CHEM 101 Answer to Homework Problems (4th edition)
Chapter 3
3.6
Using the appropriate atomic masses,
a. Li2CO3
2(6.941 amu) + 1(12.01 amu) + 3(16.00 amu) = 73.89 amu
b. C2H6
2(12.01 amu) + 6(1.008 amu) = 30.07 amu
c. NF2
1(14.01 amu) + 2(19.00 amu) = 52.01 amu
d. Al2O3
2(26.98 amu) + 3(16.00 amu) = 101.96 amu
e. Fe(NO3)3
1(55.85 amu) + 3(14.01 amu) + 9(16.00 amu) = 241.9 amu
f. PCl5
1(30.97 amu) + 5(35.45 amu) = 208.2 amu
g. Mg3N2
3(24.31 amu) + 2(14.01 amu) = 100.95 amu
3.11
a.
b.
c.
Compound
Molar mass (g)
N% by mass
(NH2)2CO
60.06
2(14.01 g)
 100% = 46.65%
60.06 g
NH4NO3
80.05
2(14.01 g)
 100% = 35.00%
80.05 g
HNC(NH2)2
59.08
3(14.01 g)
 100% = 71.14%
59.08 g
1
d.
NH3
14.01 g
100%  82.27%
17.03 g
17.03
3.21
Strategy: (1) translate each compound name into the correct chemical formula;
(2) translate "and" into "+"; this applies to compound phrases separated by commas;
(3) translate "react to form" into a reaction arrow "→"
Translation of a compound name into the correct chemical formula is the most
difficult part of the solution. If the compound name is a systematic name, it will
have Greek prefixes such as "di-" (2) which tell you what the subscript of a particular
element is in the formula. However, in most cases the name used is a common name
and you must simply look up (or learn and remember) the name/formula translation.
Solution: a. KOH + H3PO4 → K3PO4 + H2O
b. Zn + AgCl → ZnCl2 + Ag
c. NaHCO3 → Na2CO3 + H2O + CO2
d. NH4NO2 → N2 + H2O
e. CO2 + KOH → K2CO3 + H2O
f. 3KOH + H3PO4 → K3PO4 + 3H2O
Zn + 2AgCl → ZnCl2 + 2Ag
2NaHCO3 → Na2CO3 + H2O + CO2
NH4NO2 → N2 + 2H2O
CO2 + 2KOH → K2CO3 + H2O
2
3.32
In one year:
(6.5 × 109 people) ×
Total time 
365 days 24 hr 3600 s 2 particles
×
×
×
= 4.1 × 1017 particles/yr
1 yr
1 day
1 hr
1 person
6.022 × 1023 particles
4.1 × 1017 particles/yr
 1.5  106 yr
3.63
The process of combustion analysis involves the following steps:
(1) Convert the mass of CO2 to moles of CO2 - this is the same as the number of moles of C in
the sample because every C atom in the sample becomes the C atom in a CO2 molecule;
(2) Convert moles of C to grams of C;
(3) Convert the mass of H2O to moles of H2O. Twice this number is the number of moles of H in
the sample (1 mol H2O = 2 mol H) because for every two H atoms in the sample, one water
molecule is produced;
(4) Convert moles of H to grams of H;
(5) Subtract the combined mass of C and H from the sample mass; if the result is zero, then only
C and H are present in the sample; if the result is greater than zero, then this is the mass of the
other element in the sample (in this example, oxygen); convert the mass of this element to moles;
(6) The mole ratio leads to the empirical formula as in percent composition problems.
? mol C = 28.16 mg CO2 
1 mol CO2
1 mol C
1g


1000 mg 44.01 g CO2 1 mol CO2
= 0.0006399 mol C
and
0.0006399 mol C 
? mol H = 11.53 mg H2O
12.01g C
1mol C
= 0.007685 g C = 7.685 mg C
1mol H 2 O
1g
2 mol H


1000 mg 18.02g H 2O 1mol H 2O
and
3
= 0.001280 mol H
0.001280 mol H 
1.008g H
1mol H
= 0.001290 g H = 1.290 mg H
mg O = 10.00 – (7.685+1.290) = 1.025 mg O 
1g
1mol O

1000 mg 16.00 g O
= 0.0000641 mol O
C:H:O = 0.0006399 : 0.001280 : 0.0000641 = 9.98 : 19.97 : 1.00 ≈ 10 : 20 : 1
The empirical formula for menthol is C10H20O
3.70
The mole ratio from the balanced equation is 2 moles CO2 : 2 moles CO.
3.60 mol CO 
2 mol CO2
 3.60 mol CO2
2 mol CO
3.88
This is a limiting reactant problem. Calculate the moles of NO2 produced assuming complete
reaction for each reactant.
2NO(g)  O2(g)  2NO2(g)
0.886 mol NO ×
0.503 mol O2 ×
2 mol NO2
= 0.886 mol NO2
2 mol NO
2 mol NO2
= 1.01 mol NO2
1 mol O2
NO is the limiting reactant; it limits the amount of product produced. The amount of product
produced is 0.886 mole NO2.
4