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.

3.14 (a) 125 mL = 0.125 L; (0.20 mol/L)(0.125 L) = 0.025 mol NaHCO₃
(b) 650.0 mL = 0.6500 L; (2.50 mol/L)(0.650 L) = 1.62 mol H₂SO₄

3.15 (a) NaOH, 40.0 amu; 500.0 mL = 0.5000 L

(b) C₆H₁₂O₆, 180.2 amu

3.16 C₆H₁₂O₆, 180.2 amu;
25.0 g C₆H₁₂O₆ x = 0.1387 mol C₆H₁₂O₆
= 0.69 L; 0.69 L = 690 mL

3.17 C₂₇H₄₆O, 386.7 amu; 750 mL = 0.750 L

3.18 M_i x V_i = M_f x V_f; M_f = = 0.656 M

3.19 M_i x V_i = M_f x V_f;
Dilute 6.94 mL of 18.0 M H₂SO₄ with enough water to make 250.0 mL of solution. The resulting solution will be 0.500 M H₂SO₄.

3.20 50.0 mL = 0.0500 L; (0.100 mol/L)(0.0500 L) = 5.00 x 10^-3 mol NaOH
5.00 x 10^-3 mol NaOH x = 2.50 x 10^-3 mol H₂SO₄
volume = = 0.0100 L; 0.0100 L = 10.0 mL H₂SO₄

3.21 HNO₃(aq) + KOH(aq)® KNO₃(aq) + H₂O(l)
25.0 mL = 0.0250 L and 68.5 mL = 0.0685 L

3.22 From the reaction stoichiometry, moles NaOH = moles CH₃CO₂H
(0.200 mol/L)(0.0947 L) = 0.018 94 mol NaOH = 0.018 94 mol CH₃CO₂H
molarity = = 0.758 M

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

3.28 Main sources of error in calculating Avogadro's number by spreading oil on a pond are:
(i) the assumption that the oil molecules are tiny cubes
(ii) the assumption that the oil layer is one molecule thick
(iii) the assumption of a molecular mass of 200 for the oil

3.29 area of oil = 2.0 x 10⁷ cm²
volume of oil = 4.9 cm³ = area x 4l = (2.0 x 10⁷ cm²) x 4l
l = = 6.125 x 10^-8 cm
area of oil = 2.0 x 10⁷ cm² = l² x N = (6.125 x 10^-8 cm)² x N
N = = 5.33 x 10²¹ oil molecules
moles of oil = (4.9 cm³) x (0.95 g/cm³) x = 0.0233 mol oil
Avogadro's number = = 2.3 x 10²³ molecules/mole

Understanding Key Concepts

3.30 The concentration of a solution is cut in half when the volume is doubled. This is best represented by box (b).

3.31 (c) 2 A + B₂® A₂B₂

3.32 C₂H₄ + 3 O₂ 2 CO₂ + 2 H₂O

3.33 reactants, box (d), and products, box (c)

3.34 C₁₇H₁₈F₃NO 17(12.01) + 18(1.01) + 3(19.00) + 1(14.01) + 1(16.00) = 309.36 amu

3.35 Adding a subscript of "2" to the oxygen in N₂O would change nitrous oxide (N₂O) into a different compound. This is not allowed to balance an equation.

3.36 Because the two volumes are equal (let the volume = y L), the concentrations are proportional to the number of solute ions.
OH^- concentration = 1.00 M x = 0.67 M

3.37 (a) A₂ + 3 B₂ ® 2 AB₃; B₂ is the limiting reactant because it is completely consumed.
(b) For 1.0 mol of A₂, 3.0 mol of B₂ are required. Because only 1.0 mol of B₂ is available, B₂ is the limiting reactant.
1 mol B₂ x = 2/3 mol AB₃

Additional Problems

Balancing Equations

3.38 Equation (b) is balanced, (a) is not balanced.

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)

3.40 (a) Mg + 2 HNO₃ ® H₂ + Mg(NO₃)₂
(b) CaC₂ + 2 H₂O ® Ca(OH)₂ + C₂H₂
(c) 2 S + 3 O₂ ® 2 SO₃
(d) UO₂ + 4 HF ® UF₄ + 2 H₂O

