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Empirical Formula

In an ore of ...

Question

In an ore of iron, iron is present in two oxidation state. $F e^{n +}$ and $F e^{(n + 1) +}$
Number of $F e^{(n + 1) +}$ is twice the number of $F e^{n +}$ .
If empirical formula of ore is $F e_{x} O$ . Calculate value of $[x \times 100]$ .

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Solution

$^{'} F e^{'}$ is present as $+ 2$ and $+ 3$ ions. So, $n = 2$
$F e^{+ 2}$ and $F e^{+ 3}$
$F e^{+ 3}$ percentage $= F e^{+ 2}$ percentage $\times 2$
Total $= 100 %$
$⟹ F e^{+ 3} % + F e^{+ 2} % = 100$
$⟹ 2 F e^{+ 2} % + F e^{+ 2} % = 100$
$⟹ F e^{+ 2} % = 33.3$
$∴ F e^{+ 2} % = 33.3$
$∴ F e^{+ 3} % = 66.6$
$F e_{x} O \Rightarrow$ Balancing of charge.
$x [\frac{+ 2 (33.3) + 3 (66.6)}{100}] - 2 = 0$
$⟹ x [66.6 + 199.8] = 200$
$⟹ x = \frac{200}{266.4} = 0.75$
$x \times 100 = 0.75 \times 100 = 75$

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