Chapter 11 Solutions to HW
February 6, 2017 | Author: Sushant Vanmali | Category: N/A
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11
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SOLUTIONS MANUAL
The Mole Section 11.1 Measuring Matter pages 309–312
Section 11.1 Assessment
page 312
5. How is a mole similar to a dozen?
Practice Problems pages 311, 312
The mole is a unit for counting 6.02 1023 representative particles. The dozen is used to count 12 items.
1. Determine the number of atoms in 2.50 mol Zn. 6.02 1023 atoms 2.50 mol Zn 1 mol 1.51 1024 atoms of Zn
2. Given 3.25 mol AgNO3, determine the number
of formula units. 6.02 1023 formula units 3.25 mol AgNO3 1 mol 1.96 1024 formula units of AgNO3
3. Calculate the number of molecules in 11.5 mol
H2O.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
number and one mole? One mole contains Avogadro’s number (6.02 1023) of representative particles.
7. Explain how you can convert from the number
of representative particles of a substance to moles of that substance. Multiply the number of representative particles by 1 mol/6.02 1023 representative particles.
8. Explain why chemists use the mole.
6.02 1023 molecules 11.5 mol H2O 1 mol 6.92 1024 molecules of H2O
4. How many moles contain each of the following? a. 5.75
1024
atoms Al
1 mol 5.75 1024 atoms Al 6.02 1023 atoms 9.55 mol Al
b. 3.75 3.75
6. What is the relationship between Avogadro’s
1024
1024
molecules CO2
molecules CO2
1 mol 6.23 mol CO2 6.02 1023 molecules
c. 3.58 1023 formula units ZnCl2 3.58 1023 formula units ZnCl2 1 mol 0.595 mol ZnCl2 6.02 1023 formula units
d. 2.50 1020 atoms Fe 1 mol 2.50 1020 atoms Fe 6.02 1023 atoms 4.15 104 mol Fe
Chemists use the mole because it is a convenient way of knowing how many representative particle are in a sample.
9. Thinking Critically Arrange the following
from the smallest number of representative particles to the largest number of representative particles: 1.25 1025 atoms Zn; 3.56 mol Fe; 6.78 1022 molecules glucose (C6H12O6). From smallest to largest: 6.78 1022 molecules glucose, 2.14 1024 atoms Fe, 1.25 1025 atoms Zn.
10. Using Numbers Determine the number of
representative particles in each of the following and identify the representative particle: 11.5 mol Ag; 18.0 mol H2O; 0.150 mol NaCl. 6.02 1023 atoms Ag 11.5 mol Ag 1 mol Ag 6.92 1024 atoms Ag 6.02 1023 molecules H2O 18.0 mol H2O 1 mol H2O 1.08 1025 molecules H2O 6.02 1023 formula units NaCl 0.150 mol NaCl 1 mol NaCl 9.03 1022 formula units NaCl
Solutions Manual
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Section 11.2 Mass and the Mole pages 313–319
Practice Problems
SOLUTIONS MANUAL
13. How many atoms are in each of the following
samples? a. 55.2 g Li
pages 316, 318
6.02 1023 atoms 1 mol Li 55.2 g Li 6.94 g Li 1 mol
11. Determine the mass in grams of each of the
4.79 1024 atoms Li
following. a. 3.57 mol Al 26.98 g Al 3.57 mol Al 96.3 g Al 1 mol Al
b. 42.6 mol Si 28.09 g Si 42.6 mol Si 1.20 103 g Si 1 mol Si
c. 3.45 mol Co 58.93 g Co 3.45 mol Co 203 g Co 1 mol Co
d. 2.45 mol Zn 65.38 g Zn 2.45 mol Zn 1.60 102 g Zn 1 mol Zn
12. Determine the number of moles in each of the
following. a. 25.5 g Ag 1 mol Ag 25.5 g Ag 0.236 mol Ag 107.9 g Ag
b. 300.0 g S 1 mol S 300.0 g S 9.355 mol S 32.07 g S
c. 125 g Zn 1 mol Zn 125 g Zn 1.91 mol Zn 65.38 g Zn
d. 1.00 kg Fe 1000 g Fe 1 mol Fe 1.00 kg Fe 55.85 g Fe 1 kg Fe 17.9 mol Fe
b. 0.230 g Pb 1 mol Pb 0.230 g Pb 207.2 g Pb 6.02 1023 atoms 6.68 1020 atoms Pb 1 mol
c. 11.5 g Hg 1 mol Hg 11.5 g Hg 200.6 g Hg 6.02 1023 atoms 3.45 1022 atoms Hg 1 mol
d. 45.6 g Si 1 mol Si 45.6 g Si 28.09 g Si 6.02 1023 atoms 9.77 1023 atoms Si 1 mol
e. 0.120 kg Ti 1000 g Ti 1 mol Ti 0.120 kg Ti 47.87 g Ti 1 kg Ti 6.02 1023 atoms 1.51 1024 atoms Ti 1 mol
14. What is the mass in grams of each of the
following? a. 6.02 1024 atoms Bi 1 mol Bi 6.02 1024 atoms Bi 6.02 1023 atoms 209.0 g Bi 2.09 103 g Bi 1 mol Bi
b. 1.00 1024 atoms Mn 1 mol Mn 1.00 1024 atoms Mn 6.02 1023 atoms 54.94 g Mn 91.3 g Mn 1 mol Mn
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c. 3.40 1022 atoms He 3.40
1022
1 mol He atoms He 6.02 1023 atoms
4.003 g He 0.226 g He 1 mol He
d. 1.50 1015 atoms N
SOLUTIONS MANUAL
19. Sequencing Arrange the following in order of
mass from the smallest mass to the largest: 1.0 mol Ar, 3.0 1024 atoms Ne, 20 g Kr. 39.95 g Ar 1.0 mol Ar 39.95 g Ar 1 mol Ar 1 mol Ne 3.0 1024 atoms Ne 6.02 1023 atoms Ne
1 mol N 1.50 1015 atoms N 6.02 1023 atoms
20.18 g Ne 101 g Ne 1 mol Ne
14.01 g N 3.49 108 g N 1 mol N
20 g Kr, 1.0 mol Ar, 3.0 1024 atoms Ne
e. 1.50 1015 atoms U 1.50
1015
1 mol U atoms U 6.02 1023 atoms
238.0 g U 5.93 107 g U 1 mol U
Section 11.3 Moles of Compounds pages 320–327
Practice Problems pages 321–324, 326
Section 11.2 Assessment
in 2.50 mol ZnCl2.
15. Explain what is meant by molar mass.
2 mol Cl 2.50 mol ZnCl2 5.00 mol Cl 1 mol ZnCl2
Molar mass is the mass in grams of one mole of any pure substance. Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
20. Determine the number of moles of chloride ions
page 319
16. What conversion factor should be used to
convert from mass to moles? Moles to mass? Mass-to-mole conversions use the conversion factor 1 mol/number of grams. Moles-to-mass conversions use the conversion factor number of grams/1 mol.
17. Explain the steps needed to convert the mass
of an element to the number of atoms of the element. Multiply the mass by the inverse of molar mass, and then multiply by Avogadro’s number.
18. Thinking Critically The mass of a single atom
is usually given in the unit amu. Would it be possible to express the mass of a single atom in grams? Explain. Because one mole is equal to 6.02 1023 atoms and the molar mass of an element is equal to one mole, the mass of a single atom can be calculated by dividing the mass of one mole by Avogadro’s number.
Solutions Manual
21. Calculate the number of moles of each element
in 1.25 mol glucose (C6H12O6). 6 mol C 1.25 mol C6H12O6 7.50 mol C 1 mol C6H12O6 12 mol H 1.25 mol C6H12O6 15.0 mol H 1 mol C6H12O6 6 mol O 1.25 mol C6H12O6 7.50 mol O 1 mol C6H12O6
22. Determine the number of moles of sulfate ions
present in 3.00 mol iron(III) sulfate (Fe2(SO4)3). 3 mol SO42 3.00 mol Fe2(SO4)3 1 mol Fe2(SO4)3 9.00 mol SO42
23. How many moles of oxygen atoms are present
in 5.00 mol diphosphorus pentoxide (P2O5)? 5 mol O 5.00 mol P2O5 25.0 mol O 1 mol P2O5
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24. Calculate the number of moles of hydrogen
atoms in 11.5 mol water.
14.01 g N 3 mol N 42.03 g 1 mol N
2 mol H 11.5 mol H2O 23.0 mol H 1 mol H2O
1.008 g H 12 mol H 12.096 g 1 mol H
25. Determine the molar mass of each of the
following ionic compounds: NaOH, CaCl2, KC2H3O2, Sr(NO3)2, and (NH4)3PO4. NaOH 22.99 g Na 1 mol Na 22.99 g 1 mol Na
30.97 g P 1 mol P 1 mol P
16.00 g O 4 mol O 64.00 g 1 mol O molar mass (NH4)3PO4 149.10 g/mol
26. Calculate the molar mass of each of the
16.00 g O 1 mol O 1 mol O
16.00 g
1.008 g H 1 mol H 1 mol H
following molecular compounds: C2H5OH, C12H22O11, HCN, CCl4, and H2O.
1.008 g
C2H5OH
molar mass NaOH
40.00 g/mol
CaCl2
40.08 g Ca 1 mol Ca 1 mol Ca
40.08 g
35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g
molar mass CaCl2
110.98 g/mol
KC2H3O2
39.10 g K 1 mol K 1 mol K
39.10 g
12.01 g C 2 mol C 1 mol C
24.02 g
1.008 g H 3 mol H 3.024 g 1 mol H 16.00 g O 2 mol O 32.00 g 1 mol O molar mass KC2H3O2
98.14 g/mol
Sr(NO3)2
87.62 g Sr 1 mol Sr 87.62 g 1 mol Sr 14.01 g N 2 mol N 28.02 g 1 mol N 16.00 g O 6 mol O 96.00 g 1 mol O molar mass Sr(NO3)2
211.64 g/mol
(NH4)3PO4
12.01 g C 2 mol C 24.02 g 1 mol C 1.008 g H 6 mol H 6.048 g 1 mol H 16.00 g O 1 mol O 16.00 g 1 mol O molar mass C2H5OH
Chemistry: Matter and Change • Chapter 11
46.07 g/mol
C12H22O11
12.01 g C 12 mol C 1 mol C
144.12 g
1.008 g H 22 mol H 22.176 g 1 mol H 16.00 g O 11 mol O 176.00 g 1 mol O molar mass C12H22O11
342.30 g/mol
HCN 1.008 g H 1 mol H 1.008 g 1 mol H 12.01 g C 1 mol C 1 mol C
12.01 g
14.01 g N 1 mol N 14.01 g 1 mol N molar mass HCN
27.03 g/mol
CCl4
12.01 g C 1 mol C 1 mol C
12.01 g
35.45 g Cl 4 mol Cl 141.80 g 1 mol Cl molar mass CCl4
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30.97 g
153.81 g/mol Solutions Manual
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H2O
1.008 g H 2 mol H 2.016 g 1 mol H 16.00 g O 1 mol O 16.00 g 1 mol O
molar mass H2O
18.02 g/mol
(H2SO4)? Step 1: Find the molar mass of H2SO4.
98.09 g/mol
Step 2: Make mole 0 mass conversion. 98.09 g H2SO4 3.25 mol H2SO4 1 mol H2SO4 319 g H2SO4
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
158.04 g KMnO4 2.55 mol KMnO4 1 mol KMnO4 403 g KMnO4
30. Determine the number of moles present in each
107.9 g Ag 1 mol Ag 107.9 g 1 mol Ag
16.00 g O 4 mol O 64.00 g 1 mol O
28. What is the mass of 4.35
158.04 g/mol
Step 1: Find the molar mass of AgNO3.
32.07 g
102
molar mass KMnO4
of the following. a. 22.6 g AgNO3
1.008 g H 2 mol H 2.016 g 1 mol H
molar mass H2SO4
64.00 g
Step 2: Make mole 0 mass conversion.
27. What is the mass of 3.25 moles of sulfuric acid
32.07 g S 1 mol S 1 mol S
16.00 g O 4 mol O 1 mol O
moles of zinc
14.01 g N 1 mol N 1 mol N
14.01 g
16.00 g O 3 mol O 1 mol O
48.00 g
molar mass AgNO3
169.9 g/mol
Step 2: Make mass 0 mole conversion.
chloride (ZnCl2)?
1 mol AgNO 22.6 g AgNO3 169.9 g AgNO3
Step 1: Find the molar mass of ZnCl2.
