Question

A testing laboratory has been hired by a company called “Drug Company Q” to analyze a series of over-the-counter drugs that the company produces. In these over-the-counter drugs, the active ingredient is called “Active Ingredient M.” The laboratory technician collected the following data from samples with known concentrations of the Active Ingredient M. That data is shown below in Table 2. Convert %T to absorbance (A=2-log(%T)) and prepare a Beer’s law plot using this data.

Table 2. Known Concentration of M Drugs.

Sample Identification Code	Sample Concentration (M)	%T
Q5000	4.00 x 10-4	17.9
Q5001	3.20 x 10-4	25
Q5002	2.40 x 10-4	35.7
Q5003	1.60 x 10-4	50.2
Q5004	8.00 x 10-5	70.8

The technician also collected absorbance readings for the 5 over-the-counter drugs for review. The data collected for the 5 over-the-counter drugs is shown in Table 3.

Table 3. Absorbance Data for Over-the-Counter Drugs.

Sample Identification Code	%T
M21050-1	43.7
M21050-2	44.1
M21050-3	45.8
M21050-4	42.1
M21050-5	30.1

A.       Create a Beer’s law plot and best fit line for the data in Table 1.

B Use the Beer’s law plot and best fit line to determine the concentrations for samples: M21050-1, M21050-2, M21050-3, M21050-4, M21050-5.

C.       The company reported that sample M21050-2 has an M concentration of 0.0003M. Assuming that the results in Question C are 100% accurate and without error, is the company’s statement accurate? What is the percent error between the reported concentration and the concentration calculated in Question C?

E. By law, Drug Company Q must have an M concentration between 2.85 x 10^-4 M and 3.15 x 10^-4M. Do all samples analyzed meet the legal requirements? Use the information from Question C to explain your answer.

I am SUPER confused on the last two parts? *UPDATED

Answer 1

A) Prepare the following table.
Sample Identification Number   Sample Concentration (M)   %T    A = 2 – log(%T)
Q5000   4.00*10-4   17.9   0.747
Q5001   3.20*10-4   25.0   0.602
Q5002   2.40*10-4   35.7   0.447
Q5003   1.60*10-4   50.2   0.299
Q5004   8.00*10-5   70.8   0.150
Plot A vs sample concentration as below.

B) The regression equation is used to determine the concentrations of the unknown samples. The concentrations of the unknowns are obtained as below.
y = 1871.3x – 1.00*10-4
x = (y + 1.00*10-4)/(1871.3)
Put the absorbance values as y and determine x.
Sample Identification Number   %T   A = 2 – log (%T)   Concentration of the unknown sample, M
M21050-1   43.7   0.359   1.919*10-4 ≈
1.92*10-4 (rounded to 2 sig. figs)
M21050-2   44.1   0.355   1.90*10-4
M21050-3   45.8   0.339   1.81*10-4
M21050-4   42.1   0.376   2.01*10-4
M21050-5   30.1   0.521   2.78*10-4
C) The concentration of the unknown sample 3 is 1.81*10-4 M = 0.000181 M. The company claimed that the concentration of the unknown sample 3 is 0.0003 M. Since the calculated concentration is less than the claimed value, hence, the company’s claim about the concentration of the unknown sample 3 is inaccurate.
Percentage error (assuming the true concentration of the unknown sample 3 should be 0.0003 M)
= (0.0003 M – 0.000181 M)/(0.0003 M)*100
= 39.66% ≈ 39.7% (ans).
D) The legal requirement is that all the drug samples must have a concentration between 2.85*10-4 M and 3.15*10-4 M. The 5 unknown samples have concentrations lower than 2.85*10-4 M and hence, the drug samples do not meet the legal requirement.