Figure 5.14 | (a) One mole of N2(l) has a volume of approximately 35 mL and a density of 0.81 g/mL. (b) One mole of N2(g)
has a volume of 22.42 L (STP) and a density of 1.2 3 1023 g/mL. Thus the ratio of the volumes of gaseous N2 and liquid N2 is
22.42y0.035 5 640, and the spacing of the molecules is 9 times farther apart in N2(g).
A model is considered successful if it explains the observed behavior in question
and predicts correctly the results of future experiments. It is important to understand
that a model can never be proved absolutely true. In fact, any model is an approximation by its very nature and is bound to fail at some point. Models range from the simple
to the extraordinarily complex. We use simple models to predict approximate behavior
and more complicated models to account very precisely for observed quantitative behavior. In this text we will stress simple models that provide an approximate picture of
what might be happening and that fit the most important experimental results.
An example of this type of model is the kinetic molecular theory (KMT), a simple
model that attempts to explain the properties of an ideal gas. This model is based on
speculations about the behavior of the individual gas particles (atoms or molecules).
The postulates of the kinetic molecular theory as they relate to the particles of an ideal
gas can be stated as follows:
Postulates of the Kinetic Molecular Theory
1. The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero) (Fig. 5.14).
PowerLecture: Visualizing Molecular
M
­ otion: Single ­Molecule
2. The particles are in constant motion. The collisions of the particles with the walls of the
container are the cause of the pressure exerted by the gas.
3. The particles are assumed to exert no forces on each other; they are assumed neither
to attract nor to repel each other.
PowerLecture: Visualizing Molecular
M
­ otion: Many ­Molecules
4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.
Of course, the molecules in a real gas have finite volumes and do exert forces on
each other. Thus real gases do not conform to these assumptions. However, we will see
that these postulates do indeed explain ideal gas behavior.
The true test of a model is how well its predictions fit the experimental observations. The postulates of the kinetic molecular model picture an ideal gas as consisting
of particles having no volume and no attractions for each other, and the model assumes
that the gas produces pressure on its container by collisions with the walls.
Let’s consider how this model accounts for the properties of gases as summarized
by the ideal gas law: PV 5 nRT.
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216
Chapter 5
Gases
Figure 5.15 | The effects of
decreasing the volume of a sample of
gas at constant temperature.
PowerLecture: Boyle’s Law:
A Molecular-Level View
Volume is
decreased.
Pressure and Volume (Boyle’s Law)
We have seen that for a given sample of gas at a given temperature (n and T are constant) that if the volume of a gas is decreased, the pressure increases:
1
V



P 5 1nRT2
h
Constant
This makes sense based on the kinetic molecular theory because a decrease in volume
means that the gas particles will hit the wall more often, thus increasing pressure
(Fig. 5.15).
Pressure and Temperature
From the ideal gas law, we can predict that for a given sample of an ideal gas at a constant volume, the pressure will be directly proportional to the temperature:
nR
bT
V



P5a
h
Constant
The KMT accounts for this behavior because when the temperature of a gas increases,
the speeds of its particles increase, the particles hitting the wall with greater force and
greater frequency. Since the volume remains the same, this would result in increased
gas pressure (Fig. 5.16).
Critical Thinking
You have learned the postulates of the KMT. What if we could not assume the third
postulate to be true? How would this affect the measured pressure of a gas?
Temperature
is increased.
Figure 5.16 | The effects of
increasing the temperature of a
sample of gas at constant volume.
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5.6
Figure 5.17 | The effects of
increasing the temperature of a
sample of gas at constant pressure.
The Kinetic Molecular Theory of Gases
217
Temperature
is increased.
Volume and Temperature (Charles’s Law)
PowerLecture: Charles’s Law:
A Molecular-Level View
The ideal gas law indicates that for a given sample of gas at a constant pressure, the
volume of the gas is directly proportional to the temperature in kelvins:
nR
bT
P



V5a
h
Constant
This can be visualized from the KMT (Fig. 5.17). When the gas is heated to a higher
temperature, the speeds of its molecules increase and thus they hit the walls more often
and with more force. The only way to keep the pressure constant in this situation is to
increase the volume of the container. This compensates for the increased particle
speeds.
Volume and Number of Moles (Avogadro’s Law)
The ideal gas law predicts that the volume of a gas at a constant temperature and pressure depends directly on the number of gas particles present:
RT
bn
P



V5a
h
Constant
This makes sense in terms of the KMT because an increase in the number of gas particles at the same temperature would cause the pressure to increase if the volume were
held constant (Fig. 5.18). The only way to return the pressure to its original value is to
increase the volume.
Moles of gas
increase.
Figure 5.18 | The effects of
increasing the number of moles of gas
particles at constant temperature and
pressure.
Increase volume
to return to
original pressure.
Gas cylinder
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218
Chapter 5
Gases
It is important to recognize that the volume of a gas (at constant P and T) depends
only on the number of gas particles present. The individual volumes of the particles are
not a factor because the particle volumes are so small compared with the distances
between the particles (for a gas behaving ideally).
Mixture of Gases (Dalton’s Law)
The observation that the total pressure exerted by a mixture of gases is the sum of the
pressures of the individual gases is expected because the KMT assumes that all gas
particles are independent of each other and that the volumes of the individual particles
are unimportant. Thus the identities of the gas particles do not matter.
Deriving the Ideal Gas Law
We have shown qualitatively that the assumptions of the KMT successfully account
for the observed behavior of an ideal gas. We can go further. By applying the principles
of physics to the assumptions of the KMT, we can in effect derive the ideal gas law.
As shown in detail in Appendix 2, we can apply the definitions of velocity, momentum, force, and pressure to the collection of particles in an ideal gas and derive the
­following expression for pressure:
P5
2 nNA 1 12mu22
c
d
3
V
where P is the pressure of the gas, n is the number of moles of gas, NA is Avogadro’s
number, m is the mass of each particle, u2 is the average of the square of the velocities
of the particles, and V is the volume of the container.
The quantity 12mu2 represents the average kinetic energy of a gas particle. If the
average kinetic energy of an individual particle is multiplied by NA, the number of
particles in a mole, we get the average kinetic energy for a mole of gas particles:
a
b
Ken O’Donoghue
Ken O’Donoghue
Ken O’Donoghue
1KE2 avg 5 NA 1 12mu22
c
(a) A balloon filled with air at room temperature. (b) The balloon is dipped into liquid nitrogen at 77 K. (c) The balloon collapses as the molecules
inside slow down due to the decreased temperature. Slower molecules produce a lower pressure.
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5.6
Kinetic energy (KE) given by the equation
KE 5 21 mu 2 is the energy due to the
motion of a particle. We will discuss this
further in Section 6.1.
The Kinetic Molecular Theory of Gases
219
Using this definition, we can rewrite the expression for pressure as
P5
2 n 1KE2 avg
c
d
3
V
or
PV
2
5 1KE2 avg
n
3
The fourth postulate of the kinetic molecular theory is that the average kinetic energy of the particles in the gas sample is directly proportional to the temperature in
kelvin. Thus, since (KE)avg ~ T, we can write
PV
2
5 1KE2 avg ~ T or
n
3
PV
~ T
n
Note that this expression has been derived from the assumptions of the kinetic molecular theory. How does it compare to the ideal gas law—the equation obtained from experiment? Compare the ideal gas law,
PV
5 RT From experiment
n
with the result from the kinetic molecular theory,
PV
~ T From theory
n
These expressions have exactly the same form if R, the universal gas constant, is considered the proportionality constant in the second case.
The agreement between the ideal gas law and the predictions of the kinetic molecular
theory gives us confidence in the validity of the model. The characteristics we have assumed for ideal gas particles must agree, at least under certain conditions, with their actual behavior.
The Meaning of Temperature
We have seen from the kinetic molecular theory that the Kelvin temperature indicates
the average kinetic energy of the gas particles. The exact relationship between temperature and average kinetic energy can be obtained by combining the equations:
which yields the expression
PV
2
5 RT 5 1KE2 avg
n
3
3
1KE2 avg 5 RT
2
This is a very important relationship. It summarizes the meaning of the Kelvin temperature of a gas: The Kelvin temperature is an index of the random motions of the
particles of a gas, with higher temperature meaning greater motion. (As we will see in
Chapter 10, temperature is an index of the random motions in solids and liquids as well
as in gases.)
Root Mean Square Velocity
In the equation from the kinetic molecular theory, the average velocity of the gas particles is a special kind of average. The symbol u2 means the average of the squares of
the particle velocities. The square root of u2 is called the root mean square velocity
and is symbolized by urms:
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urms 5 "u2
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220
Chapter 5
Gases
We can obtain an expression for urms from the equations
1KE2 avg 5 NA 112mu22
and
Combination of these equations gives
3
1KE2 avg 5 RT
2
3
3RT
NA 112mu22 5 RT or u2 5
2
NAm
Taking the square root of both sides of the last equation produces
"u2 5 urms 5
3RT
Å NAm
In this expression m represents the mass in kilograms of a single gas particle. When
NA, the number of particles in a mole, is multiplied by m, the product is the mass of a
mole of gas particles in kilograms. We will call this quantity M. Substituting M for
NAm in the equation for urms, we obtain
urms 5
L # atm
K # mol
J
R 5 8.3145 #
K mol
R 5 0.08206
Interactive
Example 5.19
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in OWL.
3RT
Å M
Before we can use this equation, we need to consider the units for R. So far we have
used 0.08206 L # atm/K # mol as the value of R. But to obtain the desired units (meters
per second) for urms, R must be expressed in different units. As we will see in more
detail in Chapter 6, the energy unit most often used in the SI system is the joule (J). A
joule is defined as a kilogram meter squared per second squared (kg ? m2/s2). When R
is converted to include the unit of joules, it has the value 8.3145 J/K ? mol. When R in
these units is used in the expression !3RT /M, urms is obtained in the units of meters
per second as desired.
Root Mean Square Velocity
Calculate the root mean square velocity for the atoms in a sample of helium gas at
258C.
Solution
Where are we going?
To determine the root mean square velocity for the atoms of He
What do we know?
❯
❯
PowerLecture: Kinetic Molecular
Theory/Heat Transfer
T 5 258C 1 273 5 298 K
R 5 8.3145 J/K ? mol
What information do we need?
❯
Root mean square velocity is urms 5
3RT
Å M
How do we get there?
What is the mass of a mole of He in kilograms?
M 5 4.00
g
1 kg
3
5 4.00 3 1023 kg/mol
mol
1000 g
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5.6
The Kinetic Molecular Theory of Gases
221
What is the root mean square velocity for the atoms of He?
J
b 1298 K2
K # mol
J
5 1.86 3 106
kg
Å
kg
4.00 3 1023
mol
3 a8.3145
urms 5
ï
Since the units of J are kg ? m2/s2, this expression gives
j
Å
1.86 3 106
kg # m2
5 1.36 3 103 m/s
kg # s2
Reality Check | The resulting units are appropriate for velocity.
See Exercises 5.103 and 5.104
So far we have said nothing about the range of velocities actually found in a gas
sample. In a real gas, there are large numbers of collisions between particles. For example, as we will see in the next section, when an odorous gas such as ammonia is
released in a room, it takes some time for the odor to permeate the air. This delay results from collisions between the NH3 molecules and the O2 and N2 molecules in the
air, which greatly slow the mixing process.
If the path of a particular gas particle could be monitored, it would look very erratic,
something like that shown in Fig. 5.19. The average distance a particle travels between
collisions in a particular gas sample is called the mean free path. It is typically a very
small distance (1 3 1027 m for O2 at STP). One effect of the many collisions among
gas particles is to produce a large range of velocities as the particles collide and exchange kinetic energy. Although urms for oxygen gas at STP is approximately 500 m/s,
the majority of O2 molecules do not have this velocity. The actual distribution of molecular velocities for oxygen gas at STP is shown in Fig. 5.20. This figure shows the
relative number of gas molecules having each particular velocity.
We are also interested in the effect of temperature on the velocity distribution in a
gas. Figure 5.21 shows the velocity distribution (called the Maxwell-Boltzmann distribution) for nitrogen gas at three temperatures. Note that as the temperature is increased,
273 K
Relative number of N2 molecules
with given velocity
Relative number of O2 molecules
with given velocity
Figure 5.19 | Path of one particle in
a gas. Any given particle will continuously change its course as a result of
collisions with other particles, as well
as with the walls of the container.
urms 5
0
4 × 102 8 × 102
Molecular velocity (m/s)
Figure 5.20 | A plot of the relative
number of O2 molecules that have a
given velocity at STP.
1273 K
2273 K
0
1000
2000
3000
Velocity (m/s)
Figure 5.21 | A plot of the relative number
of N2 molecules that have a given velocity at
three temperatures. Note that as the temperature increases, both the average velocity and
the spread of velocities increase.
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222
Chapter 5
Gases
Figure 5.22 | The effusion of a gas
into an evacuated chamber. The rate
of effusion (the rate at which the gas
is transferred across the barrier
through the pin hole) is inversely
proportional to the square root of the
mass of the gas molecules.
Pinhole
Gas
Vacuum
the curve peak moves toward higher values and the range of velocities becomes much
larger. The peak of the curve reflects the most probable velocity (the velocity found
most often as we sample the movement of the various particles in the gas). Because the
kinetic energy increases with temperature, it makes sense that the peak of the curve
should move to higher values as the temperature of the gas is increased.
PowerLecture: Effusion
of Gas
5.7 Effusion and Diffusion
We have seen that the postulates of the kinetic molecular theory, when combined with
the appropriate physical principles, produce an equation that successfully fits the experimentally observed behavior of gases as they approach ideal behavior. Two phenomena involving gases provide further tests of this model.
Diffusion is the term used to describe the mixing of gases. When a small amount of
pungent-smelling ammonia is released at the front of a classroom, it takes some time
before everyone in the room can smell it, because time is required for the ammonia to
mix with the air. The rate of diffusion is the rate of the mixing of gases. Effusion is the
term used to describe the passage of a gas through a tiny orifice into an evacuated
chamber, as shown in Fig. 5.22. The rate of effusion measures the speed at which the
gas is transferred into the chamber.
Effusion
Thomas Graham (1805–1869), a Scottish chemist, found experimentally that the rate
of effusion of a gas is inversely proportional to the square root of the mass of its particles. Stated in another way, the relative rates of effusion of two gases at the same
temperature and pressure are given by the inverse ratio of the square roots of the
masses of the gas particles:
Rate of effusion for gas 1
"M2
5
Rate of effusion for gas 2
"M1
In Graham’s law the units for molar mass
can be g/mol or kg/mol, since the units
cancel in the ratio
"M2
"M1
Interactive
Example 5.20
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in OWL.
where M1 and M2 represent the molar masses of the gases. This equation is called
Graham’s law of effusion.
Effusion Rates
Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors
(Fig. 5.23).
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Percentage of molecules
5.7
223
Solution
0.04
0.03
First we need to compute the molar masses: Molar mass of H2 5 2.016 g/mol, and
molar mass of UF6 5 352.02 g/mol. Using Graham’s law,
UF6 at 273 K
0.02
0.01
0
Effusion and Diffusion
"MUF6
352.02
Rate of effusion for H2
5
5
5 13.2
Rate of effusion for UF6
Å 2.016
"MH2
H2 at 273 K
0
1000
2000
3000
Speed
The effusion rate of the very light H2 molecules is about 13 times that of the massive
UF6 molecules.
Figure 5.23 | Relative molecular
speed distribution of H2 and UF6.
Experiment 18: Gas Laws 2: Graham’s
Law
See Exercises 5.111 through 5.114
Does the kinetic molecular model for gases correctly predict the relative effusion
rates of gases summarized by Graham’s law? To answer this question, we must recognize that the effusion rate for a gas depends directly on the average velocity of its
particles. The faster the gas particles are moving, the more likely they are to pass
through the effusion orifice. This reasoning leads to the following prediction for two
gases at the same pressure and temperature (T ):
Effusion rate for gas 1
u for gas 1
5 rms
5
Effusion rate for gas 2
urms for gas 2
3RT
Å M1
3RT
Å M2
5
"M2
"M1
This equation is identical to Graham’s law. Thus the kinetic molecular model does fit
the experimental results for the effusion of gases.
Diffusion
NH3 1g2 1 HCl 1g2 h NH4Cl 1s2
PowerLecture: Gaseous Ammonia and
­Hydrochloric Acid
Cotton wet
with NH3(aq)
Glass tube
White solid
Air
Cotton wet
with HCl(aq)
Air
HCl
NH3
d NH3
d HCl
White ring of NH4Cl(s)
forms where the NH3
and HCl meet.
Figure 5.24 | (above left) A demonstration of the relative diffusion rates of NH3 and HCl molecules through air. Two cotton plugs, one
dipped in HCl(aq) and one dipped in NH3(aq), are simultaneously inserted into the ends of the tube. Gaseous NH3 and HCl vaporizing
from the cotton plugs diffuse toward each other and, where they meet, react to form NH4Cl(s). (above right) When HCl(g) and NH3(g)
meet in the tube, a white ring of NH4Cl(s) forms.
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Ken O’Donoghue
PowerLecture: Diffusion of Gases
Diffusion is frequently illustrated by the lecture demonstration represented in Fig.
5.24, in which two cotton plugs soaked in ammonia and hydrochloric acid are simultaneously placed at the ends of a long tube. A white ring of ammonium chloride
(NH4Cl) forms where the NH3 and HCl molecules meet several minutes later:
224
Chapter 5
Gases
As a first approximation, we might expect that the distances traveled by the two gases
are related to the relative velocities of the gas molecules:
Distance traveled by NH3
u for NH3
MHCl
36.5
5 rms
5
5
5 1.5
Distance traveled by HCl
urms for HCl
Å MNH3 Å 17
However, careful experiments produce an observed ratio of less than 1.5, indicating
that a quantitative analysis of diffusion requires a more complex analysis.
The diffusion of the gases through the tube is surprisingly slow in light of the fact
that the velocities of HCl and NH3 molecules at 258C are about 450 and 660 m/s, respectively. Why does it take several minutes for the NH3 and HCl molecules to meet?
The answer is that the tube contains air and thus the NH3 and HCl molecules undergo
many collisions with O2 and N2 molecules as they travel through the tube. Because so
many collisions occur when gases mix, diffusion is quite complicated to describe
theoretically.
5.8 Real Gases
CH4
N2
2.0
H2
PV
nRT
CO2
Ideal
gas
1.0
0
0
200 400 600 800 1000
P (atm)
Figure 5.25 | Plots of PVynRT
versus P for several gases (200 K).
Note the significant deviations from
ideal behavior (PVynRT 5 1). The
behavior is close to ideal only at low
pressures (less than 1 atm).
203 K
1.8
PV
nRT
293 K
1.4
673 K
1.0
0.6
Ideal
gas
0
200
400 600
P (atm)
800
Figure 5.26 | Plots of PVynRT
versus P for nitrogen gas at three
temperatures. Note that ­although
nonideal behavior is evident in each
case, the deviations are smaller at the
higher temperatures.
An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law, although many gases come very close at low pressures and/or high temperatures. Thus
ideal gas behavior can best be thought of as the behavior approached by real gases
under certain conditions.
We have seen that a very simple model, the kinetic molecular theory, by making
some rather drastic assumptions (no interparticle interactions and zero volume for the
gas particles), successfully explains ideal behavior. However, it is important that we
examine real gas behavior to see how it differs from that predicted by the ideal gas law
and to determine what modifications are needed in the kinetic molecular theory to
explain the observed behavior. Since a model is an approximation and will inevitably
fail, we must be ready to learn from such failures. In fact, we often learn more about
nature from the failures of our models than from their successes.
We will examine the experimentally observed behavior of real gases by measuring
the pressure, volume, temperature, and number of moles for a gas and noting how the
quantity PVynRT depends on pressure. Plots of PVynRT versus P are shown for several
gases in Fig. 5.25. For an ideal gas, PVynRT equals 1 under all conditions, but notice
that for real gases, PVynRT approaches 1 only at very low pressures (typically below
1 atm). To illustrate the effect of temperature, PVynRT is plotted versus P for nitrogen
gas at several temperatures in Fig. 5.26. Note that the behavior of the gas appears to
become more nearly ideal as the temperature is increased. The most important conclusion to be drawn from these figures is that a real gas typically exhibits behavior that is
closest to ideal behavior at low pressures and high temperatures.
