An organic compound `C_(x)H_(2y)O_(y)` was burnt with twice the amount of oxygen needed for complete combustion of `CO_(2)` and `H_(2)O`. The hot gases when cooled to `0^(@)C` and 1 atm pressure, measured 2.24 L, the water collected during cooling weighed 0.9 g The vapour pressure of pure water at `20^@C` is 17.5 mm Hg and is lowered by 0.104 mm when 50 g of the organic compound is dissolved in 1000 g of water. Give the molecular formula of the organic compound.

Explanation

Complete combustion of organic compound is as:
`C_xH_(2y)O_yrarrxCO_2+yH_2O`
Since oxygen taken is `2x` litres `x` litres `x` litres `O_2` is left at STP after reaction. Also `x` litre of `CO_4` is formed by 1 mol of organic compound.
So, `2x=2.25" litre "CO_2`
`x=1.12" litre "CO_2`
or `x=(1.12)/(22.4)molCO_2`
`=0.05molCO_2`
Moles of `H_2O` formed `(y)=(0.9)/(18)=0.05`
`x:y=0.05:0.05=1:1`
`x=1` and `y=1`
Empirical formula of organic compound `=CH_2O`
Empirical formula weight of organic compound `=30`
Now, molecular weight of compound is derived by Raoult's law:
`(p^@-p_S)/(p^@)=(w)/(m)xx(M)/(W)`
`p^@-P_S=` lowering of vapour pressure `=0.104mm`
and `p^@=` vapour pressure of pure solvent `=17.5mm`
`(0.104)/(17.5)=(50)/(m)xx(18)/(1000)`
`m=150.5`
Now, `n=("molecular weight")/("empirical formula weight")`
`=(150.5)/(30)=5`
Molecular formula`=("empirical formula")xxn`
`=(CH_2O)xx5=C_5H_(10)O_5`