CHAPTER 8
Chemical Composition
1.
100 washers
0.110 g
1 washer
= 11.0 g (assuming 100 washers is exact).
100. g
1 washer
0.110 g
= 909 washers
2.
The empirical formula is CFH from the structure given. The empirical formula represents the
smallest whole number ratio of the number and types of atoms present.
3.
The atomic mass unit (amu) is defined by scientists to more simply describe relative masses on an
atomic or molecular scale. One amu is equivalent to 1.66 10–24 g.
4.
The average atomic mass takes into account the various isotopes of an element and the relative
abundances in which those isotopes are found.
5.
a.
125 C atoms
12.01 amu
1 C atom
= 1.50 × 103 amu
b.
5 106 K atoms
39.10 amu
1 K atom
= 1.955 × 108 amu = 2 108 amu
c.
1.04 × 1022 Li atoms
6.941 amu
1 Li atom
= 7.22 1022 amu
d.
1 Mg atom
24.31 amu
1 Mg atom
= 24.31 amu
e.
3.011 1023 I atoms
126.9 amu
1 I atom
= 3.821 1025 amu
6.
a.
40.08 amu Ca ×
1 Ca atom
40.08 amu
= 1 Ca atom
b.
919.5 amu W ×
1 W atom
183.9 amu
= 5 W atoms
c.
549.4 amu Mn ×
1 Mn atom
54.94 amu
= 10 Mn atoms
d.
6345 amu I ×
1 I atom
126.9 amu
= 50 I atoms
e.
2072 amu ×
1 Pb atom
207.2 amu
= 10 Pb atoms
117
Chapter 8: Chemical Composition
7.
One “average” iron atom has a mass of 55.85 amu
299 iron atoms would weigh: 299 atoms ×
55.85 amu
1 atom
= 1.67 × 104 amu
5529.2 amu represents: 5529.2 amu ×
1 atom
55.85 amu
= 99 atoms.
8.
One bromine atom has a mass of 79.90 amu.
A sample containing 54 bromine atoms would weigh:
;
5672.9 amu of bromine would represent:
.
9.
Avogadro’s number (6.022 1023 atoms; 1.00 mol)
10.
118.7 g (1.00 mol)
11.
molar masses: Na, 22.99 g; K, 39.10 g
11.50 g Na
1 mol Na
22.99 g
= 0.5002 mol Na
0.5002 mol Na
23
6.033 10
1 mol Na
= 3.012 1023 atoms
0.5002 mol K
39.10 g
1 mol K
= 19.56 g K
12.
molar masses: Ag, 107.9 g; Cu, 63.55 g
13.
The ratio of the atomic mass of H to the atomic mass of N is (1.008 amu/14.01 amu), and the
mass of hydrogen is given by
7.00 g N
1.008 amu
14.01 amu
= 0.504 g H.
118
Chapter 8: Chemical Composition
14.
The ratio of the atomic mass of Co to the atomic mass of F is (58.93 amu Co/19.00 amu F), and
the mass of cobalt is given by
57.0 g F
= 177 g Co.
15.
mass of a magnesium atom = 3.82 × 10–23 g ×
24.31 amu Mg
22.99 amu Na
= 4.04 × 10–23 g
16.
mass of a chlorine atom = 3.16 × 10–23 g ×
= 5.90 × 10–23 g
17.
1 mol of He atoms = 4.003 g
4 mol of H atoms
1.008 g H
1 mol
= 4.032 g
4 mol of H atoms has a mass slightly larger than 1 mol of He atoms.
18.
0.25 mol Xe atoms
= 32.83 g Xe = 33 g Xe (two significant figures)
2.0 mol C atoms
= 24.02 g C = 24 g C (two significant figures)
The carbon sample weighs less.
19.
a.
4.95 g Ne
1 mol
20.18 g
= 0.245 mol Ne
b.
72.5 g Ni
1 mol
58.69 g
= 1.24 mol Ni
c.
115 mg Ag
1 g
1 mol
1000 mg
107.9 g
= 1.07 × 10–3 mol Ag
d.
6.22g U
6
1 g
1 mol
10 μg
238.0 g
= 2.61 × 10–8 mol U
e.
135 g I ×
= 1.06 mol I
20.
a.
49.2 g S ×
= 1.53 mol of S
b.
7.44 × 104 kg Pb ×
×
= 3.59 × 105 mol Pb
c.
3.27 mg Cl ×
×
= 9.22 × 10–5 mol Cl
119
Chapter 8: Chemical Composition
d.
4.01 g Li ×
= 0.578 mol Li
e.
100.0 g Cu ×
= 1.574 mol Cu
f.
82.6 mg Sr ×
×
= 9.43 × 10–4 mol Sr
21.
a.
0.251 mol Li ×
6.941 g Li
1 mol Li
= 1.74 g Li
b.
1.51 mol Al ×
26.98 g Al
1 mol Al
= 40.7 g Al
c.
8.75 × 10–2 mol Pb ×
207.2 g Pb
1 mol Pb
= 18.1 g Pb
d.
125 mol Cr ×
52.00 g Cr
1 mol Cr
= 6.50 × 103 g Cr
e.
4.25 × 103 mol Fe ×
55.85 g Fe
1 mol
= 2.37 × 105 g Fe
f.
0.000105 mol Mg ×
24.31 g Mg
1 mol Mg
= 2.55 × 10–3 g Mg
22.
a.
0.00552 mol Ca
40.08 g
1 mol
= 0.221 g Ca
b.
6.25 millimol B
3
1 mol
10.81 g
10 millmol
1 mol
= 0.0676 g B
c.
135 mol Al
26.98 g
1 mol
= 3.64 × 103 g Al
d.
1.34 × 10–7 mol Ba
137.3 g
1 mol
= 1.84 × 10–5 g Ba
e.
2.79 mol P
30.97 g
1 mol
= 86.4 g P
f.
0.0000997 mol As
74.92 g
1 mol
= 7.47 × 10–3 g As
23.
a.
1.50 g Ag
23
6.022 10 Ag atoms
107.9 g Ag
= 8.37 1021 Ag atoms
b.
0.0015 mol Cu
23
6.022 10 Cu atoms
1 mol
= 9.0 1020 Cu atoms
120
Chapter 8: Chemical Composition
c.
0.0015 g Cu
23
6.022 10 Cu atoms
63.55 g Cu
= 1.4 1019 Cu atoms
d.
2.00 kg = 2.00 103 g
2.00 103 g Mg
23
6.022 10 Mg atoms
24.31 g Mg
= 4.95 1025 Mg atoms
e.
1.000 oz = 28.35 g
2.34 oz
23
28.35 g
6.022 10 Ca atoms
1.000 oz
40.08 g Ca
= 9.97 1023 Ca atoms
f.
2.34 g Ca
23
6.022 10 Ca atoms
40.08 g Ca
= 3.52 1022 Ca atoms
g.
2.34 mol Ca
23
6.022 10 Ca atoms
1 mol Ca
= 1.41 1024 Ca atoms
24.
a.
125 Fe atoms ×
23
55.85 g Fe
6.022 10 Fe atoms
= 1.16 × 10–20 g
b.
125 Fe atoms ×
55.85 amu
1 Fe atom
= 6.98 × 103 amu
c.
125 g Fe ×
1 mol Fe
55.85 g Fe
= 2.24 mol Fe
d.
125 mol Fe ×
55.85 g Fe
1 mol Fe
= 6.98 × 103 g Fe
e.
125 g Fe ×
23
6.022 10 Fe atoms
55.85 g Fe
= 1.35 × 1024 Fe atoms
f.
125 mol Fe ×
23
6.022 10 Fe atoms
1 mol Fe
= 7.53 × 1025 Fe atoms
25.
molar mass
26.
(Answers will vary.) The molar mass is calculated by summing the individual atomic masses of
the atoms in the formula. In the compound CH4, the atomic mass of carbon and the atomic mass
of four hydrogens are summed (giving a molar mass of 16.042 g/mol).
27.
a.
H3PO4 phosphoric acid
mass of 3 mol H = 3(1.008 g) = 3.024 g
mass of 1 mol P = 30.97 g
mass of 4 mol O = 4(16.00 g) = 64.00 g
molar mass of H3PO4 = (3.024 g + 30.97 g + 64.00 g) = 97.99 g
121
Chapter 8: Chemical Composition
b.
Fe2O3 ferric oxide, iron(III) oxide
mass of 2 mol Fe = 2(55.85 g) = 111.7 g
mass of 3 mol O = 3(16.00 g) = 48.00 g
molar mass of Fe2O3 = (111.7 g + 48.00 g) = 159.7 g
c.
NaClO4 sodium perchlorate
mass of 1 mol Na = 22.99 g
mass of 1 mol Cl = 35.45 g
mass of 4 mol O = 4(16.00 g) = 64.00 g
molar mass of NaClO4 = (22.99 g + 35.45 g + 64.00 g) = 122.44 g
d.
PbCl2 plumbous chloride, lead(II) chloride
mass of 1 mol Pb = 207.2 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of PbCl2 = (207.2 g + 70.90 g) = 278.1 g
e.
