PLQ Aspirin - Post lab PDF

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Aspirin Synthesis and Analysis Post-lab Questions 1. Discuss how the following instances would affect the results of this synthesis? a. The product was washed with hot water during filtration instead of cold water. Aspirin is more soluble in warm water than in cold water which would cause a loss of product during the filtration process. Therefore, using cold water allows excess reactants to be washed away rather than losing aspirin. b. The aspirin was left out to dry for weeks, resulting in a vinegar-like smell. Write the reaction responsible for this. Allowing the aspirin to dry out for weeks, a hydrolysis reaction occurs which causes the production of acetic acid, resulting in the vinegar-like smell. C9H8O4(s) + H2O(l)  C7H6O3(s) + C2H4O2(aq) 2. We used the concept of limiting reagents in this experiment. a. Why was salicylic acid chosen as the limiting reagent? Discuss several advantages of this choice. Salicylic acid was chosen as the limiting reagent due to the many advantages of it for results and price. First off, according to the Pre-Lab Quiz, salicylic acid is more expensive at $6.56 for the 2 grams needed versus $0.26 for the 5mL of acetic anhydride. Therefore, using less salicylic acid would be more cost effective. Also, according to UMLS using acetic anhydride as the excess reactant allows the water to react with the acetic anhydride and will increase the yield of aspirin. 3. a. A student performing this experiment misread the amount of reactants and added 0.50 mL of acetic anhydride instead of 5.0 mL. Did this procedural error affect the theoretical yield? Explain and support your answer. If instead of adding 5.0mL, 0.50mL of acetic anhydride were added, then acetic anhydride becomes the limiting reactant instead of salicylic acid. Because of this, the theoretical yield of aspirin depends soley on the amount of acetic ahydride added. This changes the theoretical yield from roughly 2.165 grams that were dependent on the amount of salicylic acid added (2.005 grams in our experiment) to a theoretical yield of 0.959 grams of aspirin. 4. How many hydrogen atoms does each of the following organic molecules have?

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5. In this experiment, we used spectroscopy to analyze the purity of your aspirin.

a. The upper range for abs. measurements is around 1.0. Why is this (what is happening chemically)? What happens to the graph when the absorbance is large? When an absorbance measurement is greater than 1.0, the concentration is too great and the molecules are too packed together to get an accurate reading. According to Vernier, at an absorbance of 2, 99% of the light is being absorbed by the sample and at 3AU, 99.9% of the light is being absorbed. The distinction between the two is very hard to make and cannot accurately be made. Therefore, in order to get a more accurate reading (an absorbance level under 1.0), the solution should be diluted to a lower concentration. b. Why is it necessary to collect a calibration or standard curve when using the absorption spectroscopy technique in lab? Specifically, in this lab, having a standard curve allowed us to compare the solution’s absorbance value to the slope of the standard curve to find the concentration. c. Discuss one advantage and one disadvantage of absorption spectroscopy. According to Petro-Online, an advantage to using absorption spectroscopy the low cost. Since it provides quite accurate results, the cost being lower provides a huge advantage compared to other methods. A disadvantage, however, is that only solutions can be analyzed and solids cannot. 6. A rock containing copper is pulverized, and all copper is extracted from it with nitric acid. To make a standardization plot, you make 5 standards and measure their absorbance. The graph you obtain has a trendline of y=2.4238x + 0.0141. a. You used accidentally used a non-standard cuvet where L=1.20 cm. What is the value of epsilon? Write the Beer’s law from this plot. A=ϵLc = 2.43238x + 0.0141 Slope = ϵL = (ϵ)(1.2cm) = 2.43238  ϵ = 2.027 b. You take the absorbance of your sample. However, the absorbance you measure is too high, so you take 7.00 mL of your 10.00 mL solution and add 3.00 mL of DI water. The absorbance reading of this new solution is 0.669. How many moles of copper ion were contained in the ore? A = 0.669 = 2.027(1.2)(C) + 0.0141 C = 0.269M 0.269M (0.01L) = 0.00269 mol (7.00mL) (0.00269 mol) / (10.00mL) = 0.00188 mol Cu...