3.41 (a) 2 NH₄NO₃ ® 2 N₂ + O₂ + 4 H₂O
(b) C₂H₆O + O₂ ® C₂H₄O₂ + H₂O
(c) C₂H₈N₂ + 2 N₂O₄ ® 3 N₂ + 2 CO₂ + 4 H₂O

Molecular Masses and Moles

3.42 Hg₂Cl₂: 2(200.59) + 2(35.45) = 472.1 amu
C₄H₈O₂: 4(12.01) + 8(1.01) + 2(16.00) = 88.1 amu
CF₂Cl₂: 1(12.01) + 2(19.00) + 2(35.45) = 120.9 amu

3.43 (a) (1 x 30.97 amu) + (Y x 35.45 amu) = 137.3 amu; Solve for Y; Y = 3.
The formula is PCl_3.

(b) (10 x 12.01 amu) + (14 x 1.008 amu) + (Z x 14.01 amu) = 162.2 amu.
Solve for Z; Z = 2. The formula is C₁₀H₁₄N₂.

3.44 One mole equals the atomic mass or molecular mass in grams.
(a) Ti, 47.88 g (b) Br₂, 159.81 g (c) Hg, 200.59 g (d) H₂O, 18.02 g

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

3.46 There are 2 ions per each formula unit of NaCl. (2.5 mol)(2 mol ions/mol) = 5.0 mol ions

3.47 There are 2 K⁺ ions per each formula unit of K₂SO₄.

3.48 There are 3 ions (one Mg²⁺ and 2 Cl^-) per each formula unit of MgCl₂.
MgCl₂, 95.2 amu
27.5 g MgCl₂ x = 0.867 mol ions

3.49 There are 3 F^- anions per each formula unit of AlF₃.
AlF₃, 84.0 amu
35.6 g AlF₃ x = 1.27 mol F^-

3.50 Molar mass = ; molecular mass = 119 amu.

3.51 Molar mass = = 386.7 g/mol; molecular mass = 386.7 amu.

3.52 FeSO₄ , 151.9 amu; 300 mg = 0.300 g

3.53 0.0001 g C x = 5 x 10¹⁸ C atoms

3.54 C₈H₁₀N₄O₂, 194.2 amu; 125 mg = 0.125 g
0.125 g caffeine x 6.44 x 10^-4 mol caffeine

0.125 g caffeine x = 3.88 x 10²⁰ caffeine molecules

3.55

3.56 (a) 1.0 g Li x = 0.14 mol Li

(b) 1.0 g Au x = 0.0051 mol Au

(c) penicillin G: C₁₆H₁₇N₂O₄SK, 372.5 amu
1.0 g x = 2.7 x 10^-3 mol penicillin G

3.57 (a)
(b)
(c) C₁₆H₁₃ClN₂O, 284.7 amu

Stoichiometry Calculations

3.58 TiO₂, 79.88 amu; = 167 kg TiO₂

3.59 Fe₂O₃, 159.7 amu;
mass Fe = (0.6994)(105 kg) = 73.4 kg

3.60 (a) 2 Fe₂O₃ + 3 C ® 4 Fe + 3 CO₂

(b) Fe₂O₃, 159.7 amu; 525 g Fe₂O₃ x = 4.93 mol C

(c) 4.93 mol C x = 59.2 g C

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

(c)

3.62 (a) 2 Mg + O₂ ® 2 MgO

(b) Mg, 24.30 amu; O₂, 32.00 amu; MgO, 40.30 amu
25.0 g Mg x = 16.5 g O₂
25.0 g Mg x = 41.5 g MgO

(c) 25.0 g O₂ x = 38.0 g Mg
25.0 g O₂ x = 63.0 g MgO

3.63 C₂H₄ + H₂O ® C₂H₆O; C₂H₄, 28.05 amu; H₂O, 18.02 amu; C₂H₆O, 46.07 amu
(a)

(b)

3.64 (a) 2 HgO ® 2 Hg + O₂
(b) HgO, 216.6 amu; Hg, 200.6 amu; O₂, 32.0 amu
45.5 g HgO x = 42.1 g Hg
45.5 g HgO x = 3.36 g O₂
(c) 33.3 g O₂ x = 451 g HgO

3.65 5.60 kg = 5600 g; TiCl₄, 189.7 amu; TiO₂, 79.88 amu

3.66

Ag_0.0185Cl_0.0185 Divide both subscripts by 0.0185. The empirical formula is AgCl.

