JEE Main 2025 Jan 24 Shift 2, Chemistry Q11: Mole Concept And Equivalent Concept

JEE Main2025Jan 24, Shift 2Chemistry

The elemental composition of a compound is 54.2%, C, 9.2% H and 36.6% O. If the molar mass of the compound is 132 g mol $^{- 1}$ , the molecular formula of the compound is :

[Given : The relative atomic mass of C : H : O = 12 : 1 : 16]

A
$C_{4} H_{9} O_{3}$
B
$C_{6} H_{12} O_{6}$
C
$C_{6} H_{12} O_{3}$
D
$C_{4} H_{8} O_{2}$

Solution

Convert mass percent to moles (per 100 g of compound) by dividing each percentage by the atomic mass:

$C : H : O = \frac{54.2}{12} : \frac{9.2}{1} : \frac{36.6}{16} = 4.516 : 9.2 : 2.287$

Dividing through by the smallest value (2.287):

$C : H : O \approx 1.97 : 4.02 : 1 \approx 2 : 4 : 1$

Empirical formula $= C_{2} H_{4} O$ with empirical mass $= 24 + 4 + 16 = 44 g/mol$ .

$n = \frac{molar mass}{empirical mass} = \frac{132}{44} = 3$

$∴$ Molecular formula $= (C_{2} H_{4} O)_{3} = C_{6} H_{12} O_{3}$ .

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Chemistry Q11: Mole Concept And Equivalent Concept

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