By C.Sychov on Thursday, April 24, 2003 - 08:29 am:
Please someone explain, how can I calculate amount of compound (in injection) with molar extinction for example 14000 if I have peak surface in mV*min?
By HW Mueller on Thursday, April 24, 2003 - 11:34 pm:
There was a chain on this: Calculating peak area on basis of the absorption coefficient on May 7, 2002. Couldnīt find it, quickly. If you also have trouble, just contact me.
By bill tindall on Friday, April 25, 2003 - 03:54 pm:
I am wary of calculated results, particularly in a problem this complex. So if I faced this problem I would find a typical anthraquinone dye which would have an extinction coefficient of about 14,000 and i would calibrate on it. And then I would adjust calibration to sample based on extinction coefficient ratio.
By C.Sychov on Saturday, April 26, 2003 - 10:44 am:
To HW Mueller
C(t)= A(t)/(E*d)
C(t)=dn/dV => C(t)dV = dn =>
n= F*S/E*d, where
n- moles, F- flow rate, E-ext., S - peak surfase in AUFS*min
Am I correct?
Than question -- how I can determinate coeff. K? where mV = K*mAUFS ?
To B. Tindall
You abs. right -- anthocyanines. And (hura!) I have anthraq.! So another stupid question :) -- how can I get E for anthraq. quickly?
THANK YOU FOR HELP!
Constantine
By HW Mueller on Monday, April 28, 2003 - 11:41 pm:
Darn, my message from yesterday disappeared, here another try:
Your equation appears to be correct. I usually do the inverse, namely to calc. the area in ĩV*time from what I inject, thus the formular:
Area (in ĩV*time) = a*b*c/F*k
where
k is the ratio AUFS/ĩV
a is the absorption coefficient (ext. coeff.)
b is the length of the light path in the cell
c is the concentration of the sample
F is the flow rate
One has to whatch the units and especially what you use for c. Itīs best to check this with a analyte for which you have all the necessary data, sort of what Bill suggested, but once you know what you are doing: No problem.
By C.Sychov on Tuesday, April 29, 2003 - 11:45 am:
Yes, I`ll calibrate the device using for examle anthraquinone as a standard and I`ll find coeff.
coeff. = E*n/S, where n is moles of compound in peak.
Thank you once more for useful discussion
Constantine
By HW Mueller on Tuesday, April 29, 2003 - 11:48 pm:
If I understood your original derivatization correctly, I would get
E = F*S/n*d
or using the version given by me
a = Area*F*k/b*c
By C.Sychov on Wednesday, April 30, 2003 - 03:02 am:
Yes, you would but it seems to me the result may be significantly affected by the constuction of the cell. Is it reasonable? So that is why I decided to calibrate my chmatograph.
By Anonymous on Wednesday, April 30, 2003 - 04:16 pm:
How does the cell construction affect Lambert-Beer's law?
By HW Mueller on Friday, May 2, 2003 - 01:02 am:
One needs to calibrate to get k, thatīs all. Your last formular also does not have the flow rate in it, you would have to keep the flow rate constant (not necessarily the pumpīs speed!) for all experiments you do with a given calibration.
Anon is correct, the Beer-Lambert law does not include a term for cell dimensions except for the length of the light path through the sample.
By C.Sychov on Thursday, May 8, 2003 - 03:06 am:
Thank you one more time)))
The discussion was very informative for me.
By Sununit on Monday, March 29, 2004 - 09:27 am:
I was wondering if I could get help calculating the concentration of a substance represented by a GPC peak using a GPC machine with a UV detector. please list relevant equations.
Also, I read that the extinction coeff for a polymer named NPC is 18400 cm^-1 M^-1. but in another handbook i read that the extinction coeff for Phenylalanine is .197 molar extinctions. are the 2 units the same. If so, why the difference in magnitudes?
Thanks