Empirical Formula and Molecular Formula JEE Main PYQ

 

❓ Question

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, find its molecular formula.

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

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✍️ Solution

Assume 100 g of the compound.

Element	Percentage	Atomic Mass	Moles
C	54.2	12	$\frac{54.2}{12}=4.517$
H	9.2	1	$\frac{9.2}{1}=9.2$
O	36.6	16	$\frac{36.6}{16}=2.288$

Divide all mole values by the smallest value, $2.288$:

$$C:H:O = \frac{4.517}{2.288}: \frac{9.2}{2.288}: \frac{2.288}{2.288}$$
$$C:H:O\approx2:4:1$$

Therefore, the empirical formula is

$$\boxed{C_2H_4O}$$

Empirical formula mass:

$$=2(12)+4(1)+16$$
$$=44\text{ g mol}^{-1}$$

Now,

$$n=\frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}$$
$$n=\frac{132}{44}=3$$

Hence,

$$\text{Molecular Formula}=(C_2H_4O)_3$$

✅ Final Answer

$$\boxed{C_6H_{12}O_3}$$