Mass of $\mathrm{C}$: Mass of $\mathrm{H}=6 : 1$.
Or, for every atom of $\mathrm{C} (12 \mathrm{u}),$ we have $2$atoms of $\mathrm{H} (1\mathrm{u} \times 2=2\mathrm{u})$
Thus, the ratio of atoms of $\mathrm{C} : \mathrm{H}=\mathrm{x} : \mathrm{y}=1 : 2$
Thus, the formula is: ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}{\mathrm{O}}_{\mathrm{z}}$
Combustion of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}$:
${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}+(3\mathrm{x}/2){\mathrm{O}}_2\to {\mathrm{xCO}}_2+{\mathrm{xH}}_2\mathrm{O}$
Hence, the combustion of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}$ requires $(3\mathrm{x}/2)$ oxygen molecules.

Hence, the original molecule contains $(3\mathrm{x}/4)$molecules $=(3\mathrm{x}/2)$atoms.
Or, $\mathrm{z}=3\mathrm{x}/2$

Empirical Formula$={\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}{\mathrm{O}}_{3\mathrm{x}/2}={\mathrm{C}}_{2\mathrm{x}}{\mathrm{H}}_{4\mathrm{x}}{\mathrm{O}}_{3\mathrm{x}}={\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_3$