0.133 mol AgNO3
65.38 g Zn 1 mol Zn 65.38 g 1 mol Zn 35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g
molar mass ZnCl2
136.28 g/mol
Step 2: Make mole 0 mass conversion. 136.28 g ZnCl2 4.35 102 mol ZnCl2 1 mol ZnCl2 5.93 g ZnCl2
29. How many grams of potassium permanganate
are in 2.55 moles? Step 1: Find the molar mass of KMnO4. 39.10 g K 1 mol K 1 mol K
39.10 g
b. 6.50 g ZnSO4 Step 1: Find the molar mass of ZnSO4. 65.39 g Zn 1 mol Zn 65.39 g 1 mol Zn 32.07 g S 1 mol S 1 mol S
32.07 g
16.00 g O 4 mol O 1 mol O
64.00 g
molar mass ZnSO4
161.46 g/mol
Step 2: Make mass 0 mole conversion. 1 mol ZnSO4 6.50 g ZnSO4 161.46 g ZnSO4 0.0403 mol ZnSO4
54.94 g Mn 1 mol Mn 54.94 g 1 mol Mn
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c. 35.0 g HCl Step 1: Find the molar mass of HCl. 1.008 g H 1 mol H 1 mol H
1.008 g
35.45 g Cl 1 mol Cl 35.45 g 1 mol Cl molar mass HCl
36.46 g/mol
Step 2: Make mass 0 mole conversion. 1 mol HCl 35.0 g HCl 0.960 mol HCl 36.46 g HCl
d. 25.0 g Fe2O3
52.00 g
16.00 g O 4 mol O 1 mol O
64.00 g
molar mass Ag2CrO4
331.8 g/mol
Step 2: Make mass 0 mole conversion. 1 mol Ag2CrO4 25.8 g Ag2CrO4 331.8 g Ag2CrO4 0.0778 mol Ag2CrO4 Step 3: Make mole 0 formula unit conversion. 0.0778 mol Ag2CrO4
Step 1: Find the molar mass of Fe2O3. 55.85 g Fe 2 mol Fe 111.70 g 1 mol Fe 16.00 g O 3 mol O 1 mol O
48.00 g
molar mass Fe2O3
159.70 g/mol
Step 2: Make mass 0 mole conversion. 1 mol Fe2O3 25.0 g Fe2O3 159.70 g Fe2O3 0.157 mol Fe2O3
6.02 1023 formula units 1 mol 4.68 1022 formula units Ag2CrO4
a. How many Ag ions are present? 4.68 1022 formula units Ag2CrO4 2 Ag ions 1 formula unit Ag2CrO4 9.36 1022 Ag ions
b. How many CrO42 ions are present? 4.68 1022 formula units Ag2CrO4
e. 254 g PbCl4 Step 1: Find the molar mass of PbCl4. 207.2 g Pb 1 mol Pb 207.2 g 1 mol Pb 35.45 g Cl 4 mol Cl 1 mol Cl
141.80 g
molar mass PbCl4
349.0 g/mol
Step 2: Make mass 0 mole conversion. 1 mol PbCl4 254 g PbCl4 349.0 g PbCl4 0.728 mol PbCl4
31. A sample of silver chromate (Ag2CrO4) has a
mass of 25.8 g. Step 1: Find the molar mass of Ag2CrO4. 107.9 g Ag 2 mol Ag 215.8 g 1 mol Ag
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52.00 g Cr 1 mol Cr 1 mol Cr
Chemistry: Matter and Change • Chapter 11
1 CrO42 ion 1 formula unit Ag2CrO4 4.68 1022 CrO42 ions
c. What is the mass in grams of one formula
unit of silver chromate? 331.8 g Ag2CrO4 1 mol Ag2CrO4 1 mol Ag2CrO4 6.02 1023 formula units 5.51 1022 g Ag2CrO4/formula unit
32. What mass of sodium chloride contains
4.59 1024 formula units? Step 1: Find the number of moles of NaCl. 4.59 1024 formula units NaCl 1 mol NaCl 7.62 mol NaCl 6.02 1023 formula units
Solutions Manual
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SOLUTIONS MANUAL
Step 2: Find the molar mass of NaCl. 22.99 g Na 1 mol Na 22.99 g 1 mol Na 35.45 g Cl 1 mol Cl 1 mol Cl
35.45 g
molar mass NaCl
58.44 g/mol
Step 3: Make mole 0 mass conversion. 58.44 g NaCl 7.62 mol NaCl 445 g NaCl 1 mol NaCl
34. A sample of sodium sulfite (Na2SO3) has a
mass of 2.25 g. Step 1: Find the molar mass of Na2SO3 22.99 g Na 2 mol Na 45.98 g 1 mol Na 32.07 g S 1 mol S 1 mol S
32.07 g
45.6 g.
16.00 g O 3 mol O 1 mol O
48.00 g
Step 1: Find the molar mass of C2H5OH.
molar mass Na2SO3
126.05 g/mol
33. A sample of ethanol (C2H5OH) has a mass of
12.01 g C 2 mol C 24.02 g 1 mol C 1.008 g H 6 mol H 6.048 g 1 mol H 16.00 g O 1 mol O 16.00 g 1 mol O molar mass C2H5OH
46.07 g/mol
Step 2: Make mass 0 mole conversion. Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
1 O atom 5.96 1023 O atoms 1 molecule C2H5OH
1 mol C2H5OH 45.6 g C2H5OH 46.07 g C2H5OH 0.990 mol C2H5OH Step 3: Make mole 0 molecule conversion. 6.02 1023 molecules 0.990 mol C2H5OH 1 mol 5.96 1023 molecules C2H5OH
a. How many carbon atoms does the sample
Step 2: Make mass 0 mole conversion 1 mol Na2SO3 2.25 g Na2SO3 126.05 g Na2SO3 0.0179 mol Na2SO3 Step 3: Make mole 0 formula unit conversion 0.0179 mol Na2SO3 6.02 1023 formula units 1 mol 1.08 1022 formula units Na2SO3
a. How many Na ions are present? 1.08 1022 formula units Na2SO3 2 Na ions 1 formula unit Na2SO3 2.16 1022 Na ions
b. How many SO32 ions are present?
contain?
1.08 1022 formula units Na2SO3
5.96 1023 molecules C2H5OH
1 SO32 ions 1 formula unit Na2SO3
2 C atoms 1.19 1024 C atoms 1 molecule C2H5OH
b. How many hydrogen atoms are present? 5.96 1023 molecules C2H5OH 6 H atoms 3.58 1024 H atoms 1 molecule C2H5OH
c. How many oxygen atoms are present? 5.96 1023 molecules C2H5OH
Solutions Manual
1.08 1022 SO32 ions
c. What is the mass in grams of one formula
unit of Na2SO3? 126.08 g Na2SO3 1 mol Na2SO3 1 mol Na2SO3 6.02 1023 formula units 2.09 1022 g Na2SO3/formula unit
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35. A sample of carbon dioxide has a mass of 52.0 g. Step 1: Find the molar mass of CO2. 12.01 g C 1 mol C 12.01 g 1 mol C 16.00 g O 2 mol O 32.00 g 1 mol O molar mass CO2
44.01 g/mol
37. What three conversion factors are often used in
mole conversions? number of grams , 1 mol 6.02 1023 representative particles , 1 mol number of atoms of element 1 mol compound
Step 2: Make mass 0 mole conversion.
Step 3: Make mole 0 molecule conversion. 6.02 1023 molecules 1.18 mol CO2 1 mol 7.11 1023 molecules CO2
a. How many carbon atoms are present? 7.11 1023 molecules CO2 1 C atom 7.11 1023 C atoms 1 molecule CO2
b. How many oxygen atoms are present? 7.11 1023 molecules CO2
38. Explain how you can determine the number of
atoms or ions in a given mass of a compound. Convert the mass to moles, multiply the number of moles by the ratio of the number of atoms or ions to one mole, multiply by Avogadro’s number.
39. If you know the mass in grams of a molecule of
a substance, could you obtain the mass of a mole of that substance? Explain. Yes, multiply the mass in grams of the molecule by Avogadro’s number.
40. Thinking Critically Design a bar graph that
will show the number of moles of each element present in 500 g dioxin (C12H4Cl4O2), a powerful poison.
c. What is the mass in grams of one molecule
of CO2? 44.01 g CO2 1 mol 6.02 1023 molecules 1 mol CO2 7.31 1023 g CO2/molecule
Section 11.3 Assessment page 327
Number of moles
2 O atoms 1.42 1024 O atoms 1 molecule CO2
27 24 21 18 15 12 9 6 3 0
Moles of Elements in 500 g Dioxin
C
36. Describe how you can determine the molar
mass of a compound. Multiply the mass of one mole of each element by the ratio of that element to one mole of the compound. Add the resulting masses.
H Cl Element
O
12.01 g C 12 mol C 144.12 g C 1 mol C 1.008 g H 4 mol H 1 mol H
4.032 g H
35.45 g Cl 4 mol Cl 141.80 g Cl 1 mol Cl 16.00 g O 2 mol O 32.00 g O 1 mol O molar mass
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Chemistry: Matter and Change • Chapter 11
321.96 g/mol Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
1 mol CO2 1.18 mol CO2 52.0 g CO2 44.01 g CO2
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SOLUTIONS MANUAL
1 mol C12H4Cl4O2 500 g C12H4Cl4O2 321.96 g C12H4Cl4O2 2 mol C12H4Cl4O2
pages 328–337
12 mol C 2 mol C12H4Cl4O2 24 mol C 1 mol C12H4Cl4O2 4 mol H 2 mol C12H4Cl4O2 8 mol H 1 mol C12H4Cl4O2 4 mol Cl 2 mol C12H4Cl4O2 8 mol Cl 1 mol C12H4Cl4O2 2 mol O 2 mol C12H4Cl4O2 4 mol O 1 mol C12H4Cl4O2
41. Applying Concepts The recommended daily
allowance of calcium is 1000 mg of Ca2 ions. Calcium carbonate is used to supply the calcium in vitamin tablets. How many moles of calcium ions does 1000 mg represent? How many moles of calcium carbonate are needed to supply the required amount of calcium ions? What mass of calcium carbonate must each tablet contain?
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
40.08 g Ca 1 mol Ca 40.08 g Ca 1 mol Ca 12.01 g C 1 mol C 1 mol C
12.01 g C
16.00 g O 3 mol O 1 mol O
48.00 g O
molar mass
100.09 g/mol
1 g Ca2 1 mol Ca2 1000 mg Ca2 3 2 10 mg Ca 40.08 g Ca2 0.02 mol Ca2 1 mol CaCO3 100.09 g CaCO3 0.02 mol Ca2 1 mol CaCO3 1 mol Ca2 2 g CaCO3
Section 11.4 Empirical and Molecular Formulas Practice Problems pages 331, 333, 335, 337
42. Determine the percent by mass of each element
in calcium chloride. Steps 1 and 2: Assume 1 mole; calculate molar mass of CaCl2. 40.08 g Ca 1 mol Ca 1 mol Ca
40.08 g
35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g
molar mass CaCl2
110.98 g/mol
Step 3: Determine percent by mass of each element. 40.08 g Ca percent Ca 100 110.98 g CaCl2 36.11% Ca 70.90 g Cl percent Cl 100 110.98 g CaCl2 63.89% Cl
43. Calculate the percent composition of sodium
sulfate. Steps 1 and 2: Assume 1 mole; calculate molar mass of Na2SO4. 22.99 g Na 2 mol Na 45.98 g 1 mol Na 32.07 g S 1 mol S 1 mol S
32.07 g
16.00 g O 4 mol O 1 mol O
64.00 g
molar mass Na2SO4
142.05 g/mol
Step 3: Determine percent by mass of each element. 45.98 g Na percent Na 100 142.05 g Na2SO4 32.37% Na 32.07 g S percent S 100 142.05 g Na2SO4 22.58% S Solutions Manual
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64.00 g O percent O 100 142.05 g Na2SO4
45. What is the percent composition of phosphoric
acid (H3PO4)? Steps 1 and 2: Assume 1 mole; calculate molar mass of H3PO4.
45.05% O
44. Which has the larger percent by mass of sulfur,
H2SO3 or H2S2O8?
1.008 g H 3 mol H 3.024 g 1 mol H
Steps 1 and 2: Assume 1 mole; calculate molar mass of H2SO3.
30.97 g P 1 mol P 1 mol P
30.97 g
1.008 g H 2 mol H 1 mol H
2.016 g
16.00 g O 4 mol O 64.00 g 1 mol O
32.06 g S 1 mol S 1 mol S
32.06 g
molar mass H3PO4
16.00 g O 3 mol O 1 mol O
48.00 g
molar mass H2SO3
82.08 g/mol
Step 3: Determine percent by mass of S. 32.06 g S percent S 100 39.06% S 82.08 g H2SO3 Repeat steps 1 and 2 for H2S2O8. Assume 1 mole; calculate molar mass of H2S2O8. 1.008 g H 2 mol H 2.016 g 1 mol H 32.06 g S 2 mol S 1 mol S
64.12 g
16.00 g O 8 mol O 128.00 g 1 mol O molar mass H2S2O8
194.14 g/mol
Step 3: Determine percent by mass of S. 64.12 g S percent S 100 194.14 g H2S2O8 33.03% S H2SO3 has a larger percent by mass of S.
97.99 g/mol
Step 3: Determine percent by mass of each element. 3.024 g H percent H 100 3.08% H 97.99 g H3PO4 30.97 g P percent P 100 31.61% P 97.99 g H3PO4 64.00 g O percent O 100 65.31% O 97.99 g H3PO4
46. A blue solid is found to contain 36.84%
nitrogen and 63.16% oxygen. What is the empirical formula for this solid? Step 1: Assume 100 g sample; calculate moles of each element. 1 mol N 36.84 g N 2.630 mol N 14.01 g N 1 mol O 63.16 g O 3.948 mol O 16.00 g O Step 2: Calculate mole ratios. 2.630 mol N 1.000 mol N 1 mol N 2.630 mol N 1.000 mol N 1 mol N 1.5 mol O 3.948 mol O 1.500 mol O 2.630 mol N 1.000 mol N 1 mol N The simplest ratio is 1 mol N: 1.5 mol O. Step 3: Convert decimal fraction to whole number. In this case, multiply by 2 because 1.5 2 3. Therefore, the empirical formula is N2O3.
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47. Determine the empirical formula for a
compound that contains 35.98% aluminum and 64.02% sulfur. Step 1: Assume 100 g sample; calculate moles of each element. 1 mol Al 35.98 g Al 1.334 mol Al 26.98 g Al 1 mol S 64.02 g S 1.996 mol S 32.06 g S Step 2: Calculate mole ratios. 1.334 mol Al 1.000 mol Al 1 mol Al 1.334 mol Al 1.000 mol Al 1 mol Al 1.5 mol S 1.996 mol S 1.500 mol S 1.334 mol Al 1.000 mol Al 1 mol Al The simplest ratio is 1 mol Al: 1.5 mol S. Step 3: Convert decimal fraction to whole number. In this case, multiply by 2 because 1.5 2 3. Therefore, the empirical formula is Al2S3.
48. Propane is a hydrocarbon, a compound
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
composed only of carbon and hydrogen. It is 81.82% carbon and 18.18% hydrogen. What is the empirical formula? Step 1: Assume 100 g sample; calculate moles of each element.
