Chapter 6, Problem 31A

(a)

Interpretation Introduction

Interpretation: The mass percentage of cation with the use of overall molar mass of ammonium chloride needs to be determined.

Concept Introduction: Mass percentage is the way to calculate the percentage of each element in the given molecular compounds. The mathematical expression of mass percentage can be shown as:

  $\text{Mass percentage = }\frac{\text{Number of element × mass of element}}{\text{molar mass of compound}}\times 100$

Answer to Problem 31A

Mass percentage of cation = 39.2 %

Explanation of Solution

Cation in ammonium chloride = ${\text{NH}}_{\text{4}}^{\text{+}}$

Molar mass of ${\text{NH}}_{\text{4}}^{\text{+}}$ = 17 + (4x 1) = 21

Molar mass of ${\text{NH}}_{\text{4}}^{}\text{Cl}$ = 53.5 g/mol

Substitute the values to calculate mass percentage of cation:

  $\begin{array}{l}\text{Mass percentage = }\frac{\text{Number of cation × mass of cation}}{\text{molar mass of compound}}\times 100\\ \text{Mass percentage = }\frac{\text{1 × 21 g/mol}}{\text{53}\text{.5 g/mol}}\times 100=39.2\%\end{array}$

(b)

Interpretation Introduction

Interpretation: The mass percentage of cation with the use of overall molar mass of copper (II) sulphate needs to be determined.

Concept Introduction: Mass percentage is the way to calculate the percentage of each element in the given molecular compounds. The mathematical expression of mass percentage can be shown as:

  $\text{Mass percentage = }\frac{\text{Number of element × mass of element}}{\text{molar mass of compound}}\times 100$

Answer to Problem 31A

Mass percentage of cation = 39.8 %

Explanation of Solution

Cation in copper sulphate = ${\text{Cu}}_{}^{\text{2+}}$

Molar mass of ${\text{Cu}}_{}^{\text{2+}}$ = 63.5 g/mol

Molar mass of ${\text{CuSO}}_{\text{4}}$ = 159.6 g/mol

Substitute the values to calculate mass percentage of cation:

  $\begin{array}{l}\text{Mass percentage = }\frac{\text{Number of cation × mass of cation}}{\text{molar mass of compound}}\times 100\\ \text{Mass percentage = }\frac{\text{1 × 63}\text{.5 g/mol}}{\text{159}\text{.6 g/mol}}\times 100=39.8\%\end{array}$

(c)

Interpretation Introduction

Interpretation: The mass percentage of cation with the use of overall molar mass of gold (III) chloride needs to be determined.

Concept Introduction: Mass percentage is the way to calculate the percentage of each element in the given molecular compounds. The mathematical expression of mass percentage can be shown as:

  $\text{Mass percentage = }\frac{\text{Number of element × mass of element}}{\text{molar mass of compound}}\times 100$

Answer to Problem 31A

Mass percentage of cation = 64.9 %

Explanation of Solution

Cation in gold (III) chloride= ${\text{Au}}_{}^{\text{3+}}$

Molar mass of ${\text{Au}}_{}^{\text{3+}}$ = 196.9 g/mol

Molar mass of ${\text{AuCl}}_{\text{3}}$ = 303.3 g/mol

Substitute the values to calculate mass percentage of cation:

  $\begin{array}{l}\text{Mass percentage = }\frac{\text{Number of cation × mass of cation}}{\text{molar mass of compound}}\times 100\\ \text{Mass percentage = }\frac{\text{1 × 196}\text{.9 g/mol}}{\text{303}\text{.3 g/mol}}\times 100=64.9\%\end{array}$

(d)

Interpretation Introduction

Interpretation: The mass percentage of cation with the use of overall molar mass of silver nitrate needs to be determined.

Concept Introduction: Mass percentage is the way to calculate the percentage of each element in the given molecular compounds. The mathematical expression of mass percentage can be shown as:

  $\text{Mass percentage = }\frac{\text{Number of element × mass of element}}{\text{molar mass of compound}}\times 100$

Answer to Problem 31A

Mass percentage of cation = 63.6 %

Explanation of Solution

Cation in silver nitrate = ${\text{Ag}}_{}^{\text{+}}$

Molar mass of ${\text{Ag}}_{}^{\text{+}}$ = 108 g/mol

Molar mass of ${\text{AgNO}}_{\text{3}}$ = 169.9 g/mol

Substitute the values to calculate mass percentage of cation:

  $\begin{array}{l}\text{Mass percentage = }\frac{\text{Number of cation × mass of cation}}{\text{molar mass of compound}}\times 100\\ \text{Mass percentage = }\frac{\text{1 × 108 g/mol}}{\text{169}\text{.9 g/mol}}\times 100=63.6\%\end{array}$