Question

A piece of iron ore is found to contain a compound containing 72.3 percent iron and 27.7 percent oxygen with a molecular mass of 231.4 g/mol. What is the molecular formula of the compound?

Answer 1

Answer:

$Fe_{3} O_{4}$

Explanation:

Molecular mass of the iron ore is 231.4 u

Ore contain 72.3% of iron. Thus 72.3 % of 231.4 would be = 167.30 g

And rest is oxygen, so 231.4 - 167.30 = 64.1 g

Thus, ore contain 167.30 g of iron and 64.1 g of oxygen.

To calculate the number of iron atom, we have to divide with mass of iron ( mass of iron = 56 g mol-1)

Thus, number of iron atom = 167.3056 = 2.98 ≃ 3

Similarly, number of oxygen atom =  64.116 = 4.006 ≃ 4.

Hence, the molecular formula is $Fe_{3} O_{4}$.