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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 :-

A

`C_(8)H_(16)O_(0)`

B

`C_(4)H_(16)O_(8)`

C

`C_(2)H_(4)O`

D

`CH_(4)O_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`{:("Atom", "atomic mass", %wt, %//"Atomic wt.", "simple ratio"),(C,12,54.54,4.54,2),(H,1,9.09,9.09,4),(O,16,36.37,2.27,1):}`
Empirical formula `= C_(2)H_(4)O`
Empirical formula weight = 44
Molecular formula weight `= 88xx2=176`
`n = (176)/(44)=4`
Hence molecular formula = `C_(8)H_(16)O_(4)`
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