Determine the molecular formula of an oxide of iron in which the mass percent of iron oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is `159.8 g mol^(-1)` (Atomic mass : Fe = 55.85, O = 16.00 amu)

Explanation

Expirical formula mass of `Fe_(2)O_(3)=2xx55.85+3xx16.00=159.7gmol^(-1)`
`n=("Molar mass")/("Empirical formula mass")=(159.8)/(159.7)=1`
Hence, molecular formula is same as empirical formula, viz., `Fe_(2)O_(3)`.