One of the most important procedures in science is correcting our models as we collect more data. We will understand more clearly how gases actually behave if we can
figure out how to correct the simple model that explains the ideal gas law so that the
new model fits the behavior we actually observe for gases. So the question is: How can
we modify the assumptions of the kinetic molecular theory to fit the behavior of real
gases? The first person to do important work in this area was Johannes van der Waals
(1837–1923), a physics professor at the University of Amsterdam who in 1910 received
a Nobel Prize for his work. To follow his analysis, we start with the ideal gas law,
P5
nRT
V
Remember that this equation describes the behavior of a hypothetical gas consisting of
volumeless entities that do not interact with each other. In contrast, a real gas consists
of atoms or molecules that have finite volumes. Therefore, the volume available to a
given particle in a real gas is less than the volume of the container because the gas
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5.8
Real Gases
225
particles themselves take up some of the space. To account for this discrepancy, van
der Waals represented the actual volume as the volume of the container V minus a correction factor for the volume of the molecules nb, where n is the number of moles of
gas and b is an empirical constant (one determined by fitting the equation to the experimental results). Thus the volume actually available to a given gas molecule is given by
the difference V 2 nb.
This modification of the ideal gas equation leads to the equation
We have now corrected for both the finite
volume and the attractive forces of the
particles.
This effect can be understood using the following model. When gas particles come
close together, attractive forces occur, which cause the particles to hit the wall very
slightly less often than they would in the absence of these interactions (Fig. 5.27).
The size of the correction factor depends on the concentration of gas molecules
defined in terms of moles of gas particles per liter (nyV). The higher the concentration,
the more likely a pair of gas particles will be close enough to attract each other. For
large numbers of particles, the number of interacting pairs of particles depends on the
square of the number of particles and thus on the square of the concentration, or (nyV)2.
This can be justified as follows: In a gas sample containing N particles, there are
N 2 1 partners available for each particle (Fig. 5.28). Since the 1 c 2 pair is the same
as the 2 c 1 pair, this analysis counts each pair twice. Thus, for N particles, there are
N(N 2 1)y2 pairs. If N is a very large number, N 2 1 approximately equals N, giving
N 2y2 possible pairs. Thus the pressure, corrected for the attractions of the particles, has
the form
n 2
Pobs 5 Pr 2 a a b
V
where a is a proportionality constant (which includes the factor of 12 from N 2y2). The
value of a for a given real gas can be determined from observing the actual behavior of
that gas. Inserting the corrections for both the volume of the particles and the attractions of the particles gives the equation
Pobs 5
nRT
n 2
2 aa b
V 2 nb
V
Observed Volume
pressure
of the
container



Figure 5.27 | (a) Gas at low
concentration—relatively few
interactions between particles. The
indicated gas particle exerts a
pressure on the wall close to that
predicted for an ideal gas. (b) Gas at
high concentration—many more
interactions between particles. The
indicated gas particle exerts a much
lower pressure on the wall than would
be expected in the absence of
interactions.
nRT
2 correction factorb
V 2 nb
{
b
Pobs 5 1Pr 2 correction factor2 5 a
8n
a
Wall
The volume of the gas particles has now been taken into account.
The next step is to allow for the attractions that occur among the particles in a real
gas. The effect of these attractions is to make the observed pressure Pobs smaller than
it would be if the gas particles did not interact:
88
88
n
Wall
nRT
V 2 nb
8n
The attractive forces among molecules
will be discussed in Chapter 10.
Pr 5
Volume
correction
88n
P 9 is corrected for the finite volume of the
particles. The attractive forces have not
yet been taken into account.
Pressure
correction
Given particle
1
2
3
4
5
7
6
10
8
9
Gas sample with 10 particles
Figure 5.28 | Illustration of pairwise interactions among
gas particles. In a sample with 10 particles, each particle has
9 possible partners, to give 10(9)y2 5 45 distinct pairs. The
factor of 21 arises because when particle 1 is the particle of
2 pair, and when particle 2 is
interest, we count the 1
1 pair. However,
the particle of interest, we count the 2
2 and 2
1
1 are the same pair, which we thus have
counted twice. Therefore, we must divide by 2 to get the
correct number of pairs.
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226
Chapter 5
Gases
Figure 5.29 | The volume taken up
by the gas particles themselves is less
important at (a) large container
volume (low pressure) than at
(b) small container volume (high
pressure).
a
b
This equation can be rearranged to give the van der Waals equation:
n 2
c Pobs 1 a a b d 3 1V 2 nb2 5 nRT
V





Pobs is usually called just P.
Table 5.3 | Values of the van der
Waals Constants for
Some Common Gases
Gas
He
Ne
Ar
Kr
Xe
H2
N2
O2
Cl2
CO2
CH4
NH3
H2O
aa
atm # L2
b
mol2
0.0341
0.211
1.35
2.32
4.19
0.244
1.39
1.36
6.49
3.59
2.25
4.17
5.46
ba
L
b
mol
0.0237
0.0171
0.0322
0.0398
0.0511
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.0371
0.0305









Corrected volume









Corrected pressure
Pideal
Videal
The values of the weighting factors a and b are determined for a given gas by fitting
experimental behavior. That is, a and b are varied until the best fit of the observed pressure is obtained under all conditions. The values of a and b for various gases are given
in Table 5.3.
Experimental studies indicate that the changes van der Waals made in the basic
assumptions of the kinetic molecular theory correct the major flaws in the model.
First, consider the effects of volume. For a gas at low pressure (large volume), the
volume of the container is very large compared with the volumes of the gas particles. That is, in this case the volume available to the gas is essentially equal to the
volume of the container, and the gas behaves ideally. On the other hand, for a gas at
high pressure (small container volume), the volume of the particles becomes significant so that the volume available to the gas is significantly less than the container volume. These cases are illustrated in Fig. 5.29. Note from Table 5.3 that the
volume correction constant b generally increases with the size of the gas molecule,
which gives further support to these arguments.
The fact that a real gas tends to behave more ideally at high temperatures also can be
explained in terms of the van der Waals model. At high temperatures the particles are
moving so rapidly that the effects of interparticle interactions are not very important.
The corrections to the kinetic molecular theory that van der Waals found necessary
to explain real gas behavior make physical sense, which makes us confident that we
understand the fundamentals of gas behavior at the particle level. This is significant
because so much important chemistry takes place in the gas phase. In fact, the mixture
of gases called the atmosphere is vital to our existence. In Section 5.10 we consider
some of the important reactions that occur in the atmosphere.
Critical Thinking
You have learned that no gases behave perfectly ideally, but under conditions of high
temperature and low pressure (high volume), gases behave more ideally. What if all
gases always behaved perfectly ideally? How would the world be different?
5.9 Characteristics of Several Real Gases
We can understand gas behavior more completely if we examine the characteristics of
several common gases. Note from Fig. 5.25 that the gases H2, N2, CH4, and CO2 show
PV
different behavior when the compressibility a
b is plotted versus P. For example,
nRT
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5.10
Chemistry in the Atmosphere
227
notice that the plot for H2(g) never drops below the ideal value (1.0) in contrast to all
the other gases. What is special about H2 compared to these other gases? Recall from
Section 5.8 that the reason that the compressibility of a real gas falls below 1.0 is that
the actual (observed) pressure is lower than the pressure expected for an ideal gas due
to the intermo­lecular attractions that occur in real gases. This must mean that H2 molecules have very low attractive forces for each other. This idea is borne out by looking
at the van der Waals a value for H2 in Table 5.3. Note that H2 has the lowest value among
the gases H2, N2, CH4, and CO2. Remember that the value of a reflects how much of a
correction must be made to adjust the observed pressure up to the expected ideal
pressure:
n 2
Pideal 5 Pobserved 1 a a b
V
A low value for a reflects weak intermolecular forces among the gas molecules.
Also notice that although the compressibility for N2 dips below 1.0, it does not
show as much deviation as that for CH4, which in turn does not show as much deviation as the compressibility for CO2. Based on this behavior, we can surmise that the
importance of intermolecular interactions increases in this order:
H2 , N2 , CH4 , CO2
This order is reflected by the relative a values for these gases in Table 5.3. In Section
10.1, we will see how these variations in intermolecular interactions can be explained.
The main point to be made here is that real gas behavior can tell us about the relative
importance of intermolecular attractions among gas molecules.
5.10 Chemistry in the Atmosphere
Table 5.4 | Atmospheric
Composition Near Sea
Level (Dry Air)*
Component
Mole Fraction
N2
O2
Ar
CO2
Ne
He
CH4
Kr
H2
NO
Xe
0.78084
0.20948
0.00934
0.000345
0.00001818
0.00000524
0.00000168
0.00000114
0.0000005
0.0000005
0.000000087
*The atmosphere contains various amounts
of water vapor depending on conditions.
The most important gases to us are those in the atmosphere that surrounds the earth’s
surface. The principal components are N2 and O2, but many other important gases,
such as H2O and CO2, are also present. The average composition of the earth’s atmosphere near sea level, with the water vapor removed, is shown in Table 5.4. Because of
gravitational effects, the composition of the earth’s atmosphere is not constant; heavier
molecules tend to be near the earth’s surface, and light molecules tend to migrate to
higher altitudes, with some eventually escaping into space. The atmosphere is a highly
complex and dynamic system, but for convenience we divide it into several layers.
Fig. 5.30 shows how the temperature changes with altitude.
The chemistry occurring in the higher levels of the atmosphere is mostly determined by the effects of high-energy radiation and particles from the sun and other
sources in space. In fact, the upper atmosphere serves as an important shield to prevent this ­high-energy radiation from reaching the earth, where it would damage the
relatively fragile molecules sustaining life. In particular, the ozone in the upper atmosphere helps prevent high-energy ultraviolet radiation from penetrating to the
earth. Intensive research is in progress to determine the natural factors that control
the ozone concentration and how it is affected by chemicals released into the
atmosphere.
The chemistry occurring in the troposphere, the layer of atmosphere closest to the
earth’s surface, is strongly influenced by human activities. Millions of tons of gases
and particulates are released into the troposphere by our highly industrial civilization.
Actually, it is amazing that the atmosphere can absorb so much material with relatively
small permanent changes (so far).
Significant changes, however, are occurring. Severe air pollution is found around
many large cities, and it is probable that long-range changes in our planet’s weather are
taking place. We will discuss some of the long-range effects of pollution in Chapter 6.
In this section we will deal with short-term, localized effects of pollution.
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Chapter 5
Gases
Figure 5.30 | The variation of
temperature with altitude.
Exosphere 1000
800
600
400
Satellite
Polar lights
Spacecraft
Thermosphere 200
Altitude (km)
Mesopause
Mesosphere
Stratopause
Stratosphere
200
100
80
60
40
Meteors
Radiosonde
Supersonic
plane
Troposphere
Sea level
Commercial
aircraft
Parachute
jump
10
8
6
4
100
80
60
40
Ozone layer
20
Tropopause
1000
800
600
400
Mount Everest
20
Altitude (km)
228
10
8
6
4
2
2
1
1
0
0
–100
–60
–20 0 20 40
80
0.5
Molecules of unburned
fuel (petroleum)
0.4
Other
pollutants
0.3
0.2
NO2
O3
NO
6:00
4:00
2:00
Noon
10:00
4:00
0
8:00
0.1
6:00
Concentration (ppm)
Temperature (°C)
Time of day
Figure 5.31 | Concentration (in
molecules per million molecules of
“air”) for some smog components
versus time of day.
(Source: P.A. Leighton, “Photochemistry of Air
Pollution,” in Physical Chemistry: A Series of
Monographs, edited by Eric Hutchinson and P.
Van Rysselberghe, Vol. IX. New York: Academic
Press, 1961.)
The OH radical has no charge [it has one
fewer electron than the hydroxide ion
(OH2)].
The two main sources of pollution are transportation and the production of electricity. The combustion of petroleum in vehicles produces CO, CO2, NO, and NO2, along
with unburned molecules from petroleum. When this mixture is trapped close to the
ground in stagnant air, reactions occur, producing chemicals that are potentially irritating and harmful to living systems.
The complex chemistry of polluted air appears to center around the nitrogen oxides
(NOx). At the high temperatures found in the gasoline and diesel engines of cars and
trucks, N2 and O2 react to form a small quantity of NO that is emitted into the air with
the exhaust gases (Fig. 5.31). This NO is immediately oxidized in air to NO2, which,
in turn, absorbs energy from sunlight and breaks up into nitric oxide and free oxygen
atoms:
Radiant
energy
NO2 1g2 8888n NO 1g2 1 O 1g2
Oxygen atoms are very reactive and can combine with O2 to form ozone:
O 1g2 1 O2 1g2 h O3 1g2
Ozone is also very reactive and can react directly with other pollutants, or the ozone
can absorb light and break up to form an energetically excited O2 molecule (O2*) and
an ­energetically excited oxygen atom (O*). The latter species readily reacts with a
water molecule to form two hydroxyl radicals (OH):
O* 1 H2O h 2OH
The hydroxyl radical is a very reactive oxidizing agent. For example, OH can react
with NO2 to form nitric acid:
OH 1 NO2 h HNO3
The OH radical also can react with the unburned hydrocarbons in the polluted air to produce chemicals that cause the eyes to water and burn and are harmful to the respiratory
system.
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5.10
Chemistry in the Atmosphere
229
The end product of this whole process is often referred to as photochemical smog,
so called because light is required to initiate some of the reactions. The production of
photochemical smog can be understood more clearly by examining as a group the reactions discussed above:
NO2 1g2 h NO 1g2 1 O 1g2
O 1g2 1 O2 1g2 h O3 1g2
Although represented here as O2, the
actual oxidant for NO is OH or an organic
peroxide such as CH3COO, formed by
oxidation of organic pollutants.
Net reaction:
NO 1g2 1 12O2 1g2 h NO2 1g2
3
2 O2 1g2
h O3 1g2
Note that the NO2 molecules assist in the formation of ozone without being themselves
used up. The ozone formed then leads to the formation of OH and other pollutants.
We can observe this process by analyzing polluted air at various times during a day
(see Fig. 5.31). As people drive to work between 6 and 8 a.m., the amounts of NO,
NO2, and unburned molecules from petroleum increase. Later, as the decomposition of
NO2 occurs, the concentration of ozone and other pollutants builds up. Current efforts
to combat the formation of photochemical smog are focused on cutting down the
amounts of molecules from ­unburned fuel in automobile exhaust and designing engines that produce less nitric oxide.
The other major source of pollution results from burning coal to produce electricity.
Much of the coal found in the Midwest contains significant quantities of sulfur, which,
when burned, produces sulfur dioxide:
S 1in coal2 1 O2 1g2 h SO2 1g2
A further oxidation reaction occurs when sulfur dioxide is changed to sulfur trioxide
in the air:*
2SO2 1g2 1 O2 1g2 h 2SO3 1g2
Robert Krueger/US Forest Service/Getty Images
The production of sulfur trioxide is significant because it can combine with droplets of
water in the air to form sulfuric acid:
Figure 5.32 | Testing for acid rain in
Wilderness Lake, Colorado.
SO3 1g2 1 H2O 1l2 h H2SO4 1aq2
Sulfuric acid is very corrosive to both living things and building materials. Another
result of this type of pollution is acid rain. In many parts of the northeastern United
States and southeastern Canada, acid rain has caused some freshwater lakes to become
too acidic to support any life (Fig. 5.32).
The problem of sulfur dioxide pollution is made more complicated by the energy
crisis. As petroleum supplies dwindle and the price increases, our dependence on coal
will probably grow. As supplies of low-sulfur coal are used up, high-sulfur coal will be
utilized. One way to use high-sulfur coal without further harming the air quality is to
remove the sulfur dioxide from the exhaust gas by means of a system called a scrubber
before it is emitted from the power plant stack. A common method of scrubbing is to
blow powdered limestone (CaCO3) into the combustion chamber, where it is decomposed to lime and carbon dioxide:
CaCO3 1s2 h CaO 1s2 1 CO2 1g2
The lime then combines with the sulfur dioxide to form calcium sulfite:
CaO 1s2 1 SO2 1g2 h CaSO3 1s2
To remove the calcium sulfite and any remaining unreacted sulfur dioxide, an aqueous
suspension of lime is injected into the exhaust gases to produce a slurry (a thick suspension) (Fig. 5.33).
*This reaction is very slow unless solid particles are present. See Chapter 12 for a discussion.
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230
Chapter 5
Gases
Figure 5.33 | A schematic diagram
of the process for scrubbing sulfur
dioxide from stack gases in power
plants.
Water + CaO
CO2 + CaO
CaCO3
Coal
S + O2
SO2
CaSO3 +
Air
Scrubber
unreacted SO2
To
smokestack
Combustion chamber
CaSO3 slurry
Unfortunately, there are many problems associated with scrubbing. The systems are
complicated and expensive and consume a great deal of energy. The large quantities of
calcium sulfite produced in the process present a disposal problem. With a typical
scrubber, approximately 1 ton of calcium sulfite per year is produced per person served
by the power plant. Since no use has yet been found for this calcium sulfite, it is usually buried in a landfill. As a result of these difficulties, air pollution by sulfur dioxide
continues to be a major problem, one that is expensive in terms of damage to the environment and human health as well as in monetary terms.
For review
Key terms
State of a gas
Section 5.1
❯
barometer
manometer
mm Hg
torr
standard atmosphere
pascal
❯
1 torr 5 1 mm Hg
1 atm 5 760 torr
Section 5.2
Boyle’s law
ideal gas
Charles’s law
absolute zero
Avogadro’s law
❯
Gas laws
❯
❯
universal gas constant
ideal gas law
❯
molar volume
standard temperature and
pressure (STP)
SI unit: pascal
1 atm 5 101,325 Pa
Section 5.3
Section 5.4
The state of a gas can be described completely by specifying its pressure (P), volume (V),
temperature (T), and the amount (moles) of gas present (n)
Pressure
❯ Common units
❯
❯
❯
Discovered by observing the properties of gases
Boyle’s law: PV 5 k
Charles’s law: V 5 bT
Avogadro’s law: V 5 an
Ideal gas law: PV 5 nRT
Dalton’s law of partial pressures: Ptotal 5 P1 1 P2 1 P3 1 c, where Pn represents the
partial pressure of component n in a mixture of gases
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For Review
Key terms
Kinetic molecular theory (KMT)
Section 5.5
❯
Dalton’s law of partial
pressures
partial pressure
mole fraction
❯
Section 5.6
kinetic molecular theory
(KMT)
root mean square velocity
joule
Section 5.7
diffusion
effusion
Graham’s law of effusion
Model that accounts for ideal gas behavior
Postulates of the KMT:
❯ Volume of gas particles is zero
❯ No particle interactions
❯ Particles are in constant motion, colliding with the container walls to produce ­pressure
❯ The average kinetic energy of the gas particles is directly proportional to the ­temperature
of the gas in kelvins
Gas properties
❯
❯
The particles in any gas sample have a range of velocities
The root mean square (rms) velocity for a gas represents the average of the squares of the
particle velocities
urms 5
Section 5.8
real gas
van der Waals equation
Section 5.10
atmosphere
air pollution
photochemical smog
acid rain
❯
❯
3RT
Å M
Diffusion: the mixing of two or more gases
Effusion: the process in which a gas passes through a small hole into an empty chamber
Real gas behavior
❯
❯
❯
Review questions
231
Real gases behave ideally only at high temperatures and low pressures
Understanding how the ideal gas equation must be modified to account for real gas behavior
helps us understand how gases behave on a molecular level
Van der Waals found that to describe real gas behavior we must consider particle ­interactions
and particle volumes
Answers to the Review Questions can be found on the Student website (accessible from www.cengagebrain.com).
1. Explain how a barometer and a manometer work to
measure the pressure of the atmosphere or the pressure
of a gas in a container.
2. What are Boyle’s law, Charles’s law, and Avogadro’s
law? What plots do you make to show a linear relationship for each law?
3. Show how Boyle’s law, Charles’s law, and Avogadro’s
law are special cases of the ideal gas law. Using the
ideal gas law, determine the relationship between P and
n (at constant V and T) and between P and T (at
constant V and n).
4. Rationalize the following observations.
a. Aerosol cans will explode if heated.
b. You can drink through a soda straw.
c. A thin-walled can will collapse when the air inside
is removed by a vacuum pump.
d. Manufacturers produce different types of tennis
balls for high and low ­elevations.
5. Consider the following balanced equation in which gas
X forms gas X2:
2X 1g2 h X2 1g2
Equal moles of X are placed in two separate containers.
One container is rigid so the volume cannot change; the
other container is flexible so the volume changes to
keep the internal pressure equal to the external pressure.
The above reaction is run in each container. What
happens to the pressure and density of the gas inside
each container as reactants are converted to products?
6. Use the postulates of the kinetic molecular theory
(KMT) to explain why Boyle’s law, Charles’s law,
Avogadro’s law, and Dalton’s law of partial pressures
hold true for ideal gases. Use the KMT to explain the P
versus n (at constant V and T ) ­relationship and the P
versus T (at constant V and n) relationship.
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232
Chapter 5
Gases
Relative number of molecules
7. Consider the following velocity distribution curves A
and B.
