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
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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
.73 mol x (1 L/1000 mL) x (1 mL/1.08g) x (74.4 g NaClO/1 mol) x 100% =5.03%
There were higher chances of separating two components fully if more fractions were used. As indicated in data tables 1 and 3, approximately 91% of ethyl acetate and about 99% of butyl acetate composition indicates that the distillation was quiet an efficient and successful. The butyl acetate has a very high boiling point and longer retention time because of its more Van der Waals interactions.
During our group’s distillation process, temperature and volume were recorded. The graph above shows graphed data.
The purpose of the fractional distillation of alcohols lab was to learn a new technique to separate mixtures of our unknown compound into pure components using the specific vapor pressures of pure liquids. Before the fractional distillation lab was performed, a simple distillation lab was run. In the simple distillation setup, a boiling flask attached to an adapter holding a temperature probe. The adapter connected to a condenser into which water passed through. The condenser leads to a flask for the purified liquid. Fractional distillation was a better choice for this lab over simple distillation because
First. We weight the 10-ml graduated cylinder and record the data for 37.87g. Then put 5ml isopentyl alcohol and weight it with the 10-ml graduated cylinder for 41.94g. So, we get the isopentyl alcohol weight is 4.07g. Then, we also weight 7ml glacial acetic acid and put it into the 10-ml graduated cylinder. We add 1-ml sulfuric acid and put it into the 10-ml graduated cylinder too. After we mix all of the liquid, we put a boiling stone in the flask. Next, we good connection device like the picture. Then heating under reflux. Before reflux, we should check it. First, we should wait until the liquid began to boil and the color was changing to brown. Then, continue to heat for 75 minutes. After 75 minutes, we stop it. And we need to cool the liquid to the room temperature. Then we ready 10ml ice water to cool the liquid. After cooling it, we put the liquid in the separatory funnel. And we should add 5ml saturated aqueous sodium to separate it. After first separate it. Then add 5ml saturated aqueous sodium chloride to separate it. Then transfer the liquid to a 25-ml round-bottom flask. After one week, we first weight the sample mass. Then put our sample to another group’s graduated cylinder. And weight the combine mass. Then put the sample into a distillation unit. To begin the distillation and record the boiling point range. And we should also weight the mass after distillation. We can calculate the percentage yield.
After getting the results of the 5 different concentrations using the formula c=A/εl (Lambert-Beer Law). So a new graph was created using the concentration plotted versus the time (s). The final rates of the reactions for the five different trial were first trial was 0.5 mM/s , second trial 0.3 mM/s , third trial 0.05 mM/s , fourth trial 0.03 mM/s , and for the fifth was 0.005 mM/s. the final result of the curve line was not that accurate because of the trial 4th that was a little off.
The concentrations of 2.5% and 7.5% seem to have a direct proportion. This could occur because it had yet not had equilibrium, and the betacyanins were still being diffused. In the graph of 0% concentration I received the value of
A simple distillation was carried out involving a 30 mL pre-mixed solution of methanol and 2-pentanol in order to separate the two compounds by taking advantage of the differing boiling points. Since methanol has a lower boiling point at 64.7°C, compared to 2-pentanol’s boiling point of 119.3°C, it distilled off first. Through the use of the apparatus we set up, the solution was heated on a hot plate, the vapor that came up was cooled by the condenser into a liquid and flowed down into the receiver. Temperature was recorded for every 2 mL of distillate collected. In addition, three 10 mL distillate fractions were collected to be analyzed through gas chromatography. The purpose of gas chromatography analysis was analytical for this specific
The first peak at 5.82 minutes represents the solute and the second peak at 6.04 minutes represents the solvent. It is heterogeneous since the results show more than one component in the original liquid sample. The conclusion is that the solvent was ethanol. This was because the second peak which was a retention time of 6.04 minutes matches the standard for ethanol which is also 6.04 minutes.
A measurement of around 0.5 µL, for each compound, was inserted into a gas chromatography in such a way, a syringe holding the compound was pushed down into a column and injection occurred to then start the machine in separating the compound. A computer was right beside it, indicating the graph of multiple peaks and later, after a duration of 4 minutes, a graph was printed out and obtained to then analyze the retention times and percent composition. “The percent composition was found by inserting the following values into the following equation,
The three solvents in the experiment are one hundred percent acetone, one hundred percent isopropyl, and distilled water. Acetone is nail polish remover and isopropyl is rubbing alcohol. In the experiment, it required distilled water because it is purified and is always the same substance in the container.
the mole fraction of water was plotted. From the graph, tangent lines were drawn with respect to the curve and specified points from which the equation of these tangent lines were known. By getting the y-intercepts at x=0 for the solute and x=1 for the solvent, the partial molar volume of the ethanol-water system could be
The actual concentration of the created solutions was calculated using the equation of the line of best fit for the Beer-Lambert Law plot of the points for the created solution.
In the process of obtaining the products of an elimination reaction between 2-Methyl-2-butanol with a strong acid (sulfuric acid), it was necessary to undergo two fundamental techniques of separation and analysis: distillation and gas chromatography. Distillation is one of the oldest biotechnological techniques utilized by human kind. The history of distillation is very closely tied to the history of the production and consumption of alcohol, although, some of the oldest scientific findings on the theory and applications of distillation came from Aristotle and his observations and records which were then utilized and referenced by Alexander around the 12th century [1]. Alexandrian chemists are thought to be the major contributors to the knowledge and theory we hold today about distillation and its applications.
The three percent discrepancy could have had many causes behind it. One of the main errors that could have accounted for this was human error in reading the temperature of water due factors such as viewing angle. Because of this error, assuming we misread temperature with a lower value than the actual, our molar mass is lower than the actual because there would be more moles given our value. Also another error occurred as stated in our observations was that originally some water had escaped the graduated cylinder when the parafilm was removed, in order to compensate for this our group decided to let the water go down below the 400 mL mark unknowingly allowing more butane into the graduated cylinder, thus in our calculations 400mL was used instead