Download Ch 6 Book Answers

Chapter 6 Chemical Composition
1.
100 washers 100. g 2.
500. g 500. g 0.110 g
1 washer
1 washer
0.110 g
1 cork
1.63 g
= 11.0 g (assuming 100 washers is exact)
= 909 washers
= 306.7 = 307 corks
1 stopper
4.31 g
= 116 stoppers
1 kg (1000 g) of corks contains (1000 g 613 stoppers would weigh (613 stoppers 1 cork
1.63 g
) = 613.49 = 613 corks
4.31 g
1 stopper
) = 2644 g = 2640 g
The ratio of the mass of a stopper to the mass of a cork is (4.31 g/1.63 g). So the mass of
stoppers that contains the same number of stoppers as there are corks in 1000 g of corks is
1000 g 4.31 g
1.63 g
= 2644 g = 2640 g
3.
The atomic mass unit (amu) is a unit of mass 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.
We use the average atomic mass of an element when performing calculations because the
average mass takes into account the individual masses and relative abundance of all the
isotopes of an element.
5.
a.
635 H atoms b.
1.261 104 W atoms c.
42 K atoms d.
7.213 1023 N atoms e.
891 Fe atoms World of Chemistry
1.008 amu
= 640. amu
1 H atom
183.9 amu
39.10 amu
1 K atom
1 W atom
= 2.319 106 amu
= 1642 amu
14.01 amu
1 N atom
55.85 amu
1 Fe atom
= 1.011 1025 amu
= 4.976 104 amu
37
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38
6.
7.
Chapter 6
1 B atom
a.
10.81 amu b.
320.7 amu c.
19,691 amu d.
19,695 amu e.
3588.3 amu 8274 amu = 1.000 atom = 1 B atom
10.81 amu
1 S atom
32.07 amu
= 10 S atoms
1 Au atom
= 100.00 Au atoms = 100 Au atoms
197.97 amu
1 Xe atom
= 150.0 Xe atoms = 150 Xe atoms
131.3 amu
1 Al atom
= 133.0 Al atoms = 133 Al atoms
26.98 amu
1 S atom
= 258 S atoms
32.07 amu
5.213 1024 S atoms 32.07 amu
1 S atom
= 1.672 1026 amu
8.
1.204 1024
9.
Avogadro’s number (6.022 1023)
10.
The ratio of the atomic mass of Ca to the atomic mass of Mg is (40.08 amu/24.31 amu), and
the masses of calcium are given by
12.16 g Mg 24.31 g Mg 11.
40.08 amu
24.31 amu
= 20.05 g Ca
= 40.08 g Ca
58.93 amu
19.00 amu
= 177 g Co
1 mol O = 16.00 g O = 6.02 1023 O atoms
1 O atom 13.
24.31 amu
The ratio of the atomic mass of Co to the atomic mass of F is (58.93 amu/19.00 amu), and the
mass of cobalt is given by
57.0 g F 12.
40.08 amu
16.00 g O
6.022 x 10 23 O atoms
0.50 mol O atoms 4 mol H atoms 16.00 g O
1 mol
1.008 g
1 mol
= 2.66 10–23 g O
= 8.0 g O
=4gH
Half a mole of O atoms weighs more than 4 moles of H atoms.
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Chemical Composition
14.
15.
16.
39
1 mol Au
a.
26.2 g Au b.
41.5 g Ca c.
335 mg Ba d.
1.42 10–3 g Pd e.
3.05 10–5 g Ni f.
1.00 lb Fe g.
12.01 g C a.
2.00 mol Fe b.
0.521 mol Ni c.
1.23 10–3 mol Pt d.
72.5 mol Pb e.
0.00102 mol Mg f.
4.87 103 mol Al g.
211.5 mol Li h.
1.72 10–6 mol Na a.
0.00103 g Co b.