3.67 5.0 g Al x = 0.19 mol Al; 4.45 g O x = 0.28 mol O
Al_0.19O_0.28; divide both subscripts by the smaller, 0.19.
Al_{0.19 / 0.19}O_{0.28 / 0.19}
Al₁O_1.5; multiply both subscripts by 2 to obtain integers. The empirical formula is Al₂O₃.

Limiting Reactants and Reaction Yield

3.68 3.44 mol N₂ x = 10.3 mol H₂ required.
Because there is only 1.39 mol H₂, H₂ is the limiting reactant.
1.39 mol H₂ x = 15.8 g NH₃
1.39 mol H₂ x = 13.0 g N₂ reacted
3.44 mol N₂ x = 96.3 g N₂ initially
(96.3 g - 13.0 g) = 83.3 g N₂ left over

3.69 H₂, 2.016 amu; Cl₂, 70.91 amu; HCl 36.46 amu


Because the reaction stoichiometry between H₂ and Cl₂ is one to one, Cl₂ is the limiting reactant.

3.70 C₂H₄, 28.05 amu; Cl₂, 70.91 amu; C₂H₄Cl₂, 98.96 amu
15.4 g C₂H₄ x = 0.549 mol C₂H₄
3.74 g Cl₂ x = 0.0527 mol Cl₂
Because the reaction stoichiometry between C₂H₄ and Cl₂ is one to one, Cl₂ is the limiting reactant.
0.0527 mol Cl₂ x = 5.22 g C₂H₄Cl₂

3.71 (a) NaCl, 58.44 amu; AgNO₃, 169.9 amu; AgCl, 143.3 amu; NaNO₃, 85.00 amu
NaCl + AgNO₃ ® AgCl + NaNO₃
1.3 g NaCl x = 0.0222 mol NaCl
3.5 g AgNO₃ x = 0.0206 mol AgNO₃

Because the reaction stoichiometry between NaCl and AgNO₃ is one to one, AgNO₃ is the limiting reactant.
0.0206 mol AgNO₃ x = 3.0 g AgCl
0.0206 mol AgNO₃ x = 1.8 g NaNO₃
0.0206 mol AgNO₃ x = 1.2 g NaCl reacted
(1.3 g - 1.2 g) = 0.1 g NaCl left over

(b) BaCl₂, 208.2 amu; H₂SO₄, 98.08 amu; BaSO₄, 233.4 amu; HCl, 36.46 amu
BaCl₂ + H₂SO₄ ® BaSO₄ + 2 HCl
2.65 g BaCl₂ x = 0.0127 mol BaCl₂
6.78 g H₂SO₄ x = 0.0691 mol H₂SO₄
Because the reaction stoichiometry between BaCl₂ and H₂SO₄ is one to one, BaCl₂ is the limiting reactant.
0.0127 mol BaCl₂ x = 2.96 g BaSO₄
0.0127 mol BaCl₂ x = 0.926 g HCl
0.0127 mol BaCl₂ x = 1.25 g H₂SO₄ reacted
(6.78 g - 1.25 g) = 5.53 g H₂SO₄ left over

3.72 CaCO₃, 100.1 amu; HCl, 36.46 amu
CaCO₃ + 2 HCl ® CaCl₂ + H₂O + CO₂


The reaction stoichiometry is 1 mole of CaCO₃ for every 2 moles of HCl. For 0.0235 mol CaCO₃, we only need 2(0.0235 mol) = 0.0470 mol HCl. We have 0.0645 mol HCl; therefore CaCO₃ is the limiting reactant.

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

3.74 CH₃CO₂H + C₅H₁₂O ® C₇H₁₄O₂ + H₂O
CH₃CO₂H, 60.05 amu; C₅H₁₂O, 88.15 amu; C₇H₁₄O₂, 130.19 amu


Because the reaction stoichiometry between CH₃CO₂H and C₅H₁₂O is one to one, isopentyl alcohol (C₅H₁₂O) is the limiting reactant.

