jkk

Asked by saswati509 | 3rd May, 2020, 10:10: AM

Expert Answer:

Given:

Pecentage composition:

C= 40.68%

H= 5.08%

O = 100 - 45.76

O = 54.24%

Empirical formula is C2H3O2

Empirical formula mass : 24+ 3+ 32 = 59

Molecular mass = 2× Vapour density

                              = 2 ×59 = 118

We konw,
 
n = fraction numerator Molecular space formula space mass over denominator Empirical space formula space mass end fraction
 
   = 118 over 59
 
n = 2
 
Molecular formula = n × Empirical formula
 
                          = 2(C2H3O2)
 
Molecular formula = C4H6O 

Answered by Varsha | 4th May, 2020, 07:05: PM