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Concentration calculation from HPLC
Posted: Fri Sep 06, 2019 8:41 pm
by GarryP
Hi there,
Can someone help me confirm the calculations below please:
I extracted 120 mg of my sample (79% moisture) into 10 mL solvent. I then injected an aliquot of the extract into the HPLC and got a concentration of 7 mg/L.
41mg = 7mg/L
=0.17mg/mg of sample
% w/w=17.1
Is my calculation of the concentration in the original sample and the weight percent correct?
Thank you
Re: Concentration calculation from HPLC
Posted: Fri Sep 06, 2019 9:19 pm
by Hollow
GarryP wrote:
Hi there,
Can someone help me confirm the calculations below please:
I extracted 120 mg of my sample (79% moisture) into 10 mL solvent. I then injected an aliquot of the extract into the HPLC and got a concentration of 7 mg/L.
41mg = 7mg/L
=0.17mg/mg of sample
% w/w=17.1
Is my calculation of the concentration in the original sample and the weight percent correct?
Thank you
Where do the 41 mg come from???
w/w = [concentration of analyte] / [concentration of sample] (both at the same stage of the sample prep e.g. in the vial)
the latter one is
120 mg * (1-0.79) / 0.01 L = 2520 mg/L
So amount is:
7 mg/L / 2520 mg/L = 0.278% w/w (on dried basis)
Re: Concentration calculation from HPLC
Posted: Sun Sep 08, 2019 8:46 am
by GarryP
Thank you for your input. The 41mg was a mistake I made in deducing the dry weight content of the sample.
If I want to simply report the concentration in the sample, would that be 7mg/L or do I need to multiply it by 10 mL (the volume in which the sample was extracted) to give 70 mg/L?
Re: Concentration calculation from HPLC
Posted: Sun Sep 08, 2019 9:45 am
by Hollow
GarryP wrote:
If I want to simply report the concentration in the sample, would that be 7mg/L or do I need to multiply it by 10 mL (the volume in which the sample was extracted) to give 70 mg/L?
No,no,no, why would the concentration change if you just take more volume of it? Does a small beer have a lower alcohol conc than a big one...?
Also the unit would not be correct if you multiply mg/L * L = mg (L cancels out). Even more mg/L * ml = ug, so 70 mg/l would be totaly wrong.
So in 10ml there are 70ug of analyte, those coming from the 120 mg of sample = amount in the sample "as is" 583*10-6 w/w (=583 ppm)
Same is obtained from previous formula, if you don't correct for the moisture.
Re: Concentration calculation from HPLC
Posted: Sun Sep 08, 2019 10:05 am
by Hollow
Something I try to teach my students too:
be strict in using the right terms and check the units
A concentration is always something per volume, so the unit has to be to (something, e.g. g or mol)/L
An amount is mass per mass or unit-peace, so the unit is weight/weight (or pce), e.g g/g or g/tablet
If the unit is just like g, then call it what it is, a mass
further, if not used to it, just write the SI-prefixes as 10^-3 for milli, 10^-6 micro etc, then do the numerics and finaly convert it back to a SI-prefix; e.g mg/l*ml = 10-3g/l*10-3l = 10-6 g = ug
This may help preventing confusion.
Re: Concentration calculation from HPLC
Posted: Thu Sep 12, 2019 8:56 pm
by James_Ball
7 mg/L is concentration determined by the HPLC
7mg/L is the concentration of the 10ml final solution of your extract.
10ml of extract = 0.01L extract
10ml of extract contains (7mg/L) * 0.01L = 0.07mg of analyte
10ml of extract came from 120mg of sample therefore:
120mg sample contains 0.07mg of analyte. which is 0.07mg/120mg
amount of analyte per unit weight of sample = 0.000583mg/1mg which is 0.0583%
Corrected for moisture 0.0583/(1-0.79) = 0.2778%
Just another way to look at the problem.