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CHAPTER 7
QUANTITATIVE COMPOSITION OF COMPOUNDS
SOLUTIONS TO REVIEW QUESTIONS
1. A mole is an amount of substance containing the same number of atoms as there are atoms in exactly 12 g
of carbon-12.
It is Avogadro’s number (6.022 1023) of anything (atoms, molecules, ping-pong balls, etc.).
2. A mole of gold (197.0 g) has a higher mass than a mole of potassium (39.10 g).
3. Both samples (Au and K) contain the same number of atoms. (6.022 1023).
4. A mole of gold atoms contains more electrons than a mole of potassium atoms, as each Au atom has
79 e, while each K atom has only 19 e.
5. 6.022 1023
6. There are Avogadro’s number of particles in one mole of substance.
7. 1 mole of ozone has the greater number of oxygen atoms.
8. The molar mass of an element is the mass of one mole (or 6.022 1023 atoms) of that element.
9. No. Avogadro’s number is a constant. The mole is defined as Avogadro’s number of C-12 atoms.
Changing the atomic mass to 50 amu would change only the size of the atomic mass unit, not Avogadro’s
number.
10. (a)
(b)
(c)
(d)
(e)
A mole of oxygen atoms (O) contains 6.022 1023 atoms.
A mole of oxygen molecules (O2) contains 6.022 1023 molecules.
A mole of oxygen molecules (O2) contains 1.204 1024 atoms.
A mole of oxygen atoms (O) has a mass of 16.00 grams.
A mole of oxygen molecules (O2) has a mass of 32.00 grams.
11. 6.022 1023 molecules in one molar mass of H2SO4.
4.215 1024 atoms in one molar mass of H2SO4.
12. Either the chemical formula or experimental data giving the mass of the component elements in a sample.
13. There is 56.2% oxygen. (Remember the total percentage of all components must equal 100%.)
total mass of the element
14.
ð100Þ ¼ mass percent of the element
molar mass of the element
15. Choosing 100.0 g of a compound allows us to simply drop the % sign and use grams instead of percent.
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- Chapter 7 16. C4H6 and C8H12 both have the empirical formula C2H3.
17. The molecular formula represents the total number of atoms of each element in a molecule. The
empirical formula represents the lowest number ratio of atoms of each element in a molecule.
18. The molar mass is the most useful additional information that can be used to determine the molecular
formula of a compound from its empirical formula.
19. This formula tells us the number of units of the empirical formula found in the molecule.
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- Chapter 7 -
SOLUTIONS TO EXERCISES
1. Molar masses
(a)
KBr
1
1
K
Br
39.10 g
79.90 g
119.0
(b)
Na2SO4
2
1
4
Na
S
O
45.98 g
32.07 g
64.00 g
142.1 g
(c)
Pb(NO3)2
1
2
6
Pb
N
O
207.2 g
28.02 g
96.00 g
331.2 g
(d)
C2H5OH
2
6
1
C
H
O
24.02 g
6.048 g
16.00 g
46.07 g
(e)
HC2H3O2
4
2
2
H
C
O
4.032 g
24.02 g
32.00 g
60.05 g
(f)
Fe3O4
3
4
Fe
O
167.6 g
64.00 g
231.6 g
(g)
C12H22O11
12
22
11
C
H
O
144.1 g
22.18 g
176.0 g
342.3 g
(h)
Al2(SO4)3
2
3
12
Al
S
O
53.96 g
96.21 g
192.0 g
342.2 g
(i)
(NH4)2HPO4
9
2
1
4
H
N
P
O
9.072 g
28.02 g
30.97 g
64.00 g
132.1 g
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- Chapter 7 2. Molar masses
(a)
NaOH
1
1
1
Na
O
H
22.99 g
16.00 g
1.008 g
40.00 g
(b)
Ag2CO3
2
1
3
Ag
C
O
215.8 g
12.01 g
48.00 g
275.8 g
(c)
Cr2O3
2
3
Cr
O
104.0 g
48.00 g
152.0 g
(d)
(NH4)2CO3
2
8
1
3
N
H
C
O
28.02 g
8.064 g
12.01 g
48.00 g
96.09 g
(e)
Mg(HCO3)2
1
2
2
6
Mg
H
C
O
24.31 g
2.016 g
24.02 g
96.00 g
146.3 g
(f)
C6H5COOH
7
6
2
C
H
O
84.07 g
6.048 g
32.00 g
122.1 g
(g)
C6H12O6
6
12
6
C
H
O
72.06 g
12.10 g
96.00 g
180.2 g
(h)
K4Fe(CN)6
4
1
6
6
K
Fe
C
N
156.4 g
55.85 g
72.06 g
84.06 g
368.4 g
(i)
BaCl22 H2O
1
2
4
2
Ba
Cl
H
O
137.3
70.90
4.032
32.00
244.2
- 54 -
g
g
g
g
g
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- Chapter 7 3. Moles of atoms. 1 mol Zn
(a) ð22:5 g ZnÞ
¼ 0:344 mol Zn
65:39 g Zn
1 mol Mg
¼ 2:83 102 mol Mg
(b) ð0:688 g MgÞ
24:31 g Mg
1 mol Cu
22
¼ 7:5 102 mol Cu
(c)
4:5 10 atoms Cu
6:022 1023 atoms Cu
1 mol Co
(d) ð382 g CoÞ
¼ 6:48 mol Co
58:93 g Co
1 mol Sn
¼ 4:6 104 mol Sn
(e) ð0:055 g SnÞ
118:7 g Sn
2 atoms N
1 mol N atoms
24
¼ 28 mol N atoms
(f)
8:5 10 molecules N2
1 molecule N2 6:022 1023 atoms N
4. Number of moles. 1 mol NaOH
¼ 0:625 mol NaOH
(a) ð25:0 g NaOHÞ
40:00 g NaOH
1 mol Br2
¼ 0:275 mol Br2
(b) ð44:0 g Br2 Þ
159:8 g Br2
1 mol MgCl2
(c) ð0:684 g MgCl2 Þ
¼ 7:18 103 mol MgCl2
95:21 g MgCl2
1 mol CH3 OH
(d) ð14:8 g CH3 OHÞ
¼ 0:462 mol CH3 OH
32:04 g CH3 OH
1 mol Na2 SO4
¼ 2:03 102 mol Na2 SO4
(e) ð2:88 g Na2 SO4 Þ
142:1 g Na2 SO4
453:6 g
1 mol ZnI2
(f) ð4:20 lb ZnI2 Þ
¼ 5:97 mol ZnI2
1 lb
319:2 g ZnI2
5. Number of grams. 197:0 g Au
¼ 108 g Au
(a) ð0:550 mol AuÞ
1 mol Au
18:02 g H2 O
(b) ð15:8 mol H2 OÞ
¼ 285 g H2 O
mol H2 O
70:90 g Cl2
(c) ð12:5 mol Cl2 Þ
¼ 886 g Cl2
mol Cl2
80:05 g NH4 NO3
(d) ð3:15 mol NH4 NO3 Þ
¼ 252 g NH4 NO3
mol NH4 NO3
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- Chapter 7 6. Number of grams.
