Experiment 3B Fractional Semi-microscale Distillation Separation of Hexane and Toluene
Experiment 3B: Fractional Semi-microscale Distillation Separation of Hexane and Toluene. Objective: The main goal of this lab is to learn how separation of binary liquid mixtures is performed. Especially when the two liquids have boiling points varying by about 30° C. Hexane can be separated from toluene in this experiment because of the difference in their boiling points. Since toluene has a higher boiling point, it will left at the bottom while the hexane starts to boil out and collect in the Hickman still.
GC measurements help us in determining how accurate our data is by making a graph of the amount of hexane and toluene in each fraction. Also this lab gives experience with semi-micro distillation in order for experiments in the future that might need this purification setup. Experimental Procedure: This lab follows the experimental procedure of the book as outline in the Mayo book on pages 133 and 134. However; the following modification were made to the procedure: Use gas chromatography instead of refractive index
Mixed exact quantities in 5mL conical vial
Temperature moved really slowly but kept boiling throughout. Collected Fraction 1 at 35°C
Temperature dropped after a short while from 55°C to about 45°C. Larger amounts of vapor/liquid during Fraction 2 collected. Started collecting fraction 3 at 0°C.
Peak 1: distance from starting point = 1.2cm, therefore time = 1.2 minute. Peak 3: distance from starting point = 4cm therefore time = 4 minutes. Fraction 3:
Peak 1: distance from starting point = 1.3cm therefore time = 1.3 minute. Peak 3: distance from starting point = 4cm therefore time = 4 minutes Peak Area:
Peak 1: Peak Height measured at 116mm and 3mm width. Therefore, Peak Area = (116*4)/2 = 174 mm2. Plus the small isomers curve with height 13 mm and width 2mm should give entire peak 1 value. 174 + (13 *4)/2 = 200mm2. Peak 2: Peak height measured at 11mm and width at 5mm. Peak Area = (11*5)/2 = 27.5 mm2. Fraction 3:
Peak 1: Peak height 8.5mm and 1mm width. Peak area = (8.5 * 1)/2 = 4.5mm2. Plus the small isomer curve with height of 2mm and width 1mm should give the entire peak 1 value. 4.5 + (2*1)/2= 5.5 mm2. Peak 2: Peak height 77mm and width of 6mm. Peak area = (77 * 6)/2 = 231 mm2.
1. In the Data and Results Table, you have identified the peaks in your gas chromatogram. Explain how you made your assignments and what physical property determines the order of elution (which component elutes first, which component elutes last, and why?) a. Since Hexane is lighter and has a lower molecular weight than compared to toluene, it is expected to be faster than toluene. Therefore; when added into the GC, it is expected that the light molecules of hexanes will show up first on the charts. In the GC, the hexanes would be able to move through the system through much quicker than the toluene. 2. On a gas chromatogram, what do the peak areas represent?
The peak areas can be viewed as looking at the densities. By looking at the amount of area it covers we can approximate the amount of hexanes in gaseous form that was injected into the GC. 3. The GC data may be used to verify the results of the distillation (was there actually any separation of the hexanes/toluene mixture?). a. For each fraction collected, what do the GC data (peak identities and peak areas) tell you about the progress of the distillation?
Be sure to comment on the relative peak areas of the two peaks observed for each fraction. It helps show that the two compounds both have different temperatures under which they boil. For instance hexanes’ boiling occurs at a lower heat setting so that is why in the first fraction we can view a greater peak of hexanes than toluene. In the third fraction we notice that toluene is found in higher abundances there. This helps when trying to distill out or separate the two compounds. b. Are the GC data consistent with your expectations?
The GC data is what is expected, however there were some difference noticed with it than expected. For instance, the first fraction came out with more toluene that was originally expected and the graph showed this. Small unexpected material like this was the only thing not expected with the GC data. 4. Most of you did not have enough time to complete the redistillations of Fraction 1 and Fraction 3, but you are still equipped to anticipate the results. a. For the redistillation of Fraction 1, predict how the gas chromatogram for the initial fraction would have differed from the gas chromatogram of Fraction 1 from Distillation 1. The peaks would be more pronounced and there would be a greater variance in the size of the peaks.
Peak 1 would be much higher and steeper and peak 3 or the toluene peak would be found much smaller. b. Similarly, for the redistillation of Fraction 3, predict how the gas chromatogram for the final fraction would have differed from the gas chromatogram of Fraction 3 from Distillation 1. b.i. The difference between the re-distillation of fraction 3 and the initial distillation of fraction 3 would show some variance but overall the expected results would still be viewable. There would be a slightly lower peak for hexane and the amount of toluene would have a higher peak as compared to initial distillation of fraction 3. 5. Explain why Distillation 1 did not give two pure fractions. Suggest at least one method to improve the separation efficiency of the distillation of hexanes and toluene. a. Since we are boiling away a chemical mixture, there might be toluene that is able to heat up and boil before hexane does. No matter what we do, toluene willn still sometimes boil with the hexane at a lower temperature, but by increasing the surface area of the Hickman still or by adding more boiling stones, which also functions to increase the surface area, we can achieve greater separation efficiencies during the distillation of hexanes and toluene.