7.02 g C₇H₁₄O₂ is the theoretical yield. Actual yield = (7.02 g)(0.45) = 3.2 g.

3.75 K₂PtCl₄ + 2 NH₃ ® 2 KCl + Pt(NH₃)₂Cl₂
K₂PtCl₄, 415.1 amu; NH₃, 17.03 amu; Pt(NH₃)₂Cl₂, 300.0 amu
55.8 g K₂PtCl₄ x = 0.134 mol K₂PtCl₄
35.6 g NH₃ x = 2.09 mol NH₃
Only 2(0.134) = 0.268 mol NH₃ are needed to react with 0.134 mol K₂PtCl₄. Therefore, the NH₃ is in excess and K₂PtCl₄ is the limiting reactant.
0.134 mol K₂PtCl₄ x = 40.2 g Pt(NH₃)₂Cl₂
40.2 g Pt(NH₃)₂Cl₂ is the theoretical yield.
Actual yield = (40.2 g)(0.95) = 38 g Pt(NH₃)₂Cl₂.

3.76 CH₃CO₂H + C₅H₁₂O ® C₇H₁₄O₂ + H₂O
CH₃CO₂H, 60.05 amu; C₅H₁₂O, 88.15 amu; C₇H₁₄O₂, 130.19 amu
1.87 g CH₃CO₂H x = 0.0311 mol CH₃CO₂H
2.31 g C₅H₁₂O x = 0.0262 mol C₅H₁₂O
Because the reaction stoichiometry between CH₃CO₂H and C₅H₁₂O is one to one, isopentyl alcohol (C₅H₁₂O) is the limiting reactant.
0.0262 mol C₅H₁₂O x = 3.41 g C₇H₁₄O₂
3.41 g C₇H₁₄O₂ is the theoretical yield.
% Yield = = 86.8%

3.77 K₂PtCl₄ + 2 NH₃ ® 2 KCl + Pt(NH₃)₂Cl₂
K₂PtCl₄, 415.1 amu; NH₃, 17.03 amu; Pt(NH₃)₂Cl₂, 300.0 amu
3.42 g K₂PtCl₄ x = 0.008 24 mol K₂PtCl₄
1.61 g NH₃ x = 0.0945 mol NH₃
Only 2 x (0.008 24) = 0.0165 mol of NH₃ are needed to react with 0.008 24 mol K₂PtCl₄. Therefore, the NH₃ is in excess and K₂PtCl₄ is the limiting reactant.
0.008 24 mol K₂PtCl₄ x = 2.47 g Pt(NH₃)₂Cl₂
2.47 g Pt(NH₃)₂Cl₂ is the theoretical yield. 2.08 g Pt(NH₃)₂Cl₂ is the actual yield.
% Yield = x 100% = x 100% = 84.2%

Molarity, Solution Stoichiometry, Dilution, and Titration

3.78 (a) 35.0 mL = 0.0350 L;
(b) 175 mL = 0.175 L;

3.79 (a) C₂H₆O, 46.07 amu; 250.0 mL = 0.2500 L
= 0.150 mol C₂H₆O
(0.150 mol)(46.07 g/mol) = 6.91 g C₂H₆O

(b) H₃BO₃, 61.83 amu; 167 mL = 0.167 L
= 0.0334 mol H₃BO₃
(0.0334 mol)(61.83 g/mol) = 2.07 g H₃BO₃

3.80 BaCl₂, 208.2 amu



3.81 0.0171 mol KOH x = 0.0489 L; 0.0489 L = 48.9 mL
3.82 NaCl, 58.4 amu; 400 mg = 0.400 g; 100 mL = 0.100 L
0.400 g NaCl x = 0.006 85 mol NaCl
molarity = = 0.0685 M

3.83 C₆H₁₂O₆, 180.2 amu; 90 mg = 0.090 g; 100 mL = 0.100 L

molarity = = 0.0050 M = 5.0 x 10^-3 M

3.84 NaCl, 58.4 amu; KCl, 74.6 amu; CaCl₂, 111.0 amu; 500 mL = 0.500 L
4.30 g NaCl x = 0.0736 mol NaCl
0.150 g KCl x = 0.002 01 mol KCl
0.165 g CaCl₂ x = 0.001 49 mol CaCl₂
0.0736 mol + 0.002 01 mol + 2(0.001 49 mol) = 0.0786 mol Cl^-
Na⁺ molarity = = 0.147 M
Ca²⁺ molarity = = 0.002 98 M
K⁺ molarity = = 0.004 02 M
Cl^- molarity = = 0.157 M

