Iron oxide has formula $F e_{0.94} O .$ The fraction of $F e$ exists as $F e^{+ 3}$ is

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Solution

Iron oxide has formula $F e_{0.94} O .$
Let there be $100$ oxide ions and $94$ iron ions.
$F e^{+ x} : O^{- 2} = 94 : 100$
Total negative charge on $100$ oxide ions $= - 200$
It means as per the concept of charge neutrality, $94$ iron ions should have total charge of $+ 200$ .
Let number of $F e^{+ 3} = X$
$∴$ Number of $F e^{+ 2} = (94 - X)$
Total positive charge $= 3 X + 2 (94 - X) = 200 \Rightarrow X = 12$
$∴$ Percentage of $F e^{+ 3} = \frac{12}{94} \times 100 = 12.8 %$

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