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Grade upto college level Physical Chemistry

An organic compound CxH2yOy 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 liters. 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 are dissolved in 1000 g of water. Give the molecular formula of the organic compound.

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
Hello Student,
Please find the answer to your question
The chemical equation for the combustion of organic compound CxH2yOy can be represented as :
CxH2yOy + 2xO2 = x CO2 + y H2O + x O2
The gases obtained after cooling = x + x = 2x
∴ 2x = 2.24 litres [∵ H2O is in liquid state]
Or x = 2.24/2 = 1.12 litres
Number of moles of CO2 = 1.12 litres/222.4 litres mole [∵ 22.4 L at NTP = 1 mole]
= 1/2 mole = 0.05 mole
The empirical formula of the organic compound is C(H2O) …(i)
The mole fraction of the solute (A)
= relative decrease in vapour pressure of the solvent (B)
Po – p/po = WA/MA /WA/M­A+WB/MB
Or 0.104/17.5 = 50/MA /50/MA+1000/18 [MA = mol. wt. g A]
Or 0.104/17.5 = 50/MA(50 *18 + 100MA/18MA)
Or 104/17500 = 50 * 18/900 + 1000 MA
Or MA = 150.6
Molecular wt. of the organic compound
(CH2O)n = 150
Molecular wt. CH2O = 12 + 2 + 16 = 30
∴ 30 * n = 150 [∵ (CH2O)n = mol, formula]
Or n = 150/30 = 5
∴ Molecular formula of the given organic compound is (CH2O)or C5H10O5.

Thanks
Deepak patra
askIITians Faculty

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