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

JEE Main 2025 — 24 January, Evening Shift — Question 32

The elemental composition of a compound is 54.2%54.2\% C, 9.2%9.2\% H and 36.6%36.6\% O. If the molar mass of the compound is 132 g mol−1132\,\text{g mol}^{-1}, the molecular formula of the compound is _____\_\_\_\_\_.

(Given: Relative atomic masses of C : H : O = 12 : 1 : 16)

  1. Option A:

    C4H9O3\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{O}_{3}

  2. Option B:

    C6H12O6\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}

  3. Option C:

    C6H12O3\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}

    Correct
  4. Option D:

    C4H8O2\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}

Answer: C

Step-by-step solution

Assume 100 g100\,\text{g} of the compound.

Moles   of   C=54.212=4.52\text{Moles \;of\; C} = \frac{54.2}{12} = 4.52 Moles   of   H=9.21=9.2\text{Moles \;of \;H} = \frac{9.2}{1} = 9.2 Moles   of   O=36.616=2.29\text{Moles \;of \;O} = \frac{36.6}{16} = 2.29

On dividing by the smallest value: C : H : O≈2:4:1\text{C : H : O} \approx 2 : 4 : 1 Empirical formula =C2H4O= \mathrm{C_2H_4O}

Empirical formula mass =44= 44

Multiplier=13244=3\text{Multiplier} = \frac{132}{44} = 3

Thus, molecular formula is given as

(C2H4O)3=C6H12O3(\mathrm{C_2H_4O})_3 = \mathrm{C_6H_{12}O_3}

Thus, the correct answer is option C.

Answer key and solution verified before publishing.

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

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