98:09 g H2 SO4
4
¼ 0:0417 g H2 SO4
(a)
4:25 10 mol H2 SO4
mol H2 SO4
1 mol
153:8 g CCl4
(b)
4:5 1022 molecules CCl4
¼ 11 g CCl4
mol CCl4
6:022 1023 molecules
47:87 g Ti
(c) ð0:00255 mol TiÞ
¼ 0:122 g Ti
mol Ti
32:07 g S
¼ 8:0 107 g S
(d)
1:5 1016 atoms S
6:022 1023 atoms S
7. Number of molecules
6:022 1023 molecules
¼ 1:5 1024 molecules S8
(a) ð2:5 mol S8 Þ
mol
6:022 1023 molecules
(b) ð7:35 mol NH3 Þ
¼ 4:43 1024 molecules NH3
mol
6:022 1023 molecules
(c) ð17:5 g C2 H5 OHÞ
¼ 2:29 1023 molecules C2 H5 OH
46:07 g C2 H5 OH
6:022 1023 molecules
¼ 1:91 1024 molecules Cl2
(d) ð225 g Cl2 Þ
70:90 g Cl2
8. Number of molecules
6:022 1023 molecules
¼ 5:8 1024 molecules C2 H4
(a) ð9:6 mol C2 H4 Þ
mol
6:022 1023 molecules
(b) ð2:76 mol N2 OÞ
¼ 1:66 1024 molecules N2 O
mol
6:022 1023 molecules
(c) ð23:2 g CH3 OHÞ
¼ 4:36 1023 molecules CH3 OH
32:04 g CH3 OH
6:022 1023 molecules
¼ 1:28 1023 molecules CCl4
(d) ð32:7 g CCl4 Þ
153:8 g CCl4
9. Number of atoms
(a)
(b)
(c)
ð25 molecules P2 O5 Þ
7 atoms
1 molecule P2 O5
¼ 1:8 102 atoms
6:022 1023 molecules
2 atoms
ð3:62 mol O2 Þ
¼ 4:36 1024 atoms
mol O2
1 molecule
6:022 1023 molecules
3 atoms
ð12:2 mol CS2 Þ
¼ 2:20 1025 atoms
mol CS2
1 molecule
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- Chapter 7 6:022 1023 atoms
(d) ð1:25 g NaÞ
¼ 3:27 1022 atoms
22:99 g Na
6:022 1023 molecules
3 atoms
¼ 1:1 1023 atoms
(e) ð2:7 g CO2 Þ
44:01 g CO2
1 molecule
6:022 1023 molecules
5 atoms
¼ 4:7 1022 atoms
(f) ð0:25 g CH4 Þ
16:04 g CH4
1 molecule
10. Number of atoms.
(a)
(b)
(c)
(d)
(e)
(f)
8 atoms
ð2 molecules CH3 COOHÞ
¼ 16 atoms
1 molecule
6:022 1023 molecules CH3 COOH
8 atoms
ð0:75 mol C2 H6 Þ
¼ 3:6 1024 atoms
mol C2 H6
1 molecule
6:022 1023 molecules
3 atoms
ð25 mol H2 OÞ
¼ 4:5 1025 atoms
mol
1 molecule H2 O
6:022 1023 atoms
ð92:5 g AuÞ
¼ 2:83 1023 atoms
197:0 g Au
6:022 1023 molecules
4 atoms
ð75 g PCl3 Þ
¼ 1:3 1024 atoms
137:3 g PCl3
1 molecule
6:022 1023 molecules
24 atoms
ð15 g C6 H12 O6 Þ
¼ 1:2 1024 atoms
180:2 g C6 H12 O6
1 molecule
11. Number of grams.
(a) ð1 atom HeÞ
(b)
(c)
(d)
4:003 g
¼ 6:647 1024 g He
6:022 1023 atoms
12:01 g
¼ 2:991 1022 g C
ð15 atoms CÞ
6:022 1023 atoms
108:0 g
ð4 molecules N2 O5 Þ
¼ 7:175 1022 g N2 O5
6:022 1023 molecules
93:13 g
ð11 molecules C6 H5 NH2 Þ
¼ 1:701 1021 g C6 H5 NH2
6:022 1023 molecules
12. Number of grams.
(a) ð1 atom XeÞ
(b)
131:3 g
¼ 2:180 1022 g Xe
6:022 1023 atoms
35:45 g
¼ 1:295 1021 g Cl
ð22 atoms ClÞ
23
6:022 10 atoms
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- Chapter 7 -
(c)
(d)
13. (a)
(b)
(c)
(d)
14. (a)
(b)
(c)
(d)
ð9 molecules CH3 COOH
60:05 g
¼ 8:975 1022 g CH3 COOH
6:022 1023 molecules
116:1 g
¼ 2:892 1021 g C4 H4 O2 ðNH2 Þ2
15 molecules C4 H4 O2 ðNH2 Þ2
6:022 1023 molecules
1000 g
1 mol CO2
ð25 kg CO2 Þ
¼ 5:7 102 mol CO2
kg
44:01 g CO2
1 mol Pb
ð5 atoms PbÞ
¼ 8 1024 mol Pb
23
6:022 10 atoms Pb
6:022 1023 molecules O2
2 atoms O
ð6 mol O2 Þ
¼ 7 1024 atoms O
1 molecule O2
mol O2
123:9 g P4
ð25 molecules P4 Þ
¼ 5:1 1021 g P4
6:022 1023 molecules P4
ð275 atoms WÞ
1 mol W
¼ 4:57 1022 mol W
6:022 1023 atoms W
18:02 g H2 O
1 kg
ð95 mol H2 OÞ
¼ 1:7 kg H2 O
mol H2 O
1000 g
64:07 g SO2
ð12 molecules SO2 Þ
¼ 1:277 1021 g SO2
6:022 1023 molecules SO2
6:022 1023 molecules Cl2
2 atoms Cl
¼ 3:0 1025 atoms Cl
ð25 mol Cl2 Þ
1 molecule Cl2
mol Cl2
15. One molecule of tetraphosphorus
decoxide (P4O10) contains:
1 mol P4 O10
(a) ð1 molecule P4 O10 Þ
¼ 1:661 1024 mol P4 O10
6:022 1023 molecules P4 O10
283:9 g P4 O10
24
(b)
1:661 10 mol P4 O10
¼ 4:716 1022 g P4 O10
mol P4 O10
4 P atoms
(c) ð1 molecule P4 O10 Þ
¼ 4 atoms P
1 molecule P4 O10
10 atoms O
¼ 10 atoms O
(d) ð1 molecule P4 O10 Þ
1 molecule P4 O10
14 atoms
¼ 14 total atoms
(e) ð1 molecule P4 O10 Þ
1 molecule P4 O10
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- Chapter 7 16. 125 grams of disulfur decofluoride (S2F10) contains:
1 mol S2 F10
¼ 0:492 mol S2 F10
(a) ð125 g S2 F10 Þ
254:1 g S2 F10
6:022 1023 molecules
(b) ð0:492 mol S2 F10 Þ
¼ 2:96 1023 molecules S2 F10
mol
6:022 1023 molecules
12 atoms
(c) ð0:492 mol S2 F10 Þ
¼ 3:56 1024 total atoms
mol
1 molecule S2 F10
6:022 1023 molecules
2 S atoms
¼ 5:93 1023 atoms S
(d) ð0:492 mol S2 F10 Þ
mol
1 molecule S2 F10
6:022 1023 molecules
10 F atoms
(e) ð0:492 mol S2 F10 Þ
¼ 2:96 1024 atoms F
mol
1 molecule S2 F10
17. Atoms of hydrogen in:
(a)
(b)
(c)
ð25 molecules C6 H5 CH3 Þ
8 H atoms
¼ 2:0 102 atoms H
molecule C6 H5 CH3
6:022 1023 molecules
2 H atoms
¼ 4:2 1024 atoms H