HBr hydrogen bromide
mass of 1 mol H = 1.008 g
mass of 1 mol Br = 79.90 g
molar mass of HBr = (1.008 g + 79.90 g) = 80.91 g
f.
Al(OH)3 aluminum hydroxide
mass of 1 mol Al = 26.98 g
mass of 3 mol H = 3(1.008) = 3.024 g
mass of 3 mol O = 3(16.00 g) = 48.00 g
molar mass of Al(OH)3 = (26.98 g + 3.024 g + 48.00 g) = 78.00 g
28.
a.
KHCO3 potassium hydrogen carbonate, potassium bicarbonate
mass of 1 mol K = 39.10 g
mass of 1 mol H = 1.008 g
mass of 1 mol C = 12.01 g
mass of 3 mol O = 3(16.00 g) = 48.00 g
molar mass of KHCO3 = (39.10 + 1.008 + 12.01 + 48.00) = 100.12 g
b.
Hg2Cl2 mercurous chloride, mercury(I) chloride
mass of 2 mol Hg = 2(200.6 g) = 401.2 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of Hg2Cl2 = (401.2 g + 70.90 g) = 472.1 g
122
Chapter 8: Chemical Composition
c.
H2O2 hydrogen peroxide
mass of 2 mol H = 2(1.008 g) = 2.016 g
mass of 2 mol O = 2(16.00 g) = 32.00 g
molar mass of H2O2 = (2.016 g + 32.00 g) = 34.02 g
d.
BeCl2 beryllium chloride
mass of 1 mol Be = 9.012 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of BeCl2 = (9.012 g + 70.90 g) = 79.91 g
e.
Al2(SO4)3 aluminum sulfate
mass of 2 mol Al = 2(26.98 g) = 53.96 g
mass of 3 mol S = 3(32.07 g) = 96.21 g
mass of 12 mol O = 12(16.00 g) = 192.0 g
molar mass of Al2(SO4)3 = (53.96 g + 96.21 g + 192.0 g) = 342.2 g
f.
KClO3 potassium chlorate
mass of 1 mol K = 39.10 g
mass of 1 mol Cl = 35.45 g
mass of 3 mol O = 3(16.00 g) = 48.00 g
molar mass of KClO3 = 122.55 g
29.
a.
BaCl2
mass of 1 mol Ba = 137.3 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of BaCl2 = (137.3 g + 70.90 g) = 208.2 g
b.
Al(NO3)3
mass of 1 mol Al = 26.98 g
mass of 3 mol N = 3(14.01 g) = 42.03 g
mass of 9 mol O = 9(16.00 g) = 144.00 g
molar mass of Al(NO3)3 = (26.98 g + 42.03 g + 144.00 g) = 213.0 g
c.
FeCl2
mass of 1 mol Fe = 55.85 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of FeCl2 = (55.85 g + 70.90 g) = 126.75 g
123
Chapter 8: Chemical Composition
d.
SO2
mass of 1 mol S = 32.07 g
mass of 2 mol O = 2(16.00 g) = 32.00 g
molar mass of SO2 = (32.07 g + 32.00 g) = 64.07 g
e.
Ca(C2H3O2)2
mass of 1 mol Ca = 40.08 g
mass of 4 mol C = 4(12.01 g) = 48.04 g
mass of 6 mol H = 6(1.008 g) = 6.048 g
mass of 4 mol O = 4(16.00 g) = 64.00 g
molar mass of Ca(C2H3O2)2 = (40.08 g + 48.04 g + 6.048 g + 64.00 g) = 158.17 g
30.
a.
Ba(ClO4)2
mass of 1 mol Ba = 137.3 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
mass of 8 mol O = 8(16.00 g) = 128.00 g
molar mass of Ba(ClO4)2 = (137.3 g + 70.90 g + 128.00 g) = 336.2 g
b.
MgSO4
mass of 1 mol Mg = 24.31 g
mass of 1 mol S = 32.07 g
mass of 4 mol O = 4(16.00 g) = 64.00 g
molar mass of MgSO4 = (24.31 g + 32.07 g + 64.00 g) = 120.38 g
c.
PbCl2
mass of 1 mol Pb = 207.2 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of PbCl2 = (207.2 g + 70.90 g) = 278.1 g
d.
Cu(NO3)2
mass of 1 mol Cu = 63.55 g
mass of 2 mol N = 2(14.01 g) = 28.02 g
mass of 6 mol O = 6(16.00 g) = 96.00 g
molar mass of Cu(NO3)2 = (63.55 g + 28.02 g + 96.00 g) = 187.57 g
e.
SnCl4
mass of 1 mol Sn = 118.7 g
mass of 4 mol Cl = 4(35.45 g) = 141.8 g
molar mass of SnCl4 = (118.7 g + 141.8 g) = 260.5 g
124
Chapter 8: Chemical Composition
31.
a.
molar mass of NO2 = 46.01 g;
21.4 mg = 0.0214 g
0.0214 g NO2
1 mol
46.01 g
= 4.65 10–4 mol NO2
b.
molar mass of Cu(NO3)2 = 187.6 g
1.56 g Cu(NO3)2
1 mol
187.6 g
= 8.32 10–3 mol Cu(NO3)2
c.
molar mass of CS2 = 76.15 g
2.47 g CS2
1 mol
76.15 g
= 0.0324 mol
d.
molar mass of Al2(SO4)3 = 342.2 g
5.04 g Al2(SO4)3
1 mol
342.2 g
= 0.0147 mol Al2(SO4)3
e.
molar mass of PbCl2 = 278.1 g
2.99 g
1 mol
278.1 g
= 0.0108 mol PbCl2
f.
molar mass of CaCO3 = 100.09 g
62.4 g CaCO3
1 mol
100.09 g
= 0.623 mol CaCO3
32.
a.
molar mass Al2O3 = 101.96 g
47.2 g ×
1 mol
101.96 g
= 0.463 mol
b.
molar mass KBr = 119.00 g
1.34 kg ×
1000 g
1 mol
1 kg
119.00 g
= 11.3 mol
c.
molar mass Ge = 72.59 g
521 mg ×
1 g
1 mol
1000 mg
72.59 g
= 7.18 × 10–3 mol
d.
molar mass of U = 238.0 g
56.2 g ×
6
1 g
1 mol
10 μg
238.0 g
= 2.36 × 10–7 mol
e.
molar mass of NaC2H3O2 = 82.03 g
29.7 g ×
1 mol
82.03 g
= 1.69 mol = 0.362 mol
125
Chapter 8: Chemical Composition
f.
molar mass of SO3 = 80.07 g
1.03 g ×
1 mol
80.07 g
= 0.0129 mol
33.
a.
molar mass MgCl2 = 95.21 g
41.5 g ×
1 mol
95.21 g
= 0.436 mol
b.
molar mass Li2O = 29.88 g; 135 mg = 0.135 g
0.135 g ×
1 mol
29.88 g
= 4.52 × 10–3 mol = 4.52 mmol
c.
molar mass Cr = 52.00 g; 1.21 kg = 1210 g
1210 g ×
1 mol
52.00 g
= 23.3 mol
d.
molar mass H2SO4 = 98.09 g
62.5 g ×
1 mol
98.09 g
= 0.637 mol
e.
molar mass C6H6 = 78.11 g
42.7 g ×
1 mol
78.11 g
= 0.547 mol
f.
molar mass H2O2 = 34.02 g
135 g ×
1 mol
34.02 g
= 3.97 mol
34.
a.
molar mass of Li2CO3 = 73.89 g
1.95 10–3 g
1 mol
73.89 g
= 2.64 10–5 mol
b.
molar mass of CaCl2 = 110.98 g
4.23 kg
1000 g
1 mol
1 kg
110.98 g
= 38.1 mol
c.
molar mass of SrCl2 = 158.52 g
1.23 mg
1 g
1 mol
1000 mg 158.52 g
= 7.76 10–6 mol
d.
molar mass of CaSO4 = 136.15 g
4.75 g
1 mol
136.15 g
= 3.49 × 10–2 mol
126
Chapter 8: Chemical Composition
e.
molar mass of NO2 = 46.01 g
96.2 mg
1 g
1 mol
1000 mg
46.01 g
= 2.09 × 10–3 mol
f.
molar mass of Hg2Cl2 = 472.1 g
12.7 g ×
1 mol
472.1 g
= 0.0269 mol
35.
a.
molar mass AlCl3 = 133.3 g
1.25 mol ×
133.3 g
1 mol
= 167 g
b.
molar mass NaHCO3 = 84.01 g
3.35 mol ×
84.01 g
1 mol
= 281 g
c.
molar mass HBr = 80.91 g
4.25 mmol ×
80.91 mg
1 mmol
= 344 mg = 0.344 g
d.
molar mass U = 238.0 g
1.31 × 10–3 mol ×
238.0 g
1 mol
= 0.312 g
e.
molar mass CO2 = 44.01 g
0.00104 mol ×
44.01 g
1 mol
= 0.0458 g
f.
molar mass Fe = 55.85 g
1.49 × 102 mol Fe ×
55.85 g
1 mol
= 8.32 × 103 g
36.
a.
