quantification calculation for liquid samples

Discussions about sample preparation: extraction, cleanup, derivatization, etc.

7 posts Page 1 of 1
I have a liquid sample sample that I am extracting and quantifying an analyte in the liquid.

The extraction process is straight forward.

~3 mg of sample+ 3ml MeOH sonicate, inject

the sample matrix is essentially water, thus fully miscible in MeOH

My question is the Solvent amount in hte calculation. Do I multiply by the total volume of MeOH(3ml) or by the total volume of 6ml

I would like to find the concentration in mg/g my calculation is

analyte amount in ppm * Dilution

(((analyte amount in ppm*dilution factor)*((amount of solvent(ml)/100)/sample weight (g))/10)

example

Sx ppm result 100.00 ppm
Dilution 1
Sx weight 3.00g
solvent amount ?

the total volume is 6ml

if I plug in 6ml my result is 0.19 mg/g
If I use just the amount of MeOH the result 0.10 mg/g

Which would be correct?
You seem to be confused between amount (which has units of mole or grams) and concentration, which has units of amount/volume.

Also, you seem to have typed mg when you meant ml because you are working with a liquid sample and you end up with 6 ml after adding 3 ml of methanol, although you later refer to sample weight, which is unusual for liquids.

Peter
Peter Apps
Sorry to make it more clear I was told to report it in mg/g even though it is a liquid, but generally It can also be reported in %
that calculation would be

(((analyte amount in ppm from chromatogram*dilution factor)*(extract volume(ml)/100)/sample weight (g))/10)

it is the extract volume that im confused about. Im trying to measure the caffeine in a soda.

When I measure the caffeine in the coffee bean I use the

~1 mg of sample+ 10ml MeOH sonicate 1:20 dilution inject

calculation is

(((analyte amount in ppm*dilution factor)*((amount of solvent(ml)/100)/sample weight (g))/100)

this gives me the percent of caffeine in the bean

the trouble is now im trying the same analysis for caffeine in a liquid product and in my calculation im not sure If I just use the amount of solvent added to the liquid sample(3ml) or the total volume of solvent+sample which is 6ml

hope that makes it a little more clear
Generally speaking, you want to use the entire volume to figure out your dilution factors and your answers will come out in mg or whatever units per mL. At this point, you multiply by your original mL of sample diluted if you need units per sample (for samples that are 5mL single dose cups, for example).
For adding 3mL MeOH to a 3mL sample, your dilution is 3/6, which simplifies to 1/2, so your dilution factor is the inverse of that, 2.
Thanks,
DR
Image
You inject an extract from a specific weight of coffee and you find caffeine, its mg per g coffee.

Ideal would be, if you spike a matrix that comes close to coffee with the concentration range you need. So you spike mg in g matrix. That way your cal is in mg per g and your samples too.

If you dont spike matrix and you just have a ppm or ppb standard, you need to considerate every dilution step. You would calculate the amount of sample in your solvent and bring it back to ppm.

The liquid product has everything already dissolved. So any liquid you add, dilutes it down.

So if a coke had 35mg per L and you used 6mL as sample the concentration in these 0.035mg/mL

If you add to these 6mL, 3mL MeOH the concentration goes down to 0.023mg/mL.


I also dont understand why you have to add MeOH in an already liquid sample. If you dont hit the detector limit and the high conc. of matrix doesnt effect your measurement, i would direct inject it.
If the matrix is an issue, you should use a phase separation method.

greetings
You inject an extract from a specific weight of coffee and you find caffeine, its mg per g coffee.

Ideal would be, if you spike a matrix that comes close to coffee with the concentration range you need. So you spike mg in g matrix. That way your cal is in mg per g and your samples too.

If you dont spike matrix and you just have a ppm or ppb standard, you need to considerate every dilution step. You would calculate the amount of sample in your solvent and bring it back to ppm.

The liquid product has everything already dissolved. So any liquid you add, dilutes it down.

So if a coke had 35mg per L and you used 6mL as sample the concentration in these 0.035mg/mL

If you add to these 6mL, 3mL MeOH the concentration goes down to 0.023mg/mL.


I also dont understand why you have to add MeOH in an already liquid sample. If you dont hit the detector limit and the high conc. of matrix doesnt effect your measurement, i would direct inject it.
If the matrix is an issue, you should use a phase separation method.

greetings
We run into this problem often when someone sends us a solvent or oil and need results in a weight basis because they are going to dispose of it and the contractor charges them per ton.

If you take a volumetric flask, weigh your sample into the flask then dilute to volume you now have an extract equivalent to doing an extraction on a solid sample, so you know the initial weight and the final volume of the extract. This way you are certain to have 3g of the soda in a total of 6ml of extract. Then if your instrument result is 5mg/ml, you know that you had a total of 50mg caffeine in the volumetric flask with 6ml total volume. In that total volume you also had 3g of soda so that soda had to contain 30mg/3g so you have a final result of 10mg/g.
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