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Whats is the calculation for % Assay?

Discussions about chromatography data systems, LIMS, controllers, computer issues and related topics.

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This is causing some disagreement in my workplace, can anyone provide some clarity. So we have a test where we weigh out 100mg of sample and dissolve in 100mls of sample solvent, yielding a 1mg/mL or 1000mcg/mL solution. Standards have the same concentration and same sample solvent. So our test method states that the content of active is basically the same calculation that Empower 3 uses for amount:

Area of sample/mean area of standard * conc of standard (mcg/mL).

This just gives the actual amount in the sample. To get the % Assay, I divide this amount by the theoretical amount and * 100 to get % Assay. Example:

Weigh 100.098mg which yields a conc of 1000.98 mcg/mL. I quantify this sample and get an amount of 1000.14 Therefore 1000.14/1000.98*100 = 99.92% recovery or assay.

Any thoughts on this method?
Simplify your life by sticking with the SI units and their standard subunits - what is mcg supposed to be?
Peter Apps
Is your "standard" the actual analyte or an internal standard?

In any case, it seems that this is ok - I would not prefer it but - it's ok.

Ri = m1*Ci (for the analyte in the sample, m1 is how the detector sees the analyte)
Rstd = m1*Cstd (for a 1 point standard, response for analyte is same in sample as in the standard)

Ri/Rstd = Ci/Cstd
(Ri/Rstd)*Cstd = Ci

I'd prefer to calibrate over a range of concentrations to determine m1 (the response factor).
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