Question

A mixture of $CuS{O}_4$ and $CuS{O}_4.5H_{2}O$ has a mass of $1.245$ g. After heating, it drives off all the water, the mass is only $0.832$ g. What is the mass (in mg) of $CuS{O}_4.5H_{2}O$ in the mixture?

Answer 1

Let the mass of $CuS{O}_4=xg$

The mass of $CuS{O}_4.5H_{2}O=(1.245-x)$ g

$\underset{249.5}{\underset{}{CuS{O}_4.5H_{2}O}}⟶\underset{159.5g}{\underset{}{CuS{O}_4}}+5H_{2}O\left(g\right)\uparrow$

$CuS{O}_{4}in(1.245-x)g$ of $CuS{O}_4.5H_{2}O=\frac{159.5}{249.5}(1.245-x)g$

Thus, $\frac{159.5}{249.5}(1.245-x)+x=0.832$

$x=0.1g$

The mass of $CuS{O}_4.5H_{2}O=1.145g$.

Hence, the mass (in mg) $CuS{O}_4.5H_{2}O$ in the mixture in $1145$ mg.