preview

Gas Chromatography Lab

Decent Essays

Additionally, it is also important to understand how the percentages of each of the compound is determined. It is mainly possible due to the idea that gas chromatography is utilized. That being said, the equation used in the last lab report is again going to be used here because of the fact that the gas chromatography was used again to get the integrals of the graphs (Lab 2).

The equation for that is as follows (Lab 2):

(Area of Peak 1)/(Area of Peak 1+Area of Peak 2) x 100

Thus, it is important to note that if this equation was to be used, then the graph in each fraction needed to have two peaks and if there was only one peak then the graph does not need to have integrals which needed to get plugged into the aforementioned equation to …show more content…

30 mL of the 1:1 mixture of the acetone and 1-propanol were taken in order to perform the distillation (Landrie 45). Once the apparatus was set up, the water was run through to avoid any leakage (Landrie 45). The temperature of turned up and adjusted accordingly during the experiment (Landrie 45). The temperature was recorded for each mL of distillation (Landrie 45). The entire experiment was divided into three fraction (Landrie 45). 10 mL was accounted for the first fraction, 7 ml for the next and the last 5 ml was accounted for the last fraction (Landrie 45). Therefore, after each fraction, the distillate was transferred into a beaker and at the end of the experiment there were three beakers (Landrie 45). These three beakers were used for gas chromatography (Landrie …show more content…

That being said, a syringe was used to inject a sample into the machine (Landrie 45). Simultaneously, the "collect" button on the computer was clicked and then stopped as soon as the peak were stopped forming (Landrie 45). Therefore, the graphs were scaled and printed after analyzing the integrals (Landrie 45). This process was repeated three times for three different fractions (Landrie 45). Similar to the previous lab, the integrals were used to calculate the molar percentages using the equation above (Landrie 45).

Data:

Fraction 1 and 3 had only one peak so the molar percentage was not calculated whereas Fraction 2 had two peaks therefore it will be used to find the molar percentage:

Mf of Acetone: 1.17
Mf of Propanol: 1.2
Mole percentage = (Area of Peak 1)/(Area of Peak 1+Area of Peak 2) x 100

Mole percent of Acetone = (1601(1.17))/(1601 (1.17)+2592(1.20)) x 100 = 37.59 %
Mole percent of 1-Propanol = (2592(1.2))/(1601(1.17)+2592(1.20)) x 100 = 62.4%
Raw Table: mL versus the temperature per 1 mL during distillation (collected with the lab partner)

Volume (mL) Temperature in degree Celsius
1 54
2 54
3 54
4 55
5 55
6

Get Access