Please help me understand this calculation...

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I have recently had the opportunity of using a Beckman 126 System Gold LC pump equipped with a Rheodyne 7725i injector which has been a real treat considering most of my LC experience has involved newer instruments equipped with autosamplers such as Agilent 1260 (I like getting the "old-school" experience).

This has also posed some challenges for me as I am used to small injection volumes (less than or equal to 15 uL) and being able to easily calibrate over 11 points and 6 orders of magnitude. In the case of this Beckman system and manual injections, I am injecting 200 uL on a 20 uL loop with a mobile phase composition of Water/ACN. Due to the large injection volumes, this makes calibrating to high concentrations seem non-ideal as the standards I have access to come in MeOH. I don't like the idea of shooting 200 uL of MeOH into such a mobile phase as the one that I am using (though I never saw issues with small injection volumes '15 uL' on this same mobile phase). I haven't tried such large injections with mismatch solvent composition as the mobile phase and I would rather avoid any potential issues and waste of expensive CRM.

The requirement for large injection volumes made me realize why my prior colleague had been doing single-point calibrations... Long story short this brings me to my question. I have been completely stumped trying to understand the validity of the previously implemented calculations for % analyte in sample.

The quick and dirty:
1. 2.5 g sample is dissolved into 50 mL ACN.
2. A 500 uL aliquot of sample-containing-solution is then dissolved into 10 mL of ACN.
3. 200 uL of sample is injected.
4. The resulted Area count for the peak in question is used to back-calculate to the original concentration of analyte in sample: [(100)/(2.5 g)]*[(Response)/(RF)] ; where RF is the response factor obtained for a SINGLE calibration point, for example: Response (Area of peak) of 100 ppm standard divided by 0.1 mg/mL = RF for that analyte.

Though this may not be how I would go about setting up calibrations and back-calculations for determining analyte concentrations samples, but I sure want to understand it! I cant quite wrap my head around the above calculation. Additionally, I was hoping this factor of 100 (one of the four components to the calculation) is somehow incorporating the dilution during sample prep but I just cant seem to make any sense of it. At the end of the day this colleague of mine has decades upon decades more experience than I do and I am not doubting his technique. I am just new to doing "single-point" calibration and am still unsure of its validity, though I have observed good linearity in response factor for the analyte in question over various concentrations.

Thank you to anyone who took the time to read this, I look forward to any and all replies!
First, if you are using a 20 microliter loop, you are injecting 20 microliters, period. The rest of the 200 microliters essentially serves to wash out the loop.

Second, if the calibration standard and the sample are prepared *identically*, then the calculation is simple: Conc(sample) = Area(sample) / RF

Anything else in the formula is mathematical sleight-of-hand to account for differences in the preparation. Without knowing how the calibrator was prepared, it's hard to comment on your specific formula.

As far as your sample goes, you started with 2.5g in 50 mL, which is 50 mg/mL. You then did a 1/20 dilution, which means that your injected sample was 2.5 mg/mL, which is also 2.5 micrograms/microliter.
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
tom jupille wrote:
First, if you are using a 20 microliter loop, you are injecting 20 microliters, period. The rest of the 200 microliters essentially serves to wash out the loop.

Second, if the calibration standard and the sample are prepared *identically*, then the calculation is simple: Conc(sample) = Area(sample) / RF

Anything else in the formula is mathematical sleight-of-hand to account for differences in the preparation. Without knowing how the calibrator was prepared, it's hard to comment on your specific formula.

As far as your sample goes, you started with 2.5g in 50 mL, which is 50 mg/mL. You then did a 1/20 dilution, which means that your injected sample was 2.5 mg/mL, which is also 2.5 micrograms/microliter.


Thank you for the prompt reply! And yes you are totally right thank you for correcting that statement, I am choosing to 10x over-fill the loop as the manual that accompanied this instrument suggested over-filling to provide the best reproducibility (I've observed low deviation in response between injections even of only ~ 3-4x overfill). The statement you made makes me feel much better about including calibration points of a "high" concentration (1000 ppm) since only 20uL of the mismatched solvent will actually be injected!

