Percentage Composition & Formulae — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Percentage Composition & Formulae MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Percentage Composition & Formulae · hard · numerical
A compound contains 26.7% carbon, 2.2% hydrogen and 71.1% oxygen, and its molar mass is 90 g mol⁻¹. Its molecular formula is (C = 12, H = 1, O = 16):
A. C₃H₃O₆
B. CHO₂
C. C₂H₂O₄ ✓ Correct
D. C₄H₄O₈
Solution: In 100 g: moles C = 26.7 ÷ 12 = 2.22, H = 2.2 ÷ 1 = 2.2, O = 71.1 ÷ 16 = 4.44. Divide by smallest (2.2): C = 1, H = 1, O = 2 → empirical formula CHO₂ (mass = 12 + 1 + 32 = 45). n = 90 ÷ 45 = 2, so molecular formula = C₂H₂O₄ (oxalic acid). Choosing CHO₂ forgets to multiply by n.
Q2 — Percentage Composition & Formulae · hard · numerical
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has a molar mass of 56 g mol⁻¹. Its molecular formula is (C = 12, H = 1):
A. C₄H₈ ✓ Correct
B. CH₂
C. C₃H₆
D. C₂H₄
Solution: Moles in 100 g: C = 85.7 ÷ 12 = 7.14, H = 14.3 ÷ 1 = 14.3. Divide by 7.14: C = 1, H = 2 → empirical formula CH₂ (mass = 14). n = 56 ÷ 14 = 4, so molecular formula = C₄H₈. Using n = 3 gives C₃H₆ (molar mass 42) and n = 2 gives C₂H₄ (molar mass 28) — neither matches 56.
Q3 — Percentage Composition & Formulae · medium · numerical
An oxide of iron contains 70% iron and 30% oxygen by mass. Its empirical formula is (Fe = 56, O = 16):
A. FeO₂
B. Fe₂O₃ ✓ Correct
C. Fe₃O₄
D. FeO
Solution: Moles: Fe = 70 ÷ 56 = 1.25, O = 30 ÷ 16 = 1.875. Divide by smallest (1.25): Fe = 1, O = 1.5. To clear the fraction, multiply both by 2 → Fe = 2, O = 3, giving Fe₂O₃. Stopping at the 1 : 1.5 ratio and rounding to 1 : 1 wrongly gives FeO.
Q4 — Percentage Composition & Formulae · hard · numerical
What is the mass percentage of water of crystallisation in copper sulphate pentahydrate, CuSO₄·5H₂O? (Cu = 63.5, S = 32, O = 16, H = 1)
A. 56.4%
B. 25.5%
C. 63.9%
D. 36.1% ✓ Correct
Solution: Molar mass = 63.5 + 32 + 4(16) + 5(18) = 159.5 + 90 = 249.5. Mass of the 5 water molecules = 90. Mass % water = (90 ÷ 249.5) × 100 = 36.1%. Dividing the water mass by the anhydrous salt mass (90 ÷ 159.5) wrongly gives 56.4%; 25.5% is the mass % of copper (63.5 ÷ 249.5).
Q5 — Percentage Composition & Formulae · medium · numerical
An oxide of nitrogen is found to contain 63.6% nitrogen and 36.4% oxygen by mass. Its empirical formula is (N = 14, O = 16):
A. N₂O₃
B. NO
C. N₂O ✓ Correct
D. NO₂
Solution: Moles: N = 63.6 ÷ 14 = 4.54, O = 36.4 ÷ 16 = 2.28. Divide by smallest (2.28): N = 2, O = 1 → N₂O (nitrous oxide). Treating the ratio as 1 : 1 gives NO, and inverting the division gives NO₂; both come from mishandling the mole ratio.
Q6 — Percentage Composition & Formulae · hard · numerical
A compound has the empirical formula CH₂O and a vapour density of 30. Its molecular formula is (C = 12, H = 1, O = 16):
A. C₂H₄O₂ ✓ Correct
B. CH₂O
C. C₃H₆O₃
D. C₄H₈O₄
Solution: Molar mass = 2 × vapour density = 2 × 30 = 60 g mol⁻¹. Empirical formula mass of CH₂O = 30, so n = 60 ÷ 30 = 2. Molecular formula = C₂H₄O₂ (acetic acid). The trap CH₂O comes from using the vapour density (30) itself as the molar mass, which wrongly gives n = 1.
Q7 — Percentage Composition & Formulae · medium · numerical
A metal M of atomic mass 24 forms an oxide that contains 60% of M by mass. The empirical formula of the oxide is (O = 16):
A. M₂O
B. MO ✓ Correct
C. M₂O₃
D. MO₂
Solution: In 100 g of oxide: 60 g M and 40 g O. Moles: M = 60 ÷ 24 = 2.5, O = 40 ÷ 16 = 2.5. The ratio M : O = 1 : 1, so the empirical formula is MO (here M = 24 is magnesium, giving MgO). The equal mole values fix the 1 : 1 ratio, ruling out M₂O and MO₂.