Question

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

Answer 1

Answer:

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.