Chapter 7, Problem 43E

Questions 43 and 44: Calculate the percentage composition by mass of each compound.

a) Ammonium nitrate

b) Ammonium sulfate

c) Ammonium carbonate

d) Calcium oxide

e) Manganese $\left(\text{IV}\right)$ sulfide

Interpretation Introduction

(a)

Interpretation:

The percentage composition by mass of each component in ammonium nitrate is to be calculated.

Concept introduction:

The mass percentage of a particular element of a mixture is the ratio of the molar mass of that element and the molar mass of the mixture multiplied by $100$. The formula to calculate the mass percentage of an element A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

Answer to Problem 43E

The mass percentage of nitrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $35.00\%$.

The mass percentage of hydrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $5.037\%$.

The mass percentage of oxygen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $59.96\%$.

Explanation of Solution

The formula to calculate the mass percentage of any element, A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

The molecular formula for ammonium nitrate is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.

The molar mass of nitrogen is $14.01\text{g}/\text{mol}$.

The molar mass of hydrogen is $1.008\text{g}/\text{mol}$.

The molar mass of oxygen is $16.00\text{g}/\text{mol}$.

The molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{2}\times \text{1}4.01+\left(4\times 1.008\right)+\left(3\times 16.00\right)\\ & =& 80.05\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $80.05\text{g}/\text{mol}$.

The mass percentage of nitrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{N}\text{by}\text{mass}=\frac{\text{2}\times \text{Molar}\text{mass}\text{of}\text{N}}{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of nitrogen and ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{N}\text{by}\text{mass}& =& \frac{2\times 14.01\text{g}/\text{mol}}{80.05\text{g}/\text{mol}}\times 100\%\\ & =& 35.00\%\end{array}$

Hence, the mass percentage of nitrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $35.00\%$.

The mass percentage of hydrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{H}\text{by}\text{mass}=\frac{4\times \text{Molar}\text{mass}\text{of}\text{H}}{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of hydrogen and ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{H}\text{by}\text{mass}& =& \frac{4\times 1.008\text{g}/\text{mol}}{80.05\text{g}/\text{mol}}\times 100\%\\ & =& 5.037\%\end{array}$

Hence, the mass percentage of hydrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $5.037\%$.

The mass percentage of oxygen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{O}\text{by}\text{mass}=\frac{3\times \text{Molar}\text{mass}\text{of}\text{O}}{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of oxygen and ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{O}\text{by}\text{mass}& =& \frac{3\times 16.00\text{g}/\text{mol}}{80.05\text{g}/\text{mol}}\times 100\%\\ & =& 59.96\%\end{array}$

Hence, the mass percentage of oxygen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $59.96\%$.

Conclusion

The mass percentage of nitrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as $35.00\%$, the mass percentage of hydrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as $5.037\%$ and the mass percentage of oxygen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as $59.96\%$.

Interpretation Introduction

(b)

Interpretation:

The percentage composition by mass of each component in aluminum sulfate is to be calculated.

Concept introduction:

The mass percentage of a particular element of a mixture is the ratio of the molar mass of that element and the molar mass of the mixture multiplied by $100$. The formula to calculate the mass percentage of an element A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

Answer to Problem 43E

The mass percentage of aluminum in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $15.77\%$.

The mass percentage of sulfur in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $28.11\%$.

The mass percentage of oxygen in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $56.12\%$.

Explanation of Solution

The formula to calculate the mass percentage of any element, A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

The molecular formula for aluminum sulfate is ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$.

The molar mass of aluminum is $26.98\text{g}/\text{mol}$.

The molar mass of sulfur is $32.06\text{g}/\text{mol}$.

The molar mass of oxygen, is $16\text{g}/\text{mol}$.

The molar mass of ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{2}\times 26.98+\left(3\times 32.06\right)+\left(12\times 16\right)\\ & =& 342.14\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $342.14\text{g}/\text{mol}$.

The mass percentage of aluminum in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated as shown below.

$\text{Percentage}\text{of}\text{Al}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{Al}}{\text{Molar}\text{mass}\text{of}{\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3}\times 100\%$

Substitute the molar mass of aluminum and ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{Al}\text{by}\text{mass}& =& \frac{2\times 26.98\text{g}/\text{mol}}{342.14\text{g}/\text{mol}}\times 100\%\\ & =& 15.77\%\end{array}$

Hence, the mass percentage of aluminum in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $15.77\%$.

The mass percentage of sulfur in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated as shown below.

$\text{Percentage}\text{of}\text{S}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{S}}{\text{Molar}\text{mass}\text{of}{\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3}\times 100\%$

Substitute the molar mass of sulfur and ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{S}\text{by}\text{mass}& =& \frac{3\times 32.06\text{g}/\text{mol}}{342.14\text{g}/\text{mol}}\times 100\%\\ & =& 28.11\%\end{array}$

Hence, the mass percentage of sulfur in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $28.11\%$.

