calculate the empirical and molecular formula of a compound containing 76.6 percentage of carbon, 6.38 percentage of hydrogen and rest oxygen. it's vapour density is 47.
Answers
Answered by
4
Empirical formula = C6H6O
Va-pour density 47
∴ Molecular mass = 2 x
vapor density
= 2 x 47
= 94
Molecular formula Empirical formula x n Molecular mass x n
n = (Molecular mass) / (Empirical formula mass) = 94/94
= 1
∴ Molecular formula = C6H6O
Answered by
2
Explanation:
hope it will help you bro
Attachments:
Similar questions