Empower to calculate assay?
I have a method used to calculate assay. For standards its 20mg of active into 50mls, then a 5ml dilution of this into 50mls.
Samples exact same preparation.
The method describes % Assay as (Peak Area of Sample/Average Peak Area of Bracketing Standard) * (Standard Weight in mg/Volume of Standard Solution) * (% Potency/100)*(Volume of Sample Solution/Sample Weight of Sample) *Dilution Factor.
I understand that up to the dilution factor which is down as 100. Surely 20mg in 50mls (0.4mg/mL) * (5/50) changes this to 0.04mcg.mL, which is a dilution of 10, not 100?
Can anyone see where I'm going wrong? I want Empower to calculate % Assay. To do this I'm going to put concentration of brackets standards (nominal 40mcg/mL) in component editor and for the sampleweight of the sample line put down my sample weight obtained (nominal 20mg) and dilution factor 10. Is % Assay just amount multiplied by 100?
Samples exact same preparation.
The method describes % Assay as (Peak Area of Sample/Average Peak Area of Bracketing Standard) * (Standard Weight in mg/Volume of Standard Solution) * (% Potency/100)*(Volume of Sample Solution/Sample Weight of Sample) *Dilution Factor.
I understand that up to the dilution factor which is down as 100. Surely 20mg in 50mls (0.4mg/mL) * (5/50) changes this to 0.04mcg.mL, which is a dilution of 10, not 100?
Can anyone see where I'm going wrong? I want Empower to calculate % Assay. To do this I'm going to put concentration of brackets standards (nominal 40mcg/mL) in component editor and for the sampleweight of the sample line put down my sample weight obtained (nominal 20mg) and dilution factor 10. Is % Assay just amount multiplied by 100?
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Best Answer

The dilution factor entered into Empower would be 500...inverse of the prep: 1 / (x/50*5/50). Simple way to check: mass * sample wt / dilution...20 mg * 1 / 500 = 0.04 mg/mL for the standards.
Note that if you use the stock wt and purity in the component table, all calibration and sample calculations are effectively completed with full/max precision and won't get you into any issues with intermediate rounding like when using a concentration. It probably won't happen often, but it is seemingly inevitable that you'll have a final result that demonstrates this as problematic. I've had it happen where someone reported a result like 99.5%, but their reviewer showed that they get 99.4% and it comes down to this where full precision showed the original analyst to be correct because the result was something like 99.46% where the reviewer used rounded intermediate values and was getting something like 99.44%.
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Answers
I have made a note of your sample dilution formula.