Question

calculate empirical and molecular formula carbon-76.6%,hydrogen-6.38%and rest of o2 it's vapour density is 47​

Answer 1

Explanation:

carbon -76.6%,hydrogen-6.38%and rest of 02

76.6+6.38=82.98

100 - 82.98= 17.02

so 17.02 percentage of O2 is there.

H ---> 6.38/1 ---> 6.38

C ---->76.6/12 ---> 6.38

O --->17.02/16 --->1.38(approx)

H ---> 6.38/1.38= 6

C ----> 6.38/1.38= 6

O ---> 1.38/1.38 = 1

Empirical formula = 2× V.D

(vapour density)

Formula =C6H6O1

Empirical = 94

94(C6H6O1)

C564H564O94

C564H564O94 is your answer