3.85 3.045 g Cu x = 0.047 92 mol Cu; 50.0 mL = 0.0500 L
Cu(NO₃)₂ molarity = = 0.958 M

3.86 M_f x V_f = M_i x V_i;

3.87 M_f x V_f = M_i x V_i;

3.88 2 HBr(aq) + K₂CO₃(aq) ® 2 KBr(aq) + CO₂(g) + H₂O(l)
K₂CO₃, 138.2 amu; 450 mL = 0.450 L
= 0.225 mol HBr
0.225 mol HBr x = 15.5 g K₂CO₃

3.89 2 C₄H₁₀S + NaOCl ® C₈H₁₈S₂ + NaCl + H₂O
C₄H₁₀S, 90.19 amu; 5.00 mL = 0.005 00 L
0.0985 x 0.005 00 L = 4.925 x 10^-4 mol NaOCl
4.925 x 10^-4 mol NaOCl x = 0.0888 g C₄H₁₀S

3.90 H₂C₂O₄, 90.04 amu

= 57.2 mL

3.91 H₂C₂O₄, 90.04 amu; 400.0 mL = 0.4000 L; 25.0 mL = 0.0250 L

molarity =
H₂C₂O₄(aq) + 2 KOH(aq) ® K₂C₂O₄(aq) + 2 H₂O(l)
= 0.008 32 mol H₂C₂O₄
0.008 32 mol H₂C₂O₄ x = 0.0166 mol KOH
= 0.166 L; 0.166 L = 166 mL

Formulas and Elemental Analysis

3.92 CH₄N₂O, 60.1 amu

3.93 (a) Cu₂(OH)₂CO₃, 221.1 amu
% Cu = = 57.4%
% O = = 36.2%
% C = = 5.43%
% H = = 0.91%

(b) C₈H₉NO₂, 151.2 amu
% C = = 63.5%
% H = = 6.01%
% N = = 9.26%
% O = = 21.2%

(c) Fe₄[Fe(CN)₆]₃, 859.2 amu
% Fe = = 45.5%
% C = = 25.2%
% N = = 29.3%

3.94 Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains 24.25 g F and 75.75 g Sn.


Sn_0.6382F_1.276; divide each subscript by the smaller, 0.6382.
Sn_{0.6382 / 0.6382}F_{1.276 / 0.6382} The empirical formula is SnF₂.

3.95 (a) Assume a 100.0 g sample of ibuprofen. From the percent composition data, a 100.0 g sample contains 75.69 g C, 15.51 g O, and 8.80 g H.
75.69 g C x = 6.302 mol C
15.51 g O x = 0.9694 mol O
8.80 g H x = 8.71 mol H
C_6.302H_8.71O_0.9694, divide each subscript by the smallest, 0.9694.
C_{6.302 / 0.9694}H_{8.71 / 0.9694}O_{0.9694 / 0.9694}
C_6.5H₉O; multiply each subscript by 2 to obtain integers.
The empirical formula is C₁₃H₁₈O₂.

(b) Assume a 100.0 g sample of tetraethyllead. From the percent composition data, a 100.0 g sample contains 29.71 g C, 6.23 g H, and 64.06 g Pb.
29.71 g C x = 2.474 mol C
6.23 g H x = 6.17 mol H
64.06 g Pb x = 0.3092 mol Pb
Pb_0.3092C_2.474H_6.17; divide each subscript by the smallest, 0.3092.
Pb_{0.3092 / 0.3092}C_{2.474 / 0.3092}H_{6.17 / 0.3092} The empirical formula is PbC₈H₂₀.

