MadSci Network: Chemistry

Re: Weight of a co-distilled product?

Date: Fri Oct 5 09:51:25 2001
Posted By: Plamen Angelov, Faculty, Organic Chemistry, University of Plovdiv
Area of science: Chemistry
ID: 1001425869.Ch

The ratio of masses of two compounds in gas phase is equal to the ratio of the products of their molecular masses times their partial pressures:
m1/m2 = M1*p1 / M2*p2
The ratio of masses will remain the same after the gas mixture resulting from the distillation is condensed.

In order to calculate the weight of the natural product that codistills with each gram of water, you have to know the pressure of this product at the temperature of the distillation. If we assume that the distillation is carried out at a pressure of 760 mm Hg (which is the atmospheric pressure at standard conditions), then knowing that during this distillation the partial pressure of the water is 733 mm Hg, and knowing that boiling occurs when the total vapor pressure of the mixture equals the atmospheric pressure, we can find out the partial pressure of the natural product:
760 - 733 = 27 mm Hg.

After we have figured that out, we can use the afore mentioned formula according to which m1/m2 = 150*27 / 18*733, where m1 is the mass of the natural compound and m2 - the mass of the water.

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