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49. The chemical analysis of aspirin indicates that
the molecule is 60.00% carbon, 4.44% hydrogen, and 35.56% oxygen. Determine the empirical formula for aspirin. Step 1: Assume 100 g sample; calculate moles of each element. 1 mol C 60.00 g C 5.00 mol C 12.01 g C 1 mol H 4.44 g H 4.40 mol H 1.008 g H 1 mol O 35.56 g O 2.22 mol O 16.00 g O Step 2: Calculate mole ratios. 5.00 mol C 2.25 mol C 2.25 mol C 2.22 mol O 1.00 mol O 1 mol O 4.40 mol H 1.98 mol H 2 mol H 2.22 mol O 1.00 mol O 1 mol O 1 mol O 2.22 mol O 1.00 mol O 2.22 mol O 1.00 mol O 1 mol O The simplest ratio is 2.25 mol C: 2 mol H: 1 mol O. Step 3: Convert decimal fraction to whole number. In this case, multiply by 4 because 2.25 4 9. Therefore, the empirical formula is C9H8O4.
50. What is the empirical formula for a compound
1 mol C 81.82 g C 6.813 mol C 12.01 g C
that contains 10.89% magnesium, 31.77% chlorine, and 57.34% oxygen?
1 mol H 18.18 g H 18.04 mol H 1.008 g H
Step 1: Assume 100 g sample; calculate moles of each element.
Step 2: Calculate mole ratios. 6.813 mol C 1.000 mol C 1 mol C 6.813 mol C 1.000 mol C 1 mol C 2.65 mol H 18.04 mol H 2.649 mol H 6.813 mol C 1.000 mol C 1 mol C The simplest ratio is 1 mol: 2.65 mol H. Step 3: Convert decimal fraction to whole number. In this case, multiply by 3 because 2.65 3 7.95 8. Therefore, the empirical formula is C3H8.
1 mol Mg 10.89 g Mg 0.4480 mol Mg 24.31 g Mg 1 mol Cl 31.77 g Cl 0.8962 mol Cl 35.45 g Cl 1 mol O 57.34 g O 3.584 mol O 16.00 g O Step 2: Calculate mole ratios. 0.4480 mol Mg 1.000 mol Mg 1 mol Mg 0.4480 mol Mg 1.000 mol Mg 1 mol Mg 0.8962 mol Cl 2.000 mol Cl 2 mol Cl 0.4480 mol Mg 1.000 mol Mg 1 mol Mg 8 mol O 3.584 mol O 7.999 mol O 0.4480 mol Mg 1.000 mol Mg 1 mol Mg
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The empirical formula is MgCl2O8 or Mg(ClO4)2. The simplest ratio is 1 mol Mg: 2 mol Cl: 8 mol O.
51. Analysis of a chemical used in photographic
developing fluid indicates a chemical composition of 65.45% C, 5.45% H, and 29.09% O. The molar mass is found to be 110.0 g/mol. Determine the molecular formula. Step 1: Assume 100 g sample; calculate moles of each element. 1 mol C 65.45 g C 5.450 mol C 12.01 g C
52. A compound was found to contain 49.98 g
carbon and 10.47 g hydrogen. The molar mass of the compound is 58.12 g/mol. Determine the molecular formula. Step 1: Assume 100 g sample; calculate moles of each element. 1 mol C 49.98 g C 4.162 mol C 12.01 g C 1 mol H 10.47 g H 10.39 mol H 1.008 g H Step 2: Calculate mole ratios.
1 mol H 5.45 g H 5.41 mol H 1.008 g H
4.162 mol C 1.000 mol C 1 mol C 4.162 mol C 1.000 mol C 1 mol C
1 mol O 29.09 g O 1.818 mol O 16.00 g O
10.39 mol H 2.50 mol H 2.5 mol H 4.162 mol C 1.000 mol C 1 mol C
Step 2: Calculate mole ratios. 5.450 mol C 3.000 mol C 3 mol C 1.818 mol O 1.000 mol O 1 mol O 5.41 mol H 2.97 mol H 3 mol H 1.818 mol O 1.00 mol O 1 mol O 1 mol O 1.818 mol O 1.000 mol O 1.818 mol O 1.000 mol O 1 mol O The simplest ratio is 1 mol: 2.65 mol H. Therefore, the empirical formula is C3H3O. Step 3: Calculate the molar mass of the empirical formula. 12.01 g C 3 mol C 36.03 g 1 mol C 1.008 g H 3 mol H 3.024 g 1 mol H 16.00 g O 1 mol O 16.00 g 1 mol O molar mass C3H3O
55.05 g/mol
Step 4: Determine whole number multiplier.
The simplest ratio is 1 mol C: 2.5 mol H. Because 2.5 2 5, the empirical formula is C2H5. Step 3: Calculate the molar mass of the empirical formula. 12.01 g C 2 mol C 24.02 g 1 mol C 1.008 g H 5 mol H 5.040 g 1 mol H molar mass C2H5
29.06 g/mol
Step 4: Determine whole number multiplier. 58.12 g/mol 2.000 29.06 g/mol The molecular formula is C4H10.
53. A colorless liquid composed of 46.68% nitrogen
and 53.32% oxygen has a molar mass of 60.01 g/mol. What is the molecular formula? Step 1: Assume 100 g sample; calculate moles of each element.
110.0 g/mol 1.998, or 2 55.05 g/mol
1 mol N 46.68 g N 3.332 mol N 14.01 g N
The molecular formula is C6H6O2.
1 mol O 53.32 g O 3.333 mol O 16.00 g O Step 2: Calculate mole ratios. 3.332 mol N 1.000 mol N 1 mol N 3.332 mol N 1.000 mol N 1 mol N
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1 mol O 3.333 mol O 1.000 mol O 3.332 mol N 1.000 mol N 1 mol N The simplest ratio is 1 mol N: 1 mol O. The empirical formula is NO. Step 3: Calculate the molar mass of the empirical formula.
Step 2: Calculate mole ratios. 3.131 mol Fe 1.000 mol Fe 1 mol Fe 3.131 mol Fe 1.000 mol Fe 1 mol Fe 1.5 mol O 4.696 mol O 1.500 mol O 3.131 mol Fe 1.000 mol Fe 1 mol Fe The simplest ratio is 1 mol Fe: 1.5 mol O.
14.01 g N 1 mol N 14.01 g 1 mol N
Because 1.5 2 3, the empirical formula is Fe2O3.
16.00 g O 1 mol O 16.00 g 1 mol O
56. The pain reliever morphine contains 17.900 g C,
molar mass NO
30.01 g/mol
Step 4: Determine whole number multiplier. 60.01 g/mol 2.000 30.01 g/mol The molecular formula is N2O2.
54. When an oxide of potassium is decomposed,
19.55 g K and 4.00 g O are obtained. What is the empirical formula for the compound? Step 1: Calculate moles of each element.
1.680 g H, 4.225 g O, and 1.228 g N. Determine the empirical formula. Step 1: Calculate moles of each element. 1 mol C 17.900 g C 1.490 mol C 12.01 g C 1 mol H 1.680 g H 1.667 mol H 1.008 g H 1 mol O 4.225 g O 0.2641 mol O 16.00 g O 1 mol N 1.228 g N 0.087 65 mol N 14.01 g N
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Step 2: Calculate mole ratios. 1 mol K 19.55 g K 0.5000 mol K 39.10 g K 1 mol O 4.00 g O 0.250 mol O 16.00 g O Step 2: Calculate mole ratios. 0.5000 mol K 2.00 mol K 2 mol K 0.250 mol O 1.00 mol O 1 mol O 1 mol O 0.250 mol O 1.00 mol O 0.250 mol O 1.00 mol O 1 mol O The simplest ratio is 2 mol K: 1 mol O. The empirical formula is K2O.
55. Analysis of a compound composed of iron and
oxygen yields 174.86 g Fe and 75.14 g O. What is the empirical formula for this compound? Step 1: Calculate moles of each element
0.08765 mol N 1.000 mol N 1 mol N 0.08765 mol N 1.000 mol N 1 mol N 1.490 mol C 17.00 mol C 17 mol C 0.08765 mol N 1.000 mol N 1 mol N 1.667 mol H 19.02 mol H 19 mol H 0.08765 mol N 1.000 mol N 1 mol N 0.2641 mol O 3.013 mol O 3 mol O 0.08765 mol N 1.000 mol N 1 mol N The simplest ratio is 17 mol C: 19 mol H: 3 mol O: 1 mol N. The empirical formula is C17H19O3N.
57. An oxide of aluminum contains 0.545 g Al and
0.485 g O. Find the empirical formula for the oxide. Step 1: Calculate moles of each element.
1 mol Fe 174.86 g Fe 3.131 mol Fe 55.85 g Fe
1 mol Al 0.545 g Al 0.0202 mol Al 26.98 g Al
1 mol O 75.14 g O 4.696 mol O 16.00 g O
1 mol O 0.485 g O 0.0303 mol O 6.00 g O
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Step 2: Calculate mole ratios. 0.0202 mol Al 1.00 mol Al 1 mol Al 0.0202 mol Al 1.00 mol Al 1 mol Al 1.5 mol O 0.0303 mol O 1.50 mol O 0.0202 mol Al 1.00 mol Al 1 mol Al The simplest ratio is 1 mol Al: 1.5 mol O. Because 1.5 2 3, the empirical formula is Al2O3.
Section 11.4 Assessment page 337
58. Explain how percent composition data for a
compound are related to the masses of the elements in the compound. Percent composition is numerically equal to the mass in grams of each element in a 100.0-g sample.
SOLUTIONS MANUAL
62. Inferring Hematite (Fe2O3) and magnetite
(Fe3O4) are two ores used as sources of iron. Which ore provides the greater percent of iron per kilogram? 55.85 g Fe 2 mol Fe 111.70 g Fe 1 mol Fe 16.00 g O 3 mol O 1 mol O
48.00 g O
molar mass Fe2O3
159.70 g/mol
55.85 g Fe 3 mol Fe 167.55 g Fe 1 mol Fe 16.00 g O 4 mol O 1 mol O
64.00 g O
molar mass Fe3O4
231.55 g/mol
111.70 g Fe percent by mass 100 159.70 g Fe2O3 69.94% Fe in Fe2O3
formula and a molecular formula?
167.55 g Fe percent by mass 100 231.55 g Fe3O4
The empirical formula is the simplest wholenumber ratio of the atoms in a compound. The molecular formula is the actual ratio of atoms in the compound.
Hematite is 69.94% Fe, magnetite is 72.36% Fe. Magnetite contains a greater percentage of iron per kilogram than hematite.
59. What is the difference between an empirical
72.36% Fe in Fe3O4
60. Explain how you can find the mole ratio in a
chemical compound. The mole ratio is determined by calculating the moles of each element in the compound and dividing each number of moles by the smallest number of moles. It is sometimes necessary to multiply the ratio by an integer to obtain whole numbers.
61. Thinking Critically An analysis for copper was
performed on two pure solids. One solid was found to contain 43.0% copper; the other contained 32.0% copper. Could these solids be samples of the same copper-containing compound? Explain your answer. No, a compound will always have the same chemical analysis. Therefore, if two solid have different percents of copper, they are different compounds.
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Section 11.5 The Formula for a Hydrate pages 338–341
Practice Problems page 340
63. A hydrate is found to have the following percent
composition: 48.8% MgSO4 and 51.2% H2O. What is the formula and name for this hydrate? Step 1: Assume 100 g sample; calculate moles of each component. 1 mol MgSO4 48.8 g MgSO4 120.38 g MgSO4 0.405 mol MgSO4 1 mol H2O 51.2 g H2O 2.84 mol H2O 18.02 g H2O
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Step 2: Calculate mole ratios. 0.405 mol MgSO4 1.00 mol MgSO4 0.405 mol MgSO4 1.00 mol MgSO4 1 mol MgSO4 1 mol MgSO4 2.84 mol H2O 7.01 mol H2O 0.405 mol MgSO4 1.00 mol MgSO4 7 mol H2O 1 mol MgSO4 The formula of the hydrate is MgSO47H2O. Its name is magnesium sulfate heptahydrate.
64. Figure 11-13 shows a common hydrate of
cobalt(II) chloride. If 11.75 g of this hydrate is heated, 9.25 g of anhydrous cobalt chloride remains. What is the formula and name for this hydrate? Step 1: Calculate the mass of water driven off. mass of hydrated compound mass of anhydrous compound remaining 11.75 g CoCl2xH2O 9.25 g CoCl2 2.50 g H2O Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Step 2: Calculate moles of each component. 1 mol CoCl2 9.25 g CoCl2 129.83 g CoCl2 0.0712 mol CoCl2 1 mol H2O 2.50 g H2O 0.139 mol H2O 18.02 g H2O
SOLUTIONS MANUAL
Section 11.5 Assessment page 341
65. What is a hydrate? What does its name indicate
about its composition? A hydrate is a compound that has a specific number of water molecules associated with it. The name indicates the name of the compound and the number of water molecules associated with each formula unit of compound.
66. Describe the experimental procedure for deter-
mining the formula for a hydrate. Explain the reason for each step. Mass an empty crucible. Add some hydrate and remass. Heat the crucible to drive out the water. Cool and remass. Determine the moles of the anhydrous compound. Subtract the mass of the crucible after heating from the mass of the crucible with the hydrate. The difference is the mass of the water lost. Determine the moles of water. Determine the simplest whole-number ratio of moles of water to moles of anhydrous compound, which will yield the formula of the hydrate.
67. Name the compound having the formula
SrCl26H2O.
strontium chloride hexahydrate
68. Thinking Critically Explain how the hydrate illus-
trated in Figure 11-13 might be used as a means of roughly determining the probability of rain. The hydrate is pink in moist air.
