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

First method: In pure iron oxide, Fe and O are in the ratio of `1:1`. The compound `Fe_(0.93)O_(1.0)` as a whole should remain neutral. Suppose x atoms of `Fe^(3+)` ions are present in the compound of `Xfe^(3+)` atoms have replaced n atoms of `Fe^(2+)`.
Number of atoms of `Fe^(2+)=0.93-x`
For neutrality of compound.
Total positive charge `=` Negative charge on compound,
`implies` charge on iron atom`=`charge on O atom or charge on
`(Fe^(2+)+Fe^(2+))=2`
`2(0.93-x)+3x=2`
`1.86+x=2`
`x=0.14`
`%` of iron present as `Fe^(3+)=(0.14)/(0.93)xx100=15.05%`
Second method:
Oxidation method of Fe in `Fe_(0.093)O=0.93xx x-2xx1=0`,
`x=2.15`.
Oxidation number of Fe is an intermediate value of `Fe^(2+)` and `Fe^(3+)`
Let `%` of `Fe^(3+)=a`
`(3xxa+2(100-a))/(100)=2.15`
`thereforea=15.05%`
`%` of `Fe^(3+)=15.05`
Third method: Let x atom of `Fe^(3+)` ions be present in the compound.
`F_(0.93)` (number of `Fe^(3+)` ions of `Fe^(2+)` ions.)
Number of `Fe^(2+)` ions `=` Total `-` number of `Fe^(3+)=(0.93-x)`
Total positive charge on `Fe^(2+)+Fe^(3+)=` Total negative charge on oxygen.
`2(0.93-x)+3x=2,becausex=0.14`
`therefore% of Fe^(3+)=(0.14)/(0.93)xx100=15.05`
Third Method:
Let x atoms of `Fe^(3+)` ions be present in the compound
`F_(0.93)` (number of `Fe^(3+)` ions of `Fe^(2+)` ions) Number of `Fe^(2+)` ions=total-number of `Fe^(3+)=(0.93-x)` total positive charge on `Fe^(2+)+Fe^(3+)=` total negative charge on oxygen
`2(0.93-x)+3x=2,therefore0.14`
`therefore%` of `Fe^(3+)=0.14//0.93xx100=15.05`