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Page 54
CHAPTER 7
QUANTITATIVE COMPOSITION OF COMPOUNDS
SOLUTIONS TO REVIEW QUESTIONS
1.
A mole is an amount of substance containing the same number of particles as there are
atoms in exactly 12 g of carbon-12.
It is Avogadro’s number 16.022 * 10232 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. 16.022 * 10232.
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.
The molar mass of an element is the mass of one mole (or 6.022 * 1023 atoms) of that element.
6.
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.
7.
6.022 * 1023
8.
There are Avogadro’s number of particles in one mole of substance.
9.
(a) A mole of oxygen atoms (O) contains 6.022 : 1023 atoms.
(b) A mole of oxygen molecules (O2) contains 6.022 : 1023 molecules.
(c) A mole of oxygen molecules (O2) contains 1.204 : 1024 atoms.
(d) A mole of oxygen atoms (O) has a mass of 16.00 grams.
(e) A mole of oxygen molecules (O2) has a mass of 32.00 grams.
10.
6.022 * 1023 molecules in one molar mass of H 2SO4 .
4.215 * 1024 atoms in one molar mass of H 2SO4 .
11.
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.
12.
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
SOLUTIONS TO EXERCISES
1.
Molar masses
(a)
KBr
1
1
K
Br
39.10 g
79.90 g
119.0 g
(b)
Na 2SO4
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)
C2H 5OH
2
6
1
C
H
O
24.02 g
6.048 g
16.00 g
46.07 g
(e)
HC2H 3O2
4
2
2
H
C
O
4.032 g
24.02 g
32.00 g
60.05 g
(f)
Fe 3O4
3
4
Fe
O
167.6 g
64.00 g
231.6 g
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- Chapter 7 -
2.
(g)
C12H 22O11
12
22
11
C
H
O
144.1 g
22.18 g
176.0 g
342.3 g
(h)
Al 2(SO4)3
2
3
12
Al
S
O
53.96 g
96.21 g
192.0 g
342.2 g
(i)
(NH 4)2HPO4
9
2
1
4
H
N
P
O
9.072 g
28.02 g
30.97 g
64.00 g
132.1 g
Molar masses
(a)
NaOH
1
1
1
Na
O
H
22.99 g
16.00 g
1.008 g
40.00 g
(b)
Ag 2CO3
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)
(NH 4)2CO3
2
8
1
3
N
H
C
O
28.02 g
8.064 g
12.01 g
48.00 g
96.09 g
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- Chapter 7 -
3.
(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)
C6H 5COOH
7
6
2
C
H
O
84.07 g
6.048 g
32.00 g
122.1 g
(g)
C6H 12O6
6
12
6
(h)
K 4Fe(CN)6
(i)
BaCl 2 # 2 H 2O
C
H
O
72.06 g
12.10 g
96.00 g
180.2 g
4
1
6
6
K
Fe
C
N
156.4 g
55.85 g
72.06 g
84.06 g
368.4 g
1
2
4
2
Ba
Cl
H
O
137.3 g
70.90 g
4.032 g
32.00 g
244.2 g
Moles of atoms.
1 mol Zn
b = 0.344 mol Zn
65.39 g Zn
(a)
122.5 g Zn2a
(b)
10.688 g Mg2a
1 mol Mg
b = 2.83 * 10-2 mol Mg
24.31 g Mg
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- Chapter 7 -
(c)
14.5 * 1022 atoms Cu2a
(d)
1382 g Co2a
(e)
10.055 g Sn2a
(f)
18.5 * 1024 molecules N22a
1 mol Cu
b = 7.5 * 10-2 mol Cu
6.022 * 1023 atoms Cu
1 mol Co
b = 6.48 mol Co
58.93 g Co
1 mol Sn
b = 4.6 * 10-4 mol Sn
118.7 g Sn
2 atoms N
1 mol N atoms
b
ba
1 molecule N2 6.022 * 1023 atoms N
= 28 mol N atoms
4.
Number of moles.
(a)
(b)
(c)
(d)
5.
125.0 g NaOH2a
1 mol NaOH
b = 0.625 mol NaOH
40.00 g NaOH
1 mol Br2
144.0 g Br22a
b = 0.275 mol Br2
159.8 g Br2
1 mol MgCl2
b = 7.18 * 10-3 mol MgCl2
10.684 g MgCl22a
95.21 g MgCl2
1 mol CH3OH
114.8 g CH3OH2a
b = 0.462 mol CH3OH
32.04 g CH3OH
(e)
12.88 g Na2SO42a
(f)
14.20 lb ZnI22a
1 mol Na2SO4
b = 2.03 * 10-2 mol Na2SO4
142.1 g Na2SO4
453.6 g
1 mol ZnI2
ba
b = 5.97 mol ZnI2
1 lb
319.2 g ZnI2
Number of grams.
(a)
10.550 mol Au2a
197.0 g Au
b = 108 g Au
1 mol Au
(b)
115.8 mol H2O2a
18.02 g H2O
b = 285 g H2O
mol H2O
(c)
112.5 mol Cl22a
(d)
13.15 mol NH4NO32a
70.90 g Cl2
b = 886 g Cl2
mol Cl2
80.05 g NH4NO3
b = 252 g NH4NO3
mol NH4NO3
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- Chapter 7 6.
Number of grams.
(a)
14.25 * 10-4 mol H2SO42a
98.09 g H2SO4
b = 0.0417 g H2SO4
mol H2SO4
(b)
14.5 * 1022 molecules CCl42a
153.8 g CCl4
1 mol
ba
b
23
mol CCl4
6.022 * 10 molecules
= 11 g CCl4
(c)
(d)
7.
8.
9.
10.00255 mol Ti2a
47.87 g Ti
b = 0.122 g Ti
mol Ti
32.07 g S
b = 8.0 * 10-7 g S
11.5 * 1016 atoms S2a
6.022 * 1023 atoms S
Number of molecules.
6.022 * 1023 molecules
b = 7.59 * 1023 molecules O2
mol
(a)
11.26 mol O22a
(b)
10.56 mol C6H62a
(c)
116.0 g CH42a
(d)
11000. g HCl2a
(a)
11.75 mol Cl22a
(b)
10.27 mol C2H6O2a
(c)
112.0 g CO22a
6.022 * 1023 molecules CO2
b = 1.64 * 1023 molecules CO2
44.01 g CO2
(d)
1100. g CH42a
6.022 * 1023 molecules CH4
b = 3.75 * 1024 molecules CH4
16.04 g CH4
6.022 * 1023 molecules
b = 3.4 * 1023 molecules C6H6
mol
6.022 * 1023 molecules CH4
b = 6.01 * 1023 molecules CH4
16.04 g CH4
6.022 * 1023 molecules HCl
b = 1.652 * 1025 molecules HCl
36.46 g HCl
6.022 * 1023 molecules
b = 1.05 * 1024 molecules Cl2
mol
6.022 * 1023 molecules
b = 1.6 * 1023 molecules C2H6O
mol
Number of atoms.