This is how I would typically choose to calibrate and conduct my back-calculations: https://www.chromforum.org/viewtopic.php?f=5&t=101934

However, I have just been following the "monkey-see-monky-do" routine until I actually understand this old calculation; if it ain't broke don't fix it, right? The calibration standard is not subjected to the same prep, but is simply a neat standard in solvent at a concentration of 100 ppm. The area (response), is divided by 0.1 mg/mL (concentration in different units) to yield a response factor. This response factor is used in the calculation I mentioned in the original post: [(100)/(2.5 g)]*[(Response of sample)/(RF standard)].

Considering the dilution that the sample goes thru prior to injection (which the standard does not), I would only imagine the back-calulation from "instrument response of sample" to "% analyte in sample" to include the dilution factor and or a multiplier of some kind!

And I am sorry but I didn't quite follow where you were going with you last statement. Yes I agree the first step is a 20x dilution. Additionally, if the subsequent step involving taking a 500 uL aliquot and diluting to a final volume of 10 mL rather than diluting INTO 10 mL of ACN, this would also be a 20 x dilution; but alas, the math isnt quite as clean going from 500 uL to 10.5 mL. Overall it appears as though the sample is subject to a ~ 400x dilution, so wouldn't we want to include this somewhere in the calculation? If not, why are we able to avoid this? Thank you!
I think I see what's going on. To clear it in my own head, I tried working back from the actual concentrations and masses injected:

For the calibrator:
0.1 mg/mL = 100 ug/mL
100 ug/mL x 0.02 mL = 2 ug injected

For the sample:
2.5g/50mL = 0.05 g/mL = 50 mg/mL
50 mg/mL x 0.5 mL = 25 mg
25 mg / 10.5 mL = 2.4 mg/mL = 2400 ug/mL; that is the concentration of the injected sample*
2400 ug/mL x 0.02 mL = 48 ug injected

Since the RF was based on concentration, then the units would be something like Area/ppm, or, in different units, Area/[(mg/mL)] or, (Area*mL/mg). and then for the sample, Response/RF would give you the concentration (mg/mL) of analyte in the injected sample. Divide that by the concentration of the sample (2.4 mg/mL) to get the fraction of analyte in your sample, and then multiply by 100 to convert the fraction to % -- which is exactly what your formula does*

So, throw some hypothetical numbers in there as a "smell test".
Suppose the calibrator gave 1000 counts. Then RF = 1,000/0.1 = 10,000
Now suppose your sample gave 2000 counts for the analyte. The concentration of analyte injected would then be 2,000/10,000 = 0.2 mg/ml. The fraction of analyte in the sample would then be 0.2/2.4 = 0.083, and we multiply by 100 to get the % (8.3%)

*The 2.4 versus 2.5 is, of course the difference between the 10 and 10.5 mL in the dilution.
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
tom jupille wrote:
I think I see what's going on. To clear it in my own head, I tried working back from the actual concentrations and masses injected:

For the calibrator:
0.1 mg/mL = 100 ug/mL
100 ug/mL x 0.02 mL = 2 ug injected

For the sample:
2.5g/50mL = 0.05 g/mL = 50 mg/mL
50 mg/mL x 0.5 mL = 25 mg
25 mg / 10.5 mL = 2.4 mg/mL = 2400 ug/mL; that is the concentration of the injected sample*
2400 ug/mL x 0.02 mL = 48 ug injected

Since the RF was based on concentration, then the units would be something like Area/ppm, or, in different units, Area/[(mg/mL)] or, (Area*mL/mg). and then for the sample, Response/RF would give you the concentration (mg/mL) of analyte in the injected sample. Divide that by the concentration of the sample (2.4 mg/mL) to get the fraction of analyte in your sample, and then multiply by 100 to convert the fraction to % -- which is exactly what your formula does*

So, throw some hypothetical numbers in there as a "smell test".
Suppose the calibrator gave 1000 counts. Then RF = 1,000/0.1 = 10,000
Now suppose your sample gave 2000 counts for the analyte. The concentration of analyte injected would then be 2,000/10,000 = 0.2 mg/ml. The fraction of analyte in the sample would then be 0.2/2.4 = 0.083, and we multiply by 100 to get the % (8.3%)

*The 2.4 versus 2.5 is, of course the difference between the 10 and 10.5 mL in the dilution.


Thank you so much! The math error resulting in the factor of 2.5 rather than 2.4 made me assume he was simply plugging in the sample mass! This makes much more sense now and I feel much better knowing that the dilutions are indeed accounted for! Thank you for all of your assistance and have a great rest of your day!
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