Question

Question 1.8 Determine 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. Given that the molar mass of the oxide is 159.69 g mol–1.

Class XI Some Basic Concepts of Chemistry Page 22

Answer 1

we know , that
molecular formula = ( empirical formula ) × n
where n is integral number.


mass % of Fe = 69.9
mass % of Fe/atomic mass of Fe ( A) = 69.1/56
= 1.248

mass % of O = 30.1
mass % of O/atomic mass of O ( B) = 30.1/16
=1.88

now, dividing both ratio B to ratio A

Fe _____________ O
1.248/1.248 ________ 1.88/1.248
1________________1.5

e.g Fe : O = 1 : 1.5
Fe : O = 2 : 3

$hence,Fe_2O_3$

you can check it .
molecular mass of Fe2O3 = 2 × 56 + 3 × 16 = 160 g/mol ≈ 159.7 g/mol

hence, it's correct that molecular formula is Fe2O3