I need help with a question for a lab experiment for the sythesis of ASA. In thi

Question

I need help with a question for a lab experiment for the sythesis of ASA. In this part of the expeiment it was to test for the purity of our ASA. We weighed 0.1081g of our ASA and placed it into a 50ml volumetric flask. we added 3 ml of ethanol to dissolve. Then added 5.0ml of 5% FeCl3 solution. then added portions of distilled water to the mark on the volumetric flask.

So i need to find the molarity of SA, then convert the concentration into a weight. I must find the molarity given this equation A = (1650L/mol) x C , where A is the absorbance (measured by a spectrophotometer) My known values are A = 0.276 Molecular weight SA = 138.12 g/mol Not sure what else is needed, this experiment is the synthesis of aspirin (acetylsalicylic acid). I am trying to find the amount of salicylic acid in a volumetric flask that contains recrystallized ASA. The weight of ASA sample for the spectrophotometric analysis is 0.1081g If anyone could guide me on how to answer this question, i would really appreciate it!

Explanation / Answer

Beer-Lambert’s Law, A =e C l              - equation 1,               where,

                                                A = Absorbance

                                                e = molar absorptivity at specified wavelength

                                                l = path length (in cm)

                                                C = Molar concentration of the solute.

Assuming that your cuvette is 1 cm wide (standard path length = 1cm),

                0.276 = 1650 L mol-1 cm-1 x 1 cm x C

                Or, 0.276 = (1650 L mol-1) x C

Or, C = 1.67 x 10-4 mol L-1                              [1 molar, M = 1 mol per L = 1 mol L-1]

Thus, C = 1.67 x 10-4 M = 0.167 mM         [mM= millimolar]

Moles of ASA in the sample = molarity x volume of solution in L

                                                = 1.67 x 10 -4 x 0.050 L     [assumed: OD taken- aliquot from 50 mL flask]

                                                = 8.36 x 10-6 moles

Mass of ASA = moles x molecular mass

                                = 8.36 x 10-6 moles x 138.12 gm mol-1

                                = 1154.6832 x 10-6 gm = 1154.6832 x 10-3 mg = 1.155 mg

Mass of ASA sample taken for preparing solution = 0.1082 gram = 108.2 gm

% purity = (actual mass of ASA) / crude mass of ASA) x 100

                = (1.155 mg / 108.2 mg) x 100 = 1.067 %

Note: correct unit of e = L mol-1 cm-1. (1650 L mol-1) is a product of “e x path length”

The result is derived from the data available in the test. A variation in protocol (not mentioned in the question) may result variation in the result.

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