an organic compound an analysis : c=65%, h=3.5% n = 9.6% and o= 21.9% 0.3675 gm of the

Asked by blpradhanpradhan94137 | 1st Jun, 2020, 10:46: AM

Expert Answer:

Given:
 
C = 65%
H=3.5%
N=9.6%
O=21.9%
 
Mass of organic compound is 0.3675gm
 
 
 
Element     %    No. of moles  simple ratio
C 65 65/12= 5.42 5.42/0.68=7.97=8
H 3.5 3.5/1=3.5 3.5/0.68= 5.1=5
N 9.6 9.6/14=0.68 0.68/0.68=1
O 21.9 21.9/16=1.37 1.37/0.68= 2.01 =2
 
Empirical formula = C8H5NO2, mass of emperical formula = (8× 12)+(1×5)+(1×14)+(16×2) =168+5+14+32=219
Now, 53cc of compound = 0.3675gm22400 ml space of space compound space equals fraction numerator 0.3675 over denominator 53 end fraction cross times 22400 space equals 155.32

straight n space equals space fraction numerator Mol. space mass over denominator Emp. space formula space mass end fraction space equals space fraction numerator 219 over denominator 155.32 end fraction equals space 1.4 space tilde 1 space space
 
Therefore molecular formula= C8H5NO2
 

Answered by Ramandeep | 8th Jun, 2020, 01:19: PM