ð3:5 mol H2 CO3 Þ
mol
molecule H2 CO3
6:022 1023 molecules
6 H atoms
ð36 g CH3 CH2 OHÞ
¼ 2:8 1024 atoms H
46:07 g
molecule CH3 CH2 OH
18. Atoms of hydrogen in:
(a)
(b)
(c)
6 atoms H
¼ 138 atoms H
1 molecule CH3 CH2 COOH
6:022 1023 molecules
3 atoms H
ð7:4 mol H3 PO4 Þ
¼ 1:3 1025 atoms H
mol
1 molecule H3 PO4
7 atoms H
6:022 1023 molecules
ð57 g C6 H5 ONH2 Þ
¼ 2:2 1024 atoms H
1 molecule C6 H5 ONH2
109:1 g
ð23 molecules CH3 CH2 COOHÞ
19. The number of grams of:
(a)
silver in 25.0 g AgBr
107:9 g Ag
ð25:0 g AgBrÞ
¼ 14:4 g Ag
187:8 g AgBr
(b)
nitrogen in 6.34 mol (NH4)3PO4
42:03 g N
6:34 mol ðNH4 Þ3 PO4
¼ 266 g N
mol ðNH4 Þ3 PO4
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- Chapter 7 (c)
oxygen in 8.45 1022 molecules SO3
The conversion is: molecules SO3 ! mol SO3 ! g O
1 mol
48:00 g O
22
8:45 10 molecules SO3
¼ 6:74 g O
mol SO3
6:022 1023 molecules
20. The number of grams of:
(a) chlorine in 5.00 g PbCl2
70:90 g Cl
ð5:00 g PbCl2 Þ
¼ 1:27 g Cl
278:1 g PbCl2
(b)
hydrogen in 4.50 mol H2SO4
2:016 g H
ð4:50 mol H2 SO4 Þ
¼ 9:07 g H
1 mol H2 SO4
(c)
hydrogen in 5.45 1022 molecules NH3
The conversion is: molecules NH3 ! moles NH3 ! g H
1 mol
3:024 g H
22
5:45 10 molecules NH3
¼ 0:274 g H
mol NH3
6:022 1023 molecules
21. Percent composition
(a)
(b)
NaBr
KHCO3
Na
Br
K
H
3O
C
22.99 g
79.90 g
102.9 g
39.10 g
1.008 g
48.00 g
12.01 g
100.1 g
22:99 g
ð100Þ ¼ 22:34% Na
102:9 g
79:90 g
ð100Þ ¼ 77:65% Br
102:9 g
39:10 g
ð100Þ ¼ 39:06% K
100:1 g
1:008 g
ð100Þ ¼ 1:007% H
100:1 g
12:01 g
ð100Þ ¼ 12:00% C
100:1 g
48:00 g
ð100Þ ¼ 47:95% O
100:1 g
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- Chapter 7 (c)
FeCl3
(d)
SiCl4
(e)
Al2(SO4)3
(f)
AgNO3
Fe
3 Cl
Si
4 Cl
2 Al
3 S
12 O
Ag
N
3O
55.85 g
106.4 g
162.3 g
28.09 g
141.8 g
169.9 g
53.96 g
96.21 g
192.0 g
342.2 g
107.9
14.01
48.00
169.9
g
g
g
g
55:85 g
ð100Þ ¼ 34:41% Fe
162:3 g
106:4 g
ð100Þ ¼ 65:56% Cl
162:3 g
28:09 g
ð100Þ ¼ 16:53% Si
169:9 g
141:8 g
ð100Þ ¼ 83:46% Cl
169:9 g
53:96 g
ð100Þ ¼ 15:77% Al
342:2 g
96:21 g
ð100Þ ¼ 28:12% S
342:2 g
192:0 g
ð100Þ ¼ 56:11% O
342:2 g
107:9 g
ð100Þ ¼ 63:51% Ag
169:9 g
14:01 g
ð100Þ ¼ 8:246% N
169:9 g
48:00 g
ð100Þ ¼ 28:25% O
169:9 g
22. Percent composition
(a)
ZnCl2
(b)
NH4C2H3O2
Zn
2 Cl
N
7H
2C
2O
65.39 g
70.90 g
136.3 g
14.01
7.056
24.02
32.00
77.09
g
g
g
g
g
65:39 g
ð100Þ ¼ 47:98% Zn
136:3 g
70:90 g
ð100Þ ¼ 52:02% Cl
136:3 g
14:01 g
ð100Þ ¼ 18:17% N
77:09 g
7:056 g
ð100Þ ¼ 9:153% H
77:09 g
24:02 g
ð100Þ ¼ 31:16% C
77:09 g
32:00 g
ð100Þ ¼ 41:51% O
77:09 g
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- Chapter 7 (c)
MgP2O7
Mg
2P
24.31 g
61.94 g
7O
112.0 g
198.3 g
(d)
(NH4)2SO4
2N
8H
S
4O
28.02
8.064
32.07
64.00
132.2
g
g
g
g
g
(e)
Fe(NO3)3
Fe
3N
9O
55.85
42.03
144.0
241.9
g
g
g
g
(f)
ICl3
I
3 Cl
24:31 g
ð100Þ ¼ 12:26% Mg
198:3 g
61:94 g
ð100Þ ¼ 31:24% P
198:3 g
112:0 g
ð100Þ ¼ 56:48% O
198:3 g
28:02 g
ð100Þ ¼ 21:20% N
132:2 g
8:064 g
ð100Þ ¼ 6:100% H
132:2 g
32:07 g
ð100Þ ¼ 24:26% S
132:2 g
64:00 g
ð100Þ ¼ 48:41% O
132:2 g
55:85 g
ð100Þ ¼ 23:09% Fe
241:9 g
42:03 g
ð100Þ ¼ 17:37% N
241:9 g
144:0 g
ð100Þ ¼ 59:53% O
241:9 g
126:9 g
ð100Þ ¼ 54:39% I
233:3 g
106:4 g
ð100Þ ¼ 45:61% Cl
233:3 g
126.9 g
106.4 g
233.3 g
23. Percent of iron
55.85 g
16.00 g
71.85 g
(a)
FeO
Fe
O
(b)
Fe2O3
2 Fe
3O
111.7 g
48.00 g
159.7 g
(c)
Fe3O4
3 Fe
4O
167.6 g
64.00 g
231.6 g
(d)
K4Fe(CN)6
Fe
4K
55.85 g
156.4 g
- 62 -
55:85 g
ð100Þ ¼ 77:73% Fe
71:85 g
111:7 g
ð100Þ ¼ 69:94% Fe
159:7 g
167:6 g
ð100Þ ¼ 72:37% Fe
231:6 g
55:85 g
ð100Þ ¼ 15:16% Fe
368:4 g
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Page 63
- Chapter 7 6C
6N
72.06 g
84.06 g
368.4 g
24. Percent chlorine
39.10 g
35.45 g
74.55 g
(a)
KCl
K
Cl
(b)
BaCl2
Ba
2 Cl
137.3 g
70.90 g
208.2 g
(c)
SiCl4
Si
4 Cl
28.09 g
141.8 g
169.9 g
(d)
LiCl
Li
Cl
6.941 g
35.45 g
42.39 g
35:45 g
ð100Þ ¼ 47:55% Cl
74:55 g
70:90 g
ð100Þ ¼ 34:05% Cl
208:2 g
141:8 g
ð100Þ ¼ 83:46% Cl
169:9 g
35:45 g
ð100Þ ¼ 83:63% Cl
42:39 g
Highest % Cl is LiCl; lowest % Cl is in BaCl2
25. Percent composition
39:54 g Si
ð100Þ ¼ 54:05% Si
73:16 g
33:62 g Ba
ð100Þ ¼ 45:95% Ba
73:16 g
73.16 g barium silicide
233.62 g Ba
39.54 g Si
26. Percent composition
7.52
3.09
0.453
20.513
3.46
g
g
g
g
g
ajoene
S
H
O
C
3:09 g S
ð100Þ ¼ 41:1 % S
7:52 g
0:453 g H
ð100Þ ¼ 6:02 % H
7:52 g
0:513 g O
ð100Þ ¼ 6:82 % O
7:52 g
3:46 g C
ð100Þ ¼ 46:0 % C
7:52 g
27. (a)
(b)
(c)
H2O has the higher percent Hydrogen
N2O3 has the lower percent Nitrogen
Both have the same percent Oxygen
28. (a)
(b)
(c)
KClO3 is lower.
