Question

An impure sample of cuprite $\left(C{u}_{2}O\right)$ contains $66.6\%$ Copper by mass. Find the ratio of $\%$ of pure $C{u}_{2}O$ to $\%$ of impurity by mass in the sample. Take $=\frac{66.6\times 143}{254}=37.5$.

Answer 1

Consider the volume $V=100g$ hence $Cu=66.6g$

The number of moles $=\frac{66.6}{63.6}=1.05$ mole

$1$ mole of oxygen is per $2$ moles $Cu$.

Therefore moles of oxygen $=\frac{1.05}{2}=0.53$ moles

The weight of oxygen is then $0.53$ ml $\times 16$ g/mol $=8.42g$

The weight of ${Cu}_{2}O=$ weight of $Cu$ $+$ weight of $O$

$=66.6g+8.42g$

$=75.02g$ in $100g$ of sample

Therefore percentage of ${Cu}_{2}O$ in $100g$ of sample $=\frac{75.02}{100}\times 100$%$=75$%