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A compound on analysis gave the followin...

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.

Text Solution

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Calculation of empirical formula

Empirical formula is `C_2H_4O`.
Calculation of Moleuclar formula
Empirical formula mass `= 12 xx 2 + 1 xx 4 + 16 xx 1 = 44 `
Molecular mass `= 2 xx ` Vapour density
` =2 xx 44 =88`
`n=(" Molecular mass ")/("Empirical Formula mass ")=(88)/(44)=2`
Molecular formula = Empirical formula ` xx ` n
`=C_2H_4O xx 2 `
`= C_4 H_(8)O_2`
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