How do I account for purity?

[Nawaf]

Hello,

I have a small issue I would be grateful for help with.

I use a calibration standard that is supposed to be 100mg/L. I account for standard purity from the batch certificate of analysis, so I can report 'true' concentrations of my analytes.

E.g. if my standard purity is actually 98mg/L from the CoA and my sample gives me a result of 1mg/L from my calibration curve, I do:

(1/100)*98 = 0.98mg/L.

I routinely run one of calibration points (5mg/L) as a check throughout sequences to ensure my system is working properly (90-110% tolerance). I have recently been asked to run a sample preparation check at 5mg/L using a second 100mg/L reference standard of different origin. My calibration standards are bought in pre-derivatised but this second standard will be derivatised inhouse.

If this second standard batch has a purity of say 101mg/L I am worried that the acceptance may fail on paper due to the differences in purity, but in reality is within specification.

My question is - how do I account for the 101mg/L purity compared to the calibration standard purity of 98mg/L?

Thanks you so much for any help.

[vmu]

X mg/L is the certified ("true") concentration of the standard rather than its "purity".

If the check standard is certified as 101 mg/L and your result for this standard is 101 mg/L, the recovery is 100*101/101 = 100 %.

If the check standard is certified as 101 mg/L and your result for this standard is 96 mg/L, the recovery is 100*96/101 = 95 %.

If the check standard is certified as 102 mg/L and your result for this standard is 96 mg/L, the recovery is 100*96/102 = 94 %.
...

[Nawaf]

X mg/L is the certified ("true") concentration of the standard rather than its "purity".

If the check standard is certified as 101 mg/L and your result for this standard is 101 mg/L, the recovery is 100*101/101 = 100 %.

If the check standard is certified as 101 mg/L and your result for this standard is 96 mg/L, the recovery is 100*96/101 = 95 %.

If the check standard is certified as 102 mg/L and your result for this standard is 96 mg/L, the recovery is 100*96/102 = 94 %.
...

Thank you for the correction.

Thank you for the examples, using your first example if my calibration standard material is 98mg/L but my check standard is 101mg/L my check standard result output will be higher than true. I’m trying to account for the difference between that 98mg/L and the 101mg/L.

[vmu]

You are making a problem from nothing. Use the certified concentration value of the calibration material to calculate the true concentrations of the calibration solutions and use these true concentrations for the calibration (e.g. 0.98, 9.8 and 49 instead of 1, 10 and 50). In this case, the calibration will give you correct results without the need for any additional corrections. Divide the obtained true result for the check standard by the true concentration of the check standard to calculate the recovery.

Alternatively, if you prefer to use the rounded concentrations for calibration (1, 10 and 50), you have to correct the initial results for the certified concentration value of the standard material to obtain the true results (e.g. correct a result of 40 as 40*98/100 = 39.2). Then, divide the obtained true result (e.g. 39.2) for the check standard by the true concentration of the check standard [e.g., if the rounded concentration of the check standard is 40, its true concentration is 40*101/100 = 40.4] to calculate the recovery (39.2/40.4 = 0.97 or 97 %).

[Nawaf]

You are making a problem from nothing. Use the certified concentration value of the calibration material to calculate the true concentrations of the calibration solutions and use these true concentrations for the calibration (e.g. 0.98, 9.8 and 49 instead of 1, 10 and 50). In this case, the calibration will give you correct results without the need for any additional corrections. Divide the obtained true result for the check standard by the true concentration of the check standard to calculate the recovery.

Alternatively, if you prefer to use the rounded concentrations for calibration (1, 10 and 50), you have to correct the initial results for the certified concentration value of the standard material to obtain the true results (e.g. correct a result of 40 as 40*98/100 = 39.2). Then, divide the obtained true result (e.g. 39.2) for the check standard by the true concentration of the check standard [e.g., if the rounded concentration of the check standard is 40, its true concentration is 40*101/100 = 40.4] to calculate the recovery (39.2/40.4 = 0.97 or 97 %).

Thank you.

Yes we use rounded calibration values.

I have been doing for a result of 40:

(40*98)/101 = 38.8 true result.

Where 40 is the check standard result, 98 is the calibration concentration and 101 is the check standard concentration. My method and your method seem to produce different results.

[vmu]

I have been doing for a result of 40:

(40*98)/101 = 38.8 true result.

This is wrong.

If the rounded check std. conc. is 40, its true conc. is 40*101/100 = 40.4. This is the value that should be obtained ideally from the chromatography of this check std.

With your rounded calibration, you only have to correct the result calculated from this calibration for the certified conc. of the calib. material. If the initial result of the calculation from the check std peak area and the cal. curve is X, then the corrected (true) result is X*98/100.

If X*98/100 is 40.4 for the 40.4 check std, the recovery is 100*40.4/40.4 = 100 %.

If X*98/100 is, e.g. 35.0 (i.e. X = 35.7) for the 40.4 check std, the recovery is 100*35.0/40.4 = 86.6 %.

[vmu]

(40*98)/101 = 38.8 true result.

Here, you are recalculating the true result 40*98/100 = 39.2 to the "rounded value" for the check std. The value 38.8 (not a true value) can be considered useful only if you use it to calculate the recovery relative to the rounded conc. of the check std, i.e. recovery = 100*38.8/40 = 97 %. This way of recovery calculation is too artificial and may lead to confusions and mistakes.

[Nawaf]

(40*98)/101 = 38.8 true result.

Here, you are recalculating the true result 40*98/100 = 39.2 to the "rounded value" for the check std. The value 38.8 (not a true value) can be considered useful only if you use it to calculate the recovery relative to the rounded conc. of the check std, i.e. recovery = 100*38.8/40 = 97 %. This way of recovery calculation is too artificial and may lead to confusions and mistakes.

Thank you so much for your help.

Yes I think this what we need to do. We recently had a 1mg/L check standard JUST fail on another method, I think it was 1.150mg/L or something (0.8mg/L - 1.1mg/L tolerance or 80-110%) but then I realised that the calibration standard certified concentration was 97mg/L and the check standard certified concentration was about 102mg/L so the check standard was always going to come back higher and if we could account for the difference in concentration it would actually be a pass.