Step 3: Calculate mole ratios. 0.0712 mol CoCl2 1.00 mol CoCl2 0.0712 mol CoCl2 1.00 mol CoCl2 1 mol CoCl2 1 mol CoCl2
69. Sequencing Arrange these hydrates in order of
increasing percent water content: MgSO47H2O, Ba(OH)28H2O, CoCl26H2O. 24.31 g Mg 1 mol Mg 24.31 g Mg 1 mol Mg
0.139 mol H2O 1.95 mol H2O 0.0712 mol CoCl2 1.00 mol CoCl2
32.07 g S 1 mol S 1 mol S
32.07 g S
2 mol H2O 1 mol CoCl2
16.00 g O 11 mol O 1 mol O
176.00 g O
1.008 g H 14 mol H 1 mol H
14.112 g H
The formula of the hydrate is CoCl22H2O. Its name is cobalt(II) chloride dihydrate.
molar mass MgSO47H2O 246.49 g
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137.33 g Ba 1 mol Ba 1 mol Ba
137.33 g Ba
16.00 g O 10 mol O 1 mol O
160.00 g O
1.008 g H 18 mol H 1 mol H
18.144 g H
molar mass Ba(OH)28H2O 315.47 g
Chapter 11 Assessment pages 346–350 Concept Mapping 70. Complete this concept map by placing in each box the conversion factor needed to convert from each measure of matter to the next. Mass 1.
4.
Moles
Moles
2.
3.
58.93 g Co 1 mol Co 1 mol Co
58.93 g Co
35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g Cl
16.00 g O 6 mol O 1 mol O
96.00 g O
1.008 g H 12 mol H 1 mol H
12.096 g H
molar mass CoCl26H2O
237.93 g
16.00 g O 1 mol O 1 mol O
16.00 g O
6.02 1023 representative particles 2. 1 mol
1.008 g H 2 mol H 1 mol H
2.016 g H
1 mol 3. 6.02 1023 representative particles
molar mass H2O
18.02 g
7(18.02 g H2O) 100 246.49 g MgSO47H2O 51.17% H2O in MgSO47H2O 8(18.02 g H2O) 100 315.47 g Ba(OH)28H2O
45.70% H2O in Ba(OH)28H2O
Number of particles 1 mol 1. number of grams
number of grams 4. 1 mol
Mastering Concepts 71. Why is the mole an important unit to chemists? (11.1) A mole allows a chemist to accurately measure the number of atoms, molecules, or formula units in a substance.
6(18.02 g H2O) 100 237.93 g CoCl26H2O
72. How is a mole similar to a dozen? (11.1)
CoCl26H2O; Ba(OH)28H2O; MgSO47H2O
73. What is the numerical value of Avogadro’s
45.44% H2O in CoCl26H2O
A mole, like a dozen, contains a specific number of items.
number? (11.1) 6.02 1023
74. What is molar mass? (11.2) Molar mass is the mass in grams of one mole of any element or compound.
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75. Which contains more atoms, a mole of silver
atoms or a mole of gold atoms? Explain your answer. (11.2) They both contain the same number of atoms because a mole of anything contains 6.02 1023 representative particles.
76. Discuss the relationships that exist between the
mole, molar mass, and Avogadro’s number. (11.2) Molar mass is the mass in grams of one mole of any pure substance. Avogadro’s number is the number of representative particles in one mole. The mass of 6.02 1023 representative particles of a substance is the molar mass of the substance.
77. Which has a greater mass, a mole of silver
atoms or a mole of gold atoms? Explain your answer. (11.2) The molar mass of Au is 197.0 g/mol; the molar mass of Ag is 107.9 g/mol. Thus, a mole of Au has a greater mass.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
78. Explain the difference between atomic mass
SOLUTIONS MANUAL
82. Which of the following molecules contains the
most moles of carbon atoms per mole of the compound: ascorbic acid (C6H8O6), glycerin (C3H8O3), or vanillin (C8H8O3)? Explain. (11.3) The formula for vanillin (C8H8O3) shows there are eight carbon atoms per molecule, more than ascorbic acid or glycerin.
83. Explain what is meant by percent composition.
(11.4) Percent composition is the percent by mass of each element in a compound.
84. What is the difference between an empirical
formula and a molecular formula? Use an example to illustrate your answer. (11.4) An empirical formula is the smallest wholenumber ratio of elements that make up a compound (CH). A molecular formula specifies the actual number of atoms of each element in one molecule or formula unit of the substance (C6H6).
85. Do all pure samples of a given compound have
(amu) and molar mass (gram). (11.2)
the same percent composition? Explain. (11.4)
Atomic mass (amu) is the mass of an individual particle (atom, molecule). Molar mass (grams) is the mass of a mole of particles.
Yes, for every pure substance, the percent by mass of each element is the same regardless of the size of the sample.
79. If you divide the molar mass of an element by
86. What information must a chemist obtain in
Avogadro’s number, what is the meaning of the quotient? (11.2)
order to determine the empirical formula of an unknown compound? (11.4)
the mass of one atom of the element
the percent composition of the compound
80. List three conversion factors used in molar
87. What is a hydrated compound? Use an example
conversions. (11.3)
to illustrate your answer. (11.5)
1 mol , number of grams
A hydrated compound is a compound that has a specific number of water molecules associated with its atoms, for example, Na2CO310H2O and CuSO45H2O.
6.02 1023 representative particles , 1 mol 1 mol , 6.02 1023 representative particles number of grams 1 mol
88. Explain how hydrates are named. (11.5) First, name the compound. Then, add a prefix (mono- di- tri-) that indicates how many water molecules are associated with one mole of compound.
81. What information is provided by the formula for
potassium chromate (K2CrO4)? (11.3) One mole of K2CrO4 contains two moles of K ions and one mole of CrO42 ions. Solutions Manual
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c. 1.56 1023 formula units sodium hydroxide
Mastering Problems Mole-Particle Conversions (11.1) Level 1 89. Determine the number of representative particles in each of the following. a. 0.250 mol silver 6.02 atoms Ag 0.250 mol Ag 1 mol Ag 1023
1.51 1023 atoms Ag
b. 8.56 103 mol sodium chloride 8.56 103 mol NaCl 6.02 1023 formula units NaCl 1 mol NaCl 5.15 1021 formula units NaCl
c. 35.3 mol carbon dioxide 6.02 molecules CO2 35.3 mol CO2 1 mol CO2 1023
2.13 1025 molecules CO2
d. 0.425 mol nitrogen (N2) molecules N2 6.02 0.425 mol N2 1 mol N2 1023
2.56
1023
molecules N2
90. Determine the number of moles in each of the
following. a. 3.25 1020 atoms lead 1 mol Pb 3.25 1020 atoms Pb 6.02 1023 atoms Pb 5.39 104 mol Pb
b. 4.96 1024 molecules glucose 4.96 1024 molecules glucose 1 mol glucose 6.02 1023 molecules glucose 8.24 mol of glucose
1.56 1023 formula units NaOH 1 mol NaOH 6.02 1023 formula units NaOH 2.59 101 mol NaOH
d. 1.25 1025 copper(II) ions 1.25 1025 Cu2 ions 1 mol Cu2 ions 6.02 1023 Cu2 ions 2.08 101 mol Cu2 ions
91. Make the following conversions. a. 1.51 1015 atoms Si to mol Si 1 mol Si 1.51 1015 atoms Si 6.02 1023 atoms Si 2.51 109 mol Si
b. 4.25 102 mol H2SO4 to molecules
H2SO4 4.25 102 mol H2SO4 6.02 1023 molecules H2SO4 1 mol H2SO4 2.56 1022 molecules H2SO4
c. 8.95 1025 molecules CCl4 to mol CCl4 8.95 1025 molecules CCl4 1 mol CCl4 6.02 1023 molecules CCl4 1.49 102 mol CCl4
d. 5.90 mol Ca to atoms Ca 6.02 1023 atoms Ca 5.90 mol Ca 1 mol Ca 3.55 1024 atoms Ca
92. How many molecules are contained in each of
the following? a. 1.35 mol carbon disulfide (CS2) 6.02 1023 molecules CS2 1.35 mol CS2 1 mol CS2 8.13 1023 molecules CS2
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b. 0.254 mol diarsenic trioxide (As2O3) 0.254 mol As2O3 6.02 1023 molecules As2O3 1 mol As2O3 1.53 1023 molecules As2O3
c. 1.25 mol water
added to another solution containing 0.130 mol Ca2 ions. What is the total number of metal ions in the combined solution? 0.250 mol Ca2 ions 0.130 mol Ca2 ion 0.380 mol metal ions 0.380 mol metal ions
6.02 molecules H2O 1.25 mol H2O 1 mol H2O
6.02 1023 mol metal ions 1 mol metal ions
7.53 1023 molecules H2O
2.29 1023 Cu2 and Ca2 ions
1023
d. 150.0 mol HCl
96. If a snowflake contains 1.9 1018 molecules
6.02 1023 molecules HCl 150.0 mol HCl 1 mol HCl
of water, how many moles of water does it contain?
9.030 1025 molecules HCl
1.9 1018 molecules H2O
93. How many moles contain each of the following? a. 1.25 1015 molecules carbon dioxide 1.25 1015 molecules CO2 1 mol CO2 6.02 1023 molecules CO2 2.08 109 mol CO2
b. 3.59 1021 formula units sodium nitrate Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
95. A solution containing 0.250 mol Cu2 ions is
3.59 1021 formula units NaNO3
1 mol H2O 6.02 1023 molecules H2O 3.2 106 mol H2O
97. If you could count two atoms every second,
how long would it take you to count a mole of atoms? Assume that you counted continually 24 hours every day. How does the time you calculated compare with the age of Earth, which is estimated to be 4.5 109 years old?
1 mol NaNO3 6.02 1023 formula units NaNO3
1s 6.02 1023 particles 3.01 1023 s 2 particles
5.96 103 mol NaNO3
1y 1 day 1h 3.01 1023 s 3600 s 24 h 365 day
c. 2.89 1027 formula units calcium carbonate 2.89 1027 formula units CaCO3 1 mol CaCO3 6.02 1023 formula units CaCO3 4.80 103 mol CaCO3
Level 2 94. A bracelet containing 0.200 mol of metal atoms is 75% gold. How many particles of gold atoms are in the bracelet? 0.200 mol Au 0.75 0.150 mol Au 6.02 atoms Au 0.150 mol Au 1 mol Au 1023
9.54 1015 y 2.1 106 4.5 109 y 9.5 1015 yr; about 2 million times longer
Mole-Mass Conversions (11.2) Level 1 98. Calculate the mass of the following. a. 5.22 mol He 4.00 g He 5.22 mol He 20.9 g He 1 mol He
b. 0.0455 mol Ni 58.69 g Ni 0.0455 mol Ni 2.67 g Ni 1 mol Ni
9.03 1022 atoms Au
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c. 2.22 mol Ti 47.87 g Ti 2.22 mol Ti 106 g Ti 1 mol Ti
d. 0.00566 mol Ge 72.61 g Ge 0.00566 mol Ge 0.411 g Ge 1 mol Ge
99. Make the following conversions. a. 3.50 mol Li to g Li 6.94 g Li 3.50 mol Li 24.3 g Li 1 mol Li
b. 7.65 g Co to mol Co 1 mol Co 7.65 g Co 0.130 mol Co 58.93 g Co
c. 5.62 g Kr to mol Kr 1 mol Kr 5.62 g Kr 0.0671 mol Kr 83.80 g Kr
d. 0.0550 mol As to g As 74.92 g As 0.0550 mol As 4.12 g As 1 mol As
Level 2 100. Determine the mass in grams of the following. a. 1.33 1022 mol Sb 121.76 g Sb 1.33 1022 mol Sb 1 mol Sb 1.62
1024
g Sb
b. 4.75 1014 mol Pt 195.08 g Pt 4.75 1014 mol Pt 1 mol Pt 9.27 1016 g Pt
c. 1.22 1023 mol Ag 1.22 1023
107.87 g Ag mol Ag 1 mol Ag
1.32 1025 g Ag
d. 9.85 1024 mol Cr 52.00 g Cr 9.85 1024 mol Cr 1 mol Cr
Particle-Mass Conversions (11.2) Level 1 101. Convert the following to mass in grams. a. 4.22 1015 atoms U 1 mol U 4.22 1015 atoms U 6.02 1023 atoms U 283.03 g U 1.67 106 g U 1 mol U
b. 8.65 1025 atoms H 8.65 1025 atoms H 1.008 g H 1 mol H 1 mol H 6.02 1023 atoms H 145 g H
c. l.25 1022 atoms O 1.25 1022 atoms O 16.00 g O 1 mol O 1 mol O 6.02 1023 atoms O 0.332 g O
d. 4.44 1023 atoms Pb 4.44 1023 atoms Pb 207.2 g Pb 1 mol Pb 1 mol Pb 6.02 1023 atoms Pb 153 g Pb
102. A sensitive balance can detect masses of
1 108 g. How many atoms of silver would be in a sample having this mass? 1 mol Ag 1 108 g Ag 107.87 g Ag 6.02 1023 atoms Ag 5 1013 atoms Ag 1 mol Ag
103. Calculate the number of atoms in each of the
following. a. 25.8 g Hg 1 mol Hg 25.8 g Hg 200.59 g Hg 6.02 1023 atoms Hg 1 mol Hg 7.74 1022 atoms Hg
5.12 1026 g Cr
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b. 0.0340 g Zn 1 mol Zn 0.0340 g Zn 63.59 g Zn 6.02 1023 atoms Zn 3.13 1020 atoms Zn 1 mol Zn
c. 150 g Ar
Level 2 106. A mixture contains 0.250 mol Fe and 1.20 g C. What is the total number of atoms in the mixture? 6.02 1023 atoms Fe 0.250 mol Fe 1 mol Fe 1.51 1023 atoms Fe
1 mol Ar 150 g Ar 39.95 g Ar
6.02 1023 atoms C 1 mol C 1.20 g C 12.01 g C 1 mol C
6.02 1023 atoms Ar 2.3 1024 atoms Ar 1 mol Ar
0.601 1023 atoms C
d. 0.124 g Mg 1 mol Mg 0.124 g Mg 24.31 g Mg 6.02 1023 atoms Mg 1 mol Mg 3.07 1021 atoms Mg
104. Which has more atoms, 10.0 g of carbon or
10.0 g of calcium? How many atoms does each have? 1 mol C 6.02 1023 atoms C 10.0 g C 12.01 g C 1 mol C Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
SOLUTIONS MANUAL
5.01 1023 atoms C 1 mol Ca 6.02 1023 atoms Ca 10.0 g Ca 40.08 g Ca 1 mol Ca 1.50 1023 atoms Ca 10.0 g C contains more atoms.
