Chapter 3, Problem 3.78QP

(a)

Interpretation Introduction

Interpretation: The percent composition of the given compounds is to be stated.

Concept introduction: Molar mass of a compound is sum of the atomic mass of each elements present in the compound.

To determine: The percent composition of sodium sulfate $\left({\text{Na}}_{\text{2}}{\text{SO}}_4\right)$.

Answer to Problem 3.78QP

Solution

Percent composition of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is $\underset{\_}{32.36\%Na}$, $\underset{\_}{22.57\%S}$ and $\underset{\_}{45.04\%O}$.

Explanation of Solution

Explanation

The molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is calculated as,

$\text{Molar mass}=\left(\begin{array}{c}\left(\text{Number of Na atoms}\times \text{Molar mass of Na}\right)+\\ \left(\text{Number of S atoms}\times \text{Molar mass of S}\right)\\ \left(\text{Number of O atoms}\times \text{Molar mass of O}\right)\end{array}\right)$

In ${\text{Na}}_{\text{2}}{\text{SO}}_4$, number of Na atoms $=2$, number of $\text{S}$ atoms $=1$ and number of O atoms $=4$.

Molar mass of $\text{Na}=22.\text{99 g/mol}$, molar mass of $\text{S}=32.07\text{ g/mol}$, molar mass of $\text{O}=16\text{ g/mol}$ and molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is $142.1\text{ g/mol}$

Substitute the number of sodium, sulfur and oxygen atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=\text{2}\times \text{22}\text{.99}+\text{1}\times 32.07+4\times 16\\ \text{=142}\text{.1 g/mol}\end{array}$

The percent composition of sodium (Na) is calculated by the formula,

$\text{Percent composition of Na}=\frac{\text{Mass of Na}}{{\text{Mass of Na}}_{\text{2}}{\text{SO}}_4}\times 100$

Mass of $2$ atoms of Na $=2\times 22.9\text{9 g}=45.9\text{8 g}$.

Substitute the value of mass of Na and mass of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ in the above equation.

$\begin{array}{c}\text{Percent composition of Na}=\frac{\text{45}\text{.98 g}}{\text{142}\text{.1 g}}\times 100\\ =\underset{\_}{32.36\%Na}\end{array}$

Therefore, the percent composition of Na is $\underset{\_}{32.36\%Na}$.

The percent composition of sulfur ( $\text{S}$) is calculated by the formula,

$\text{Percent composition of S}=\frac{\text{Mass of S}}{{\text{Mass of Na}}_{\text{2}}{\text{SO}}_4}\times 100$

Substitute the value of mass of $\text{S}$ and mass of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ in the above equation.

$\begin{array}{c}\text{Percent composition of S}=\frac{\text{32}\text{.07 g}}{\text{142}\text{.1 g}}\times 100\\ =\underset{\_}{22.57\%S}\end{array}$

Therefore, the percent composition of $\text{S}$ is $\underset{\_}{22.57\%S}$.

The percent composition of oxygen (O) is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of Na}}_{\text{2}}{\text{SO}}_4}\times 100$

Mass of $4$ atoms of O $=4\times 16\text{ g}=64\text{ g}$.

Substitute the value of mass of O and mass of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{64 g}}{\text{142}\text{.1 g}}\times 100\\ =\underset{\_}{45.04\%O}\end{array}$

Therefore, the percent composition of O is $\underset{\_}{45.04\%O}$.

Percent composition of ${\text{Na}}_{\text{2}}{\text{SO}}_4$ is $\underset{\_}{32.36\%Na}$, $\underset{\_}{22.57\%S}$ and $\underset{\_}{45.04\%O}$.

(b)

Interpretation Introduction

To determine: The percent composition of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$.

Answer to Problem 3.78QP

Solution

Percent composition of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\underset{\_}{30.45\%N}$ and $\underset{\_}{69.55\%O}$.