0.00103 mol Co World of Chemistry
= 0.133 mol Au
197.0 g
1 mol Ca
= 1.04 mol Ca
40.08 g
1g
1 mol Ba
3
10 mg
137.3 g
1 mol Pd
= 1.33 10–5 mol Pd
106.4 g
1g
6
10 g
453.59 g
1 lb
1 mol C
= 2.44 10–3 mol Ba
1 mol Ni
58.70 g
1 mol Fe
55.85 g
= 5.20 10–13 mol Ni
= 8.12 mol Fe
= 1.000 mol C
12.01 g
55.85 g
= 112 g Fe
1 mol
58.70 g
1 mol
= 30.6 g Ni
195.1 g
1 mol
207.2 g
= 1.50 104 g Pb
1 mol
24.31 g
1 mol
= 0.0248 g Mg
26.98 g
1 mol
6.941 g
1 mol
= 0.240 g Pt
= 1.31 105 g Al
= 1468 g Li
22.99 g
1 mol
= 3.95 10–5 g Na
6.022 x 10 23 Co atoms
58.93 g Co
= 1.05 1019 Co atoms
6.022 x 10 23 Co atoms
1 mol
= 6.20 1020 Co atoms
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40
Chapter 6
1 mol
c.
2.75 g Co d.
5.99 1021 Co atoms e.
4.23 mol Co f.
4.23 mol Co g.
4.23 g Co 58.93 g Co
= 0.0467 mol Co
1 mol
6.022 x 10 23 Co atoms
58.93 g Co
1 mol Co
= 0.00995 mol Co
= 249 g Co
6.022 x 10 23 Co atoms
1 mol Co
6.022 x 10 23 Co atoms
58.93 g Co
= 2.55 1024 Co atoms
= 4.32 1022 Co atoms
17.
molar mass
18.
adding together (summing)
19.
a. mass of 3 mol Na = 3 (22.99 g) =
mass of 1 mol N = 1 (14.01 g) =
molar mass of Na3N =
68.97 g
14.01 g
82.98 g
b. mass of 1 mol C =
12.01 g =
mass of 2 mol S = 2 (32.07 g) =
molar mass of CS2 =
12.01 g
64.14 g
76.15 g
c. mass of 1 mol N = 14.01 g =
mass of 4 mol H = 4 (1.008 g) =
mass of 1 mol Br = 79.90 g =
molar mass of NH4Br =
14.01 g
4.032 g
79.90 g
97.942 g = 97.94 g
d. mass of 2 mol C = 2 (12.01 g) =
mass of 6 mol H = 6 (1.008 g) =
mass of 1 mol O =
16.00 g =
molar mass of C2H6O =
24.02 g
6.048 g
16.00 g
46.068 g = 46.07 g
e. mass of 2 mol H = 2 (1.008 g) = 2.016 g
mass of 1 mol S =
32.07 g =
32.07 g
mass of 3 mol O = 3 (16.00 g) = 48.00 g
82.086 g = 82.09 g
molar mass of H2SO3 =
f.
mass of 2 mol H = 2 (1.008 g) = 2.016 g
mass of 1 mol S =
32.07 g =
32.07 g
mass of 4 mol O = 4 (16.00 g) = 64.00 g
98.086 g = 98.09 g
molar mass of H2SO4 =
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Chemical Composition
20.
21.
41
a.
mass of 1 mol Ba =
137.3 g =
mass of 2 mol Cl =
2 (35.45 g) =
mass of 8 mol O =
8 (16.00 g) =
molar mass of Ba(ClO4)2 =
137.3 g
70.90 g
128.0 g
336.2 g
b.
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) =
molar mass of MgSO4 =
24.31 g
32.07 g
64.00 g
120.38 g
c.
mass of 1 mol Pb =
207.2 g =
mass of 2 mol Cl =
2 (35.45 g) =
molar mass of PbCl2 =
207.2 g
79.90 g
278.1 g
d.
mass of 1 mol Cu =
63.55 g =
mass of 2 mol N =
2 (14.01 g) =
mass of 6 mol O =
6 (16.00 g) =
molar mass of Cu(NO3)2 =
63.55 g
28.02 g
96.00 g
187.57 g
e.
mass of 1 mol Sn =
118.7 g =
mass of 4 mol Cl =
4 (35.45 g) =
molar mass of SnCl4 =
118.7 g
141.80 g
260.5 g
f.
mass of 6 mol C =
6 (12.01 g) = 72.06 g
mass of 6 mol H =
6 (1.008 g) = 6.048 g
mass of 1 mol O =
16.00 g =
16.00 g
94.11 g
molar mass of C6H6O =
a.
molar mass of SO3 = 80.07 g
49.2 mg SO3 b.
