Question

I am working on a HPLC lab where we are to determine the concentration of caffeine in two energy drinks. I am trying to figure out the concentrations but don't think I am doing it right. Below is the info from our chromatograms:

Standard Caffeine peak area
blank 0 mg/100mL 0
Std 1 2.5 mg/100mL 3,039,775
Std 2 5 mg/100mL 7,913,698
Std 3 10mg/100mL 14,790,962
Unknwon 1 unknwon 36,474,730
Unknown 2 unknown 9,784,763

unknown 1 was prepared by diluting 2 mL of the unknown in 8 mL of a 20/80 water/methanol mix. Unknown 2 was prepared by diluting 5 mL of the unknown in 5 mL of the same mix.

When I plot the data in excel I get a equation for the line of y = 753789x - 913332. If I try and figure out the concentrations of the 2 unknowns just by using the graph I am getting a totally different answer then if I plug the peak areas for the unknowns in to the equation for the line. I don't know what I am doing wrong.

I am assuming the better way to go would be to use the equation for the line. If not please let me know. If so should I be doing something else beyond just plugging the peak areas for the unknown in to the equation for the line? If so, what and how do I do it? HELP!! I have spend severl hours trying to figure this out.

Answer 1

I didn't plot the data and I can't verify the equation you have for the line; however, just looking at the data suggests that unknown 2 must have a concn of about 7 mg/100 mL [I used 5 mg/100 x (about 15/about 8) = about 9 mg/100 mL) and 10 mg/100 mL x (about 8/about 15) = about 5 mg/100mL with the average between the two of about 7 mg/100 mL). That should be in the ball park anyway. That's the concn on the column since the standards were made from the column. Now unknown 2 was diluted by a factor of 2 (from 5 mL to 10 mL; therefore, the final concn is about 7mg/100 x (10/5) = about 14 mg/100 mL.
For unknown 1, I note that NO standard is as high as unknown 1; therefore, you must extrapolate the line to more than double your highest standard. Frankly, I wouldn't like to do that. If I had a choice I would run another unknown 1 of a different dilution. Since the first dilution gave a value more than double, I would make a new dilution at least 1:10 and perhaps 0.5:10(1 mL up to 20 mL of the mix) would be better.

Answer 2

Awesome. Thank you! I can't run another unknown but when I plug the peak area for unknown 2 in to my equation for the line I get a concentration of 14.19 mg/100mL so I will just plug in the peak area for unknown 1 to get my concentration for that unknown....unless I shouldn't. Please advise if I shouldn't