JEE Chemistry 2025 Question with Solution

Given: Percentage composition of the compound is C = $14.5\%$, Cl = $64.46\%$, and H = $1.8\%$.

Find: The empirical formula mass of compound $X$.

From the given working:

$$\text{Moles of C} = \frac{14.5}{12} = 1.21, \quad \text{Moles of Cl} = \frac{64.46}{35.5} = 1.81, \quad \text{Moles of H} = \frac{1.8}{1} = 1.8$$

Now divide by the smallest value, $1.21$:

$$\text{C} : \frac{1.21}{1.21} = 1, \quad \text{Cl} : \frac{1.81}{1.21} = 1.5, \quad \text{H} : \frac{1.8}{1.21} = 1.49$$

This gives the approximate ratio:

$$\text{C} : \text{Cl} : \text{H} = 1 : 1.5 : 1.5$$

Multiplying throughout by $2$:

$$\text{C} : \text{Cl} : \text{H} = 2 : 3 : 3$$

So the empirical formula is:

$$\text{C}_2\text{Cl}_3\text{H}_3$$

Hence, the empirical formula mass is:

$$2(12) + 3(35.5) + 3(1) = 146.5 \, \text{g/mol}$$

the solution shows 4.85 as the listed answer, but the worked calculation clearly gives $146.5$. Therefore, the empirical formula mass is $146.5$.

Given: Elemental percentages of C, Cl, and H.

Find: The empirical formula mass.

The key idea is to convert percentages into moles and then reduce them to the simplest whole-number ratio. After calculating moles, the ratios become approximately $1 : 1.5 : 1.49$, which is effectively:

$$1 : 1.5 : 1.5$$

To remove the fraction, multiply by $2$:

$$2 : 3 : 3$$

Thus the empirical formula is:

$$\text{C}_2\text{Cl}_3\text{H}_3$$

Now add the atomic masses:

$$2 \times 12 = 24, \quad 3 \times 35.5 = 106.5, \quad 3 \times 1 = 3$$

Therefore:

$$24 + 106.5 + 3 = 146.5$$

So, the empirical formula mass is $146.5 \, \text{g/mol}$.