Question

Determine the moleculer formula of an oxide of iron in which the mass of iron & oxygen are 69.9 & 30 % respectively​.

Answer 1

Given:

The iron oxide has 69.9% iron and 30.1% dioxygen by mass.

Solution:

Thus, 100 g of iron oxide contains 69.9 g iron and 30.1 g dioxygen.

The number of moles of iron present in 100 g of iron oxide are= $\sf\large\ \frac{69.9}{55.8}$

$\underline{\boxed{\bf{\red{ =1.25.}}}}$

The number of moles of dioxygen present in 100 g of iron oxide are=$\sf\large\ \frac{30.1}{32}$

$\underline{\boxed{\bf{\orange{ =0.94}}}}$

The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is = $\sf\large\ \frac{2×0.94}{1.25}$

$\underline{\boxed{\bf{\purple{ =1.5:1=3:2.}}}}$

Hence, the formula of the iron oxide is Fe2O3.