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.