(c) Assume a 100.0 g sample of zircon. From the percent composition data, a 100.0 g sample contains 34.91 g O, 15.32 g Si, and 49.76 g Zr.
34.91 g O x = 2.182 mol O
15.32 g Si x = 0.5454 mol Si
49.76 g Zr x = 0.5455 mol Zr
Zr_0.5455Si_0.5454O_2.182; divide each subscript by the smallest, 0.5454.
Zr_{0.5455 / 0.5454}Si_{0.5454 / 0.5454}O_{2.182 / 0.5454} The empirical formula is ZrSiO₄.

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.

3.97 5.024 mg = 0.005 024 g; 13.90 mg = 0.013 90 g; 6.048 mg = 0.006 048 g
0.013 90 g CO₂ x = 3.158 x 10^-4 mol C
0.006 048 g H₂O x = 6.713 x 10^-4 mol H
3.158 x 10^-4 mol C x = 0.003 793 g C
6.713 x 10^-4 mol H x = 0.000 676 7 g H
mass N = 0.005 024 g - (0.003 793 g + 0.000 676 7 g) = 0.000 554 g N
0.000 554 g N x = 3.95 x 10^-5 mol N
Scale each mol quantity to eliminate exponents.
C_3.158H_6.713N_0.395; divide each subscript by the smallest, 0.395.
C_{3.158 / 0.395}H_{6.713 / 0.395}N_{0.395 / 0.395} The empirical formula is C₈H₁₇N.

3.98 Let X equal the molecular mass of cytochrome c.
;

3.99 Let X equal the molecular mass of nitrogenase.
0.000 872 = ; X = = 220,000 amu

3.100 Let X equal the molecular mass of disilane.
;
62.23 amu - 2(Si atomic mass) = 62.23 amu - 2(28.09 amu) = 6.05 amu
6.05 amu is the total mass of H atoms.
; Disilane is Si₂H₆.

3.101 Let X equal the molecular mass of MS₂.
0.4006 = ; X = = 160.1 amu
Atomic mass of M = 160.1 amu - 2(S atomic mass)
= 160.1 amu - 2(32.07 amu) = 95.96 amu
M is Mo.

General Problems

3.102 (a) C₆H₁₂O₆, 180.2 amu

(b) H₂SO₄, 98.08 amu

(c) KMnO₄, 158.0 amu




(d) C₇H₅NO₃S, 183.2 amu

3.103 (a) Assume a 100.0 g sample of aspirin. From the percent composition data, a 100.0 g sample contains 60.00 g C, 35.52 g O, and 4.48 g H.
60.00 g C x = 5.00 mol C
35.52 g O x = 2.22 mol O
4.48 g H x = 4.44 mol H
C_5.00H_4.44O_2.22; divide each subscript by the smallest, 2.22.
C_{5.00 / 2.22}H_{4.44 / 2.22}O_{2.22 / 2.22}
C_2.25H₂O₁; multiply each subscript by 4 to obtain integers.
The empirical formula is C₉H₈O₄.

(b) Assume a 100.0 g sample of ilmenite. From the percent composition data, a 100.0 g sample contains 31.63 g O, 31.56 g Ti, and 36.81 g Fe.
31.63 g O x = 1.98 mol O
31.56 g Ti x = 0.659 mol Ti
36.81 g Fe x = 0.659 mol Fe
Fe_0.659Ti_0.659O_1.98; divide each subscript by the smallest, 0.659.
Fe_{0.659 / 0.659}Ti_{0.659 / 0.659}O_{1.98 / 0.659} The empirical formula is FeTiO₃.

(c) Assume a 100.0 g sample of sodium thiosulfate. From the percent composition data, a 100.0 g sample contains 30.36 g O, 29.08 g Na, and 40.56 g S.
30.36 g O x = 1.90 mol O
29.08 g Na x = 1.26 mol Na
40.56 g S x = 1.26 mol S
Na_1.26S_1.26O_1.90; divide each subscript by the smallest, 1.26.
Na_{1.26 / 1.26}S_{1.26 / 1.26}O_{1.90 / 1.26}
NaSO_1.5; multiply each subscript by 2 to obtain integers.
The empirical formula is Na₂S₂O₃.

