Calculation of pesticide residue after QuEChERS extraction

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I am new to pesticide residue analysis in food commodities.

Just to give a little background, we do a QuEChERS extraction and then inject the samples on LC/MS and GC/MS. Our calibration range is 5-500 ppb, with standards spiked in a matrix matched solution. We incorporate QC’s where we spike and extract negative matrix exactly the same way we extract the samples.
Problems:

1. How do we calculate the concentration in the 10g sample from the concentration read from the instrument (in the vial)? Is there some sort of dilution and conversion factors that one has to consider when calculating? That after the QuEChERS extraction.

2. We spike the QC with 300ppb analyte and then extract exactly like our samples. Then the instrument will read the QC at approximately 300ppb. Does that mean there is no dilution in the whole extraction? (QuEChERS)

Summary of extraction:
10g sample + 10ml Acetonitrile > 1ml aliquot in dSPE > Dry the 1ml aliquot and make up in 200ul > Inject 2ul of the 200ul extract.

Please help...
Let's start with the low standard you have on the instrument. If you make that up by adding 5ng of analyte to 1ml of blank matrix you have 5ppb. If you calibrate with that injecting the same amount as you do the sample then you can ignore the injection volume.

Your final volume is 200ul so we figure the mass in the extract from a result given by an instrument reading of 5ppb

5ng/ml*0.2ml = 1ng of analyte in the final extract of 200ul.

That 1ng came from 1ml of the extraction process so you have 1ng in the final 1ml after dSPE so that aliquot represents the solvent portion of the extraction with a concentration of 1ng/ml. There were 10ml of solvent so you have

1ng/ml * 10ml Acetonitrile = 10ng in the extract.

You extracted 10g of sample to get the 10ng in the extract so you have

10ng/10g sample or 1ng/g sample. ng/g=ug/Kg=ppb therefore you have 1ppb in the sample.

The QC will be

((300ng/ml(instrument reading)*0.2ml)*10ml extract volume)/10g sample=60ppb in sample, which would indicate a 20% recovery

Essentially the first part of the extraction is a direct gram to ml extraction so no correction factor, the final step is a concentration of 1ml to 0.2ml which is a 5 fold concentration factor so your results are the instrument reading divided by 5 if you have exactly 10g sample. You can replace the 10g with the exact weight for more accuracy.

On the other hand if you prepare your standard in the extracted matrix then process through the dSPE and evaporation process and bring to final 200ul volume before injecting in the instrument then there is no conversion factor since you are using a procedural standard. I wasn't sure which way you are preparing the standard for injection.
The past is there to guide us into the future, not to dwell in.
James_Ball wrote:
Let's start with the low standard you have on the instrument. If you make that up by adding 5ng of analyte to 1ml of blank matrix you have 5ppb. If you calibrate with that injecting the same amount as you do the sample then you can ignore the injection volume.

Your final volume is 200ul so we figure the mass in the extract from a result given by an instrument reading of 5ppb

5ng/ml*0.2ml = 1ng of analyte in the final extract of 200ul.

That 1ng came from 1ml of the extraction process so you have 1ng in the final 1ml after dSPE so that aliquot represents the solvent portion of the extraction with a concentration of 1ng/ml. There were 10ml of solvent so you have

1ng/ml * 10ml Acetonitrile = 10ng in the extract.

You extracted 10g of sample to get the 10ng in the extract so you have

10ng/10g sample or 1ng/g sample. ng/g=ug/Kg=ppb therefore you have 1ppb in the sample.

The QC will be

((300ng/ml(instrument reading)*0.2ml)*10ml extract volume)/10g sample=60ppb in sample, which would indicate a 20% recovery

Essentially the first part of the extraction is a direct gram to ml extraction so no correction factor, the final step is a concentration of 1ml to 0.2ml which is a 5 fold concentration factor so your results are the instrument reading divided by 5 if you have exactly 10g sample. You can replace the 10g with the exact weight for more accuracy.

On the other hand if you prepare your standard in the extracted matrix then process through the dSPE and evaporation process and bring to final 200ul volume before injecting in the instrument then there is no conversion factor since you are using a procedural standard. I wasn't sure which way you are preparing the standard for injection.



Thank you very much James, this will help me very much! I really appreciate it... let me set up and see how far I can go.
Extraction volume/Sample weight x Final volume/Analysis volume = Your multiplication factor.

For you that's
10 ml/10 g x 0.2 ml/1 ml = 0.2

Multiply the value you obtain from your calibration curve by 0.2 to give you the concentration of analyte in your sample. Most software will do this calculation automatically if you enter your multiplication factor in the appropriate place.

On a side note I believe your QC is too high. Assuming you get close to 100% recovery what you would see in your vial is 300 ppb/0.2 = 1500 ppb. That's outside of your calibration range.
Thanks, this thread was really helpful. I am quite new to these calculations as I didn't use pesticides in my garden. This year I decided to use some from Food Rapid Test as my garden got overwhelmed by parasite grass. The manufacturer was recommended by a friend of mine who is into gardening. According to him, these are the best pesticides in terms of quality. Still, he recommended that I run regular tests as it is very easy to mess up the proportions.
for Quechers extraction in peanuts for every 5 grams of sample, 7.5 mL of water should be added, after which weigh 10 g of this mixture and carry out the steps. my question is: how is the final concentration calculated? having mixed 5 grams with water and added 10 acetonitrile was a dilution made?

can someone help me?
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