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

First method
Oxidation number `(ON)` of `Fe_(0.93)O`
`= 0.93 xx X - 2 xx 1 = 0 implies X = 2.15`
Oxidation number `(ON)` of `Fe` is an intermediate value of `Fe^(2+)` and `Fe^(3+)`
Let `%` of `F^(3+) = a`
`(3 xx a + 2(100 - a))/(100) = 2.15 implies a = 15.05%`
`:.` Percent of `Fe^(3+) = 15.05, %` of `Fe^(2+) = 84.95`
Second method
Let `X` atom of `Fe^(3+)` ions are present in the compounds.
`:. Fe_(0.93)` (number of `Fe^(3+)` ions `+` Number of `Fe^(2+)` ions)
Number of `Fe^(2+)` inos = Total - Number of `Fe^(3+)`
`= (0.93 -X)`
Total positive charge (i.e., on `Fe^(2+) + Fe^(3+)`) `= Total`
(negative0 charge on oxygen
`2(0.93 - X) + 3X = 2 implies X = 0.14`
`:.` Precent of `Fe^(3+) = 0.14//0.93 xx 100 = 15.05`