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EXPERIMENT 5 THE ASSAY OF ASPIRIN Aspirin is made by the reaction of salicylic acid and acetic acid. As a consequence of this aspirin has two acid portions, each of which can be neutralized by base. One mole of pure aspirin is therefore able to consume two moles of base. The base first neutralizes the acetic acid but releases a new acid, salicylic acid. Salicylic acid cannot be neutralized until the acetic acid is released. This new acid also has to be neutralized so a second portion of base must be added. It is therefore possible to titrate aspirin in two portions, the first neutralizing the acetic acid and the second neutralizing the salicylic acid. The reactions occur relatively quickly so speed is important when titrating the acetic acid (the first part of the titration). In theory, both titrations should take equal amounts of base since one mole of aspirin is made from one mole of acetic acid (neutralized in the first titration), and one mole of salicylic acid (neutralized in the second titration). Unfortunately this is not always the case. If aspirin is stored for long periods of time, water in the air can react with the aspirin and reform the salicylic acid and acetic acid from which it was made. If all of the aspirin was decomposed in this manner then there would be no second titration since the salicylic acid would already be released. It would be titrated along with the acetic acid in the first titration. Suppose for example that you have 1.0 mole of aspirin, but 0.2 mole of it has decomposed. What you would have left is 0.8 mole of aspirin, 0.2 mole of acetic acid, and 0.2 mole of salicylic acid. There would therefore be 1.2 moles of acid present in the first titration and not 1.0 as one might assume. In the first titration you would be neutralizing the acetic acid, the salicylic acid and the aspirin (1.2 moles of acids total), but you would be releasing 0.8 mole of salicylic acid (from the aspirin) during the titration process. You would therefore still have 0.8 mole of the released salicylic acid left to be titrated in the second titration. So even though it should take one mole of base for the first titration and one mole of base for the second titration it may take more moles for the first than for the second if the aspirin has decomposed. In this experiment you will be given a sample of aspirin to titrate and then determine whether any of this aspirin has decomposed. In addition you will calculate the random error involved in each step and the overall error in your measurements. Measurement of Errors Every procedure has an error associated with it. These errors have three possible sources, 1) Personal error 2) Method error 3) Random errors Personal errors are never reported because they can be fixed. If you make a mistake, correct the mistake and continue the experiment. Method errors are errors inherent in the way in which an experiment is done. For example, it is not possible to scrape all of the solid off of a piece of filter paper to weigh the solid. This would be an example of method error. Method errors are non-random, they are always either positive or negative (In the case of the solid, the real mass of the solid will always be larger than the amount you scrape off. Therefore this error is always positive). Random Errors Unlike personal and method errors, random errors do not have a particular value nor can they be fixed. Random errors arise because of the equipment we use is not perfect. You have noted by now that the balances "flicker" between two mass values when you read them. This is due to random error. When you read a buret sometimes your eye will be a little low, and sometimes a bit high. If your are always low or high this is personal error for which you can account. But if you have done everything you can to assure that you are reading the buret correctly, you still have to decide what that last digit is in your measurement and if you are doing everything right, sometimes you will guess high, and sometimes low. This is random error. Random error is always written as a plus and a minus. For example you might read the mass of a sample as 2.456 +/- 0.002 grams. This means that the real mass of the sample is somewhere between 2.454 and 2.458 grams, but due to fluctuations in the sensitivity of the balance you do not know which value exactly. The plus/minus values are determined by the best estimate that one can make when READING an instrument (not when adding or subtracting errors). Our burets can be read to within +/- 0.02 mL and the balances can be read