Choose the molecular formula of an oxide of iron in which, the mass percent of iron and oxygen are 69.9 and 30.1 respectively and its molecular mass is 160.

For element Fe, moles of atoms $=\frac{69.9}{56}=1.25$
For element O, moles of atoms $=\frac{30.1}{16}=1.88$
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 = Fe₂O₃
Molecular mass of Fe₂O₃ = 160
$\mathrm{n}=\frac{ \mathrm{Molecular} \mathrm{mass} }{ \mathrm{Empirical} \mathrm{formula} \mathrm{mass} }=\frac{160}{160}=1$
Molecular formula = Fe₂O₃