You visited us 1 times! Enjoying our articles? Unlock Full Access!

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 $(x + 100)$ .

Open in App

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.
$x [\frac{+ 2 (33.3) + 3 (66.6)}{100}] - 2 = 0$
$⟹ x [66.6 + 199.8] = 200$
$⟹ x = \frac{200}{266.4} = 0.75$
$x + 100 = 100 + 0.75 = 100.75$

Suggest Corrections

Similar questions

F e^{n +}

F e^{(n + 1) +}

F e^{(n + 1) +}

F e^{n +}

F e_{x} O

[x \times 100]

F e^{n +}

3 d

F e^{n +}

F e^{n +}

3 d

F e^{n +}

m

n

K^{+}

{F e}^{m +}

[{F e}^{n +} {(C N)}_{6}^{4 -}] \to prussian blue (colloidal solution)

m^{2}

Join BYJU'S Learning Program