HPLC column efficiency

[txxplc110]

I have a question about HPLC. My peak is very small (it’s more like a little line). It is definitely not like Gauss peak, but I have to find column efficiency. Can I calculate it using the formula Ne=16*(t’r/Wb)^2? Or what else should I do? I found the information that I can only calculate it when the peak is big.

I got a six-digit result :/

[tom jupille]

If your peak looks like a vertical line, re-plot it with an expanded x-axis to stretch it out so that you can see its actual shape. Assuming your data acquisition rate is fast enough (so that you collect at least 20 or so data points across the peak), you should be able to get a width measurement to plug into the formula.

A caveat here: if your separation was a gradient (i.e., the mobile phase composition changed during the chromatogram) then that formula is invalid; it applies only to isocratic (constant mobile phase) separations. In fact, you cannot measure the plate number directly from a single gradient chromatogram. At least two chromatograms with different gradient steepness are required, and even then the math is messy.

[txxplc110]

If your peak looks like a vertical line

But that peak is actually a horizontal line. I got this chromatogram as online project, it is saved as png so I can't do anything with it...

it applies only to isocratic (constant mobile phase) separations.

Yes, it is isocratic.

[DR]

If you open your png in Windows Paint, save it as a jpg and find a free photo hosting program online, and upload it there, you can post it here via image link.

[tom jupille]

Yes, one picture is worth many words!

The Forum should support png files. If you downloaded the chromatogram as a png file, all you need to do to display it in the Forum is to use the tag (insert the link to your file as the url).

Here's an example of a png image:

[txxplc110]

Yes, one picture is worth many words!

Yes, I know! Here is my chromatogram.

The first one is a clear chromatogram. On the second one, I marked that peak and calculated it with that Ne=16(t'r/W)^2 formula (only this formula is in my script). I'm also not sure if my dead time is fine because there is also a small mark on 0,400min (I just found it now). I don't know if this matters or not.

[tom jupille]

That isn't a peak; it's the tip of a peak. Most of the peak is below the limit of detection.

Think of it this way: how can you measure the width of an iceberg when all you can see is the little bit showing above the waterline? The answer is "you can't!". Your peak is like that iceberg: you're only seeing a bit of the top. If you had access to the original file (and the data system, of course) you could "zoom' the chromatogram to see what the peak actually looks like -- assuming it's not buried in the baseline noise.

All of that said, isocratic chromatograms usually show approximately the same plate number for all peaks, so why not use one of the other peaks?

As to the dead time, you can generally make a reasonable estimate of the dead volume (V0) as 0.5 * L * (dc^2) where L is the column length and dc is the diameter. The dead volume divided by the flow gives the dead time.

[txxplc110]

If you had access to the original file (and the data system, of course) you could "zoom' the chromatogram to see what the peak actually looks like -- assuming it's not buried in the baseline noise.

Unfortunately, I don't have access to the original file. But I zoomed my picture to 500% and now I can see that peak. Thank You for helping me!