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Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 8 April, Evening Shift — Question 12

On combustion of 0.210 g of an organic compound containing C,H\mathrm{C}, \mathrm{H} and O gave 0.127 gH2O0.127 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} and 0.307 gCO2\mathrm{g} \mathrm{CO}_{2}. The percentage of hydrogen and oxygen in the given organic compound respectively are:

  1. Option A:

    53.41,39.653.41,39.6

  2. Option B:

    6.72,39.876.72,39.87

  3. Option C:

    6.72,53.416.72,53.41

    Correct
  4. Option D:

    7.55,43.857.55,43.85

Answer: C

Step-by-step solution

Mass of organic compound =0.210 g=0.210 \mathrm{~g}

Mass of water formed =0.127 g=0.127 \mathrm{~g}

Mass of CO2\mathrm{CO}_{2} formed =0.307 g=0.307 \mathrm{~g}

Mass of hydrogen =0.127×218=0.014 g=\frac{0.127 \times 2}{18}=0.014 \mathrm{~g}

Percentage of hydrogen =0.014×1000.210=6.72%=\frac{0.014 \times 100}{0.210}=6.72 \%

Mass of carbon =0.307×1244=0.084 g=\frac{0.307 \times 12}{44}=0.084 \mathrm{~g}

Percentage of carbon ==0.084×1000.210=39.87%==\frac{0.084 \times 100}{0.210}=39.87 \%

Percentage of oxygen =53.41%=53.41 \%

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic
Percentage composition & Empirical Formula

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