Question

A student performed the synthesis of aspirin. After weighing their product, they had 0.501g of product. They took 0.250g of this product and diluted it to 250mL as instructed, and then used 10ml of that stock solution and diluted it to 100mL with 0.025 M Fe(NO₃)_3.Upon analysis with a Spectrophotometer, that solution was found to have an absorbance of .150. Using a Beer's Law plot with the equation A=0.250xC where A is absorbance and C is concentration (1x10^-5M), determine the percent purity of their sample.

Answer 1

Ans. Part 1:

Given, Absorbance = 0.150

            Beer’s law plot equation = 0.250 x C                   [where, C has unit of 1x 10^-5 M]

Putting absorbance value in above equation,

            0.150 = 0.250 x C

            Or, C = 0.150 / 0.250 = 0.6

Since, C has a unit of 1x 10^-5 M, the concentration of final solution = 0.6 x 10^-5 M

Part 2: Number of moles of aspirin in final 100 mL solution =

            Molarity x volume of solution in Liter

            = 0.6 x 10^-5 M x 0.100 L              [1L = 1000 mL]

            = 0.6 x 10^-6 moles

Since, this solution is prepared from 10 mL of previous solution (250 mL, stock solution), the 10 mL volume of the stock solution also contains 0.6 x 10^-6 moles of aspirin.

Part 3: 10 mL of the stock solution has 0.6 x 10^-6 moles of aspirin.

Now,

            Since, 10 mL of stock solution has 0.6 x 10^-6 moles aspiring

            Or,       1 mL     -           -           - [ (0.6 x 10^-6) / 10 ] moles

            Or,      250 mL    -           -           - [ (0.6 x 10^-6) / 10 ] x 250 moles

                                                            = 1.5 x 10^-4 moles

Thus, total number of moles of aspirin in 250 mL stock solution = 1.5 x 10^-4 moles

Part 4: Since stock solution is prepared using 0.250 g of the sample, the total number of moles of aspirin in 0.250 g sample = 1.5 x 10^-4 moles

Mass of pure aspirin in 0.250 g sample = moles of aspirin x molar mass of aspirin

                                                            = 1.5 x 10^-4 moles x 180.157 g mol^-1

                                                            = 2.702355 x 10^-2 g

                                                            = 0.02702355 g

% purity of aspirin = (Actual mass of aspirin / Mass of sample taken) x 100

                                                            = (0.02702355 g / 0.250) x 100

                                                            = 10.81 %

Thus, % purity of aspirin = 10.81 %