Well, the amount of IS on the column was 10 ng. So the amount in the sample can be calculated by simple equation C analyte = (10/A) IS *A analyte. However, this amount will be divided by the sample weight. Do you use initial weight taken for the extraction or on column weight in calculating ng/mg of sample?
For example, if you take 10 mg of tissue spiked with 1000 nmol/L internal standard, performed the extraction and the final elution volume was 100 uL and injection volume is 10uL, then concentration is calculated as:
Concentration of IS will be: 1000 nmol/L * 10 uL = 10 pmol.
Conc. of analyte = (Conc. of IS/Area of IS) * Area of analyte
weight of the tissue
My question is the weight of the sample is 10 mg or 1 mg? For data normalization.
First, calculate amount of the analyte in the solution (either with external or internal standard method) in wt/vol unit. For instance ng/mL or ug/mL. Then calculate the wt of matrix or sample in the same solution you injected, such as g/mL. Therefore the concentration of the analyte in the sample would be (ng/mL)/(g/mL) and the final result is ng/g or ppb
Other than during the preparation of the internal standard, the concentration of the internal standard is a moot point. You are using the ISTD peak area as an indicator of interferences or losses during sample prep and or between injections. With that in mind you calculate the uncompensated concentration of the sample then multiply that value by the value obtained by dividing the area of the ISTD in the standard by that of the area of the ISTD in the sample.
Dr. Narong thanks for your reply. I understand your explanation. My point is this: We use say 10 mg of the tissue for lipid extraction and then the final elution volume is 100 ul and injection volume is 10 uL. In this case, when normalization of the analyte concentration (say in pmol or ng) by tissue weight do you use 10 mg or 1mg?
I am don't understand your question but can give you an example. As you implied, the final solution has 10 mg of the tissue in 100 uL so the concentration of tissue is 10 mg/100 uL or 100 mg/1 mL. The injection volume is the same for all samples so it will not be used in the calculation. If you find the concentration of the analyte by comparing it with the standard curve and you get 10 ug/mL, the final concentration of the analyte in the tissue would be 10 ug/100 mg or 100 ug/1000 mg or 100 ug/g or 100 ppm.
Perhaps you need to talk to someone on your staff to guide you through this experiment.
Dear Michael
Thanks for your advice and suggestions. I got the answer.
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