Empirical Formula Mass from Percentage Composition JEE Chemistry

 

❓ Question

Quantitative analysis of an organic compound $X$ gives the following composition:

• Carbon = 14.5%
• Chlorine = 64.46%
• Hydrogen = 1.8%

Find the empirical formula mass of $X$.

(Atomic masses: C = 12, H = 1, O = 16, Cl = 35.5)

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✍️ Solution

Assume the mass of the compound is 100 g.

Percentage of oxygen:

$$100-(14.5+64.46+1.8)=19.24\%$$

Now calculate the moles of each element.

Element	Percentage	Atomic Mass	Moles
C	14.5	12	$\frac{14.5}{12}=1.208$
Cl	64.46	35.5	$\frac{64.46}{35.5}=1.816$
H	1.8	1	$\frac{1.8}{1}=1.8$
O	19.24	16	$\frac{19.24}{16}=1.2025$

Divide all mole values by the smallest value ($1.2025$):

$$C:Cl:H:O = 1:1.5:1.5:1$$

Multiply by 2 to obtain whole numbers:

$$2:3:3:2$$

Hence, the empirical formula is

$$\boxed{C_2H_3Cl_3O_2}$$

Empirical formula mass:

$$=2(12)+3(1)+3(35.5)+2(16)$$
$$=24+3+106.5+32$$
$$=165.5\ \text{g mol}^{-1}$$

✅ Final Answer

$$\boxed{165.5\ \text{g mol}^{-1}}$$

Note: The handwritten solution in the image incorrectly adds the masses as 130. The correct sum is 24 + 3 + 106.5 + 32 = 165.5 g mol⁻¹.