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CORNERSHOT : NATURAL DISTILLATION IN OUR BACKYARD  Photo : Gilles

THIS SECTION IS A LITTLE HEAVY IN THEORY AND APPLIED FORMULAS, BUT NECESSARY TO FULLY UNDERSTAND THE DISTILLATION PROCESS.
 
FOR THOSE OF YOU CAPABLE OF WHIZZING THROUGH THESE THINGS, YOU HAVE MY DEEPEST RESPECT,
BUT FOR MOST OF US, IT'S PURE MARTIAN GRAFITIS, THEREFORE, IT'S NICE TO FIND WAYS TO EXPLAIN SUCH A COMPLEX YET SO SIMPLE PROCESS.
 

 >> ExploreFractional Distillation Tutorial  

AND ENJOY THE VIDEO DEMONSTRATING DISTILLATION OF CRUDE OIL ( PRINCIPLE IS THE SAME AS ETHANOL EXCEPT THAT ETHANOL SEPERATES FROM WATER '2 ELEMENTS' AS OPPOSED TO CRUDE OIL WICH  HAS MANY ELEMENTS TO SEPERATE .)

 
















CHM 1046
General Chemistry II
Dr. Michael Blaber


  • Properties of Solutions 
    • Distillation and an application of Raoult's Law

We will now take a foray into an important application of Raoult's Law regarding mole fractions of components in a mixture and their related contribution to the overall pressure of the sample. The mixture we will consider will involve ethanol and water.

       Vapor pressures @ 20C : H2O = 20 mm Hg         Ethanol = 50 mm Hg

The above comparison of vapor pressures immediately informs us that ethanol is "more volatile" than water (i.e. requires less kinetic energy for molecules on the surface of the liquid to escape into the vapor phase)

Yeast produce ethanol as a waste product as they extract energy from glucose (this actually requires the absence of oxygen also). If a sugar solution is allowed to ferment we can obtain an aqueous solution that will be approximately 15% by mass ethanol (mass fraction = 0.15, mass % = 15). Let us start our calculations…

In a 100g sample of this brew we would therefore have:  15g ethanol  AND  85g H2O

With regard to mole fractions, these amounts would represent:

  • Ethanol, C2H5OH, is (2*12)+(6*1.0)+(1*16) = 46g/mole
    15g * (1mole/46g) = 0.326 moles
  • Water, H2O is (2*1.0)+(1*16.0) = 18g/mole
    85g * (1mole/18g) = 4.72 moles
  • Total moles = (0.326 + 4.72) = 5.05 moles
  • Mole fraction of ethanol, XEtoh = 0.326/5.05 = 0.0646
  • Mole fraction of H2O, XH2O = 4.72/5.05 = 0.935

Now that we have worked out all this data, we can ask the question, "what is the concentration of the components in the vapor above this solution"? This question is an application of Raoult's law regarding mole fractions of components and their contribution to overall pressure of the sample:

Pcomponent = Xcomponent * P0component

Pressure = mole fraction * Pressure for pure substance

  • PEtoh = XEtoh * P0Etoh = 0.0646 * 50mmHg = 3.23mmHg
  • PH2O = XH2O * P0H2O = 0.935 * 20mmHg = 18.7mmHg
  • Ptotal of vapor = PEtoh + PH2O = (3.23 mmHg + 18.7mmHg) = 21.9mmHg

 

What is the mole fraction of the water and ethanol components in this vapor based on the above partial pressures? From the ideal gas law we can relate the number of moles, n, to the pressure:       PV = Nrt    and        n = PV/RT     Since the ethanol and water in the vapor share the same volume and temperature, the value V/RT is a constant, and n P:       Xcomponent = ncomponent/ntotal = Pcomponent/Ptotal

Therefore, the mole fractions can be determined from the ratio of the partial pressure to total pressure:    XEtoh = PEtoh/Ptotal = 3.23/21.9 = 0.147

                           XH2O = PH2O/Ptotal = 18.7/21.9 = 0.853

 Now, if we take this vapor and condense it to a liquid, what will be the mass fraction of the ethanol component? In a 1.0 mole sample of condensed liquid we would have 0.147 moles of ethanol and 0.853 moles of H2O (from the mole fractions determined above).

  • 0.147 moles ethanol * (46g/mole) = 6.76g ethanol
  • 0.853 moles H2O * (18g/mole) = 15.4g H2O
  • Total mass of such a sample would be 6.76 + 15.4 = 22.2g
  • Mass% ethanol = (6.76/22.2) * 100 = 30.5%
  • Mass% H2O = (15.4/22.2) * 100 = 69.5%

Thus, we started with a sample that was 15% by mass ethanol. The vapor, however, has a higher ethanol concentration because ethanol has a higher vapor pressure. The vapor of a mixture will always have a relatively higher percentage of the more volatile component. If we condense this vapor, we have a solution with a higher percentage of the more volatile component (in this case, we have effectively doubled the concentration of the ethanol). This is the basis of distillation (heating a liquid to produce a vapor that is enriched in the more volatile component, and then condensing this vapor to produce a liquid that will correspondingly be enriched in the volatile component).

This is also the principle behind the isolation of volatile (i.e. smaller molecular mass) hydrocarbons from crude oil.

2000 Dr. Michael Blaber
















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