JEE Chemistry 2023 Question with Solution

Given: Amount of compound = $0.01 \, \text{mol}$, carbon content = $60\%$, and $4.4 \, \text{g}$ of $CO_2$ is produced on complete combustion.

Find: The molar mass of the compound.

From the solution:

So, moles of carbon atoms present in the burnt sample = $0.1 \, \text{mol}$.

Hence mass of carbon in the sample:

Since carbon is $60\%$ by mass, if the sample mass is $m$ g, then

Therefore,

This mass corresponds to $0.01 \, \text{mol}$ of the compound. Therefore molar mass is

Therefore, the molar mass of the compound is $200 \, \text{g mol}^{-1}$.

The provided the solution contains an inconsistent line: it states "Mass of carbon in the compound: $0.01 \times \frac{60}{100} = 0.006 \, \text{g}$", which is dimensionally incorrect because $0.01$ is the amount in moles, not mass in grams.

Using the rest of the provided working and the combustion data:

1. Convert $4.4 \, \text{g}$ of $CO_2$ to moles.
2. Use those moles to get mass of carbon in the sample.
3. Since carbon is $60\%$ of the sample mass, obtain the sample mass.
4. Divide sample mass by $0.01 \, \text{mol}$ to get molar mass.

This gives the consistent final result $200 \, \text{g mol}^{-1}$, matching the listed correct answer.