An organic compound on analysis gave the following percentage composition : C = 57.8 %, H= 3.6 % and the rest is oxygen. The vapour density of the compound was found to be 83. Find the molecular formula of the compound.

An organic compound on analysis gave the following percentage composit

1.	An organic compound on analysis gave the following percentage composition : C = 57.8 %, H= 3.6 % and the rest is oxygen. The vapour density of the compound was found to be 83. Find the molecular formula of the compound.
Answer» Solution :STEP I. Percentage of oxygen `100 - (57.8 + 3.6) = 100 - 61.4 = 38.6` Step II. Empirical formula of organic compound `{:("Element","Percentage","Atomic mass","GRAM atoms (Moles)","Atomic ratio (Molar ratio)","Simplest WHOLE no. ratio"),("C",57.8,12,(57.8)/(1)=4.82,(4.82)/(2.41)=2.0,4),("H",3.6,1,(3.6)/(1)=3.6,(3.60)/(2.41)=1.5,3),("O",38.6,16,(38.6)/(16)=2.41,(2.41)/(2.41)=1.0,2):}` Empirical formula of the compound `= C_(4)H_(3)O_(2)` Step II. Molecular formula of the compound Empirical formula mass `= 4 xx 12 + 3xx 1+ 2 xx 16 = 83 u` Molecular mass `= 2 xx V.D = 2 xx 83 = 166u` `n=("Molecular mass")/("Empirical formula mass")=((166u))/((83u))=2` `:.` Molecular formula of compound `= 2 xx C_(4)H_(3)O_(2) = C_(8)H_(6)O_(4)`.