105. Which has more atoms, 10.0 moles of carbon
or 10.0 moles of calcium? How many does each have? One mole of any substance contains 6.02 1023 representative particles. Thus, 10.0 moles of carbon and 10.0 moles of calcium contain the same number or atoms.
1.51 1023 atoms Fe 0.601 1023 atoms C 2.11 1023 total atoms
Chemical Formulas (11.3) Level 1 107. In the formula for sodium phosphate (Na3PO4), how many moles of sodium are represented? How many moles of phosphorus? How many moles of oxygen? 3 mol Na, 1 mol P, 4 mol O
108. How many moles of oxygen atoms are
contained in the following? a. 2.50 mol KMnO4 4 mol O 2.50 mol KMnO4 1 mol KMnO4 10.0 mol O
b. 45.9 mol CO2 2 mol O 45.9 mol CO2 91.8 mol O 1 mol CO2
c. 1.25 102 mol CuSO45H2O 1.25 102 mol CuSO45H2O
9 mol O 0.113 mol O 1 mol CuSO45H2O
6.02 1023 atoms 10.0 mol 1 mol 6.02 1024 atoms
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The Composition of a Compound 6 5 4 3 2 1
Molar Mass (11.3) Level 1 111. Determine the molar mass of each of the following. a. nitric acid (HNO3) 1.008 g H 1 mol H 1.008 g H 1 mol H 14.01 g N 1 mol N 14.01 g N 1 mol N 16.00 g O 3 mol O 48.00 g O 1 mol O
Ca
C H Atoms
O
CaC4H6O4 (calcium acetate) 40.08 g Ca 1 mol Ca 40.08 g Ca 1 mol Ca
molar mass
63.02 g/mol HNO3
b. ammonium nitrate (NH4NO3) 14.01 g N 2 mol N 28.02 g N 1 mol N
12.01 g C 4 mol C 1 mol C
48.04 g C
1.008 g H 4 mol H 4.032 g H 1 mol H
1.008 g H 6 mol H 1 mol H
6.048 g H
16.00 g O 3 mol O 48.00 g O 1 mol O
16.00 g O 4 mol O 1 mol O
64.00 g O
molar mass
158.18 g/mol
molar mass
110. How many carbon tetrachloride molecules are
in 3.00 mol carbon tetrachloride (CCl4)? How many carbon atoms? How many chlorine atoms? How many total atoms? 6.02 1023 molecules CCl4 3.00 mol CCl4 1 mol CCl4
80.05 g/mol NH4NO3
c. zinc oxide (ZnO) 65.39 g Zn 1 mol Zn 65.39 g Zn 1 mol Zn 16.00 g O 1 mol O 1 mol O
16.00 g O
molar mass
81.39 g/mol ZnO
d. cobalt chloride (CoCl2)
1.81 1024 molecules CCl4
58.93 g Co 1 mol Co 58.93 g Co 1 mol Co
1 mol atoms C 1.81 1024 molecules CCl4 1 molecule CCl4
35.45 g Cl 2 mol Cl 70.90 g Cl 1 mol Cl
1.81 1024 atoms C 1.81 1024
molar mass
129.83 g/mol CoCl2
4 mol atoms Cl molecules CCl4 1 molecule CCl4
7.24 1024 atoms Cl 1.81 1024 atoms C 7.24 1024 atoms Cl 9.05 1024 total atoms
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Atoms per formula unit
Level 2 109. The graph shows the numbers of atoms of each element in a compound. What is its formula? What is its molar mass?
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112. Calculate the molar mass of each of the
following. a. ascorbic acid (C6H8O6) 12.01 g C 6 mol C 72.06 g C 1 mol C 1.008 g H 8 mol H 8.064 g H 1 mol H 16.00 g O 6 mol O 96.00 g O 1 mol O 176.12 g/mol C6H8O6
molar mass
113. Determine the molar mass of allyl sulfide, the
compound responsible for the smell of garlic. The chemical formula of allyl sulfide is (C3H5)2S. 12.01 g C 6 mol C 1 mol C
72.06 g C
1.008 g H 10 mol H 10.080 g H 1 mol H 32.07 g S 1 mol S 1 mol S
32.07 g S
molar mass
114.21 g/mol (C3H5)2S
b. sulfuric acid (H2SO4) 1.008 g H 2 mol H 2.016 g H 1 mol H 32.07 g S 1 mol S 32.07 g S 1 mol S 16.00 g O 4 mol O 64.00 g O 1 mol O 98.09 g/mol H2SO4
molar mass
107.87 g Ag 1 mol Ag 107.87 g Ag 1 mol Ag Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
14.01 g N 2 mol N 28.02 g N 1 mol N 16.00 g O 1 mol O 16.00 g O 1 mol O
c. silver nitrate (AgNO3)
14.01 g N 1 mol N 1 mol N
14.01 g N
16.00 g O 3 mol O 1 mol O
48.00 g O
molar mass
Mass-Mole Conversions (11.3) Level 1 114. How many moles are in 100.0 g of each of the following compounds? a. dinitrogen oxide (N2O)
169.88 g/mol AgNO3
d. saccharin (C7H5NO3S)
molar mass
44.02 g/mol
1 mol N2O 2.27 mol N2O 100.0 g N2O 44.02 g N2O
b. methanol (CH3OH) 12.01 g C 1 mol C 12.01 g C 1 mol C 1.008 g H 4 mol H 4.032 g H 1 mol H
12.01 g C 7 mol C 84.07 g C 1 mol C
16.00 g O 1 mol O 16.00 g O 1 mol O
1.008 g H 5 mol H 5.040 g H 1 mol H
molar mass
14.01 g N 1 mol N 14.01 g N 1 mol N
32.04 g/mol
1 mol CH3OH 100.0 g CH3OH 32.04 g CH3OH 3.12 mol CH3OH
16.00 g O 3 mol O 48.00 g O 1 mol O 32.07 g S 1 mol S 32.07 g S 1 mol S molar mass
Solutions Manual
183.19 g/mol C7H5NO3S
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115. What is the mass of each of the following? a. 4.50
102
mol CuCl2
63.55 g Cu 1 mol Cu 63.55 g Cu 1 mol Cu 35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g Cl
molar mass
134.45 g/mol
134.35 g CuCl2 4.50 102 mol CuCl2 1 mol CuCl2 6.05 g CuCl2
b. 1.25 102 mol Ca(OH)2
Level 2 117. Benzoyl peroxide is a substance used as an acne medicine. What is the mass in grams of 3.50 102 moles of benzoyl peroxide (C14H10O4)? 12.01 g C 14 mol C 168.14 g C 1 mol C 1.008 g H 10 mol H 10.080 g H 1 mol H 16.00 g O 4 mol O 1 mol O
64.00 g O
molar mass
242.22 g/mol
40.08 g Ca 1 mol Ca 40.08 g Ca 1 mol Ca
242.22 g 3.50 102 mol 1 mol
1.008 g H 2 mol H 1 mol H
8.48 g benzoyl peroxide
2.016 g H
118. Hydrofluoric acid is a substance used to etch
16.00 g O 2 mol O 1 mol O
32.00 g O
molar mass
74.10 g/mol
1.25
102
74.10 g Ca(OH)2 mol Ca(OH)2 1 mol Ca(OH)2
9.26 102 g Ca(OH)2
116. Determine the number of moles in each of the
following. a. 1.25 102 g Na2S
glass. Determine the mass of 4.95 1025 HF molecules. 1.008 g H 1 mol H 1.008 g H 1 mol H 19.00 g F 1 mol F 1 mol F
19.00 g F
molar mass
20.01 g/mol
4.95 1025 molecules HF
22.99 g Na 2 mol Na 45.98 g Na 1 mol Na 32.07 g S 1 mol S 1 mol S
32.07 g S
molar mass
78.05 g/mol
1 mol Na2S 1.25 102 g Na2S 78.05 g Na2S 1.60 mol Na2S
b. 0.145 g H2S
20.01 g HF 1 mol HF 1 mol HF 6.02 1023 molecules HF 1650 g HF
119. How many moles of aluminum ions are in
45.0 g of aluminum oxide? 26.98 g Al 2 mol Al 53.96 g Al 1 mol Al 16.00 g O 3 mol O 1 mol O
48.00 g O
molar mass
101.96 g/mol
1.008 g H 2 mol H 2.016 g H 1 mol H
1 mol Al2O3 45.0 g Al2O3 101.96 g Al2O3
32.07 g S 1 mol S 32.07 g S 1 mol S
2 mol Al3 ions 0.883 mol Al3 ions 1 mol Al2O3
molar mass
34.09 g/mol
1 mol H2S 0.145 g H2S 34.09 g H2S 4.25 103 mol H2S
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120. How many moles of ions are in the following? a. 0.0200 g AgNO3 107.87 g Ag 1 mol Ag 107.87 g Ag 1 mol Ag 14.01 g N 1 mol N 1 mol N
14.01 g N
16.00 g O 3 mol O 1 mol O
48.00 g O
molar mass
169.88 g/mol
Table 11-2 Moles, Mass, and Representative Particles Compound
Number Mass(g) of moles
Representative particles
Silver acetate Ag(C2H3O2)
2.50
417
1.51 1024
1 mol AgNO3 0.0200 g AgNO3 169.88 g AgNO3
Glucose C6H12O6
1.798
324.0
1.082 1024
2 mol ions 2.35 104 mol ions 1 mol AgNO3
Benzene C6H6
0.009 39
0.733
5.65 1021
0.4178
100.0
2.516 1023
Lead(II) sulfide PbS
b. 0.100 mol K2CrO4 3 mol ions 0.100 mol K2CrO4 1 mol K2CrO4 0.300 mol ions
137.33 g Ba 1 mol Ba 137.33 g Ba 1 mol Ba 16.00 g O 2 mol O 1 mol O
32.00 g O
1.008 g H 2 mol H 1 mol H
2.016 g H
molar mass
171.35 g/mol
1 mol Ba(OH)2 0.500 g Ba(OH)2 171.35 g Ba(OH)2 3 mol ions 8.75 103 mol ions 1 mol Ba(OH)2
d. 1.00 109 mol Na2CO3 3 mol ions 1.00 109 mol Na2CO3 1 mol Na2CO3 3.00 109 mol ions
Silver acetate: 107.87 g Ag 1 mol Ag 107.87 g Ag 1 mol Ag
c. 0.500 g Ba(OH)2
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Mass-Particle Conversions (11.3) Level 1 121. Calculate the values that will complete the table.
12.01 g C 2 mol C 1 mol C
24.02 g C
1.008 g H 3 mol H 1 mol H
3.024 g H
16.00 g O 2 mol O 1 mol O
32.00 g O
molar mass
166.91 g/mol
166.91 g AgC2H3O2 2.50 mol AgC2H3O2 1 mol AgC2H3O2 417 g AgC2H3O2 6.02 1023 formula units 2.50 mol AgC2H3O2 1 mol AgC2H3O2 1.51 1024 formula units Glucose: 12.01 g C 6 mol C 1 mol C
72.06 g C
1.008 g H 12 mol H 12.10 g H 1 mol H 16.00 g O 6 mol O 96.00 g O 1 mol O molar mass
Solutions Manual
180.16 g/mol
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1 mol glucose 324.0 g glucose 180.16 g glucose 1.798 mol glucose
1 mol Au 3.50 g Au 196.97 g Au
6.02 1023 molecules 1.798 mol glucose 1 mol glucose
6.02 1023 atoms Au 1.07 1022 atoms Au 1 mol Au
1.082 1024 molecules Benzene:
124. Calculate the mass of 3.62 1024 molecules
of glucose (C6H12O6).
12.01 g C 6 mol C 72.06 g C 1 mol C
From solution #121, the molar mass of glucose equals 180.16 g/mol.
1.008 g H 6 mol H 6.048 g H 1 mol H
3.62 1024 molecules glucose 1 mol glucose 6.02 1023 molecules glucose
molar mass
78.11 g/mol
5.65 1021 molecules benzene 1 mol benzene 6.02 1023 molecules benzene 9.39 103 mol benzene 78.11 g benzene mol benzene 9.39 1 mol benzene 0.733 g benzene 103
180.16 g glucose 1080 g glucose 1 mol glucose
125. Determine the number of molecules of ethanol
(C2H5OH) in 47.0 g. 12.01 g C 2 mol C 24.02 g C 1 mol C 1.008 g H 6 mol H 6.048 g H 1 mol H
Lead(II) sulfide: 207.2 g Pb 1 mol Pb 207.2 g Pb 1 mol Pb
16.00 g O 1 mol O 16.00 g O 1 mol O
32.07 g S 1 mol S 1 mol S
32.07 g S
molar mass
molar mass
239.3 g/mol
1 mol ethanol 47.0 g ethanol 46.07 g ethanol
1 mol PbS 100.0 g PbS 0.4178 mol PbS 239.3 g PbS 6.02 1023 formula units 0.4178 mol PbS 1 mol PbS 2.516 1023 formula units
122. How many formula units are present in 500.0 g
lead(II) chloride? 207.2 g Pb 1 mol Pb 207.2 g Pb 1 mol Pb 35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g Cl
molar mass
278.1 g/mol
1 mol PbCl2 1.798 mol PbCl2 500.0 g PbCl2 278.1 g PbCl2 6.02 1023 formula units 1.798 mol PbCl2 1 mol PbCl2 1.082 1024 formula units PbCl2
160
123. Determine the number of atoms in 3.50 g gold.
Chemistry: Matter and Change • Chapter 11
46.07 g/mol
6.02 1023 molecules ethanol 1 mol ethanol 6.14 1023 molecules ethanol
126. What mass of iron(III) chloride contains
2.35 1023 chloride ions? 55.85 g Fe 1 mol Fe 55.85 g Fe 1 mol Fe 35.45 g Cl 3 mol Cl 1 mol Cl
106.35 g Cl
molar mass
162.20 g/mol
2.35 1023 chloride ions 1 mol chloride ions 6.02 1023 chloride ions 1 mol FeCl3 162.20 g FeCl3 3 mol chloride ions 1 mol FeCl3 21.1 g FeCl3
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127. How many moles of iron can be recovered
in 250.0 g aluminum oxide (Al2O3).