molar mass of SO3 = 80.07 g
6.14 10–4 mol
= 0.0492 g
b.
molar mass of PbO2 = 239.2 g
3.11 105 mol
= 7.44 107 g
c.
molar mass of CHCl3 = 119.368 g
0.495 mol
= 59.1 g
d.
molar mass of C2H3Cl3 = 133.394 g
127
Chapter 8: Chemical Composition
2.45 10–8 mol
= 3.27 10–6 g
e.
molar mass of LiOH = 23.949 g
0.167 mol
= 4.00 g
f.
molar mass of CuCl = 99.00 g
5.26 mol
= 521 g
37.
a.
molar mass of C2H6O = 46.07 g
0.251 mol
46.07 g
1 mol
= 11.6 g C2H6O
b.
molar mass of CO2 = 44.01 g
1.26 mol
44.01 g
1 mol
= 55.5 g CO2
c.
molar mass of AuCl3 = 303.4 g
9.31 10–4 mol
303.4 g
1 mol
= 0.282 g AuCl3
d.
molar mass of NaNO3 = 85.00 g
7.74 mol
85.00 g
1 mol
= 658 g NaNO3
e.
molar mass of Fe = 55.85 g
0.000357 mol
55.85 g
1 mol
= 0.0199 g Fe
38.
a.
molar mass C6H6 = 78.11 g
0.994 mol ×
78.11 g
1 mol
= 77.6 g
b.
molar mass CaH2 = 42.10 g
4.21 mol ×
42.10 g
1 mol
= 177 g
c.
molar mass H2O2 = 34.02 g
1.79 × 10–4 mol ×
34.02 g
1 mol
= 6.09 × 10–3 g
d.
molar mass C6H12O6 = 180.16 g
1.22 mmol ×
= 0.220 g
128
Chapter 8: Chemical Composition
e.
molar mass Sn = 118.7 g
10.6 mol ×
118.7 g
1 mol
= 1.26 × 103 g
f.
molar mass SrF2 = 125.62 g
0.000301 mol ×
125.62 g
1 mol
= 0.0378 g
39.
a.
4.75 millimol = 0.00475 mol
0.00475 mol
23
6.022 10 molecules
1 mol
= 2.86 1021 molecules
b.
molar mass of PH3 = 33.99 g
4.75 g
23
6.022 10 molecules
33.99 g
= 8.42 1022 molecules
c.
molar mass of Pb(C2H3O2)2 = 325.3 g
1.25 10–2 g
23
6.022 10 formula units
325.3 g
= 2.31 1019 formula units
d.
1.25 10–2 mol
23
6.022 10 formula units
1 mol
= 7.53 1021 formula units
e.
If the sample contains a total of 5.40 mol of carbon, then because each benzene contains
six carbons, there must be (5.40/6) = 0.900 mol of benzene present.
0.900 mol
23
6.022 10 molecules
1 mol
= 5.42 1023 molecules
40.
a.
3.54 mol SO2
23
6.022 10 molecules
1 mol
= 2.13 1024 molecules SO2
b.
molar mass of SO2 = 64.07 g
3.54 g
= 3.33 1022 molecules SO2
c.
molar mass of NH3 = 17.034 g
4.46 10–5 g
= 1.58 1018 molecules NH3
d.
4.46 10–5 mol NH3
23
6.022 10 molecules
1 mol
= 2.69 1019 molecules NH3
e.
molar mass of C2H6 = 30.068 g
1.96 mg
= 3.93 1019 molecules
C2H6
129
Chapter 8: Chemical Composition
41.
a.
molar mass of C2H6O = 46.07 g
1.271 g
1 mol
46.07 g
= 0.02759 mol C2H6O
0.02759 mol C2H6O
2
6
2 mol C
1 mol C H O
= 0.05518 mol C
b.
molar mass of C6H4Cl2 = 147.0 g
3.982 g
1 mol
147.0 g
= 0.027088 mol C6H4Cl2
0.027088 mol C6H4Cl2
6
4
2
6 mol C
1 mol C H Cl
= 0.1625 mol C
c.
molar mass of C3O2 = 68.03 g
0.4438 g
1 mol
68.03 g
= 0.0065236 mol C3O2
0.0065236 mol C3O2
3
2
3 mol C
1 mol C O
= 0.01957 mol C
d.
molar mass of CH2Cl2 = 84.93 g
2.910 g
1 mol
84.93 g
= 0.034264 mol CH2Cl2
0.034264 mol CH2Cl2
2
2
1 mol C
1 mol CH Cl
= 0.03426 mol C
42.
a.
molar mass of Na2SO4 = 142.1 g
2.01 g Na2SO4
2
4
2
4
1 mol Na SO
1 mol S
142.1 g
1 mol Na SO
= 0.0141 mol S
b.
molar mass of Na2SO3 = 126.1 g
2.01 g Na2SO3
2
3
2
3
1 mol Na SO
1 mol S
126.1 g
1 mol Na SO
= 0.0159 mol S
c.
molar mass of Na2S = 78.05 g
2.01 g Na2S
2
2
1 mol Na S
1 mol S
78.05 g
1 mol Na S
= 0.0258 mol S
d.
molar mass of Na2S2O3 = 158.1 g
2.01 g Na2S2O3
= 0.0254 mol S
43.
molar
130
Chapter 8: Chemical Composition
44.
The percent composition of each element in the compound does not change because a compound
consists of the same percent composition by mass regardless of the starting amount in the sample.
45.
a.
mass of H present = 1.008 g = 1.008 g
mass of Cl present = 35.45 g = 35.45 g
mass of O present = 3(16.00 g) = 48.00 g
molar mass of HClO3 = 84.46 g
% H =
1.008 g H
84.46 g
100 = 1.193 %H
% Cl =
35.45 g Cl
84.46 g
100 = 41.97 %Cl
% O =
48.00 g O
84.46 g
100 = 56.83 %O
b.
mass of U present = 238.0 g = 238.0 g
mass of F present = 4(19.00 g) = 76.00 g
molar mass of UF4 =
314.0 g
% U =
238.0 g U
314.0 g
100 = 75.80% U
% F =
76.00 g F
314.0 g
100 = 24.20% F
c.
mass of Ca present = 40.08 g = 40.08 g
mass of H present = 2(1.008 g) = 2.016 g
molar mass of CaH2 = 42.10 g
% Ca =
40.08 g C
42.10 g
100 = 95.21% Ca
% H =
2.016 g H
42.10 g
100 = 4.789% H
d.
mass of Ag present =
2(107.9 g) = 215.8 g
mass of S present = 32.07 g =
32.07 g
molar mass of Ag2S = 247.9 g
% Ag =
215.8 g Ag
247.9 g
100 = 87.06% Ag
% S =
32.07 g S
247.9 g
100 = 12.94% S
131
Chapter 8: Chemical Composition
e.
mass of Na present = 22.99 g = 22.99 g
mass of H present = 1.008 g = 1.008 g
mass of S present = 32.07 g = 32.07 g
mass of O present = 3(16.00 g) = 48.00 g
molar mass of NaHSO3 = 104.07 g
% Na =
22.99 g Na
104.07 g
100 = 22.09% Na
% H =
1.008 g H
104.07 g
100 = 0.9686% H
% S =
32.07 g S
104.07 g
100 = 30.82% S
% O =
48.00 g O
104.07 g
100 = 46.12% O
f.
mass of Mn present = 54.94 g = 54.94 g
mass of O present = 2(16.00 g) = 32.00 g
molar mass of MnO2 = 86.94 g
% Mn =
54.94 g Mn
86.94 g
100 = 63.19% Mn
% S =
32.00 g O
86.94 g
100 = 36.81% O
46.
a.
mass of Zn present = 65.38 g
mass of O present = 16.00 g
molar mass of ZnO = 81.38 g
%Zn =
65.38 g Zn
100
81.38 g
= 80.34% Zn
%O =
16.00 g O
81.38 g
× 100 = 19.66% O
b.
mass of Na present = 2(22.99 g) = 45.98 g
mass of S present = 32.07 g
molar mass of Na2S = 78.05 g
%Na =
45.98 g Na
78.05 g
× 100 = 58.91% Na
%S =
32.07 g S
78.05 g
× 100 = 41.09% S
132
Chapter 8: Chemical Composition
c.
mass of Mg present = 24.31 g
mass of O present = 2(16.00 g) = 32.00 g
mass of H present = 2(1.008 g) = 2.016 g
molar mass of Mg(OH)2 = 58.33 g
%Mg =
24.31 g Mg
58.33 g
× 100 = 41.68% Mg
%O =
32.00 g O
58.33 g
× 100 = 54.86% O
%H =
2.016 g H
58.33 g
× 100 = 3.456% H
d.
mass of H present = 2(1.008 g) = 2.016 g
mass of O present = 2(16.00 g) = 32.00 g
molar mass of H2O2 = 34.02 g
%H =
2.016 g H
34.02 g
× 100 = 5.926% H
%O =
32.00 g O
34.02 g
× 100 = 94.06% O
e.
mass of Ca present = 40.08 g
mass of H present = 2(1.008 g) = 2.016 g
molar mass of CaH2 = 42.10 g
%Ca =
40.08 g Ca
42.10 g
× 100 = 95.20% Ca
%H =
2.016 g H
42.10 g
× 100 = 4.789% H
f.
