Question

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. Given that the molar mass of the oxide is 159.69 g mol¹.

Answer 1

Calculation of the chemical formula for iron oxide-

Elements	Atomic mass	Percentage Composition	Moles of atoms	Mole ratio or atomic ratio	Simples wholes number ratio
Fe	56	69.9	$\frac{69.9}{56}=$1.25	$\frac{1.25}{1.25}=1$	2
O	16	30.1	$\frac{30.1}{16}=$1.88	$\frac{1.88}{1.25}=$1.50	3

Empirical formula- Fe₂O₃
Empirical formula mass = 2(55.84) + 3(16) = 159.68
Molar mass given = 159.69g/mol
n = $\frac{159.69}{159.68}=1$

Hence chemical formula of iron oxide = Fe₂O₃

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