Percentage Composition & Formulae — NEET Chemistry MCQs with Solutions
Free NEET 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 · easy · numerical
The mass percentage of oxygen in water (H₂O), taking H = 1 and O = 16, is:
A. 11.11%
B. 88.89% ✓ Correct
C. 50%
D. 80%
Solution: Mass % of an element = (mass of that element in the compound ÷ molar mass) × 100. Molar mass of H₂O = 2(1) + 16 = 18. Mass % O = (16 ÷ 18) × 100 = 88.89%. The value 11.11% is the mass % of hydrogen (2 ÷ 18 × 100), not oxygen.
Q2 — Percentage Composition & Formulae · easy · numerical
What is the mass percentage of carbon in carbon dioxide (CO₂)? (C = 12, O = 16)
A. 42.86%
B. 72.73%
C. 12%
D. 27.27% ✓ Correct
Solution: Molar mass of CO₂ = 12 + 2(16) = 44. Mass % C = (12 ÷ 44) × 100 = 27.27%. The value 72.73% is the mass % of oxygen (32 ÷ 44 × 100); 42.86% is the carbon percentage in CO (12 ÷ 28), a wrong compound.
Q3 — Percentage Composition & Formulae · easy · numerical
The mass percentage of nitrogen in ammonia (NH₃) is (N = 14, H = 1):
A. 82.35% ✓ Correct
B. 14%
C. 17.65%
D. 46.67%
Solution: Molar mass of NH₃ = 14 + 3(1) = 17. Mass % N = (14 ÷ 17) × 100 = 82.35%. The value 17.65% is the mass % of hydrogen (3 ÷ 17 × 100); 46.67% is the nitrogen percentage in urea, a different compound.
Q4 — Percentage Composition & Formulae · medium · numerical
A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its empirical formula is (C = 12, H = 1, O = 16):
A. C₂H₄O
B. CHO
C. CH₂O ✓ Correct
D. CH₃O
Solution: Take 100 g of compound: 40 g C, 6.7 g H, 53.3 g O. Moles → C = 40 ÷ 12 = 3.33, H = 6.7 ÷ 1 = 6.7, O = 53.3 ÷ 16 = 3.33. Divide by the smallest (3.33): C = 1, H = 2, O = 1. So the simplest whole-number ratio gives the empirical formula CH₂O.
Q5 — Percentage Composition & Formulae · easy · numerical
The mass percentage of oxygen in calcium carbonate (CaCO₃) is (Ca = 40, C = 12, O = 16):
A. 12%
B. 48% ✓ Correct
C. 60%
D. 40%
Solution: Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100. Mass % O = (48 ÷ 100) × 100 = 48%. The value 40% is the mass % of calcium and 12% is the mass % of carbon; only the three oxygen atoms (mass 48) give 48%.
Q6 — Percentage Composition & Formulae · medium · numerical
A compound has the empirical formula CH₂O and a molar mass of 180 g mol⁻¹. Its molecular formula is (C = 12, H = 1, O = 16):
A. C₅H₁₀O₅
B. C₂H₄O₂
C. C₆H₁₂O₆ ✓ Correct
D. C₃H₆O₃
Solution: Empirical formula mass of CH₂O = 12 + 2(1) + 16 = 30. n = molar mass ÷ empirical formula mass = 180 ÷ 30 = 6. Molecular formula = n × empirical formula = 6 × CH₂O = C₆H₁₂O₆ (glucose). Using n = 3 or n = 2 would wrongly give C₃H₆O₃ or C₂H₄O₂.
Q7 — Percentage Composition & Formulae · medium · numerical
How many grams of carbon are present in 44 g of carbon dioxide (CO₂)? (C = 12, O = 16)
A. 12 g ✓ Correct
B. 27.3 g
C. 32 g
D. 24 g
Solution: Molar mass of CO₂ = 44, so 44 g of CO₂ is exactly 1 mole and contains 1 mole of carbon = 12 g. Equivalently, mass of C = 44 × (12 ÷ 44) = 12 g. The value 32 g is the mass of oxygen in the sample; 27.3 g is the percentage of carbon, not its mass.