QJEEM20Ch55f6: JEE Main 2020 Chemistry - Class 11 JEEM20Ch55f6 Solution: The ratio of the mass percentages of ‘C & H’ and ‘C & O’ of a…

Let the organic compound X is = C_xH_yO_z

Here moles of C = x, moles of H = y, moles of O = z

Given, ${{{W_C}} \over {{W_H}}} = {4 \over 1}$

$\therefore$ ${x \over y} = {{{{{W_C}} \over {12}}} \over {{{{W_H}} \over 1}}}$ = ${4 \over 1} \times {1 \over {12}}$ = ${1 \over 3}$

Also Given, ${{{W_C}} \over {{W_O}}} = {3 \over 4}$

$\therefore$ ${x \over z} = {{{{{W_C}} \over {12}}} \over {{{{W_O}} \over {16}}}}$ = ${3 \over 4} \times {{16} \over {12}}$ = 1

$\Rightarrow$ x = z

$\therefore$ Empirical formula = C_xH_3xO_x = CH₃O

As compound is saturated acyclic organic
compound, so molecular formula = C₂H₆O₂

C₂H₆O₂ + ${5 \over 2}$ O₂ $\to$ 2CO₂ + 3H₂O

For 1 mole of C₂H₆O₂ number of moles of O₂ required = ${5 \over 2}$

$\therefore$ For 2 mole of C₂H₆O₂ number of moles
of O₂ required = ${5 \over 2}$ $\times$ 2 = 5