(a)
(11 molecules C2H5OH) a
(b)
(25.0 mol Ag) a
9 atoms
b = 99 atoms C2H5OH
1 molecule
6.022 * 1023atoms
b = 1.5 * 1025 atoms Ag
mol
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- Chapter 7 -
6.022 * 1023atoms
b = 4.52 * 1020 atoms Xe
131.3 g
(c)
(0.0986 g Xe) a
(d)
(72.5 g CHCl3) a
6.022 * 1023 molecules
5 atoms
ba
b
119.4
1 molecule
= 1.83 * 1024 atom CHCl3
10.
11.
Number of atoms.
(a)
(18 molecules N2O5) a
(b)
(10.0 mol Au) a
(c)
(75.2 g BF3) a
(d)
(15.2 g U) a
7 atoms
b = 1.3 * 102 atoms N2O5
1 molecule
6.022 * 1023atoms
b = 6.02 * 1024 atoms Au
mol
6.022 * 1023 molecules BF3
4 atoms
ba
b = 2.67 * 1024 atoms BF3
67.81 g BF3
1 molecules
6.022 * 1023atoms U
b = 3.85 * 1022 atoms U
238.0 g U
Number of grams.
(a)
11 atom Pb2a
(b)
11 atom Ag2a
(c)
11 molecule H2O2a
(d)
[1 molecule C3H5(NO3)3]a
207.2 g Pb
6.022 * 1023 atoms Pb
b = 3.441 * 10-22 g Pb
107.9 g Ag
6.022 * 1023 atoms Ag
b = 1.792 * 10-22 g Ag
18.02 g H2O
23
6.022 * 10 molecules H2O
b = 2.992 * 10-23 g H2O
227.1 g C3H5(NO3)3
6.022 * 1023 molecules C3H5(NO3)3
b
= 3.771 * 10 - 22 g C3H5(NO3)3
12.
(a)
11 atom Au2a
(b)
11 atom U2a
(c)
197.0 g Au
6.022 * 1023 atoms Au
b = 3.271 * 10-22 g Au
b = 3.952 * 10-22 g U
6.022 * 1023 atoms U
17.03 g NH3
11 molecule NH32a
b = 2.828 * 10-23 g NH3
6.022 * 1023 molecules NH3
238.0 g U
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- Chapter 7 -
(d)
[1 molecule C6H4(NH2)2]a
108.1 g C6H4(NH2)2
6.022 * 1023 molecules C6H4(NH2)2
b
= 1.795 * 10 - 22 g C6H4(NH2)2
13.
14.
15.
(a)
18.66 mol Cu2a
(b)
1125 mol Au2a
197.0 g Au
1 kg
ba
b = 24.6 kg Au
mol Au
1000 g
(c)
110. atoms C2a
1 mol
b = 1.7 * 10-23 mol C
23
6.022 * 10 atoms
(d)
15000 molecules CO22a
(a)
128.4 g S2a
(b)
12.50 kg NaCl2a
(c)
142.4 g Mg2a
(d)
1485 mL Br22a
63.55 g Cu
b = 550. g Cu
mol Cu
1 mol
b = 8 * 10-21 mol CO2
6.022 * 1023 molecules
1 mol S
b = 0.886 mol S
32.07 g S
1000 g
1 mol NaCl
ba
b = 42.8 mol NaCl
kg
58.44 g NaCl
6.022 * 1023 atoms
b = 1.05 * 1024 atoms Mg
24.31 g
3.12 g
1 mol Br2
ba
b = 9.47 mol Br2
mL
159.8 g Br2
One mole of carbon disulfide (CS2) contains:
(a)
6.022 * 1023 molecules of CS2
(b)
16.022 * 1023 molecules of CS22 ¢
1 C atom
≤ = 6.022 * 1023 C atoms
1 molecule CS2
(c)
16.022 * 1023 molecules of CS22 ¢
2 S atoms
≤ = 1.204 * 1024 S atoms
1 molecule CS2
(d)
16.022 * 1023 atoms2 + 11.204 * 1024 atoms2 = 1.806 * 1024 total atoms
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- Chapter 7 16.
17.
18.
One mole of ammonia (NH 3) contains
(a)
6.022 * 1023 molecules of NH 3
(b)
16.022 * 1023 molecules of NH 32 ¢
1 N atom
≤ = 6.022 * 1023 N atoms
molecule NH 3
(c)
16.022 * 1023 molecules of NH 32 ¢
3 H atoms
≤ = 1.807 * 1024 H atoms
molecule NH 3
(d)
16.022 * 1023 atoms2 + 11.807 * 1024 atoms2 = 2.409 * 1024 total atoms
Atoms of oxygen in:
1 mol O2
2 mol O
6.022 * 1023 atoms
b¢
≤¢
≤ = 6.02 * 1023 atoms O
32.00 g O2 1 mol O2
mol
(a)
116.0 g O22a
(b)
10.622 mol MgO2a
(c)
16.00 * 1022 molecules C6H 12O62 ¢
6.022 * 1023 atoms
1 mol O
b¢
≤ = 3.75 * 1023 atoms O
mol MgO
mol
6 atoms O
≤ = 3.60 * 1023 atoms O
molecule C6H 12O6
Atoms of oxygen in:
(a)
15.0 mol MnO22 ¢
2 mol O
6.022 * 1023 atoms
≤¢
≤ = 6.0 * 1024 atoms O
mol MnO2
mol
(b)
1255 g MgCO32a
1 mol MgCO3
3 mol O
6.022 * 1023 atoms
b¢
≤¢
≤
84.32 g MgCO3 mol MgCO3
mol
= 5.46 * 1024 atoms O
(c)
19.