1g
1000 mg
80.07 g
= 6.14 10–4 mol SO3
1000 g
1 kg
1 mol
239.2 g
= 3.11 105 mol PbO2
molar mass of CHCl3 = 119.37 g
59.1 g CHCl3 d.
1 mol
molar mass of PbO2 = 239.2 g
7.44 x 104 kg PbO2 c.
1 mol
119.37 g
= 0.495 mol CHCl3
molar mass of C2H3Cl3 = 133.39 g
3.27 g C2H3Cl3 1g
6
10 g
1 mol
133.39 g
= 2.45 10–8 mol C2H3Cl3
e. molar mass of LiOH = 23.95 g
4.01 g LiOH World of Chemistry
1 mol
23.95 g
= 0.167 mol LiOH
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42
22.
Chapter 6
a.
molar mass of NaH2PO4 = 120.0 g
4.26 10–3 g NaH2PO4 b.
1 mol
a.
1 mol
= 0.0697 mol SrF2
125.6 g
molar mass of Al = 26.98 g
1 mol
26.98 g
407.7 g
1 mol
= 612 g AlI3
molar mass of C6H6 = 78.11 g
78.11 g
1 mol
180.2 g
1 mol
= 721 g C6H12O6
molar mass of C2H5OH = 46.07 g
4.56 105 mol C2H5OH e.
= 0.149 g C6H6
molar mass of C6H12O6 = 180.2 g
4.00 mol C6H12O6 d.
= 467 mol Al
molar mass of AlI3 = 407.7 g
1.91 10–3 mol C6H6 c.
= 2.70 103 mol Fe
55.85 g
molar mass of SrF2 = 125.6 g
1.50 mol AlI3 b.
1 mol
1 kg
1.26 104 g Al 23.
= 5.26 mol CuCl
99.00 g
1000 g
8.76 g SrF2 e.
= 3.55 10–5 mol NaH2PO4
molar mass of Fe = 55.85 g
151 kg Fe d.
120.0 g
molar mass of CuCl = 99.00 g
521 g CuCl c.
1 mol
46.07 g
1 mol
= 2.10 107 g C2H5OH
molar mass of Ca(NO3)2 = 164.1 g
2.27 mol Ca(NO3)2 World of Chemistry
164.1 g
1 mol
= 373 g Ca(NO3)2
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Chemical Composition
24.
a.
43
molar mass of CO2 = 44.01 g
1.27 mmol b.
1 mol
10 mmol
1 mol
80.05 g
1 mol
18.02 g
1 mol
= 0.361 g NH4NO3
= 324 g H2O
molar mass of CuSO4 = 159.6 g
159.6 g
1 mol
= 1.00 104 g CuSO4
6.022 x 10 23 molecules
a.
6.37 mol CO b.
molar mass of CO = 28.01 g
6.37 g c.
= 4.96 105 g NCl3
molar mass of H2O = 18.02 g
62.7 mol CuSO4 25.
= 0.0559 g CO2
molar mass of NH4NO3 = 80.05 g
18.0 mol H2O e.
1 mol
120.4 g
0.00451 mol NH4NO3 d.
44.01 g
molar mass of NCl3 = 120.4 g
4.12 103 mol NCl3 c.
3
1 mol
= 3.84 1024 molecules CO
1 mol
6.022 x 10 23 molecules
= 1.37 1023 molecules CO
28.01 g
1 mol
molar mass of H2O = 18.02 g
2.62 10–6 g 6.022 x 10 23 molecules
18.02 g
6.022 x 10 23 molecules
d.
2.62 10–6 mol e.
molar mass of C6H6 = 78.11 g
5.23 g World of Chemistry
= 8.76 1016 molecules H2O
1 mol
6.022 x 10 23 molecules
78.11 g
= 1.58 1018 molecules H2O
= 4.03 1022 molecules C6H6
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44
26.
