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 @ 20°C

• 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
• 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, X_Etoh = 0.326/5.05 = 0.0646
• Mole fraction of H₂O, X_H2O = 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:

P_component = X_component * P⁰_component

Pressure = mole fraction * Pressure for pure substance

• P_Etoh = X_Etoh * P⁰_Etoh = 0.0646 * 50mmHg = 3.23mmHg
• P_H2O = X_H2O * P⁰_H2O = 0.935 * 20mmHg = 18.7mmHg
• P_total of vapor = P_Etoh + P_H2O = (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

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 µ P:

X_component = n_component/n_total = P_component/P_total

Therefore, the mole fractions can be determined from the ratio of the partial pressure to total pressure:

• X_Etoh = P_Etoh/P_total = 3.23/21.9 = 0.147
• X_H2O = P_H2O/P_total = 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.