15.0 * 1018 molecules H 2O2 ¢
1 atom O
≤ = 5.0 * 1018 atoms O
molecule H 2O
The number of grams of:
(a)
silver in 25.0 g AgBr
125.0 g AgBr2a
107.9 g Ag
b = 14.4 g Ag
187.8 g AgBr
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- Chapter 7 (b)
nitrogen in 6.34 mol (NH 4)3PO4
16.34 mol (NH 4)3PO42 ¢
(c)
42.03 g N
≤ = 266 g N
mol (NH 4)3PO4
oxygen in 8.45 * 1022 molecules SO3
The conversion is: molecules SO3 ¡ mol SO3 ¡ g O
18.45 * 1022 molecules SO32 ¢
20.
48.00 g O
1 mol
≤¢
≤ = 6.74 g O
23
mol SO3
6.022 * 10 molecules
The number of grams of:
(a)
chlorine in 5.00 g PbCl 2
15.00 g PbCl 22 ¢
(b)
70.90 g Cl
≤ = 1.27 g Cl
278.1 g PbCl 2
hydrogen in 4.50 g H 2SO4
14.50 g H 2SO42 ¢
(c)
2.016 g H
≤ = 9.25 * 10-2 g H
98.09 g H 2SO4
hydrogen in 5.45 * 1022 molecules NH 3
The conversion is: molecules NH 3 ¡ moles NH 3 ¡ g H
15.45 * 1022 molecules NH 32 ¢
21.
3.024 g H
1 mol
≤¢
≤ = 0.274 g H
23
mol NH 3
6.022 * 10 molecules
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
a
22.99 g
b 11002 = 22.34% Na
102.9 g
a
79.90 g
b 11002 = 77.65% Br
102.9 g
a
39.10 g
b 11002 = 39.06% K
100.1 g
a
1.008 g
b11002 = 1.007% H
100.1 g
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- Chapter 7 -
(c)
(d)
(e)
(f)
FeCl 3
SiCl 4
Fe
3 Cl
Si
4 Cl
Al 2(SO4)3 2 Al
3S
12 O
AgNO3
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 g
14.01 g
48.00 g
169.9 g
- 64 -
a
12.01 g
b11002 = 12.00% C
100.1 g
a
48.00 g
b11002 = 47.95% O
100.1 g
a
55.85 g
b11002 = 34.41% Fe
162.3 g
a
106.4 g
b11002 = 65.56% Cl
162.3 g
a
28.09 g
b11002 = 16.53% Si
169.9 g
a
141.8 g
b11002 = 83.46% Cl
169.9 g
a
53.96 g
b11002 = 15.77% Al
342.2 g
a
96.21 g
b11002 = 28.12% S
342.2 g
a
192.0 g
b11002 = 56.11% O
342.2 g
a
107.9 g
b11002 = 63.51% Ag
169.9 g
a
14.01 g
b11002 = 8.246% N
169.9 g
a
48.00 g
b11002 = 28.25% O
169.9 g
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Page 65
- Chapter 7 22.
Percent composition
(a)
(b)
(c)
(d)
ZnCl 2
Zn
2 Cl
NH 4C2H 3O2
MgP2O7
(NH 4)2SO4
N
7H
2C
2O
Mg
2P
7O
2N
8H
S
4O
65.39 g
70.90 g
136.3 g
14.01 g
7.056 g
24.02 g
32.00 g
77.09 g
24.31 g
61.94 g
112.0 g
198.3 g
28.02 g
8.064 g
32.07 g
64.00 g
132.2 g
- 65 -
a
65.39 g
b11002 = 47.98% Zn
136.3 g
a
70.90 g
b11002 = 52.02% Cl
136.3 g
a
14.01 g
b11002 = 18.17% N
77.09 g
a
7.056 g
b11002 = 9.153% H
77.09 g
a
24.02 g
b11002 = 31.16% C
77.09 g
a
32.00 g
b11002 = 41.51% O
77.09 g
a
24.31 g
b11002 = 12.26% Mg
198.3 g
a
61.94 g
b11002 = 31.24% P
198.3 g
a
112.0 g
b11002 = 56.48% O
198.3 g
a
28.02 g
b11002 = 21.20% N
132.2 g
a
8.064 g
b11002 = 6.100% H
132.2 g
a
32.07 g
b11002 = 24.26% S
132.2 g
a
64.00 g
b11002 = 48.41% O
132.2 g
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- Chapter 7 (e)
(f)
23.
Fe(NO3)3
ICI 3
Fe
3N
9O
I
3 Cl
55.85 g
42.03 g
144.0 g
241.9 g
a
55.85 g
b11002 = 23.09% Fe
241.9 g
a
42.03 g
b11002 = 17.37% N
241.9 g
a
144.0 g
b11002 = 59.53% O
241.9 g
a
126.9 g
b11002 = 54.39% I
233.3 g
a
106.4 g
b11002 = 45.61% Cl
233.3 g
55.85 g
16.00 g
71.85 g
a
55.85 g
b11002 = 77.73% Fe
71.85 g
126.9 g
106.4 g
233.3 g
Percent of iron
(a)
FeO
Fe
O
(b)
Fe 2O3
2 Fe
3O
111.7 g
48.00 g
159.7 g
a
111.7 g
b11002 = 69.94% Fe
159.7 g
(c)
Fe 3O4
3 Fe
4O
167.6 g
64.00 g
231.6 g
a
167.6 g
b11002 = 72.37% Fe
231.6 g
(d)
K 4Fe(CN)6
Fe
4K
6C
6N
55.85 g
156.4 g
72.06 g
84.06 g
368.4 g
a
55.85 g
b11002 = 15.16% Fe
368.4 g
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- Chapter 7 24.
Percent chlorine
(a)
KCl
K
Cl
39.10 g
35.45 g
74.55 g
a
35.45 g
b11002 = 47.55% Cl
74.55 g
(b)
BaCl 2
Ba
2 Cl
137.3 g
70.90 g
208.2 g
a
70.90 g
b11002 = 34.05% Cl
208.2 g
(c)
SiCl 4
Si
4 Cl
28.09 g
141.8 g
169.9 g
a
141.8 g
b11002 = 83.46% Cl
169.9 g
(d)
LiCl
Li
Cl
6.941 g
35.45 g
42.39 g
a
35.45 g
b11002 = 83.63% Cl
42.39 g
Highest % Cl is in LiCl; lowest % Cl is in BaCl 2
25.
Percent composition of an oxide
14.20 g oxide
-6.20 g P
8.00 g oxygen
26.