KHSO4 is higher.
Na2CrO4 is lower.
(Because a K atom has more mass than a Na atom.)
(Because a H atom has less mass than a K atom.)
(Because only one Cr atom is present.)
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- Chapter 7 29. Empirical formulas from percent composition.
(a) Step 1. Express each element as grams/100 g material.
63:6% N ¼ 63:6 g N=100 g material
36:4% O ¼ 36:4 g O=100 g material
Step 2. Calculate the relative moles of each element.
1 mol N
ð63:6 g NÞ
¼ 4:54 mol N
14:01 g N
1 mol O
ð36:4 g OÞ
¼ 2:28 mol O
16:00 g O
Step 3. Change these moles to whole numbers by dividing each by the smaller number.
4:54 mol N
¼ 1:99 mol N
2:28
2:28 mol O
¼ 1:00 mol O
2:28
The simplest ratio of N:O is 2:1. The empirical formula, therefore, is N2O.
(b)
46.7% N, 53.3% O
1 mol N
ð46:7 g NÞ
¼ 3:33 mol N
14:01 g N
1 mol O
ð53:3 g NÞ
¼ 3:33 mol O
16:00 g O
3:33 mol N
¼ 1:00 mol N
3:33
3:33 mol O
¼ 1:00 mol O
3:33
The empirical formula is NO.
(c)
25.9% N, 74.1% O
1 mol N
ð25:9 g NÞ
¼ 1:85 mol N
14:01 g N
1 mol O
ð74:1 g OÞ
¼ 4:63 mol O
16:00 g O
1:85 mol N
¼ 1:00 mol N
1:85
4:63 mol O
¼ 2:50 mol O
1:85
Since these values are not whole numbers, multiply each by 2 to change them to whole numbers.
ð1:00 mol NÞð2Þ ¼ 2:00 mol N; ð2:5 mol OÞð2Þ ¼ 5:00 mol O
The empirical formula is N2O5.
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- Chapter 7 (d)
43.4% Na, 11.3% C, 45.3% O
1 mol Na
ð43:4 g NaÞ
¼ 1:89 mol Na
22:99 g Na
1 mol C
ð11:3 g CÞ
¼ 0:941 mol C
12:01 g C
1 mol O
ð45:3 g OÞ
¼ 2:83 mol O
16:00 g O
1:89 mol Na
¼ 2:01 mol Na
0:941
0:941 mol C
¼ 1:00 mol C
0:941
2:83 mol O
¼ 3:00 mol O
0:941
The empirical formula is Na2CO3.
(e)
18.8% Na, 29.0% Cl, 52.3% O
1 mol Na
ð18:8 g NaÞ
¼ 0:818 mol Na
22:99 g Na
1 mol Cl
ð29:0 g ClÞ
¼ 0:818 mol Cl
35:45 g Cl
1 mol O
ð52:3 g OÞ
¼ 3:27 mol O
16:00 g O
0:818 mol Na
¼ 1:00 mol Na
0:818
0:818 mol Cl
¼ 1:00 mol Cl
0:818
3:27 mol O
¼ 4:00 mol O
0:818
The empirical formula is NaClO4.
(f)
72.02% Mn, 27.98% O
1 mol Mn
ð72:02 g MnÞ
¼ 1:311 mol Mn
54:94 g Mn
1 mol O
ð27:98 g OÞ
¼ 1:749 mol O
16:00 g O
1:311 mol Mn
¼ 1:000 mol Mn
1:311
1:749 mol O
¼ 1:334 mol O
1:311
Multiply both values by 3 to give whole numbers.
ð1:000 mol MnÞð3Þ ¼ 3:000 mol Mn; ð1:334 mol OÞð3Þ ¼ 4:002 mol O
The empirical formula is Mn3O4.
30. Empirical formulas from percent composition.
(a)
64.1% Cu, 35.9% Cl
1 mol Cu
ð64:1 g CuÞ
¼ 1:01 mol Cu
63:55 g Cu
1 mol Cl
ð35:9 g ClÞ
¼ 1:01 mol Cl
35:45 g Cl
The empirical formula is CuCl.
- 65 -
1:01 mol Cu
¼ 1:00 mol Cu
1:01
1:01 mol Cl
¼ 1:00 mol Cl
1:01
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- Chapter 7 (b)
47.2% Cu, 52.8% Cl
1 mol Cu
ð47:2 g CuÞ
¼ 0:743 mol Cu
63:55 g Cu
1 mol Cl
ð52:8 g ClÞ
¼ 1:49 mol Cl
35:45 g Cl
0:743 mol Cu
¼ 1:00 mol Cu
0:743
1:49 mol Cl
0:743
¼ 2:01 mol Cl
The empirical formula is CuCl2.
(c)
51.9% Cr, 48.1% S
1 mol Cr
ð51:9 g CrÞ
¼ 0:998 mol Cr
52:00 g Cr
1 mol S
ð48:1 g SÞ
¼ 1:50 mol S
32:07 g S
0:998 mol Cr
¼ 1:00 mol Cr
0:998
1:50 mol S
0:998
¼ 1:50 mol S
Multiply both values by 2 to give whole numbers.
ð1:00 mol CrÞð2Þ ¼ 2:00 mol Cr; ð1:50 mol SÞð2Þ ¼ 3:00 mol S
The empirical formula is Cr2S3.
(d)
55.3% K, 14.6% P, 30.1% O
1 mol K
ð55:3 g KÞ
¼ 1:41 mol K
39:10 g K
1 mol P
ð14:6 g PÞ
¼ 0:471 mol P
30:97 g P
1 mol O
ð30:1 g OÞ
¼ 1:88 mol O
16:00 g O
1:41 mol K
¼ 2:99 mol K
0:471
1:471 mol P
¼ 1:00 mol P
0:471
1:88 mol O
¼ 3:99 mol O
0:471
The empirical formula is K3PO4.
(e)
38.9% Ba, 29.4% Cr, 31.7% O
1 mol Ba
ð38:9 g BaÞ
¼ 0:283 mol Ba
137:3 g Ba
1 mol Cr
ð29:4 g CrÞ
¼ 0:565 mol Cr
52:00 g Cr
1 mol O
ð31:7 g OÞ
¼ 1:98 mol O
16:00 g O
The empirical formula is BaCr2O7.
- 66 -
0:283 mol Ba
¼ 1:00 mol Ba
0:283
0:565 mol Cr
¼ 2:00 mol Cr
0:283
1:98 mol O
¼ 7:00 mol O
0:283
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- Chapter 7 (f)
3.99% P, 82.3% Br, 13.7% Cl
1 mol P
ð3:99 g PÞ
¼ 0:129 mol P
30:97 g P
1 mol Br
ð82:3 g BrÞ
¼ 1:03 mol Br
79:90 g Br
1 mol Cl
ð13:7 g ClÞ
¼ 0:386 mol Cl
35:45 g Cl
0:129 mol P
¼ 1:00 mol P
0:129
1:03 mol Br
¼ 7:98 mol Br
0:129
0:386 mol Cl
¼ 2:99 mol Cl
0:129
The empirical formula is PBr8Cl3.
