QJEEM18Cha70e: JEE Main 2018 Chemistry - Class 11 JEEM18Cha70e Solution: The ratio of mass percent of C and H of an organic compound (C…

$\therefore\,\,\,$ The ratio of no of atoms of C and H in one molecule of C_x H_y O_z = 1 : 2

$\therefore\,\,\,$ y = 2x

In one molecule of C_x H_y O_z compound contain z atoms of oxygen.

According to the question, no of atoms of oxygen required to burn C_xH_y completely should be twice of z atoms of oxygen.

C_xH_y + (x + ${y \over 4}$ ) O₂ $\to$ HCO₂ + ${y \over 2}$ H₂O

No. of O₂ molecules required = (x + ${y \over 4}$ )

$\therefore\,\,\,$ No. of O atoms required = 2 (x + ${y \over 4}$ )

According to question,

$2\left( {x + {y \over 4}} \right) = 2z$

$\Rightarrow \,\,\,2\left( {x + {{2x} \over 4}} \right) = 2z\,\,\,$ [ Putting y = 2x ]

$\Rightarrow \,\,\,\,{{3x} \over 2} = z$

$\therefore\,\,\,$ x : y : z

= x : 2x : ${{3x} \over 2}$

= 2 : 4 : 3

$\therefore\,\,\,$ Empirical formula = C₂H₄O₃