6.20 g
b 11002 = 43.7% P
14.20 g
a
8.00 g
b 11002 = 56.3% O
14.20 g
Percent composition of ethylene chloride
6.00 g C
1.00 g H
17.75 g Cl
24.75 g total
27.
a
(a)
(b)
(c)
H 2O
N2O3
equal
a
6.00 g
b11002 = 24.2% C
24.75 g
a
1.00 g
b11002 = 4.04% H
24.75 g
a
17.75 g
b11002 = 71.72% Cl
24.75 g
(It has 2 H atoms/O atom)
(It has only 0.67 N atom/O atom)
(Both have 20 atoms/N atom)
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- Chapter 7 28.
(a)
(b)
(c)
29.
(Because a K atom has mare mass than a Na atom.)
(Because a H atom has less mass then a K atom.)
(Because only one Cr atom is present.)
KClO3
KHSO4
Na 2CrO4
Empirical Formulas. Change all percents to grams.
(a)
5.94 gH
= 5.89 mol H
1.008 g>mol
94.06 g O
= 5.879 mol O
16.00 g>mol
This is a ratio of 1 mol H to 1 mol O.
The empirical formula is HO
(b)
80.34 g Zn
= 1.228 mol Zn
65.39 g>mol
19.66 g O
= 1.229 mol O
16.00 g>mol
This is a ratio of 1 mol Zn to 1 mol O.
The empirical formula is ZnO
(c)
35.18 g Fe
= 0.6299 mol Fe
55.85 g>mol
44.66 g Cl
= 1.260 mol Cl
35.45 g>mol
20.16 g O
= 1.260 mol O
16.00 g O>mol
mol ratios =
0.6299 Fe
= 1.000 mol Fe
0.6299
1.260 mol O
= 2.000 mol O
0.6299
1.260 mol Cl
= 2.000 mol Cl
0.6299
The empirical formula is Fe(ClO)2
(d)
26.19 g N
= 1.869 mol N
14.01 g>mol
7.55 g H
= 7.49 mol H
1.008 g>mol
66.26 g Cl
= 1.869 mol Cl
35.45 g>mol
mol ratios:
1.869 mol N
= 1 mol N
1.869
7.49 mol H
= 4.00 mol H
1.869
1.869 mol Cl
= 1.000 mol Cl
1.869
The empirical formula is NH4Cl
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- Chapter 7 30.
Empirical formulas: Change all percents to grams.
(a)
32.86 g K
= 0.8404 mol K
39.10 g>mol
67.14 g Br
= 0.8403 mol Br
79.90 g>mol
This is a ratio of 1 mol K to 1 mol Br.
The empirical formula is KBr
(b)
63.50 g Ag
= 0.5885 mol Ag
107.9 g>mol
8.25 g N
= 0.5889 mol N
14.01 g>mol
28.25 g O
= 1.766 mol O
16.00 g>mol
mole ratios:
0.5885 mol Ag
= 1.000 mol Ag
0.5885
0.5889 mol N
= 1.001 mol N
0.5885
1.766 mol O
= 3.000 mol O
0.5885
The empirical formula is Ag NO3
(c)
54.09 g Ca
= 1.350 mol Ca
40.08 g>mol
2.72 g H
= 2.70 mol H
1.008 g>mol
43.18 g O
= 2.699 mol O
16.00 g>mol
mole ratios:
1.350 mol Ca
= 1.000 mol Ca
1.350
2.70 mol H
= 2.00 mol H
1.350
2.699 mol O
= 1.999 mol O
1.35
The empirical formula is CaO2H 2 or Ca(OH)2
(d)
2.06 g H
= 2.07 mol H
1.008 g>mol
32.69 g S
= 1.019 mol S
32.07 g>mol
65.25 g O
= 4.078 mol O
16.00 g>mol
mole ratios:
2.04 mol H
= 2.00 mol H
1.019
4.078 mol O
= 4.002 mol O
1.019
The empirical formula is H2SO4
- 69 -
1.019 mol S
= 1.000 mol S
1.019
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- Chapter 7 31.
Empirical formulas from percent composition.
(a)
Step 1. Express each element as grams> 100 g material.
63.6% N = 63.6 g N>100 g material
36.4% O = 36.4 g O>100 g material
Step 2. Calculate the relative moles of each element.
163.6 g N2a
1 mol N
b = 4.54 mol N
14.01 g N
136.4 g O2a
1 mol O
b = 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
= 1.99 mol N
2.28
2.28 mol O
= 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
146.7 g N2a
1 mol N
b = 3.33 mol N
14.01 g N
3.33 mol N
= 1.00 mol N
3.33
153.3 g O2a
1 mol O
b = 3.33 mol O
16.00 g O
3.33 mol O
= 1.00 mol O
3.33
The empirical formula is NO.
(c)
25.9% N, 71.4% O
125.9 g N2a
1 mol N
b = 1.85 mol N
14.01 g N
1.85 mol N
= 1.00 mol N
1.85
174.1 g O2a
1 mol O
b = 4.63 mol O
16.00 g O
4.63 mol O
= 2.5 mol O
1.85
Since these values are not whole numbers, multiply each by 2 to change them to
whole numbers.
11.00 mol N2122 = 2.00 mol N; 12.5 mol O2122 = 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
143.4 g Na2a
b = 1.89 mol Na
22.99 g Na
111.3 g C2a
1 mol C
b = 0.941 mol C
12.01 g C
145.3 g O2a
1 mol O
b = 2.83 mol O
16.00 g O
1.89 mol Na
= 2.01 mol Na
0.941
0.941 mol C
= 1.00 mol C
0.941
2.83 mol O
= 3.00 mol O
0.941
The empirical formula is Na 2CO3 .
(e)
18.8% Na, 29.0% Cl, 52.3% O
118.8 g Na2a
1 mol Na
b = 0.818 mol Na
22.99 g Na
1 mol Cl
b = 0.818 mol Cl
35.45 g Cl
1 mol O
152.3 g O2a
b = 3.27 mol O
16.00 g O
The empirical formula is NaClO4 .
129.0 g Cl2a
(f)
0.818 mol Na
= 1.00 mol Na
0.818
0.818 mol Cl
= 1.00 mol Cl
0.818
3.27 mol O
= 4.00 mol O
0.818
72.02% Mn, 27.98% O
172.02 g Mn2a
1 mol Mn
b = 1.311 mol Mn
54.94 g Mn
1.311 mol Mn
= 1.000 mol Mn
1.311
1 mol O
b = 1.749 mol O
16.00 g O
1.749 mol O
= 1.334 mol O
1.311
127.98 g O2a
Multiply both values by 3 to give whole numbers.
