Question

The ratio of mass percent of C and H of an organic compound (C,H,O,) is 6:1. If one molecule of the above compound (C, 40.) contains half as much oxygen as required to burn one molecule of compound C,H, completely to CO, and H,O. The empirical formula of compound C,H,O, is:
(A) C,H,O, (B) C,H,O,
(C) C,H,0 (D) C,H,O,​

Answer 1

Explanation:

Cx Hu + nO2 react with x CO2 + y/2 H2O

n= 4x+y/4 and amount of O required m = n*2

given that z= m/2 = n

therefore , mass of % ratio of C:H::6:1 = 12:2

therefore for each C and H are present in the molecule for x = 1 , y = 2

therefore, Z = 4× 1+2/4

Z= 1.5

Empirical Formula == C2H4O3