1.

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

Answer»

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

SOLUTION :`C_(x)H_(y) + (x+(y)/(4))O_(2) RARR xCO_(2) + (y)/(2) H_(2)O`
Moles of `H_(2)O = (0.72)/(18) = 0.04`
Moles of `CO_(2)= (3.08)/(44) = 0.07`
`-""x : (y)/(2) = 0.07 : 0.04`
or `x : y = 0.07 : 0.08 or 7 : 8`
`therefore""C : H = 7 : 8`
Therefore, the empirical formula is `C_(7)H_(8)`.


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