Question

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 out the molecular formula of the compound.

Answer 1

We have,

An organic compound $C=57.8\%$

$H=3.6\%$

Thus the oxygen is $38.6\%$ Since the total $\%is100$

So,

Molecular weight i$=2\times vapordensity$

$=2\times 83=166$

So, Mass of C=\dfrac{57.8}{100}\times166$

$=95.948g$

No pof carbon atoms $=\frac{95.948}{12}=7.99\sim 8atoms$ Since

$12=mass$ of the carbon

Mass of $H=\frac{3.6}{100}\times 166=5.976\sim 6$ atoms

Mass of $O=\frac{38.6}{100}\times 166=64.076g$

No of atoms $=\frac{64.076}{16}=4atoms$

Thus. The formulaa will be $C_8{H}_6{O}_4$