31. Empirical formula:
1 mol Zn
(a) ð26:08 g ZnÞ
¼ 0:3988 mol Zn
65:39 g
1 mol C
ð4:79 g CÞ
¼ 0:399 mol C
12:01 g
1 mol O
ð19:14 g OÞ
¼ 1:196 mol O
16:00 g
0:3988 mol Zn
¼ 1:00 mol Zn
0:3988
0:399 mol C
0:3988
1:196 mol O
0:3988
¼ 1:00 mol C
¼ 2:999 mol O
The empirical formula is ZnCO3
(b)
150.0 g compound
57:66 g C
7:26 g H
85:1 g Cl
1 mol C
ð57:66 g CÞ
¼ 4:801 mol C
12:01 g C
1 mol H
ð7:26 g HÞ
¼ 7:20 mol H
1:008 g H
1 mol Cl
ð85:1 g ClÞ
¼ 2:40 mol Cl
35:45 g
4:801 mol C
¼ 2:000 mol C
2:40
7:20 mol H
¼ 3:00 mol H
2:40
2:40 mol Cl
¼ 1:00 mol Cl
2:40
The empirical formula is C2H3Cl
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- Chapter 7 (c)
75.0 g Oxide 42.0 gV ¼ 33.0 g O
1 mol V
ð42:0 g VÞ
¼ 0:824 mol V
50:94 g V
1 mol O
ð33:0 g OÞ
¼ 2:06 mol O
16:00 g O
0:824 mol V
¼ 1:00 mol V
0:824
2:06 mol O
¼ 2:50 mol O
0:824
Multiplying both by 2 gives the empirical formula V2O5
(d)
1 mol Ni
ð67:35 g NiÞ
¼ 1:148 mol Ni
58:69 g Ni
1 mol O
ð48:96 g OÞ
¼ 3:060 mol O
16:00 g O
1 mol P
ð23:69 g PÞ
¼ 0:7649 mol P
30:97 g P
1:148 mol Ni
¼ 1:501 mol Ni
0:7649
3:060 mol O
¼ 4:001 mol O
0:7649
0:7649 mol P
¼ 1:000 mol P
0:7649
Multiplying all by 2 gives the empirical formula Ni3O8P2
32. Empirical formula
1 mol C
(a)
ð55:08 g CÞ
¼ 4:586 mol C
12:01 g C
1 mol H
ð3:85 g HÞ
¼ 3:82 mol H
1:008 g H
1 mol Br
ð61:07 g BrÞ
¼ 0:7643 mol Br
79:90 g Br
4:586 mol C
0:7643 mol
¼ 6:000 mol C
3:82 mol H
¼ 5:00 mol H
0:7643
0:7643 mol Br
¼ 1:000 mol Br
0:7643
The empirical formula is C6H5Br
(b)
65.2 g compound 36.8 g Ag 12.1 g Cl ¼ 16.3 g O
0:341 mol Ag
1 mol Ag
ð36:8 g AgÞ
¼ 1:00 mol Ag
¼ 0:341 mol Ag
0:341
107:9 g Ag
1 mol Cl
0:341mol Cl
ð12:1 g ClÞ
¼ 0:341 mol Cl
¼ 1:00 mol Cl
35:45 g Cl
0:341
1 mol O
1:02 mol O
¼ 2:99 mol O
ð16:3 g OÞ
¼ 1:02 mol O
16:00 g O
0:341
The empirical formula is AgClO3
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- Chapter 7 (c)
(d)
25.25 g sulfide 12.99 gV ¼ 12.26 g S
1 mol V
ð12:99 g VÞ
¼ 0:2550 mol V
50:94 g V
1 mol S
ð12:26 g SÞ
¼ 0:3823 mol S
32:07 g S
0:2550 mol V
¼ 1:000 mol V
0:2550
0:3823 mol S
¼ 1:499 mol S
0:2550
Multiplying both by 2 gives the empirical formula V2S3
1 mol Zn
0:581 mol Zn
¼ 1:50 mol Zn
ð38:0 g ZnÞ
¼ 0:581 mol Zn
65:39 g
0:387
1 mol P
0:387 mol P
ð12:0 g PÞ
¼ 0:387 mol P
¼ 1:00 mol P
30:97 g
0:387
Multiplying both by 2 gives the empirical formula Zn3P2
33. 15.267 g sulfide 12.272 g Au ¼ 2.995 g S
1 mol Au
ð12:272 g AuÞ
¼ 0:06229 mol Au
197:0 g Au
1 mol S
ð2:995 g SÞ
¼ 0:09339 mol S
32:07 g S
0:06229 mol Au
¼ 1:000 mol Au
0:06229
0:09339 mol S
0:06229
¼ 1:499 mol S
Multiplying both by 2 gives the empirical formula Au2S3
34. 10.724 g oxide 7.143 g Ti ¼ 3.581 g O
1 mol Ti
ð7:143 g TiÞ
¼ 0:1492 mol Ti
47:88 g Ti
1 mol O
ð3:581 g OÞ
¼ 0:2238 mol O
16:00 g O
0:1492 mol Ti
¼ 1:000 mol Ti
0:1492
0:2238 mol O
¼ 1:500 mol O
0:1492
Multiplying both by 2 gives the empirical formula Ti2O3
35. 5:000 g compound ð0:6375 g C þ 0:1070 g HÞ ¼ 4:256 g S
1 mol C
0:05308 mol C
ð0:6375 g CÞ
¼ 0:05308 mol C
¼ 1:000 mol C
12:01 g C
0:05308 mol
1 mol H
0:1062 mol H
ð0:1070 g HÞ
¼ 0:1062 mol H
¼ 2:001 mol H
1:008 g H
0:05308 mol
1 mol S
0:1327 mol S
¼ 0:1327 mol S
¼ 2:500 mol S
ð4:256 g SÞ
32:07 g S
0:05308 mol
Multiplying by 2 gives the empirical formula C2H4S5
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- Chapter 7 36. Empirical formula
5.276 g compound 3.898 g Hg ¼ 1.378 g Cl
1 mol Hg
ð3:898 g HgÞ
¼ 0:01943 mol Hg
200:6 g Hg
1 mol Cl
ð1:378 g ClÞ
¼ 0:03887 mol Cl
35:45 g Cl
0:01943 mol Hg
¼ 1:000 mol Hg
0:01943
0:03887 mol Cl
¼ 2:001 mol Cl
0:01943
The empirical formula is HgCl2.
37. Empirical and molecular formula of traumatic acid.
63.13% C, 8.830% H, 28.03% O; molar mass ¼ 228 g
1 mol C
5:256 mol C
¼ 5:256 mol C
¼ 3:000 mol C
ð63:13 g CÞ
12:01 g C
1:752
1 mol H
8:760 mol H
ð8:830 g HÞ
¼ 8:760 mol H
¼ 5:000 mol H
1:008 g H
1:752
1 mol O
1:752 mol O
¼ 1:752 mol O
¼ 1:000 mol O
ð28:03 g OÞ
16:00 g O
1:752 mol
The empirical formula for traumatic acid is C3H5O. The empirical formula mass is 57 g.
molar mass
228
¼
¼4
empirical formula mass
57
The molecular formula is four times that of the empirical formula.
Molecular formula is (C3H5O)4 ¼ C12H20O4.