11.000 mol Mn2132 = 3.000 mol Mn; 11.334 mol O2132 = 4.002 mol O
The empirical formula is Mn 3O4 .
32.
Empirical formulas from percent composition.
(a)
64.1% Cu, 35.9% Cl
1 mol Cu
b = 1.01 mol Cu
164.1 g Cu2a
63.55 g Cu
1 mol Cl
b = 1.01 mol Cl
135.9 g Cl2a
35.45 g Cl
The empirical formula is CuCl.
- 71 -
1.01 mol Cu
= 1.00 mol Cu
1.01
1.01 mol Cl
= 1.00 mol Cl
1.01
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- Chapter 7 (b)
47.2% Cu, 52.8% Cl
147.2 g Cu2a
1 mol Cu
b = 0.743 mol Cu
63.55 g Cu
152.8 g Cl2a
1 mol Cl
b = 1.49 mol Cl
35.45 g Cl
The empirical formula is CuCl 2 .
(c)
0.743 mol Cu
= 1.00 mol Cu
0.743
1.49 mol Cl
= 2.01 mol Cl
0.743
51.9% Cr, 48.1% S
151.9 g Cr2a
1 mol Cr
b = 0.998 mol Cr
52.00 g Cr
0.998 mol Cr
= 1.00 mol Cr
0.998
148.1 g S2a
1 mol S
1.50 mol S
b = 1.50 mol S
= 1.50 mol S
32.07 g S
0.998
Multiply both values by 2 to give whole numbers.
11.50 mol S2122 = 3.00 mol S
11.00 mol Cr2122 = 2.00 mol Cr;
The empirical formula is Cr2S3 .
(d)
55.3% K, 14.6% P, 30.1% O
155.3 g K2a
1 mol K
b = 1.41 mol K
39.10 g K
1.41 mol K
= 2.99 mol K
0.471
1 mol P
b = 0.471 mol P
30.97 g P
1 mol O
130.1 g O2a
b = 1.88 mol O
16.00 g O
0.471 mol P
= 1.00 mol P
0.471
1.88 mol O
= 3.99 mol O
0.471
114.6 g P2a
The empirical formula is K 3PO4 .
(e)
38.9% Ba, 29.4% Cr, 31.7% O
138.9 g Ba2a
1 mol Ba
b = 0.283 mol Ba
137.3 g Ba
129.4 g Cr2a
1 mol Cr
b = 0.565 mol Cr
52.00 g Cr
1 mol O
131.7 g O2a
b = 1.98 mol O
16.00 g O
The empirical formula is BaCr2O7 .
- 72 -
0.283 mol Ba
= 1.00 mol Ba
0.283
0.565 mol Cr
= 2.00 mol Cr
0.283
1.98 mol O
= 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
b = 0.129 mol P
30.97 g P
1 mol Br
182.3 g Br2a
b = 1.03 mol Br
79.90 g Br
1 mol Cl
113.7 g Cl2a
b = 0.386 mol Cl
35.45 g Cl
13.99 g P2a
0.129 mol P
= 1.00 mol P
0.129
1.03 mol Br
= 7.98 mol Br
0.129
0.386 mol Cl
= 2.99 mol Cl
0.129
The empirical formula is PBr8Cl 3 .
33.
Empirical formula
13.996 g Sn2a
1 mol Sn
b = 0.0337 mol Sn
118.7 g Sn
11.077 g O2a
1 mol O
b = 0.0673 mol O
16.00 g O
The empirical formula is SnO2 .
0.0337 mol Sn
= 1.00 mol Sn
0.0337
0.0673 mol O
= 2.00 mol O
0.0337
34.
Empirical formula
5.454 g product - 3.054 g V = 2.400 g O
1 mol V
0.0600 mol V
13.054 g V2a
b = 0.0600 mol V
= 1.00 mol V
50.94 g V
0.0600
1 mol O
0.1500 mol O
12.400 g O2a
b = 0.1500 mol O
= 2.50 mol O
16.00 g O
0.0600
Multiplying both by 2 gives the empirical formula V2O5 .
35.
Empirical formula
2.775 g oxide - 2.465 g Cu = 0.310 g O
1 mol Cu
b = 0.03879 mol Cu
12.465 g Cu2a
63.55 g Cu
1 mol O
b = 0.0194 mol O
16.00 g O
The empirical formula is Cu2O.
10.310 g O2a
36.
Empirical formula
5.276 g compound - 3.898 g Hg = 1.378 g Cl
1 mol Hg
b = 0.01943 mol Hg
13.898 g Hg2a
200.6 g Hg
- 73 -
0.03879 mol Cu
= 2.00 mol Cu
0.0194
0.0194 mol O
= 1.00 mol O
0.0194
0.01943 mol Hg
= 1.000 mol Hg
0.01943
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- Chapter 7 -
11.378 g Cl2a
1 mol Cl
b = 0.03887 mol Cl
35.45 g Cl
0.03887 mol Cl
= 2.001 mol Cl
0.01943
The empirical formula is HgCl2 .
37.
Molecular formula of hydroquinone
65.45% C, 5.45% H, 29.09% O; molar mass = 110.1
1 mol C
5.450 mol C
b = 5.450 mol C
= 2.998 mol C
165.45 g C2a
12.01 g C
1.818
1 mol H
5.41 mol H
b = 5.41 mol H
= 2.98 mol H
15.45 g H2a
1.008 g H
1.818
1.818 mol O
1 mol O
b = 1.818 mol O
= 1.000 mol O
129.09 g O2a
16.00 g O
1.818
The empirical formula is C3H 3O making the empirical formula mass 55.05 g.
110.1 g
molar mass
=
= 2
mass of empirical formula
55.05 g
The molecular formula is twice that of the empirical formula.
Molecular formula = (C3H 3O)2 = C6H 6O2
38.
Molecular formula of fructose
40.0% C, 6.7% H, 53.3% O; molar mass = 180.1
1 mol C
3.33 mol C
b = 3.33 mol C
= 1.00 mol C
140.0 g C2a
12.01 g C
3.33
6.6 mol H
1 mol H
b = 6.6 mol H
= 2.0 mol H
16.7 g H2a
1.008 g H
3.33
1 mol O
3.33 mol O
b = 3.33 mol O
= 1.00 mol O
153.3 g O2a
16.00 g O
3.33
The empirical formula is CH 2O making the empirical formula mass 33.03 g.
180.1 g
molar mass
=
= 5.994
mass of empirical formula
33.03 g
The molecular formula is six times that of the empirical formula.
