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.
