A non stoichiometric compound $F e_{7} S_{8}$ consist of iron in both $F e^{+ 2}$ and $F e^{+ 3}$ form and sulphur is present as sulphide ions. Calculate cation vacancies as a percentage of $F e^{+ 2}$ initially present in the sample. (Express your answer as $\frac{100}{x}$ and you should write $x$ as answer)

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Solution

The given non-stoichiometric compound is $F e_{7} S_{8}$ . The compound contains $F e^{2 +}$ and $F e^{3 +}$ . So the given non-stoichiometric compound $F e_{7} S_{8}$ has 5, 2 and 8 number of $F e^{2 +}$ , $F e^{3 +}$ and $S^{2 -}$ respectively
Therefore cationvacancies $= \frac{(8 - 5) X 100}{8} \Rightarrow x = 2.67$ .

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