38. Empirical and molecular formulas of dixanthogen.
29.73% C, 4.16% H, 13.20% O, 52.91% S;
1 mol C
ð29:73 g CÞ
¼ 2:475 mol C
12:01 g C
1 mol H
ð4:16 g HÞ
¼ 4:13 mol H
1:008 g H
1 mol O
ð13:20 g OÞ
¼ 0:8250 mol O
16:00 g O
1 mol S
ð52:91 g SÞ
¼ 1:650 mol S
32:07 g S
molar mass ¼ 242.4 g
2:475 mol C
¼ 3:000 mol C
0:8250
4:13 mol H
¼ 5:01 mol H
0:8250
0:8250 mol O
¼ 1:000 mol O
0:8250
1:650 mol S
¼ 2:000 mol S
0:8250
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- Chapter 7 The empirical formula is C3H5OS2. The empirical formula mass is 121.2 g.
molar mass
242:4 g
¼
¼2
empirical formula mass 121:2 g
The molecular formula is twice that of the empirical formula
Molecular formula is (C3H5OS2)2 ¼ C6H10O2S4
39. Molecular formula of oxalic acid (ethanedioic acid)
26.7% C, 2.24% H, 71.1% O; molar mass ¼ 90.04
1 mol C
2:22 mol C
¼ 1:0 mol C
26:7 g C
¼ 2:22 mol C
12:01 g C
2:2
1 mol H
2:2 mol H
2:2 g H
¼ 2:2 mol H
¼ 1:0 mol H
1:008 g H
2:2
1 mol O
4:44 mol O
¼ 2:0 mol O
71:1 g O
¼ 4:44 mol O
16:00 g O
2:2
The empirical formula is CHO2, making the empirical formula mass 45.02 g.
molar mass
90:04 g
¼
¼2
mass of empirical formula 45:02 g
The molecular formula is twice that of the empirical formula.
Molecular formula ¼ (CHO2)2 ¼ C2H2O4
40. Molecular formula of butyric acid
54.5% C, 9.2% H, 36.3% O; molar mass ¼ 88.11
1 mol C
4:54 mol C
ð54:5 g CÞ
¼ 4:54 mol C
¼ 2:00 mol C
12:01 g C
2:27
1 mol H
9:1 mol H
ð9:2 g HÞ
¼ 9:1 mol H
¼ 4:0 mol H
1:008 g H
2:27
1 mol O
2:27 mol O
ð36:3 g OÞ
¼ 2:27 mol O
¼ 1:0 mol O
16:00 g O
2:27
The empirical formula is C2H4O, making the empirical formula mass 44.05 g.
molar mass
88:11 g
¼
¼2
mass of empirical formula 44:05 g
The molecular formula is twice that of the empirical formula.
Molecular formula ¼ (C2H4O)2 ¼ C4H8O2
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- Chapter 7 12:04 g
ð100Þ ¼ 30:45%
39:54 g
39:54 g 12:04 g
ð100Þ ¼ 69:55%
% oxygen ¼
39:54 g
41. % nitrogen ¼
12:04 g N
¼ 0:8594 mol N
14:01 g=mol
27:50 g O
moles of oxygen ¼
¼ 1:719 mol O
16:00 g=mol O
0:8594 mol
¼ 1:000
relative number of nitrogen atoms ¼
0:8594 mol
1:719 mol
relative number of oxygen atoms ¼
¼ 2:000
0:8594 mol
empirical formula ¼ NO2
molecular formula: (molar mass of NO2) x ¼ 92.02 g, 46.01 x ¼ 92.02, x ¼ 2
The molecular formula is twice the empirical formula.
molecular formula ¼ N2O4
empirical formula: moles of nitrogen ¼
42.
Total mass of C þ H þ O
30:21 g
ð100Þ
% carbon ¼
75:53 g
5:08 g
% hydrogen ¼
ð100Þ
75:53 g
40:24 g
ð100Þ
% oxygen ¼
75:53 g
¼ 30:21 g þ 5:08 g þ 40:24 g ¼ 75:53 g
¼ 40:0%
¼ 6:73%
¼ 53:3%
30:21 g C
12:01 g=mol
5:080 g H
moles of hydrogen ¼
1:008 g=mol
40:24 g O
moles of oxygen ¼
16:00 g=mol
2:515 mol
¼
relative number of carbon atoms ¼
2:515 mol
5:03 mol
relative number of hydrogen atoms ¼
¼
2:515 mol
2:515 mol
¼
relative number of oxygen atoms ¼
2:515 mol
empirical formula ¼ CH2 O
empirical formula: moles of carbon ¼
¼ 2:515 mol C
¼ 5:03 mol H
¼ 2:515 mol O
1:000
2:00
1:000
molecular formula: (molar mass of CH2O) x ¼ 180.18 g=mol,
ð30:03 g=molÞx ¼ 180:18 g=mol,
x¼
180:18 g=mol
¼6
30:03 g=mol
The molecular formula is six times the empirical formula.
molecular formula ¼ C6H12O6
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- Chapter 7 43. What is compound XYZ3
X:
ð0:4004Þð100:09 gÞ ¼ 40:08 g ðcalciumÞ
ð0:1200Þð100:09 gÞ ¼ 12:01 g ðcarbonÞ
48:00 g
Z:
ð0:4796Þð100:09 gÞ ¼ 48:00 g;
¼ 16:00 g ðoxygenÞ
3
Elements determined from atomic masses in the periodic table.
XYZ3 ¼ CaCO3
Y:
44. What is compound X2(YZ3)3
53:96 g
¼ 26:98 g ðaluminumÞ
2
84:27 g
Y: ð0:2986Þð282:23 gÞ ¼
¼ 28:09 g ðsiliconÞ
3
143:99 g
ð0:5102Þð282:23 gÞ ¼
¼ 16:00 g ðoxygenÞ
Z:
9
Elements determined from atomic masses in the periodic table.
X2(YZ3)3 ¼ Al2(SiO3)3
X:
ð0:1912Þð282:23 gÞ ¼
6:022 1023 molecules
4 atoms P
45. ð0:350 mol P4 Þ
¼ 8:43 1023 atoms P
mol
molecule P4
1 mol K
1 mol Na 22:99 g Na
46. ð10:0 g KÞ
¼ 5:88 g Na
39:10 g K
1 mol K
mol Na
47.
3:27 1022 g
1 molecule
6:022 1023 molecules
197 g
¼
1 mol
mol
453:6 g 6:022 1023 molecules
48. ð5 lb C12 H22 O11 Þ
¼ 4 1024 molecules C12 H22 O11
1 lb
342:3 g
4:60 cm
1m
23
49. 6:022 10 sheets
¼ 5:54 1019 m
500 sheets 100 cm
50.
6:022 1023 dollars
¼ 8:6 1013 dollars=person
7:0 109 people
51. The conversion is: mi3 ! ft3 ! in:3 ! cm3 ! drops
5280 ft 3 12:0 in: 3 2:54 cm 3 20 drops
(a)
1 mi3
¼ 8 1016 drops
mile
ft
inch
1:0 cm3
1 mi3
23
(b)
6:022 10 drops
¼ 8 106 mi3
8 1016 drops
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- Chapter 7 52. 1 mol Ag ¼ 107.9 g Ag
1 cm3
(a) ð107:9 g AgÞ
¼ 10:3 cm3 ðvolume of cubeÞ
10:5 g
(b) 10:3 cm3 ¼ volume of cube ¼ ðsideÞ3
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
side ¼ 3 10:3 cm3 ¼ 2:18 cm
53. (a)
(b)
Determine the molar mass of each compound.
CO2, 44.01 g; O2, 32.00 g; H2O, 18.02 g; CH3OH, 32.04 g. The 1.00 gram sample with the lowest
molar mass will contain the most molecules. Thus, H2O will contain the most molecules.
ð3Þ 6:022 1023 atoms
1 mol
¼ 1:00 1023 atoms
ð1:00 g H2 OÞ
mol
18:02 g
ð6Þ 6:022 1023 atoms
1 mol
ð1:00 g CH3 OHÞ
¼ 1:13 1023 atoms
mol
32:04 g
ð3Þ 6:022 1023 atoms
1 mol
¼ 4:10 1022 atoms
ð1:00 g CO2 Þ
mol
44:01 g
ð2Þ 6:022 1023 atoms
1 mol
ð1:00 g O2 Þ
¼ 3:76 1022 atoms
mol
32:00 g
The 1.00 g sample of CH3OH contains the most atoms
54. 1 mol Fe2S3 ¼ 207.9 g Fe2S3 ¼ 6.022 1023 formula units
1 formula unit
207:9 g Fe2 S3
23
¼ 41:58 g Fe2 S3
6:022 10 atoms
5 atoms
6:022 1023 formula units
55. The conversion is g P ! mol P ! mol Ca ! g Ca
1 mol P
3 mol Ca 40:08 g Ca
¼ 1:94 g Ca
ð1:00 g PÞ
30:97 g P
2 mol P
1 mol Ca
1.94 g Ca combines with 1.00 g P.
