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

Explanation

The composition is `Fe_0.93 O_1.00` instead and FeO because some `Fe^(2+)` ions have been replaced by `Fe^(+)` ions. Let us first calculate the number of `Fe^(2+)` and `Fe^(3+)` ions present. The formula `Fe_0.93 O_1.00` implies that 93 Fe atoms are combined with 100 O-atoms. Out of 93 Fe atoms , suppoe Fe atoms present as `Fe^(3+)=x`. Then `Fe^(2+)=93-x`.As the compound is neutral, total charge on `Fe^(2+)` and `Fe^(3+)` ions =total charge on `O^(2-)` ions. Thus,
`3 xx x +2 (93-x)=2 xx 100` or 3x+ 186-2x =200 or x=14 , i.e., `Fe^(3+)=14`
Hence, `Fe^(2+)` =93-14=79
Thus, out of 93 Fe atoms , Fe present as `Fe^(3+)`=14 `therefore` % age of Fe present of Fe (III) =`14/93xx100=15%`