The composition of a sample of wustite is `Fe_(0.93)O_(1.00)`. What percentage of iron is present in the form of `Fe(III)`?

Answer

`11.5%`

Explanation

`Fe_(0.93) O_(1.0)` is non - stoichiometric and is a mixture of `Fe^(2+)` and `Fe^(3+)` ions. Let x atoms of `Fe^(3+)` ions are present in the compound. This means that `x Fe^(3+)` ions have been replaced by `Fe^(2+)` ions.
No. of `Fe^(2+)` ions = 0.93 - x
For electrical neutrality, positive charge on the compound = Negative charge on the compound
`2(0.93 - x) + 3x = 2`
`1.86 + x = 2`
or `x = 0.14`
`:.` Fraction of `Fe^(3+) = 0.14, Fe^(2+) = 0.93 - 0.14 = 0.79`
Thus, the given formula may be represented as
`Fe_(0.79)^(2+) Fe_(0.14)^(3+) O_(1.0)^(2-)`
Total molar mass = `0.93 xx 56 + 1 xx 16 = 68.08 g `
% of irons present as Fe (III) `= (0.14 xx 56)/(68.08) xx 100 = 11.5%.