[d_link]

Hi, so I've read all the other HPLC topics and available literature but still have some uncertainties which bug me =/ so please help

So.... we have been trying to analyze a component (vanillin) in our sample. As a standard we used vanillic acid (not the same I know but we didn't have anything else, since it was for practice only it's ok  ) with only one concentration (yeah I know but life's hard).

Relevant info:
Injection volume = 20 ul
Flow rate = 1 ml/min
c (vanillic acid) = 0,1 mg/ml

After we identified our component in the sampe we gathered a fraction which was supposed to contain vanillin. Then we started another run with the 20 ul of the fraction from where we obtained the peak area.

Standard area = 2955
Sample (fraction) area = 2069

Ok, from here I calculated the concentration of our component via:
Conc of unknown component =conc. of it corresponding component of standard x Peak area(sample)/Peak area(standard)

c (vanillin) = 0,07 mg/ml

Okay... but to what does it respond to? Is it the concentration of vanillin in the 20 ul of the gathered fraction? How do I get to the concentration of vanillin in my starting sample?

Obviously we diluted our "sample" when we made a run with the gathered fraction? I mean the absorbance of our peak of interest isn't the same...

Gathering the fraction run(5,8-6,5 min):
http://img685.imageshack.us/img685/2286/gatheringfraction.png

Fraction run:
http://img691.imageshack.us/img691/1553/fraction.png

Can you calculate the dilution of the sample with the flow rate or how? I'm confused  

[Mitch]

To figure out your concentration, you will need a 3-point calibration curve. You can not figure it out with 1-point.

[d_link]

To figure out your concentration, you will need a 3-point calibration curve. You can not figure it out with 1-point.

I know but let's pretend it's been properly calibrated and has a corresponding standard component. As said it was just for practice, at first we didn't have a standard at all but then they found the vanillic acid.

I know I would never do it with only one standard concentration (like any other experiment) when doing some actual work.

Please help me with the calculation =/

[Mitch]

Why did you collect a fraction? It makes more sense to just use the multiple component system and subtract your background which looks like it goes from about 400 to 300 across your peak of interest.

[d_link]

You're taking it way too serious  

We had to do everything by ourselves, the assistant told us only that they had a C18 column (not even the dimensions). So we had to find/come up with a protocol for identifying and separating a component from some plant material. The method from an article we've used doesn't even have the same dimensions of the column and it all doesn't matter since it was just and introduction to see how HPLC is working (some of us have just seen it for the first time in their life). Therefore... the point of the exercise was to just see how it all looks like. Yes I know it doesn't make sense and is wrong but that's how they do it at our university. Working conditions are really bad.

Why we collected a fraction? I suppose to determine if it really is the component we are interested it, or at least it's one of the components that the give the peak. So by running the standard if it came up on the same place we could confirm the presence of the component. I know there are more sophisticated and reasonable methods but since I didn't design this "exercise" and since materials are limited that's how we've done it. So, please bare with me. Any suggestions how it could have been done better just don't make any sense since the "exercise" is over and we have to work with what we have got. I mean to start with... using an untested method on a different column doesn't make any sense but that's how it is, unfortunately 

=/

[Mitch]

Fine.

The concentration you calculated is for the 2nd 20 uL fraction. Since this is for 'fun' there are probably a bunch of different ways to calculate your sample's concentration. Use the flow rate provided and assume you collected in the center of the multicomponent peak. Also remember to include the different dilutions you performed.

[d_link]

Yeah and that's where I don't know how to progress =/

We are diluting the "samples" by passing them through the column right? Shouldn't the peak area and the corresponding concentration I calculated respond to 20 ul diluted to the volume passing 5,8-6,5 mins (700 ul) and not the actual 20 ul of 2nd injection? That's what confuses me and what I don't understand =/

By obtaining the concentration from the 2nd injection I have the concentration of the 700 ul fraction which was diluted from the 20 ul of 1st injection, my starting undiluted sample volume?

[peterliulei]

BY THE WAY, do you need the columns...

[RobbinB]

Assuming you used the same injection volume for all injections, then the concentration in the 2nd sample you injected is just as you stated = c = (peak area sample / peak area standard) * conc std = 0.07 mg/mL

So all you really need to do is figure out what the dilution factor is in going from your first injection to your second injection.  If you indeed collected the entire peak in the first chromatogram, then you in fact got all your 20 uL injection back so your dilution factor is just flow (mL /min)* peak width (min) / 0.020 mL = DF.

(or as you stated 700 uL / 20 uL = 35)

Then just multiply that DF by the calculated concentration from your subsequent injection, ie 35 * 0.07 mg/mL = 2.45 mg/mL.

Now if you didn't collect the entire peak from the first run, but only a fraction of that peak, then you didn't get the entire 20 uL you injected back, and you will have to guesttimate what percentage of the initial 20 uL injected you did recover.