1. Assume 100 g of compound: 50.0 g O, 37.5 g C, 12.5 g H.
Moles: O = 50.0 / 16.00 = 3.125 mol; C = 37.5 / 12.01 ≈ 3.122 mol; H = 12.5 / 1.008 ≈ 12.40 mol.
Divide by smallest (≈3.122): O ≈ 1.00, C ≈ 1.00, H ≈ 3.97 ≈ 4.
Empirical formula: CH₄O.
2. Assume 100 g of compound: 40.0 g C, 6.7 g H, 53.5 g O.
Moles: C = 40.0 / 12.01 ≈ 3.33 mol; H = 6.7 / 1.008 ≈ 6.65 mol; O = 53.5 / 16.00 ≈ 3.34 mol.
Divide by smallest (≈3.33): C ≈ 1.00, H ≈ 2.00, O ≈ 1.00.
Empirical formula: CH₂O.
3. Moles N = 4.04 / 14.01 ≈ 0.288 mol; moles O = 11.46 / 16.00 ≈ 0.716 mol.
Ratio O/N = 0.716 / 0.288 ≈ 2.49 ≈ 2.5 → multiply by 2: N₂O₅.
Empirical formula mass = 2×14.01 + 5×16.00 = 108.02 g/mol.
Molar mass given is 108.0 g/mol → molecular formula = empirical formula = N₂O₅.
4. Moles N = 0.606 / 14.01 ≈ 0.04326 mol; moles O = 1.390 / 16.00 ≈ 0.08688 mol.
Ratio O/N = 0.08688 / 0.04326 ≈ 2.008 ≈ 2 → empirical formula: NO₂.
Empirical formula mass = 14.01 + 2×16.00 = 46.01 g/mol.
Molar mass = 92 g/mol → multiplier = 92 / 46.01 ≈ 2.
Molecular formula: N₂O₄.
Parent Tip: Review the logic above to help your child master the concept of molecular formulas worksheet.
