Empirical formula from percentage composition

Steps:

• Assume 100 g sample: 63 g N and 36.4 g O.
• Calculate moles: N = 63 / 14 = 4.5 mol; O = 36.4 / 16 = 2.275 mol.
• Find ratio: divide by smallest (2.275): N = 4.5 / 2.275 ≈ 1.98 (≈2); O = 1.
• Empirical formula is N₂O (simplest whole-number ratio).

Why A is correct:

• N₂O gives N:O mass ratio of (2×14):(16) = 28:16 = 63.6%:36.4%, matching the composition by the law of definite proportions.

Why the others are wrong:

• B. NO: 1:1 ratio yields 46.3% N and 53.7% O, not matching.
• C. NO₂: 1:2 ratio yields 30.4% N and 69.6% O, not matching.
• D. N₂O₄: 1:2 ratio (simplified) yields 30.4% N and 69.6% O, not matching.

Final answer: A