3.104 (a) SiCl₄ + 2 H₂O ® SiO₂ + 4 HCl
(b) P₄O₁₀ + 6 H₂O ® 4 H₃PO₄
(c) CaCN₂ + 3 H₂O ® CaCO₃ + 2 NH₃
(d) 3 NO₂ + H₂O ® 2 HNO₃ + NO

3.105 NaH, 24.00 amu; B₂H₆, 27.67 amu; NaBH₄, 37.83 amu
2 NaH + B₂H₆ 2 NaBH₄
8.55 g NaH x = 0.356 mol NaH
6.75 g B₂H₆ x = 0.244 mol B₂H₆
For 0.244 mol B₂H₆, 2 x (0.244) = 0.488 mol NaH are needed. Because only 0.356 mol of NaH is available, NaH is the limiting reactant.
0.356 mol NaH x = 13.5 g NaBH₄ produced
0.356 mol NaH x = 4.93 g B₂H₆ reacted
B₂H₆ left over = 6.75 g - 4.93 g = 1.82 g B₂H₆

3.106 Assume a 100.0 g sample of ferrocene. From the percent composition data, a 100.0 g sample contains 5.42 g H, 64.56 g C, and 30.02 g Fe.
5.42 g H x = 5.37 mol H
64.56 g C x = 5.38 mol C
30.02 g Fe x = 0.538 mol Fe
C_5.38H_5.37Fe_0.538; divide each subscript by the smallest, 0.538.
C_{5.38 / 0.538}H_{5.37 / 0.538}Fe_{0.538 / 0.538} The empirical formula is C₁₀H₁₀Fe.

3.107 Mass of 1 HCl molecule = (36.5 )(1.6605 x 10^-24 ) = 6.06 x 10^-23 g/molecule
Avogadro's number = = 6.02 x 10²³ molecules/mole

3.108 Na₂SO₄, 142.04 amu; Na₃PO₄, 163.94 amu; Li₂SO₄, 109.95 amu; 100.00 mL = 0.10000 L
0.550 g Na₂SO₄ x = 0.003 872 mol Na₂SO₄
1.188 g Na₃PO₄ x = 0.007 247 mol Na₃PO₄
0.223 g Li₂SO₄ x = 0.002 028 mol Li₂SO₄
Na⁺ molarity = = 0.295 M
Li⁺ molarity = = 0.0406 M
SO₄^2- molarity = = 0.0590 M
PO₄^3- molarity = = 0.0725 M

3.109 23.46 mg = 0.023 46 g; 20.42 mg = 0.02042 g; 33.27 mg = 0.033 27 g




mass O = 0.023 46 g - (0.009 080 g + 0.002 284 g) = 0.012 10 g O

Scale each mol quantity to eliminate exponents.
C_0.7560H_2.266O_0.7563; divide each subscript by the smallest, 0.7560.
C_{0.7560 / 0.7560}H_{2.266 / 0.7560}O_{0.7563 / 0.7560} The empirical formula is CH₃O, 31.0 amu.
62.0 amu / 31.0 amu = 2; molecular formula = C_{(2 x 1)}H_{(2 x 3)}O_{(2 x 1)} = C₂H₆O₂

3.110 High resolution mass spectrometry is capable of measuring the mass of molecules with a particular isotopic composition.

3.111 AgCl, 143.32 amu; %Cl in AgCl = x 100% = 24.73%
Find the mass of Cl in 1.68 g of AgCl. Mass Cl = 1.68 g x 0.2473 = 0.4155 g Cl

All of the Cl in AgCl came from XCl₃.
Find the mass of X in 0.634 g of XCl₃. Mass of X = 0.634 g - 0.4155 g = 0.2185 g X
0.4155 g Cl x = 0.011 72 mol Cl
0.011 72 mol Cl x = 0.003 907 mol X
molar mass of X = = 55.93 g/mol; X = Fe

3.112 C₆H₁₂O₆ + 6 O₂ ® 6 CO₂ + 6 H₂O; C₆H₁₂O₆, 180.16 amu; CO₂, 44.01 amu
66.3 g C₆H₁₂O₆ x = 97.2 g CO₂
66.3 g C₆H₁₂O₆ x = 56.1 L CO₂

3.113 H₂C₂O₄, 90.04 amu; 22.35 mL = 0.02235 L
0.5170 g H₂C₂O₂ x = 0.002 297 mol KMnO₄
KMnO₄ molarity = = 0.1028 M

3.114 Mass of Cu = 2.196 g; mass of S = 2.748 g - 2.196 g = 0.552 g S
(a) %Cu = x 100% = 79.9%
%S = x 100% = 20.1%
(b) 2.196 g Cu x = 0.0346 mol Cu
0.552 g S x = 0.0172 mol S
Cu_0.0346S_0.0172; divide each subscript by the smaller, 0.0172.
Cu_{0.0346 / 0.0172}S_{0.0172 / 0.0172} The empirical formula is Cu₂S.

