Question

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?
a. 6.4%.
b. 10%.
c. 7.8%.
d. 92%.
e. none of the above.

Answer 1

Answer:

7.8%

Explanation:

Given that:

The initial mass amount of aspirin = 0.020g

The standard molar mass of aspirin = 240 g/mol

Thus, the number of moles = mass/molar mass

= 0.020/240

= 0.0000833 moles

Now, the molarity of aspirin in the solution(diluted) $C_1$ = $\dfrac{8.33 \times 10^{-5}}{0.01}$

= $= 8.33 \times 10^{-3} \ M$ (provided the volume v = 0.01 L)

The absorbance of the sample solution A =1.07

The path length (b) = 1 cm

From the standard value of salicylic acid, the coefficient (e)= 139.322 /M/cm

Now; according to Beer's law, the concentration of aspirin is:

A = e×b×c

$c = \dfrac{A}{eb}$

$c = \dfrac{1.07}{139.322 \times 1}$

c = 0.00768 M

Finally, relating the concentration of the aspirin, the percentage of salicylic acid the product  $= \dfrac{0.00833-0.00768}{0.00833} \times 100$

= 7.8%