Molecular formula = (CH 2O)6 = C6H 12O6
39.
Molecular formula of ethanedioic acid
26.7% C, 2.24% H, 71.1% O; molar mass = 90.04
26.7 g C a
1 mol C
b = 2.22 mol C
12.01 g C
2.22 mol C
= 1.0 mol C
2.2
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- Chapter 7 -
2.2 g Ha
1 mol H
b = 2.2 mol H
1.008 g H
2.2 mol H
= 1.0 mol H
2.2
1 mol O
4.44 mol O
b = 4.44 mol O
= 2.0 mol O
16.00 g O
2.2
The empirical formula is CHO2, making the empirical formula mass 45.02 g.
90.04 g
molar mass
=
= 2
mass of empirical formula
45.02 g
The molecular formula is twice that of the empirical formula.
71.1 g Oa
Molecular formula = (CHO2)2 = C2H2O4
40.
Molecular formula of butyric acid
54.5% C, 9.2% H, 36.3% O; molar mass = 88.11
154.5 g C2a
1 mol C
b = 4.54 mol C
12.01 g C
4.54 mol C
= 2.00 mol C
2.27
19.2 g H2a
1 mol H
b = 9.1 mol H
1.008 g H
9.1 mol H
= 4.0 mol H
2.27
136.3 g O2a
1 mol O
b = 2.27 mol O
16.00 g O
2.27 mol O
= 1.0 mol O
2.27
The empirical formula is C2H4O, making the empirical formula mass 44.05 g.
88.11 g
molar mass
=
= 2
mass of empirical formula
44.05 g
The molecular formula is twice that of the empirical formula.
Molecular formula = (C2H4O)2 = C4H8O2
41.
12.04 g
11002 = 30.45%
39.54 g
39.54 g - 12.04 g
% oxygen =
11002 = 69.55%
39.54 g
12.04 g N
empirical formula: moles of nitrogen =
= 0.8594 mol N
14.01 g>mol
27.50 g O
moles of oxygen =
= 1.719 mol O
16.00 g>mol O
% nitrogen =
relative number of nitrogen atoms =
- 75 -
0.8594 mol
= 1.000
0.8594 mol
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- Chapter 7 -
relative number of oxygen atoms =
1.719 mol
= 2.000
0.8594 mol
empirical formula = NO2
molecular formula: 1molar mass of NO22x = 92.02 g, 46.01x = 92.02, x = 2
The molecular formula is twice the empirical formula.
molecular formula = N2O4
42.
Total mass of C + H + O =
30.21 g
% carbon =
11002
75.53 g
5.08 g
% hydrogen =
11002
75.53 g
40.24 g
% oxygen =
11002
75.53 g
30.21 g + 5.08 g + 40.24 g = 75.53 g
= 40.0%
= 6.73%
= 53.3%
30.21 g C
= 2.515 mol C
12.01 g>mol
5.080 g H
moles of hydrogen =
= 5.03 mol H
1.008 g>mol
40.24 g O
moles of oxygen =
= 2.515 mol O
16.00 g>mol
empirical formula: moles of carbon =
relative number of carbon atoms =
relative number of hydrogen atoms =
relative number of oxygen atoms =
empirical formula = CH 2O
2.515 mol
= 1.000
2.515 mol
5.03 mol
= 2.00
2.515 mol
2.515 mol
= 1.000
2.515 mol
molecular formula: 1molar mass of CH 2O2x = 180.18 g>mol,
130.03 g>mol2x = 180.18 g>mol,
x =
180.18 g>mol
= 6
30.03 g>mol
The molecular formula is six times the empirical formula.
molecular formula = C6H 12O6
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- Chapter 7 43.
What is compound XYZ 3
X: 10.400421100.09 g2 = 40.08 g (calcium)
10.120021100.09 g2 = 12.01 g (carbon)
48.00 g
Z: 10.479621100.09 g2 = 48.00 g;
= 16.00 g (oxygen)
3
Elements determined from atomic masses in the periodic table.
Y:
XYZ 3 = CaCO3
44.
What is compound X 2(YZ 3)3
53.96 g
= 26.98 g (aluminum)
X: 10.191221282.23 g2 =
2
84.27 g
= 28.09 g (silicon)
Y: 10.298621282.23 g2 =
3
143.99 g
Z: 10.510221282.23 g2 =
= 16.00 g (oxygen)
9
Elements determined from atomic masses in the periodic table.
X 2(YZ 3)3 = Al 2(SiO3)3
6.022 * 1023 molecules
4 atoms P
≤¢
≤ = 8.43 * 1023 atoms P
mol
molecule P4
45.
10.350 mol P42 ¢
46.
110.0 g K2a
47.
11.79 * 10-23 g>atom216.022 * 1023 atoms>molar mass2 = 10.8 g>molar mass
48.
15 lb C12H22O112 ¢
49.
16.022 * 1023 sheets2a
50.
¢
1 mol K
1 mol Na 22.99 g Na
ba
ba
b = 5.88 g Na
39.10 g K
1 mol K
mol Na
453.6 g
6.022 * 1023 molecules
≤¢
≤ = 4 * 1024 molecules
1 lb
342.3 g
1m
4.60 cm
ba
b = 5.54 * 1019 m
500 sheets 100 cm
6.022 * 1023 dollars
≤ = 9.9 * 1013 dollars>person
6.1 * 109 people
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- Chapter 7 51.
52.
The conversion is: mi3 ¡ ft 3 ¡ in.3 ¡ cm3 ¡ drops
5280 ft 3 12.0 in. 3 2.54 cm 3 20 drops
b a
b a
b a
b = 8 * 1016 drops
mile
ft
inch
1.0 cm3
(a)
11 mi32a
(b)
16.022 * 1023 drops2 ¢
1 mi3
≤ = 8 * 106 mi3
8 * 1016 drops
1 mol Ag = 107.9 g Ag
1 cm3
≤ = 10.3 cm3 (volume of cube)
10.5 g
(a)
1107.9 g Ag2 ¢
(b)
10.3 cm3 = volume of cube = (side)3
side = 2
3 10.3 cm3 = 2.18 cm
53.
(a)
Determine the molar mass of each compound.
CO2 , 44.01 g; O2 , 32.00 g; H 2O, 18.02 g; CH 3OH, 32.04 g. The 1.00 gram sample
with the lowest molar mass will contain the most molecules. Thus, H 2O will
contain the most molecules.
