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