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)`?

Explanation

`Fe_(0.93)O_(1.0)` is a non - stoichiometric compound. It is a mixture of `Fe^(2+)` ions
Let, x atoms of `Fe^(3+)` ions are present in the compound
So, (0.93 - x) atoms of `Fe^(2+)` ions are present in the compound In the compound,
Total positive charge = Total negative charge
`x xx(+3)+(0.93-x)xx(+2)=+2`
or `3x +0.93xx2-2x = +2`
or `x + 2xx0.93 = 2`
`x = 0.14`
`therefore` % of `Fe^(3+)` ions present `= (0.14)/(0.93)xx100 = 15.053`.