An organic compound contains carbon, hydrogen and oxygen. If the ratio percentage of C and H is `6:1` calculate the simplest formula of the compound, given that one molecule of the compound contains half as much oxygen as would be required to burn all the carbon and hydrogen atoms in it to `CO_2` and `H_2O`

Explanation

Percentage of C: Percentage of `H=6:1`
Percentage of molar ratio
`=(6)/(12):(1)/(1)-=(1)/(2):1` `{:[("Atomic weight of "C=12),("Atmic weight of "H=1)]:}`
`-=1:2`
Molar ratio `C:H::1:2`
General formula `C_(x)H_(2x)O_(y)+O_(2)toxCO_2+xH_2O`
oxygen atom in product `=3x`
According to question.
`(3x)/(2)=y`
`therefore3x=2y`
`y=(3x)/(2)`
Formula is `C_(x)H_(2x)O_((3x)/(2))`. Simplifying, we get `C_(2)H_(4)O_(3)`.