The ratio of mass percent of $\mathrm{C}\mathrm{and}\mathrm{H}$ of an organic compound $\left({\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}{\mathrm{O}}_{\mathrm{Z}}\right)\mathrm{is}6:1$. If one molecule of the above compound $\left({\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}{\mathrm{O}}_{\mathrm{Z}}\right)$ contains half as much oxygen as required to burn one molecule of compound ${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}$ completely to ${\mathrm{CO}}_2\mathrm{and}{\mathrm{H}}_2\mathrm{O}$. The empirical formula of compound ${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}{\mathrm{O}}_{\mathrm{Z}}$ is

The correct option is C

${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_3$

The explanation for the correct option:

C: ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_3$

Element	Relative mass	Relative mole	Simplest whole number ratio
$\mathrm{C}$	$6$	$6/12=0.5$	$1$
$\mathrm{H}$	$1$	$1/1=1$	$2$

So, $\mathrm{X}=1,\mathrm{Y}=2$

Equation for combustion of ${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}$

$$\begin{gathered} {\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}+(\mathrm{X}+\mathrm{Y}/4){\mathrm{O}}_2->{\mathrm{XCO}}_2+(\mathrm{Y}/2){\mathrm{H}}_2\mathrm{O} \\ \mathrm{Oxygen}\mathrm{atoms}\mathrm{required}=2(\mathrm{X}+\mathrm{Y}/4) \\ \mathrm{Given}2(\mathrm{X}+\mathrm{Y}/4)=2\mathrm{Z} \\ \Rightarrow (1+2/4)=\mathrm{Z} \\ \Rightarrow \mathrm{Z}=1.5 \\ \mathrm{Molecule}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}{\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}{\mathrm{O}}_{\mathrm{Z}} \\ \Rightarrow {\mathrm{C}}_1{\mathrm{H}}_2{\mathrm{O}}_{3/2} \\ \Rightarrow {\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_3 \end{gathered}$$

Final Answer: Therefore, option C is the correct option.