What fraction of Fe exists as Fe(III) in $Fe_{0.96}O$? (Consider $Fe_{0.96}$ to be made up of Fe(II) and Fe(II

Question

What fraction of Fe exists as Fe(III) in $Fe_{0.96}O$? (Consider $Fe_{0.96}$ to be made up of Fe(II) and Fe(III) only)

Accepted Answer

1. **Establish Charge Neutrality:** The compound $Fe_{0.96}O$ is neutral. The oxide ion ($O^{2-}$) has a charge of $-2$. Therefore, the total positive charge contributed by $0.96$ moles of Iron atoms must be $+2$.
2. **Set up Equations:**
   - Let the fraction of $Fe^{3+}$ be $x$ and the fraction of $Fe^{2+}$ be $(1-x)$.
   - Alternatively, consider 100 formula units of $Fe_{0.96}O$ contains 96 Fe atoms and 100 O atoms.
   - Total charge of Fe atoms = Total charge of O atoms = $200$.
   - Let number of $Fe^{3+}$ be $a$ and $Fe^{2+}$ be $b$.
   - Eq 1 (Atom balance): $a + b = 96$
   - Eq 2 (Charge balance): $3a + 2b = 200$
3. **Solve for 'a' ($Fe^{3+}$):**
   - From Eq 1, $b = 96 - a$.
   - Substitute into Eq 2: $3a + 2(96 - a) = 200$
   - $3a + 192 - 2a = 200$
   - $a = 8$
4. **Calculate Fraction:**
   - The number of $Fe(III)$ ions is 8.
   - The total number of Fe ions is 96.
   - Fraction = $\frac{8}{96} = \frac{1}{12}$.

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