The mass percentage of oxygen in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated as shown below.

$\text{Percentage}\text{of}\text{O}\text{by}\text{mass}=\frac{\text{12}\times \text{Molar}\text{mass}\text{of}\text{O}}{\text{Molar}\text{mass}\text{of}{\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3}\times 100\%$

Substitute the molar mass of oxygen and ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{O}\text{by}\text{mass}& =& \frac{12\times 16\text{g}/\text{mol}}{342.14\text{g}/\text{mol}}\times 100\%\\ & =& 56.12\%\end{array}$

Hence, the mass percentage of oxygen in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is $9.75\%$.

Conclusion

The mass percentage of aluminum in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated as $15.77\%$, the mass percentage of sulfur in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated as $28.11\%$ and the mass percentage of oxygen in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is calculated as $56.12\%$.

Interpretation Introduction

(c)

Interpretation:

The percentage composition by mass of each component in ammonium carbonate is to be calculated.

Concept introduction:

The mass percentage of a particular element of a mixture is the ratio of the molar mass of that element and the molar mass of the mixture multiplied by $100$. The formula to calculate the mass percentage of an element A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

Answer to Problem 43E

The mass percentage of nitrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $29.16\%$.

The mass percentage of hydrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $8.392\%$.

The mass percentage of carbon in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $12.50\%$.

The mass percentage of oxygen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $49.95\%$.

Explanation of Solution

The formula to calculate the mass percentage of any element, A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

The molecular formula for ammonium carbonate is ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$.

The molar mass of nitrogen is $14.01\text{g}/\text{mol}$.

The molar mass of hydrogen is $1.008\text{g}/\text{mol}$.

The molar mass of carbon is $12.01\text{g}/\text{mol}$.

The molar mass of oxygen is $16\text{g}/\text{mol}$.

The molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{2}\times \text{14}\text{.01}+\left(8\times 1.008\right)+\left(12.01\right)+\left(3\times 16\right)\\ & =& 96.09\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $96.09\text{g}/\text{mol}$.

The mass percentage of nitrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{N}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{N}}{\text{Molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of nitrogen and ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{N}\text{by}\text{mass}& =& \frac{2\times 14.01\text{g}/\text{mol}}{96.09\text{g}/\text{mol}}\times 100\%\\ & =& 29.16\%\end{array}$

Hence, the mass percentage of nitrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $29.16\%$.

The mass percentage of hydrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{H}\text{by}\text{mass}=\frac{\text{8}\times \text{Molar}\text{mass}\text{of}\text{H}}{\text{Molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of hydrogen and ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{H}\text{by}\text{mass}& =& \frac{8\times 1.008\text{g}/\text{mol}}{96.09\text{g}/\text{mol}}\times 100\%\\ & =& 8.392\%\end{array}$

Hence, the mass percentage of hydrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $8.392\%$.

The mass percentage of oxygen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{O}\text{by}\text{mass}=\frac{\text{3}\times \text{Molar}\text{mass}\text{of}\text{O}}{\text{Molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of oxygen and ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{O}\text{by}\text{mass}& =& \frac{3\times 16\text{g}/\text{mol}}{96.09\text{g}/\text{mol}}\times 100\%\\ & =& 49.95\%\end{array}$

Hence, the mass percentage of oxygen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $49.95\%$.

The mass percentage of carbon in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{C}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{C}}{\text{Molar}\text{mass}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}}\times 100\%$

Substitute the molar mass of carbon and ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{C}\text{by}\text{mass}& =& \frac{12.01\text{g}/\text{mol}}{96.09\text{g}/\text{mol}}\times 100\%\\ & =& 12.50\%\end{array}$

Hence, the mass percentage of carbon in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is $12.50\%$.

Conclusion

The mass percentage of nitrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as $29.16\%$.

The mass percentage of hydrogen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as $8.392\%$.

The mass percentage of carbon in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as $12.50\%$.

The mass percentage of oxygen in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as $49.95\%$.

Interpretation Introduction

(d)

Interpretation:

The percentage composition by mass of each component in calcium oxide is to be calculated.

Concept introduction:

The mass percentage of a particular element of a mixture is the ratio of the molar mass of that element and the molar mass of the mixture multiplied by $100$. The formula to calculate the mass percentage of an element A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

Answer to Problem 43E

The mass percentage of calcium in $\text{CaO}$ is $71.47\%$.

The mass percentage of oxygen in $\text{CaO}$ is $28.53\%$.

Explanation of Solution

The formula to calculate the mass percentage of any element, A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

The molecular formula for calcium oxide is $\text{CaO}$.

