Stumped by HPLC

[ravendragonwing]

Hello all,
I have a problem and I am feeling extremely stupid, because I cannot solve it.
Background: boss decides I am to perform an HPLC analysis. Fine, have never run Agilent software or instrument. Whatever, figured this out, ran my sample.......now what?
I am supposed to calculate the % of the analyte in the sample. I am also to calculate the % recovery for the spike sample I prepared.
Never have done these before and I have NO formula. And the only procedure I have is an AOAC method. I am also using 2 older sample runs (just what the analyst wrote up, not the actual files). Both analysts are no longer here and they are both different. *head, desk*
Sample: filtered into HPLC vial
Spike: 10mL of filtered sample + 910uL of a 10mg/ml standard, diluted to 50mL. 550uL of this solution was then diluted to 1 mL and injected. (no idea why I did this, it is what a prior analyst did)
Results:
Sample peak area: 6400
Spike peak area: 950
????????
how did that happen? and yes, I did check vial placement and my sequence table to ensure I did not switch the vials. I think I need to add a dilution factor to the formula....
Any info I am missing to calculate the amount of analyte or % recovery of the spike?
Help appreciated greatly. And please keep it simple, as I do not run HPLC ever and have little to no background.
Thanks!

[tom jupille]

In order to explain this appropriately, it would help to know what your general experience/skill level is.

[ravendragonwing]

In order to explain this appropriately, it would help to know what your general experience/skill level is.

little to no background......I understand the concept of HPLC, what to do with the answers.....no idea.
I have run HPLC in the past (10+ years ago) and only for qualitative, not quantitative.

[WillyOne]

Lets suppose that your sample is 1mg/ml;
It means that you have 10 mg (from sample: 1mg/ml x 10 ml) plus 9.1 mg (from standard: 0.91ml x 10mg/ml) in 50 ml.
It means a conc of 0.38 mg/ml (19.1mg/50 ml)
later dissolution: 0.55 ml to 1 ml, so 0.38 x 0.55/1 means that your spiked sample is 0.21 mg/l.

That I think is that, depending of your sample conc you may have diluted to five times (or more) your spiked sample.

From your results: 6400/950= 6.74, you have diluted your sample 6.74 times, so your sample should be lower than 1 mg/ml.

Sorry I will continue in a second post in a few minutes

[WillyOne]

2nd part:
So your spiking procedure makes no sense.
If you are using an AOAC method look for it. It should explain how calibrate the method (10 mg/ml seems too high)

Prepare calibration standard(s) as per AOAC method and build up a calibration straight line representing Response in front of concentration (mg/l) of Std.
With an Excel sheet calculate the equation of that straight line: Response = a(conc) + b ( y= ax+b)
Prepare your sample and inject it.
Then calculate conc = (response-b)/a
Spike your sample in a way that you increase the concentration around 10 to 20 % and inject it.
Calculate the concentration of your spiked sample.
Calculate recovery as Conc found in spiked sample- concentration in sample x 100/ Conc added

Or something like that...