What is the percentage by mass of nitrogen in ammonium trioxonitrate(V), NH₄NO₃? [H = 1, N = 14, O = 16]

Molar mass of NH₄NO₃ = 14 + 4(1) + 14 + 3(16) = 14 + 4 + 14 + 48 = 80 g/mol. There are 2 nitrogen atoms, so total mass of N = 2 × 14 = 28 g. Percentage = (28/80) × 100 = 35.0%. Option D (28%) uses the mass of nitrogen but divides by 100 instead of 80; option B considers only one nitrogen atom (14/80 × 100 ≈ 17.5%, which is option A); option B is another common miscalculation.