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.(molecular mass is 159.8).

Explanation

Mass percent of iron (Fe) = 69.9 % (Given)
Mass percent of oxygen (O) = 30.1% (Given )
number of moles of iron present in the oxide ` (69.90)/(55.85)`
=1.25
Number ofmoles of oxygen present in the oxide `=(30.1)/(16.0)`
=1.88
Ratio of iron to oxygen in the oxide ,
=1.25 :1.88
`=(1.25)/(1.25):(1.88)/(1.25)`
=1:1.5
=2:3
` therefore` The empirical formula of the oxide is `Fe_(2)O_(3)`.
Empirical formula mass of `Fe_(20O_(3) =[2(55.85)+3(16.00)]g`
Molar mass of `Fe_(2)O_(3) = 159.69 g`
`therefore n=("molar mass")/("Emprical Formula Mass ") =(159.96 g)/(159.7 g)`
=.0999
=1 ( approx)
Molecular formula of a compound i9s obtined by mutiplying the empircal From with n thus , the empirical formula of the given oxide is `Fe_(2)O_(3)` and n is 1.
Hence , the molecular formula of the oxide is `Fe_(2)O_(3)`