Effect of flow rate on peak area

[Rocnex1]

Theoretically when injecting the same sample, if reduce the flow rate by half while keep everything else same, the peak area for this compound should be unchanged, the peak is wider and lower.

We observed with a mAb, both in SEC and CEX, PA increased almost linearly when reducing flow rate.

Is that unusual?

Thanks!

[tom jupille]

I really need to add this one to the FAQ ("Frequently Asked Questions") section, because it comes up with fair regularity.

Assuming a concentration-sensitive detector (such as UV), the peak area is obtained by reading the signal (absorbance) at regular time intervals, and adding up all the readings from the beginning to the end of the peak.

To a first approximation, if you cut the flow rate in half, the sample spends twice as much time in the flow cell, you get twice as many readings, and the area becomes twice as large. What you are seeing is perfectly normal.

[Rocnex1]

Tom,
Thank you for your input. I had this theory and was not very sure if it stands well or not:

Cutting FR half and your sample stays twice as long in the flow cell. however, at any given momnet (a reading happens), the absorbance should be less (elute comes out slower) than that at normal flow rate. So when you adding up all the signal during that time period, the total may not be twice as large.

I guess it also depends on how fast the detector reads compared to the dwell time of the elute in the flow cell.

I really need to add this one to the FAQ ("Frequently Asked Questions") section, because it comes up with fair regularity.

Assuming a concentration-sensitive detector (such as UV), the peak area is obtained by reading the signal (absorbance) at regular time intervals, and adding up all the readings from the beginning to the end of the peak.

To a first approximation, if you cut the flow rate in half, the sample spends twice as much time in the flow cell, you get twice as many readings, and the area becomes twice as large. What you are seeing is perfectly normal.

[tom jupille]

the absorbance should be less (elute comes out slower) than that at normal flow rate

I see where you're coming from (but trust me, you've got it wrong).

The analyte comes out more slowly, but so does the solvent; the concentration therefore stays the same. Peak area is essentially the product of concentration and residence time in the flow cell.


Imagine the following experiment: Inject a sample under your normal conditions and wait until one of your peaks starts to elute. At the very top of the peak, instantly stop the flow (I know you can't really instantly stop the flow, that's why it's an imaginary experiment). What will happen to the signal?
*
*
*
*
*
*
*
pause to let the suspense build!
*
*
*
*
*
*
If you were using a mass-sensitive detector, the signal would drop back down to zero (think of a flame ionization detector in GC: once you burn all the sample, it's gone).

In a concentration-sensitive detector (like UV), the sample is still in the flow cell, still absorbing light. If you waited 5 minutes and then instantly turned the flow back on, the signal would start dropping again; you would end up with a huge, flat-topped peak; something like this (with apologies for the crude artwork!):




The area in this case would bear no relationship to the amount of sample injected; by changing the duration of the stop-flow interval, you could get any area you wanted.

[koen_shimadzu]

Tom is absolutely right in this.

Sometimes it's difficult to make a picture in your mind , but the illustration shown above is clear. To add it to FAQ is a good idea.

If you want to have the equations

Concentration-sensitive detectors,
S=A x F/W

M-ass-Flow sensitive detectors
S=A /W

As you can see the peak area depends on the flow when using a concentration sensitive detector.

[Rocnex1]

Thanks a lot, Tom. The art work is actually great, it explains!

Really glad I asked here.