mass of K present = 2(39.10 g) = 78.20 g
mass of O present = 16.00 g
molar mass of K2O = 94.20 g
%K =
78.20 g K
94.20 g
× 100 =83.01% K
%O =
16.00 g O
94.20 g
× 100 = 16.99% O
47.
a.
molar mass of CH4 = 16.04 g
% C =
12.01 g C
16.04 g
100 = 74.88% C
133
Chapter 8: Chemical Composition
b.
molar mass NaNO3 = 85.00 g
% Na =
22.99 g Na
85.00 g
100 = 27.05% Na
c.
molar mass of CO = 28.01 g
% C =
12.01 g C
28.01 g
100 = 42.88% C
d.
molar mass of NO2 = 46.01 g
%N =
14.01 g N
46.01 g
× 100 =30.45% N
e.
molar mass of C8H18O = 130.2 g
% C =
96.08 g C
130.2 g
100 = 73.79% C
f.
molar mass of Ca3(PO4)2 = 310.1 g
% Ca =
120.2 g Ca
310.1 g
100 = 38.76% Ca
g.
molar mass of C12H10O = 170.2 g
% C =
144.1 g C
170.2 g
100 = 84.67% C
h.
molar mass of Al(C2H3O2)3 = 204.1 g
% Al =
26.98 g Al
204.1 g
100 = 13.22% Al
48.
a.
molar mass of CuBr2 = 223.35 g
% Cu =
100 = 28.45% Cu
b.
molar mass of CuBr = 143.45 g
% Cu =
100 = 44.30% Cu
c.
molar mass of FeCl2 = 126.75 g
% Fe =
100 = 44.06% Fe
d.
molar mass of FeCl3 = 162.2 g
% Fe =
100 = 34.43% Fe
134
Chapter 8: Chemical Composition
e.
molar mass of CoI2 = 312.73 g
% Co =
100 = 18.84% Co
f.
molar mass of CoI3 = 439.63 g
% Co =
100 = 13.40% Co
g.
molar mass of SnO = 134.7 g
% Sn =
100 = 88.12% Sn
h.
molar mass of SnO2 = 150.7 g
% Sn =
100 = 78.77% Sn
49.
a.
molar mass of C6H10O4 = 146.1 g
% C =
72.06 g C
146.1 g
100 = 49.32% C
b.
molar mass of NH4NO3 = 80.05 g
% N =
28.02 g N
80.05 g
100 = 35.00% N
c.
molar mass of C8H10N4O2 = 194.2 g
% C =
96.09 g C
194.2 g
100 = 49.47% C
d.
molar mass of ClO2 = 67.45 g
% Cl =
35.45 g Cl
67.45 g
100 = 52.56% Cl
e.
molar mass of C6H11OH = 100.2 g
% C =
72.06 g C
100.2 g
100 = 71.92% C
f.
molar mass of C6H12O6 = 180.2 g
% C =
72.06 g C
180.2 g
100 = 39.99% C
g.
molar mass of C20H42 = 282.5 g
% C =
240.2 g C
282.5 g
100 = 85.03% C
135
Chapter 8: Chemical Composition
h.
molar mass of C2H5OH = 46.07 g
% C =
24.02 g C
46.07 g
100 = 52.14% C
50.
a.
molar mass of CO = 28.01 g
% O =
100 = 57.12% O
b.
molar mass of MnO2 = 86.94 g
% O =
100 = 36.81% O
c.
molar mass of KClO3 = 122.55 g
% O =
100 = 39.17% O
d.
molar mass of FeO = 71.85 g
% O =
100 = 22.27% O
e.
molar mass of Ca(OH)2 = 74.096 g
% O =
100 = 43.19% O
51.
a.
molar mass of NH4I = 144.94 g
4.25 g
+
4
4
4
1 mol NH I 1 mol NH
144.94 g
1 mol NH I
= 0.0293 mol NH4+
molar mass of NH4+ = 18.04 g
0.0293 mol NH4+
+
4
+
4
18.04 g NH
1 mol NH
= 0.529 g NH4+
b.
6.31 mol (NH4)2S
+
4
4 2
2 mol NH
1 mol (NH ) S
= 12.6 mol NH4+
molar mass of NH4+ = 18.04 g
12.6 mol NH4+
+
4
+
4
18.04 g NH
1 mol NH
= 227 g NH4+ ion
c.
molar mass of Ba3P2 = 473.8 g
9.71 g
2+
3 2
3 2
1 mol Ba P
3 mol Ba
473.8 g
1 mol Ba P
= 0.0615 mol Ba2+
molar mass of Ba2+ = 137.3 g
136
Chapter 8: Chemical Composition
0.0615 mol Ba2+
2+
2+
137.3 g Ba
1 mol Ba
= 8.44 g Ba2+
d.
7.63 mol Ca3(PO4)2
2+
3
4 2
3 mol Ca
1 mol Ca (PO )
= 22.9 mol Ca2+
molar mass of Ca2+ = 40.08 g
22.9 mol Ca2+
2+
2+
40.08 g Ca
1 mol Ca
= 918 g Ca2+
52.
a.
molar mass of (NH4)2S = 68.15 g; molar mass of S2– ion = 32.07 g
% S2– =
2
4
32.07 g S
68.15 g NH S
100 = 47.06% S2–
b.
molar mass of CaCl2 = 110.98 g; molar mass of Cl– = 35.45 g
% Cl– =
2
70.90 g Cl
110.98 g CaCl
100 = 63.89% Cl–
c.
molar mass of BaO = 153.3 g; molar mass of O2– ion = 16.00 g
% O2– =
2
16.00 g O
153.3 g BaO
100 = 10.44% O2–
d.
molar mass of NiSO4 = 154.76 g; molar mass of SO42– ion = 96.07 g
% SO42– =
2
4
4
96.07 g SO
154.76 g NiSO
100 = 62.08% SO42–
53.
To determine the empirical formula of a new compound, the composition of the compound by
mass must be known. To determine the molecular formula of the compound, the molar mass of
the compound must also be known.
54.
The empirical formula indicates the smallest whole number ratio of the number and type of atoms
present in a molecule. For example, NO2 and N2O4 both have two oxygen atoms for every
nitrogen atom and therefore have the same empirical formula
55.
a.
NaO
b.
C4H3O2
c.
C12H12N2O3 is already the empirical formula.
d.
C2H3Cl
56.
a.
yes (each of these has empirical formula CH)
b.
no (the number of hydrogen atoms is wrong)
c.
yes (both have empirical formula NO2)
d.
no (the number of hydrogen and oxygen atoms is wrong)
137
Chapter 8: Chemical Composition
57.
Assume we have 100.0 g of the compound so that the percentages become masses.
89.56 g Ba ×
= 0.6253 mol Ba
10.44 g O ×
1 mol
16.00 g
= 0.6525 mol O
Dividing both of these numbers of moles by the smaller number of moles (0.6523 mol Ba) gives
The empirical formula is BaO.
58.
Assume we have 100.0 g of the compound so that the percentages become masses.
11.64 g N ×
1 mol
14.01 g
= 0.8308 mol N
88.36 g Cl ×
1 mol
35.45 g
= 2.493 mol Cl
Dividing both of these numbers of moles by the smaller number of moles gives
0.8308 mol N
0.8308
= 1.000 mol N
2.493 mol Cl
0.8308 mol
= 3.001 mol Cl
The empirical formula is NCl3.
59.
0.2322 g C
1 mol
12.01 g
= 0.01933 mol C
0.05848 g H
1 mol
1.008 g
= 0.05802 mol H
0.3091 g O
1 mol
16.00 g
= 0.01932 mol O
Dividing each number of moles by the smallest number of moles (0.01932 mol C) gives
0.01933 mol C
0.01932 mol
= 1.001 mol C
0.05802 mol H
0.01932 mol
= 3.003 mol H
138
Chapter 8: Chemical Composition
0.01932 mol O
0.01932 mol
= 1.000 mol O
The empirical formula is CH3O.
60.
Assume we have 100.0 g of the compound, so that the percentages become masses.
78.14 g B ×
1 mol
10.81 g
= 7.228 mol B
21.86 g H ×
1 mol
1.008 g
= 21.69 mol H
Dividing each number of moles by the smaller number of moles gives
7.228 mol B
7.228 mol
= 1.000 mol B
21.69 mol H
7.228 mol
= 3.000 mol H
The empirical formula is BH3.
61.
The mass of chlorine in the reaction is 6.280 – 1.271 = 5.009 g Cl.
1.271 g Al
1 mol
26.98 g
= 0.04711 mol Al
5.009 g Cl
1 mol
35.45 g
= 0.1413 mol Cl
Dividing each of these number of moles by the smaller (0.04711 mol Al) gives
0.04711 mol Al
0.04711 mol
= 1.000 mol Al
0.1413 mol Cl
0.04711 mol
= 3.000 mol Cl
The empirical formula is AlCl3.
62.
Consider 100.0 g of the compound so that percentages become masses.
