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 weighed 0.9 g. The vapour pressure of pure water at 20° C is 1 7.5 mm Hg and is lowered by 0.104 mm when 50 g of the organic compound are dissolved in 1000 g of water. Give the molecular formula of the organic compound.

Answer 1

 Complete combustion of organic compound is as :

Since, oxygen taken is 2x litre and thus, x litre O₂ is left at STP after reaction. Also x litre of CO₂ is formed by 1 mole of organic compound.

∴ Empirical formula of organic compound = CH₂O

Empirical formula wt. of organic compound = 30

Now, mol. wt. of compound is derived by Raoult's law:

∴ p° - p_s = lowering of vapour pressure = 0.104 mm

and p°  = vapour pressure of pure solvent = 17.5 mm