Chapter 6
a.
molar mass of Na2SO4 = 142.1 g
1 mol Na 2SO 4
2.01 g Na2SO4 b.
142.1 g
1 mol Na 2SO 3
a.
= 0.0141 mol S
1 mol S
1 mol Na 2SO 3
= 0.0159 mol S
1 mol Na 2S
78.05 g
1 mol S
1 mol Na 2S
= 0.0258 mol S
molar mass of Na2S2O3 = 158.1 g
1 mol Na 2S2 O 3
158.1 g
2 mol S
1 mol Na 2S2 O 3
= 0.0254 mol S
mass of Na present = 2 (22.99 g) = 45.98 g
mass of S present =
32.07 g =
32.07 g
mass of O present =
4 (16.00 g) = 64.00 g
molar mass of Na2SO4 =
142.05 g
% Na =
%S=
45.98 g Na
142.05 g
32.07 g S
142.05 g
100 = 32.37% Na
100 = 22.58% S
64.00 g O
%O=
b.
126.1 g
2.01 g Na2S2O3 27.
1 mol Na 2SO 4
molar mass of Na2S = 78.05 g
2.01 g Na2S d.
1 mol S
molar mass of Na2SO3 = 126.1 g
2.01 g Na2SO3 c.
100 = 45.05% O
142.05 g
mass of Na present = 2 (22.99 g) = 45.98 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 Na2SO3 =
126.05 g
% Na =
%S=
%O=
World of Chemistry
45.98 g Na
126.05 g
32.07 g S
126.05 g
48.00 g O
126.05 g
100 = 36.48% Na
100 = 25.44% S
100 = 38.08% O
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Chemical Composition
c.
45
mass of Na present =
mass of S present =
molar mass of Na2S =
% Na =
%S=
d.
78.05 g
45.98 g
32.07 g
78.05 g
100 = 58.91% Na
100 = 41.09% S
mass of Na present = 2 (22.99 g) = 45.98 g
mass of S present =
2 (32.07 g) = 64.14 g
mass of O present =
3 (16.00 g) = 48.00 g
158.12 g
molar mass of Na2S2O3 =
%S=
%O=
45.98 g Na
158.12 g
64.14 g S
158.12 g
48.00 g O
158.12 g
100 = 29.08% Na
100 = 40.56% S
100 = 30.36% O
mass of K present =
3 (39.10 g) = 117.3 g
mass of P present =
30.97 g =
30.97 g
mass of O present =
4 (16.00 g) = 64.00 g
212.3 g
molar mass of K3PO4 =
%K=
%P=
%O=
f.
78.05 g
32.07 g S
% Na =
e.
45.98 g Na
2 (22.99 g) =
32.07 g =
117.3 g K
212.3 g
30.97 g P
212.3 g
64.00 g O
212.3 g
100 = 55.25% K
100 = 14.59% P
100 = 30.15% O
mass of K present =
2 (39.10 g) = 78.20 g
mass of H present =
1.008 g =
1.008 g
mass of P present =
30.97 g =
30.97 g
mass of O present =
4 (16.00 g) = 64.00 g
174.178 g = 174.18 g
molar mass of K2HPO4 =
%K=
%H=
World of Chemistry
78.20 g K
174.18 g
1.008 g H
174.18 g
100 = 44.90% K
100 = 0.5787% H
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46
Chapter 6
%P=
30.97 g P
174.18 g
64.00 g O
%O=
g.
mass of K present =
39.10 g =
mass of H present =
2 (1.008 g) =
mass of P present =
30.97 g =
mass of O present =
4 (16.00 g) =
molar mass of KH2PO4 =
39.10 g K
%P=
2.016 g H
136.09 g
%P=
117.3 g K
30.97 g P
148.3 g
3 (39.10) g =
30.97 g =
148.27 g =
117.3 g
30.97 g
148.3 g
100 = 79.10% K
148.3 g
100 = 20.88% P
molar mass of CuBr2 = 223.4 g
% Cu =
63.55 g Cu
223.4 g
100 = 28.45% Cu
molar mass of CuBr = 143.5 g
% Cu =
c.