A
B
Velocity (m/s)
a. If the plots represent the velocity distribution of
1.0 L of He(g) at STP versus 1.0 L of Cl2(g) at
STP, which plot corresponds to each gas? Explain
your reasoning.
b. If the plots represent the velocity distribution of
1.0 L of O2(g) at temperatures of 273 K versus
1273 K, which plot corresponds to each temperature? Explain your reasoning. Under which temperature condition would the O2(g) sample ­behave
most ideally? Explain.
8. Briefly describe two methods one might use to find the
molar mass of a newly synthesized gas for which a
molecular formula was not known.
9. In the van der Waals equation, why is a term added to
the observed pressure and why is a term subtracted
from the container volume to correct for nonideal gas
behavior?
10. Why do real gases not always behave ideally? Under
what conditions does a real gas behave most ideally?
Why?
A discussion of the Active Learning ­Questions can be found online in the ­Instructor’s Resource Guide and on PowerLecture. The questions
allow students to explore their understanding of concepts through discussion and peer teaching. The real value of these questions is the
learning that occurs while students talk to each other about chemical concepts.
Active Learning Questions
These questions are designed to be used by groups of students in
class.
1. Consider the following apparatus: a test tube covered with a
nonpermeable elastic membrane inside a container that is
closed with a cork. A syringe goes through the cork.
Syringe
d. Capillary action of the mercury causes the mercury to go
up the tube.
e. The vacuum that is formed at the top of the tube holds up
the mercury.
Justify your choice, and for the choices you did not pick, explain what is wrong with them. Pictures help!
3. The barometer below shows the level of mercury at a given atmospheric pressure. Fill all the other barometers with mercury
for that same atmospheric pressure. Explain your answer.
Cork
Membrane
Hg(l )
a. As you push down on the syringe, how does the membrane covering the test tube change?
b. You stop pushing the syringe but continue to hold it
down. In a few seconds, what happens to the membrane?
2. Figure 5.2 shows a picture of a barometer. Which of the following statements is the best explanation of how this barometer works?
a. Air pressure outside the tube causes the mercury to move in
the tube until the air pressure inside and outside the tube is
equal.
b. Air pressure inside the tube causes the mercury to move in
the tube until the air pressure inside and outside the tube is
equal.
c. Air pressure outside the tube counterbalances the weight
of the mercury in the tube.
4. As you increase the temperature of a gas in a sealed, rigid
container, what happens to the density of the gas? Would the
results be the same if you did the same experiment in a container with a piston at constant pressure? (See Fig. 5.17.)
5. A diagram in a chemistry book shows a magnified view of a
flask of air as follows:
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For Review
What do you suppose is between the dots (the dots represent
air molecules)?
a. air
b. dust
c. pollutants
d. oxygen
e. nothing
6. If you put a drinking straw in water, place your finger over the
opening, and lift the straw out of the water, some water stays
in the straw. Explain.
7. A chemistry student relates the following story: I noticed my
tires were a bit low and went to the gas station. As I was filling
the tires, I thought about the kinetic molecular theory (KMT).
I noticed the tires because the volume was low, and I realized
that I was increasing both the pressure and volume of the tires.
“Hmmm,” I thought, “that goes against what I learned in
chemistry, where I was told pressure and volume are inversely
proportional.” What is the fault in the logic of the chemistry
student in this situation? Explain why we think pressure and
volume to be inversely related (draw pictures and use the
KMT).
8. Chemicals X and Y (both gases) react to form the gas XY, but
it takes a bit of time for the reaction to occur. Both X and Y are
placed in a container with a piston (free to move), and you
note the volume. As the reaction occurs, what happens to the
volume of the container? (See Fig. 5.18.)
9. Which statement best explains why a hot-air balloon rises
when the air in the balloon is heated?
a. According to Charles’s law, the temperature of a gas is
directly related to its volume. Thus the volume of the balloon increases, making the density smaller. This lifts the
balloon.
b. Hot air rises inside the balloon, and this lifts the balloon.
c. The temperature of a gas is directly related to its pressure.
The pressure therefore increases, and this lifts the balloon.
d. Some of the gas escapes from the bottom of the balloon,
thus decreasing the mass of gas in the balloon. This
decreases the density of the gas in the balloon, which lifts
the balloon.
e. Temperature is related to the root mean square velocity of
the gas molecules. Thus the molecules are moving faster,
hitting the balloon more, and thus lifting the balloon.
Justify your choice, and for the choices you did not pick, explain what is wrong with them.
10. Draw a highly magnified view of a sealed, rigid container
filled with a gas. Then draw what it would look like if you
cooled the gas significantly but kept the temperature above the
boiling point of the substance in the container. Also draw what
it would look like if you heated the gas significantly. Finally,
draw what each situation would look like if you evacuated
enough of the gas to decrease the pressure by a factor of 2.
11. If you release a helium balloon, it soars upward and eventually
pops. Explain this behavior.
12. If you have any two gases in different containers that are the
same size at the same pressure and same temperature, what is
true about the moles of each gas? Why is this true?
233
13. Explain the following seeming contradiction: You have two
gases, A and B, in two separate containers of equal volume and
at equal pressure and temperature. Therefore, you must have
the same number of moles of each gas. Because the two temperatures are equal, the average kinetic energies of the two
samples are equal. Therefore, since the energy given such a
system will be converted to translational motion (that is, move
the molecules), the root mean square velocities of the two are
equal, and thus the particles in each sample move, on average,
with the same relative speed. Since A and B are different gases,
they each must have a different molar mass. If A has a higher
molar mass than B, the particles of A must be hitting the sides
of the container with more force. Thus the pressure in the container of gas A must be higher than that in the container with
gas B. However, one of our initial assumptions was that the
pressures were equal.
14. You have a balloon covering the mouth of a flask filled with air
at 1 atm. You apply heat to the bottom of the flask until the
volume of the balloon is equal to that of the flask.
a. Which has more air in it, the balloon or the flask? Or do
both have the same amount? Explain.
b. In which is the pressure greater, the balloon or the flask?
Or is the pressure the same? Explain.
15. How does Dalton’s law of partial pressures help us with our
model of ideal gases? That is, what postulates of the kinetic
mo­lecular theory does it support?
16. At the same conditions of pressure and temperature, ammonia
gas is less dense than air. Why is this true?
17. For each of the quantities listed below, explain which of the
following properties (mass of the molecule, density of the
gas sample, temperature of the gas sample, size of the molecule, and number of moles of gas) must be known to calculate the quantity.
a. average kinetic energy
b. average number of collisions per second with other gas
molecules
­
c. average force of each impact with the wall of the
container
d. root mean square velocity
e. average number of collisions with a given area of the
container
f. distance between collisions
18. You have two containers each with 1 mole of xenon gas at
158C. Container A has a volume of 3.0 L, and container B has
a volume of 1.0 L. Explain how the following quantities compare ­between the two containers.
a. the average kinetic energy of the Xe atoms
b. the force with which the Xe atoms collide with the container walls
c. the root mean square velocity of the Xe atoms
d. the collision frequency of the Xe atoms (with other
atoms)
e. the pressure of the Xe sample
19. Draw molecular-level views that show the differences among
solids, liquids, and gases.
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234
Chapter 5
Gases
A blue question or exercise number indicates that the answer to
that question or exercise appears at the back of the book and a
solution appears in the Solutions Guide, as found on PowerLecture.
28. Consider the flasks in the following diagrams.
Questions
20. At room temperature, water is a liquid with a molar volume of
18 mL. At 1058C and 1 atm pressure, water is a gas and has a
molar volume of over 30 L. Explain the large difference in
molar volumes.
21. If a barometer were built using water (d 5 1.0 g/cm3) instead
of mercury (d 5 13.6 g/cm3), would the column of water be
higher than, lower than, or the same as the column of mercury
at 1.00 atm? If the level is different, by what factor? Explain.
22. A bag of potato chips is packed and sealed in Los Angeles,
­California, and then shipped to Lake Tahoe, Nevada, during
ski season. It is noticed that the volume of the bag of potato
chips has increased upon its arrival in Lake Tahoe. What external conditions would most likely cause the volume increase?
23. Boyle’s law can be represented graphically in several ways.
Which of the following plots does not correctly represent
Boyle’s law (assuming constant T and n)? Explain.
PV
P
P
V
V
P
1/P
1/V
24. As weather balloons rise from the earth’s surface, the pressure
of the atmosphere becomes less, tending to cause the volume
of the balloons to expand. However, the temperature is much
lower in the upper atmosphere than at sea level. Would this
temperature effect tend to make such a balloon expand or contract? Weather balloons do, in fact, expand as they rise. What
does this tell you?
25. Which noble gas has the smallest density at STP? Explain.
26. Consider two different containers, each filled with 2 moles of
Ne(g). One of the containers is rigid and has constant volume.
The other container is flexible (like a balloon) and is capable
of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in
both containers, what happens to the pressure and density of
the gas inside each container? Assume a constant external
pressure.
27. In Example 5.11 of the text, the molar volume of N2(g) at STP
is given as 22.42 L/mol N2. How is this number calculated?
How does the molar volume of He(g) at STP compare to the
molar volume of N2(g) at STP (assuming ideal gas behavior)?
Is the molar volume of N2(g) at 1.000 atm and 25.0°C equal to,
less than, or greater than 22.42 L/mol? Explain. Is the molar
volume of N2(g) collected over water at a total pressure of
1.000 atm and 0.0°C equal to, less than, or greater than 22.42
L/mol? Explain.
volume = 2X
volume = X
volume = X
volume = X
Assuming the connecting tube has negligible volume, draw
what each diagram will look like after the stopcock between
the two flasks is opened. Also, solve for the final pressure in
each case, in terms of the original pressure. Assume temperature is constant.
29. Do all the molecules in a 1-mole sample of CH4(g) have the
same kinetic energy at 273 K? Do all molecules in a 1-mole
sample of N2(g) have the same velocity at 546 K? Explain.
30. Consider the following samples of gases at the same
temperature.
Ne
Ar
i
ii
iii
iv
v
vi
vii
viii
Arrange each of these samples in order from lowest to highest:
a. pressure
b. average kinetic energy
c. density
d. root mean square velocity
Note: Some samples of gases may have equal values for these
attributes. Assume the larger containers have a volume twice
the volume of the smaller containers, and assume the mass of
an argon atom is twice the mass of a neon atom.
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For Review
Exercises
39. A sealed-tube manometer (as shown below) can be used to
measure pressures below atmospheric pressure. The tube
above the mercury is evacuated. When there is a vacuum in the
flask, the mercury levels in both arms of the U-tube are equal.
If a gaseous sample is introduced into the flask, the mercury
levels are different. The difference h is a measure of the pressure of the gas inside the flask. If h is equal to 6.5 cm, calculate
the pressure in the flask in torr, pascals, and atmospheres.
h
Gas
40. If the sealed-tube manometer in Exercise 39 had a height difference of 20.0 inches between the mercury levels, what is the
pressure in the flask in torr and atmospheres?
41. A diagram for an open-tube manometer is shown below.
Atmosphere
If the flask is open to the atmosphere, the mercury levels are
equal. For each of the following situations where a gas is contained in the flask, calculate the pressure in the flask in torr,
atmospheres, and pascals.
Atmosphere
(760. torr)
In this section similar exercises are paired.
Flask
Flask
118 mm
Pressure
37. Freon-12 (CF2Cl2) is commonly used as the refrigerant in central home air conditioners. The system is initially charged to a
pressure of 4.8 atm. Express this pressure in each of the following units (1 atm 5 14.7 psi).
a. mm Hg
c. Pa
b. torr
d. psi
38. A gauge on a compressed gas cylinder reads 2200 psi (pounds
per square inch; 1 atm 5 14.7 psi). Express this pressure in
each of the following units.
a. standard atmospheres
b. megapascals (MPa)
c. torr
Atmosphere
(760. torr)
215 mm
31. As NH3(g) is decomposed into nitrogen gas and hydrogen gas
at constant pressure and temperature, the volume of the product gases collected is twice the volume of NH3 reacted. Explain. As NH3(g) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure
increases by some factor. Why the increase in pressure and by
what factor does the total pressure increase when reactants are
completely converted into products? How do the partial pressures of the product gases compare to each other and to the
initial pressure of NH3?
32. Which of the following statements is(are) true? For the false
statements, correct them.
a. At constant temperature, the lighter the gas molecules, the
faster the average velocity of the gas molecules.
b. At constant temperature, the heavier the gas molecules,
the larger the average kinetic energy of the gas molecules.
c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving
relatively quickly.
d. As temperature increases, the effect of interparticle interactions on gas behavior is increased.
e. At constant V and T, as gas molecules are added into a
container, the number of collisions per unit area increases
resulting in a higher pressure.
f. The kinetic molecular theory predicts that pressure is
inversely proportional to temperature at constant volume
and moles of gas.
33. From the values in Table 5.3 for the van der Waals constant a
for the gases H2, CO2, N2, and CH4, predict which of these gas
molecules show the strongest intermolecular attractions.
34. Without looking at a table of values, which of the following
gases would you expect to have the largest value of the van der
Waals constant b: H2, N2, CH4, C2H6, or C3H8?
35. Figure 5.6 shows the PV versus P plot for three different gases.
Which gas behaves most ideally? Explain.
36. Ideal gas particles are assumed to be volumeless and to neither
attract nor repel each other. Why are these assumptions crucial
to the validity of Dalton’s law of partial pressures?
235
a.
b.
c. Calculate the pressures in the flask in parts a and b (in
torr) if the atmospheric pressure is 635 torr.
42. a. If the open-tube manometer in Exercise 41 contains a nonvolatile silicone oil (density 5 1.30 g/cm3) instead of mercury (density 5 13.6 g/cm3), what are the pressures in the
flask as shown in parts a and b in torr, atmospheres, and
pascals?
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236
Chapter 5
Gases
b. What advantage would there be in using a less dense fluid
than mercury in a manometer used to measure relatively
small differences in pressure?
Gas Laws
43. A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L. If the balloon is filled
with 2.0 L helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only 500. mm Hg,
will the balloon burst? (Assume temperature is constant.)
44. A balloon is filled to a volume of 7.00 3 102 mL at a temperature of 20.08C. The balloon is then cooled at constant pressure
to a temperature of 1.00 3 102 K. What is the final volume of
the balloon?
45. An 11.2-L sample of gas is determined to contain 0.50 mole of
N2. At the same temperature and pressure, how many moles of
gas would there be in a 20.-L sample?
46. Consider the following chemical equation.
2NO2 1g2 h N2O4 1g2
If 25.0 mL NO2 gas is completely converted to N2O4 gas under
the same conditions, what volume will the N2O4 occupy?
47. Complete the following table for an ideal gas.
P (atm)
a.
5.00
b.
0.300
c.
4.47
V (L)
n (mol)
T
2.00
1558C
2.00
2.25
10.5
P
V
7.74 3 103 Pa
12.2 mL
b.
43.0 mL
c.
455 torr
d.
745 mm Hg
57. Consider two separate gas containers at the following
conditions:
Container A
Container B
Contents: SO2(g)
Pressure 5 PA
Moles of gas 5 1.0 mol
Volume 5 1.0 L
Temperature 5 78C
Contents: unknown gas
Pressure 5 PB
Moles of gas 5 2.0 mol
Volume 5 2.0 L
Temperature 5 2878C
How is the pressure in container B related to the pressure in
container A?
58. What will be the effect on the volume of an ideal gas if the
pressure is doubled and the absolute temperature is halved?
2.01
758C
48. Complete the following table for an ideal gas.
a.
55. A gas sample containing 1.50 moles at 258C exerts a pressure
of 400. torr. Some gas is added to the same container and the
temperature is increased to 50.8C. If the pressure increases to
800. torr, how many moles of gas were added to the container?
Assume a constant-volume container.
56. A bicycle tire is filled with air to a pressure of 75 psi at a temperature of 198C. Riding the bike on asphalt on a hot day increases the temperature of the tire to 588C. The volume of the tire
increases by 4.0%. What is the new pressure in the bicycle tire?
155 K
25.0
d.
53. A 2.50-L container is filled with 175 g argon.
a. If the pressure is 10.0 atm, what is the temperature?
b. If the temperature is 225 K, what is the pressure?
54. A person accidentally swallows a drop of liquid oxygen, O2(l),
which has a density of 1.149 g/mL. Assuming the drop has a
volume of 0.050 mL, what volume of gas will be produced in
the person’s stomach at body temperature (378C) and a pressure of 1.0 atm?
11.2 L
n
T
258C
0.421 mol
223 K
4.4 3 1022 mol
3318C
0.401 mol
49. Suppose two 200.0-L tanks are to be filled separately with the
gases helium and hydrogen. What mass of each gas is needed
to produce a pressure of 2.70 atm in its respective tank at
248C?
50. The average lung capacity of a human is 6.0 L. How many
moles of air are in your lungs when you are in the following
situations?
a. At sea level (T 5 298 K, P 5 1.00 atm).
b. 10. m below water (T 5 298 K, P 5 1.97 atm).
c. At the top of Mount Everest (T 5 200. K, P 5 0.296 atm).
51. The steel reaction vessel of a bomb calorimeter, which has a
volume of 75.0 mL, is charged with oxygen gas to a pressure
of 14.5 atm at 228C. Calculate the moles of oxygen in the reaction vessel.
52. A 5.0-L flask contains 0.60 g O2 at a temperature of 228C.
What is the pressure (in atm) inside the flask?
59. A container is filled with an ideal gas to a pressure of 11.0 atm
at 08C.
a. What will be the pressure in the container if it is heated to
458C?
b. At what temperature would the pressure be 6.50 atm?
c. At what temperature would the pressure be 25.0 atm?
60. An ideal gas at 78C is in a spherical flexible container having
a radius of 1.00 cm. The gas is heated at constant pressure to
888C. Determine the radius of the spherical container after the
gas is heated. [Volume of a sphere 5 (4y3)pr 3.]
61. An ideal gas is contained in a cylinder with a volume of 5.0 3
102 mL at a temperature of 30.8C and a pressure of 710. torr.
The gas is then compressed to a volume of 25 mL, and the
temperature is raised to 820.8C. What is the new pressure of
the gas?
62. A compressed gas cylinder contains 1.00 3 103 g argon gas. The
pressure inside the cylinder is 2050. psi (pounds per square inch)
at a temperature of 188C. How much gas remains in the cylinder
if the pressure is decreased to 650. psi at a temperature of 268C?
63. A sealed balloon is filled with 1.00 L helium at 238C and 1.00
atm. The balloon rises to a point in the atmosphere where the
pressure is 220. torr and the temperature is 2318C. What is the
change in volume of the balloon as it ascends from 1.00 atm to
a pressure of 220. torr?
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For Review
64. A hot-air balloon is filled with air to a volume of 4.00 3
103 m3 at 745 torr and 218C. The air in the balloon is then
heated to 628C, causing the balloon to expand to a volume of
4.20 3 103 m3. What is the ratio of the number of moles of air
in the heated balloon to the original number of moles of air in
the balloon? (Hint: Openings in the balloon allow air to flow
in and out. Thus the pressure in the balloon is always the same
as that of the atmosphere.)
Gas Density, Molar Mass, and Reaction Stoichiometry
65. Consider the following reaction:
4Al 1s2 1 3O2 1g2 h 2Al2O3 1s2
It takes 2.00 L of pure oxygen gas at STP to react completely
with a certain sample of aluminum. What is the mass of aluminum reacted?
66. A student adds 4.00 g of dry ice (solid CO2) to an empty balloon. What will be the volume of the balloon at STP after all
the dry ice sublimes (converts to gaseous CO2)?
67. Air bags are activated when a severe impact causes a steel ball
to compress a spring and electrically ignite a detonator cap.
This causes sodium azide (NaN3) to decompose explosively
according to the following reaction:
2NaN3 1s2 h 2Na 1s2 1 3N2 1g2
What mass of NaN3(s) must be reacted to inflate an air bag to
70.0 L at STP?
68. Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe):
2H2O2 1aq2 h 2H2O 1l2 1 O2 1g2
What volume of pure O2(g), collected at 278C and 746 torr,
would be generated by decomposition of 125 g of a 50.0% by
mass hydrogen peroxide solution? Ignore any water vapor that
may be present.
69. In 1897 the Swedish explorer Andreé tried to reach the North
Pole in a balloon. The balloon was filled with hydrogen gas.
The hydrogen gas was prepared from iron splints and diluted
sulfuric acid. The reaction is
Fe 1s2 1 H2SO4 1aq2 h FeSO4 1aq2 1 H2 1g2
The volume of the balloon was 4800 m3 and the loss of hydrogen gas during filling was estimated at 20.%. What mass of
iron splints and 98% (by mass) H2SO4 were needed to ensure
the complete filling of the balloon? Assume a temperature of
08C, a pressure of 1.0 atm during filling, and 100% yield.