56. Grams of Fe per ton of ore that contains 5% FeSO4.
The conversion is: ton ! lb ! g ! g FeSO4 ! g Fe
2000 lb 453:6 g
55:85 g Fe
ð0:05 FeSO4 Þ
¼ 2 104 g Fe
ð1:0 tonÞ
ton
lb
151:9 g FeSO4
1.0 ton of iron ore contains 2 104 g Fe.
57. From the formula, 2 Li (13.88 g) combine with 1 S (32.07 g).
13:88 g Li
ð20:0 g SÞ ¼ 8:66 g Li
32:07 g S
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- Chapter 7 HgCO3
Hg
C
3O
200.6 g
12.01 g
48.00 g
260.6 g
(b)
Ca(ClO3)2
6O
2 Cl
Ca
96.00 g
70.90 g
40.08 g
207.0 g
(c)
C10H14N2
2N
10 C
14 H
28.02 g
120.1 g
14.11 g
162.2 g
(d)
C55H72MgN4O5
Mg
55 C
72 H
4N
5O
24.31 g
660.55 g
72.58 g
56.04 g
80.00 g
893.5 g
58. (a)
200:6 g Hg
ð100Þ ¼ 76:98% Hg
260:6 g
96:00 g O
ð100Þ ¼ 46:38% O
207:0 g
28:02 g N
ð100Þ ¼ 17:27% N
162:6 g
24:31 g Mg
ð100Þ ¼ 2:721% Mg
893:5 g
59. According to the formula, 1 mol (65.39 g) Zn combines with 1 mol (32.07 g) S.
32:07 g S
¼ 9:56 g S
ð19:5 g ZnÞ
65:39 g Zn
19.5 g Zn require 9.56 g S for complete reaction. Therefore, there is not sufficient S present (9.40 g) to
react with the Zn.
60. Percent composition of C15H20O6
15 C
20 H
6O
180.2
20.16
96.00
296.4
g
g
g
g
180:2 g C
ð100Þ ¼ 60:80 % C
296:4 g
20:16 g H
ð100Þ ¼ 6:802 % H
296:4 g
96:00 g O
ð100Þ ¼ 32:39 % O
296:4 g
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- Chapter 7 61. Percent composition of C17 H21 NO HCl
17 C
22 H
N
O
Cl
204.2
22.18
14.01
16.00
35.45
291.8
g
g
g
g
g
g
62. Percent composition of sucrose
12 C
22 H
11 O
144.1
22.18
176.0
342.3
g
g
g
g
63. Molecular formula of aspirin
60.0% C, 4.48% H, 35.5% O; molar mass
1 mol C
¼ 5:00 mol C
ð60:0 g CÞ
12:01 g C
1 mol H
ð4:48 g HÞ
¼ 4:44 mol H
1:008 g H
1 mol O
ð35:5 g OÞ
¼ 2:22 mol O
16:00 g O
204:2 g C
ð100Þ ¼ 69:98% C
291:8 g
22:18 g H
ð100Þ ¼ 7:60% H
291:8 g
14:01 g N
ð100Þ ¼ 4:80% N
291:8 g
16:00 g O
ð100Þ ¼ 5:48% O
291:8 g
35:45 g Cl
ð100Þ ¼ 12:15% Cl
291:8 g
144:1 g C
ð100Þ ¼ 42:10% C
342:3 g
22:18 g H
ð100Þ ¼ 6:480% H
342:3 g
176:0 g O
ð100Þ ¼ 51:42% O
342:3 g
of aspirin ¼ 180.2
5:00 mol C
¼ 2:25 mol C
2:22
4:44 mol H
¼ 2:00 mol H
2:22
2:22 mol O
¼ 1:00 mol O
2:22
Multiplying each by 4 give the empirical formula C9H8O4. The empirical formula mass is 180.2 g. Since
the empirical formula mass equals the molar mass, the molecular formula is the same as the empirical
formula, C9H8O4.
64. First calculate the percent oxygen in Al2(SO4)3.
53.96 g
2 Al
192:0 g
ð100Þ ¼ 56:11% O
96.21 g
3S
342:2 g
192.0 g
12 O
342.2 g
Second calculate grams of oxygen in 8.50 g of Al2(SO4)3.
Now take 56:11% of
8:50 g
8:50 g Al2 ðSO4 Þ3 ð0:5611Þ ¼ 4:77 g O
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Page 77
- Chapter 7 65. Empirical formula of gallium arsenide; 48.2% Ga, 51.8% As
1 mol Ga
0:691 mol Ga
ð48:2 g GaÞ
¼ 0:691 mol Ga
¼ 1:00 mol Ga
69:72 g Ga
0:691
1 mol As
0:691 mol As
ð51:8 g AsÞ
¼ 0:691 mol As
¼ 1:00 mol As
74:92 g As
0:691
The empirical formula is GaAs.
66. Empirical formula of calcium tartrate; 25.5%
1 mol C
¼ 2:12 mol C
ð25:5 g CÞ
12:01 g C
1 mol H
ð2:1 g HÞ
¼ 2:1 mol H
1:008 g H
1 mol C
ð21:2 g CaÞ
¼ 0:531 mol Ca
40:08 g Ca
1 mol O
ð51:0 g OÞ
¼ 3:19 mol O
16:00 g O
C, 2.1% H, 21.3% Ca, 51.0% O.
2:212 mol C
¼ 3:99 mol C
0:531
2:1 mol H
¼ 4:0 mol H
0:531
0:529 mol Ca
¼ 1:00 mol Ca
0:531
3:19 mol O
¼ 6:01 mol O
0:531
The empirical formula is C4H4CaO6
67. (a)
7.79% C, 92.21% Cl
1 mol C
¼ 0:649 mol C
ð7:79 g CÞ
12:01 g C
1 mol Cl
ð92:21 g ClÞ
¼ 2:601 mol Cl
35:45 g Cl
0:649 mol C
¼ 1:00 mol C
0:649
2:601 mol Cl
¼ 4:01 mol Cl
0:649
The empirical formula is CCl4. The empirical formula mass is 153.8 which equals the molar mass,
therefore the molecular formula is CCl4.
(b)
10.13% C, 89.87% Cl
1 mol C
ð10:13 g CÞ
¼ 0:8435 mol C
12:01 g C
1 mol Cl
ð89:87 g ClÞ
¼ 2:535 mol Cl
35:45 g Cl
0:8435 mol C
¼ 1:000 mol C
0:8435
2:535 mol Cl
¼ 3:005 mol Cl
0:8435
The empirical formula is CCl3. The empirical formula mass is 118.4 g.
molar mass
236:7 g
¼
¼ 1:999
empirical formula mass
118:4 g
The molecular formula is twice that of the empirical formula.
Molecular formula ¼ C2Cl6.
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Page 78
- Chapter 7 (c)
25.26% C, 74.74% Cl
1 mol C
ð25:26 g CÞ
¼ 2:103 mol C
12:01 g C
1 mol Cl
ð74:74 g ClÞ
¼ 2:108 mol Cl
35:45 g Cl
2:103 mol C
¼ 1:000 mol C
2:103
2:108 mol Cl
¼ 1:002 mol Cl
2:103
The empirical formula is CCl. The empirical formula mass is 47.46 g.
molar mass
284:8 g
¼
¼ 6:000
empirical formula mass
47:46 g
The molecular formula is six times that of the empirical formula.