100 = 47.03% O
136.09 g
mass of K present =
mass of P present =
molar mass of K3P =
%K=
b.
100 = 22.76% P
64.00 g O
%O=
a.
100 = 1.481% H
136.09 g
30.97 g P
39.10 g
2.016 g
30.97 g
64.00 g
136.09 g
100 = 28.73% K
136.09 g
%H=
28.
100 = 36.74% O
174.18 g
%K=
h.
100 = 17.78% P
63.55 g Cu
143.5 g
100 = 44.29% Cu
molar mass of FeCl2 = 126.75 g
% Fe =
World of Chemistry
55.85 g Fe
126.75 g
100 = 44.06% Fe
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Chemical Composition
d.
47
molar mass of FeCl3 = 162.2 g
55.85 g Fe
% Fe =
e.
molar mass of CoI2 = 312.7 g
% Co =
f.
a.
134.7 g
100 = 88.12% Sn
118.7 g Sn
150.7 g
100 = 78.77% Sn
72.06 g C
146.1 g
100 = 49.32% C
28.02 g N
80.05 g
100 = 35.00% N
96.08 g C
194.2 g
100 = 49.47% C
35.45 g Cl
67.45 g
100 = 52.56% Cl
molar mass of C6H11OH = 100.2 g
%C=
f.
118.7 g Sn
molar mass of ClO2 = 67.45 g
% Cl =
e.
100 = 13.41% Co
molar mass of C8H10N4O2 = 194.2 g
%C=
d.
439.6 g
molar mass of NH4NO3 = 80.05 g
%N=
c.
58.93 g Co
molar mass of C6H10O4 = 146.1 g
%C=
b.
100 = 18.85% Co
molar mass of SnO2 = 150.7 g
% Sn =
29.
312.7 g
molar mass of SnO = 134.7 g
% Sn =
h.
58.93 g Co
molar mass of CoI3 = 439.6 g
% Co =
g.
100 = 34.43% Fe
162.2 g
72.06 g C
100.2 g
100 = 71.92% C
molar mass of C6H12O6 = 180.2 g
%C=
World of Chemistry
72.06 g C
180.2 g
100 = 39.99% C
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48
Chapter 6
g.
molar mass of C20H42 = 282.5 g
%C=
h.
a.
100 = 85.03% C
24.02 g C
46.07 g
100 = 52.14% C
molar mass of NH4Cl = 53.49 g
molar mass of NH4+ ion = 18.04 g
%
b.
282.5 g
molar mass of C2H5OH = 46.07 g
%C=
30.
240.2 g C
NH4+ =
18.04 g NH +4
53.49 g
100 = 33.73% NH4+
molar mass of CuSO4 = 159.62
molar mass of Cu2+ ion = 63.55 g
% Cu2+ =
c.
63.55 g Cu 2+
159.62 g
100 = 39.81% Cu2+
molar mass of AuCl3 = 303.4 g
molar mass of Au3+ ion = 197.0 g
% Au3+ =
d.
197.0 g Au 3+
303.4 g
100 = 64.93% Au3+
molar mass of AgNO3 = 169.9 g
molar mass of Ag+ ion = 107.9 g
% Ag+ =
107.9 g Ag +
169.9 g
100 = 63.51% Ag+
31.
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.
32.
The empirical formula represents the smallest whole number ratio of the elements present in a
compound. The molecular formula indicates the actual number of atoms of each element
found in a molecule of the substance.
33.
a.
NaO
b.
C4H3O2
c.
C12H12N2O3 is already the empirical formula
d.
C2H3Cl
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Chemical Composition
34.
35.
49
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)
1 mol C
0.1929 g C = 0.01606 mol C
12.01 g C
1 mol H
0.01079 g H 1.008 g H
1 mol O
0.08566 g O 16.00 g O
1 mol Cl
0.1898 g Cl 35.45 g Cl
= 0.01070 mol H
= 0.005354 mol O
= 0.005354 mol Cl
Dividing each number of moles by the smallest number of moles gives:
0.01606 mol C
= 3.000 mol C
0.005354
0.01070 mol H
0.005354
0.005354 mol O
0.005354
0.005354 Cl
= 1.999 mol H
= 1.000 mol O
= 1.000 mol Cl
0.005354
The empirical formula is C3H2OCl
36.