(b)
11.00 g H 2O2a
13216.022 * 1023 atoms2
1 mol
b¢
≤ = 1.00 * 1023 atoms
18.02 g
mol
11.00 g CH 3OH2a
16216.022 * 1023 atoms2
1 mol
b¢
≤ = 1.13 * 1023 atoms
32.04 g
mol
13216.022 * 1023 atoms2
1 mol
b¢
11.00 g CO22a
≤ = 4.10 * 1022 atoms
44.01 g
mol
12216.022 * 1023 atoms2
1 mol
11.00 g O22a
b¢
≤ = 3.76 * 1022 atoms
32.00 g
mol
The 1.00 g sample of CH 3OH contains the most atoms
54.
1 mol Fe 2S3 = 207.9 g Fe 2S3 = 6.022 * 1023 formula units
16.022 * 1023 atoms2a
207.9 g Fe 2S3
1 formula unit
b¢
≤ = 41.58 g Fe 2S3
5 atoms
6.022 * 1023 formula units
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- Chapter 7 55.
The conversion is g P ¡ mol P ¡ mol Ca ¡ g Ca
11.00 g P2a
1 mol P
3 mol Ca 40.08 g Ca
ba
ba
b = 1.94 g Ca
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
11.0 ton2a
55.85 g Fe
2000 lb 453.6 g
ba
b 10.05 FeSO42 ¢
≤ = 2 * 104 g Fe
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).
a
58.
13.88 g Li
b120.0 g S2 = 8.66 g Li
32.07 g S
(a)
HgCO3
(b)
(c)
200.6 g
12.01 g
48.00 g
260.6 g
a
200.6 g Hg
b11002 = 76.98% Hg
260.6 g
Ca(ClO3)2 6 O
2 Cl
Ca
96.00 g
70.90 g
40.08 g
207.0 g
a
96.00 g O
b 11002 = 46.38% O
207.0 g
C10H 14N2
28.02 g
120.1 g
14.11 g
162.2 g
a
28.02 g N
b11002 = 17.27% N
162.2 g
Hg
C
3O
2N
10 C
14 H
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- Chapter 7 (d)
C55H 72MgN4O5
24.31 g
660.55 g
72.58 g
56.04 g
80.00 g
893.5 g
Mg
55 C
72 H
4N
5O
a
24.31 g Mg
b 11002 = 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
119.5 g Zn2a
b = 9.56 g S
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 C21H28O3
21 C
28 H
3O
61.
252.2 g
28.22 g
48.00 g
328.4 g
a
252.2 g C
b11002 = 76.80% C
328.4 g
28.22 g H
a
b(100) = 8.593% H
328.4 g
48.00 g O
a
b (100) = 14.62% O
328.4 g
Percent composition of C17H21NO•HCl
17 C
22 H
N
O
Cl
204.2 g
22.18 g
14.01 g
16.00 g
35.45 g
291.8 g
204.2 g C
b11002
291.8 g
22.18 g H
a
b(100)
291.8 g
14.01 g N
a
b(100)
291.8 g
16.00 g O
a
b(100)
291.8 g
a
a
= 69.98% C
= 7.60% H
= 4.80% N
= 5.48% O
35.45g Cl
b(100) = 12.15% Cl
291.8 g
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- Chapter 7 62.
Percent composition of sucrose
144.1 g
22.18 g
176.0 g
342.3 g
12 C
22 H
11 O
63.
a
144.1 g C
b11002 = 42.10% C
342.3 g
22.18 g H
a
b(100) = 6.480% H
342.3 g
176.0 g O
b (100) = 51.42% O
a
342.3 g
Molecular formula of aspirin
60.0% C, 4.48% H, 35.5% O; molar mass of aspirin = 180.2
1 mol C
5.00 mol C
b = 5.00 mol C
= 2.25 mol C
160.0 g C2a
12.01 g C
2.22
4.44 mol H
1 mol H
14.48 g H2a
b = 4.44 mol H
= 2.00 mol H
1.008 g H
2.22
135.5 g O2a
1 mol O
b = 2.22 mol O
16.00 g O
2.22 mol O
= 1.00 mol O
2.22
Multiplying each by 4 give the empirical formula C9H 8O4 . 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, C9H 8O4 .
64.
Calculate the percent oxygen in Al 2(SO4)3 .
2 Al
3S
12 O
65.
53.96 g
96.21 g
192.0 g
342.2 g
a
192.0 g
b11002 = 56.11% O
342.2 g
Now take 56.11% of 8.50 g
18.50 g Al2(SO4)3210.56112 = 4.77 g O
Empirical formula of gallium arsenide; 48.2% Ga, 51.8% As
1 mol Ga
b = 0.691 mol Ga
69.72 g Ga
1 mol As
151.8 g As2a
b = 0.691 mol As
74.92 g As
148.2 g Ga2a
The empirical formula is GaAs.
- 81 -
0.691 mol Ga
= 1.00 mol Ga
0.691
0.691 mol As
= 1.00 mol As
0.691
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- Chapter 7 66.
Empirical formula of calcium tartrate; 25.5% C, 2.1% H, 21.3% Ca, 51.0% O.
125.5 g C2a
1 mol C
b = 2.12 mol C
12.01 g C
2.212 mol C
= 3.99 mol C
0.531
12.1 g H2a
1 mol H
2.1 mol H
b = 2.1 mol H
= 4.0 mol H
1.008 g H
0.531
1 mol Ca
0.529 mol Ca
121.2 g Ca2a
b = 0.531 mol Ca
= 1.00 mol Ca
40.08 g Ca
0.531
151.0 g O2a
1 mol O
b = 3.19 mol O
16.00 g O
3.19 mol O
= 6.01 mol O
0.531
The empirical formula is C4H4CaO6
67.
(a)
7.79% C, 92.21% Cl
17.79 g C2a
1 mol C
0.649 mol C
b = 0.649 mol C
= 1.00 mol C
12.01 g C
0.649
1 mol Cl
2.601 mol Cl
192.21 g Cl2a
b = 2.601 mol Cl
= 4.01 mol Cl
35.45 g Cl
0.649
The empirical formula is CCl 4 . The empirical formula mass is 153.8 which equals
the molar mass, therefore the molecular formula is CCl 4 .
(b)
10.13% C, 89.87% Cl
1 mol C
0.8435 mol C
110.13 g C2a
b = 0.8435 mol C
= 1.000 mol C
12.01 g C
0.8435
2.535 mol Cl
1 mol Cl
189.87 g Cl2a
b = 2.535 mol Cl
= 3.005 mol Cl
35.45 g Cl
0.8435
The empirical formula is CCl 3 . The empirical formula mass is 118.4 g.