45.56 g Sn
1 mol
118.7 g
= 0.3838 mol Sn
54.43 g Cl
1 mol
35.45 g
= 1.535 mol Cl
Dividing each number of moles by the smaller number of moles gives
0.3838 mol Sn
0.3838 mol
= 1.000 mol Sn
139
Chapter 8: Chemical Composition
1.535 mol Cl
0.3838 mol
= 3.999 mol Cl
The empirical formula is SnCl4.
63.
3.269 g Zn
1 mol
65.38 g
= 0.05000 mol Zn
0.800 g O
1 mol
16.00 g
= 0.0500 mol O
Because the two components are present in equal amounts on a molar basis, the empirical
formula must be simply ZnO.
64.
Consider 100.0 g of the compound.
55.06 g Co
1 mol
58.93 g
= 0.9343 mol Co
If the sulfide of cobalt is 55.06% Co, then it is 44.94% S by mass.
44.94 g S
1 mol
32.07 g
= 1.401 mol S
Dividing each number of moles by the smaller (0.9343 mol Co) gives
0.09343 mol Co
0.9343
= 1.000 mol Co
1.401 mol S
0.9343 mol
= 1.500 mol S
Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for the
compound as Co2S3.
65.
The amount of fluorine that reacted with the aluminum sample must be (3.89 g – 1.25 g) = 2.64 g
of fluorine
1.25 g Al ×
1 mol
26.98 g
= 0.04633 mol Al
2.64 g F ×
1 mol
19.00 g
= 0.1389 mol F
Dividing each number of moles by the smaller number of moles gives
0.04633 mol Al
0.04633 mol
= 1.000 mol Al
0.1389 mol F
0.04633 mol
= 2.999 mol F
The empirical formula is just AlF3
140
Chapter 8: Chemical Composition
66.
2.50 g Al ×
1 mol
26.98 g
= 0.09266 mol Al
5.28 g F ×
1 mol
19.00 g
= 0.2779 mol F
Dividing each number of moles by the smaller number of moles gives
0.09266 mol Al
0.09266 mol
= 1.000 mol Al
0.2779 mol F
0.09266 mol
= 2.999 mol F
The empirical formula is just AlF3. Note the similarity between this problem and question 65:
they differ in the way the data is given. In question 65, you were given the mass of the product,
and first had to calculate how much fluorine had reacted.
67.
Consider 100.0 g of the compound.
67.61 g U
1 mol
238.0 g
= 0.2841 mol U
32.39 g F
1 mol
19.00 g
= 1.705 mol F
Dividing each number of moles by the smaller number of moles (0.2841 mol U) gives
0.2841 mol U
0.2841 mol
= 1.000 mol U
1.705 mol F
0.2841 mol
= 6.000 mol F
The empirical formula is UF6.
68.
Consider 100.0 g of the compound so that percentages become masses.
46.46 g Li
1 mol
6.941 g
= 6.694 mol Li
53.54 g O
1 mol
16.00 g
= 3.346 mol O
Dividing each number of moles by the smaller number of moles gives
6.694 mol Li
3.346 mol
= 2.001 mol Li
3.346 mol O
3.346 mol
= 1.000 mol O
The empirical formula is Li2O
141
Chapter 8: Chemical Composition
69.
Consider 100.0 g of the compound.
33.88 g Cu
1 mol
63.55 g
= 0.5331 mol Cu
14.94 g N
1 mol
14.01 g
= 1.066 mol N
51.18 g O
1 mol
16.00 g
= 3.199 mol O
Dividing each number of moles by the smallest number of moles (0.5331 mol Cu) gives
0.5331 mol Cu
0.5331 mol
= 1.000 mol Cu
1.066 mol N
0.5331 mol
= 2.000 mol N
3.199 mol O
0.5331 mol
= 6.001 mol O
The empirical formula is CuN2O6 [i.e., Cu(NO3)2].
70.
Consider 100.0 g of the compound.
59.78 g Li
1 mol
6.941 g
= 8.613 mol Li
40.22 g N
1 mol
14.01 g
= 2.871 mol N
Dividing each number of moles by the smaller number of moles (2.871 mol N) gives
8.613 mol Li
2.871 mol
= 3.000 mol Li
2.871 mol N
2.871 mol
= 1.000 mol N
The empirical formula is Li3N.
71.
Consider 100.0 g of the compound.
66.75 g Cu
1 mol
63.55 g
= 1.050 mol Cu
10.84 g P
1 mol
30.97 g
= 0.3500 mol P
22.41 g O
1 mol
16.00 g
= 1.401 mol O
Dividing each number of moles by the smallest number of moles (0.3500 mol P) gives
142
Chapter 8: Chemical Composition
1.050 mol Cu
0.3500 mol
= 3.000 mol Cu
0.3500 mol P
0.3500
= 1.000 mol P
1.401 mol O
0.3500 mol
= 4.003 mol O
The empirical formula is thus Cu3PO4.
72.
Consider 100.0 g of the compound so that percentages become masses.
48.64 g C
= 4.050 mol C
8.16 g H
= 8.095 mol H
(100.0 g – 48.64 g – 8.16 g = 43.2 g O)
1 mol
16.00 g
= 2.70 mol O
Dividing each number of moles by the smaller number of moles gives
= 1.500 mol C
= 2.998 mol H
= 1.000 mol O
Multiplying these relative numbers of moles by 2 to give whole numbers gives the empirical
formula as C3H6O2.
73.
The compound must contain 1.00 mg of lithium and 2.73 mg of fluorine.
1.00 mg Li
1 mmol
6.941 mg
= 0.144 mmol Li
2.73 mg F
1 mmol
19.00 mg
= 0.144 mol F
The empirical formula of the compound is LiF.
74.
Compound 1: Assume 100.0 g of the compound.
22.55 g P
1 mol
30.97 g
= 0.7281 mol P
77.45 g Cl
1 mol
35.45 g
= 2.185 mol Cl
143
Chapter 8: Chemical Composition
Dividing each number of moles by the smaller (0.7281 mol P) indicates that the formula of
Compound 1 is PCl3.
Compound 2: Assume 100.0 g of the compound.
14.87 g P
1 mol
30.97 g
= 0.4801 mol P
85.13 g Cl
1 mol
35.45 g
= 2.401 mol Cl
Dividing each number of moles by the smaller (0.4801 mol P) indicates that the formula of
Compound 2 is PCl5.
75.
The empirical formula of a compound represents only the smallest whole number relationship
between the number and type of atoms in a compound, whereas the molecular formula represents
the actual number of atoms of each type in a true molecule of the substance. Many compounds
(for example, H2O) may have the same empirical and molecular formulas.
76.
molecular formula: C6H12O6 (count the number of each type of element in the molecule
represented); empirical formula: CH2O (simplest whole-number ratio of C6H12O6; divisible by 6)
77.
Assume that we have 100.00 g of the compound: then 78.14 g will be boron and 21.86 g will be
hydrogen.
78.14 g B
1 mol
10.81 g
= 7.228 mol B
21.86 g H
1 mol
1.008 g
= 21.69 mol H
Dividing each number of moles by the smaller number (7.228 mol B) gives the empirical formula
as BH3. The empirical molar mass of BH3 would be [10.81 g + 3(1.008 g)] = 13.83 g. This is
approximately half of the indicated actual molar mass, and therefore the molecular formula must
be B2H6.
78.
empirical formula mass of CH = 13 g
n =
molar mass
empirical formula mass
=
78 g
13 g
= 6
The molecular formula is (CH)6 or C6H6.
79.
empirical formula mass of CH2 = 14
n =
molar mass
empirical formula mass
=
84 g
14 g
= 6
molecular formula is (CH2)6 = C6H12.
144
Chapter 8: Chemical Composition
80.
empirical formula mass of C2H5O = 46 g
n =
molar mass
empirical formula mass
=
90 g
~ 2
46 g
molecular formula is (C2H5O)2 = C4H10O2
81.
Consider 100.0 g of the compound.
42.87 g C
1 mol
12.01 g
= 3.570 mol C
3.598 g H
1 mol
1.008 g
= 3.569 mol H
28.55 g O
1 mol
16.00 g
= 1.784 mol O
25.00 g N
1 mol
14.01 g
= 1.784 mol N
Dividing each number of moles by the smallest number of moles (1.784 mol O or N) gives
3.570 mol C
1.784 mol
= 2.001 mol C
3.569 mol H
1.784 mol
= 2.001 mol H
1.784 mol O
1.784 mol
= 1.000 mol O
1.784 mol N
1.784 mol
= 1.000 mol N
The empirical formula of the compound is C2H2ON and the empirical formula mass of
C2H2ON = 56.
n =
molar mass
empirical formula mass
=
168 g
56 g
= 3
The molecular formula is (C2H2ON)3 = C6H6O3N3.
82.
The empirical formula is therefore C3H8, with a molar mass of 44.094 g/mol. Since the molar
mass of the compound is 44.1 g/mol, the molecular formula and the empirical formula are the
same. The molecular formula is C3H8.
145
Chapter 8: Chemical Composition
83.
[1]
c
[6]
d
[2]
e
[7]
a
[3]
j
[8]
g
[4]
h
[9]
i
[5]
b
[10] f
84.
