post lab 9

post lab 9 - value of 135 We suspect that any impurities in...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Michele Chan CHEM 1212L Lisa James Results Salicylic acid- .200g salicylic acid Acetic Anhydride- .5ml .200g SA X (1 mol SA/138.12g) X (1 mol product/1 mol SA) X ( 180.16g product/ 1 mol product) = 0.261g .5ml AA X ( 1.58g AA/1 ml AA) X ( 1 mol AA/ 102.09gAA) X (1 mol product/ 1mol AA) X ( 180.16g product/1mol product) =1.39g Limiting reagent is Salicylic acid Mass of product is 0.169g Melting point 134-138 Rf value of Salicylic acid =.66 Rf value of Acetic Anhydride= .8 Rf value of product= one at .8 and one at .5 Percentage yield = .169/.261 X 100 = 64.7% Looking at the percent yield for the product, we got 64.7%. I believe that it is an accurate measurement as we had difficulties dissolving our reaction. We had to break up the mixture with stirring rods before it would melt and dissolve, so we might not have allowed enough time for all the reactants to react, which would account for our low percent yield. The melting point for our product , however, was right on par at 134-138 degrees Celsius which is close to the literature
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: value of 135. We suspect that any impurities in the sample are reminiscent reactants. The Rf value we got for our product showed two spots one at .8 and one at .5. We believe that the one at .8 contains some of the reactant as it has the same Rf value as that of Acetic Anhydride. We think that the Rf at .5 is representative of either un reactive Salicylic acid. The IR spectrum for the reactant salicylic acid and our product had some significant differences. There are a couple of peaks from the 3000to 3200cm-1 range for the salicylic acid, which is not present in the product, indicating that the hydroxyl group from salicylic acid is no longer present in the product. The peak at 1749cm -1 indicates the presence of an ester group, which is also not visible in the reactant. Therefore, I conclude that we successfully created Aspirin...
View Full Document

This note was uploaded on 10/05/2011 for the course CHEM 2211 taught by Professor Hubbard during the Fall '07 term at UGA.

Ask a homework question - tutors are online