Question

A $\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{g}$ sample consisting of a mixture of sodium chloride and potassium sulphate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead nitrate to form $\mathbf{21}\mathbf{.}\mathbf{75}\mathbf{g}$of solid. Determine the mass percent of sodium chloride in the original mixture.

Short answer

The initial sodium chloride and potassium sulfate combination had a mass percentage of$68\%$ sodium chloride.

Step-by-step solution

Introduction

Mixtures are always formed when two or more chemicals are combined. Each chemical in the combination has a mass that determines its mass fraction. The sum of all mass percentages of all substances in the combination equals $100\%$. One efficient way to expose the chemical makeup of a mixture is to use mass percentage.

The balanced chemical equation

The total mass of solid precipitate$=21.75\mathrm{g}$

The given mass mixture of and Potassium sulfate$=10.0\mathrm{g}$

The mass of $\mathrm{NaCl}$is $\mathrm{x}$.

Thus,

Mass of ${\mathrm{K}}_2{\mathrm{SO}}_4$ is $(10.0-\mathrm{x})\mathrm{g}$.

The reaction of $\mathrm{NaCl}$with$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ is:

$\mathbf{2}\mathbf{NaCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Pb}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{PbCl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{NaNO}}_{\mathbf{3}}$

The reaction of ${\mathrm{K}}_2{\mathrm{SO}}_4$with $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$ is:

${\mathbf{K}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{Pb}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{PbSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{KNO}}_{\mathbf{3}}$

Calculating the number of moles

The number of moles can be calculated as:

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

Where $\mathrm{n}$is the number of moles.

$\mathrm{m}$ is the given mass.

$\mathrm{M}$is the molar mass.

The molar mass of $\mathrm{NaCl}$is $58.44\mathrm{g}/\mathrm{mol}$.

The molar mass of ${\mathrm{K}}_2{\mathrm{SO}}_4$is $174.26\mathrm{g}/\mathrm{mol}$.

On calculating,

$\begin{array}{c}{\mathrm{n}}_{\mathrm{NaCl}}=\frac{\mathrm{x}\mathrm{g}}{58.44\mathrm{g}/\mathrm{mol}}\\ =\frac{\mathrm{x}}{58.44}\mathrm{mol}\\ {\mathrm{n}}_{{\mathrm{K}}_2{\mathrm{SO}}_4}=\frac{(10.0-\mathrm{x})\mathrm{g}}{174.26\mathrm{g}/\mathrm{mol}}\\ =\frac{(10.0-\mathrm{x})}{174.26}\mathrm{mol}\end{array}$

Calculating the mass of both solids

The mass of both solids can be calculated by:

$\mathrm{Mass}=\mathrm{Number}\mathrm{of}\mathrm{moles}\times \mathrm{Molar}\mathrm{mass}$

The molar mass of${\mathrm{PbCl}}_2$ is $278.1\mathrm{g}/\mathrm{mol}$.

The molar mass of${\mathrm{PbSO}}_4$ is $303.26\mathrm{g}/\mathrm{mol}$.

On substituting,

$\begin{array}{c}\mathrm{Mass}\mathrm{of}{\mathrm{PbCl}}_2=\frac{\mathrm{x}}{116.88}\mathrm{mol}\times 278.1\mathrm{g}/\mathrm{mol}\\ =2.379\mathrm{x}\\ \mathrm{Mass}\mathrm{of}{\mathrm{PbSO}}_4=\frac{(10.0-\mathrm{x})}{174.26}\mathrm{mol}\times 303.26\mathrm{g}/\mathrm{mol}\\ =17.4-1.74\mathrm{x}\end{array}$

The given total mass of solid is:

$\begin{array}{c}2.379\mathrm{x}+17.4-1.74\mathrm{x}=21.75\\ 0.639\mathrm{x}=4.35\\ \mathrm{x}=6.8\end{array}$

Calculating the mass percentage

The mass percentage is:

$\begin{array}{c}\mathrm{Mass}\mathrm{percentage}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{NaCl}}{\mathrm{Mass}\mathrm{of}\mathrm{mixture}}\times 100\mathrm{\%}\\ =\frac{6.8\mathrm{g}}{10.0\mathrm{g}}\times 100\mathrm{\%}\\ =68\mathrm{\%}\end{array}$

Interpretation

Thus, the mass percentage was computed based on the sodium chloride and potassium sulfate combination.