5.00 g Al
0.185 mol
1.12 1023 atoms
0.140 g Fe
0.00250 mol
1.51 1021 atoms
2.7 102 g Cu
4.3 mol
2.6 1024 atoms
0.00250 g Mg
1.03 10–4 mol
6.19 1019 atoms
0.062 g Na
2.7 10–3 mol
1.6 1021 atoms
3.95 10–18g U 1.66 10–20 mol 1.00 104 atoms
85.
4.24 g
0.0543 mol
3.27 1022 molec. 3.92 1023 atoms
4.04 g
0.224 mol
1.35 1023 molec. 4.05 1023 atoms
1.98 g
0.0450 mol
2.71 1022 molec. 8.13 1022 atoms
45.9 g
1.26 mol
7.59 1023 molec. 1.52 1024 atoms
126 g
6.99 mol
4.21 1024 molec. 1.26 1025 atoms
0.297 g 0.00927 mol
5.58 1021 molec. 3.35 1022 atoms
86.
mass of 2 mol X = 2(41.2 g) = 82.4 g
mass of 1 mol Y = 57.7 g = 57.7 g
mass of 3 mol Z = 3(63.9 g) = 191.7 g
molar mass of X2YZ3 = 331.8 g
%X =
82.4 g
100
331.8 g
= 24.8 %X
% Y =
57.7 g
100
331.8 g
= 17.4% Y
% Z =
191.7 g
100
331.8 g
= 57.8% Z
If the molecular formula were actually X4Y2Z6, the percentage composition would be the same,
and the relative mass of each element present would not change. The molecular formula is always
a whole number multiple of the empirical formula.
87.
magnesium/nitrogen compound: mass of nitrogen = 1.2791 g – 0.9240 = 0.3551 g N
0.9240 g Mg
1 mol
24.31 g
= 0.03801 mol Mg
0.3551 g N
1 mol
14.01 g
= 0.02535 mol N
Dividing each number of moles by the smaller number of moles gives
146
Chapter 8: Chemical Composition
0.03801 mol Mg
0.02535 mol
= 1.499 mol Mg
0.02535 mol N
0.02535 mol
= 1.000 mol N
Multiplying by two, to convert to whole numbers, gives the empirical formula as Mg3N2.
magnesium/oxygen compound: Consider 100.0 g of the compound.
60.31 g Mg
1 mol
24.31 g
= 2.481 mol Mg
39.69 g O
1 mol
16.00 g
= 2.481 mol O
The empirical formula is MgO.
88.
For the first compound (restricted amount of oxygen)
2.118 g Cu
1 mol
63.55 g
= 0.03333 mol Cu
0.2666 g O
1 mol
16.00 g
= 0.01666 mol O
Since the number of moles of Cu (0.03333 mol) is twice the number of moles of O (0.01666
mol), the empirical formula is Cu2O.
For the second compound (stream of pure oxygen)
2.118 g Cu
1 mol
63.55 g
= 0.03333 mol Cu
0.5332 g O
1 mol
16.00 g
= 0.03333 mol O
Since the numbers of moles are the same, the empirical formula is CuO.
89.
Compound
Molar Mass
% H
HF
20.01 g
5.037%
HCl
36.46 g
2.765%
HBr
80.91 g
1.246%
HI
127.9 g
0.7881%
90.
a.
molar mass H2O = 18.02 g
4.21 g
23
1 mol
6.022 10 molecules
18.02 g
1 mol
= 1.41 1023 molecules
The sample contains 1.41 1023 oxygen atoms and 2(1.41 1023) = 2.82 1023
hydrogen
atoms.
147
Chapter 8: Chemical Composition
b.
molar mass CO2 = 44.01 g
6.81 g
23
1 mol
6.022 10 molecules
44.01 g
1 mol
= 9.32 1022 molecules
The sample contains 9.32 1022 carbon atoms and 2(9.32 1022) = 1.86 1023 oxygen
atoms.
c.
molar mass C6H6 = 78.11 g
0.000221 g
23
1 mol
6.022 10 molecules
78.11 g
1 mol
= 1.70 1018 molec.
The sample contains 6(1.70 1018) = 1.02 1019 atoms of each element.
d.
2.26 mol
23
6.022 10 molecules
1 mol
= 1.36 1024 molecules
atoms C = 12(1.36 1024) = 1.63 1025 atoms
atoms H = 22(1.36 1024) = 2.99 1025 atoms
atoms O = 11(1.36 1024) = 1.50 1025 atoms
91.
a.
Assuming 10,000,000,000 = 1.0 1010
1.0 1010 molecules
2
23
28.02 g N
6.022 10 molecules
= 4.7 10–13 g N2
b.
2.49 1020 molecules
2
23
44.01 g CO
6.022 10 molecules
= 0.0182 g CO2
c.
7.0983 mol NaCl
58.44 g
1 mol
= 414.8 g NaCl
d.
9.012 10–6 mol C2H4Cl2
98.95 g
1 mol
= 8.918 10–4 g C2H4Cl2
92.
a.
molar mass of C3O2 = 3(12.01 g) + 2(16.00 g) = 68.03 g
% C =
36.03 g C
68.03 g
× 100 = 52.96% C
7.819 g C3O2
3
2
52.96 g C
100.0 g C O
= 4.141 g C
4.141 g C
23
6.022 10 molecules
12.01 g C
= 2.076 1023 C atoms
b.
molar mass of CO = 12.01 g + 16.00 g = 28.01 g
% C =
12.01 g C
28.01 g
100 = 42.88% C
1.53 1021 molecules CO
1 C atom
1 molecule CO
= 1.53 1021 C atoms
148
Chapter 8: Chemical Composition
1.53 1021 C atoms
23
12.01 g C
6.022 10 C atoms
= 0.0305 g C
c.
molar mass of C6H6O = 6(12.01 g) + 6(1.008 g) + 16.00 g = 94.11 g
% C =
72.06 g C
94.11 g
100 = 76.57% C
0.200 mol C6H6O
6
6
6 mol C
1 mol C H O
= 1.20 mol C
1.20 mol C
12.01 g
1 mol
= 14.4 g C
14.4 g C
23
6.022 10 C atoms
12.01 g C
= 7.22 1023 C atoms
93.
[1]
g
[6]
i
[2]
c
[7]
f
[3]
b
[8]
h
[4]
a
[9]
e
[5]
j
[10] d
94.
All are true (a, b, c, d, e). 6.022 1023 = 1 mol. 18.016 g/mol is the molar mass of H2O. There are
3 atoms in every one molecule of H2O, thus 3 (6.022 1023) = 1.807 1024 atoms. There are 2
moles of H in every 1 mole of H2O.
95.
2.24 g Fe
58.93 g Co
55.85 g Fe
= 2.36 g Co
96.
(a); Na has the smallest molar mass (22.99 g/mol). Since the same mass is used for each sample,
Na contains the largest number of moles and thus atoms (because the mass is divided by the
molar mass to get moles).
97.
1.00 kg = 1.00 103 g
1.00 103 g Zr
6.941 g Li
91.22 g Zr
= 76.1 g Li
98.
153.8 g CCl4 = 6.022 1023 molecules CCl4
1 molecule
4
23
153.8 g CCl
6.022 10 molecules
= 2.554 10–22 g
149
Chapter 8: Chemical Composition
99.
a.
molar mass of C6H6 = 78.11 g
2.500 g
6.048 g H
78.11 g
= 0.1936 g H
b.
molar mass of CaH2 = 42.10 g
2.500 g
2.016 g H
42.10 g
= 0.1197 g H
c.
molar mass of C2H5OH = 46.07 g
2.500 g
6.048 g H
46.07 g
= 0.3282 g H
d.
molar mass of C3H7O3N = 105.1 g
2.500 g
7.056 g H
105.1 g
= 0.1678 g H
100.
(a); Both NO2 and F2 are molecules and since you have equal moles of each (but not necessarily 1
mole of each), the number of molecules must be the same (6.022 1023 = 1 mol). However, the
number of atoms is different since NO2 contains 3 atoms and F2 contains 2. Furthermore, the
masses are different since NO2 and F2 have different molar masses.
101.
Consider 100.0 g of the compound.
25.45 g Cu
1 mol
63.55 g
= 0.4005 mol Cu
12.84 g S
1 mol
32.07 g
= 0.4004 mol S
4.036 g H
1 mol
1.008 g
= 4.004 mol H
57.67 g O
1 mol
16.00 g
= 3.604 mol O
Dividing each number of moles by the smallest number of moles (0.4004 mol) gives
0.4005 mol Cu
0.4004 mol
= 1.000 mol Cu
0.4004 mol S
0.4004 mol
= 1.000 mol S
4.004 mol H
0.4004 mol
= 10.00 mol H
3.604 mol O
0.4004 mol
= 9.001 mol O
150
Chapter 8: Chemical Composition
The empirical formula is CuSH10O9 (which is usually written as CuSO45H2O).
102.
Consider 100.0 g of the compound so that percentages become masses.
