Chemistry - Aromatic Hydrocarbon Question with Solution | TestHub

Ratio of mass % of C and H in${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}{\mathrm{O}}_{\mathrm{Z}}$is 6 : 1.

Therefore, Ratio of mole % of C and H in${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}{\mathrm{O}}_{\mathrm{Z}}$will be 1 : 2.

Therefore x : y = 1 : 2, which is possible in options 1, 2 and 3.

Now oxygen required to burn${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}$

${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right){\mathrm{O}}_2\to \mathrm{x}\mathrm{C}{\mathrm{O}}_2+\frac{\mathrm{y}}{2}{\mathrm{H}}_2\mathrm{O}$

Now z is half of oxygen atoms required to burn${\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}}$

So$\mathrm{Z}=\frac{\left(2\mathrm{X}+\frac{\mathrm{Y}}{2}\right)}{2}=\left(\mathrm{X}+\frac{\mathrm{Y}}{4}\right)$

Now putting values of x and y from the given options

Option A,$\mathrm{x}=2,\mathrm{y}=4$

$\mathrm{Z}=\left(2+\frac{4}{4}\right)=3$

Option B,$\mathrm{x}=3,\mathrm{y}=6$

$\mathrm{Z}=\left(3+\frac{6}{4}\right)=4.5$

Therefore correct option is 1$\left({\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_3\right)$