Question

Choose the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are $69.9$ and $30.1$ respectively and its molecular mass is $160$

• A. $FeO$
• B. $Fe_{3}O_{4}$
• C. $Fe_{2}O_{3}$
• D. $Fe O_{2}$

Answer 1

Answer: (C)

For element $Fe$, mole of atoms $=\frac {69.9}{56} = 1.25$
For element $O$, mole of atoms $={30.1}{16} = 1.88$
Mole ratio of $Fe = \frac{1.25}{1.25} = 1$,
Mole ratio of $O = \frac{1.88}{1.25} = 1.5$
Simplest whole number ratio of $Fe$ and $O = 2,3 $
Empirical formula of compound $= Fe_{2}O_{3}$
Molecular mass of $Fe_{2}O_{3} = 160$
$ n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{160}{160} = 1$
Molecular formula $ =Fe_{2}O_{3}$