60.87 g C
= 5.068 mol C
4.38 g H
= 4.345 mol H
(100.0 g – 60.87 g – 4.38 g = 34.75 g O)
1 mol
16.00 g
= 2.172 mol O
Dividing each number of moles by the smaller number of moles gives
= 2.333 mol C
= 2.000 mol H
= 1.000 mol O
Multiplying these relative numbers of moles by 3 to give whole numbers gives the empirical
formula as C7H6O3.
103.
atomic mass unit (amu)
104.
We use the average mass because this average is a weighted average and takes into account both
the masses and the relative abundances of the various isotopes.
105.
a.
160,000 amu
1 O atom
16.00 amu
= 1.0 104 O atoms (assuming exact)
b.
8139.81 amu
1 N atom
14.01 amu
= 581 N atoms
c.
13,490 amu
1 Al atom
26.98 amu
= 500 Al atoms
d.
5040 amu
1 H atom
1.008 amu
= 5.00 103 H atoms
e.
367,495.15 amu
1 Na atom
22.99 amu
= 1.599 104 Na atoms
106.
1.98 1013 amu
1 Na atom
22.99 amu
= 8.61 1011 Na atoms
3.01 1023 Na atoms
22.99 amu
1 Na atom
= 6.92 1024 amu
151
Chapter 8: Chemical Composition
107.
a.
1.5 mg = 0.0015 g
0.0015 g Cr
1 mol
52.00 g
= 2.9 10–5 mol Cr
b.
2.0 10–3 g Sr
1 mol
87.62 g
= 2.3 10–5 mol Sr
c.
4.84 104 g B
1 mol
10.81 g
= 4.48 103 mol B
d.
3.6 10–6 g = 3.6 10–12 g
3.6 10–12 g Cf
1 mol
251 g
= 1.4 10–14 mol Cf
e.
2000 lb
453.59 g
1 lb
= 9.1 105 g
9.1 105 g Fe
1 mol
55.85 g
= 1.6 104 mol Fe
f.
20.4 g Ba
1 mol
137.3 g
= 0.149 mol Ba
g.
62.8 g Co
1 mol
58.93 g
= 1.07 mol Co
108.
a.
5.0 mol K
39.10 g
1 mol
= 195 g = 2.0 102 g K
b.
0.000305 mol Hg
200.6 g
1 mol
= 0.0612 g Hg
c.
2.31 10–5 mol Mn
54.94 g
1 mol
= 1.27 10–3 g Mn
d.
10.5 mol P
30.97 g
1 mol
= 325 g P
e.
4.9 104 mol Fe
55.85 g
1 mol
= 2.7 106 g Fe
f.
125 mol Li
6.941 g
1 mol
= 868 g Li
g.
0.01205 mol F
19.00 g
1 mol
= 0.2290 g F
109.
a.
2.89 g Au
23
6.022 10 Au atoms
197.0 g Au
= 8.83 1021 Au atoms
152
Chapter 8: Chemical Composition
b.
0.000259 mol Pt
23
6.022 10 Pt atoms
1 mol
= 1.56 1020 Pt atoms
c.
0.000259 g Pt
23
6.022 10 Pt atoms
195.1 g Pt
= 7.99 1017 Pt atoms
d.
2.0 lb
453.59 g
1 lb
= 908 g
908 g Mg
23
6.022 10 Mg atoms
24.31 g Mg
= 2.2 1025 Mg atoms
e.
1.90 mL
13.6 g
1 mL
= 25.8 g Hg
25.8 g Hg
23
6.022 10 Hg atoms
200.6 g Hg
= 7.75 1022 Hg atoms
f.
4.30 mol W
23
6.022 10 W atoms
1 mol
= 2.59 1024 W atoms
g.
4.30 g W
23
6.022 10 W atoms
183.9 g W
= 1.41 1022 W atoms
110.
a.
mass of 1 mol Fe = 1(55.85 g) = 55.85 g
mass of 1 mol S = 1(32.07 g) = 32.07 g
mass of 4 mol O = 4(16.00 g) = 64.00 g
molar mass of FeSO4 = 151.92 g
b.
mass of 1 mol Hg = 1(200.6 g) = 200.6 g
mass of 2 mol I = 2(126.9 g) = 253.8 g
molar mass of HgI2 =
454.4 g
c.
mass of 1 mol Sn = 1(118.7 g) = 118.7 g
mass of 2 mol O = 2(16.00 g) = 32.00 g
molar mass of SnO2 = 150.7 g
d.
mass of 1 mol Co = 1(58.93 g) = 58.93 g
mass of 2 mol Cl = 2(35.45 g) = 70.90 g
molar mass of CoCl2 = 129.83 g
e.
mass of 1 mol Cu = 1(63.55 g) = 63.55 g
mass of 2 mol N = 2(14.01 g) = 28.02 g
mass of 6 mol O = 6(16.00 g) = 96.00 g
molar mass of Cu(NO3)2 = 187.57 g
153
Chapter 8: Chemical Composition
111.
a.
mass of 6 mol C = 6(12.01 g) = 72.06 g
mass of 10 mol H = 10(1.008 g) = 10.08 g
mass of 4 mol O = 4(16.00 g) = 64.00 g
molar mass of C6H10O4 = 146.14 g
b.
mass of 8 mol C = 8(12.01 g) = 96.08 g
mass of 10 mol H = 10(1.008 g) = 10.08 g
mass of 4 mol N = 4(14.01 g) = 56.04 g
mass of 2 mol O = 1(16.00 g) = 32.00 g
molar mass of C8H10N4O2 = 194.20 g
c.
mass of 20 mol C = 20(12.01 g) = 240.2 g
mass of 42 mol H = 42(1.008 g) = 42.34 g
molar mass of C20H42 = 282.5 g
d.
mass of 6 mol C = 6(12.01 g) = 72.06 g
mass of 12 mol H = 12(1.008 g) = 12.10 g
mass of 1 mol O = 16.00 g = 16.00 g
molar mass of C6H11OH = 100.16 g
e.
mass of 4 mol C = 4(12.01 g) = 48.04 g
mass of 6 mol H = 6(1.008 g) = 6.048 g
mass of 2 mol O = 2(16.00 g) = 32.00 g
molar mass of C4H6O2 = 86.09 g
f.
mass of 6 mol C = 6(12.01 g) = 72.06 g
mass of 12 mol H = 12(1.008 g) = 12.10 g
mass of 6 mol O = 6(16.00 g) = 96.00 g
molar mass of C6H12O6 = 180.16 g
112.
a.
molar mass of (NH4)2S = 68.15 g
21.2 g
1 mol
68.15 g
= 0.311 mol (NH4)2S
b.
molar mass of Ca(NO3)2 = 164.1 g
44.3 g
1 mol
164.1 g
= 0.270 mol Ca(NO3)2
c.
molar mass of Cl2O = 86.9 g
4.35 g
1 mol
86.9 g
= 0.0501 mol Cl2O
154
Chapter 8: Chemical Composition
d.
1.0 lb = 454 g; molar mass of FeCl3 = 162.2
454 g
1 mol
162.2 g
= 2.8 mol FeCl3
e.
1.0 kg = 1.0 103 g;
molar mass of FeCl3 = 162.2 g
1.0 103 g
1 mol
162.2 g
= 6.2 mol FeCl3
113.
a.
molar mass of FeSO4 = 151.92 g
1.28 g
1 mol
151.92 g
= 8.43 10–3 mol FeSO4
b.
5.14 mg = 0.00514 g; molar mass of HgI2 = 454.4 g
0.00514 g
1 mol
454.4 g
= 1.13 10–5 mol HgI2
c.
9.21 g = 9.21 10–6 g; molar mass of SnO2 = 150.7 g
9.21 10–6 g
1 mol
150.7 g
= 6.11 10–8 mol SnO2
d.
1.26 lb = 1.26(453.59 g) = 572 g; molar mass of CoCl2 = 129.83 g
572 g
1 mol
129.83 g
= 4.41 mol CoCl2
e.
molar mass of Cu(NO3)2 = 187.57 g
4.25 g
1 mol
187.57 g
= 2.27 10–2 mol Cu(NO3)2
114.
CuCO3, Na3PO4, P4O10; Assuming 1 mole of each, rank in order of increasing molar mass: 123.56
g/mol for CuCO3, 163.94 g/mol for Na3PO4, and 283.88 g/mol for P4O10.
115.
a.
molar mass of (NH4)2CO3 = 96.09 g
3.09 mol
96.09 g
1 mol
= 297 g (NH4)2CO3
b.
molar mass of NaHCO3 = 84.01 g
4.01 10–6 mol
84.01 g
1 mol
= 3.37 10–4 g NaHCO3
c.
molar mass of CO2 = 44.01 g
88.02 mol
44.01 g
1 mol
= 3874 g CO2
155
Chapter 8: Chemical Composition
d.
1.29 mmol = 0.00129 mol; molar mass of AgNO3 = 169.9 g
0.00129 mol
169.9 g
1 mol
= 0.219 g AgNO3
e.
molar mass of CrCl3 = 158.4 g
0.0024 mol
158.4 g
1 mol
= 0.38 g CrCl3
116.
a.
molar mass of C6H12O6 = 180.2 g
3.45 g
23
6.022 10 molecules
180.2 g
= 1.15 1022 molecules C6H12O6
b.
