Chapter 6

(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

(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