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.

Answer

`C_(5)H_(10)O_(5)`

Explanation

`C_xH_(2y)O_y+2xO_2 hArrxCO_2+yH_2O+xO_2`
After cooling only `CO_(2)` and `O_(2)` are present because water is in liquid form.
Thus, moles of gases after cooling =`x + x =2x`
Volume of gases after cooling =`2.24 L`
`2x=2.24 L`
`x=1.12 L`
Number of moles of `CO_(2)=1.12/2.24 =0.5 mol`
Hence, the empirical formula of organic compound is `C(H_(2)O)`
Given, Vapour pressure of pure water `(P^(@))=17.5 mm Hg`
Lowering of vapour pressure `(P^(@)-P)=0.104 mm Hg`
weight of solute `(W_(2)) =50 g`
weight of solvent `(W_(1)) =1000 g`
`(P^(@)-P)/ P^(@)=(W_(2) xx Mw_(1))/(Mw_(1) xx W_(1))`
`0.104/17.5=(50 xx 18)/(Mw_(2) xx 1000)`
`Mw_(2)=(900 xx 17.5)/104=151.4 g`
Empirical formula weight =`C(H_(2)O)`
=`12 +2 +16 =30`
Molecular weight (m)=151.4 g`
`n=("Molecular weight")/("Empirical formula weight")`
=`151.4/30 =5.04 =5`
Thus, molecular formula =`[C(H_(2)O]_(5)=C_(5)H_(10)O_(5)`