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A compound on analysis gave the follwing...

A compound on analysis gave the follwing result C=54.54%,H=9.09% and vapour density of compound = 88.Determine the molecular formula of the compound :-

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Percentage of Oxygen =100 - (54.54 + 9.09) = 36.37%

Empirical formula `C_(2)H_(4)O`
Empirical formula weight =44
`"Molecular weight" = 2 xx 88 =176, n= (176)/(44)= 4`
So, `"molecular formula" = 4 xx E.F = 4 C_(2)H_(4)O = C_(6)H_(16)O_(4)`
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