Explanation of Solution

Explanation

The molar mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is calculated as,

$\text{Molar mass}=\left(\begin{array}{c}\left(\text{Number of N atoms}\times \text{Molar mass of N}\right)+\\ \left(\text{Number of O atoms}\times \text{Molar mass of O}\right)\end{array}\right)$

In ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$, number of $\text{N}$ atoms $=2$, number of $\text{O}$ atoms $=4$.

Molar mass of $\text{N}=14.01\text{ g/mol}$ and molar mass of O $=16\text{ g/mol}$.

Substitute the number of nitrogen and oxygen atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=2\times 14.01+4\times 16.00\\ =92.02\text{ g/mol}\end{array}$

The percent composition of nitrogen ( $\text{N}$) is calculated by the formula,

$\text{Percent composition of N}=\frac{\text{Mass of N}}{{\text{Mass of N}}_{\text{2}}{\text{O}}_{\text{4}}}\times 100$

Mass of $\text{2}\text{N}$ atoms $=2\times 14.01=28.02\text{g}$.

Substitute the value of mass of $\text{N}$ and mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of N}=\frac{\text{28}\text{.02 g}}{\text{92}\text{.02 g}}\times 100\\ =\underset{\_}{30.45\%N}\end{array}$

Therefore, the percent composition of $\text{N}$ is $\underset{\_}{30.45\%N}$.

The percent composition of oxygen (O) is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{{\text{Mass of N}}_{\text{2}}{\text{O}}_{\text{4}}}\times 100$

Mass of $4$ atoms of O $=4\times 16\text{ g}=64\text{ g}$.

Substitute the value of mass of O and mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{64 g}}{\text{92}\text{.02 g}}\times 100\\ =\underset{\_}{69.55\%O}\end{array}$

Therefore, the percent composition of O is $\underset{\_}{69.55\%O}$.

Percent composition of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\underset{\_}{30.45\%N}$ and $\underset{\_}{69.55\%O}$.

(c)

Interpretation Introduction

To determine: The percent composition of strontium nitrate ( $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$).

Answer to Problem 3.78QP

Solution

Percent composition of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\underset{\_}{13.24\%N}$, $\underset{\_}{45.36\%O}$ and $\underset{\_}{41.40\%Sr}$.

Explanation of Solution

Explanation

The molar mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as,

$\text{Molar mass}=\left(\begin{array}{c}\left(\text{Number of Sr atoms}\times \text{Molar mass of Sr}\right)+\\ \left(\text{Number of N atoms}\times \text{Molar mass of N}\right)+\\ \left(\text{Number of O atoms}\times \text{Molar mass of O}\right)\end{array}\right)$

In $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$, number of $\text{Sr}$ atoms $=1$, number of $\text{N}$ atoms $=2$, number of $\text{O}$ atoms $=6$.

Molar mass of $\text{Sr}=87.62\text{g/mol}$, molar mass of $\text{N}=14.01\text{ g/mol}$ and molar mass of O $=16\text{ g/mol}$.

Substitute the number of strontium, nitrogen and oxygen atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=1\times 87.62+2\times 14.01+6\times 16.00\\ =211.64\text{ g/mol}\end{array}$

The percent composition of nitrogen ( $\text{N}$) is calculated by the formula,

$\text{Percent composition of N}=\frac{\text{Mass of N}}{\text{Mass of Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times 100$

Mass of $\text{2}\text{N}$ atoms $=2\times 14.01=28.02\text{g}$.

Substitute the value of mass of $\text{N}$ and mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of N}=\frac{\text{28}\text{.02 g}}{\text{211}\text{.64 g}}\times 100\\ =\underset{\_}{13.24\%N}\end{array}$

Therefore, the percent composition of $\text{N}$ is $\underset{\_}{13.24\%N}$.

The percent composition of oxygen (O) is calculated by the formula,

$\text{Percent composition of O}=\frac{\text{Mass of O}}{\text{Mass of Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times 100$

Mass of $6$ atoms of O $=6\times 16\text{ g}=96\text{ g}$.

