On complete combustion 1.0 g of an organic compound $(\mathrm{X})$ gave 1.46 g of $\mathrm{CO}_2$ and 0.567 g of $\mathrm{H}_2 \mathrm{O}$. The empirical formula mass of compound $(\mathrm{X})$ is __________ g. (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16$ )

<p><strong>Calculate the Moles of Carbon:</strong></p> <p><p>Carbon is fully converted to $ \mathrm{CO}_2 $.</p></p> <p><p>Moles of $ \mathrm{CO}_2 = \frac{1.46\, \text{g}}{44\, \text{g/mol}} = 0.033 \text{ mol} $.</p></p> <p><p>Moles of Carbon ($ \text{C} $) = Moles of $ \mathrm{CO}_2 $ = 0.033 mol.</p></p> <p><p>Mass of Carbon = $ 0.033 \times 12\, \text{g/mol} = 0.396\, \text{g} $.</p></p> <p><strong>Calculate the Moles of Hydrogen:</strong></p> <p><p>Hydrogen is fully converted to $ \mathrm{H}_2\mathrm{O} $.</p></p> <p><p>Moles of $ \mathrm{H}_2\mathrm{O} = \frac{0.567\, \text{g}}{18\, \text{g/mol}} = 0.0315 \text{ mol} $.</p></p> <p><p>Moles of Hydrogen ($ \text{H} $) = $ 2 \times 0.0315 = 0.063 \text{ mol} $.</p></p> <p><p>Mass of Hydrogen = $ 0.063 \times 1\, \text{g/mol} = 0.063\, \text{g} $.</p></p> <p><strong>Calculate the Mass of Oxygen:</strong></p> <p><p>Total mass of compound $ \mathrm{X} $ = 1.0 g.</p></p> <p><p>Mass of Oxygen = Total mass - (Mass of Carbon + Mass of Hydrogen) </p></p> <p><p>Mass of Oxygen = $ 1.0\, \text{g} - (0.396\, \text{g} + 0.063\, \text{g}) = 0.541\, \text{g} $.</p></p> <p><p>Moles of Oxygen ($ \text{O} $) = $ \frac{0.541}{16} = 0.0338 \approx 0.033 \text{ mol} $.</p></p> <p><strong>Determine the Empirical Formula:</strong></p> <p><p>The ratio of moles: $ \text{C} = 0.033 $, $ \text{H} = 0.063 $, $ \text{O} = 0.033 $.</p></p> <p><p>Simplified, this ratio is approximately 1:2:1.</p></p> <p><p>Empirical formula: $ \mathrm{CH}_2\mathrm{O} $.</p></p> <p><strong>Calculate the Empirical Formula Mass:</strong></p> <p>Empirical formula mass = $ 12 \times 1 + 1 \times 2 + 16 \times 1 = 30\, \text{g/mol} $.</p> <p>Therefore, the empirical formula mass of compound $ \mathrm{X} $ is 30 g/mol.</p>