Question

A certain sample of cuprous sulphide is found to have the composition $C{u}_{1.8}S$ due to incorporation of $C{u}^{2+}$ ions in the lattice. What is the mole % of $C{u}^{2+}$ in total copper content in this crystal? (Cu is present as $C{u}^{+}$ and $C{u}^{2+}$).

Answer 1

Two $C{u}^{+}$ ions can be replaced by one $C{u}^{2+}$.
The compound should have been $C{u}_{2}S$ as ideally all the copper ions should have been $C{u}^{+}$.
$C{u}_{1.8}S$ - for every mole of $S^{2-}$, x moles are $C{u}^{2+}$ and the rest i.e. (1.8-x) moles of $C{u}^{+}$ are present.
Using the idea of charge conservation:
Net charge on cations =Net charge on anions
⇒Charge on ( $C{u}^{+}+C{u}^{2+}$ )= Charge on $S^{2-}\Rightarrow x\times 2+(1.8-x)\times 1=2\Rightarrow x+1.8=2\Rightarrow x=0.2$
0.2 moles of $C{u}^{2+}$ have replaced 0.4 moles of $C{u}^{+}$ in 1 mole of compound.
∴ % of $C{u}^{2+}$ in the total copper content
$=\frac{0.2}{1.8}\times 100=11.11$ %