Question

An organic compound (C_xH_2yO_y) was burnt with twice the amount of oxygen needed for complete combustion to CO₂ and H₂O. The hot gases when cooled to 0°C and 1 atm pressure, measured 2.24 L. The water collected during cooling weigh 0.9 g. The vapour pressure of pure water at 20°C is 17.5 mm Hg and is lowered by 0.104 mm Hg when 50g of the organic compound are dissolved in 1000g of water. Give the molecular formula of the organic compound.

Answer 1

The chemical equation for the combustion of organic compound C_xH_2yO_y can be represented as : C_xH_2yO_y + 2xO₂ = x CO₂ + y H₂O + xO₂
As oxygen taken is 2X litre and thus X litre O_{2 is left at STP after reaction and also X litre of CO2 is formed by 1 mole of organic compound.}
So the gases obtained after cooling = x + x = 2x
∴ 2x = 2.24 litre
Or x = 2.24/2 = 1.12 litres CO₂
Number of moles of CO₂ = 1.12 /22.4 mole [∵ 22.4 L at NTP = 1 mole]
= 0.05 mole CO_{2
Moles of} H₂O formed (Y) = 0.9/18 = 0.05

therefore X:Y= 0.05 : 0.05 = 1 : 1
X=1 and Y=1
So the empirical formula of the organic compound is C(H₂O)
Emperical formula wt. of organic compound = 30

According to Raoult's law :
(Po - Ps)/Ps = (w/m)*(M)/(W)

where (Po - Ps) = lowering of V.P. = 0.104 mm, Po = V.P. of pure solvent = 17.5 mm

or 0.104/17.396 = (50/m)/(18/1000)
on solving we get m = 150.5

n = Molecular wt./Emperical formula wt. = 150.5/30 = 5

Molecular formula = (Emperical formula)n
= (CH₂O)5 = C₅H₁₀O₅