70. Sulfur trioxide, SO3, is produced in enormous quantities each
year for use in the synthesis of sulfuric acid.
S 1s2 1 O2 1g2 h SO2 1g2
2SO2 1g2 1 O2 1g2 h 2SO3 1g2
What volume of O2(g) at 350.8C and a pressure of 5.25 atm is
needed to completely convert 5.00 g sulfur to sulfur trioxide?
71. A 15.0-L rigid container was charged with 0.500 atm of krypton gas and 1.50 atm of chlorine gas at 350.8C. The krypton
and chlorine react to form krypton tetrachloride. What mass of
krypton tetrachloride can be produced assuming 100% yield?
237
72. An important process for the production of acrylonitrile
(C3H3N) is given by the following equation:
2C3H6 1g2 1 2NH3 1g2 1 3O2 1g2 h 2C3H3N 1g2 1 6H2O 1g2
A 150.-L reactor is charged to the following partial pressures
at 258C:
PC3H6 5 0.500 MPa
PNH3 5 0.800 MPa
PO2 5 1.500 MPa
What mass of acrylonitrile can be produced from this mixture
(MPa 5 106 Pa)?
73. Consider the reaction between 50.0 mL liquid methanol,
CH3OH (density 5 0.850 g/mL), and 22.8 L O2 at 278C and a
pressure of 2.00 atm. The products of the reaction are CO2(g)
and H2O(g). Calculate the number of moles of H2O formed if
the reaction goes to completion.
74. Urea (H2NCONH2) is used extensively as a nitrogen source in
fertilizers. It is produced commercially from the reaction of
ammonia and carbon dioxide:
Heat
2NH3 1g2 1 CO2 1g2 8888n
H2NCONH2 1s2 1 H2O 1g2
Pressure
Ammonia gas at 2238C and 90. atm flows into a reactor at a
rate of 500. L/min. Carbon dioxide at 2238C and 45 atm flows
into the reactor at a rate of 600. L/min. What mass of urea is
produced per minute by this reaction assuming 100% yield?
75. Hydrogen cyanide is prepared commercially by the reaction of
methane, CH4(g), ammonia, NH3(g), and oxygen, O2(g), at
high temperature. The other product is gaseous water.
a. Write a chemical equation for the reaction.
b. What volume of HCN(g) can be obtained from the reaction
of 20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g)? The
volumes of all gases are measured at the same temperature
and pressure.
76. Ethene is converted to ethane by the reaction
Catalyst
C2H4 1g2 1 H2 1g2 8888n C2H6 1g2
C2H4 flows into a catalytic reactor at 25.0 atm and 300.8C with
a flow rate of 1000. L/min. Hydrogen at 25.0 atm and 300.8C
flows into the reactor at a flow rate of 1500. L/min. If 15.0 kg
C2H6 is collected per minute, what is the percent yield of the
reaction?
77. An unknown diatomic gas has a density of 3.164 g/L at STP.
What is the identity of the gas?
78. A compound has the empirical formula CHCl. A 256-mL
flask, at 373 K and 750. torr, contains 0.800 g of the gaseous
compound. Give the molecular formula.
79. Uranium hexafluoride is a solid at room temperature, but it
boils at 568C. Determine the density of uranium hexafluoride
at 60.8C and 745 torr.
80. Given that a sample of air is made up of nitrogen, oxygen, and
argon in the mole fractions 0.78 N2, 0.21 O2, and 0.010 Ar, what
is the density of air at standard temperature and pressure?
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238
Chapter 5
Gases
Partial Pressure
81. Determine the partial pressure of each gas as shown in the
figure below. Note: The relative numbers of each type of gas
are depicted in the figure.
1 atm
86. Consider the flask apparatus in Exercise 85, which now contains 2.00 L H2 at a pressure of 360. torr and 1.00 L N2 at an
unknown pressure. If the total pressure in the flasks is 320. torr
after the stopcock is opened, determine the initial pressure of
N2 in the 1.00-L flask.
87. Consider the three flasks in the diagram below. Assuming the
connecting tubes have negligible volume, what is the partial
pressure of each gas and the total pressure after all the stopcocks are opened?
He
Ne
Ar
82. Consider the flasks in the following diagrams.
He
Ne
He
Ne
1.00 L
200. torr
volume = X
volume = X
a. Which is greater, the initial pressure of helium or the initial pressure of neon? How much greater?
b. Assuming the connecting tube has negligible volume,
draw what each diagram will look like after the stopcock
between the two flasks is opened.
c. Solve for the final pressure in terms of the original pressures of helium and neon. Assume temperature is constant.
d. Solve for the final partial pressures of helium and neon in
terms of their original pressures. Assume the temperature
is constant.
83. A piece of solid carbon dioxide, with a mass of 7.8 g, is placed
in a 4.0-L otherwise empty container at 278C. What is the
pressure in the container after all the carbon dioxide vaporizes? If 7.8 g solid carbon dioxide were placed in the same
container but it already contained air at 740 torr, what would
be the partial pressure of carbon dioxide and the total pressure
in the container after the carbon dioxide vaporizes?
84. A mixture of 1.00 g H2 and 1.00 g He is placed in a 1.00-L
container at 278C. Calculate the partial pressure of each gas
and the total pressure.
85. Consider the flasks in the following diagram. What are the final partial pressures of H2 and N2 after the stopcock between
the two flasks is opened? (Assume the final volume is 3.00 L.)
What is the total pressure (in torr)?
2.00 L H2
475 torr
1.00 L N2
0.200 atm
1.00 L
0.400 atm
Ar
2.00 L
24.0 kPa
88. At 08C a 1.0-L flask contains 5.0 3 1022 mole of N2, 1.5 3 102
mg O2, and 5.0 3 1021 molecules of NH3. What is the partial
pressure of each gas, and what is the total pressure in the
flask?
89. The partial pressure of CH4(g) is 0.175 atm and that of O2(g)
is 0.250 atm in a mixture of the two gases.
a. What is the mole fraction of each gas in the mixture?
b. If the mixture occupies a volume of 10.5 L at 658C, calculate the total number of moles of gas in the mixture.
c. Calculate the number of grams of each gas in the mixture.
90. A tank contains a mixture of 52.5 g oxygen gas and 65.1 g
carbon dioxide gas at 27°C. The total pressure in the tank is
9.21 atm. Calculate the partial pressures of each gas in the
container.
91. Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic
zinc.
Zn 1s2 1 2HCl 1aq2 h ZnCl2 1aq2 1 H2 1g2
Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose
240. mL of hydrogen gas is collected at 30.8C and has a total
pressure of 1.032 atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc
must have reacted to produce this quantity of hydrogen? (The
vapor pressure of water is 32 torr at 308C.)
92. Helium is collected over water at 258C and 1.00 atm total pressure. What total volume of gas must be collected to obtain
0.586 g helium? (At 258C the vapor pressure of water is
23.8 torr.)
93. At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8765-g sample of
impure sodium chlorate was heated until the production of
oxygen gas ceased. The oxygen gas collected over water occupied 57.2 mL at a temperature of 228C and a pressure of
734 torr. Calculate the mass percent of NaClO3 in the original
sample. (At 228C the vapor pressure of water is 19.8 torr.)
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For Review
94. Xenon and fluorine will react to form binary compounds when
a mixture of these two gases is heated to 4008C in a nickel reaction vessel. A 100.0-mL nickel container is filled with xenon
and fluorine, giving partial pressures of 1.24 atm and 10.10 atm,
respectively, at a temperature of 258C. The reaction vessel is
heated to 4008C to cause a reaction to occur and then cooled to
a temperature at which F2 is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining F2 gas is
transferred to another 100.0-mL nickel container, where the
pressure of F2 at 258C is 7.62 atm. Assuming all of the xenon
has reacted, what is the formula of the product?
95. Methanol (CH3OH) can be produced by the following reaction:
CO 1g2 1 2H2 1g2 h CH3OH 1g2
Hydrogen at STP flows into a reactor at a rate of 16.0 L/min.
Carbon monoxide at STP flows into the reactor at a rate of
25.0 L/min. If 5.30 g methanol is produced per minute, what
is the percent yield of the reaction?
96. In the “Méthode Champenoise,” grape juice is fermented in a
wine bottle to produce sparkling wine. The reaction is
C6H12O6 1aq2 h 2C2H5OH 1aq2 1 2CO2 1g2
Fermentation of 750. mL grape juice (density 5 1.0 g/cm3) is
allowed to take place in a bottle with a total volume of 825 mL
until 12% by volume is ethanol (C2H5OH). Assuming that the
CO2 is insoluble in H2O (actually, a wrong assumption), what
would be the pressure of CO2 inside the wine bottle at 258C?
(The density of ethanol is 0.79 g/cm3.)
239
of 740. torr at 20.8C. After the reaction has gone to completion,
the pressure inside the flask is 390. torr at 20.8C. What is the
mass percent of MgO in the mixture? Assume that only the MgO
reacts with CO2.
Kinetic Molecular Theory and Real Gases
101. Calculate the average kinetic energies of CH4(g) and N2(g)
­mo­lecules at 273 K and 546 K.
102. A 100.-L flask contains a mixture of methane (CH4) and argon
gases at 258C. The mass of argon present is 228 g and the mole
fraction of methane in the mixture is 0.650. Calculate the total
kinetic energy of the gaseous mixture.
103. Calculate the root mean square velocities of CH4(g) and N2(g)
molecules at 273 K and 546 K.
104. Consider separate 1.0-L samples of He(g) and UF6(g), both at
1.00 atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the
same root mean square velocity?
105. You have a gas in a container fitted with a piston and you
change one of the conditions of the gas such that a change
takes place, as shown below:
1.00 atm
97. Hydrogen azide, HN3, decomposes on heating by the following unbalanced equation:
HN3 1g2 h N2 1g2 1 H2 1g2
If 3.0 atm of pure HN3(g) is decomposed initially, what is the
final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.
98. Equal moles of sulfur dioxide gas and oxygen gas are mixed in
a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction
goes to completion, what is the ratio of the final volume of the
gas mixture to the initial volume of the gas mixture if both
volumes are measured at the same temperature and pressure?
99. Some very effective rocket fuels are composed of lightweight
liquids. The fuel composed of dimethylhydrazine [(CH3)2N2H2]
mixed with dinitrogen tetroxide was used to power the Lunar
Lander in its missions to the moon. The two components react
according to the following equation:
1CH32 2N2H2 1l2 1 2N2O4 1l2 h 3N2 1g2 1 4H2O 1g2 1 2CO2 1g2
If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 1278C in an
evacuated 250-L tank, what is the partial pressure of nitrogen
gas produced and what is the total pressure in the tank assuming the reaction has 100% yield?
100. The oxides of Group 2A metals (symbolized by M here) react
with carbon dioxide according to the following reaction:
MO 1s2 1 CO2 1g2 h MCO3 1s2
A 2.85-g sample containing only MgO and CuO is placed in a
3.00-L container. The container is filled with CO2 to a pressure
State two distinct changes you can make to accomplish this,
and explain why each would work.
106. You have a gas in a container fitted with a piston and you
change one of the conditions of the gas such that a change
takes place, as shown below:
volume = X
volume = 2X
State three distinct changes you can make to accomplish this,
and explain why each would work.
107. Consider a 1.0-L container of neon gas at STP. Will the average kinetic energy, average velocity, and frequency of collisions of gas molecules with the walls of the container increase,
decrease, or remain the same under each of the following
conditions?
a. The temperature is increased to 1008C.
b. The temperature is decreased to 2508C.
c. The volume is decreased to 0.5 L.
d. The number of moles of neon is doubled.
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240
Chapter 5
Gases
108. Consider two gases, A and B, each in a 1.0-L container with
both gases at the same temperature and pressure. The mass of
gas A in the container is 0.34 g and the mass of gas B in the
container is 0.48 g.
A
B
0.34 g
0.48 g
a. Which gas sample has the most molecules present?
Explain.
b. Which gas sample has the largest average kinetic energy?
Explain.
c. Which gas sample has the fastest average velocity?
Explain.
d. How can the pressure in the two containers be equal to
each other since the larger gas B molecules collide with
the container walls more forcefully?
109.
Consider three identical flasks filled with different gases.
Flask A: CO at 760 torr and 08C
Flask B: N2 at 250 torr and 08C
Flask C: H2 at 100 torr and 08C
a. In which flask will the molecules have the greatest average kinetic energy?
b. In which flask will the molecules have the greatest average velocity?
110. Consider separate 1.0-L gaseous samples of H2, Xe, Cl2, and
O2 all at STP.
a. Rank the gases in order of increasing average kinetic
­energy.
b. Rank the gases in order of increasing average velocity.
c. How can separate 1.0-L samples of O2 and H2 each have
the same average velocity?
111. Freon-12 is used as a refrigerant in central home air conditioners. The rate of effusion of Freon-12 to Freon-11 (molar mass
5 137.4 g/mol) is 1.07:1. The formula of Freon-12 is one of
the following: CF4, CF3Cl, CF2Cl2, CFCl3, or CCl4. Which
formula is correct for Freon-12?
112. The rate of effusion of a particular gas was measured and
found to be 24.0 mL/min. Under the same conditions, the rate
of effusion of pure methane (CH4) gas is 47.8 mL/min. What
is the molar mass of the unknown gas?
113. One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of
effusion of 12C16O, 12C17O, and 12C18O. Name some advantages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.
114. It took 4.5 minutes for 1.0 L helium to effuse through a porous
barrier. How long will it take for 1.0 L Cl2 gas to effuse under
identical conditions?
115. Calculate the pressure exerted by 0.5000 mole of N2 in a
1.0000-L container at 25.08C
a. using the ideal gas law.
b. using the van der Waals equation.
c. Compare the results.
116. Calculate the pressure exerted by 0.5000 mole of N2 in a
10.000-L container at 25.08C
a. using the ideal gas law.
b. using the van der Waals equation.
c. Compare the results.
d. Compare the results with those in Exercise 115.
Atmosphere Chemistry
117. Use the data in Table 5.4 to calculate the partial pressure of He
in dry air assuming that the total pressure is 1.0 atm. Assuming
a temperature of 258C, calculate the number of He atoms per
cubic centimeter.
118. A 1.0-L sample of air is collected at 258C at sea level
(1.00 atm). Estimate the volume this sample of air would have
at an altitude of 15 km (see Fig. 5.30). At 15 km, the pressure
is about 0.1 atm.
119. Write an equation to show how sulfuric acid is produced in the
atmosphere.
120. Write an equation to show how sulfuric acids in acid rain reacts with marble and limestone. (Both marble and limestone
are primarily calcium carbonate.)
121. Atmospheric scientists often use mixing ratios to express the
concentrations of trace compounds in air. Mixing ratios are
often expressed as ppmv (parts per million volume):
ppmv of X 5
vol of X at STP
3 106
total vol of air at STP
On a certain November day, the concentration of carbon
monoxide in the air in downtown Denver, Colorado, reached
3.0 3 102 ppmv. The atmospheric pressure at that time was
628 torr and the temperature was 08C.
a. What was the partial pressure of CO?
b. What was the concentration of CO in molecules per
cubic meter?
c. What was the concentration of CO in molecules per cubic
centimeter?
122. Trace organic compounds in the atmosphere are first concentrated and then measured by gas chromatography. In the concentration step, several liters of air are pumped through a
tube containing a porous substance that traps organic compounds. The tube is then connected to a gas chromatograph
and heated to release the trapped compounds. The organic
compounds are separated in the column and the amounts are
measured. In an analysis for benzene and toluene in air, a
3.00-L sample of air at 748 torr and 238C was passed through
the trap. The gas chromatography analysis showed that this
air sample contained 89.6 ng benzene (C6H6) and 153 ng
toluene (C7H8). Calculate the mixing ratio (see Exercise 121)
and number of molecules per cubic centimeter for both benzene and toluene.
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For Review
Additional Exercises
123. Draw a qualitative graph to show how the first property varies
with the second in each of the following (assume 1 mole of an
ideal gas and T in kelvin).
a. PV versus V with constant T
b. P versus T with constant V
c. T versus V with constant P
d. P versus V with constant T
e. P versus 1yV with constant T
f. PVyT versus P
124. At STP, 1.0 L Br2 reacts completely with 3.0 L F2, producing
2.0 L of a product. What is the formula of the product? (All
substances are gases.)
125. A form of Boyle’s law is PV 5 k (at constant T and n). Table
5.1 contains actual data from pressure–volume experiments
conducted by Robert Boyle. The value of k in most experiments is 14.1 3 102 in Hg ? in3. Express k in units of atm ? L.
In Example 5.3, k was determined for NH3 at various pressures and volumes. Give some reasons why the k values differ
so dramatically between Example 5.3 and Table 5.1.
126. A 2.747-g sample of manganese metal is reacted with excess HCl gas to produce 3.22 L H2(g) at 373 K and 0.951
atm and a manganese chloride compound (MnClx). What is
the ­formula of the manganese chloride compound produced
in the reaction?
127. A 1.00-L gas sample at 100.8C and 600. torr contains 50.0%
helium and 50.0% xenon by mass. What are the partial pressures of the individual gases?
128. Cyclopropane, a gas that when mixed with oxygen is used as
a general anesthetic, is composed of 85.7% C and 14.3% H by
mass. If the density of cyclopropane is 1.88 g/L at STP, what
is the molecular formula of cyclopropane?
129. The nitrogen content of organic compounds can be determined
by the Dumas method. The compound in question is first reacted by passage over hot CuO(s):
132.
133.
134.
135.
241
of 1.00 atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below
1.00 atm ­pressure.
A spherical glass container of unknown volume contains helium gas at 258C and 1.960 atm. When a portion of the helium
is withdrawn and adjusted to 1.00 atm at 258C, it is found to
have a volume of 1.75 cm3. The gas remaining in the first container shows a pressure of 1.710 atm. Calculate the volume of
the spherical container.
A 2.00-L sample of O2(g) was collected over water at a total
pressure of 785 torr and 258C. When the O2(g) was dried (water vapor removed), the gas had a volume of 1.94 L at 258C
and 785 torr. Calculate the vapor pressure of water at 258C.
A 20.0-L stainless steel container at 258C was charged with
2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark
ignited the mixture, producing water. What is the pressure in
the tank at 258C? If the exact same experiment were performed, but the temperature was 1258C instead of 258C, what
would be the pressure in the tank?
Metallic molybdenum can be produced from the mineral molybdenite, MoS2. The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is
then reduced to metallic molybdenum using hydrogen gas.
The balanced equations are
MoS2 1s2 1 72 O2 1g2 h MoO3 1s2 1 2SO2 1g2
MoO3 1s2 1 3H2 1g2 h Mo 1s2 1 3H2O 1l2
Calculate the volumes of air and hydrogen gas at 178C and
1.00 atm that are necessary to produce 1.00 3 103 kg pure
molyb­denum from MoS2. Assume air contains 21% oxygen
by volume, and assume 100% yield for each reaction.
136. Nitric acid is produced commercially by the Ostwald process.
In the first step ammonia is oxidized to nitric oxide:
4NH3 1g2 1 5O2 1g2 h 4NO 1g2 1 6H2O 1g2
Assume this reaction is carried out in the apparatus diagramed
below.
Hot
Compound 8888n
N2 1g2 1 CO2 1g2 1 H2O 1g2
CuO(s)
The product gas is then passed through a concentrated solution of KOH to remove the CO2. After passage through the
KOH solution, the gas contains N2 and is saturated with water
vapor. In a given experiment a 0.253-g sample of a compound
produced 31.8 mL N2 saturated with water vapor at 258C and
726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at 258C is 23.8 torr.)
130. An organic compound containing only C, H, and N yields the
following data.
i. C
omplete combustion of 35.0 mg of the compound produced 33.5 mg CO2 and 41.1 mg H2O.
ii.A 65.2-mg sample of the compound was analyzed for
­nitrogen by the Dumas method (see Exercise 129),
giving 35.6 mL of dry N2 at 740. torr and 258C.
iii.The effusion rate of the compound as a gas was measured and found to be 24.6 mL/min. The effusion rate of
argon gas, under identical conditions, is 26.4 mL/min.
What is the molecular formula of the compound?
131. A 15.0-L tank is filled with H2 to a pressure of 2.00 3 102 atm.
How many balloons (each 2.00 L) can be inflated to a pressure
2.00 L NH3
0.500 atm
1.00 L O2
1.50 atm
The stopcock between the two reaction containers is opened,
and the reaction proceeds using proper catalysts. Calculate the
partial pressure of NO after the reaction is complete. Assume
100% yield for the reaction, assume the final container volume
is 3.00 L, and assume the temperature is constant.
