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The combustion of one mole of a compound...

The combustion of one mole of a compound yields 3.0 moles of `CO_(2)`, and 4.0 moles of `H_(2)`O. The empirical formula of this compound is.

A

`C_(3)H_(3)`

B

`CH_(4)`

C

`C_(3)H_(4)`

D

`C_(3)H_(8)`

Text Solution

Verified by Experts

The correct Answer is:
D
`C_(x)H_(y) + (x + (y)/(4)) O_(2) to xCO_(2) + (y)/(2) H_(2)O`
`x = 3, (y)/(2) H_(2)O`
` therefore " " E.F = C_(3)H_(6) = M.F`
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