❓ Question:
The complete combustion of 1.80 g of an oxygen-containing compound (CxHyOz) gave 2.64 g of CO₂ and 1.08 g of H₂O. What is the percentage of oxygen in the original organic compound?
📄 Solution Image:
🔍 Concept Behind the Question:
This is a classic combustion analysis problem where you calculate the amount of C, H, and O in a compound based on the masses of CO₂ and H₂O produced.
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All the carbon in the compound ends up in CO₂
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All the hydrogen ends up in H₂O
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The oxygen is found by difference (mass balance)
Such problems are crucial in elemental analysis, often asked in JEE to test your stoichiometry and logical deduction skills.
🧠 Step-by-Step Explanation:
Given:
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Mass of compound = 1.80 g
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CO₂ formed = 2.64 g
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H₂O formed = 1.08 g
Step 1: Find moles of Carbon
Molar mass of CO₂ = 44 g/mol
Moles of CO₂ = 2.64 / 44 = 0.06 mol
⇒ Moles of C = 0.06 mol
⇒ Mass of C = 0.06 × 12 = 0.72 g
Step 2: Find moles of Hydrogen
Molar mass of H₂O = 18 g/mol
Moles of H₂O = 1.08 / 18 = 0.06 mol
Each H₂O has 2 H atoms ⇒ Moles of H = 0.06 × 2 = 0.12 mol
⇒ Mass of H = 0.12 × 1 = 0.12 g
Step 3: Find mass of Oxygen
Total mass = 1.80 g
Mass of O = 1.80 – (0.72 + 0.12) = 0.96 g
Step 4: Find percentage of oxygen
% O = (0.96 / 1.80) × 100 = 53.33%
✅ Final Answer:
53.33% oxygen is present in the compound.
📽️ Video Solution:
💡 Why This Question is Important:
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Applies stoichiometry + basic chemistry logic
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Tests your grip on combustion reactions and mass relationships
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Helps you prepare for both empirical formula and percentage composition questions in JEE
This type of question has appeared repeatedly in JEE Mains and Advanced papers, especially in the organic and general chemistry sections.
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