JEE Main 2025 — Some Basic Concepts Of Chemistry Question with Solution

Calculate the Moles of Carbon:

Carbon is fully converted to ${\mathrm{CO}}_2$.

Moles of ${\mathrm{CO}}_2=\frac{1.46\text{g}}{44\text{g/mol}}=0.033\text{ mol}$.

Moles of Carbon ($\text{C}$) = Moles of ${\mathrm{CO}}_2$ = 0.033 mol.

Mass of Carbon = $0.033\times 12\text{g/mol}=0.396\text{g}$.

Calculate the Moles of Hydrogen:

Hydrogen is fully converted to ${\mathrm{H}}_2\mathrm{O}$.

Moles of ${\mathrm{H}}_2\mathrm{O}=\frac{0.567\text{g}}{18\text{g/mol}}=0.0315\text{ mol}$.

Moles of Hydrogen ($\text{H}$) = $2\times 0.0315=0.063\text{ mol}$.

Mass of Hydrogen = $0.063\times 1\text{g/mol}=0.063\text{g}$.

Calculate the Mass of Oxygen:

Total mass of compound $\mathrm{X}$ = 1.0 g.

Mass of Oxygen = Total mass - (Mass of Carbon + Mass of Hydrogen)

Mass of Oxygen = $1.0\text{g}-(0.396\text{g}+0.063\text{g})=0.541\text{g}$.

Moles of Oxygen ($\text{O}$) = $\frac{0.541}{16}=0.0338\approx 0.033\text{ mol}$.

Determine the Empirical Formula:

The ratio of moles: $\text{C}=0.033$, $\text{H}=0.063$, $\text{O}=0.033$.

Simplified, this ratio is approximately 1:2:1.

Empirical formula: ${\mathrm{CH}}_2\mathrm{O}$.

Calculate the Empirical Formula Mass:

Empirical formula mass = $12\times 1+1\times 2+16\times 1=30\text{g/mol}$.

Therefore, the empirical formula mass of compound $\mathrm{X}$ is 30 g/mol.