An oxide of iron contains 69.9% iron, its empirical formula, is:

Question

(Given: Molar mass of Fe and O are 56 and 16 $g$ ${mol}^{- 1}$ respectively.) [2026]

Smart Achievers · Accepted Answer

(2)

Element	Mass percentage (%w/w)	Number of gram atoms per 100 g compound = (%w/w)/gram atomic mass	Relative ratio	Simple ratio
Fe	69.9	$\frac{69.9}{56} = 1.25$	$\frac{1.25}{1.25} = 1$	2
O	100 − 69.9 = 30.1	1.88	$\frac{1.88}{1.25} = 1.5$	3

Empirical formula = ${Fe}_{2} O_{3}$

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