A fuel has 30.43% nitrogen and 69.57% oxygen. Find the molecular formula of the compound if it has a mass of 92 grams per mole

A. NO

B. N2O4

C. NO2

D. N4O8

B. N₂O₄.

Explanation:

We will suppose the fuel has the formula NxOy. The molar mass of the fuel = (x. atomic mass of N) + (y. atomic mass of O) = 92.0 g/mole.

The % composition of N = 30.43 %

The % composition of N = (x. atomic mass of N) / (molar mass of fuel) x 100. then x = (% composition of N) (molar mass of fuel) / 100 (atomic mass of N) = (30.43 %) (92.0 g/mole) / 100 (14.00 g/mole) = 1.9996 ≅ 2.0.

By the same way;

The % composition of O = 69.57 %

The % composition of O = (y. atomic mass of O) / (molar mass of fuel) x 100 then y = (% composition of O) (molar mass of fuel) / 100 (atomic mass of O) = (69.57 %) (92.0 g/mole) / 100 (16.00 g/mole) = 4.0.

So, the formula of the compound is N₂O₄.