Question

FLUORIMETER LAB:

A 0.1019 g portion of an aspirin tablet was dissolved in sufficient water to prepare 1.000 L of solution. A pipet was used to transfer 10.00 mL of the solution to a 100.00 mL volumetric flask. A 2.00 mL portion of 4 M NaOH (aq) was added to the flask, and the resulting solution was diluted to the mark with water. The fluorescence of the diluted sample solution was found, by the working curve method, to correspond to a salicylate ion concentration of 4.80 x 10^-5 M. Calculate the % acetylsalicylic acid in the aspirin tablet.

Answer 1

Ans. The steps involved in serial dilution to prepare final aliquot (whose fluorescence was recorded) are as follow-

I. 0.1019 g aspirin tablet is diluted to 1.0 L. Let the concertation of aspirin in this solution be X molar- and let’s label it as solution 1.

II. 10.0 mL of solution 1 mixed with 2.0 mL of 4M NaOH is diluted to 100.0 mL. Let’s label this solution as solution 2.

Now, using

            C1V1 (solution 1) = C2V2 (solution 2)

            Or, X M x 10.0 mL = C2 x 100.0 mL

            Or, C2 = (X M x 10.0 mL) / 100.0 mL = 0.1X M

Hence, concentration of aspirin in solution 2 = 0.1X M

# The fluorescence of solution 2 is equivalent to the salicylate ion concertation of 4.80 x 10^-5 M.

So,

            0.1X M = 4.80 x 10^-5 M

            Or, X = (4.80 x 10^-5 M) / 0.1M = 4.80 x 10^-4 M

Therefore, [Salicylate ion] in solution = 4.80 x 10^-4 M

1 mol acetylsalicylic acid (aspirin) yield 1 mol salicylate ion. Thus, concertation of salicylate ion is equivalent to the concentration of acetylsalicylic acid (aspirin).

Therefore, [Salicylate ion] = [Aspirin] or [Acetylsalicylic acid] = 4.80 x 10^-4 M

# Moles of Acetylsalicylic acid in solution 1 = Molarity x Volume of solution in liters

                                                = 4.80 x 10^-4 M x 1.0 L

                                                = 4.80 x 10^-4 mol

Mass of Acetylsalicylic acid in solution 1 = Moles of Acetylsalicylic acid x Molar mass

                                                = 4.80 x 10^-4 mol x (180.157 g/ mol)

                                                = 0.08648 g

Since solution 1 is prepared from 0.1019 g aspirin tablet, the total mass of Acetylsalicylic acid in 0.1019 g tablet is equal to 0.08648 g.

That is, 0.1019 g tablet consists of 0.08648 g Acetylsalicylic acid.

Now,

            % Acetylsalicylic acid = (Mass of Acetylsalicylic acid / Mass of tablet sample) x 100

                                                = (0.08648 g / 0.1019 g) x 100

                                                = 84.86 %