The molar mass of calcium is $40.08\text{g}/\text{mol}$.

The molar mass of oxygen, is $16\text{g}/\text{mol}$.

The molar mass of $\text{CaO}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{40}\text{.08}+16\\ & =& 56.08\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of $\text{CaO}$ is $56.08\text{g}/\text{mol}$.

The mass percentage of calcium in $\text{CaO}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{Ca}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{Ca}}{\text{Molar}\text{mass}\text{of}\text{CaO}}\times 100\%$

Substitute the molar mass of calcium and $\text{CaO}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{Ca}\text{by}\text{mass}& =& \frac{40.08\text{g}/\text{mol}}{56.08\text{g}/\text{mol}}\times 100\%\\ & =& 71.47\%\end{array}$

Hence, the mass percentage of calcium in $\text{CaO}$ is $71.47\%$.

The mass percentage of oxygen in $\text{CaO}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{O}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{O}}{\text{Molar}\text{mass}\text{of}\text{CaO}}\times 100\%$

Substitute the molar mass of oxygen and $\text{CaO}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{O}\text{by}\text{mass}& =& \frac{16\text{g}/\text{mol}}{56.08\text{g}/\text{mol}}\times 100\%\\ & =& 28.53\%\end{array}$

Hence, the mass percentage of oxygen in $\text{CaO}$ is $28.53\%$.

Conclusion

The mass percentage of calcium in $\text{CaO}$ is calculated as $71.47\%$ and the mass percentage of oxygen in $\text{CaO}$ is calculated as $28.53\%$.

Interpretation Introduction

(e)

Interpretation:

The percentage composition by mass of each component in $\text{manganese}\left(\text{IV}\right)\text{sulfide}$ is to be calculated.

Concept introduction:

The mass percentage of a particular element of a mixture is the ratio of the molar mass of that element and the molar mass of the mixture multiplied by $100$. The formula to calculate the mass percentage of an element A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

Answer to Problem 43E

The mass percentage of manganese in ${\text{MnS}}_{\text{2}}$ is $46.14\%$.

The mass percentage of sulfur in ${\text{MnS}}_{\text{2}}$ is $53.86\%$.

Explanation of Solution

The formula to calculate the mass percentage of any element, A is shown below.

$\text{Percentage}\text{of}\text{A}\text{by}\text{mass}=\frac{\text{Parts}\text{of}\text{A}}{\text{Total}\text{parts}}\times 100\%$

The molecular formula for $\text{manganese}\left(\text{IV}\right)\text{sulfide}$ is ${\text{MnS}}_{\text{2}}$.

The molar mass of manganese is $54.94\text{g}/\text{mol}$.

The molar mass of sulfur is $32.06\text{g}/\text{mol}$.

The molar mass of ${\text{MnS}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{54}\text{.94}+\left(2\times 32.06\right)\\ & =& 119.06\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{MnS}}_{\text{2}}$ is $119.06\text{g}/\text{mol}$.

The mass percentage of manganese in ${\text{MnS}}_{\text{2}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{Mn}\text{by}\text{mass}=\frac{\text{Molar}\text{mass}\text{of}\text{Mn}}{\text{Molar}\text{mass}\text{of}{\text{MnS}}_{\text{2}}}\times 100\%$

Substitute the molar mass of manganese and ${\text{MnS}}_{\text{2}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{Mn}\text{by}\text{mass}& =& \frac{54.94\text{g}/\text{mol}}{119.06\text{g}/\text{mol}}\times 100\%\\ & =& 46.14\%\end{array}$

Hence, the mass percentage of manganese in ${\text{MnS}}_{\text{2}}$ is $46.14\%$.

The mass percentage of sulfur in ${\text{MnS}}_{\text{2}}$ is calculated as shown below.

$\text{Percentage}\text{of}\text{S}\text{by}\text{mass}=2\times \frac{\text{Molar}\text{mass}\text{of}\text{S}}{\text{Molar}\text{mass}\text{of}{\text{MnS}}_{\text{2}}}\times 100\%$

Substitute the molar mass of sulfur and ${\text{MnS}}_{\text{2}}$ in the above expression.

$\begin{array}{rll}\text{Percentage}\text{of}\text{S}\text{by}\text{mass}& =& 2\times \frac{32.06\text{g}/\text{mol}}{119.06\text{g}/\text{mol}}\times 100\%\\ & =& 53.86\%\end{array}$

Hence, the mass percentage of sulfur in ${\text{MnS}}_{\text{2}}$ is $53.86\%$.

Conclusion

The mass percentage of manganese in ${\text{MnS}}_{\text{2}}$ is calculated as $46.14\%$ and the mass percentage of sulfur in ${\text{MnS}}_{\text{2}}$ is calculated as $53.86\%$.