2.514 g Ca 1 mol
40.08 g Ca
= 0.06272 mol Ca
The increase in mass represents the oxygen with which the calcium reacted:
1 mol O
1.004 g O 16.00 g O
= 0.06275 mol O
Since we have effectively the same number of moles of Ca and O, the empirical formula must
be CaO.
37.
Consider having 100.0 g of the compound. Then the percentages of the elements present are
numerically equal to their masses in grams.
58.84 g Ba 13.74 g S World of Chemistry
1 mol
137.3 g Ba
1 mol
32.07 g S
= 0.4286 mol Ba
= 0.4284 mol S
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50
Chapter 6
1 mol O
27.43 g O = 1.714 mol O
16.00 g O
Dividing each number of moles by the smallest number of moles (0.4284 mol S) gives
0.4286 mol Ba
= 1.000 mol Ba
0.4284
0.4284 mol S
= 1.000 mol S
0.4284
1.714 mol O
= 4.001 mol O
0.4284
The empirical formula is BaSO4.
38.
The mass of chlorine involved in the reaction is 6.280 – 1.271 = 5.009 g Cl
1 mol Al
1.271 g Al 26.98 g Al
1 mol Cl
5.009 g Cl 34.45 g Cl
= 0.04711 mol Al
= 0.1413 mol Cl
Dividing each of these numbers of moles by the smaller (0.04711 mol Al) shows that the
empirical formula is AlCl3.
39.
Consider 100.0 g of the compound.
1 mol
55.06 g Co 58.93 g Co
= 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 S
= 1.401 mol S
Dividing each number of moles by the smaller (0.9343 mol Co) gives
0.9343 mol Co
0.9343
1.401 mol S
0.9343
= 1.000 mol Co
= 1.500 mol S
Multiplying by two, to convert to whole numbers of moles, gives the empirical formula for
the compound as Co2S3.
40.
2.461 g Ca 4.353 g Cl 1 mol Ca
40.08 g Ca
1 mol Cl
35.45 g Cl
= 0.06140 mol Ca
= 0.1228 mol Cl
Dividing each of the number of moles by the smaller (0.06140 mol Ca) shows that the
empirical formula is CaCl2.
World of Chemistry
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Chemical Composition
41.
1 mol
10.00 g Cu 2.52 g O 51
63.55 g Cu
1 mol O
16.00 g O
= 0.1574 mol Cu
= 0.158 mol O
The numbers are almost equal: the empirical formula is CuO.
42.
Consider 100.0 g of the compound.
1 mol Cu
33.88 g Cu 14.94 g N 51.18 g O 63.55 g Cu
1 mol N
14.01 g N
1 mol O
16.00 g O
= 0.5331 mol Cu
= 1.066 mol N
= 3.199 mol O
Dividing each number of moles by the smaller number of moles (0.5331 mol Cu) gives
0.5331 mol Cu
0.5331
1.066 mol N
= 1.000 mol Cu
= 2.000 mol N
0.5331
3.199 mol O
= 6.001 mol O
0.5331
The empirical formula is CuN2O6 [i.e., Cu(NO3)2]
43.
Compound 1: Assume 100.0 g of the compound.
83.12 g Na 16.88 g N 1 mol Na
22.99 g Na
1 mol N
14.01 g N
= 3.615 mol Na
= 1.205 mol Na
Dividing each number of moles by the smaller (1.205 mol Na) indicates that the formula of
Compound 1 is Na3N.
Compound 2: Assume 100.0 g of the compound.
35.36 g Na 64.64 g N 1 mol Na
22.99 g Na
1 mol N
14.01 g N
= 1.538 mol Na
= 4.614 mol N
Dividing each number of moles by the smaller (1.538 mol Na) indicates that the formula of
Compound 2 is NaN3.
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52
Chapter 6
44.
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.
45.
If only the empirical formula is known, the molar mass of the substance must be determined
before the molecular formula can be calculated.
