Question

A mixture of $CuS{O}_4.5H_{2}O$ and $MgS{O}_4.7H_{2}O$ was heated until all the water was driven off. If $5.0$ g of mixture gave $3$ g of anhydrous salt, what was the percentage by mass of $CuS{O}_4.5H_{2}O$ in the original mixture?

• A. $74.4$
• B. $70$
• C. $80$
• D. $90$

Answer 1

The correct option is A $74.4$

Let the mixture is X% by mass of $CuS{O}_4.5H_{2}O$ and 100 - X % by mass of $MgS{O}_4.7H_{2}O$. 5.0 g of mixture will contain 0.05X g $CuS{O}_4.5H_{2}O$ and 5.0 - 0.05X g $MgS{O}_4.7H_{2}O$

The molar masses of $CuS{O}_4.5H_{2}O$ and $MgS{O}_4.7H_{2}O$ are 249.7 g/mol and 246.5 g/mol respectively.

The number of moles of $CuS{O}_4.5H_{2}O=\frac{0.05X}{249.7}=2.00\times {10}^{-4}X$ moles.

The number of moles of $MgS{O}_4.7H_{2}O=\frac{5.0-0.05X}{246.5}$
Total number of moles of water obtained $=5\times 2.00\times {10}^{-4}X+7\times \frac{5.0-0.05X}{246.5}=1.00\times {10}^{-3}X+\frac{5.0-0.05X}{35.22}$

Mass of water obtained $=5.0-3.0=2.0$ g

Moles of water obtained $=\frac{2.0}{18}=0.11111$
Hence,
$1.00\times {10}^{-3}X+\frac{5.0-0.05X}{35.22}=0.11111$

$3.522\times {10}^{-2}X+5.0-0.05X=3.9133$

$0.01478X=1.0867$

$X=74.4$ %

Hence, the percentage by mass of $CuS{O}_4.5H_{2}O$ in the original mixture was 74.4 %