JEE Chemistry 2025 Question with Solution

Given: The hydrocarbon $X$ has molar mass $80 \, \text{g mol}^{-1}$ and contains $90\%$ carbon.

Find: The degree of unsaturation.

First determine the masses of carbon and hydrogen in $80 \, \text{g}$ of the compound:

$$\text{Mass of carbon} = 0.90 \times 80 = 72 \, \text{g}$$ $$\text{Mass of hydrogen} = 80 - 72 = 8 \, \text{g}$$

Now calculate moles of carbon and hydrogen:

$$\text{Moles of carbon} = \frac{72 \, \text{g}}{12 \, \text{g/mol}} = 6$$ $$\text{Moles of hydrogen} = \frac{8 \, \text{g}}{1 \, \text{g/mol}} = 8$$

So the molecular composition is $\mathrm{C_6H_8}$.

Using the degree of unsaturation formula:

$$\text{DU} = \frac{2C + 2 - H}{2}$$

Substitute $C = 6$ and $H = 8$:

$$\text{DU} = \frac{2(6) + 2 - 8}{2} = \frac{12 + 2 - 8}{2} = \frac{6}{2} = 3$$

Therefore, the degree of unsaturation is $3$.

Given: Molar mass of the hydrocarbon is $80 \, \text{g mol}^{-1}$ and carbon content is $90\%$.

Find: Degree of unsaturation.

From $80 \, \text{g}$ of compound:

$$\text{Carbon mass} = 72 \, \text{g}, \qquad \text{Hydrogen mass} = 8 \, \text{g}$$

Hence,

$$\text{Moles of C} = \frac{72}{12} = 6, \qquad \text{Moles of H} = \frac{8}{1} = 8$$

The ratio $6:8$ simplifies to $3:4$, so the empirical formula is $\mathrm{C_3H_4}$.

Its empirical formula mass is:

$$3 \times 12 + 4 \times 1 = 40 \, \text{g mol}^{-1}$$

Now,

$$\frac{80}{40} = 2$$

So the molecular formula is:

$$\mathrm{C_6H_8}$$

Now apply:

$$\text{DU} = \frac{2C + 2 - H}{2}$$ $$\text{DU} = \frac{2(6) + 2 - 8}{2} = 3$$

Therefore, the hydrocarbon has degree of unsaturation $3$.