Banana Oil Lab Report

Topics: Ester, Oxygen, Acetic acid Pages: 5 (924 words) Published: November 18, 2014

Banana Oil Lab Report

Jesse Bradford


MTWR Section

In the banana oil lab we began with isopentyl alcohol + acetic acid  isopentyl acetate + Water. We needed for this experiment a hot plate, clamps, pipette, 5mL vial, caps, hoses and a thermometer. Upon starting, our group set up an open system experiment that allowed gases to be released to avoid pressure build up. We mixed together to molecules, 1.0mL of isopentyl alcohol, 1.5mL of acetic acid and added three drops of sulphuric acid. The acetic acid was used as a catalyst to speed up the reaction. Once all the needed chemicals were added we waited for about 70-75mintues for the reaction to take place. The desired temperature for the reaction was 150oC. We also had the solution at a constant stir.

After the reaction was done taking place, we began to purification process. We used a pipette to remove the excess water and impurities that were underneath the banana oil. We removed all that was available and then began to add sodium carbonate to help wash and dry the mixture. Slowly shaking the banana oil inside the 5mL side to side, allowing CO2 to escape the 5mL vial. We did this twice making sure all the excess impurities were removed. As we had our final solution of banana oil, we used the I.R. spectra to conclude our results. The I.R. spectra showed us that the compound we produced had no peak at 3300cm-1. The banana oil peaked at roughly 1750cm-1. We concluded from these results that all the alcohol was extracted and the final product was banana oil. Purpose

The purpose of our lab was to reflux an isopentyl alcohol with carboxylic acid (acetic acid), adding three drops of sulphuric acid acting as a catalyst to produce an ester (isopentyl acetate) and water. We verified our answers using infrared spectroscopy.

Calculations for Percent Yield
To find the percent yield we had to first find moles of our limiting reagent. 1) Weight of 5mL vial + cap =28.2526g
2) Weight of 5mL vial + cap +1.0mL of isopentyl alcohol= 28.9852g 3) Weight of 5mL vial + cap + 1.0mL of isopentyl alcohol + 1.5mL acetic acid= 30.5185g
First we calculated the difference of the 1 and 2 (2-1= difference)= 0.7326g. Then, we found the difference of 1 and 3, =1.5333g. Using the molecular weight of the molecules of isopentyl alcohol and acetic acid we found the moles of each molecule. For isopentyl alcohol we calculated, (0.7326g/1) / (88.2g/mol=).0083 moles of isopentyl alcohol. The moles for acetic acid were calculated at 0.0255 mol= (1.5333g/1) / (60.1g/mol). Once we found that isopentyl alcohol was the limiting reagent, we found how many grams of banana oil were possible to produce. This number came out to be, (0,0083mol) x (130.2g/mol)= 1.08066g= theoretical yield. Our final product (actual yield) we produce 0.4127g of banana oil. We then took this number to find our percent yield, (0.4127g)/(1.08066g) x 100= 38.2%

Purification Process
For the purification process we needed to rid the solution of any excess chemicals besides the banana oil. To do this, we needed to understand a little about what makes a solution. Two factors that affect the ability of a solution to form are: 1) the natural tendency of substances to mix and spread into larger volumes when not restrained in some way 2) the types of intermolecular forces in the solution.

In the case of the alcohol and the carboxyl acid, the two molecules reacted to form isopentyl acetate and water. All of the isopentyl alcohol was reacted with acetic acid and due to the solubility between the two molecules we were able to extract the water and acetic acid. This was done using a pipette to remove the bottom layer of the solution and it was possible because the alcohol was not soluble in the water. The second extraction required the use of one mL of sodium carbonate to wash the mixture by slowly shaking it until the bubbling...
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