137. A compound contains only C, H, and N. It is 58.51% C and
7.37% H by mass. Helium effuses through a porous frit 3.20
times as fast as the compound does. Determine the empirical
and molecular formulas of this compound.
138. One of the chemical controversies of the nineteenth century
concerned the element beryllium (Be). Berzelius originally
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242
Chapter 5
Gases
claimed that beryllium was a trivalent element (forming Be31
ions) and that it gave an oxide with the formula Be2O3. This
resulted in a calculated atomic mass of 13.5 for beryllium. In
formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming Be21 ions) and that it gave an
oxide with the formula BeO. This assumption gives an atomic
mass of 9.0. In 1894, A. Combes (Comptes Rendus 1894,
p. 1221) reacted beryllium with the anion C5H7O22 and measured the density of the gaseous product. Combes’s data for
two different experiments are as follows:
Mass
Volume
Temperature
Pressure
I
II
0.2022 g
22.6 cm3
138C
765.2 mm Hg
0.2224 g
26.0 cm3
178C
764.6 mm
If beryllium is a divalent metal, the molecular formula of the
product will be Be(C5H7O2)2; if it is trivalent, the formula will
be Be(C5H7O2)3. Show how Combes’s data help to confirm
that beryllium is a divalent metal.
139. An organic compound contains C, H, N, and O. Combustion
of 0.1023 g of the compound in excess oxygen yielded
0.2766 g CO2 and 0.0991 g H2O. A sample of 0.4831 g of the
compound was analyzed for nitrogen by the Dumas method
(see Exercise 129). At STP, 27.6 mL of dry N2 was obtained.
In a third ­experiment, the density of the compound as a gas
was found to be 4.02 g/L at 1278C and 256 torr. What are the
empirical and molecular formulas of the compound?
140. Consider the following diagram:
B
H2
A
Container A (with porous walls) is filled with air at STP. It is
then inserted into a large enclosed container (B), which is then
flushed with H2(g). What will happen to the pressure inside
container A? Explain your answer.
ChemWork Problems
These multiconcept problems (and additional ones) are found interactively online with the same type of assistance a student would get
from an instructor.
141. A glass vessel contains 28 g of nitrogen gas. Assuming ideal
behavior, which of the processes listed below would double
the pressure exerted on the walls of the vessel?
a. Adding 28 g of oxygen gas
b. Raising the temperature of the container from 2738C to
1278C
c. Adding enough mercury to fill one-half the container
d. Adding 32 g of oxygen gas
142.
143.
144.
145.
146.
e. Raising the temperature of the container from 30.8C to
60.8C
A steel cylinder contains 150.0 moles of argon gas at a temperature of 258C and a pressure of 8.93 MPa. After some argon has been used, the pressure is 2.00 MPa at a temperature
of 198C. What mass of argon remains in the cylinder?
A certain flexible weather balloon contains helium gas at a
volume of 855 L. Initially, the balloon is at sea level where the
temperature is 258C and the barometric pressure is 730 torr.
The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is 158C. What is the
change in volume of the balloon as it ascends from sea level to
6000 ft?
A large flask with a volume of 936 mL is evacuated and found
to have a mass of 134.66 g. It is then filled to a pressure of
0.967 atm at 318C with a gas of unknown molar mass and then
reweighed to give a new mass of 135.87 g. What is the molar
mass of this gas?
A 20.0-L nickel container was charged with 0.859 atm of xenon gas and 1.37 atm of fluorine gas at 4008C. The xenon and
fluorine react to form xenon tetrafluoride. What mass of xenon
tetrafluoride can be produced assuming 100% yield?
Consider the unbalanced chemical equation below:
CaSiO3 1s2 1 HF 1g2 h CaF2 1aq2 1 SiF4 1g2 1 H2O 1l2
Suppose a 32.9-g sample of CaSiO3 is reacted with 31.8 L of
HF at 27.08C and 1.00 atm. Assuming the reaction goes to
completion, calculate the mass of the SiF4 and H2O produced
in the reaction.
147. Consider separate 1.0-L gaseous samples of He, N2, and O2,
all at STP and all acting ideally. Rank the gases in order of
increasing average kinetic energy and in order of increasing
average velocity.
148. Which of the following statements is(are) true?
a. If the number of moles of a gas is doubled, the volume
will double, assuming the pressure and temperature of the
gas remain constant.
b. If the temperature of a gas increases from 258C to 508C,
the volume of the gas would double, assuming that the
pressure and the number of moles of gas remain constant.
c. The device that measures atmospheric pressure is called a
barometer.
d. If the volume of a gas decreases by one half, then the
pressure would double, assuming that the number of
moles and the temperature of the gas remain constant.
Challenge Problems
149. A chemist weighed out 5.14 g of a mixture containing unknown amounts of BaO(s) and CaO(s) and placed the sample in
a 1.50-L flask containing CO2(g) at 30.08C and 750. torr. After
the reaction to form BaCO3(s) and CaCO3(s) was completed,
the pressure of CO2(g) remaining was 230. torr. Calculate the
mass percentages of CaO(s) and BaO(s) in the mixture.
150. A mixture of chromium and zinc weighing 0.362 g was reacted with an excess of hydrochloric acid. After all the metals
in the mixture reacted, 225 mL dry of hydrogen gas was collected at 278C and 750. torr. Determine the mass percent of Zn
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For Review
151.
152.
153.
154.
155.
in the metal sample. [Zinc reacts with hydrochloric acid to
produce zinc chloride and hydrogen gas; chromium reacts
with hydrochloric acid to produce chromium(III) chloride and
hydrogen gas.]
Consider a sample of a hydrocarbon (a compound consisting
of only carbon and hydrogen) at 0.959 atm and 298 K. Upon
combusting the entire sample in oxygen, you collect a mixture
of gaseous carbon dioxide and water vapor at 1.51 atm and
375 K. This mixture has a density of 1.391 g/L and occupies a
volume four times as large as that of the pure hydrocarbon.
Determine the molecular formula of the hydrocarbon.
You have an equimolar mixture of the gases SO2 and O2, along
with some He, in a container fitted with a piston. The density
of this mixture at STP is 1.924 g/L. Assume ideal behavior and
constant temperature and pressure.
a. What is the mole fraction of He in the original mixture?
b. The SO2 and O2 react to completion to form SO3. What is
the density of the gas mixture after the reaction is complete?
Methane (CH4) gas flows into a combustion chamber at a rate
of 200. L/min at 1.50 atm and ambient temperature. Air is
added to the chamber at 1.00 atm and the same temperature,
and the gases are ignited.
a. To ensure complete combustion of CH4 to CO2(g) and
H2O(g), three times as much oxygen as is necessary is
reacted. Assuming air is 21 mole percent O2 and 79 mole
percent N2, calculate the flow rate of air necessary to
deliver the required amount of oxygen.
b. Under the conditions in part a, combustion of methane
was not complete as a mixture of CO2(g) and CO(g) was
produced. It was determined that 95.0% of the carbon in
the exhaust gas was present in CO2. The remainder was
present as carbon in CO. Calculate the composition of the
exhaust gas in terms of mole fraction of CO, CO2, O2, N2,
and H2O. Assume CH4 is completely reacted and N2 is
unreacted.
A steel cylinder contains 5.00 mole of graphite (pure carbon)
and 5.00 moles of O2. The mixture is ignited and all the graphite reacts. Combustion produces a mixture of CO gas and CO2
gas. After the cylinder has cooled to its original temperature, it
is found that the pressure of the cylinder has increased by
17.0%. Calculate the mole fractions of CO, CO2, and O2 in the
final gaseous mixture.
The total mass that can be lifted by a balloon is given by the
difference between the mass of air displaced by the balloon
and the mass of the gas inside the balloon. Consider a hot-air
balloon that approximates a sphere 5.00 m in diameter and
contains air heated to 658C. The surrounding air temperature
is 218C. The pressure in the balloon is equal to the atmospheric
pressure, which is 745 torr.
a. What total mass can the balloon lift? Assume that the
average molar mass of air is 29.0 g/mol. (Hint: Heated air
is less dense than cool air.)
b. If the balloon is filled with enough helium at 218C and
745 torr to achieve the same volume as in part a, what
total mass can the balloon lift?
c. What mass could the hot-air balloon in part a lift if it were
on the ground in Denver, Colorado, where a typical atmospheric pressure is 630. torr?
243
156. You have a sealed, flexible balloon filled with argon gas. The
­atmospheric pressure is 1.00 atm and the temperature is 258C.
­Assume that air has a mole fraction of nitrogen of 0.790, the rest
being oxygen.
a. Explain why the balloon would float when heated. Make
sure to discuss which factors change and which remain
constant, and why this matters.
b. Above what temperature would you heat the balloon so
that it would float?
157. You have a helium balloon at 1.00 atm and 258C. You want to
make a hot-air balloon with the same volume and same lift as
the helium balloon. Assume air is 79.0% nitrogen and 21.0%
oxygen by volume. The “lift” of a balloon is given by the difference between the mass of air displaced by the balloon and
the mass of gas inside the balloon.
a. Will the temperature in the hot-air balloon have to be
higher or lower than 258C? Explain.
b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon
at 1.00 atm and 258C. Assume atmospheric conditions are
1.00 atm and 258C.
158. We state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.
159. You are given an unknown gaseous binary compound (that is,
a compound consisting of two different elements). When 10.0
g of the compound is burned in excess oxygen, 16.3 g of water
is ­produced. The compound has a density 1.38 times that of
oxygen gas at the same conditions of temperature and pressure. Give a possible identity for the unknown compound.
160. Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia gas (NH3). You have nitrogen and hydrogen gases in a
15.0-L container fitted with a movable piston (the piston allows the container volume to change so as to keep the pressure constant inside the container). Initially the partial pressure of each reactant gas is 1.00 atm. Assume the temperature
is constant and that the reaction goes to completion.
a. Calculate the partial pressure of ammonia in the container
after the reaction has reached completion.
b. Calculate the volume of the container after the reaction
has reached completion.
Integrative Problems
These problems require the integration of multiple concepts to find
the solutions.
161. In the presence of nitric acid, UO21 undergoes a redox process. It is converted to UO221 and nitric oxide (NO) gas is
produced according to the following unbalanced equation:
H1 1aq2 1 NO32 1aq2 1 UO21 1aq2 h
NO 1g2 1 UO221 1aq2 1 H2O 1l2
If 2.55 3 102 mL NO(g) is isolated at 298C and 1.5 atm, what
amount (moles) of UO21 was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)
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244
Chapter 5
Gases
162. Silane, SiH4, is the silicon analogue of methane, CH4. It is
prepared industrially according to the following equations:
Si 1s2 1 3HCl 1g2 h HSiCl3 1l2 1 H2 1g2
4HSiCl3 1l2 h SiH4 1g2 1 3SiCl4 1l2
a. If 156 mL HSiCl3 (d 5 1.34 g/mL) is isolated when 15.0 L
HCl at 10.0 atm and 358C is used, what is the percent yield
of HSiCl3?
b. When 156 mL HSiCl3 is heated, what volume of SiH4 at
10.0 atm and 358C will be obtained if the percent yield of
the reaction is 93.1%?
163. Solid thorium(IV) fluoride has a boiling point of 16808C.
What is the density of a sample of gaseous thorium(IV) fluoride at its boiling point under a pressure of 2.5 atm in a 1.7-L
container? Which gas will effuse faster at 16808C, thorium(IV)
fluoride or uranium(III) fluoride? How much faster?
164. Natural gas is a mixture of hydrocarbons, primarily methane
(CH4) and ethane (C2H6). A typical mixture might have xmethane 5
0.915 and xethane 5 0.085. What are the partial pressures of the
two gases in a 15.00-L container of natural gas at 20.8C and
1.44 atm? Assuming complete combustion of both gases in the
natural gas sample, what is the total mass of water formed?
Marathon Problem
This problem is designed to incorporate several concepts and techniques into one situation.
165. Consider an equimolar mixture (equal number of moles) of
two diatomic gases (A2 and B2) in a container fitted with a
piston. The gases react to form one product (which is also a
gas) with the formula AxBy. The density of the sample after the
reaction is complete (and the temperature returns to its original state) is 1.50 times greater than the density of the reactant
mixture.
a. Specify the formula of the product, and explain if more
than one answer is possible based on the given data.
b. Can you determine the molecular formula of the product
with the information given or only the empirical formula?
Marathon Problems can be used in class by groups of students to
help facilitate problem-solving skills.
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Chapter 6
Thermochemistry
6.1
The Nature of Energy
6.3
Chemical Energy
6.2
Hess’s Law
6.6
New Energy Sources
Characteristics of Enthalpy Changes
Coal Conversion
Enthalpy and Calorimetry
6.4
Standard Enthalpies of Formation
Hydrogen as a Fuel
Enthalpy
6.5
Present Sources of Energy
Other Energy Alternatives
Calorimetry
Petroleum and Natural Gas
Coal
Effects of Carbon Dioxide on Climate
A burning match is an example of exothermic reaction.This double exposure shows the match lit and the
match blown out. (© Caren Brinkema/Science Faction/Corbis)
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245
E
nergy is the essence of our very existence as individuals and as a society. The
food that we eat furnishes the energy to live, work, and play, just as the coal and
oil consumed by manufacturing and transportation systems power our modern industrialized civilization.
In the past, huge quantities of carbon-based fossil fuels have been available for the
taking. This abundance of fuels has led to a world society with a voracious appetite for
energy, consuming millions of barrels of petroleum every day. We are now dangerously dependent on the dwindling supplies of oil, and this dependence is an important
source of tension among nations in today’s world. In an incredibly short time, we have
moved from a period of ample and cheap supplies of petroleum to one of high prices
and uncertain supplies. If our present standard of living is to be maintained, we must
find alternatives to petroleum. To do this, we need to know the relationship between
chemistry and energy, which we explore in this chapter.
There are additional problems with fossil fuels. The waste products from burning
fossil fuels significantly affect our environment. For example, when a carbon-based
fuel is burned, the carbon reacts with oxygen to form carbon dioxide, which is released
into the atmosphere. Although much of this carbon dioxide is consumed in various
natural processes such as photosynthesis and the formation of carbonate materials, the
amount of carbon dioxide in the atmosphere is steadily increasing. This increase is
significant because atmospheric carbon dioxide absorbs heat radiated from the earth’s
surface and radiates it back toward the earth. Since this is an important mechanism for
controlling the earth’s temperature, many scientists fear that an increase in the concentration of carbon dioxide will warm the earth, causing significant changes in climate.
In addition, impurities in the fossil fuels react with components of the air to produce
air pollution. We discussed some aspects of this problem in Chapter 5.
Just as energy is important to our society on a macroscopic scale, it is critically
­important to each living organism on a microscopic scale. The living cell is a miniature
chemical factory powered by energy from chemical reactions. The process of cellular
respiration extracts the energy stored in sugars and other nutrients to drive the various
tasks of the cell. Although the extraction process is more complex and more subtle, the
energy obtained from “fuel” molecules by the cell is the same as would be obtained from
burning the fuel to power an internal combustion engine.
Whether it is an engine or a cell that is converting energy from one form to another, the processes are all governed by the same principles, which we will begin to
explore in this chapter. Additional aspects of energy transformation will be covered
in Chapter 17.
IBLG: See questions from
“The Nature of Energy”
6.1 The Nature of Energy
The total energy content of the universe
is constant.
246
Although the concept of energy is quite familiar, energy itself is rather difficult to define
precisely. We will define energy as the capacity to do work or to produce heat. In this
chapter we will concentrate specifically on the heat transfer that accompanies chemical
processes.
One of the most important characteristics of energy is that it is conserved. The law
of conservation of energy states that energy can be converted from one form to
another but can be neither created nor destroyed. That is, the energy of the universe
is constant. Energy can be classified as either potential or kinetic energy. Potential
energy is energy due to position or composition. For example, water behind a dam has
potential energy that can be converted to work when the water flows down through
turbines, thereby creating electricity. Attractive and repulsive forces also lead to
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6.1
A
Held in place
B
a
Initial
B
A
b
Final
Figure 6.1 | (a) In the initial
positions, ball A has a higher potential
energy than ball B. (b) After A has
rolled down the hill, the potential
energy lost by A has been ­converted
to random motions of the components of the hill (frictional heating)
and to the increase in the potential
energy of B.
Courtesy, Sierra Pacific Innovations
Heat involves a transfer of energy.
This infrared photo of a house shows
where energy leaks occur. The more
red the color, the more energy (heat) is
leaving the house.
PowerLecture: Surface Area and
­Reaction Rate: Coffee Creamer
Flammability
The Nature of Energy
247
potential energy. The energy released when gasoline is burned results from differences
in attractive forces between the nuclei and electrons in the reactants and products. The
kinetic energy of an object is ­energy due to the motion of the object and depends on
the mass of the object m and its velocity v: KE 5 12mv2 .
Energy can be converted from one form to another. For example, consider the two
balls in Fig. 6.1(a). Ball A, because of its higher position initially, has more potential
energy than ball B. When A is released, it moves down the hill and strikes B. Eventually, the arrangement shown in Fig. 6.1(b) is achieved. What has happened in going
from the initial to the final arrangement? The potential energy of A has decreased,
but since energy is conserved, all the energy lost by A must be accounted for. How
is this energy distributed?
Initially, the potential energy of A is changed to kinetic energy as the ball rolls
down the hill. Part of this kinetic energy is then transferred to B, causing it to be raised
to a higher final position. Thus the potential energy of B has been increased. However,
since the final position of B is lower than the original position of A, some of the energy
is still unaccounted for. Both balls in their final positions are at rest, so the missing
energy cannot be due to their motions. What has happened to the remaining energy?
The answer lies in the interaction between the hill’s surface and the ball. As ball A
rolls down the hill, some of its kinetic energy is transferred to the surface of the hill as
heat. This transfer of energy is called frictional heating. The temperature of the hill
increases very slightly as the ball rolls down.
Before we proceed further, it is important to recognize that heat and temperature are
decidedly different. As we saw in Chapter 5, temperature is a property that reflects the
random motions of the particles in a particular substance. Heat, on the other hand, involves the transfer of energy between two objects due to a temperature difference. Heat is
not a substance contained by an object, although we often talk of heat as if this were true.
Note that in going from the initial to the final arrangements in Fig. 6.1, ball B gains
potential energy because work was done by ball A on B. Work is defined as force acting over a distance. Work is required to raise B from its original position to its final
one. Part of the original energy stored as potential energy in A has been transferred
through work to B, thereby increasing B’s potential energy. Thus there are two ways to
transfer energy: through work and through heat.
In rolling to the bottom of the hill shown in Fig. 6.1, ball A will always lose the
same amount of potential energy. However, the way that this energy transfer is divided
between work and heat depends on the specific conditions—the pathway. For example, the surface of the hill might be so rough that the energy of A is expended completely through frictional heating; A is moving so slowly when it hits B that it cannot
move B to the next level. In this case, no work is done. Regardless of the condition of
the hill’s surface, the total energy transferred will be constant. However, the amounts
of heat and work will ­differ. Energy change is independent of the pathway; however,
work and heat are both ­dependent on the pathway.
This brings us to a very important concept: the state function or state property. A
state function refers to a property of the system that depends only on its present state.
A state function (property) does not depend in any way on the system’s past (or future). In other words, the value of a state function does not depend on how the system
arrived at the present state; it depends only on the characteristics of the present state.
This leads to a very important characteristic of a state function: A change in this function (property) in going from one state to another state is independent of the particular
pathway taken between the two states.
A nonscientific analogy that illustrates the difference between a state function and
a nonstate function is elevation on the earth’s surface and distance between two points.
In traveling from Chicago (elevation 674 ft) to Denver (elevation 5280 ft), the change
in elevation is always 5280 2 674 5 4606 ft regardless of the route taken between the
two cities. The distance traveled, however, depends on how you make the trip. Thus
elevation is a function that does not depend on the route (pathway), but distance is
pathway dependent. Elevation is a state function and distance is not.
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248
Chapter 6
Thermochemistry
Energy is a state function; work and heat
are not.
Of the functions considered in our present example, energy is a state function, but
work and heat are not state functions.
Chemical Energy
The ideas we have just illustrated using mechanical examples also apply to chemical
systems. The combustion of methane, for example, is used to heat many homes in
the United States:
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1g2 1 energy 1heat2
PowerLecture: Combustion Reaction:
Sugar and ­Potassium Chlorate
To discuss this reaction, we divide the universe into two parts: the system and the surroundings. The system is the part of the universe on which we wish to focus attention;
the surroundings include everything else in the universe. In this case we define the
system as the reactants and products of the reaction. The surroundings consist of the
reaction container (a furnace, for example), the room, and anything else other than
the reactants and products.
