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Selectivity Calculation by FID area%

Discussions about GC and other "gas phase" separation techniques.

5 posts Page 1 of 1
Hello,

I have a quick question. For calculating Selectivity of Hydrocarbon, can I use FID area% directly or I need to calibrate it by standard gas? I am using TCD-Methanizer-FID in a series connection with a single column HayeSep D.

Thanks in advance.
I think that you are confusing your terminology. Selectivity speaks to how well a certain stationary phase can differentiate between 2 molecules. For instance, under a particular set of conditions, benzene and ethanol coelute on a wax phase but the same set of conditions, they are easily separable on a 1-phase (DB-1, Rtx-1, ZB-1, etc.).

You seem to be interested in sensitivity. Detection limit - in some way shape and form - is dependent on how well your instrument can see a real signal above the normal, hopefully random, noise of your detector. Area% is not the way to go with that. The magnitude of the signal above the noise is what's important.

This might be a good starting point for you.

http://www.chromatographyonline.com/limit-detection
Thanks for your reply but I am trying to make my question simple:

"Let's say I have got CH4,C2H6 and C6H14 from FID. I will get area% for individual component from FID. Can I use this area% directly as the CH4,C2H6 and C6H14 concentration in the product? or I need calibration for getting the product concentration?"

Thanks.
I think I understand you better now. To get accurate concentrations of the analytes in your sample, you will need to get (or make) a calibration standard. All "Area Percent" tells you is what fraction of the total material that provides a response is that component. For instance, if you have 3 components and the chromatogram only shows 3 peaks:

A = 1000 cts
B = 2000 cts
C = 40,000 cts

The total area is 43,000 cts. The %A = 1,000/43,000 x 100 = 2.3%, %B = 2,000/43,000 x 100 = 4.7% and %C = 40,000/43,000 x 100 = 93.0%

If your chromatogram would show your 3 peaks plus something that is unknown, the Area% for your analytes will change because of that additional peak. The concentrations of your analytes might be the same. This scenario doesn't even account for part of your sample that might be invisible to your detector. If your sample has air or a noble gas or CO2 in it, the FID will not respond to them.

You really need a standard of known concentrations of your analytes for good quantitation.
Yes, this is exactly what I want to know. Thanks a lot for your quick response.
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