3.45 mol
23
6.022 10 molecules
1 mol
= 2.08 1024 molecules C6H12O6
c.
molar mass of ICl5 = 304.2 g
25.0 g
23
6.022 10 molecules
304.2 g
= 4.95 1022 molecules ICl5
d.
molar mass of B2H6 = 27.67 g
1.00 g
23
6.022 10 molecules
27.67 g
= 2.18 1022 molecules B2H6
e.
1.05 mmol = 0.00105 mol
0.00105 mol
23
6.022 10 formula units
1 mol
= 6.32 1020 formula units
117.
a.
molar mass of NH3 = 17.03 g
2.71 g
1 mol
17.03 g
= 0.159 mol NH3
mol H = 3(0.159 mol) = 0.477 mol H
b.
mol H = 2(0.824 mol) = 1.648 mol H = 1.65 mol H
c.
6.25 mg = 0.00625 g; molar mass of H2SO4 = 98.09 g
0.00625 g
1 mol
98.09 g
= 6.37 10–5 mol H2SO4
mol H = 2(6.37 10–5 mol) = 1.27 10–4 mol H
d.
molar mass of (NH4)2CO3 = 96.09 g
451 g
1 mol
96.09 g
= 4.69 mol (NH4)2CO3
mol H = 8(4.69 mol) = 37.5 mol H
156
Chapter 8: Chemical Composition
118.
molar mass of Mg3N2 = 100.95 g
5.00 g Mg3N2
= 5.97 × 1022 N atoms
119.
a.
molar mass of NaN3 = 65.02 g
% Na =
22.99 g Na
65.02 g
100 = 35.36% Na
b.
molar mass of CuSO4 = 159.62 g
% Cu =
63.55 g Cu
159.62 g
100 = 39.81% Cu
c.
molar mass of AuCl3 = 303.4 g
% Au =
197.0 g Au
303.4 g
100 = 64.93% Au
d.
molar mass of AgNO3 = 169.9 g
% Ag =
107.9 g Ag
169.9 g
100 = 63.51% Ag
e.
molar mass of Rb2SO4 = 267.0 g
% Rb =
170.9 g Rb
267.0 g
100 = 64.01% Rb
f.
molar mass of NaClO3 = 106.44 g
% Na =
22.99 g Na
106.44 g
100 = 21.60% Na
g.
molar mass of NI3 = 394.7 g
% N =
14.01 g N
394.7 g
100 = 3.550% N
h.
molar mass of CsBr = 212.8 g
% Cs =
132.9 g Cs
212.8 g
100 = 62.45% Cs
120.
(d); The percent mass of an element in a compound is independent of the amount of compound
present.
121.
0.7238 g C
1 mol
12.01 g
= 0.06027 mol C
0.07088 g H
1 mol
1.008 g
= 0.07032 mol H
157
Chapter 8: Chemical Composition
0.1407 g N
1 mol
14.01 g
= 0.01004 mol N
0.3214 g O
1 mol
16.00 g
= 0.02009 mol O
Dividing each number of moles by the smallest number of moles (0.01004 mol) gives
0.06027 mol C
0.01004
= 6.003 mol C
0.07032 mol H
0.01004
= 7.004 mol H
0.01004 mol N
0.01004
= 1.000 mol N
0.02009 mol O
0.01004
= 2.001 mol O
The empirical formula is C6H7NO2.
122.
123.
2.004 g Ca
1 mol
40.08 g
= 0.05000 mol Ca
0.4670 g N
1 mol
14.01 g
= 0.03333 mol N
Dividing each number of moles by the smaller number of moles gives
0.05000 mol Ca
0.03333 mol
= 1.500 mol Ca
0.03333 mol N
1.000 mol N
0.03333 mol
Multiplying these relative numbers of moles by 2 to give whole numbers gives the empirical
formula as Ca3N2.
124.
Assume 100.000 grams. 100.000 g – 13.102 g = 86.898 g Cl
125.
Mass of chlorine in compound = 3.045 g – 1.00 g = 2.045 g Cl
1.00 g Cr
1 mol
52.00 g
= 0.0192 mol Cr
2.045 g Cl
1 mol
35.45 g
= 0.0577 mol Cl
Dividing each number of moles by the smaller number of moles
158
Chapter 8: Chemical Composition
0.0192 mol Cr
0.0192
= 1.00 mol Cr
0.0577 mol Cl
3.01 mol Cl
0.0192 mol
The empirical formula is CrCl3.
126.
Assume we have 100.0 g of the compound.
65.95 g Ba
1 mol
137.3 g
= 0.4803 mol Ba
34.05 g Cl
1 mol
35.45 g
= 0.9605 mol Cl
Dividing each of these number of moles by the smaller number gives
0.4803 mol Ba
0.4803 mol
= 1.000 mol Ba
0.9605 mol Cl
0.4803 mol
= 2.000 mol Cl
The empirical formula is then BaCl2.
127.
a.
H2O water
mass of 2 mol H = 2(1.008 g) = 2.016 g
mass of 1 mol O = 16.00 g
molar mass of H2O = (2.016 g + 16.00 g) = 18.02 g
b.
FeCl3 iron(III) chloride
mass of 1 mol Fe = 55.85 g
mass of 3 mol Cl = 3(35.45 g) = 106.35 g
molar mass of FeCl3 = (55.85 g + 106.35 g) = 162.2 g
c.
KBr potassium bromide
mass of 1 mol K = 39.1 g
mass of 1 mol Br = 79.90 g
molar mass of KBr = (39.1 g + 79.90 g) = 119.0 g
d.
NH4NO3 ammonium nitrate
mass of 2 mol N = 2(14.01 g) = 28.02 g
mass of 4 mol H = 4(1.008 g) = 4.032 g
mass of 3 mol O = 3(16.00 g) = 48.00 g
molar mass of NH4NO3 = (28.02 g + 4.032 g + 48.00 g) = 80.05 g
159
Chapter 8: Chemical Composition
e.
NaOH sodium hydroxide
mass of 1 mol Na = 22.99 g
mass of 1 mol O = 16.00 g
mass of 1 mol H = 1.008 g
molar mass of NaOH = (22.99 g + 16.00 g + 1.008 g) = 40.00 g
128.
a.
C63H88CoN14O14P
mass of 63 mol C = 63(12.01 g) = 756.63 g
mass of 88 mol H = 88(1.008 g) = 88.704 g
mass of 1 mol Co = 58.93 g
mass of 14 mol N = 14(14.01 g) = 196.14 g
mass of 14 mol O = 14(16.00 g) = 224.00 g
mass of 1 mol P = 30.97 g
molar mass of C63H88CoN14O14P = (756.63+88.704+58.93+196.14+224.00+30.97) = 1355.37 g
b.
250 mg C63H88CoN14O14P
= 1.8 × 10–4 mol
c.
0.60 mol C63H88CoN14O14P
= 810 g
d.
1.0 mol C63H88CoN14O14P
=
= 5.3 × 1025 H atoms
e.
1.0 × 107 molecules C63H88CoN14O14P
= 2.3 × 10–14 g
f.
1.0 molecule C63H88CoN14O14P
= 2.3 × 10–21 g
129.
326.4 g Mg3(PO4)2
= 1.242 mol
303.0 g Ca(NO3)2
= 1.846 mol
141.6 g K2CrO4
= 0.7291 mol
406.3 g N2O5
= 3.761 mol
160
Chapter 8: Chemical Composition
130.
a.
1.0 g CH4O
= 1.9 × 1022 C
atoms
b.
1.0 g CH3CH2OH
=
= 2.6 × 1022 C atoms
c.
25.0 g CO(NH2)2
=
= 5.01 × 1023 N atoms
131.
Least to greatest number of H atoms: HF, H2S, PH3, H2O
119 g HF
= 3.58 × 1024 H atoms
119 g H2S
= 4.20 × 1024 H atoms
119 g PH3
= 6.32 × 1024 H atoms
119 g H2O
= 7.96 × 1024 H atoms
132.
molar mass of C9H8O4 = 180.154 g
% C =
100 = 60.00% C
% H =
100 = 4.476% H
% O =
100 = 35.53% O
133.
SNH, NO, N2O, NH3
SNH: % N =
100 = 29.75% N
NO: % N =
100 = 46.68% N
N2O: % N =
100 = 63.65% N
161
Chapter 8: Chemical Composition
NH3: % N =
100 = 82.25% N
162
Chapter 8: Chemical Composition
134.
Consider 100.0 g of the compound so that percentages become masses.
40.0 g C
= 3.33 mol C
6.70 g H
= 6.65 mol H
53.3 g O
1 mol
16.00 g
= 3.33 mol O
Dividing each number of moles by the smaller number of moles gives
= 1.00 mol C
= 2.00 mol H
= 1.000 mol O
The empirical formula is therefore CH2O (with a molar mass of 30.026 g/mol). The molar mass
of the compound is 180.1 g/mol, thus the empirical formula mass goes into the molecular formula
mass 6 times (180.1/30.026 = 6). The molecular formula is 6 (CH2O) = C6H12O6.
163