When a reaction results in the evolution of heat, it is said to be exothermic (exo- is
a prefix meaning “out of”); that is, energy flows out of the system. For example, in the
combustion of methane, energy flows out of the system as heat. Reactions that absorb
energy from the surroundings are said to be endothermic. When the heat flow is into
a system, the process is endothermic. For example, the formation of nitric oxide from
nitrogen and oxygen is endothermic:
N2 1g2 1 O2 1g2 1 energy 1heat2 h 2NO 1g2
Where does the energy, released as heat, come from in an exothermic reaction? The
answer lies in the difference in potential energies between the products and the reactants. Which has lower potential energy, the reactants or the products? We know that
total energy is conserved and that energy flows from the system into the surroundings
in an exothermic reaction. This means that the energy gained by the surroundings must
be equal to the energy lost by the system. In the combustion of methane, the energy
content of the system decreases, which means that 1 mole of CO2 and 2 moles of H2O
molecules (the products) possess less potential energy than do 1 mole of CH4 and
2 moles of O2 molecules (the reactants). The heat flow into the surroundings results
from a lowering of the potential energy of the reaction system. This always holds true.
In any exothermic reaction, some of the potential energy stored in the chemical bonds is
being converted to thermal ­energy (random kinetic energy) via heat.
The energy diagram for the combustion of methane is shown in Fig. 6.2, where
D(PE) represents the change in potential energy stored in the bonds of the products as
compared with the bonds of the reactants. In other words, this quantity represents the
difference between the energy required to break the bonds in the reactants and the
Figure 6.2 | The combustion of
methane releases the quantity of
energy D(PE) to the surroundings via
heat flow. This is an exothermic
process.
Potential energy
System
Surroundings
2 mol O2
1 mol CH4
( Reactants)
Δ(PE)
Energy released to the surroundings as heat
Exothermic reaction
2 mol H2O
1 mol CO2
( Products)
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6.1
The Nature of Energy
249
Figure 6.3 | The energy diagram for
the reaction of nitrogen and oxygen to
form nitric oxide. This is an endothermic process: Heat [equal in magnitude
to D(PE)] flows into the system from
the surroundings.
System
2 mol NO
( Products)
Potential energy
PowerLecture: Work versus Energy Flow
Surroundings
Δ(PE)
Heat absorbed from the surroundings
Endothermic reaction
1 mol N2
1 mol O2
( Reactants)
energy released when the bonds in the products are formed. In an exothermic process,
the bonds in the products are stronger (on average) than those of the reactants. That is,
more energy is released by forming the new bonds in the products than is consumed to
break the bonds in the reactants. The net result is that the quantity of energy D(PE) is
transferred to the surroundings through heat.
For an endothermic reaction, the situation is reversed, as shown in Fig. 6.3. Energy
that flows into the system as heat is used to increase the potential energy of the system.
In this case the products have higher potential energy (weaker bonds on average) than
the reactants.
The study of energy and its interconversions is called thermodynamics. The law of
conservation of energy is often called the first law of thermodynamics and is stated
as follows: The energy of the universe is constant.
The internal energy E of a system can be defined most precisely as the sum of the
kinetic and potential energies of all the “particles” in the system. The internal energy
of a system can be changed by a flow of work, heat, or both. That is,
DE 5 q 1 w
where DE represents the change in the system’s internal energy, q represents heat, and
w represents work.
Thermodynamic quantities always consist of two parts: a number, giving the magnitude of the change, and a sign, indicating the direction of the flow. The sign reflects
the system’s point of view. For example, if a quantity of energy flows into the system
via heat (an endothermic process), q is equal to 1x, where the positive sign indicates
that the system’s energy is increasing. On the other hand, when energy flows out of the
system via heat (an exothermic process), q is equal to 2x, where the negative sign indicates that the system’s energy is decreasing.
Surroundings
Surroundings
Energy
Energy
System
System
ΔE < 0
Exothermic
ΔE > 0
Endothermic
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250
Chapter 6
Thermochemistry
The convention in this text is to take the
system’s point of view; q 5 2x denotes an
exothermic process, and q 5 1x denotes
an endothermic one.
Interactive
Example 6.1
In this text the same conventions also apply to the flow of work. If the system does
work on the surroundings (energy flows out of the system), w is negative. If the surroundings do work on the system (energy flows into the system), w is positive. We
define work from the system’s point of view to be consistent for all thermodynamic
quantities. That is, in this convention the signs of both q and w reflect what happens to
the system; thus we use DE 5 q 1 w.
In this text we always take the system’s point of view. This convention is not followed in every area of science. For example, engineers are in the business of designing
machines to do work, that is, to make the system (the machine) transfer energy to its
surroundings through work. Consequently, engineers define work from the surroundings’ point of view. In their convention, work that flows out of the system is treated as
positive because the energy of the surroundings has increased. The first law of thermodynamics is then written DE 5 q 2 w9, where w9 signifies work from the surroundings’ point of view.
Internal Energy
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.com to try this Interactive Example
in OWL.
The joule (J) is the fundamental SI unit for
energy:
kg # m2
J5
s2
Calculate DE for a system undergoing an endothermic process in which 15.6 kJ of heat
flows and where 1.4 kJ of work is done on the system.
Solution
We use the equation
DE 5 q 1 w
where q 5 115.6 kJ, since the process is endothermic, and w 5 11.4 kJ, since work
is done on the system. Thus
One kilojoule (kJ) 5 103 J.
DE 5 15.6 kJ 1 1.4 kJ 5 17.0 kJ
The system has gained 17.0 kJ of energy.
See Exercises 6.29 through 6.32
PowerLecture: Work versus Energy Flow
A common type of work associated with chemical processes is work done by a gas
(through expansion) or work done to a gas (through compression). For example, in an
automobile engine, the heat from the combustion of the gasoline expands the gases in
the cylinder to push back the piston, and this motion is then translated into the motion
of the car.
Suppose we have a gas confined to a cylindrical container with a movable piston as
shown in Fig. 6.4, where F is the force acting on a piston of area A. Since pressure is
defined as force per unit area, the pressure of the gas is
P= F
A
P= F
A
P5
Area = A
F
A
Work is defined as force applied over a distance, so if the piston moves a distance
Dh, as shown in Fig. 6.4, then the work done is
Work 5 force 3 distance 5 F 3 Dh
Δh
Δh
ΔV
Since P 5 FyA or F 5 P 3 A, then
Work 5 F 3 Dh 5 P 3 A 3 Dh
a
b
Figure 6.4 | (a) The piston, moving a distance Dh against a pressure P, does work on the
Initial
state
Final
state
surroundings. (b) Since the volume of a cylinder is the area of the base times its height, the
change in volume of the gas is given by Dh 3 A 5 DV.
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6.1
The Nature of Energy
251
Since the volume of a cylinder equals the area of the piston times the height of the cylinder
(see Fig. 6.4), the change in volume DV resulting from the piston moving a distance Dh is
DV 5 final volume 2 initial volume 5 A 3 Dh
Substituting DV 5 A 3 Dh into the expression for work gives
Work 5 P 3 A 3 Dh 5 PDV
w and PDV have opposite signs because
when the gas expands (DV is positive),
work flows into the surroundings (w is
negative).
This gives us the magnitude (size) of the work required to expand a gas DV against a
pressure P.
What about the sign of the work? The gas (the system) is expanding, moving the
piston against the pressure. Thus the system is doing work on the surroundings, so
from the system’s point of view the sign of the work should be negative.
For an expanding gas, DV is a positive quantity because the volume is increasing.
Thus DV and w must have opposite signs, which leads to the equation
w 5 2PDV
Note that for a gas expanding against an external pressure P, w is a negative quantity
as required, since work flows out of the system. When a gas is compressed, DV is a
negative quantity (the volume decreases), which makes w a positive quantity (work
flows into the system).
Critical Thinking
You are calculating DE in a chemistry problem. What if you confuse the system and
the surroundings? How would this affect the magnitude of the answer you calculate?
The sign?
Interactive
Example 6.2
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in OWL.
PV Work
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.
Solution
For a gas at constant pressure,
w 5 2PDV
For an ideal gas, work can occur only
when its volume changes. Thus, if a gas is
heated at constant volume, the ­pressure
increases but no work occurs.
In this case P 5 15 atm and DV 5 64 2 46 5 18 L. Hence
w 5 215 atm 3 18 L 5 2270 L # atm
Note that since the gas expands, it does work on its surroundings.
Reality Check | Energy flows out of the gas, so w is a negative quantity.
See Exercises 6.35 through 6.37
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Interactive
Example 6.3
In dealing with “PV work,” keep in mind that the P in PDV always refers to the external pressure—the pressure that causes a compression or that resists an expansion.
Internal Energy, Heat, and Work
A balloon is being inflated to its full extent by heating the air inside it. In the final
stages of this process, the volume of the balloon changes from 4.00 3 106 L to
4.50 3 106 L by the addition of 1.3 3 108 J of energy as heat. Assuming that the
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252
Chapter 6
Thermochemistry
balloon expands against a constant pressure of 1.0 atm, calculate DE for the process.
(To convert between L ? atm and J, use 1 L ? atm 5 101.3 J.)
Solution
Where are we going?
To calculate DE
What do we know?
❯
❯
❯
P 5 1.0 atm
1 L # atm 5 101.3 J
V2 5 4.50 3 106 L
❯
DE 5 q 1 w
Carlos Caetano/Shutterstock.com
❯
A propane burner is used to heat the air
in a hot-air balloon.
V1 5 4.00 3 106 L
q 5 11.3 3 108 J
❯
What do we need?
How do we get there?
What is the work done on the gas?
w 5 2PDV
What is DV?
DV 5 V2 2 V1 5 4.50 3 106 L 2 4.00 3 106 L 5 5.0 3 105 L
What is the work?
w 5 2PDV 5 21.0 atm 3 5.0 3 105 L 5 25.0 3 105 L # atm
The negative sign makes sense because the gas is expanding and doing work on the
surroundings. To calculate DE, we must sum q and w. However, since q is given in
units of J and w is given in units of L ? atm, we must change the work to units of joules:
w 5 25.0 3 105 L # atm 3
101.3 J
5 25.1 3 107 J
L # atm
Then
❯
DE 5 q 1 w 5 111.3 3 108 J2 1 125.1 3 107 J2 5 8 3 107 J
Reality Check | Since more energy is added through heating than the gas expends doing work, there is a net increase in the internal energy of the gas in the balloon. Hence
DE is positive.
See Exercises 6.38 through 6.40
IBLG: See questions from
“Enthalpy and Calorimetry”
6.2 Enthalpy and Calorimetry
Enthalpy
So far we have discussed the internal energy of a system. A less familiar property of a
system is its enthalpy H, which is defined as
H 5 E 1 PV
where E is the internal energy of the system, P is the pressure of the system, and V is
the volume of the system.
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6.2
Enthalpy is a state function. A change in
enthalpy does not depend on the pathway
between two states.
Enthalpy and Calorimetry
253
Since internal energy, pressure, and volume are all state functions, enthalpy is also
a state function. But what exactly is enthalpy? To help answer this question, consider
a process carried out at constant pressure and where the only work allowed is pressure–
volume work (w 5 2PDV ). Under these conditions, the expression
DE 5 qP 1 w
becomes
Recall from the previous section that w
and PDV have opposite signs:
w 5 2PDV
DE 5 qP 2 PDV
or
qP 5 DE 1 PDV
where qP is the heat at constant pressure.
We will now relate qP to a change in enthalpy. The definition of enthalpy is
H 5 E 1 PV. Therefore, we can say
or
Change in H 5 1change in E2 1 1change in PV2
DH 5 DE 1 D 1PV2
Since P is constant, the change in PV is due only to a change in volume. Thus
and
D 1PV2 5 PDV
DH 5 DE 1 PDV
This expression is identical to the one we obtained for qP:
qP 5 DE 1 PDV
Thus, for a process carried out at constant pressure and where the only work allowed
is that from a volume change, we have
DH 5 qP
DH 5 q only at constant pressure.
The change in enthalpy of a system has
no easily interpreted meaning except at
constant pressure, where DH 5 heat.
At constant pressure (where only PV work is allowed), the change in enthalpy DH of
the system is equal to the energy flow as heat. This means that for a reaction studied at
constant pressure, the flow of heat is a measure of the change in enthalpy for the system. For this reason, the terms heat of reaction and change in enthalpy are used interchangeably for reactions studied at constant pressure.
For a chemical reaction, the enthalpy change is given by the equation
DH 5 Hproducts 2 Hreactants
At constant pressure, exothermic means
DH is negative; endothermic means DH is
positive.
In a case in which the products of a reaction have a greater enthalpy than the reactants, DH will be positive. Thus heat will be absorbed by the system, and the reaction
is endothermic. On the other hand, if the enthalpy of the products is less than that of
the ­reactants, DH will be negative. In this case the overall decrease in enthalpy is
achieved by the generation of heat, and the reaction is exothermic.
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in OWL.
Interactive
Example 6.4
Enthalpy
When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is
released as heat. Calculate DH for a process in which a 5.8-g sample of methane is
burned at constant pressure.
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254
Chapter 6
Thermochemistry
Solution
Where are we going?
To calculate DH
What do we know?
qP 5 DH 5 2890 kJ /mol CH4
Mass 5 5.8 g CH4
Molar mass CH4 5 16.04 g
❯
❯
❯
How do we get there?
What are the moles of CH4 burned?
5.8 g CH4 3
1 mol CH4
5 0.36 mol CH4
16.04 g CH4
What is DH?
DH 5 0.36 mol CH4 3
2890 kJ
5 2320 kJ
mol CH4
Thus, when a 5.8-g sample of CH4 is burned at constant pressure,
❯
DH 5 heat flow 5 2320 kJ
Reality Check | In this case, a 5.8-g sample of CH4 is burned. Since this amount is
smaller than 1 mole, less than 890 kJ will be released as heat.
See Exercises 6.45 through 6.48
Calorimetry
Table 6.1 | The Specific Heat
Capacities of Some
Common Substances
Substance
Specific Heat
Capacity (J/8C ? g)
H2O(l)
H2O(s)
Al(s)
Fe(s)
Hg(l)
C(s)
4.18
2.03
0.89
0.45
0.14
0.71
Specific heat capacity: the energy
required to raise the temperature of
one gram of a substance by one degree
Celsius.
Molar heat capacity: the energy required
to raise the temperature of one mole of a
substance by one degree Celsius.
The device used experimentally to determine the heat associated with a chemical reaction is called a calorimeter. Calorimetry, the science of measuring heat, is based on
observing the temperature change when a body absorbs or discharges energy as heat.
Substances respond differently to being heated. One substance might require a great
deal of heat energy to raise its temperature by one degree, whereas another will exhibit
the same temperature change after absorbing relatively little heat. The heat capacity
C of a substance, which is a measure of this property, is defined as
C5
heat absorbed
increase in temperature
When an element or a compound is heated, the energy required will depend on the
amount of the substance present (for example, it takes twice as much energy to raise
the temperature of 2 g of water by one degree than it takes to raise the temperature of
1 g of water by one degree). Thus, in defining the heat capacity of a substance, the
amount of substance must be specified. If the heat capacity is given per gram of substance, it is called the specific heat capacity, and its units are J/8C ? g or J/K ? g. If the
heat capacity is given per mole of the substance, it is called the molar heat capacity,
and it has the units J/8C ? mol or J/K ? mol. The specific heat capacities of some common substances are given in Table 6.1. Note from this table that the heat capacities of
metals are very different from that of water. It takes much less energy to change the
temperature of a gram of a metal by 18C than for a gram of water.
Although the calorimeters used for highly accurate work are precision instruments,
a very simple calorimeter can be used to examine the fundamentals of calorimetry. All
we need are two nested Styrofoam cups with a cover through which a stirrer and thermometer can be inserted (Fig. 6.5). This device is called a “coffee-cup calorimeter.”
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6.2
Enthalpy and Calorimetry
255
The outer cup is used to provide extra insulation. The inner cup holds the solution in
which the reaction occurs.
The measurement of heat using a simple calorimeter such as that shown in Fig. 6.5
is an example of constant-pressure calorimetry, since the pressure (atmospheric
pressure) remains constant during the process. Constant-pressure calorimetry is used
in ­determining the changes in enthalpy (heats of reactions) for reactions occurring in
solution. Recall that under these conditions, the change in enthalpy equals the heat;
that is, DH 5 qP .
For example, suppose we mix 50.0 mL of 1.0 M HCl at 25.08C with 50.0 mL of 1.0 M
NaOH also at 258C in a calorimeter. After the reactants are mixed by stirring, the temperature is observed to increase to 31.98C. As we saw in Section 4.8, the net ionic
equation for this reaction is
If two reactants at the same temperature
are mixed and the resulting solution gets
warmer, this means the reaction taking
place is exothermic. An endothermic
reaction cools the solution.
Thermometer
Styrofoam
cover
Styrofoam
cups
Stirrer
H 1 1aq2 1 OH 2 1aq2 h H2O 1l2
When these reactants (each originally at the same temperature) are mixed, the temperature of the mixed solution is observed to increase. Therefore, the chemical reaction must be releasing energy as heat. This released energy increases the random motions of the solution components, which in turn increases the temperature. The quantity
of energy released can be determined from the temperature increase, the mass of solution, and the specific heat capacity of the solution. For an approximate result, we will
assume that the calorimeter does not absorb or leak any heat and that the solution can
be treated as if it were pure water with a density of 1.0 g/mL.
We also need to know the heat required to raise the temperature of a given amount
of water by 18C. Table 6.1 lists the specific heat capacity of water as 4.18 J/8C ? g. This
means that 4.18 J of energy is required to raise the temperature of 1 g of water by 18C.
From these assumptions and definitions, we can calculate the heat (change in enthalpy) for the neutralization reaction:
Energy 1as heat2 released by the reaction
5 energy 1as heat2 absorbed by the solution
5 specific heat capacity 3 mass of solution 3 increase in temperature
5 s 3 m 3 DT
In this case the increase in temperature (DT ) 5 31.98C 2 25.08C 5 6.98C, and the
mass of solution (m) 5 100.0 mL 3 1.0 g/mL 5 1.0 3 102 g. Thus
Energy released 1as heat2 5 s 3 m 3 DT
J
5 a4.18
b 11.0 3 102 g2 16.9°C2
°C # g
5 2.9 3 103 J
Figure 6.5 | A coffee-cup calorimeter made of two Styrofoam cups.
How much energy would have been released if twice these amounts of solutions
had been mixed? The answer is that twice as much heat would have been produced.
The heat of a reaction is an extensive property; it depends directly on the amount of
substance, in this case on the amounts of reactants. In contrast, an intensive property is
not related to the amount of a substance. For example, temperature is an intensive
property.
Enthalpies of reaction are often expressed in terms of moles of reacting substances.
The number of moles of H1 ions consumed in the preceding experiment is
50.0 mL 3
1L
1.0 mol H 1
3
5 5.0 3 1022 mol H 1
1000 mL
L
Thus 2.9 3 103 J heat was released when 5.0 3 1022 mole of H1 ions reacted, or
2.9 3 103 J
5 5.8 3 104 J /mol
5.0 3 1022 mol H 1
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256
Chapter 6
Thermochemistry
Chemical connections
Nature Has Hot Plants
The voodoo lily is a beautiful,
seductive—and foul-smelling—plant.
The exotic-looking lily features an
elaborate reproductive mechanism—
a purple spike that can reach nearly
3 feet in length and is cloaked by a
hoodlike leaf. But approach to the plant
reveals bad news—it smells terrible!
Despite its antisocial odor, this
putrid plant has fascinated biologists
for many years because of its ability to
generate heat. At the peak of its
Notice that in this example we mentally
keep track of the direction of the energy
flow and assign the correct sign at the end
of the calculation.
metabolic activity, the plant’s blossom
can be as much as 158C above its
ambient temperature. To generate this
much heat, the metabolic rate of the
plant must be close to that of a flying
hummingbird!
What’s the purpose of this intense
heat production? For a plant faced with
limited food supplies in the very
competitive tropical climate where it
grows, heat production seems like a
great waste of energy. The answer to
this mystery is that the voodoo lily is
pollinated mainly by carrion-loving
insects. Thus the lily prepares a
malodorous mixture of chemicals
characteristic of rotting meat, which it
then “cooks” off into the surrounding
air to attract flesh-feeding beetles and
flies. Then, once the insects enter the
pollination chamber, the high temperatures there (as high as 1108F) cause the
insects to remain very active to better
carry out their pollination duties.
of heat released per 1.0 mole of H1 ions neutralized. Thus the magnitude of the
enthalpy change per mole for the reaction
H 1 1aq2 1 OH 2 1aq2 h H2O 1l2
is 58 kJ/mol. Since heat is evolved, DH 5 258 kJ/mol.
To summarize, we have found that when an object changes temperature, the heat
can be calculated from the equation
q 5 s 3 m 3 DT
where s is the specific heat capacity, m is the mass, and DT is the change in temperature. Note that when DT is positive, q also has a positive sign. The object has absorbed
heat, so its temperature increases.
