( 0.30 ~g ) of an organic compound containing ( C , H ) and Oxygen on combustion yields ( 0.44 ~g CO ₂ ) and ( 0.18 ~g H ₂ O ). If one mol of compound weighs ( 60 ), then molecular formula of the compound is

Step-by-step solution

Percentage of C= 12 44 0.44 0.30 100=40 % Percentage of H = 2 18 0.18 0.30 100=6.6 % Percentage of O=100-(40+6.6)=53.4 % array |l|l|l|l| Element & % & Molar ratio & Simplest ratio C & 40 & 40 12 =3.3 & 3.3 3.3 =1 H & 6.6 & 6.6 1 =6.6 & 6.6 3.3 =2 O & 53.4 & 53.4 16 =3.3 & 3.3 3.3 =1 array hence, empirical formula = CH ₂ O We know n= Molecular mass Empirial formula mass = 60 30 =2 Molecular formula of compound = ( CH ₂ O ₂ )= C ₂ H ₄ O ₂ Hence, molar mass of C ₂ H ₄ O ₂=2 12+1 4+16 2=60 gmol ⁻¹