to within +/- 0.002 grams. Propagation of Error Once an error is known for an individual measurement, these measurements can be added, subtracted, multiplied, and divided and the errors associated with the results of these mathematical manipulations can be reported. Adding and Subtracting Errors When two numbers are added or subtracted and each of them has an error associated with them then you take the square root of the sum of the squares of the errors. Example: Addition Add 1.2 +/- 0.02 and 2.5+/- 0.03 Answer: 1.2 + 2.5 = 3.7 +/- ? To get the error you do the following, ? = ((0.02)2 + (0.03)2)1/2 = 0.0361 = 0.04 (sig.figs) You would therefore report your answer as 3.7 +/- 0.04. Example: Subtraction Subtract 1.2 +/- 0.02 from 2.5+/- 0.03 Answer: 2.5 - 1.2 = 1.3 +/- ? To get the error you do the following, ? = ((0.02)2 + (0.03)2)1/2 = 0.0361 = 0.04 (sig.figs) You would therefore report your answer as 1.3 +/- 0.04. Multiplying and Dividing Errors When two numbers are multiplied or divided then you take the square root of the sum of the squares of the relative errors. The answer is then multiplied by the result of the multiplication or division. Example: Multiplication Multiply 3.60 +/- 0.020 by 4.00 +/- 0.030 Answer 3.60 x 4.00 = 14.4 +/- ? To get the error do the following, ? = ((0.02/3.6)2 + (0.03/4)2)1/2 = 0.00933 0.00933 x 14.4 = 0.134 You would therefore report your answer as 14.4 +/- 0.134 = 14.4 +/- 0.1 (sig.figs) Example: Division Divide 3.60 +/- 0.020 by 4.00 +/- 0.030 Answer 3.60/4.00 = 0.90 +/- ? To get the error do the following, ? = ((0.02/3.6)2 + (0.03/4)2)1/2 = 0.00933 0.00933 x 0.90 = 0.0084 You would therefore report your answer as 0.90 +/- 0.0084= 0.900 +/- 0.008(sig.figs) EXPERIMENT: Get a sample of powdered aspirin from the instructional assistant or the instructor and accurately weigh three samples of 0.40 - 0.45 gm each into 150 mL Erlenmeyer flasks. Do not dry the samples, Dissolve each sample in turn in 15 mL of absolute alcohol, add 4 drops phenolphthalein indicator, and titrate each sample quickly to the first persistent faint pink color with standard 0.1 M NaOH. Record this volume and then add, from your buret, that same volume again + 5 mL excess. Place the flasks on the steam bath for 15 minutes to allow reaction (2) to proceed to completion. Then back-titrate the excess base with your standard 0.1 M HCl. From the total titration volumes you may calculate the percentage aspirin in your sample, and a comparison of the first and second titration values will give you a qualitative measure of the decomposition which has occurred. Perform a complete error analysis for this experiment. Name Date THE ASSAY OF ASPIRIN (WORKSHEET) OBJECTIVE: PROCEDURE: Quickly titrate a sample of aspirin with a portion of NaOH using phenolphthalein as the indicator. Add an equivalent amount of NaOH plus excess. Titrate the excess with standardized HCl. Use these values to calculate the percentage of aspirin in the sample, percent decomposition, and propagate the errors involved in the titration. DATA: Molarity of NaOH = Molarity of HCl = Mass of Trial #1 Volume Volume Volume HCl Aspirin Used NaOH NaOH Start Finish Total Mass of Trial #2 Volume Volume Volume HCl Aspirin Used NaOH NaOH Start Finish Total Mass of Trial #3 Volume Volume Volume HCl Aspirin Used NaOH NaOH Start Finish Total Aspirin Assay: Sample Calculations 1) Sample mass sample wt. ±0. (= A) 2) First base titration final reading ±0.02 initial reading ±0.02 first volume ±0. 3) Second base increment final reading ±0.02 total added base ±0. ( = B - convert to liters) 4) Back titration final reading ±0.02 initial reading ±0.02 volume of HCl ±0. ( = C - convert to liters) 5) Molarity of NaOH 0. ± 0. (from bottles) ( = D) 6) Molarity of HCl 0. ± 0. (from bottles) ( = E) 7) Total moles of base consumed (B x D) - (C x E ) = ± ( = F) 8) % Aspirin = [½ x 180.15 x 100] x F / A = ± 9) Second sample: Repeat calculations in step 8 ± 10) Third sample: Repeat calculations in step 8 ± 11) Average % aspirin = (8 + 9 + 10) / 3 = RESULTS: The percent of aspirin is +/- . CONCLUSION: SOURCES OF ERROR: PROBLEMS: 1) If a portion of your aspirin sample had broken down, how would that effect the total amount of base used? What would the effect be on the percentage of aspirin in your sample? 2) Proprogate the error for the following: a) (1.51 ± 0.03) + (4.93 ± 0.05) + (2.47 ± 0.02) b) (1.51 ± 0.03) (4.93 ± 0.05) (2.47 ± 0.02) 3) In your own words describe the difference between random and method error. Include at least one example of each. Your examples do not need to be from the aspirin experiment.