Question

The composition of a sample of wustite is fe0.93o what percentage of fe in present in form of fe(3)

Answer 1

In pure iron oxide (FeO), iron and oxygen are present in the ratio 1 : 1.

However, here number of Fe2+ present = 0.93

Or No. of Fe2+ ions missing = 0.07

Since each Fe2+ ion has 2 positive charge, the total number of charge due to missing (0.07) Fe2+ ions = 0.07 * 2 = 0.14

To maintain electrical neutrality, 0.14 positive charge is compensated by the presence of Fe3+ ions. Now since, replacement of one Fe2+ ion by one Fe3+ ion increases one positive charge, 0.14 positive charge must be compensated by the presence of 0.14 Fe3+ ions.

In short, 0.93 Fe2+ ions have 0.14 Fe3+ ions

100 Fe2+ ions have = 0.14/0.93 * 100 = 15.05%

Answer 2

Answer:

The percentage of Fe²⁺ present in the $Fe_{0.93}O$ sample is found to be 15%.

Explanation:

Let assume that the charge on Fe²⁺ is x

Then the charge on Fe³⁺ $=0.93-x$

Now we know the overall compound is neutral

∴   $(+2)x+(0.93-x)(+3)+(-2)(1)=0$

      $2x+2.79-3x-2=0$

             $-x-0.79=0\\ x=0.79$

The composition of $Fe^{3+}=0.93-0.79=0.19$

The % of  $Fe^{3+}=(\frac{0.14}{0.93})*100$

Therefore percentage of Fe³⁺ in the given sample of $Fe_{0.93}O$ = 15%