46.
empirical formula mass of CH2O = 30 g
n=
molar mass
empirical formula mass
=
90 g
30 g
=3
molecular formula is (CH2O)3 = C3H6O3.
47.
empirical formula mass of CH2 = 14
n=
molar mass
empirical formula mass
=
84 g
14 g
=6
molecular formula is (CH2)6 = C6H12.
48.
empirical formula mass of CH4O = 32.04 g
n=
molar mass
empirical formula mass
=
192 g
32.04 g
=6
molecular formula is (CH4O)6 = C6H24O6.
49.
Consider 100.0 g of the compound.
42.87 g C 3.598 g H 28.55 g O 25.00 g N 1 mol C
12.01 g C
1 mol H
1.008 g H
1 mol O
16.00 g O
1 mol N
14.01 g N
= 3.570 mol C
= 3.569 mol H
= 1.784 mol O
= 1.784 mol N
Dividing each number of moles by the smallest number of moles (1.784 mol O) gives
3.570 mol C
1.784
3.569 mol H
1.784
1.784 mol O
1.784
World of Chemistry
= 2.001 mol C
= 2.001 mol H
= 1.000 mol O
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Chemical Composition
1.784 mol N
1.784
53
= 1.000 mol N
The empirical formula of the compound is C2H2ON, empirical formula mass of
C2H2ON = 56
n=
molar mass
empirical formula mass
=
168 g
56 g
=3
The molecular formula is (C2H2ON)3 = C6H6O3N3.
50.
Consider 100.0 g of the compound.
65.45 g C 5.492 g H 29.06 g O 1 mol C
12.01 g C
1 mol H
1.008 g H
1 mol O
16.00 g O
= 5.450 mol C
= 5.448 mol H
= 1.816 mol O
Dividing each number of moles by the smallest number of moles (1.816 mol O) gives
5.450 mol C
1.816
5.448 mol H
1.816
1.816 mol O
1.816
= 3.001 mol C
= 3.000 mol H
= 1.000 mol O
The empirical formula is C3H3O, and the empirical formula mass is approximately 55 g.
n=
molar mass
empirical formula mass
=
110 g
55 g
=2
The molecular formula is (C3H3O)2 = C6H6O2.
51.
52.
[1]
c
[6]
d
[2]
e
[7]
a
[3]
j
[8]
g
[4]
h
[9]
i
[5]
b
[10]
f
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
3.95 10–18 g U
2.7 10–3 mol
1.66 10–20 mol
1.6 1021 atoms
1.00 104 atoms
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54
53.
54.
Chapter 6
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
126 g
1.26 mol
6.99 mol
7.59 1023 molec.
4.21 1024 molec.
1.52 1024 atoms
1.26 1025 atoms
0.267 g
0.00927 mol
5.58 x 1021 molec.
3.35 x 1022 atoms
magnesium/nitrogen compound:
mass of nitrogen contained = 1.2791 g – 0.9240 g = 0.3551 g N
1 mol Mg
0.9240 g Mg 24.31 g Mg
1 mol N
0.3551 g N = 0.03801 mol Mg
= 0.02535 mol N
14.01 g N
Dividing each number of moles by the smaller number of moles gives
0.03801 mol Mg
0.02535
0.02535 mol N
0.02535
= 1.499 mol Mg
= 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 this compound.
60.31 g Mg 39.69 g O 1 mol Mg
24.31 g Mg
1 mol O
16.00 g O
= 2.481 mol Mg
= 2.481 mol O
Since the numbers of moles are the same, the compound contains the same relative number of
Mg and O atoms: the empirical formula is MgO.
55.
For the first compound (restricted amount of oxygen):
2.118 g Cu 0.2666 g O 1 mol Cu
63.54 g Cu
1 mol O
16.00 g O
= 0.03333 mol Cu
= 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.
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Chemical Composition
55
For the second compound (stream of pure oxygen):
1 mol Cu
2.118 g Cu 63.54 g Cu
0.5332 g O 1 mol O
= 0.03333 mol Cu
= 0.03333 mol O
16.00 g O
Since the numbers of moles are the same, the empirical formula is CuO.
56.
