Question

Question on solving concentration using area and y=mx+b. I know that y=area x= concentration, but I’m not sure what to do from there. If their is another way to calculate concentration from area, feel free to mention.

finding the concentration using y=mx tlə I have a project that had us us a HPLC machine. Iran my known stock Solution (looppm Caffeine) and diluted soluticos at 75, so, and 25ppm made from the same solution. The area that was given to me are listed below: Conc. area 25 ppm 742 Area soppn 1,5664) 75ppm 1,110.7 100ppm 1,488.9 When graphed, soppm has a very obvious spine I would like to go in and Calculate the actual concentration of the Solutions above in order to see if I should just re-run the whole sample or make another soppm solution from my stocks

Answer 1

HPLC is not an absolute method. This means that there is no way te relate areas to concentrations in a direct manner. In order to calculate concentrations, the procedure you are attempting is usually the method: obtaining the value of the signal (area in this case) for KNOWN concentrations, then graphing them and, if the relationship is linear (as stated by your y = mx + b expression, which is usually known as a calibration curve), the proportionality constant between area and concentration can be determined. This means that there is no way to calculate the "actual" concentration of your samples, since you actually need to know those concentrations to create your calibration curve, which you can then use to determine the unknown concentration of other samples.

The most you could do is to draw a calibration curve using the 25, 75 and 100 ppm values and, if that relationship is linear, use it to calculate the actual concentration of the 25 ppm sample. Also, if the 25, 75, 100 calibration curve is linear, it means you don't have to prepare them all again and you only have to make another 50 ppm solution. Your area vs concentration data (25, 75 and 100) is plotted here (commas are decimal separators):

1600 1400 1200 Area y = 9,5894x + 474,57 R2 = 0,9615 20 100 120 40 60 80 Concentration (ppm)

It's a decent linear correlation. I would try to re-do the 50 ppm sample and measure ir again, to see if it fits in the linear trend.