Question

A 0.649-g sample containing only ${\mathbf{K}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{(}\mathbf{FM}\mathbf{174}\mathbf{.}\mathbf{27}\mathbf{)}$and $\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\right(}\mathbf{FM}\mathbf{132}\mathbf{.}\mathbf{14}\mathbf{)}$was dissolved in water and treated with$\mathbf{Ba}\mathbf{(}{\mathbf{NO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$to precipitate all${\mathbf{SO}}_{\mathbf{4}}^{\mathbf{-}\mathbf{2}}{\mathbf{asBaSO}}_{\mathbf{4}}\mathbf{(}\mathbf{FM}\mathbf{233}\mathbf{.}\mathbf{39}\mathbf{)}$. Find the weight percent of${\mathbf{K}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$in the sample if 0.977 g of precipitate was formed.

Short answer

The solution for the mass percent of potassium sulfate from the given sample is 61.1 %.

Step-by-step solution

Calculate the mass percent of potassium sulfate from the given sample

Given

$0.977\mathrm{gof}{\mathrm{K}}_2{\mathrm{SO}}_4$

Consider x is mass of potassium sulfate and y is mass of$\left({\mathrm{NH}}_4\right){\mathrm{H}}_2{\mathrm{SO}}_4$

$x+y=0.649g$

$\frac{\mathrm{xg}}{174.27\mathrm{g}/\mathrm{mol}}+\frac{\mathrm{yg}}{132.14\mathrm{g}/\mathrm{mol}}=\frac{0.977\mathrm{g}}{233.39\mathrm{g}/\mathrm{mol}}$

$\frac{x}{174.27}$is moles of ${\mathrm{K}}_2{\mathrm{SO}}_4$

$\frac{y}{132.14}$is moles of$\left({\mathrm{NH}}_4\right){\mathrm{H}}_2{\mathrm{SO}}_4$

Substituting $\mathrm{y}=(0.649\mathrm{g}-\mathrm{x})$we get,

$\mathrm{x}=0.397\mathrm{g}$

Calculation of Mass percent.

The mass percent is calculated as follows,

$$\begin{gathered} \mathrm{mass}\%=\frac{0.397\mathrm{g}}{0.649\mathrm{g}}\times 100\% \\ =61.1\% \end{gathered}$$

The mass percent of potassium sulfate is 61.1%