1.

A gaseous hydrocarbon given upon combustion, 0.72g water and 3.08g of CO_(2). The empirical formula of the hydrocarbon is:

Answer»

`C_(6)H_(5)`
`C_(7)H_(8)`
`C_(2)H_(4)`
`C_(3)H_(4)`

SOLUTION :Number of moles of `H_(2)O=(0.72)/(18)=0.04`
Number of moles of `CO_(2)=(3.08)/(44)=0.07`
`n_(H)=2n_(H_(2))=2xx0.04=0.08`
`n_(C)=n_(CO_(2))=0.07`
`C:H=0.07:0.08=7:8`
`THEREFORE` Empirical =`C_(7)H_(8)`.


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