We need to find the subscripts for CaHbOcSd, where a:b:c:d is the mole ratio of C:H:O:S
atoms. We are given the following relationships:
b = 2a
a=c
b = 8d
We can see that d will be the smallest subscript. Thus, let d = x and we get
b = 8x
2a = 8x or
a = 4x
since a = c,
c = 4x
Thus, we have C4xH8xO4xSx. The empirical formula is C4H8O4S, which has a molar mass of
about 152 g/mol [4(12.01) + 8(1.008) + 4(16.00) + 1(32.07)]. We are given that the molar
mass of the compound is 152 g/mol. Thus, the molecular formula is C4H8O4S.
55.85 g Fe
57.
2.24 g Co 58.
2.24 g Fe 59.
Consider 100.0 g of the compound.
58.93 g Co
58.93 g Co
12.84 g S 4.036 g H 57.67 g O World of Chemistry
= 2.36 g Co
55.85 g Fe
25.45 g Cu 1 mol Cu
63.55 g Cu
1 mol S
32.07 g S
1 mol H
1.008 g H
1 mol O
16.00 g O
= 2.12 g Fe
= 0.4005 mol Cu
= 0.4004 mol S
= 4.004 mol H
= 3.604 mol O
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56
Chapter 6
Dividing each number of moles by the smallest number of moles gives
0.4005 mol Cu
= 1.000 mol Cu
0.4004
0.4004 mol S
= 1.000 mol S
0.4004
4.004 mol H
= 10.00 mol H
0.4004
3.604 mol O
= 9.001 mol O
0.4004
The empirical formula is CuSH10O9 (which is usually written as CuSO45H2O).
60.
1 mol C
0.2990 g C 12.01 g C
1 mol H
0.05849 g H 1.008 g H
1 mol N
0. 2318 g N 14.01 g N
1 mol O
0.1328 g O = 0.02490 mol C
16.00 g O
= 0.05803 mol H
= 0.01655 mol N
= 0.008300 mol O
Dividing each number of moles by the smallest number of moles (0.008300 mol O) gives
0.02490 mol C
0.008300
0.05803 mol H
0.008300
0.01655 mol N
0.008300
0.008300 mol O
0.008300
= 3.000 mol C
= 6.992 mol H
= 1.994 mol N
= 1.000 mol O
The empirical formula is C3H7N2O.
61.
Mass of oxygen in compound = 4.33 g – 4.01 g = 0.32 g O
4.01 g Hg 0.32 g O 1 mol Hg
200.6 g Hg
1 mol O
16.00 g O
= 0.0200 mol Hg
= 0.020 mol O
Since the numbers of moles are equal, the empirical formula is HgO.
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Chemical Composition
62.
57
Assume we have 100.0 g of the compound.
65.95 g Ba 1 mol Ba
1 mol Cl
34.05 g Cl = 0.4803 mol Ba
137.3 g Ba
35.45 g Cl
= 0.9605 mol Cl
Dividing each of these numbers of moles by the smaller number gives
0.4803 mol Ba
0.4803
0.9605 mol Cl
0.4803
= 1.000 mol Ba
= 2.000 mol Cl
The empirical formula is then BaCl2.
63.
We need to find the subscripts for HxNyOz, where x:y:z is the mole ratio of H:N:O atoms. We
are given that x = 4.0 moles H.
To solve for y, we use
56.0 g N 1 mol N
14.01 g N
= 4.00 moles N
To solve for z, we use
7.2 x 1024 atoms O 1 mol O
6.022 x 10 23 atoms
= 12 moles O
The mole ratio is 4:4:12 or 1:1:3. The empirical formula is HNO3.
64.
For every 100.0 g of A2O, we have 63.7 g A and 36.3 g O.
36.3 g O 1 mol O
16.00 g O
= 2.27 mol O
The ratio between A and O is 2:1; with 2.27 mol O we must have 4.54 mol A.
63.7 g A
4.54 mol A
= 14.0 g/mol
Thus, A must be nitrogen. The compound is N2O.
65.
[1]
[2]
[3]
[4]
[5]
g
c
b
a
j
World of Chemistry
[6]
[7]
[8]
[9]
[10]
i
f
h
e
d
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