Substitute the value of mass of O and mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of O}=\frac{\text{96 g}}{\text{211}\text{.64 g}}\times 100\\ =\underset{\_}{45.36\%O}\end{array}$

Therefore, the percent composition of O is $\underset{\_}{45.36\%O}$.

The percent composition of strontium ( $\text{Sr}$) is calculated by the formula,

$\text{Percent composition of Sr}=\frac{\text{Mass of Sr}}{\text{Mass of Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times 100$

Substitute the value of mass of $\text{Sr}$ and mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of Sr}=\frac{\text{87}\text{.62 g}}{\text{211}\text{.64 g}}\times 100\\ =\underset{\_}{41.40\%Sr}\end{array}$

Therefore, the percent composition of $\text{Sr}$ is $\underset{\_}{41.40\%Sr}$.

Percent composition of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\underset{\_}{13.24\%N}$, $\underset{\_}{45.36\%O}$ and $\underset{\_}{41.40\%Sr}$.

(d)

Interpretation Introduction

To determine: The percent composition of aluminum sulfide ( ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$).

Answer to Problem 3.78QP

Solution

Percent composition of ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ is $\underset{\_}{35.93\%Al}$ and $\underset{\_}{64.07\%S}$.

Explanation of Solution

Explanation

The molar mass of ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ is calculated as,

$\text{Molar mass}=\left(\begin{array}{c}\left(\text{Number of Al atoms}\times \text{Molar mass of Al}\right)+\\ \left(\text{Number of S atoms}\times \text{Molar mass of S}\right)\end{array}\right)$

In ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$, number of $\text{Al}$ atoms $=2$, and number of $\text{S}$ atoms $=3$.

Molar mass of $\text{Al}=26.98\text{ g/mol}$, and molar mass of $\text{S}=32.07\text{ g/mol}$.

Substitute the number of $\text{Al}$ and $\text{S}$ atoms and their molar masses in the above equation.

$\begin{array}{c}\text{Molar mass}=2\times \text{26}\text{.98}+3\times 32.07\\ =150.17\text{ g/mol}\end{array}$

The percent composition of aluminum ( $\text{Al}$) is calculated by the formula,

$\text{Percent composition of Al}=\frac{\text{Mass of Al}}{{\text{Mass of Al}}_{\text{2}}{\text{S}}_{\text{3}}}\times 100$

Mass of $\text{2 Al}$ atoms $=2\times 26.98\text{g}=56.96\text{ g}$.

Substitute the value of mass of $\text{Al}$ and mass of ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of Al}=\frac{\text{56}\text{.96 g}}{\text{150}\text{.17 g}}\times 100\\ =\underset{\_}{35.93\%Al}\end{array}$

Therefore, the percent composition of $\text{Al}$ is $\underset{\_}{35.93\%Al}$.

The percent composition of sulfur ( $\text{S}$) is calculated by the formula,

$\text{Percent composition of S}=\frac{\text{Mass of S}}{{\text{Mass of Al}}_{\text{2}}{\text{S}}_{\text{3}}}\times 100$

Mass of $3$ atom of $\text{S}$ $=3\times 32.07\text{ g}=96.21\text{ g}$.

Substitute the value of mass of $\text{S}$ and mass of ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ in the above equation.

$\begin{array}{c}\text{Percent composition of S}=\frac{\text{96}\text{.21 g}}{\text{150}\text{.17 g}}\times 100\\ =\underset{\_}{64.07\%S}\end{array}$

Therefore, the percent composition of $\text{S}$ is $\underset{\_}{64.07\%S}$.

Percent composition of ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ is $\underset{\_}{35.93\%Al}$ and $\underset{\_}{64.07\%S}$.

Conclusion

Percent composition of ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ is $\underset{\_}{35.93\%Al}$ and $\underset{\_}{64.07\%S}$