Question

A compound on analysis gave the following percentage
composition C = 54.54%, H, 9.09% 0 = 36.36. The vapour density of the
compound was found to be 44. Find out the molecular formula of the
compound.

Answer 1

atomic mass        percentage     atomic ratio( %/a.m.)  simple ratio
1)carbon is 12            54.54%          4.545                       2
2)hydrogen is 1          09.09%          9.09                         4
3)oxygen is 16           36.36%          2.2725                     1

:. the empirical formula is C₂H₄O

empirical formula weight =44

vapour density is 44
molecular mass is 2(44)=88

'n' factor = molecular formula weight/Empicical formula weight

'n' factor= 88/44=2
Molecular formula = (Empirical Formula)n
M.F = C₄H₈O₂