I realise this topic has been discussed alot, but I am having difficulty understanding why peak area decreases as flow rate increases. All else being equal the same mass passes through the detector. I realise that UV response is dependent on concentration, but am having difficulties conceptualising the effect. For a relatively higher flow rate a tighter sample slug of a sample passes thru the detector and at a slower flow rate the same mass will pass thru the detector but as a wider sample slug. In both cases the detector sees the same amount of sample just more concentrated in one case than the other. I believe UV is additive, in that if you take two UV spectra of a 0.01M solution (at one wavelength) and add then together you will get the same UV spectrum as the 0.02M solution. I liken that to adding the wider sample slug slices (I am refering to the sample slices created by the chosen data rate) together and the narrower sample slug slices together - this is why I am having trouble understanding the reason the peak area/flow rate relationship. Hope all that makes sense. Someone may say I have answered the question myself - mass vs. concentration but that is not a complete enough answer for me.

Imagine a system with no column and no band broadaning and a zero volume detector cell. When we inject a sample we get a perfect square wave output. Now imagine we inject 10ul of a solution that produces a 1000mAU response at a flow rate of 0.6 ml/min (600uL/60sec = 10 uL/sec). This injection will produce a response of 1000mAU that lasts 1 sec or an area of 1000mAU*sec. Now if you lower the flow rate to 5ul/sec you will get the same response because the sample hasn't changed, however, the signal will last for 2 seconds. Thus the area will be 2000mAU*sec.

You are right, you did answer it yourself! Just forget about the width of your "sample slug", sampling rate, and all of the other complexities. In practice it works out like our imaginary system and peak area is directly proportional to flow rate.

How about this idea.... peak shape will cause variations in area (ie tailing, fronting) so if you inject the same concentration and have different peak shapes you will change the area. Same idea can be applied to flow rates....

Here's another way to look at it:

Suppose you have a separation at 1 ml per minute with a retention time for the component of interest at five minutes and a peak width at baseline of one minute. Since area is expressed in units of detector response multiplied by the peak duration in units of time, the area is a function of peak response and the time value of the peak. Assume this particular component gives an area of 100,000 area counts. Now if you cut the flow rate in half, the retention time will increase to 10 minutes, the peak width will double (to a first approximation), doubling the area counts even though the peak height will be approximately the same (to a first approximation) making area counts double to: ~200,000.

So, this doubling of peak area is nothing more than an artifact of our use of time in the area calculation. If you do the same mental experiment above but use area based on detector response multiplied by the peak duration in units of volume, the area would the approximately the same regardless of the flow rate.

Thanks everybody for the explanation. I guess the time factor is key.

12:03,

Not exactly, to a first approximation changes in peak shape have no effect on area assuming you keep the flow rate constant. The area is fixed, and it just gets distributed differently. Any change in area due to peak shape is the result of not being able to accurately define the begining and end of the peak.

You can replace your column with a piece of capillary tubing and inject a standard and you will get area counts remarkably close to those with a column.

Sorry, I meant Anon at 1:59. It would be easier to have a discussion if people chose to be pseudonymous (make up a name) instead of anonymous.

Tom: I purposely ran a std under 3 separate isocratic conditions to create different peak shapes to test your theory. I was amazed that the areas were all the same. Thanks for the input

Check the chain "Molar extinction" of April 24, 03. Couldnīt find the original chain anymore, where Uwe and Tom helped me overcome a mental block to arrive at the formular in the chain of April 24. One can use that for predicting peak areas, etc.
Another simple mental prop of "what gives" here: If you stop the flow during a peak, the area will approach infinity as stop time approaches infinity.