Complete combustion of 1.18 g of organic compound gives 2.64 g of `CO_(2) & 1.26 g " of " H_(2)O`. Find emprical formula of compound ?

Options

A. `C_(3)H_(7)O`
B. `C_(2)H_(4)`
C. `C_(3)H_(6)O`
D. `C_(3)H_(7)`

Answer

A

Explanation

n of x C = `(2.64)/(44)xx 1 = 0.06`
mass of `C=0.06 xx 12 = 0.72 g`
n of H = `(1.26)/(18) xx 2 = 0.14`
mass of `O=1.18-(0.72+0.14) = 0.32 g`
`{:(n_(C)=0.06,,,0.06//0.02=3),(n_(H)=0.14,,,0.14//0.02=7),(n_(O)=0.02,,,0.02//0.02=1):}`
`EFimplies C_(3)H_(7)O_(1)`