JEE Chemistry 2025 Question with Solution

Given: Mass of organic compound = $500 \, \text{mg}$; mass of $\mathrm{CO_2}$ produced = $220 \, \text{mg}$.

Find: Percentage composition of carbon in the compound.

On complete combustion, all the carbon in the compound appears in $\mathrm{CO_2}$.

Molar mass of $\mathrm{CO_2}$ is

$$12 + 2 \times 16 = 44 \, \text{g mol}^{-1}$$

So, in $44 \, \text{g}$ of $\mathrm{CO_2}$, the mass of carbon is $12 \, \text{g}$.

Therefore, carbon present in $220 \, \text{mg}$ of $\mathrm{CO_2}$ is

$$\text{Mass of C} = \left(\frac{12}{44}\right) \times 220 = 60 \, \text{mg}$$

Now, percentage of carbon in the compound is

$$\text{Percentage of C} = \left(\frac{60}{500}\right) \times 100 = 12\%$$

Therefore, the percentage composition of carbon in the compound is $12\%$.

Given: The compound has mass $500 \, \text{mg}$ and gives $220 \, \text{mg}$ of $\mathrm{CO_2}$ on complete combustion.

Find: The percentage of carbon in the original compound.

Step 1: Use the composition of $\mathrm{CO_2}$.

$$\text{Molar mass of } \mathrm{CO_2} = 12 + 32 = 44 \, \text{g mol}^{-1}$$

Out of this $44$, carbon contributes $12$.

Step 2: Find the carbon mass in $220 \, \text{mg}$ of $\mathrm{CO_2}$.

$$\text{Mass of carbon} = \frac{12}{44} \times 220 = 60 \, \text{mg}$$

Step 3: Compare this carbon mass with the original sample mass.

$$\text{Percentage of carbon} = \frac{60}{500} \times 100 = 12\%$$

Thus, the nearest integer value is 12.