Chromatography Forum

Hi,

manufacturer of the compound gives several results on the CoA:
water by KF 5%
solvent (acetone) by GC 2.9 %
sodium by ICP 0.2%
purity by HPLC 95 % (corrected for the water and solvent content)

Is the correct way to calculate the content of the compound by this equation (100 % - water - acetone)*(HPLC purity-sodium) =
(100 % - 5 - 2.9)*(95 %-0.2 %)= 87.3 %

Thanks!

In 100g you have 5g water and 2.9 g solvent. Of the remaining 92.1 g, 95% is target compound by HPLC, so the purity is 87.5 %.

Does the difference really make a difference ??

Peter

I'ts early and my brain is on strike (OK, I'm at work and my brain is on strike) but I am having problems understanding how 95 % purity is correcting for the water and acetone content when the sum of the water and acetone is greater than 5. Did you mean uncorrected? As far as the sodium content, since you are probably not dealing with atomic sodium, there is likely a counterion involved. For example, if the sodium contribution is from NaCl, for 0.2 % (w/w) Na you would have 0.5 % (w/w) NaCl present. If the sodium is from some other salt, the correction would be different.

It means that, chromatographically, your standard is 95% pure by area.

Figure your standard conc. based on 0.95*(standard mass)(100-%water-%Acetone)/100%

Ah, yes, the old smoke and mirrors game. Even the USP falls prey to this, where 99+% USP glycerin can contain up to 5% water.

Also remember that the HPLC may see the acetone, so the 95% purity can be either of all peaks excluding any acetone peak, or by comparison of the compound peak area with the area of a known standard. You need to know the method used if you want to replicate it.

Messy, messy. Typically, unless explicit stated, I'd ignore metals, ash etc. in any calculation, just use the solvents and water, as shown by Peter.

Bruce Hamilton