You weighed out 0.020 g of your crude aspirin product in order to determine the amount of salicylic acid impurity. Following the procedure outlined in the manual, you dissolved the solid and diluted the solution to a final volume of 10.0 mL. If the absorbance of your sample solution was 1.07, what was the percent salicylic acid in your product

The correct answer is 7.8 percent.

Explanation:

As mentioned in the given question, the absorbance (A) of the sample solution is 1.07. To find the concentration of aspirin, Beer's law is used, that is, A = ebc

Here, e is the extinction coefficient, which is equal to 139.322 M^-1cm^-1 as per the standard value for salicylic acid, b is the pathlength, which is equivalent to 1 cm. Now putting the values we get,

A = ebc

c = A / (eb)

c = 1.07 / (139.322 * 1)

c = 0.00768 M

Now to determine the percent salicylic acid in the sample, there is a need to compare the value of concentration determined with the concentration of aspirin given initially.

0.02 grams is the initial concentration of aspirin mentioned in the question. The molar mass of aspirin is 240 g/mol.

Therefore, the moles of aspirin will be,

0.02 / 240 = 8.33 * 10^-5 moles

The final volume of the diluted solution given is 10 ml or 0.01 liters.

The molarity of aspirin in the diluted solution will be,

c1 = 8.33 * 10^-5 / 0.01 = 8.33 * 10^-3 M or 0.00833 M

Now, the percent of salicylic acid in the product will be,

c1 - c / c1 * 100

(0.00833 - 0.00768) / 0.00833 * 100 = 7.8 %