Chapter F, Problem F.27E

Interpretation Introduction

Interpretation:

Percentage by mass of ${\text{NaNO}}_{\text{3}}$ in the mixture has to be calculated.

Answer to Problem F.27E

Mass percentage of ${\text{NaNO}}_{\text{3}}$ is $44.88\%$.

Explanation of Solution

The mixture that is given is said to contain ${\text{NaNO}}_{\text{3}}$ and ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$.  The mass of mixture is $\text{5}\text{.37}\text{g}$.  Molar mass of the individual components is calculated as follows;

Molar mass of ${\text{NaNO}}_{\text{3}}$:

    $\begin{array}{c}\text{Molar}\text{mass}=1\times (22.98\text{g}\cdot {\text{mol}}^{-1})+1\times (14.01\text{g}\cdot {\text{mol}}^{-1})+3\times (16.00\text{g}\cdot {\text{mol}}^{-1})\\ =22.98\text{g}\cdot {\text{mol}}^{-1}+14.01\text{g}\cdot {\text{mol}}^{-1}+48.00\text{g}\cdot {\text{mol}}^{-1}\\ =84.99\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$:

    $\begin{array}{c}\text{Molar}\text{mass}=2\times (22.98\text{g}\cdot {\text{mol}}^{-1})+1\times (32.06\text{g}\cdot {\text{mol}}^{-1})+4\times (16.00\text{g}\cdot {\text{mol}}^{-1})\\ =45.96\text{g}\cdot {\text{mol}}^{-1}+32.06\text{g}\cdot {\text{mol}}^{-1}+64.00\text{g}\cdot {\text{mol}}^{-1}\\ =142.02\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Mass percentage of sodium in individual component in the mixture is calculated as shown below;

Mass percentage of $\text{Na}$ in ${\text{NaNO}}_{\text{3}}$:

    $\begin{array}{c}\text{Mass}\text{percentage}=\frac{\text{Mass}\text{of}\text{Na}\text{atom}\text{in}\text{1}\text{mol}\text{of}{\text{NaNO}}_{\text{3}}}{\text{molar}\text{mass}\text{of}{\text{NaNO}}_{\text{3}}}\times 100\\ =\frac{22.98\text{g}\cdot {\text{mol}}^{-1}}{84.99\text{g}\cdot {\text{mol}}^{-1}}\times 100\\ =27.04\%\end{array}$

Mass percentage of $\text{Na}$ in ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$:

    $\begin{array}{c}\text{Mass}\text{percentage}=\frac{\text{Mass}\text{of}\text{Na}\text{atom}\text{in}\text{1}\text{mol}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{molar}\text{mass}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}\times 100\\ =\frac{2(22.98\text{g}\cdot {\text{mol}}^{-1})}{142.02\text{g}\cdot {\text{mol}}^{-1}}\times 100\\ =32.36\%\end{array}$

The total mass of the mixture is given as $\text{5}\text{.37}\text{g}$.  Let the mass of ${\text{NaNO}}_{\text{3}}$ be considered as $x\text{g}$.  Therefore, the mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be expressed as follows;

    $\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}=(5.37-x)\text{g}$

Mass of sodium in $x\text{g}$ of ${\text{NaNO}}_{\text{3}}$ is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}\text{Na}\text{in}x\text{g}\text{of}{\text{NaNO}}_{\text{3}}=\frac{\text{Mass}\text{percentage}\text{of}\text{Na}\text{in}{\text{NaNO}}_{\text{3}}}{\text{100}}\times x\text{g}\\ \text{=}\frac{27.04}{100}x\text{g}\\ \text{=0}\text{.2704}x\text{g}\end{array}$

Mass of sodium in $(5.37-x)\text{g}$ of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as shown below;

$\begin{array}{c}\text{Mass}\text{of}\text{Na}\text{in}(5.37-x)\text{g}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}=\frac{\text{Mass}\text{percentage}\text{of}\text{Na}\text{in}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{100}}\times (5.37-x)\text{g}\\ \text{=}\frac{32.36}{100}(5.37-x)\text{g}\\ \text{=1}\text{.738}\text{g}-0.3236x\text{g}\end{array}$

Total mass of sodium that is present in the mixture is given as $\text{1}\text{.61}\text{g}$.  Therefore, the equation obtained can be given as shown below;

    $\begin{array}{c}\text{0}\text{.2704}x\text{g + 1}\text{.738}\text{g}-0.3236x\text{g = 1}\text{.61}\text{g}\\ \text{0}\text{.0532}x\text{g}=0.128\\ x\text{g=}\frac{0.128}{0.0532}\\ =2.41\text{g}\end{array}$

Therefore, the mass of ${\text{NaNO}}_{\text{3}}$ in the mixture is found to be $\text{2}\text{.41}\text{g}$.  The mass percentage of ${\text{NaNO}}_{\text{3}}$ is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{percentage}\text{of}{\text{NaNO}}_{\text{3}}=\frac{\text{Mass}\text{of}{\text{NaNO}}_{\text{3}}}{\text{Total}\text{mass}\text{of}\text{mixture}}\times 100\\ =\frac{\text{2}\text{.41}\text{g}}{\text{5}\text{.37}\text{g}}\times 100\\ =44.88\%\end{array}$

Thus the mass percentage of ${\text{NaNO}}_{\text{3}}$ in the mixture is $44.88\%$.