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.

Answer 1

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 of moles 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

∴ The empirical formula of the oxide is Fe₂O_3.

Empirical formula mass of Fe₂O₃ = [2(55.85) + 3(16.00)] g

Molar mass of Fe₂O₃ = 159.69 g

`therefore "n" = "Molar Mass"/"Emprical formula mass"`

=(159.69  "g")/(159.7  "g")`

= 0.999

= 1 (approx)

The molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is Fe₂O₃ and n is 1.

Hence, the molecular formula of the oxide is Fe₂O_3.