JEE Main 2025 Jan 28 Shift 1, Chemistry Q22: Mole Concept And Equivalent Concept

JEE Main2025Jan 28, Shift 1Chemistry

Quantitative analysis of an organic compound (X) shows following % composition.

C : 14.5% Cl : 64.46%
H : 1.8%

Empirical formula mass of the compound (X) is ______ $\times 1 0^{- 1}$

(Given molar mass in g mol $^{- 1}$ of C : 12, H : 1, O : 16, Cl : 35.5)

Your answer

Solution

Oxygen accounts for the remaining mass:

%O $= 100 - 14.5 - 64.46 - 1.8 = 19.24%$

Compute mole ratios by dividing each percentage by the atomic mass:

Element	% mass	mol ratio	÷ smallest	$\times 5/3$
C	14.5	14.5/12 = 1.208	1.0	2
H	1.8	1.8/1 = 1.80	1.49	3
Cl	64.46	64.46/35.5 = 1.816	1.50	3
O	19.24	19.24/16 = 1.203	1.0	2

Empirical formula: C $_{2}$ H $_{3}$ Cl $_{3}$ O $_{2}$ .

Empirical formula mass $= 2 (12) + 3 (1) + 3 (35.5) + 2 (16) = 24 + 3 + 106.5 + 32 = 165.5$ g mol $^{- 1}$ .

Expressed as $\times 1 0^{- 1}$ : $165.5 = 1655 \times 1 0^{- 1}$ .

$∴$ Answer = $1655$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.