55.85 g Fe 3 mol Fe 167.55 g Fe 1 mol Fe
From solution # 119, the molar mass of aluminum oxide equals 101.96 g/mol.
16.00 g O 4 mol O 1 mol O
64.00 g O
1 mol Al2O3 2 mol Al3 250.0 g Al2O3 1 mol Al2O3 101.96 g Al2O3
molar mass
231.55 g/mol
4.90 mol Al3
103 g Fe3O4 100.0 kg Fe3O4 1 kg Fe3O4 1.000 105 g Fe3O4 1 mol Fe3O4 1.000 105 g Fe3O4 231.55 g Fe3O4 3 mol Fe 1296 mol Fe 1 mol Fe3O4
128. The mass of an electron is 9.11 1028 g.
What is the mass of a mole of electrons? 9.11 1028 g 6.02 1023 electrons 1 mol electrons 1 electron 5.48 104 g/mol electrons
129. Vinegar is 5.0% acetic acid (CH3COOH). How
many molecules of acetic acid are present in 25.0 g vinegar? Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
131. Calculate the moles of aluminum ions present
from 100.0 kg Fe3O4?
Level 2 132. Determine the number of chloride ions in 10.75 g of magnesium chloride. 24.31 g Mg 1 mol Mg 24.31 g 1 mol Mg 35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g
molar mass
94.31 g/mol
1 mol MgCl2 10.75 g MgCl2 94.31 g MgCl2 6.02 1023 ions MgCl2 2 mol Cl 1 mol MgCl2 1 mol Cl 1.37 1023 ions Cl
133. Acetaminophen, a common aspirin substitute,
12.01 g C 2 mol C 24.02 g C 1 mol C
has the formula C8H9NO2. Determine the number of molecules of acetaminophen in a 500 mg tablet.
1.008 g H 4 mol H 4.032 g H 1 mol H
12.01 g C 8 mol C 96.08 g C 1 mol C
16.00 g O 2 mol O 32.00 g O 1 mol O
1.008 g H 9 mol H 9.072 g H 1 mol H
molar mass
60.05 g/mol
0.050 25.0 g vinegar 1.25 g CH3COOH
14.01 g N 1 mol N 14.01 g N 1 mol N
1 mol CH3COOH 1.25 g CH3COOH 60.05 g CH3COOH
16.00 g O 2 mol O 32.00 g O 1 mol O
6.02 1023 molecules CH3COOH 1 mol CH3COOH
molar mass
1.25
1022
molecules CH3COOH
130. The density of lead is 11.3 g/cm3. Calculate
the volume of one mole lead. 207.2 g Pb 1 cm3 1 mol Pb 18.3 cm3 11.3 g Pb 1 mol Pb
151.16 g/mol
1 g C8H9NO2 500 mg C8H9NO2 103 mg C8H9NO2 1 mol C8H9NO2 0.003 mol C8H9NO2 151.16 g C8H9NO2 0.003 mol C8H9NO2 6.02 1023 molecules C8H9NO2 1 mol C8H9NO2 2 1021 molecules C8H9NO2
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134. Calculate the number of sodium ions present in
25.0 g sodium chloride. 22.99 g Na 1 mol Na 22.99 g Na 1 mol Na 35.45 g Cl 1 mol Cl 1 mol Cl
35.45 g Cl
molar mass
58.44 g/mol
1 mol NaCl 25.0 g NaCl 58.44 g NaCl 6.02 1023 ions Na 2.58 1023 ions Na 1 mol NaCl
135. Determine the number of oxygen atoms
present in 25.0 g carbon dioxide. 12.01 g C 1 mol C 12.01 g C 1 mol C 16.00 g O 2 mol O 32.00 g O 1 mol O molar mass
44.01 g/mol
1 mol CO2 0.568 mol CO2 25.0 g CO2 44.01 g CO2 2(6.02 atoms O) 0.568 mol CO2 1 mol CO2 1023
6.84 1023 atoms O
176.00 g percent by mass 100 342.30 g 51.42% O
b. magnetite (Fe3O4) 55.85 g Fe 3 mol Fe 167.55 g Fe 1 mol Fe 16.00 g O 4 mol O 64.00 g O 1 mol O
167.55 g percent by mass 100 231.55 g 72.360% Fe 64.00 g percent by mass 100 231.55 g 27.64% O
c. aluminum sulfate (Al2(SO4)3) 26.98 g Al 2 mol Al 53.96 g Al 1 mol Al 32.07 g S 3 mol S 1 mol S
12.01 g C 12 mol C 144.12 g C 1 mol C 1.008 g H 22 mol H 22.18 g H 1 mol H 16.00 g O 11 mol O 176.00 g O 1 mol O molar mass
342.30 g/mol
144.12 g percent by mass 100 342.30 g 42.10% C 22.18 g percent by mass 100 342.30 g 6.480% H
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Chemistry: Matter and Change • Chapter 11
96.21 g S
16.00 g O 12 mol O 192.00 g O 1 mol O molar mass
Percent Composition (11.4) 136. Express the composition of each of the following as the mass percent of its elements (percent composition). a. sucrose (C12H22O11)
231.55 g/mol
molar mass
342.17 g/mol
53.96 g percent by mass 100 342.17 g 15.77% Al 96.21 g percent by mass 100 342.17 g 28.12% S 96.21 g percent by mass = 100 342.17 g 56.11% O
137. Which of the following iron compounds
contain the greatest percentage of iron: pyrite (FeS2), hematite (Fe2O3), or siderite (FeCO3)? 55.85 g Fe 1 mol Fe 55.85 g Fe 1 mol Fe 32.07 g S 2 mol S 1 mol S
64.14 g S
molar mass FeS2
119.99 g/mol
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55.85 g Fe 2 mol Fe 111.70 g Fe 1 mol Fe 16.00 g O 3 mol O 48.00 g O 1 mol O molar mass Fe2O3
159.70 g/mol
chemical formula C8H10N4O2. a. Calculate the molar mass of caffeine. 12.01 g C 8 mol C 1 mol C
96.08 g C
55.85 g Fe 1 mol Fe 55.85 g Fe 1 mol Fe
1.008 g H 10 mol H 10.080 g H 1 mol H
12.01 g C 1 mol C 1 mol C
12.01 g C
14.01 g N 4 mol N 56.04 g N 1 mol N
16.00 g O 3 mol O 96.00 g O 1 mol O
16.00 g O 2 mol O 32.00 g O 1 mol O
molar mass FeCO3
152.95 g/mol
55.85 g percent mass 100 119.99 g 46.55% in FeS2 111.70 g percent mass 100 159.70 g 69.944% in Fe2O3 55.85 g percent mass 100 152.95 g 36.51% in FeCO3
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
139. Caffeine, a stimulant found in coffee, has the
Hematite, with 69.944% Fe, has the greatest percentage of iron.
138. Determine the empirical formula for each of
the following compounds. a. ethylene (C2H4) Both subscripts can be divided by two. The empirical formula is CH2.
b. ascorbic acid (C6H8O6) All subscripts can be divided by two. The empirical formula is C3H4O3.
c. naphthalene (C10H8) Both subscripts can be divided by two. The empirical formula is C5H4.
molar mass
194.20 g/mol
b. Determine the percent composition of
caffeine. 96.08 g percent by mass 100 194.20 g 49.47% C 10.08 g percent by mass 100 194.20 g 5.191% H 56.04 g percent by mass 100 194.20 g 28.86% N 32.00 g percent by mass 100 194.20 g 16.48% O
140. Which of the titanium-containing minerals
rutile (TiO2) or ilmenite (FeTiO3) has the larger percent of titanium? 47.87 g Ti 1 mol Ti 47.87 g Ti 1 mol Ti 16.00 g O 2 mol O 1 mol O
32.00 g O
molar mass TiO2
79.87 g/mol
55.85 g Fe 1 mol Fe 55.85 g Fe 1 mol Fe 47.87 g Ti 1 mol Ti 47.87 g Ti 1 mol Ti
Solutions Manual
16.00 g O 3 mol O 1 mol O
48.00 g O
molar mass FeTiO3
151.72 g/mol
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47.87 g percent by mass 100 79.87 g 59.93% Ti in TiO2 47.87 g percent by mass 100 151.72 g 31.55% Ti in FeTiO3 TiO2 has the greater percent by mass of titanium.
141. Vitamin E, found in many plants, is thought to
retard the aging process in humans. The formula for vitamin E is C29H50O2. What is the percent composition of vitamin E?
80.00 g percent by mass 100 27.18% O 294.30 g
Empirical and Molecular Formulas (11.4) 143. The hydrocarbon used in the manufacture of foam plastics is called styrene. Analysis of styrene indicates the compound is 92.25% C and 7.75% H and has a molar mass of 104 g/mol. Determine the molecular formula for styrene.
12.01 g C 29 mol C 348.29 g C 1 mol C
1 mol C 92.25 g C 7.681 mol C 12.01 g C
1.008 g H 50 mol H 50.400 g H 1 mol H
1 mol H 7.75 g H 7.69 mol H 1.008 g H
16.00 g O 2 mol O 32.00 g O 1 mol O
7.681 mol C 1.00 7.69
molar mass
430.69 g/mol
7.69 mol H 1.00 7.69
348.29 g percent by mass 100 80.87% C 430.69 g
The empirical formula is CH.
50.40 g percent by mass 100 11.70% H 430.69 g
12.01 g C 1 mol C 12.01 g C 1 mol C
32.00 g percent by mass 100 7.430% O 430.69 g
1.008 g H 1 mol H 1.008 g H 1 mol H
142. Aspartame, an artificial sweetener, has the
molar mass CH
13.02 g/mol
formula C14H18N2O5. Determine the percent composition of aspartame.
104 g/mol n 8.00 13.02 g/mol
12.01 g C 14 mol C 168.14 g C 1 mol C
Multiply the subscripts in the empirical formula by 8 to obtain the molecular formula C8H8.
1.008 g H 18 mol H 18.144 g H 1 mol H
144. Monosodium glutamate (MSG) is sometimes
14.01 g N 2 mol N 28.02 g N 1 mol N 16.00 g O 5 mol O 80.00 g O 1 mol O
added to food to enhance flavor. Analysis determined this compound to be 35.5% C, 4.77% H, 8.29% N, 13.6% Na, and 37.9% O. What is the empirical formula for MSG? 1 mol C 35.5 g C 12.01 g C
2.96 mol C
168.14 g percent by mass 100 57.13% C 294.30 g
1 mol H 4.77 g H 1.008 g H
4.73 mol H
18.144 g percent by mass 100 6.165% H 294.30 g
1 mol N 8.29 g N 14.01 g N
0.592 mol N
molar mass
164
28.02 g percent by mass 100 9.521% N 294.30 g
294.30 g/mol
Chemistry: Matter and Change • Chapter 11
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1 mol Na 13.6 g Na 0.592 mol Na 22.99 g Na
12.01 g C 13 mol C 156.13 g C 1 mol C
1 mol O 37.9 g O 16.00 g O
1.008 g H 18 mol H 18.144 g H 1 mol H
2.37 mol O
2.96 mol C 5.00 mol C 0.592 4.73 mol H 8.00 mol H 0.592 0.592 mol N 1.00 mol N 0.592 0.592 mol Na 1.00 mol Na 0.592 2.37 mol O 4.00 mol O 0.592 The simplest whole-number ratio is 5.00 mol C: 8.00 mol H: 1.00 mol N: 1.00 mol Na: 4.00 mol O. The empirical formula is C5H8NO4Na.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
145. Determine the molecular formula for
16.00 g O 2 mol O 32.00 g O 1 mol O molar mass C13H18O2 206.27 g/mol The molar mass C13H18O2 equals the molar mass of ibuprofen. Thus, the molecular formula is the same as the empirical formula C13H18O2.
146. Vanadium oxide is used as an industrial cata-
lyst. The percent composition of this oxide is 56.0% vanadium and 44.0% oxygen. Determine the empirical formula for vanadium oxide. 1 mol V 56.0 g V 1.10 mol V 50.94 g V 1 mol O 44.0 g O 2.75 mol O 16.00 g O
ibuprofen, a common headache remedy. Analysis of ibuprofen yields a molar mass of 206 g/mol and a percent composition of 75.7% C, 8.80% H and 15.5% O.
1.10 mol V 1.00 mol V 1.10
1 mol C 75.7 g C 6.30 mol C 12.01 g C
The simplest ratio is 1.00 mol V: 2.50 mol O. Multiply the simplest ratio by two to obtain the simplest whole-number ratio. The empirical formula is V2O5.
1 mol H 8.80 g H 8.73 mol H 1.008 g H 1 mol O 15.5 g O 0.969 mol O 16.00 g O 6.30 mol C = 6.50 mol C 0.969 8.73 mol H = 9.01 mol H 0.969 0.969 mol O = 1.00 mol O 0.969 The simplest ratio is 6.50 mol C: 9.01 mol H: 1.00 mol O. Multiply the simplest ratio by two to obtain the simplest whole-number ratio. The empirical formula is C13H18O2.
2.75 mol O 2.50 mol O 1.10
147. What is the empirical formula of a compound
that contains 10.52 g Ni, 4.38 g C, and 5.10 g N? 1 mol Ni 10.52 g Ni 0.1792 mol Ni 58.69 g Ni 1 mol C 4.38 g C 0.3470 mol C 12.01 g C 1 mol N 5.10 g N 0.3640 mol N 14.01 g N 0.1792 mol Ni 1.000 mol Ni 0.1792 0.3470 mol C 1.936 mol C 0.1792 0.3640 mol N 2.031 mol N 0.1792 The simplest ratio is 1.00 mol Ni: 1.936 mol C: 2.031 mol N. The empirical formula is Ni(CN)2.
