# How to Calculate Mole Fraction of Vapor in a Distillation Fraction

Written by brian baer
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A distillation fraction is the breakdown of multiple components' concentration at a certain temperature in a distillation column. The components concentration can be listed in mass terms (weight fraction) or mole terms (mole fraction). A mole is a unit of measure that relates atoms/elements to mass. For instance, water has a molecular weight of 8.17 Kilogram per mole. Fifty pounds of water is 2.78 moles (50/18.01).

Skill level:
Moderate

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## Instructions

1. 1

Set up a simple distillation model with water, ethanol and isopropyl alcohol. The molecular weight of water is 18.01, ethanol is 46.07 and isopropyl alcohol is 60.1. The boiling point of water is 100 degrees C, ethanol is 77.8 degrees C and isopropyl alcohol is 82.8 degrees C. Assume the liquid broth to be distilled has 50 per cent water by weight, 30 per cent ethanol and 20 per cent isopropyl alcohol and the broth is 0.454kg.

2. 2

Determine a temperature that will boil two components at the same time. A temperature of 85 degrees C would boil off the ethanol and isopropyl alcohol but not the water.

3. 3

Determine the total number of moles that are boiled up at 85 degrees C. This would assume an ideal boil-up. There are 136kg. of ethanol which is 6.5 moles (136kg. or 2090kg. per mole). There are 90.7kg. of isopropyl alcohol which is 3.3 moles (90.7kg. or 273kg. per mole) for a total of 9.8 total moles of vapour.

4. 4

Determine the mole fraction of each component in the vapour fraction. This is done by dividing the moles of each component by the total number of moles. The mole fraction of ethanol is 6.5 / 9.8 or 0.66. The mole fraction of isopropyl alcohol is 3.3/9.8 or 0.34.

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