Ncert Exercises & Solutions

1.8 Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively.

Mass percent of iron (Fe) = 69.9%
Mass percent of oxygen (O) = 30.1%
Number of moles of iron present in the oxide = $\frac{69.90}{55.85}$ = 1.25

A number of moles of oxygen present in the oxide = 30.1/16.0
= 1.88
Ratio of iron to oxygen in the oxide,
= 1.25:1.88
= $\frac{1.25}{1.25} : \frac{1.88}{1.25}$
= 1 : 1.5
= 2 : 3

$∴$ The empirical formula of the oxide is ${Fe}_{2} O_{3}$ .

Empirical formula mass of Fe2O3 = 159.69 g

$∴ n = \frac{Molar mass}{Emprical formula mass} = \frac{159.69 g}{159.7 g}$
$= 0.999 = 1 (approx)$

Molecular formula of a compound is obtained by multiplying the empirical formula with n.
Thus, the empirical formula of the given oxide is ${Fe}_{2} O_{3}$ and n is 1.
Hence, the molecular formula of the oxide is ${Fe}_{2} O_{3}$ .

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