Chapter 3

(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu

(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu

(a) 56.107 amu;
(b) 54.091 amu;
(c) 199.9976 amu;
(d) 97.9950 amu

Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.

Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.

The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 $\times$ 10²³ molecules.

(a) 256.528 g/mol; (b) 72.150 g mol⁻¹; (c) 378.103 g mol⁻¹; (d) 58.080 g mol⁻¹; (e) 180.158 g mol⁻¹

(a) 197.382 g mol⁻¹; (b) 257.163 g mol⁻¹; (c) 194.193 g mol⁻¹; (d) 60.056 g mol⁻¹; (e) 306.464 g mol⁻¹

(a) 0.819 g;
(b) 307 g;
(c) 0.23 g;
(d) 1.235 $\times$ 10⁶ g (1235 kg);
(e) 765 g

(a) 99.41 g;
(b) 2.27 g;
(c) 3.5 g;
(d) 222 kg;
(e) 160.1 g

(a) 9.60 g; (b) 19.2 g; (c) 28.8 g

zirconium: 2.038 $\times$ 10²³ atoms; 30.87 g; silicon: 2.038 $\times$ 10²³ atoms; 9.504 g; oxygen: 8.151 $\times$ 10²³ atoms; 21.66 g

AlPO₄: 1.000 mol, or 26.98 g Al
Al₂Cl₆: 1.994 mol, or 53.74 g Al
Al₂S₃: 3.00 mol, or 80.94 g Al
The Al2S3 sample thus contains the greatest mass of Al.

3.113 $\times$ 10²⁵ C atoms

0.865 servings, or about 1 serving.

20.0 g H₂O represents the least number of molecules since it has the least number of moles.

(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca²⁺ = 38.76%

% NH₃ = 38.2%

(a) CS₂
(b) CH₂O

C₆H₆

Mg₃Si₂H₃O₈ (empirical formula), Mg₆Si₄H₆O₁₆ (molecular formula)

C₁₅H₁₅N₃

We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

(a) 0.679 M;
(b) 1.00 M;
(c) 0.06998 M;
(d) 1.75 M;
(e) 0.070 M;
(f) 6.6 M

(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g

(a) 37.0 mol H₂SO₄;
3.63 $\times$ 10³ g H₂SO₄;
(b) 3.8 $\times$ 10⁻⁶ mol NaCN;
1.9 $\times$ 10⁻⁴ g NaCN;
(c) 73.2 mol H₂CO;
2.20 kg H₂CO;
(d) 5.9 $\times$ 10⁻⁷ mol FeSO₄;
8.9 $\times$ 10⁻⁵ g FeSO₄

(a) Determine the molar mass of KMnO₄; determine the number of moles of KMnO₄ in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 $\times$ 10⁻³ M

(a) 5.04 $\times$ 10⁻³ M;
(b) 0.499 M;
(c) 9.92 M;
(d) 1.1 $\times$ 10⁻³ M

0.025 M

0.5000 L

1.9 mL

(a) 0.125 M;
(b) 0.04888 M;
(c) 0.206 M;
(e) 0.0056 M

11.9 M

1.6 L

(a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:
$\%{\text{mass}}_1\times {\text{mass}}_1=\%{\text{mass}}_2\times {\text{mass}}_2$
This equation can be rearranged to isolate mass₁ and the given quantities substituted into this equation.
(b) 58.8 g

114 g

1.75 $\times$ 10⁻³ M

95 mg/dL

2.38 $\times$ 10⁻⁴ mol

0.29 mol