Interactive
Example 6.5
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in OWL.
Constant-Pressure Calorimetry
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.08C is mixed with 1.00 L of 1.00 M
Na2SO4 solution at 25.08C in a calorimeter, the white solid BaSO4 forms, and the temperature of the mixture increases to 28.18C. Assuming that the calorimeter absorbs
only a negligible quantity of heat, the specific heat capacity of the solution is 4.18 J/8C
? g, and the density of the final solution is 1.0 g/mL, calculate the enthalpy change per
mole of BaSO4 formed.
Solution
Experiment 20: Calorimetry
Where are we going?
To calculate DH per mole of BaSO4 formed
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6.2
For example, very precise calorimeters
have been designed that can be used
to study the heat produced, and thus
the metabolic activities, of clumps of
cells no larger than a bread crumb.
Several scientists have suggested that
a single calorimetric measurement
taking just a few minutes on a tiny
plant might be useful in predicting the
growth rate of the mature plant
throughout its lifetime. If true, this
would provide a very efficient method
for selecting the plants most likely to
thrive as adults.
Because the study of the heat
production by plants is an excellent
way to learn about plant metabolism,
this continues to be a “hot” area of
research.
257
Neil Lucas/Nature Picture Library
The voodoo lily is only one of many
such thermogenic (heat-producing)
plants. Another interesting example is
the eastern skunk cabbage, which
produces enough heat to bloom inside
of a snow bank by creating its own ice
caves. These plants are of special
interest to biologists because they
provide opportunities to study
metabolic reactions that are quite
subtle in “normal” plants. For example,
recent studies have shown that
salicylic acid, the active form of
aspirin, is probably very important in
producing the metabolic bursts in
thermogenic plants.
Besides studying the dramatic heat
effects in thermogenic plants,
biologists are also interested in
calorimetric studies of regular plants.
Enthalpy and Calorimetry
The voodoo lily attracts pollinating insects
with its foul odor.
What do we know?
❯ 1.00 L of 1.00 M Ba(NO3)2
❯ 1.00 L of 1.00 M Na2SO4
❯ Tinitial 5 25.08C
❯ Tfinal 5 28.18C
❯ Heat capacity of solution 5 4.18 J/8C ? g
❯ Density of final solution 5 1.0 g/mL
What do we need?
❯ Net ionic equation for the reaction
The ions present before any reaction occurs are Ba21, NO32, Na1, and SO422. The Na1
and NO32 ions are spectator ions, since NaNO3 is very soluble in water and will not
precipitate under these conditions. The net ionic equation for the reaction is therefore
Ba21 1aq2 1 SO422 1aq2 h BaSO4 1s2
How do we get there?
What is DH?
Since the temperature increases, formation of solid BaSO4 must be exothermic; DH
is negative.
Heat evolved by the reaction
5 heat absorbed by the solution
5 specific heat capacity 3 mass of solution 3 increase in temperature
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258
Chapter 6
Thermochemistry
What is the mass of the final solution?
Mass of solution 5 2.00 L 3
1000 mL
1.0 g
3
5 2.0 3 103 g
1L
mL
What is the temperature increase?
DT 5 Tfinal 2 Tinitial 5 28.1°C 2 25.0°C 5 3.1°C
How much heat is evolved by the reaction?
Thus
Heat evolved 5 14.18 J /°C # g2 12.0 3 103 g2 13.1°C2 5 2.6 3 104 J
q 5 qP 5 DH 5 22.6 3 104 J
What is DH per mole of BaSO4 formed?
Since 1.0 L of 1.0 M Ba(NO3)2 contains 1 mole of Ba21 ions and 1.0 L of 1.0 M
Na2SO4 ­contains 1.0 mole of SO422 ions, 1.0 mole of solid BaSO4 is formed in this
experiment. Thus the enthalpy change per mole of BaSO4 formed is
❯
DH 5 22.6 3 104 J /mol 5 226 kJ /mol
See Exercises 6.61 through 6.64
Constant-volume calorimetry experiments can also be performed. For example,
when a photographic flashbulb flashes, the bulb becomes very hot, because the reaction of the zirconium or magnesium wire with the oxygen inside the bulb is exothermic. The reaction occurs inside the flashbulb, which is rigid and does not change volume. Under these conditions, no work is done (because the volume must change for
pressure–volume work to be performed). To study the energy changes in reactions
under conditions of constant volume, a “bomb calorimeter” (Fig. 6.6) is used. Weighed
reactants are placed inside a rigid steel container (the “bomb”) and ignited. The energy
change is determined by measuring the increase in the temperature of the water and
Stirrer
Ignition
wires
Thermometer
Charles D. Winters/Photo Researchers, Inc.
Insulating
container
Steel
bomb
Water
Reactants
in sample
cup
Figure 6.6 | A bomb calorimeter. The reaction is carried out inside a rigid steel “bomb” (photo of actual
disassembled “bomb’’ shown on right), and the heat evolved is absorbed by the surrounding water and other
calorimeter parts. The quantity of energy produced by the reaction can be calculated from the temperature
increase.
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6.2
Enthalpy and Calorimetry
259
other calorimeter parts. For a constant-volume process, the change in volume DV is
equal to zero, so work (which is 2PDV ) is also equal to zero. Therefore,
DE 5 q 1 w 5 q 5 qV (constant volume)
Suppose we wish to measure the energy of combustion of octane (C8H18), a component of gasoline. A 0.5269-g sample of octane is placed in a bomb calorimeter known
to have a heat capacity of 11.3 kJ/8C. This means that 11.3 kJ of energy is required to
raise the temperature of the water and other parts of the calorimeter by 18C. The octane
is ­ignited in the presence of excess oxygen, and the temperature increase of the calorimeter is 2.258C. The amount of energy released is calculated as follows:
Energy released by the reaction
5 temperature increase 3 energy required to change the temperature by 1°C
5 DT 3 heat capacity of calorimeter
5 2.25°C 3 11.3 kJ /°C 5 25.4 kJ
This means that 25.4 kJ of energy was released by the combustion of 0.5269 g octane.
The number of moles of octane is
0.5269 g octane 3
1 mol octane
5 4.614 3 1023 mol octane
114.2 g octane
Since 25.4 kJ of energy was released for 4.614 3 1023 mole of octane, the energy released per mole is
25.4 kJ
5 5.50 3 103 kJ /mol
4.614 3 1023 mol
Since the reaction is exothermic, DE is negative:
DEcombustion 5 25.50 3 103 kJ /mol
Note that since no work is done in this case, DE is equal to the heat.
DE 5 q 1 w 5 q
since w 5 0
Thus q 5 25.50 3 103 kJ/mol.
Example 6.6
Hydrogen’s potential as a fuel is ­discussed
in Section 6.6.
Constant-Volume Calorimetry
It has been suggested that hydrogen gas obtained by the decomposition of water might be
a substitute for natural gas (principally methane). To compare the energies of combustion
of these fuels, the following experiment was carried out using a bomb calorimeter with a
heat capacity of 11.3 kJ/8C. When a 1.50-g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.38C. When a 1.15-g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.38C.
Compare the energies of combustion (per gram) for hydrogen and methane.
Solution
Where are we going?
To calculate DH of combustion per gram for H2 and CH4
What do we know?
❯
❯
❯
1.50 g CH4 1 DT 5 7.3°C
1.15 g H2 1 DT 5 14.3°C
Heat capacity of calorimeter 5 11.3 kJ /°C
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260
Chapter 6
Thermochemistry
What do we need?
❯ DE 5 DT 3 heat capacity of calorimeter
How do we get there?
What is the energy released for each combustion?
For CH4, we calculate the energy of combustion for methane using the heat capacity of
the calorimeter (11.3 kJ/8C) and the observed temperature increase of 7.38C:
Energy released in the combustion of 1.5 g CH4 5 111.3 kJ /°C2 17.3°C2
5 83 kJ
83 kJ
Energy released in the combustion of 1 g CH4 5
5 55 kJ /g
1.5 g
For H2,
The direction of energy flow is indicated
by words in this example. Using signs to
designate the direction of energy flow:
DEcombustion 5 255 kJ/g
for methane and
DEcombustion 5 2141 kJ/g
Energy released in the combustion of 1.15 g H2 5 111.3 kJ /°C2 114.3°C2
5 162 kJ
162 kJ
Energy released in the combustion of 1 g H2 5
5 141 kJ /g
1.15 g
How do the energies of combustion compare?
❯ The energy released in the combustion of 1 g hydrogen is approximately 2.5 times
that for 1 g methane, indicating that hydrogen gas is a potentially useful fuel.
See Exercises 6.67 and 6.68
for hydrogen.
IBLG: See questions from
“Hess’s Law and Standard
Enthalpies of ­Formation”
6.3 Hess’s Law
DH is not dependent on the reaction
pathway.
Since enthalpy is a state function, the change in enthalpy in going from some initial
state to some final state is independent of the pathway. This means that in going from
a particular set of reactants to a particular set of products, the change in enthalpy is
the same whether the reaction takes place in one step or in a series of steps. This principle is known as Hess’s law and can be illustrated by examining the oxidation of nitrogen to produce nitrogen dioxide. The overall reaction can be written in one step,
where the enthalpy change is represented by DH1.
N2 1g2 1 2O2 1g2 h 2NO2 1g2
DH1 5 68 kJ
This reaction also can be carried out in two distinct steps, with enthalpy changes designated by DH2 and DH3:
PowerLecture: Hess’s Law
N2 1g2 1 O2 1g2 h 2NO 1g2
2NO 1g2 1 O2 1g2 h 2NO2 1g2
Net reaction: N2 1g2 1 2O2 1g2 h 2NO2 1g2
DH2 5 180 kJ
DH3 5 2112 kJ
DH2 1 DH3 5 68 kJ
Note that the sum of the two steps gives the net, or overall, reaction and that
DH1 5 DH2 1 DH3 5 68 kJ
The principle of Hess’s law is shown schematically in Fig. 6.7.
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6.3
Figure 6.7 | The principle of Hess’s
Hess’s Law
261
Two-step reaction
law. The same change in enthalpy
occurs when nitrogen and oxygen
react to form nitrogen dioxide,
regardless of whether the reaction
occurs in one (red) or two (blue) steps.
O2(g), 2NO(g)
O2(g), 2NO(g)
H (kJ)
ΔH3 = −112 kJ
ΔH2 = 180 kJ
2NO2(g)
ΔH
N2(g), 2O2(g)
2NO2(g)
ΔH1 = 68 kJ
= ΔH2 + ΔH3 = 180 kJ − 112 kJ
N2(g), 2O2(g)
One-step reaction
Characteristics of Enthalpy Changes
To use Hess’s law to compute enthalpy changes for reactions, it is important to understand two characteristics of DH for a reaction:
Reversing the direction of a reaction
changes the sign of DH.
Characteristics of DH for a Reaction
❯ If a reaction is reversed, the sign of DH is also reversed.
❯ The magnitude of D H is directly proportional to the quantities of reactants and
products in a reaction. If the coefficients in a balanced reaction are multiplied by an
integer, the value of DH is multiplied by the same integer.
Both these rules follow in a straightforward way from the properties of enthalpy
changes. The first rule can be explained by recalling that the sign of DH indicates the
direction of the heat flow at constant pressure. If the direction of the reaction is reversed, the direction of the heat flow also will be reversed. To see this, consider the
preparation of xenon tetrafluoride, which was the first binary compound made from a
noble gas:
Courtesy, Argonne National Laboratory
Xe 1g2 1 2F2 1g2 h XeF4 1s2
Crystals of xenon tetrafluoride, the first
reported binary compound containing a
noble gas element.
DH 5 2251 kJ
This reaction is exothermic, and 251 kJ of energy flows into the surroundings as heat.
On the other hand, if the colorless XeF4 crystals are decomposed into the elements,
according to the equation
XeF4 1s2 h Xe 1g2 1 2F2 1g2
the opposite energy flow occurs because 251 kJ of energy must be added to the system
to produce this endothermic reaction. Thus, for this reaction, DH 5 1251 kJ.
The second rule comes from the fact that DH is an extensive property, depending
on the amount of substances reacting. For example, since 251 kJ of energy is evolved
for the reaction
Xe 1g2 1 2F2 1g2 h XeF4 1s2
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262
Chapter 6
Thermochemistry
then for a preparation involving twice the quantities of reactants and products, or
2Xe 1g2 1 4F2 1g2 h 2XeF4 1s2
twice as much heat would be evolved:
DH 5 2 12251 kJ2 5 2502 kJ
Critical Thinking
What if Hess’s law were not true? What are some possible repercussions this would
have?
Interactive
Example 6.7
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in OWL.
Hess’s Law I
Two forms of carbon are graphite, the soft, black, slippery material used in “lead”
pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using
the ­enthalpies of combustion for graphite (2394 kJ/mol) and diamond (2396 kJ/mol),
calculate DH for the conversion of graphite to diamond:
Cgraphite 1s2 h Cdiamond 1s2
Solution
Where are we going?
To calculate DH for the conversion of graphite to diamond
What do we know?
The combustion reactions are
❯
❯
Cgraphite 1s2 1 O2 1g2 h CO2 1g2
Cdiamond 1s2 1 O2 1g2 h CO2 1g2
DH 5 2394 kJ
DH 5 2396 kJ
How do we get there?
How do we combine the combustion equations to obtain the equation for the
conversion of graphite to diamond?
Note that if we reverse the second reaction (which means we must change the sign of
DH) and sum the two reactions, we obtain the desired reaction:
Cgraphite 1s2 1 O2 1g2 h CO2 1g2
CO2 1g2 h Cdiamond 1s2 1 O2 1g2
❯
Cgraphite 1s2 h Cdiamond 1s2
DH 5 2394 kJ
DH 5 2 12396 kJ2
DH 5 2 kJ
The DH for the conversion of graphite to diamond is
DH 5 2 kJ /mol graphite
We obtain this value by summing the DH values for the equations as shown above.
This process is endothermic since the sign of DH is positive.
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Hess’s Law
Graphite
263
Comstock/Jupiter Images
Rich Treptow/Visuals Unlimited
6.3
Diamond
See Exercises 6.69 and 6.70
Interactive
Example 6.8
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in OWL.
Hess’s Law II
Diborane (B2H6) is a highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program. Calculate DH for the synthesis of diborane from its elements, according to the equation
using the following data:
2B 1s2 1 3H2 1g2 h B2H6 1g2
Reaction
(a)
(b)
(c)
(d)
DH
3
2 O2 1g2
2B 1s2 1 1
h B2O3 1s2
B2H6 1g2 1 3O2 1g2 h B2O3 1s2 1 3H2O 1g2
H2 1g2 1 1 12 O2 1g2 h H2O 1l2
H2O 1l2 h H2O 1g2
21273 kJ
22035 kJ
2286 kJ
44 kJ
Solution
To obtain DH for the required reaction, we must somehow combine equations (a), (b),
(c), and (d) to produce that reaction and add the corresponding DH values. This can
best be done by focusing on the reactants and products of the required reaction. The
reactants are B(s) and H2(g), and the product is B2H6(g). How can we obtain the correct equation? Reaction (a) has B(s) as a reactant, as needed in the required equation.
Thus reaction (a) will be used as it is. Reaction (b) has B2H6(g) as a reactant, but this
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264
Chapter 6
Thermochemistry
substance is needed as a product. Thus reaction (b) must be reversed, and the sign of
DH must be changed accordingly. Up to this point we have
2B 1s2 1 32O2 1g2 h B2O3 1s2
B2O3 1s2 1 3H2O 1g2 h B2H6 1g2 1 3O2 1g2
Sum: B2O3 1s2 1 2B 1s2 1 32O2 1g2 1 3H2O 1g2 h B2O3 1s2 1 B2H6 1g2 1 3O2 1g2
1a2
2 1b2
DH 5 21273 kJ
DH 5 2 122035 kJ2
DH 5 762 kJ
Deleting the species that occur on both sides gives
2B 1s2 1 3H2O 1g2 h B2H6 1g2 1 32O2 1g2
DH 5 762 kJ
We are closer to the required reaction, but we still need to remove H2O(g) and O2(g)
and introduce H2(g) as a reactant. We can do this using reactions (c) and (d). If we
multiply reaction (c) and its DH value by 3 and add the result to the preceding equation, we have
3 3 1c2
2B 1s2 1 3H2O 1g2 h B2H6 1g2 1 32O2 1g2
3 3 H2 1g2 1 12O2 1g2 h H2O 1l2 4
Sum: 2B 1s2 1 3H2 1g2 1 32O2 1g2 1 3H2O 1g2 h B2H6 1g2 1 32O2 1g2 1 3H2O 1l2
DH 5 762 kJ
DH 5 3 12286 kJ2
DH 5 296 kJ
3
2 O2(g)
We can cancel the
on both sides, but we cannot cancel the H2O because it is
gaseous on one side and liquid on the other. This can be solved by adding reaction (d),
multiplied by 3:
2B 1s2 1 3H2 1g2 1 3H2O 1g2 h B2H6 1g2 1 3H2O 1l2
3 3 1d2
3 3 H2O 1l 2 h H2O 1g2 4
2B 1s2 1 3H2 1g2 1 3H2O 1g2 1 3H2O 1l 2 h B2H6 1g2 1 3H2O 1l2 1 3H2O 1g2
DH 5 296 kJ
DH 5 3 144 kJ2
DH 5 136 kJ
This gives the reaction required by the problem:
2B 1s2 1 3H2 1g2 h B2H6 1g2
DH 5 136 kJ
Thus DH for the synthesis of 1 mole of diborane from the elements is 136 kJ.
See Exercises 6.71 through 6.76
Problem-Solving Strategy
Hess’s Law
Calculations involving Hess’s law typically require that several reactions be
manipulated and combined to finally give the reaction of interest. In doing this
procedure you should
❯ Work backward from the required reaction, using the reactants and products
to decide how to manipulate the other given reactions at your disposal
❯ Reverse any reactions as needed to give the required reactants and products
❯ Multiply reactions to give the correct numbers of reactants and products
This process involves some trial and error, but it can be very systematic if you
always allow the final reaction to guide you.
6.4 Standard Enthalpies of Formation
For a reaction studied under conditions of constant pressure, we can obtain the enthalpy change using a calorimeter. However, this process can be very difficult. In fact,
in some cases it is impossible, since certain reactions do not lend themselves to such
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6.4
Standard Enthalpies of Formation
265
study. An example is the conversion of solid carbon from its graphite form to its diamond form:
Cgraphite 1s2 h Cdiamond 1s2
The value of DH for this process cannot be obtained by direct measurement in a calorimeter because the process is much too slow under normal conditions. However, as
we saw in Example 6.7, DH for this process can be calculated from heats of combustion. This is only one example of how useful it is to be able to calculate DH values for
chemical ­reactions. We will next show how to do this using standard enthalpies of
formation.
The standard enthalpy of formation (DHf8) of a compound is defined as the
change in enthalpy that accompanies the formation of 1 mole of a compound from its
elements with all substances in their standard states.
A degree symbol on a thermodynamic function, for example, DH8, indicates that the
corresponding process has been carried out under standard conditions. The standard
state for a substance is a precisely defined reference state. Because thermodynamic
functions often depend on the concentrations (or pressures) of the substances involved,
we must use a common reference state to properly compare the thermodynamic properties of two substances. This is especially important because, for most thermodynamic properties, we can measure only changes in the property. For example, we have
no method for determining absolute values of enthalpy. We can measure enthalpy
changes (DH values) only by performing heat-flow experiments.
Conventional Definitions of Standard States
For a Compound
The International Union of Pure and
­Applied Chemists (IUPAC) has adopted ­
1 bar (100,000 Pa) as the standard
­pressure instead of 1 atm (101,305 Pa).
Both standards are now in wide use.
❯ The standard state of a gaseous substance is a pressure of exactly 1 atmosphere.
❯ For a pure substance in a condensed state (liquid or solid), the standard state is the
pure liquid or solid.
❯ For a substance present in a solution, the standard state is a concentration of
exactly 1 M.
For an Element
Standard state is not the same as the
standard temperature and pressure (STP)
for a gas (discussed in Section 5.4).
❯ The standard state of an element is the form in which the element exists under
conditions of 1 atmosphere and 258C. (The standard state for oxygen is O2(g) at a
pressure of 1 atmosphere; the standard state for sodium is Na(s); the standard state
for mercury is Hg(l); and so on.)
Photo © Cengage Learning. All rights reserved.
Several important characteristics of the definition of the enthalpy of formation will
become clearer if we again consider the formation of nitrogen dioxide from the elements in their standard states:
Brown nitrogen dioxide gas.