236.7 g
molar mass
=
= 1.999
empirical formula mass
118.4 g
The molecular formula is twice that of the empirical formula.
Molecular formula = C2Cl 6 .
(c)
25.26% C, 74.74% Cl
1 mol C
125.26 g C2a
b = 2.103 mol C
12.01 g C
1 mol Cl
b = 2.108 mol Cl
174.74 g Cl2a
35.45 g Cl
- 82 -
2.103 mol C
= 1.000 mol C
2.103
2.103 mol Cl
= 1.002 mol Cl
2.108
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The empirical formula is CCl. The empirical formula mass is 47.46 g.
284.8 g
molar mass
=
= 6.000
empirical formula mass
47.46 g
The molecular formula is six times that of the empirical formula.
Molecular formula = C6Cl 6 .
(d)
11.25% C, 88.75% Cl
1 mol C
111.25 g C2a
b = 0.9367 mol C
12.01 g C
1 mol Cl
b = 2.504 mol Cl
188.75 g Cl2a
35.45 g Cl
0.9367 mol C
= 1.000 mol C
0.9367
2.504 mol Cl
= 2.673 mol Cl
0.9367
Multiplying each by 3 give the empirical formula C3Cl 8 . The empirical formula
mass is 319.6. Since the molar mass is also 319.6 the molecular formula is C3Cl 8 .
68.
The conversion is: s ¡ min ¡ hr ¡ day ¡ yr
16.022 * 1023 s2a
69.
The conversion is: g ¡ mol ¡ atom
12.5 g Cu2a
70.
1 day
1 year
1 min
1 hr
ba
ba
ba
b = 1.910 * 1016 years
60 s
60 min
24 hr
365 days
1 mol Cu
6.022 * 1023 atoms
b¢
≤ = 2.4 * 1022 atoms Cu
63.55 g Cu
mol
The conversion is: molecules ¡ mol ¡ g
11000. * 1012 molecules C3H8O32 ¢
1 trillion = 1012
92.09 g C3H8O3
1 mol
b
≤a
23
mol C3H8O3
6.022 * 10 molecules
= 1.529 * 10 - 7 g C3H 8O3
71.
16.1 * 109 people2 ¢
72.
Empirical formula
1 mol people
6.022 * 1023 people
≤ = 1.0 * 10-14 mol of people
23.3% Co, 25.3% Mo, 51.4% Cl
123.3 g Co2a
1 mol Co
b = 0.395 mol Co
58.93 g Co
- 83 -
0.395 mol Co
= 1.50 mol Co
0.264
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- Chapter 7 -
125.3 g Mo2a
1 mol Mo
0.264 mol Mo
b = 0.264 mol Mo
= 1.00 mol Mo
95.94 g Mo
0.264
1 mol Cl
1.45 mol Cl
151.4 g Cl2a
b = 1.45 mol Cl
= 5.49 mol Cl
35.45 g Cl
0.264
Multiplying by 2 gives the empirical formula Co 3Mo 2Cl 11 .
73.
74.
The conversion is: g Al ¡ mol Al ¡ mol Mg ¡ g Mg
2 mol Mg 24.31 g Mg
1 mol Al
ba
ba
b = 32 g Mg
118 g Al2a
26.98 g Al
1 mol Al
mol Mg
110.0 g compound210.1772 = 1.77 g N
1 mol N
b = 0.126 mol N
11.77 g N2a
14.01 g N
13.8 * 1023 atoms H2 ¢
1 mol
≤ = 0.63 mol H
6.022 * 1023 atoms
To determine the mol C, first find grams H and subtract the grams of H and N from the
grams of the sample.
1.008 g H
b = 0.64 g H
mol H
10.0 g sample
-1.77 g N
-0.64 g H
7.6 g C
10.63 mol H2a
17.6 g C2a
1 mol C
b = 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
= 1.00 mol N
N
0.126
0.63 mol H
= 5.0 mol H
H
0.126
0.63 mol C
C
= 5.0 mol C
0.126
The empirical formula is C5H 5N
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- Chapter 7 75.
Let x = molar mass of A 2O
0.400x = 16.00 g O (Since A 2O has only one mol of O atoms)
x = 40.0 g O>mol A 2O
40.0 = 16.00 + 2y
y = molar mass of A
40.0 - 16.00 = 2y
g
12.0
= y
mol
Look in the periodic table for the element that has 12.0 g>mol.
The element is carbon. The mystery element is carbon.
CH 2O
C4H 9
CH 2O
C25H 52
C6H 2Cl 2O
(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)
76.
(a)
(b)
(c)
(d)
(e)
77.
First determine the elements in compound A(BC)3 :
A: 10.34592178.01 g2 = 26.98 g (aluminum)
48.00 g
= 16.00 g (oxygen)
B: (0.6153)(78.01 g) =
3
3.03 g
C: 10.03882178.01 g2 =
= 1.01 g (hydrogen)
3
Elements determined from atomic masses in the periodic table.
A(BC)3 = Al(OH)3
Then compound A 2B3 = Al 2O3 with a molar mass of
2126.98 g2 + 3116.00 g2 = 102.0 g
2126.98 g2
11002 = 52.90%
102.0 g
3116.002
%O =
11002 = 47.06%
103.0
% Al =
78.
(a)
Percent composition of the original unknown compound.
Convert g CO2 to g C and g H2O to g H
12.01 g C
b = 1.303 g C
(4.776 g CO2)a
44.01 g CO2
(2.934 g H2O)a
2.016 g H
b = 0.3282 g H
18.02 g H2O
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2.500 g compound -1.303 g C -0.3282 g H = 0.869 g O
(b)
a
1.303 g C
b(100) = 52.12% C
2.500 g
a
0.328 g H
b(100) = 13.13% H
2.500 g
a
0.869 g O
b(100) = 34.76% O
2.500 g
Empirical formula of unknown compound; 52.12% C, 13.13% H, 34.76% O.
(52.12 g C)a
1 mol C
b = 4.340
12.01 g
(13.13 g H)a
1 mol H
b = 13.03
1.008 g H
13.03 mol H
= 5.996 mol H
2.173
(34.76 g O)a
1 mol O
b = 2.173
16.00 g O
2.173 mol O
= 1.000 mol O
2.173
4.340 mol C
= 1.997 mol C
2.173
The empirical formula is C2H6O
- 86 -