[groover]

Hi, I am an undergraduate Chemistry student who has taken on a major HPLC experiment independent study and I need a bit of guidance from those who... well... know what they're doing!!!

I have created a standard solutions and have created a calibration plot for my experiment. I am now trying to reflux solids into a sample solution. Now here is my problem:

For instance, I'll take 30g of a solid and solvate it with 100mL of a liquid. I'll then filter and sequence it on the HPLC and get a nice, clean peak. From my standards, I know exactly which peak it is in the chromatogram. However, when it comes to calculations, I'm having a problem determining concentration in going from the chromatogram of my sample and plotting it on the calibration curve.

What calculation do I need to be doing to correct for the dilution of the solid to create the sample solution???

Thanks so much, in advance.

[Borek]

Give more details. Seems like either you are missing the most basic things, or you have not posted everything.

Or both

[groover]

I'm working on a method to determine capsaicin in a solid sample.
So my standard has an area of about 15mAU and when I run a sample, I'll get an area of about 9mAU. However, I know I can't calculate a concentration of the capsaicin in the sample because there is a dilution that I'm doing when I put the solid into a solution. I'm so lost!

Long story short, I know there is a formula for HPLC with the peak area, volume and mass of solution components, and dilution factor for sample and standard when calculating  HPLC concentration with a standard. I believe its a %weight formula.  Is this what I'm looking for?

[groover]

Okay, Let me re-ask this in a diffierent way.
I am using an official HPLC method. At the "calculation" section, it shows a calculation to determince concentration, but its based upon the weights and volumes outlined in the method, which don't quite serve the purpose that I have here.
(peak area sample/ peak area standard)*(concentration of standards solution/weight of solid used in the standard) *(200/0.98)*9300

Now, I know that the end numbers are specific to the method and this is just an altered form of an original equation that just states volumes, masses, dilution factors, peak areas, etc.

Does this make more sense?