Question

The composition of a sample of wustite is $\text{F}{\text{e}}_{0.93}\text{O}$.What percentage of the iron is present in the Fe(III)?

Answer 1

We know that in pure iron oxide(FeO),iron and oxygen are present in the ratio 1: 1

In wustite (Fe0.93O1.00)(Fe0.93O1.00),some of the Fe2+Fe2+ ions are missing and the number of Fe2+Fe2+ ions present in 0.93 instead of 1.

From here we find the number of missing Fe2+Fe2+ ions

⇒1.0−0.93=0.07⇒1.0−0.93=0.07

Since each Fe2+Fe2+ ion carries two units of positive charge so the total positiv charge missing

⇒0.07×2=0.14⇒0.07×2=0.14

For maintenance of electrical neutrality this much (ie 0.14) positive charge has to be compensated by the presence of Fe3+Fe3+ ions.

If one Fe3+Fe3+ ion replaces one Fe2+Fe2+ ion then there is an increase of one unit positive charge.So to compensate 0.14 unit positive charge we require 0.14 Fe3+Fe3+ ions to replace Fe2+Fe2+ ion.

So,100Fe2+100Fe2+ ions have Fe3+Fe3+ ions=0.140.930.140.93×100=15.05%×100=15.05%