Question

An organic compound contain 61% carbon and 11.8% hydrogen if the vapour density of compound be 59 then calculate its empirical formula and molecular formula

Answer 1

Explanation:

O%=100-(61+11.8)=27.5

Now

C=61/12=5.08

H=11.8/1=11.8

O=27/16=1.68

Lowest ratio value multiple to all value

C=5.08/1.68=3.02

H=11.8/1.68=7.2

O=1.68/1.68=1.68

Empirical formula is =C3H7O1

Empirical formula mass =3×12+7×1+1×16=59

n=density /emyrical formula mass =59/59=1

Molecular formula =n×emyrical formula =1×C3H7O1=C3H7O