Chapter 3
Formulas, Equations, and Moles

3.1 2 KClO₃ ® 2 KCl + 3 O₂
3.2 (a) C₆H₁₂O₆ ® 2 C₂H₆O + 2 CO₂
(b) 4 Fe + 3 O₂ ® 2 Fe₂O₃
(c) 4 NH₃ + Cl₂ ® N₂H₄ + 2 NH₄Cl

3.3 3 A₂ + 2 B ® 2 BA₃

3.4 (a) Fe₂O₃: 2(55.85) + 3(16.00) = 159.7 amu
(b) H₂SO₄: 2(1.01) + 1(32.07) + 4(16.00) = 98.1 amu
(c) C₆H₈O₇: 6(12.01) + 8(1.01) + 7(16.00) = 192.1 amu
(d) C₁₆H₁₈N₂O₄S: 16(12.01) + 18(1.01) + 2(14.01) + 4(16.00) + 1(32.07) = 334.4 amu

3.5 Fe₂O₃(s) + 3 CO(g) ® 2 Fe(s) + 3 CO₂(g)

0.500 mol

3.6 C₅H₁₁NO₂S: 5(12.01) + 11(1.01) + 1(14.01) + 2(16.00) + 1(32.07) = 149.24 amu

3.7 C₉H₈O₄, 180.2 amu; 500 mg = 500 x 10^-3 g = 0.500 g
0.500 g x
2.77 x 10^-3 mol x

3.8 salicylic acid, C₇H₆O₃, 138.1 amu; acetic anhydride, C₄H₆O₃, 102.1 amu
aspirin, C₉H₈O₄, 180.2 amu; acetic acid, C₂H₄O₂, 60.1 amu
4.50 g C₇H₆O₃ x = 3.33 g C₄H₆O₃
4.50 g C₇H₆O₃ x = 5.87 g C₉H₈O₄
4.50 g C₇H₆O₃ x = 1.96 g C₂H₄O₂

3.9 C₂H₄, 28.1 amu; C₂H₆O, 46.1 amu
4.6 g = 7.5 g C₂H₆O (theoretical yield)

3.10 CH₄, 16.04 amu; CH₂Cl₂, 84.93 amu; 1.85 kg = 1850 g
850 g CH₄ x = 9800 g CH₂Cl₂ (theoretical yield)
Actual yield = (9800 g)(0.431) = 4220 g CH₂Cl₂

3.11 Li₂O, 29.9 amu: 65 kg = 65,000 g; H₂O, 18.0 amu: 80.0 kg = 80,000 g
65,000 g Li₂O x = 2.17 x 10³ mol Li₂O
80,000 g H₂O x = 4.44 x 10³ mol H₂O
The reaction stoichiometry between Li₂O and H₂O is one to one. There are twice as many moles of H₂O as there are moles of Li₂O. Therefore, Li₂O is the limiting reactant.
(4.44 x 10³ mol - 2.17 x 10³ mol) = 2.27 x 10³ mol H₂O remaining
2.27 x 10³ mol H₂O x = 40,860 g H₂O = 40.9 kg = 41 kg H₂O

3.12 LiOH, 23.9 amu; CO₂, 44.0 amu

3.13 (a) A + B₂ ® AB₂
There is a 1:1 stoichiometry between the two reactants. A is the limiting reactant because there are fewer reactant A's than there are reactant B₂'s.
(b) 1.0 mol of AB₂ can be made from 1.0 mol of A and 1.0 mol of B₂.

3.23 For dimethylhydrazine, C₂H₈N₂, divide each subscript by 2 to obtain the empirical formula. The empirical formula is CH₄N. C₂H₈N₂, 60.1 amu or 60.1 g/mol

3.24 Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains 14.25 g C, 56.93 g O, and 28.83 g Mg.
14.25 g C x = 1.19 mol C
56.93 g O x = 3.56 mol O
28.83 g Mg x = 1.19 mol Mg
Mg_1.19C_1.19O_3.56; divide each subscript by the smallest, 1.19.
Mg_{1.19 / 1.19}C_{1.19 / 1.19}O_{3.56 / 1.19}
The empirical formula is MgCO₃.

3.25 1.161 g H₂O x = 0.129 mol H
2.818 g CO₂ x = 0.0640 mol C
0.129 mol H x = 0.130 g H
0.0640 mol C x = 0.768 g C
1.00 g total - (0.130 g H + 0.768 g C) = 0.102 g O
0.102 g O x = 0.006 38 mol O
C_0.0640H_0.129O_{0.006 38}; divide each subscript by the smallest, 0.006 38.
C_{0.0640 / 0.006 38}H_{0.129 / 0.006 38}O_{0.006 38 / 0.006 38}
C_10.03H_20.22O₁ The empirical formula is C₁₀H₂₀O.

3.26 The empirical formula is CH₂O, 30 amu: molecular mass = 150 amu.
; therefore
molecular formula = 5 x empirical formula = C_{(5 x 1)}H_{(5 x 2)}O_{(5 x 1)} = C₅H₁₀O₅

3.27 (a) Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains 21.86 g H and 78.14 g B.


B_7.24 H_21.6; divide each subscript by the smaller, 7.24.
B_{7.24 / 7.24} H_{21.6 / 7.24} The empirical formula is BH₃, 13.8 amu.
27.7 amu / 13.8 amu = 2; molecular formula = B_{(2 x 1)}H_{(2 x 3)} = B₂H₆.

(b) Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains 6.71 g H, 40.00 g C, and 53.28 g O.



C_3.33 H_6.64 O_3.33; divide each subscript by the smallest, 3.33.
C_{3.33 / 3.33} H_{6.64 / 3.33} O_{3.33 / 3.33} The empirical formula is CH₂O, 30.0 amu.
90.08 amu / 30.0 amu = 3; molecular formula = C_{(3 x 1)}H_{(3 x 2)}O_{(3 x 1)} = C₃H₆O₃

3.32 C₂H₄ + 3 O₂ 2 CO₂ + 2 H₂O

3.39 (a) and (c) are not balanced, (b) is balanced.
(a) 2 Al + Fe₂O₃ Al₂O₃ + 2 Fe (balanced)
(c) 4 Au + 8 NaCN + O₂ + 2 H₂O 4 NaAu(CN)₂ + 4 NaOH (balanced)

3.45 (a)
(b)
(c)
(d)

3.61 (a) Fe₂O₃ + 3 CO ® 2 Fe + 3 CO₂
(b) Fe₂O₃, 159.7 amu; CO, 28.01 amu

(c)

3.73 2 NaN₃ ® 3 N₂ + 2 Na; NaN₃, 65.01 amu; N₂, 28.01 amu
38.5 g NaN₃ x = 41.8 L

3.96 Toluene contains only C and H.
152.5 mg = 0.1525 g and 35.67 mg = 0.035 67 g


C_{0.003 465}H_{0.003 959}; divide each subscript by the smaller, 0.003 465.
C_{0.003 465 / 0.003 465}H_{0.003 959 / 0.003 465}
CH_1.14; multiply each subscript by 7 to obtain integers.
The empirical formula is C₇H₈.

3.115 Mass of added Cl = mass of XCl₅ - mass of XCl₃ = 13.233 g - 8.729 g = 4.504 g
mass of Cl in XCl₅ = 5 Cl's x = 11.26 g Cl
mass of X in XCl₅ = 13.233 g - 11.26 g = 1.973 g X
11.26 g Cl x = 0.3176 mol Cl
0.3176 mol Cl x = 0.063 52 mol X
molar mass of X = = 31.1 g/mol; atomic mass =31.1 amu, X = P