It should be possible to predict peak areas detected via UV/Vis from a knowledge of the molar absorption coefficient of an analyte and the flow rate, etc... Now, LR Snyder, JL Glajch, JJ Kirkland, in "Practical HPLC Method Development", Wiley, 1988 describe an equation to estimate the minimum detectable mass (to be injected), based on the absorption coefficient, plate hight, etc. Rearrangement of this equation can lead to a formular for the prediction of peak hight. Converting it so that peak area could be calculated proved a bit "hairy", mildly put. As a matter of fact, I am not sure that everything I tried was legitimate. The same holds for an attempt to derive a formular for the area using various forms of the equation describing a normal (Gaussian) distribution, etc.
Is there anybody out there with a better mathematical acuity than I can muster, who has successfully done this or who knows where to find a derivatization? A derivatization of the equation by Snyder, et al, would be very instructive as well.

Incidentally, this has implications for the former discussion "Problem with absorption coefficients" started on Feb. 26. (For instance: One can derive the absorption coefficient using HPLC detectors in a normal chromatographic run. The calculation of specific rotation from HPLC data has some similarities: http://www.pdr-chiral.com/specificrotation.htm .)

Note: this was erroneously (better: stupidly) posted in the news board, I will try to better myself)

Hans

Hans,

Here is a short version of what I think you want. If there are any difficulties with it, let me know.
The signal is determined by Beer's law:
S = E*d*c
E = epsilon = absorption coefficient, d is the cell length and c is the concentration
The mass can be obtained by integration over the concentration with respect to the volume. I cna simplify this by assuming that the peak is a triangle with a height c(max) and a width w = 4*sigma (the standard deviation of the peak).
m = 1/2*c*w
Plugging the c from the first equation into this equation yields:
m = 1/2 * w * S /(E*d)
the detector gives you the signal in absorption units (if not, you simply convert it from mV), you can measure the width, you know the cell length, and you can look up the extinction coefficient. Everything should work. Or you take the peak area, which commonly comes in mV*sec, and convert it in an analogous way.

Uwe: Actually I am more after predicting the area before performing the HPLC, knowing the mass injected, your E, and the flow rate, and not knowing sigma or the peak hight. That is, I am after your S in an equation which must also include the flow rate.

Better yet: A formular which can be plucked into a spread sheet that closely approximates the peak one should get on the HPLC. VR Meyer in LC.GC Int, October (1994), p. 590 does this with hypothetical peaks, I can´t find the handle to convert this to predicting a peak knowing the factors mentioned above + maybe the plate hight of the column.

Hans,

I not sure what you are after is possible. Wouldn't peak area be a function of cell volume in the same way that it is a function of flow rate. Not to mention detector time constant, etc.

I think any formula will have to be empirical.

Hans, this is not impossible, but there are a lot of steps in it. First, you got to know what the sample is. More specifically, you got to know its molecular weight. Then you need to define the mobile phase, more specifically, the viscosity of the mobile phase. From both of these things, you can get the diffusion coefficient. Now you need the flow rate, the particle size, and the column length. With this, and the equation for the reduced plate height, you can calculate the plate count. From this, the column dimensions and the retention factor, you can calculate the peak width in volume units as the peak leaves the column. To this, you may need to add extra-column bandspreading from injection, detector and tubing. All of these things are well understood, it just takes some time to put it all together.
I did this a while ago for the more complicated situation of gradient elution. It is all actually very straightforward. It may sound complicated, but in principle it is no more difficult than adding 1+1.
I won't have time at the moment, but I can do this for you if you don't need it for a month.

Rereading your question, I think I went overboard with my answer. Let me try again:

mass = concentration * volume = (AUFS/(E*d)) * (F*t)

F is flow rate, t is the time, AUFS = Absorption units full scale, E = the absorption cofficient and d = cell length.
AUFS = R x mV where R is the conversion factor from millivolts to AUFS.
I think that should do...

Uwe
Time is not a problem.
Are you talking about a formular for predicting area or predicting the peak in a spread sheet? For predicting only area it seems to be a bit too involved?
Do you, or anybody else see how Snyder´s, et al, equation is derived? I will try to give it here:

W = (1000M(k+1)(S/N)(No)L^0.5)/E(Lc)d^2N^0.5

w is the minimum mass needed to get a specified S/N the signal to noise ratio, M is the mol. wt. No is the detector baseline noise (so the same as N in S/N) k the retention factor, L the column length, E the abs. coeff., Lc is the flow cell length, dc is the column diameter, N is the column plate number (the last N).
If one sets S = expected peak hight and W = the injected mass one can use this equ. to predict peak hight.

Ups, our contributions crossed, Uwe. I still have a problem with your concentration term, AUFS/Ed. The AUFS (better µV) are the area of the peak and AUFS/Ed are then the area in conc.... makes sense.
I modified a form of a Gaussian equation and got your equation with the t missing, that´s what made me suspicious.
Too bad, if books gave more mathematical derivs one could understand things much more easily.