Molecular formula ¼ C6Cl6.
(d)
11.25% C, 88.75% Cl
1 mol C
ð11:25 g CÞ
¼ 0:9367 mol C
12:01 g C
1 mol Cl
ð88:75 g ClÞ
¼ 2:504 mol Cl
35:45 g Cl
0:9367 mol C
¼ 1:000 mol C
0:9367
2:504 mol Cl
¼ 2:673 mol Cl
0:9367
Multiplying each by 3 gives the empirical formula C3Cl8. The empirical formula mass is 319.6.
Since the molar mass is also 319.6 the molecular formula is C3Cl8.
68. The conversion is: s ! min ! hr ! day ! yr
1 min
1 hr
1 day
1 year
23
6:022 10 s
¼ 1:910 1016 years
60 s
60 min
24 hr
365 days
69. The conversion is: g ! mol ! atom
1 mol Cu
6:022 1023 atoms
ð2:5 g CuÞ
¼ 2:4 1022 atoms Cu
63:55 g Cu
mol
70. The conversion is: molecules ! mol ! g 1 trillion ¼ 1012
1 mol
92:09 g C3 H8 O3
12
1000: 10 molecules C3 H8 O3
mol C3 H8 O3
6:022 1023 molecules
7
¼ 1:529 10 g C3 H8 O3
71.
7:0 10 people
9
1 mol people
¼ 1:2 1014 mol of people
6:022 1023 people
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Page 79
- Chapter 7 72. Empirical formula 23.3% Co, 25.3% Mo, 51.4%
1 mol Co
ð23:3 g CoÞ
¼ 0:935 mol Co
58:93 g Co
1 mol Mo
ð25:3 g MoÞ
¼ 0:264 mol Mo
95:94 g Mo
1 mol Cl
ð51:4 g ClÞ
¼ 1:45 mol Cl
35:45 g Cl
Cl
0:395 mol Co
¼ 1:50 mol Co
0:264
0:264 mol Mo
¼ 1:00 mol Mo
0:264
1:45 mol Cl
¼ 5:49 mol Cl
0:264
Multiplying by 2 gives the empirical formula Co3Mo2Cl11.
73. The conversion is: g Al ! mol Al ! mol Mg ! g Mg
1 mol Al
2 mol Mg 24:31 g Mg
¼ 32 g Mg
ð18 g AlÞ
26:98 g Al
1 mol Al
mol Mg
74. (10.0 g compound) (0.177) ¼ 1.77 g N
1 mol N
ð1:77 g NÞ
¼ 0:126 mol N
14:01 g N
1 mol
23
3:8 10 atoms H
¼ 0:63 mol H
6:022 1023 atoms
To determine the mol C, first find grams of H and subtract the grams of H and N from the grams of the
sample.
1:008 g H
¼ 0:64 g H
ð0:63 mol HÞ
mol H
10:0 g sample
1:77 g N
0:64 g H
7:6 g C
1 mol C
ð7:6 g CÞ
¼ 0:63 mol C
12:01 g C
Now determine the empirical formula from the moles of C, H, and N.
0:126 mol N
¼ 1:00 mol N
0:126
0:63 mol H
H
¼ 5:0 mol H
0:126
0:63 mol C
¼ 5:0 mol C
C
0:126
The empirical formula is C5H5N
N
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Page 80
- Chapter 7 75. Let x ¼ molar mass of A2O
0:400x ¼ 16:00 g O ðsince A2 O has only one mol of O atomsÞ
x ¼ 40:0 g O=mol A2 O
40:0 ¼ 16:00 þ 2y
y ¼ molar mass of A
40:0 16:00 ¼ 2y
g
¼y
mol
Look in the periodic table for the element that has 12.0 g=mol.
12:0
The element is carbon. The mystery element is carbon.
76. (a)
(b)
(c)
(d)
(e)
CH2O
C4H9
CH2O
C25H52
C6H2Cl2O
(divide the molecular formula by 6)
(divide the molecular formula by 2)
(divide the molecular formula by 3)
(divide the molecular formula by 1)
(divide the molecular formula by 2)
9:0 mg Cu2þ ions
1g
1 mol Cu2þ ions
6:022 1023 Cu2þ ions
77. ð1 L waterÞ
1 L water
1 mol Cu2þ ions
106 mg 63:55 g Cu2þ ions
¼ 8:5 1016 Cu2þ ions
1 mol H2
6:022 1023 molecules H2
1 molecule O2
78. ð3:0 g H2 Þ
2:016 g H2
1 mol H2
1:0 106 molecule H2
¼ 9:0 1017 molecules O2
10:75 hr 60 min
60 s
250:0 kg H2 O 1000 g H2 O
1 mol H2 O
79. ð1 dayÞ
1 day
1 hr
1 min
1s
1 kg H2 O
18:02 g H2 O
¼ 5:369 108 mol H2 O
80. Empirical formula of 38.65% C, 9.74% H, 51.61% S
1 mol C
3:218 mol C
ð38:65 g CÞ
¼ 3:218 mol C
¼ 2:000 mol C
12:01 g C
1:609
1 mol H
9:66 mol H
¼ 9:66 mol H
¼ 6:00 mol H
ð9:74 g HÞ
1:008 g H
1:609
1 mol S
1:609 mol S
¼ 1:609 mol S
¼ 1:000 mol S
ð51:61 g SÞ
32:07 g S
1:609
The empirical formula is C2H6S
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Page 81
- Chapter 7 81. First determine the elements in compound A(BC)3:
A: ð0:3459Þð78:01 gÞ ¼ 26:98 g ðaluminumÞ
48:00 g
B: ð0:6153Þð78:01 gÞ ¼
¼ 16:00 g ðoxygenÞ
3
3:03 g
C: ð0:0388Þð78:01 gÞ ¼
¼ 1:01 g ðhydrogenÞ
3
Element determined from atomic masses in the periodic table.
A(BC)3 ¼ Al(OH)3
Then compound A2B3 ¼ Al2O3 with a molar mass of
2ð26:98 gÞ þ 3ð16:00 gÞ ¼ 102:0 g
2ð26:98 gÞ
ð100Þ ¼ 52:90%
% Al ¼
102:0 g
3ð16:00Þ
%O ¼
ð100Þ ¼ 47:06%
102:0
82. (a) Percent composition of the original unknown compound.
Convert g CO2 to g C and g H2O to g H
12:01 g C
¼ 1:303 g C
ð4:776 g CO2 Þ
44:01 g CO2
2:016 g H
ð2:934 g H2 OÞ
¼ 0:3282 g H
18:02 g H2 O
2.500 g compound 1.303 g C 0.3282 g H ¼ 0.8688 g O
1:303 g C
ð100Þ ¼ 52:12% C
2:500 g
0:3282 g H
ð100Þ ¼ 13:13% H
2:500 g
0:8688 g O
ð100Þ ¼ 34:75% O
2:500 g
(b)
Empirical formula of unknown compound; 52.12% C, 13.13% H, 34.76% O.
1 mol C
4:340 mol C
ð52:12 g CÞ
¼ 1:997 mol C
¼ 4:340
12:01 g
2:173
1 mol H
13:03 mol H
ð13:13 g HÞ
¼ 13:03
¼ 5:996 mol H
1:008 g H
2:173
1 mol C
2:173 mol O
ð34:76 g OÞ
¼ 2:173
¼ 1:000 mol O
16:00 g
2:173
The empirical formula is C2H6O
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Page 82
- Chapter 7 83. 1:00 pg C10 H16
1g
1 mol C10 H16
6:022 1023 molecules
1 mol
1012 pg 136:2 g C10 H16
1 photon
1 molecule C10 H16
1s
¼ 1:67 109 s
2:64 1018 photons
1 gram ¼ 1012 picogram (pg)
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