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148. The Statue of Liberty turns green in air
because of the formation of two copper compounds, Cu3(OH)4SO4 and Cu4(OH)6SO4. Determine the mass percent of copper in these compounds.
0.2865 mol Pb 1.000 mol Pb 0.2865 1.146 mol Cl 4.000 mol Cl 0.2865
Cu3(OH)4SO4
The simplest ratio is 1.000 mol Pb: 4.000 mol Cl. The empirical formula is PbCl4.
63.55 g Cu 3 mol Cu 190.65 g Cu 1 mol Cu
207.2 g Pb 1 mol Pb 207.2 g Pb 1 mol Pb
16.00 g O 8 mol O 1 mol O
128.00 g O
35.45 g Cl 4 mol Cl 141.80 g Cl 1 mol Cl
1.008 g H 4 mol H 1 mol H
4.032 g H
molar mass PbCl4
32.07 g S 1 mol S 1 mol S
32.07 g S
The molar mass based on the empirical formula equals the molar mass of the compound. The molecular formula is PbCl4.
molar mass Cu3(OH)4SO4 354.75 g/mol 190.65 g percent by mass 100 53.74% 338.75 g Cu4(OH)6SO4 63.55 g Cu 4 mol Cu 254.20 g Cu 1 mol Cu 16.00 g O 10 mol O 1 mol O
160.00 g O
1.008 g H 6 mol H 1 mol H
6.048 g H
32.07 g S 1 mol S 1 mol S
32.07 g S
molar mass Cu4(OH)6SO4 452.32 g/mol
349.00 g/mol
150. Glycerol is a thick, sweet liquid obtained as a
byproduct of the manufacture of soap. Its percent composition is 39.12% carbon, 8.75% hydrogen, and 52.12% oxygen. The molar mass is 92.11 g/mol. What is the molecular formula for glycerol? 1 mol C 39.12 g C 3.257 mol C 12.01 g C 1 mol H 8.75 g H 8.681 mol H 1.008 g H 1 mol O 52.12 g O 3.258 mol O 16.00 g O 3.257 mol C 1.000 mol C 3.257
254.20 g percent by mass 100 56.20% 452.32 g
8.681 mol H 2.665 mol H 3.257
149. Analysis of a compound containing chlorine
3.258 mol O 1.000 mol O 3.257
and lead reveals that the compound is 59.37% lead. The molar mass of the compound is 349.0 g/mol. What is the empirical formula for the chloride? What is the molecular formula? percent Pb percent Cl 100% percent Cl 100% percent Pb 100% 59.37% 40.63% 1 mol Pb 59.37 g Pb 0.2865 mol Pb 207.2 g Pb 1 mol Cl 40.63 g Cl 1.146 mol Cl 35.45 g Cl
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The simplest whole-number ratio is 1.000 mol C: 2.655 mol H: 1.000 mol O. To obtain the simplest whole-number ratio, multiply each number by three. The empirical formula is C3H8O3. 12.01 g C 3 mol C 36.03 g C 1 mol C 1.008 g H 8 mol H 8.064 g H 1 mol H 16.00 g O 3 mol O 48.00 g O 1 mol O molar mass
92.094 g/mol
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The molar mass based on the empirical formula equals the molar mass of the compound. The molecular formula is C3H8O3.
The Formula for a Hydrate (11.5) Level 1 151. Determine the mass percent of anhydrous sodium carbonate (Na2CO3) and water in sodium carbonate decahydrate (Na2CO310H2O).
1.008 g H 2 mol H 2.016 g H 1 mol H molar mass H2O
18.02 g/mol
1 mol BaCl2 0.410 mol BaCl2 85.3 g BaCl2 208.23 g BaCl2 1 mol H2O 14.7 g H2O 0.816 mol H2O 18.02 g H2O
22.99 g Na 2 mol Na 1 mol Na
45.98 g Na
0.816 mol n 2.00 0.410 mol
12.01 g C 1 mol C 1 mol C
12.01 g C
BaCl22H2O, barium chloride dihydrate
16.00 g O 13 mol O 1 mol O
208.00 g O
1.008 g H 20 mol H 1 mol H
20.16 g H
molar mass Na2CO310H2O 286.15 g/mol
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
16.00 g O 1 mol O 16.00 g O 1 mol O
153. Gypsum is hydrated calcium sulfate. A 4.89-g
sample of this hydrate was heated, and after the water was driven off, 3.87 g anhydrous calcium sulfate remained. Determine the formula of this hydrate and name the compound.
22.99 g Na 2 mol Na 45.98 g Na 1 mol Na
mass of hydrated CaSO4 mass of CaSO4 mass of H2O
12.01 g C 1 mol C 1 mol C
12.01 g C
mass of H2O mass of hydrated CaSO4 mass of H2O 4.89 g 3.87 g 1.02 g
16.00 g O 3 mol O 1 mol O
48.00 g O
40.08 g Ca 1 mol Ca 40.08 g Ca 1 mol Ca
molar mass Na2CO3
105.99 g/mol
16.00 g O 1 mol O 1 mol O
16.00 g O
1.008 g H 2 mol H 1 mol H
2.016 g H
molar mass H2O
18.02 g/mol
105.99 g 100 37.03% Na2CO3 286.15 g 10(18.02 g) 100 62.97% H2O 286.15 g
152. What is the formula and name for a hydrate that
is 85.3% barium chloride and 14.7% water? 137.33 g Ba 1 mol Ba 137.33 g Ba 1 mol Ba 35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g Cl
molar mass BaCl2
208.23 g/mol
32.07 g S 1 mol S 1 mol S
32.07 g S
16.00 g O 4 mol O 1 mol O
64.00 g O
molar mass CaSO4
136.15 g/mol
16.00 g O 1 mol O 1 mol O
16.00 g O
1.008 g H 2 mol H 1 mol H
2.016 g H
molar mass H2O
18.02 g/mol
1 mol CaSO4 3.87 g CaSO4 136.15 g CaSO4 0.0284 mol CaSO4 1 mol H2O 1.02 g H2O 0.0566 mol H2O 18.02 g H2O 0.0566 mol n 2.00 0.0284 mol CaSO42H2O, calcium sulfate dihydrate
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154. The table shows data from an experiment to
determine the formulas of hydrated barium chloride. Determine the formula for the hydrate and its name. Table 11-3 Data for BaCl2xH20
155. A 1.628-g sample of a hydrate of magnesium
iodide is heated until its mass is reduced to 1.072 g and all water has been removed. What is the formula of the hydrate? mass of water mass of hydrate mass of anhydrous solid 1.628 g 1.072 g 0.556 g 24.31 g Mg 1 mol Mg 24.31 g Mg 1 mol Mg
Mass of empty crucible
21.30 g
Mass of hydrate crucible
31.35 g
Initial mass of hydrate
10.05 g
126.90 g I 2 mol I 1 mol I
253.80 g I
Mass after heating 5 min
29.87 g
molar mass MgI2
278.11 g/mol
Mass of anhydrous solid
8.57 g
16.00 g O 1 mol O 16.00 g O 1 mol O
initial mass of hydrate (mass of hydrate crucible) (mass of empty crucible) 31.35 g 21.30 g 10.05 g
1.008 g H 2 mol H 2.016 g H 1 mol H molar mass H2O
mass of water (initial mass of hydrate) (mass of anhydrous solid) 10.05 g 8.57 g 1.48 g
0.003855 mol MgI2
137.33 g B 1 mol Ba 137.33 g Ba 1 mol Ba
0.0309 mol n 8.02 0.003855 mol
35.45 g Cl 2 mol Cl 70.90 g Cl 1 mol Cl
MgI28H2O
molar mass BaCl2
208.23 g/mol
16.00 g O 1 mol O 1 mol O
16.00 g O
1.008 g H 2 mol H 1 mol H
2.016 g H
molar mass H2O
18.02 g/mol
1 mol BaCl2 8.57 g BaCl2 208.23 g BaCl2 0.0412 mol BaCl2 1 mol H2O 1.48 g H2O 0.0821 mol H2O 18.02 g H2O 0.0821 mol n 2.00 0.0412 mol BaCl22H2O, barium chloride dihydrate
1 mol MgI2 1.072 g MgI2 278.11 g MgI2 1 mol H2O 0.556 g H2O 0.0309 mol H2O 18.02 g H2O
156. Hydrated sodium tetraborate (Na2B4O7xH2O)
is commonly called borax. Chemical analysis indicates that this hydrate is 52.8% sodium tetraborate and 47.2% water. Determine the formula and name the hydrate.
22.99 g Na 2 mol Na 45.98 g Na 1 mol Na 12.01 g B 4 mol B 1 mol B
48.04 g B
16.00 g O 7 mol O 1 mol O
112.00 g O
molar mass Na2B4O7
206.02 g/mol
16.00 g O 1 mol O 16.00 g O 1 mol O 1.008 g H 2 mol H 2.016 g H 1 mol H molar mass H2O
168
18.02 g/mol
mass of anhydrous solid (mass after heating 5 min) (mass of empty crucible) 29.87 g 21.30 g 8.57 g
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1 mol Na2B4O7 52.8 g Na2B4O7 206.02 g Na2B4O7 0.256 mol Na2B4O7 1 mol H2O 47.2 g H2O 2.61 mol H2O 18.02 g H2O 2.61 mol n 10.2 0.256 mol Na2B4O710H2O, sodium tetraborate decahydrate
2 mol O 0.4008 mol CO2 0.8016 mol O 1 mol CO2 6.02 1023 atoms O 0.8016 mol O 1 mol O 4.82 1023 atoms O
b. 3.21 1022 molecules CH3OH
Mixed Review
3.21 1022 molecules CH3OH
Sharpen your problem-solving skills by answering the following. 157. Determine the following: a. the number of representative particles in 3.75 g Zn
1 atom O 3.21 1022 atoms O 1 molecule CH3OH
1 mol Zn 3.75 g Zn 65.39 g Zn 6.02 1023 atoms Zn 3.45 1022 atoms Zn 1 mol Zn
b. the mass of 4.32 1022 atoms Ag 4.32 1022 atoms Ag Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
1 mol CO2 17.64 g CO2 0.4008 mol CO2 44.01 g CO2
1 mol Ag 107.87 g Ag 1 mol Ag 6.02 1023 atoms Ag 7.74 g Ag
c. the number of sodium ions is 25.0 g of
c. 0.250 mol C6H12O6 6 mol O 0.250 mol C6H12O6 1 mol C6H12O6 1.5 mol O 6.02 1023 atoms O 1.5 mol O 1 mol O 9.0 1023 atoms O
159. Which of the following compounds has the
greatest percent of oxygen by mass: TiO2, Fe2O3, Al2O3? 47.87 g Ti 1 mol Ti 47.87 g Ti 1 mol Ti 16.00 g O 2 mol O 32.00 g O 1 mol O
Na2O.
molar mass TiO2
1 mol Na2O 25.0 g Na2O 61.98 g Na2O
55.85 g Fe 2 mol Fe 111.70 g Fe 1 mol Fe
6.02 1023 ions Na 2 mol Na 1 mol Na2O 1 mol Na
16.00 g O 3 mol O 48.00 g O 1 mol O
4.86 1023 ions Na
molar mass Fe2O3
158. Which of the following has the greatest
number of oxygen atoms? a. 17.63 g CO2 12.01 g C 1 mol C 12.01 g C 1 mol C 16.00 g O 2 mol O 32.00 g O 1 mol O molar mass
Solutions Manual
44.01 g/mol
79.87 g/mol
159.70 g/mol
26.98 g Al 2 mol Al 53.96 g Al 1 mol Al 16.00 g O 3 mol O 48.00 g O 1 mol O molar mass Al2O3
54.44 g/mol
32.00 g O percent by mass 100 79.87 g TiO2 40.07% O in TiO2
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48.00 g O percent by mass 100 159.70 g Fe2O3 30.06% O in Fe2O3 48.00 g O percent by mass 100 54.44 g Al2O3 87.99% O in Al2O3 Al2O3 has the greatest percent of oxygen by mass.
160. Naphthalene, commonly known as moth
balls, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of napthalene is 128 g/mol. Determine the empirical and molecular formulas for naphthalene. 1 mol C 93.7 g C 7.80 mol C 12.01 g C 1 mol H 6.3 g H 6.2 mol H 1.008 g H 7.80 mol C 1.25 mol C 6.2 6.2 mol H 1.0 mol H 6.2
162. Calculate each of the following: a. the number of moles in 15.5 g Na2SO4 22.99 g Na 2 mol Na 45.98 g Na 1 mol Na 32.07 g S 1 mol S 1 mol S
32.07 g S
16.00 g O 4 mol O 1 mol O
64.00 g O
molar mass
142.05 g/mol
1 mol Na2SO4 15.5 g Na2SO4 142.05 g Na2SO4 0.109 mol Na2SO4
b. the number of formula units in 0.255 mol
NaCl 0.255 mol NaCl 6.02 1023 formula units NaCl 1 mol NaCl 1.54 1023 formula units NaCl
c. the mass in grams of 0.775 mol SF6
The simplest ratio is 1.25 mol C : 1.0 mol H. Multiply both subscripts by four to obtain the simplest whole-number ratio. The empirical formula is C5H4.
32.07 g S 1 mol S 32.07 g S 1 mol S
12.01 g C 5 mol C 60.05 g C 1 mol C
molar mass
1.008 g H 4 mol H 4.032 g H 1 mol H molar mass
64.09 g/mol
19.00 g F 6 mol F 114.00 g F 1 mol F 146.07 g/mol
146.07 g SF6 113 g SF6 0.775 mol SF6 1 mol SF6
d. the number of Cl ions in 14.5 g MgCl2.
