Using peak area to estimate concentration

[niba]

I have been told that it is possible to estimate the concentration of a compound in a solution using the HPLC peak area if you know the compound's extinction coefficient. Based on Beer's Law I have the following calculation:
Concentration of sample = (peak area * flow rate) / (ext. coeff * path length * volume injected)

However when I look at the units in this calculation I arrive at the following

Concentration of sample = ((Au*s)*(L*s-1))/((L*g-1*cm-1)*(cm)*(L))
= (Au*g)/L

rather than

Concentration of sample = g/L

Have I misunderstood this calculation or have I misunderstood the units? I was hoping to be able to estimate the concentration of compounds for which I do not have standards but do have the published extinction coefficients.

[lmh]

Yes, we've got into this knotty area, too, and frankly it's a bit hard.

May I suggest an alternative non-mathematical approach.

Pick a simple, preferably related standard whose extinction coefficient is known, and which is reasonably cheap and runs OK (i.e. as similar to your target compound as possible). Make a calibration curve using this standard. Now "quantify" your compound using this calibration curve, but correct the final values by using the ratio of extinction coefficients of the standard and the target compound.

Even if your standard is unrelated, if it runs with approximately the same peak shape, it will save you a lot of hassle worrying about the geometry of a flow cell.

This may sound a bit feeble (and it is!), but it's been used before in some areas. For instance, glucosinolates in plants and food are rarely available as reliable, clean standards, but I believe someone somewhere published a set of extinction coefficients relative to that of Sinigrin, the one glucosinolate that is reasonably easy to get hold of; these values can be used to convert relative peak areas into amounts.

[Uwe Neue]

I assume that you want to know the concentration in the sample.

The peak area is proportional to the mass injected, and by knowing the extinction coefficient, you can calculate this mass as follows:

Au*sec = E*d*c*sec

Au*sec is the measured peak area, E is the extinction coefficient, d is the cell length, c is the concentration, and sec are seconds. The concentration is mass/volume or m/V. Substituting this results in:

Au*sec = E*d*m/V*sec or E*d*m/F

where F is the flow rate (in mL/sec). Now the mass injected is:

m = Au/sec*F/(E*d)

Since you know, how many microL you have injected and with the mass calculated as above, you get directly the concentration in the sample. In what units you get it depends on the dimension of the extinction coefficient. If you use the molar extinction coefficient, you get a molar concentration.

[niba]

Thank you both for your replies. I have two questions for Uwe, firstly you stated:

" Au*sec = E*d*m/V*sec or E*d*m/F

where F is the flow rate (in mL/sec). Now the mass injected is:

m = Au/sec*F/(E*d) "

Should the last line read m = Au*sec*F/(E*d) ? Was this a typing error or have I missed something?

Second question, if the mass injected is calculated using

mass = Au*sec*F/(E*d)

and the dimension of the Extinction coefficient is in the units L g-1 cm-1

then does it matter that

mass = (Au*s*ml*s-1)/(L*g-1*cm-1*cm)
mass = Au*g

Any clarification would be appreciated,
Many thanks,
Nicole

[Uwe Neue]

You are correct *this was a typo):

m = Au*sec*F/(E*d)

"Au" does not have a dimension, so the outcome of this calculation is just in grams or moles, depending on the extinction coefficient.

[lmh]

Nice! Thanks Uwe...

[HW Mueller]

Another aspect of this has been discussed some years ago. This is to use a known absorption (extinction) coefficient to estimate the peak area in order to have an idea about possible material loss.
Starting with Uwe´s equation,

Au*sec = E*b*m/F

one can see that

Au*sec = E*b*c*(injected volume)/F

(c is the concentration of analyte)
or

Area of peak = E*b*c*(injected volume)/F

If your software puts out µV*sec instead of Au*sec you have to correct with the factor µV/Au, or its reciprocal, depending..... (µV is micro Volts).
Careful, the units are mixed up quite easily.

[lmh]

I have to ask a naive question:

given that using extinction coefficients in hplc isn't half as difficult as I thought, why are we all running standard curves? After all, in straightforward UV spectrometry, it's perfectly acceptable to use extinction coefficients, and indeed in some quarters frowned upon to use calibration curves, as the preparation of standards is just another step where errors can occur.

[HW Mueller]

Remember that we all talked about "estimations". My post hints at one problem: I used that calculation to check on what the absolute area should be, that is, I checked whether there is a considerable loss of analyte in the HPLC. Note that this is a more absolute (again "absolute" is used in the sense of what it should be) check than injecting the anaylyte without a column. As far as the analysis is concerned one could be way off by using this calculation. Chromatography is a relative method, not absolute. Note all the questions about some strange things going on during chromatography!
"Absolute" UV depends on high or known optical purity of the substances measured, if you have an unknown mixture, even of known substances, you can usually forget about it (unless the spectra are very unique).

[lmh]

Thanks for very clear reply.

[Jade.Barker]

why are we all running standard curves? ... the preparation of standards is just another step where errors can occur.

For our lab the standard is used for both creating the calibration curve, and to QC the instrument. If the standard looks bad - something is wrong. We purchase our standard already prepared so we minimize the risk of tech error. I would be more worried about a manual calculation as a source of error, at least in our lab.