Iron oxide has formula $F e_{0.96} O_{1.00}$ . What fraction of Fe exists as $F e^{+ 2} and F e^{+ 3}$ ions?

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Solution

Let's assume that there are $x$ molecule fraction of $F e^{+ 2}$ & $0.93 - x$ molecular fraction of $F e^{+ 3}$ .
Now, using charge conservation formula for $F e_{0.96} O_{1.00}$ , we get
$\Rightarrow x (+ 2) + (0.99 - x) (+ 3) - 2 = 0$
$\Rightarrow 2 x - 3 x + 2.79 - 2 = 0$
$\Rightarrow x = 0.79$
Fraction of $F e^{+ 2} = 0.79$ & frcation of $F e^{+ 3} = 0.14$
Percentage of $F e^{+ 2} = \frac{0.79}{0.93} \times 100 = 84.9$ %
Percentage of $F e^{+ 3} = \frac{0.14}{0.93} \times 100 = 15.05$ %

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