The ratio of mass percent of CandH of an organic compound ( C _ X H...

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

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

Empirical Formula$= {C}_{x}{H}_{2x}{O}_{3x / 2} = {C}_{2x}{H}_{4x}{O}_{3x} = {C}_{2}{H}_{4}{O}_{3}$