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Experiment 6, Acetone / n-Hexane Azeotrope, November 28, 2012, Landon Mutch
Introduction
In this experiment, a solution of incrementally-varied composition of acetone and n-hexane was rapidly
refluxed, and its boiling temperature and refractive index were measured at various compositions.
These boiling temperatures and refractive indexes were then graphed, and the plots were used to
determine the presence and location of the azeotrope of the mixture.
Procedure
The procedure followed in this experiment is outlined in the Lab Manual [1]. L. Mutch obtained the data
in Table 1, while E. Jones [2] obtained the data in Table 2.
Data
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Table 3 - Calibration of Thermocouple
Thermometer used
Calibrated thermometer
Temperature difference
L. Mutch Temp. /oC
57.5
56.5
-1.0
E. Jones Temp. /oC
61.2
63.0
+1.8
density of n-hexane = 0.659 g/mL (at 25oC) [3]
density of acetone = 0.791 g/mL (at 25oC) [4]
nD pure acetone = 1.3587
nD pure n-hexane = 1.3771
nD pure water = 1.3328
Calculations
corrected temp = boiling temp – 1.0 oC = 58.8 oC – 1.0 oC = 57.8 oC
corrected temp = boiling temp + 1.8 oC = 70.1 oC + 1.8 oC = 71.9 oC
molarity of n-hexane = density/molar weight = (0.659 g/mL)/86.18 g/mol = 0.007646785mol/mL
molarity of acetone = density/molar weight = (0.791 g/mL)/58.08 g/mol = 0.01361915mol/mL
Table 1:
mole fraction of n-hexane = molarityn-hex*volumen-hex/(molarityacet*volumeacet) + mole fractionprevious
= 0.007646785mol/mL*0.25mL/(0.01361915mol/mL*20mL) + 0mol/mL = 0.007018414 = 0.00702
Table 2:
mole fraction of n-hexane = mole fractionprevious - [molarityn-hex*volumen-hex/(molarityacet*volumeacet)]
= 1.00 - [0.007646785mol/mL*1.0mL/(0.01361915mol/mL*40mL)] = 0.9859632 = 0.986
mole fraction of condensate determined by comparing RI’s to Table 1 in Lab Manual [1]
Results
From Graph 1, the azeotrophic mixture can be seen to have a boiling point of 52oC and a 0.65 mole
fraction of n-hexane. The azeotrope is at a minimum, so we can say that acetone/n-hexane interactions
are weaker than acetone/acetone or n-hexane/n-hexane interactions [1].
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Graph 1 -
Discussion
At the azeotrope, the compositions of the vapour and the liquid are the same, unlike the other areas of
the graph, where the compositions of the liquid and vapour differ. However, over all the temperatures,
the vapour and liquid compositions remain in equilibrium. Discovery of the azeotrope is important
because here fractional distillation is no longer able to separate the two components, as they are
present in the vapour in constant proportions. As discussed in the results, the presence and location of
the azeotrope can also tell us about the interactions between the components. We can conclude that
the substances are dissimilar because of their differing interactions, and also that the interactions
between acetone/n-hexane are weaker than those between acetone/acetone or n-hexane/n-hexane.
Note that two data points have been excluded from the best fit curves. As the liquid was refluxing
rapidly, the sample of condensation should have been the same as the vapour; so the errors are likely
due to Refractive Index misreading’s.
In this experiment, a mixture of acetone and n-hexane was continuously refluxed at various
compositions, and the resulting data was plotted to determine the presence and location of the
azeotrope. This allowed us to draw the discussed conclusions about the mixture.
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Works Cited
[1] A. Taylor, Chemistry 245 Laboratory Manual, Victoria, BC: University of Victoria, Fall 2012.
[2] E. Jones, Lab Partner.
[3] "Sigma Aldrich," 2012. [Online]. Available:
http://www.sigmaaldrich.com/catalog/product/sial/650501?lang=en&region=CA. [Accessed
November 28 2012].
[4] "Sigma Aldrich," 2012. [Online]. Available:
http://www.sigmaaldrich.com/catalog/product/sial/296090?lang=en&region=CA. [Accessed
November 28 2012].