Hi,
I am a research associate at LSU. I am fairly new at HPLC analysis and am having a problem
with quantification. My units do not cancel to give me the desired results. Can someone there
please help?
Here are my conditions:
I have a 25 umole/l solution of an amino acid standard.
In the vial that sample is injected from I have:
10 ul of 25 umole/l std
10 ul of 150 umole/l of norvaline as the internal standard
40 ul of OPA
100 ul of buffer
for a total of 160 ul.
From this vial I inject 50 ul onto the column.
Do I develop response factors using the concentration of the std sol or do I calculate what
actually goes on the column?
This is the calculation I was using, but something is not correct.
=[area peak (uv) / RF (area(uv)/Conc(umoles/L)) * (area IS in std/Area Is sample) / V inj (ul)
X (Vol (ul in vial) / Vol (ul sample der) X Dilution Factor X ?????
I want to end up with umoles/L or grams/umole.
I feel like an idiot, but I can't seem to figure out what I am doing wrong. Of course, very few of
the papers I have read on amino acid analysis give the actual quantitation calculations.
If you could be of any help, I would really appreciate it.
Thanks,
Millie B Williams
![]()
![]()
![]()
![]()
By Anonymous on Monday, May 10, 1999 - 10:24 am:
Conc of unk= (area unk/(1/(area std/conc std))*(Instd area STD/Instd area UNK)
Use the actual concentration of compound in the vial. In your example (25 * 10/160)=1.56 umole/L
If injection volume is constant it does not need to be used in the calculations.
Good Luck
![]()
![]()
![]()
![]()
By Anonymous on Thursday, June 10, 1999 - 08:25 am:
I too had the same ideas when moving into analytical chemistry from biological sciences.
Providing the injection vol. for the std. and the sample are the same, and both are treated in the same manner, there is no need for the equation, as your peak areas can be expressed proportionally to the concentration injected (in your case 25uM/L).
Hope this helps!
Posting is currently disabled in this topic. Contact your discussion moderator for more information.