128 g/mol n 2.00 64.09 g/mol
24.31 g Mg 1 mol Mg 24.31 g Mg 1 mol Mg
Molecular formula is C10H8.
35.45 g Cl 2 mol Cl 1 mol Cl
70.90 g Cl
molar mass
95.21 g/mol
161. Which of the following molecular formulas are
also empirical formulas: ethyl ether (C4H10O), aspirin (C9H8O4), butyl dichloride (C4H8Cl2), glucose (C6H12O6). The subscripts in C4H10O and C9H8O4 represent simplest whole-number ratios.
1 mol MgCl2 14.5 g MgCl2 95.21 g MgCl2 0.152 mol MgCl2 2(6.02 1023 ion Cl) 0.152 mol MgCl2 1 mol MgCl2 1.83 1023 ions Cl
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163. The graph shows the percent composition
of a compound containing carbon, hydrogen, oxygen, and nitrogen. How many grams of each element are present in 100 g of the compound?
C 19.68%
H 4.96% 52.42 g O, 22.95 g N, 19.68 g C, 4.96 g H
164. A party balloon was filled with 9.80 1022
atoms of helium. After 24 hours, 45% of the helium had escaped. How many atoms of helium remained? percent remaining 100% percent that escaped 100% 45% 55% 0.55 9.80 1022 atoms He 5.39 1022 atoms He Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
38.00 g F 2 mol F 38.00 g F 1 mol F molar mass CF2
50.01 g/mol
100.0 g/mol n 2.000 50.01 g/mol
N 22.95% O 52.42%
12.01 g C 1 mol C 12.01 g C 1 mol C
165. Tetrafluoroethylene, which is used in the
production of Teflon, is composed of 24.0% carbon and 76.0% fluorine and has a molar mass of 100.0 g/mol. Determine the empirical and molecular formulas of this compound. 1 mol C 24.0 g C 2.00 mol C 12.01 g C
The molar mass of the compound is two times greater than the molar mass of the empirical formula. Multiply the subscripts in the empirical formula by two to obtain the molecular formula C2F4.
166. Calculate the mass in grams of one atom of
lead. 1 mol Pb 1 atom Pb 6.02 1023 atoms Pb 207.02 g Pb 3.44 1022 g Pb 1 mol Pb
167. Diamond is a naturally occurring form of
carbon. If you have a 0.25-carat diamond, how many carbon atoms are present? (1 carat 0.200 g) 0.200 g C 0.25 carat 5 102 g C 1 carat 1 mol C 5 102 g C 12.01 g C 6.02 1023 atoms C 2.5 1021 atoms C 1 mol C
168. How many molecules of isooctane (C8H18)
1 mol F 76.0 g F 4.00 mol F 19.00 g F
are present in 1.00 L? (density of isooctane 0.680 g/mL)
2.00 mol C 1.00 mol C 2.00
12.01 g C 8 mol C 1 mol C
4.00 mol F 2.00 mol F 2.00
1.008 g H 18 mol H 18.18 g H 1 mol H
The simplest ratio is 1.00 mol C : 2.00 mol F. The empirical formula is CF2.
molar mass
96.08 g C
114.22 g/mol
0.680 g isooctane 103 mL 1.00 L 1L 1 mL 6.80 102 g isooctane
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1 mol isooctane 6.80 102 g isooctane 114.26 g isooctane 5.95 mol isooctane 5.95 mol isooctane 6.02 1023 molecules isooctane 1 mol isooctane 3.58 1024 molecules isooctane
169. Calculate the number of molecules of water in
a swimming pool that is 40.0 m in length, 20.0 m in width, and 5.00 m in depth. Assume that the density of water is 1.00 g/cm3. volume length width height volume 40.0 m 20.0 m 5.00 m 4.00 103 m3 1 g H2O (102 cm)3 4.00 103 m3 3 1m 1 cm3 4.00 109 g H2O 1 mol H2O 4.00 1032 g H2O 18.02 g H2O 6.02 1023 molecules H2O 1 mol H2O 1.34
1032
molecules H2O
Thinking Critically 170. Analyze and Conclude A mining company has two possible sources of copper: chalcopyrite (CuFeS2) and chalcocite (Cu2S). If the mining conditions and the extraction of copper from the ore were identical for each of the ores, which ore would yield the greater quantity of copper? Explain your answer. Chalcopyrite (CuFeS2) is 34.6% copper by mass (determined from percent composition) and chalcocite (Cu2S) is 79.9% copper by mass. Chalcocite would yield the greater quantity of copper because the ore has the greater percentage copper by mass.
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171. Designing an Experiment Design an experi-
ment that can be used to determine the amount of water in alum (KAl(SO4)2xH2O). Determine and record the mass of an empty evaporating dish. Add about 2 g of the hydrate. Measure and record the mass. Heat the evaporating dish gently for 5 minutes, and strongly for another 5 minutes to evaporate all the water. Allow the dish to cool, and measure and record the mass. Determine the masses of the anhydrous solid and the water lost. Calculate the number of moles of anhydrous compound and water. Determine the ratio of moles of water to moles of anhydrous compound. Use the wholenumber ratio of the moles as the coefficient of H2O in the formula.
172. Concept Mapping Design a concept map that
illustrates the mole concept. Include moles, Avogadro’s number, molar mass, number of particles, percent composition, empirical formula, and molecular formula. Concept maps will vary.
173. Communicating If you use the expression
“a mole of nitrogen,” is it perfectly clear what you mean, or is there more than one way to interpret the expression? Explain. How could you change the expression to make it more precise? The phrase could be interpreted to mean a mole of nitrogen atoms or a mole of nitrogen molecules. Specify atoms or molecules.
Writing in Chemistry 174. Octane ratings are used to identify certain grades of gasoline. Research octane rating and prepare a pamphlet for consumers identifying the different types of gasoline, the advantages of each, and when each grade is used. Student pamphlets will vary but should discuss how octane ratings are found. Octane rating of 100% means pure octane. Octane rating of 90% means 90% octane and 10% heptane.
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175. Research the life of the Italian chemist
c. Aqueous solutions of sulfuric acid and
Amedeo Avogadro (1776–1856) and how his work led scientists to the number of particles in a mole. Students should mention Avogadro’s hypothesis. Avogadro formulated his hypothesis as an explanation for earlier works by Gay-Lussac and Ritter. His ideas were rejected by chemists of his day but were revived later by the Italian chemist Stanislao Cannizzaro. Avogadro died before seeing his ideas accepted.
Cumulative Review Refresh your understanding of previous chapters by answering the following. 176. Express the following answers with the correct number of significant figures. (Chapter 2) a. 18.23 456.7
potassium hydroxide undergo a double replacement reaction. H2SO4(aq) 2KOH(aq) 0 K2SO4(aq) 2H2O(l)
179. How can you tell if a chemical equation is
balanced? (Chapter 10) There are equal numbers of each kind of atom on both sides of the equation.
Standardized Test Practice Chapter 11 page 351
Interpreting Graphs Use the graph to answer questions 1–4. Percent Composition of Some Organic Compounds
438.5
323
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
c. (82.44 4.92) 0.125 406
Percent by mass
b. 4.233 0.0131
60 50 52.2 40
number. How do these two numbers compare for isotopes of an element? (Chapter 4) Atomic number equals the number of protons. Mass number equals the number of protons plus the number of neutrons. Two isotopes of an element will have the same atomic number but different mass numbers.
178. Write balanced equations for the following
reactions. (Chapter 10) a. Magnesium metal and water combine to form solid magnesium hydroxide and hydrogen gas. Mg(s) 2H2O(l) 0 Mg(OH)2(s) H2(g)
b. Dinitrogen tetroxide gas decomposes into
nitrogen dioxide gas.
53.3
54.5
54.5
40.0
36.4
36.4
34.8
30 20 10
177. Distinguish between atomic number and mass
%C %H %O
0
13.0 6.7 Ethanol
9.1
9.1
Formaldehyde Acetaldehyde Butanoic acid
Compound name
1. Acetaldehyde and butanoic acid must have the
same _____ . a. molecular formula b. empirical formula c. molar mass d. chemical properties Acetaldehyde and butanoic acid must have the same empirical formula because they have the same number of grams and therefore the same number of moles of each element in a 100.0 g sample. b
N2O4(g) 0 2NO2(g)
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2. If the molar mass of butanoic acid is 88.1 g/mol,
then what is its molecular formula? a. CH2O b. C2H4O c. C6HO2 d. C4H8O2 Determine the empirical formula. Assume a 100.0 g sample. 1 mol C 54.5 g C 4.54 mol C 12.01 g C 1 mol H 9.1 g H 9.03 mol H 1.008 g H 1 mol O 36.4 g O 2.28 mol O 16.00 g O Determine the mole ratios. 4.54 mol C 1.99 2 mol C 2.28 9.03 mol H 3.96 4 mol H 2.28 2.28 mol O 1.00 1 mol O 2.28 The empirical formula for butanoic acid is C2H4O. The mass of the empirical formula is 44.06 g/mol. 88.1 g/mol 2.00 44.06 g/mol The molecular formula for butanoic acid (C2H4O)2 C4H8O2. d
3. What is the empirical formula of ethanol? a. b. c. d.
C4HO3 C52H13O35 C2H6O C4H13O2
1 mol C 52.2 g C 4.35 mol C 12.01 g C
SOLUTIONS MANUAL
Determine the mole ratios. 4.35 mol C 2.00 mol C 2 mol C 2.18 12.9 mol H 5.92 mol H 6 mol H 2.18 2.18 mol O 1.00 mol O 1 mol O 2.18 The empirical formula for ethanol is C2H6O. c
4. The empirical formula of formaldehyde is the
same as its molecular formula. How many grams are in 2.000 moles of formaldehyde? a. 30.00 g b. 60.06 g c. 182.0 g d. 200.0 g Determine the empirical formula for formaldahyde. 1 mol C 40.0 g C 3.33 mol C 12.01 g C 1 mol H 6.7 g H 6.65 mol H 1.008 g H 1 mol O 53.3 g O 3.33 mol O 16.00 g O 3.33 mol C 1.00 mol C 1 mol C 3.33 6.65 mol H 2.00 mol H 2 mol H 3.33 3.33 mol O 1.00 mol O 1 mol O 3.33 The empirical formula for formaldahyde is CH2O. Molar mass of CH2O 30.03 g/mol 30.03 g 2.000 mol 60.06 g formaldahyde 1 mol b
1 mol H 13.0 g H 12.9 mol H 1.008 g H 1 mol O 34.8 g O 1 mol O 16.00 g O
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5. A mole is all of the following EXCEPT _____ . a. the atomic or molar mass of an element or
compound b. Avogadro’s number of molecules of a compound c. the number of atoms in exactly 12 g of pure 12C d. the SI measurement unit for the amount of a substance A mole is not a mass so a is the answer. a
6. How many atoms are in 0.625 moles of Ge
(atomic mass 72.59 amu)? a. 2.73 1025 b. 6.99 1025 c. 3.76 1023 d. 9.63 1023 atoms Ge 6.02 0.625 mol Ge 1 mol Ge 1023
3.76 1023 atoms Ge
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
c
7. The molar mass of fluorapatite (Ca5(PO4)3F)
is _____ . a. 314 g/mol b. 344 g/mol c. 442 g/mol d. 504 g/mol 40.08 g Ca 5 mol Ca 200.40 g Ca 1 mol Ca 30.97 g P 3 mol P 1 mol P
92.91 g P
16.00 g O 12 mol O 192.00 g O 1 mol O 19.00 g F 1 mol F 1 mol F
19.00 g F
molar mass 504.31 g/mol d
8. How many moles of cobalt(III) titanate
(Co2TiO4) are in 7.13 g of the compound? a. 2.39 101 mol b. 3.14 102 mol c. 3.22 101 mol d. 4.17 102 mol Find the molar mass of Co2TiO4. 58.93 g Co 2 mol Co 117.86 g Co 1 mol Co 44.88 g Ti 1 mol Ti 44.88 g Ti 1 mol Ti 16.00 g O 4 mol O 64.00 g O 1 mol O Molar mass Co2TiO4 226.74 g/mol 1 mol Co2TiO4 7.13 g Co2TiO4 227 g Co2TiO4 3.14 102 mol b
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9. Magnesium sulfate (MgSO4) is often added to
water-insoluble liquid products of chemical reactions to remove any unwanted water. MgSO4 readily absorbs water to form two different hydrates. One of these hydrates is found to contain 13.0% H2O and 87.0% MgSO4. What is the name of this hydrate? a. magnesium sulfate monohydrate b. magnesium sulfate dihydrate c. magnesium sulfate hexahydrate d. magnesium sulfate heptahydrate
SOLUTIONS MANUAL
10. The mass of one molecule of barium hexafluo-
rosilicate (BaSiF6) is _____ . a. 1.68 1026 g b. 2.16 1021 g c. 4.64 1022 g d. 6.02 1023 g Determine the molar mass of BaSiF6. 137.33 g Ba 1 mol Ba 137.33 g Ba 1 mol Ba 28.09 g Si 1 mol Si 1 mol Si
28.09 g Si
Molar mass MgSO4 120.38 g/mol
19.00 g F 6 mol F 1 mol F
114.00 g F
Molar mass H2O 18.02 g/mol
Molar mass
279.42 g/mol BaSiF6
1 mol MgSO4 87.0 g MgSO4 120.38 g MgSO4
279.42 g BaSiF6 1 mol BaSiF6 1 mol BaSiF6 6.02 1023 molecules BaSiF6
Assume a 100.0 g sample of the hydrate. Then, the sample contains 87.0 g MgSO4 and 13.0 g H2O.
0.723 mol MgSO4
4.64 1022 g/molecule BaSiF6
1 mol H2O 13.0 g H2O 0.721 mol H2O 18.02 g H2O
c
0.721 mol H2O 0.997 1 1 1 0.723 mol MgSO4 Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
The hydrate is MgSO4H2O, magnesium sulfate monohydrate a
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Chemistry: Matter and Change • Chapter 11
Solutions Manual
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