(c) Cu₂S, 159.16 amu

= 4.2 x 10²² Cu⁺ ions/cm³

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

3.116 PCl₃, 137.33 amu; PCl₅, 208.24 amu
Let Y = mass of PCl₃ in the mixture, and (10.00 - Y) = mass of PCl₅ in the mixture.
fraction Cl in PCl₃ = = 0.774 48
fraction Cl in PCl₅ = = 0.851 25
(mass of Cl in PCl₃) + (mass of Cl in PCl₅) = mass of Cl in the mixture
0.774 48Y + 0.851 25(10.00 g - Y) = (0.8104)(10.00 g)
Y = 5.32 g PCl₃ and 10.00 - Y = 4.68 g PCl₅

3.117 100.00 mL = 0.100 00 L; 71.02 mL = 0.071 02 L
mol H₂SO₄ = 0.1083 x 0.100 00 L = 0.010 83 mol H₂SO₄
mol NaOH = 0.1241 x 0.071 02 L = 0.008 814 mol NaOH
H₂SO₄ + 2 NaOH ® Na₂SO₄ + 2 H₂O
mol H₂SO₄ reacted with NaOH = 0.008 814 mol NaOH x = 0.004 407 mol H₂SO₄
mol H₂SO₄ reacted with MCO₃ = 0.010 83 mol - 0.004 407 mol = 0.006 423 mol H₂SO₄
mol H₂SO₄ reacted with MCO₃ = mol CO₃^2- in MCO₃ = mol CO₂ produced = 0.006 423 mol CO₂
(a) CO₃^2-, 60.01 amu; 0.006 423 mol CO₃^2- x = 0.3854 g CO₃^2-
mass of M = 1.268 g - 0.3854 g = 0.8826 g M
molar mass of M = = 137.4 g/mol; M is Ba
(b) 0.006 423 mol CO₂ x = 0.1571 L CO₂

3.118 NH₄NO₃, 80.04 amu; (NH₄)₂HPO₄, 132.06 amu
Assume you have a 100.0 g sample of the mixture.
Let X = grams of NH₄NO₃ and (100.0 - X) = grams of (NH₄)₂HPO₄.
Both compounds contain 2 nitrogen atoms per formula unit.
Because the mass % N in the sample is 30.43%, the 100.0 g sample contains 30.43 g N.
mol NH₄NO₃ =
mol (NH₄)₂HPO₄ =
mass N = x
= 30.43 g
Solve for X.
= 30.43
= 1.08627
= 1.08627
(132.06)(X) + (100.0 - X)(80.04) = (1.08627)(80.04)(132.06)
132.06X + 8004 - 80.04X = 11481.96
132.06X - 80.04X = 11481.96 - 8004
52.02X = 3477.96
X = = 66.86 g NH₄NO₃
(100.0 - X) = (100.0 - 66.86) = 33.14 g (NH₄)₂HPO₄

The mass ratio of NH₄NO₃ to (NH₄)₂HPO₄ in the mixture is 2 to 1.

3.119 (a) 56.0 mL = 0.0560 L
mol X₂ = (0.0560 L X₂)= 0.00250 mol X₂
mass X₂ = 1.12 g MX₂ - 0.720 g MX = 0.40 g X₂
molar mass X₂ = = 160 g/mol
atomic mass of X = 160/2 = 80 amu; X is Br.

(b) mol MX = 0.00250 mol X₂ x = 0.00500 mol MX
mass of X in MX = 0.00500 mol MX x = 0.40 g X
mass of M in MX = 0.720 g MX - 0.40 g X = 0.32 g M
molar mass M = = 64 g/mol
atomic mass of X = 64 amu; M is Cu.