Chromatography Forum

I was wondering if someone could help me here, I've got myself really confused using multiple internal standard calibrations.......

I recently developed a method to extract acids from my samples as follows - I take 10mL of my sample, extract into 20mL of hexane and then dry down. The extracted acids are then reconstituted in 5mL of MeOH ready for HPLC analysis. To this 5mL I also add an internal standard (200uL of a 200mg/L stock solution)
As for my standards, I made up a serial dilution ranging from 10-100mg/L from which I also take 5mL each and add the same amount of internal standard as above (200uL of a 200mg/L stock solution).
I have now ran all my samples but the numbers I've worked out don't make sense (I don't think), the numbers I am getting are higher than what I expect so I was wondering if someone could enlighten me on the following -

1) Does the ratio of IS to sample and standard matter for quantification? because in my case I'm adding 0.2mL/10mL sample while it's 0.2mL/2mL in my standards
2) Ignoring the IS, if say I get a value of 45mg/L by using unnormalized area, is this the amount of acids in the 5mL sample from which an HPLC injection was made or the amount of acids in the original 10mL sample from which the acids were extracted?
3) Finally, if I wanted to determine the total concentration of acids in the original volume of say 100mL from which 10mL was taken for extraction and as in the second question the calculated acid concentration was 45mg/L, do I just work out the equivalence of 45mg/L in 100mL or I also have to take into consideration the extraction and reconstitution step as a dilution factor type calculation?

1) Does the ratio of IS to sample and standard matter for quantification? because in my case I'm adding 0.2mL/10mL sample while it's 0.2mL/2mL in my standards. I think in the standards you are adding 0.2mL to 5mL right? Which is the point at which you are adding in the sample because you've reconstituted back up to 5mL right before the addition.

2) Ignoring the IS, if say I get a value of 45mg/L by using unnormalized area, is this the amount of acids in the 5mL sample from which an HPLC injection was made or the amount of acids in the original 10mL sample from which the acids were extracted?
This is the concentration of the sample as injected. Since by your drying and reconstituting, you concentrated your sample you need to account for a dilution factor, in your case 0.5. So if you return a value as read of 45mg/L, that is actually 22.5mg/L for your original sample.

3) Finally, if I wanted to determine the total concentration of acids in the original volume of say 100mL from which 10mL was taken for extraction and as in the second question the calculated acid concentration was 45mg/L, do I just work out the equivalence of 45mg/L in 100mL or I also have to take into consideration the extraction and reconstitution step as a dilution factor type calculation? You take the concentration of your original sample as calculated in response to question 2, which is 22.5mg/L, then take the amount you had originally of 100mL (0.1L), and multiply to get a final value of 2.25mg/100mL.

Paul

Hi,

Thank you very much for your reply. Yes indeed, I add the 0.2mL to 5mL of my standards. I read somewhere that the amount of IS added is per sample which in this case I took to mean 0.2mL/10mL of sample so I thought that the amount of IS in the sample and standard will be different so this might affect my calculations. In any case, your other two answers have clarified my major issues and although this was how I originally worked it out I wasn't 100% sure about the numbers but I am now.
Thank you very much once again!