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 \((\mathrm{X})\) is _____ \(\times 10^{-1}\) (Given molar mass in \(\mathrm{g} \mathrm{~mol}^{-1}\) of \(\mathrm{C}: 12, \mathrm{H}: 1\), \(\mathrm{O}: 16, \mathrm{Cl}: 35.5)\)

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Some Basic Concepts of Chemistry

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}

C : 12, H : 1

O : 16, Cl : 35.5)

Answer: 1655

Solution

With C 14.5%, H 1.8%, Cl 64.46% and the balance as oxygen,

100 - 14.5 - 1.8 - 64.46 = 19.24%

, convert each to moles: C

= 14.5/12 = 1.208

, H

= 1.8/1 = 1.8

, Cl

= 64.46/35.5 = 1.816

, O

= 19.24/16 = 1.203

. Dividing through by the smallest value gives approximately

1 : 1.5 : 1.5 : 1

for C:H:Cl:O, and doubling clears the halves to the whole-number ratio

C_{2} H_{3} C l_{3} O_{2}

. The empirical formula mass is

2 (12) + 3 (1) + 3 (35.5) + 2 (16) = 165.5

, i.e.

1655 \times 1 0^{- 1}

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer

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