1
2 N2 1g2
1 O2 1g2 h NO2 1g2
DH°f 5 34 kJ /mol
Note that the reaction is written so that both elements are in their standard states, and
1 mole of product is formed. Enthalpies of formation are always given per mole of
product with the product in its standard state.
The formation reaction for methanol is written as
C 1s2 1 2H2 1g2 1 12O2 1g2 h CH3OH 1l2
DH°f 5 2239 kJ /mol
The standard state of carbon is graphite, the standard states for oxygen and hydrogen
are the diatomic gases, and the standard state for methanol is the liquid.
The DH f8 values for some common substances are shown in Table 6.2. More values
are found in Appendix 4. The importance of the tabulated DH f8 values is that enthalpies
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266
Chapter 6
Thermochemistry
Table 6.2 | Standard Enthalpies of
Formation for Several
Compounds at 258C
Compound
DH8f (kJ/mol)
NH3(g)
NO2(g)
H2O(l)
Al2O3(s)
Fe2O3(s)
CO2(g)
CH3OH(l)
C8H18(l)
246
34
2286
21676
2826
2394
2239
2269
for many reactions can be calculated using these numbers. To see how this is done, we
will calculate the standard enthalpy change for the combustion of methane:
CH4 1g2 1 2O2 1g2 h CO2 1g2 1 2H2O 1l2
Enthalpy is a state function, so we can invoke Hess’s law and choose any convenient
pathway from reactants to products and then sum the enthalpy changes along the chosen pathway. A convenient pathway, shown in Fig. 6.8, involves taking the reactants
apart to the respective elements in their standard states in reactions (a) and (b) and then
forming the products from these elements in reactions (c) and (d). This general pathway will work for any reaction, since atoms are conserved in a chemical reaction.
Note from Fig. 6.8 that reaction (a), where methane is taken apart into its
elements,
CH4 1g2 h C 1s2 1 2H2 1g2
is just the reverse of the formation reaction for methane:
C 1s2 1 2H2 1g2 h CH4 1g2
DH°f 5 275 kJ /mol
C 1s2 1 O2 1g2 h CO2 1g2
DH°f 5 2394 kJ /mol
Since reversing a reaction means changing the sign of DH but keeping the magnitude
the same, DH for reaction (a) is 2DH f8, or 75 kJ. Thus DH8(a) 5 75 kJ.
Next we consider reaction (b). Here oxygen is already an element in its standard
state, so no change is needed. Thus DH8(b) 5 0.
The next steps, reactions (c) and (d), use the elements formed in reactions (a) and
(b) to form the products. Note that reaction (c) is simply the formation reaction for
carbon dioxide:
and
DH°1c2 5 DH°f for CO2 1g2 5 2394 kJ
Reaction (d) is the formation reaction for water:
H2 1g2 1 12 O2 1g2 h H2O 1l2
DH°f 5 2286 kJ /mol
However, since 2 moles of water are required in the balanced equation, we must form
2 moles of water from the elements:
2H2 1g2 1 O2 1g2 h 2H2O 1l2
Thus
DH°1d2 5 2 3 DH°f for H2O 1l2 5 2 12286 kJ2 5 2572 kJ
Reactants
Elements
Products
C(s)
CH4(g)
(c)
(a)
CO2(g)
2H2(g)
Figure 6.8 | In this pathway for the
combustion of methane, the reactants
are first taken apart in reactions
(a) and (b) to form the constituent
elements in their standard states,
which are then used to assemble the
products in reactions (c) and (d).
(d)
2O2(g)
(b)
2H2O(l)
2O2(g)
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6.4
Figure 6.9 | A schematic diagram of
Reactants
the energy changes for the reaction
CH4(g) 1 2O2(g) n CO2(g) 1 2H2O(l).
Standard Enthalpies of Formation
Elements
267
Products
Step 1
(a)
Step 2
(c)
ΔHa = 75 kJ
ΔHc = −394 kJ
(b)
(d)
ΔHb = 0 kJ
ΔHd = −572 kJ
We have now completed the pathway from the reactants to the products. The change
in enthalpy for the reaction is the sum of the DH values (including their signs) for
the steps:
DH°reaction
5 DH°1a2 1 DH°1b2 1 DH°1c2 1 DH°1d2
5 3 2DH °f for CH4 1g2 4 1 0 1 3 DH °f for CO2 1g2 4 1 3 2 3 DH °f for H2O 1l2 4
5 2 1275 kJ2 1 0 1 12394 kJ2 1 12572 kJ2
5 2891 kJ
Subtraction means to reverse the sign
and add.
This process is diagramed in Fig. 6.9. Notice that the reactants are taken apart and
converted to elements [not necessary for O2(g)] that are then used to form products.
You can see that this is a very exothermic reaction because very little energy is required to convert the reactants to the respective elements but a great deal of energy is
released when these elements form the products. This is why this reaction is so useful
for producing heat to warm homes and offices.
Let’s examine carefully the pathway we used in this example. First, the reactants
were broken down into the elements in their standard states. This process involved
reversing the formation reactions and thus switching the signs of the enthalpies of
formation. The products were then constructed from these elements. This involved
formation reactions and thus enthalpies of formation. We can summarize this entire
process as follows: The enthalpy change for a given reaction can be calculated by
subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products. Remember to multiply the enthalpies of formation by integers as
required by the balanced equation. This statement can be represented symbolically as
follows:
DH°reaction 5 SnpDH°f 1products2 2 SnrDH°f 1reactants2 Elements in their standard states are not
included in enthalpy calculations using
DHf8 values.
(6.1)
where the symbol S (sigma) means “to take the sum of the terms,” and np and nr represent the moles of each product or reactant, respectively.
Elements are not included in the calculation because elements require no change in
form. We have in effect defined the enthalpy of formation of an element in its standard
state as zero, since we have chosen this as our reference point for calculating enthalpy
changes in reactions.
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268
Chapter 6
Thermochemistry
Problem-Solving Strategy
Enthalpy Calculations
❯
❯
PowerLecture: Reduction of Iron:
Thermite ­Reaction
❯
❯
Interactive
Example 6.9
Sign in at http://login.cengagebrain
.com to try this Interactive Example
in OWL.
When a reaction is reversed, the magnitude of DH remains the same, but its
sign changes.
When the balanced equation for a reaction is multiplied by an integer, the
value of DH for that reaction must be multiplied by the same integer.
The change in enthalpy for a given reaction can be calculated from the
enthalpies of formation of the reactants and products:
DH°reaction 5 SnpDH°f 1products2 2 SnrDH°f 1reactants2
Elements in their standard states are not included in the DHreaction calculations. That is, DH f8 for an element in its standard state is zero.
Enthalpies from Standard Enthalpies of Formation I
Using the standard enthalpies of formation listed in Table 6.2, calculate the standard
enthalpy change for the overall reaction that occurs when ammonia is burned in air to
form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid.
4NH3 1g2 1 7O2 1g2 h 4NO2 1g2 1 6H2O 1l2
Solution
We will use the pathway in which the reactants are broken down into elements in their
standard states, which are then used to form the products (Fig. 6.10).
❯ Decomposition of NH3(g) into elements [reaction (a) in Fig. 6.10]. The first
step is to decompose 4 moles of NH3 into N2 and H2:
4NH3 1g2 h 2N2 1g2 1 6H2 1g2
The preceding reaction is 4 times the reverse of the formation reaction for NH3:
Thus
1
2 N2 1g2
1 32H2 1g2 h NH3 1g2
DH°f 5 246 kJ /mol
DH°1a2 5 4 mol 3 2 1246 kJ /mol2 4 5 184 kJ
Reactants
Elements
Products
2N2(g)
4NH3(g)
(c)
(a)
4NO2(g)
6H2(g)
(d)
7O2(g)
(b)
6H2O(l)
7O2(g)
Figure 6.10 | A pathway for the
combustion of ammonia.
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6.4
❯
❯
Standard Enthalpies of Formation
269
Elemental oxygen [reaction (b) in Fig. 6.10]. Since O2(g) is an element in its
standard state, DH8(b) 5 0.
We now have the elements N2(g), H2(g), and O2(g), which can be combined
to form the products of the overall reaction.
Synthesis of NO2(g) from elements [reaction (c) in Fig. 6.10]. The overall
reaction equation
has 4 moles of NO2. Thus the required reaction is 4 times the
­
formation reaction for NO2:
4 3 3 12N2 1g2 1 O2 1g2 h NO2 1g2 4
and
DH°1c2 5 4 3 DH°f for NO2 1g2
From Table 6.2, DH f8 for NO2(g) 5 34 kJ/mol and
❯
DH°1c2 5 4 mol 3 34 kJ /mol 5 136 kJ
Synthesis of H2O(l) from elements [reaction (d) in Fig. 6.10]. Since the overall
equation for the reaction has 6 moles of H2O(l), the required reaction is 6 times
the formation reaction for H2O(l):
6 3 3 H2 1g2 1 12O2 1g2 h H2O 1l 2 4
and
DH°1d2 5 6 3 DH°f for H2O 1l 2
From Table 6.2, DH f8 for H2O(l) 5 2286 kJ/mol and
DH°1d2 5 6 mol 12286 kJ /mol2 5 21716 kJ
To summarize, we have done the following:
7O2 1g2





4NH3 1g2
DH8(a)
DH8(c)
 2N2 1g2 1 6H2 1g2 8888n
8888n
4NO2 1g2

DH8(b) 
DH8(d)
 7O2 1g2
8888n
8888n
6H2O 1l2

Elements in their
standard states
We add the DH8 values for the steps to get DH8 for the overall reaction:
DH°reaction 5 DH°1a2 1 DH°1b2 1 DH°1c2 1 DH°1d2
5 3 4 3 2DH°f for NH3 1g2 4 1 0 1 3 4 3 DH°f for NO2 1g2 4
1 3 6 3 DH°f for H2O 1l 2 4
5 3 4 3 DH°f for NO2 1g2 4 1 3 6 3 DH°f for H2O 1l2 4
2 3 4 3 DH°f for NH3 1g2 4
5 SnpDH °f 1products2 2 SnrDH °f 1reactants2
Remember that elemental reactants and products do not need to be included, since
DH f8 for an element in its standard state is zero. Note that we have again obtained
Equation (6.1). The final solution is
DH°reaction 5 3 4 3 134 kJ2 4 1 3 6 3 12286 kJ2 4 2 3 4 3 1246 kJ2 4
5 21396 kJ
See Exercises 6.79 and 6.80
Now that we have shown the basis for Equation (6.1), we will make direct use of it
to calculate DH for reactions in succeeding exercises.
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270
Chapter 6
Thermochemistry
Interactive
Example 6.10
Enthalpies from Standard Enthalpies of Formation II
Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction:
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.com to try this Interactive Example
in OWL.
2Al 1s2 1 Fe2O3 1s2 h Al2O3 1s2 1 2Fe 1s2
This reaction occurs when a mixture of powdered aluminum and iron(III) oxide is ignited with a magnesium fuse.
Solution
Where are we going?
To calculate DH for the reaction
What do we know?
❯
❯
Photo © Cengage Learning. All rights reserved.
❯
The thermite reaction is one of the most
energetic chemical reactions known.
DH°f for Fe2O3 1s2 5 2826 kJ /mol
DH°f for Al2O3 1s2 5 21676 kJ /mol
DH°f for Al 1s2 5 DH°f for Fe 1s2 5 0
What do we need?
❯ We use Equation (6.1):
DH° 5 SnpDHf° 1products2 2 SnrDHf° 1reactants2
How do we get there?
❯
DH°reaction 5 DH°f for Al2O3 1s2 2 DH°f for Fe2O3 1s2
5 21676 kJ 2 12826 kJ2 5 2850. kJ
This reaction is so highly exothermic that the iron produced is initially molten. This
process is often used as a lecture demonstration and also has been used in welding
massive steel objects such as ships’ propellers.
See Exercises 6.83 and 6.84
Critical Thinking
For DHreaction calculations, we define DH8f for an element in its standard state as zero.
What if we define DH8f for an element in its standard state as 10 kJ/mol? How would
this affect your determination of DHreaction? Provide support for your answer with a
sample calculation.
Example 6.11
Enthalpies from Standard Enthalpies of Formation III
Until recently, methanol (CH3OH) was used as a fuel in high-performance engines in
race cars. Using the data in Table 6.2, compare the standard enthalpy of combustion
per gram of methanol with that per gram of gasoline. Gasoline is actually a mixture of
compounds, but assume for this problem that gasoline is pure liquid octane (C8H18).
Solution
Where are we going?
To compare DH of combustion for methanol and octane
What do we know?
❯ Standard enthalpies of formation from Table 6.2
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6.5
Present Sources of Energy
271
How do we get there?
For methanol:
What is the combustion reaction?
2CH3OH 1l2 1 3O2 1g2 h 2CO2 1g2 1 4H2O 1l2
What is the DH8reaction?
Using the standard enthalpies of formation from Table 6.2 and Equation (6.1), we have
DH°reaction 5 2 3 DH°f for CO2 1g2 1 4 3 DH°f for H2O 1l2
2 2 3 DH°f for CH3OH 1l2
5 2 3 12394 kJ2 1 4 3 12286 kJ2 2 2 3 12239 kJ2
5 21.45 3 103 kJ
hat is the enthalpy of combustion per gram?
W
Thus 1.45 3 103 kJ of heat is evolved when 2 moles of methanol burn. The molar mass
of methanol is 32.04 g/mol. This means that 1.45 3 103 kJ of energy is produced when
64.08 g methanol burns. The enthalpy of combustion per gram of methanol is
21.45 3 103 kJ
5 222.6 kJ /g
64.08 g
For octane:
What is the combustion reaction?
2C8H18 1l2 1 25O2 1g2 h 16CO2 1g2 1 18H2O 1l2
hat is the DH8reaction?
W
Using the standard enthalpies of formation from Table 6.2 and Equation (6.1), we have
DH°reaction 5 16 3 DH°f for CO2 1g2 1 18 3 DH°f for H2O 1l2
2 2 3 DH°f for C8H18 1l 2
5 16 3 12394 kJ2 1 18 3 12286 kJ2 2 2 3 12269 kJ2
5 21.09 3 104 kJ
What is the enthalpy of combustion per gram?
This is the amount of heat evolved when 2 moles of octane burn. Since the molar mass
of octane is 114.22 g/mol, the enthalpy of combustion per gram of octane is
❯
In the cars used in the Indianapolis 500,
ethanol is now used instead of methanol.
21.09 3 104 kJ
5 247.7 kJ /g
2 1114.22 g2
The enthalpy of combustion per gram of octane is approximately twice that per
gram of methanol. On this basis, gasoline appears to be superior to methanol
for use in a racing car, where weight considerations are usually very important.
Why, then, was methanol used in racing cars? The answer is that methanol
burns much more smoothly than gasoline in high-performance engines, and
this advantage more than compensates for its weight disadvantage.
See Exercise 6.91
6.5 Present Sources of Energy
Woody plants, coal, petroleum, and natural gas hold a vast amount of energy that originally came from the sun. By the process of photosynthesis, plants store energy that can
be claimed by burning the plants themselves or the decay products that have been
converted over millions of years to fossil fuels. Although the United States currently
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272
Chapter 6
Thermochemistry
Figure 6.11 | Energy sources used
in the United States.
91%
73%
71%
62%
52%
36%
21%
9%
1850
Wood
23%
18%
5% 3%
1900
Coal
6%
6%
6%
3%
1950
1975
Petroleum/natural gas
11%
4%
2000
Hydro and nuclear
depends heavily on petroleum for energy, this dependency is a relatively recent phenomenon, as shown in Fig. 6.11. In this section we will discuss some sources of energy
and their effects on the environment.
Keith Wood/Stone/Getty Images
Petroleum and Natural Gas
Oil rig in Gulf of Mexico.
Table 6.3 | Names and Formulas
for Some Common
Hydrocarbons
Formula
Name
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
C7H16
C8H18
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Although how they were produced is not completely understood, petroleum and natural gas were most likely formed from the remains of marine organisms that lived approximately 500 million years ago. Petroleum is a thick, dark liquid composed mostly
of compounds called hydrocarbons that contain carbon and hydrogen. (Carbon is
unique among elements in the extent to which it can bond to itself to form chains of
various lengths.) Table 6.3 gives the formulas and names for several common hydrocarbons. Natural gas, usually associated with petroleum deposits, consists mostly of
methane, but it also contains significant amounts of ethane, propane, and butane. Over
the last several years it has become clear that there are tremendous reserves of natural
gas deep in shale deposits. Estimates indicate that there may be as much as 200 trillion
cubic meters of recoverable natural gas in these deposits around the globe. In the eastern United States, the 1-mile-deep Marsellus Shale deposit may contain as much as
two trillion cubic meters of recoverable gas. The problem with these gas deposits is
that the shale is very impermeable, and the gas does not flow out into wells on its own.
A technique called hydraulic fracturing, or “fracking,” is now being used to access
these gas deposits. Fracking involves injecting a slurry of water, sand, and chemical
additives under pressure through a well bore drilled into the shale. This produces fractures in the rock that allow the gas to flow out into wells. With the increased use of
fracking has come environmental concerns, including potential contamination of
ground water, risks to air quality, and possible mishandling of wastes associated with
the process. However, even in view of these potential hazards, it is expected that natural gas obtained from fracking may supply as much as half of the U.S. natural gas
supply by the year 2020.
The composition of petroleum varies somewhat, but it consists mostly of hydrocarbons having chains that contain from 5 to more than 25 carbons. To be used efficiently,
the petroleum must be separated into fractions by boiling. The lighter molecules (having the lowest boiling points) can be boiled off, leaving the heavier ones behind. The
commercial uses of various petroleum fractions are shown in Table 6.4.
The petroleum era began when the demand for lamp oil during the Industrial Revolution outstripped the traditional sources: animal fats and whale oil. In response to this
increased demand, Edwin Drake drilled the first oil well in 1859 at Titusville, Pennsylvania. The petroleum from this well was refined to produce kerosene (fraction C10–C18),
which served as an excellent lamp oil. Gasoline (fraction C5–C10) had limited use and
was ­often discarded. However, this situation soon changed. The development of the
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6.5
Table 6.4 | Uses of the Various
Petroleum Fractions
Petroleum
Fraction in Terms
of Numbers of
Carbon Atoms
C5–C10
C10–C18
C15–C25
.C25
Major Uses
Gasoline
Kerosene
Jet fuel
Diesel fuel
Heating oil
Lubricating oil
Asphalt
Present Sources of Energy
273
electric light ­decreased the need for kerosene, and the advent of the “horseless carriage” with its ­gasoline-powered engine signaled the birth of the gasoline age.
As gasoline became more important, new ways were sought to increase the yield
of gasoline obtained from each barrel of petroleum. William Burton invented a process at Standard Oil of Indiana called pyrolytic (high-temperature) cracking. In this
process, the heavier molecules of the kerosene fraction are heated to about 7008C,
causing them to break (crack) into the smaller molecules of hydrocarbons in the gasoline fraction. As cars became larger, more efficient internal combustion engines were
designed. Because of the uneven burning of the gasoline then available, these engines
“knocked,” producing unwanted noise and even engine damage. Intensive research to
find additives that would promote smoother burning produced tetraethyl lead,
(C2H5)4Pb, a very effective “antiknock” agent.
The addition of tetraethyl lead to gasoline became a common practice, and by 1960,
gasoline contained as much as 3 grams of lead per gallon. As we have discovered so
often in recent years, technological advances can produce environmental problems. To
prevent air pollution from automobile exhaust, catalytic converters have been added to
car exhaust systems. The effectiveness of these converters, however, is destroyed by
lead. The use of leaded gasoline also greatly increased the amount of lead in the environment, where it can be ingested by animals and humans. For these reasons, the use
of lead in gasoline has been phased out, requiring extensive (and expensive) modifications of engines and of the gasoline refining process.
Coal
Coal has variable composition depending
on both its age and location.
Coal was formed from the remains of plants that were buried and subjected to high
pressure and heat over long periods of time. Plant materials have a high content of
cellulose, a complex molecule whose empirical formula is CH2O but whose molar
mass is around 500,000 g/mol. After the plants and trees that flourished on the earth
at various times and places died and were buried, chemical changes gradually lowered the oxygen and hydrogen content of the cellulose molecules. Coal “matures”
through four stages: lignite, subbituminous, bituminous, and anthracite. Each stage has
a higher carbon-to-oxygen and carbon-to-hydrogen ratio; that is, the relative carbon
content gradually increases. Typical elemental compositions of the various coals
are given in Table 6.5. The energy available from the combustion of a given mass of
coal increases as the ­carbon content increases. Therefore, anthracite is the most
valuable coal and lignite the least valuable.
Coal is an important and plentiful fuel in the United States, currently furnishing
approximately 23% of our energy. As the supply of petroleum dwindles, the share of
the e