(The mentioned Gaussian distr.:

C = (M/Fs(2pi)^0.5)e^-1/2((tr-t)/s)^2

from WJ Lough, et al, HPLC, Fundamental Principals a. Practice, Blackie Academic Professional, London, 1995. The s should be sigma.) Note that the e term will = 1 for t being equal to the retention time, tr. The F, M (mol wt.) etc., must then come from the reasoning behind your equation.

You really don't need all the Gaussian mumbo jumbo. This is why I had approximated the peak by a triangle. As the volume, take half of the peak width. AUFS is still in absortion units full scale:
mass = concentration * volume = (AUFS/(E*d)) * (F*t)
To get it into mV, this is where the conversion factor from 1 AUFS to 1 mV is coming from:
mass = concentration * volume = (AUFS/(E*d)) * (F*t) = (mV * R/ (E*d)) * (F*t)
The time t is half the peak width at the bottom of the peak.
PS: I doubt that I will be at a computer for several days.

Letting it simmer over the weekend my problem with this became clear:
Wanting to predict the peak area one does not know t. The rest in your formular, Uwe, is known.

Also, I seem to have a mental block as to why the volume is based on the area within the peak (1/2tF), rather than the whole volume (tF) of mobile phase in which the eluate emerges (t here is the peak width at peak bottom). This 1/2tF was also used in the link which I gave above.
Also, I agree that one shouldn´t complicate things visually, here with the normal distribution instead of triangulaion, but if one sticks the formular in a PC one might as well do away (or almost) with approximations.

Tom, since peak hight can be predicted
(using Snyder at al, or the link above) area should be even easier to predict. Now whether this is theoretical, empirical, semiempirical, etc., is not possible to know if one does not know, in detail, how such formulars are derived.

Hans,

I think I understand your question now. For a fixed flow rate I think area will be independent of all of that plate count and Gaussian distribution stuff.

For a thought experiment imagine a system with no column no band broadening and a zero volume detector cell. Let’s say you prepare a 1mg/ml solution of a compound and use a syringe and fill the detector flow cell and it produces a response of 1AU or 1000 mAU. Now we make a 10 microliter injection of this solution at 0.6 ml/min (10 uL/sec). Our hypothetical system will have a square wave response of 1000 mAU that lasts 1 sec. This will be the peak area (1000 mAU*sec) regardless of how you distribute it.

Area = Response * Time

Response = ext. coeff * path length * concentration of sample

Time = volume of sample/flow rate

Thanks Tom for reasserting that area depends on only simple factors. (Strange, I suggested similar reasoning in an earlier chain. . . .) Now Uwe´s reasoning also appears to be correct, the diff. being that he used AUFS and you used AUFS*t.
Also, to use your example, but inject the same amount in 100µL would mean that the peak is 10 times longer, but only 1/10 as high and the area is still 1000.... Extending this it seems reasonable? to just use a unit volume, 1mL, for the sample and get (partially using Uwe´s terms for consistency):

Area (in AU*time) = Ecd*1mL/F

This is it!? if one gets the units straight?
Now I am also beginning to grasp why I was hung up on Gauss, etc.: I found only Gaussian type formulars in the lit. which I used to arrive at Area, finally substituted Gaussian stuff back into these equations and came up with the above, but didn´t belive it.

Once again in more easily remembered form:

Area = AUFS*time = k*µV = abc/F

where the area given by the PC is µV, k is the ratio of AUFS*time/µV
a is the absorption coefficient
b is the length of the light path in the cell
c is the concentration of the sample
F is the flow rate.

Have to try this.

But: Anybody see how the formular by Snyder, et al, was derived?

does anyone know the equation used in Fluorescence detection to determine concentrations. I am thinking along the lines of the Beer Lambert law but have been told that it is called the Kubelke-Muncke equation. I have searched everywhere for this but have not had any success. The name may have been distorted with time but I am sure that it should be freely avaialble somewhere

colin.crowley, there is indeed an equation, though I never saw a name attached to it. One simply gets it by combining Beer-Lambert (for the excitation part) with the formular of fluorescence efficiency (fluo. quantum yield, for the emission part). Like the equation for transmittance it has this log function in it. A trick uses a polynomial expansion (power series) to get rid of the log function, but then one has an approximation (why all these power series, for what does one need them in the days of PC? Do the PC electronics require them as well??)
Anyway in the present context this equation is of theoretical use only, neither the light intensity nor the fluo. efficiency (depends on all kinds of things) will be known before you do an experiment.
I am planning to try to use the above area formular to predict fluorescence peak areas by the following:

replace a*b in the last formular above with my own Fluo efficiency (Phe), determined by direct injection (no column) into the detector of a known solution of the analyte in mobile phase.
The formualur should then be:

Area = I*time = (Phe)c/F

I is the fluorescence intensity on my detector.