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Question

A certain sample of cuprous sulphide is found to have the composition Cu1.8S due to incorporation of Cu2+ ions in the lattice. What is the mole % of Cu2+ in total copper content in this crystal? (Cu is present as Cu+ and Cu2+).

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Solution

Two Cu+ ions can be replaced by one Cu2+.
The compound should have been Cu2S as ideally all the copper ions should have been Cu+.
Cu1.8S - for every mole of S2, x moles are Cu2+ and the rest i.e. (1.8-x) moles of Cu+ are present.
Using the idea of charge conservation:
Net charge on cations =Net charge on anions
⇒Charge on ( Cu++Cu2+ )= Charge on S2x×2+(1.8x)×1=2x+1.8=2x=0.2
0.2 moles of Cu2+ have replaced 0.4 moles of Cu+ in 1 mole of compound.
∴ % of Cu2+ in the total copper content
=0.21.8×100=11.11 %

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