Bioequivalence and Bioavailability Forum

Dear sesame!

❝ I have to weight equal to 44.4mg of atorvastatin from atorvastatin calcium trihydrated working standard.

Let's see whether I get the stoichiometry right...
The formula of atorvastatin is C₃₃H₃₅FN₂O₅ giving a molecular mass of 558.64 g·mol^-1.
Since atorvastation is a monovalent acid and Ca(OH)₂ a bivalent base the equation looks like:

2 RCOOH + Ca(OH)₂ → (RCOO)₂Ca + 2 H₂O

In other words each mole of the calcium salt equals two moles of atorvastatin acid.

The formula of anhydrous atorvastatin calcium is Ca(C₃₃H₃₄FN₂O₅)₂ with a molecular mass of 1155.36 g·mol^-1. For the trihydrate Ca(C₃₃H₃₄FN₂O₅)₂ · 3 H₂O we get 1209.41 g·mol^-1.The factors to the free acid based on the molecular masses are 2.06813 (calcium salt) and 2.16488 (trihydrate).
44.4 mg of atorvastatin are equivalent to 44.4 mg / 2.06813 = 21.47 mg anhydrous Ca salt or 44.4 mg / 2.16488 = 20.51 mg trihydrate.

❝ potency as atorvastatin is 94.6%

❝ potency as atorvastatin calcium is 98.7%

❝ Please anyone can show me how to calculate for the weight... I'm confused.

Me too. The factor of the calcium salt to the trihydrate is 1155.36/1209.41 = 0.95531, but 94.6%/98.7% = 0.95846 - even if we use just three significant digits 0.955 # 0.958...


I would ask the supplier to explain this discrepancy. Since you have the trihydrate (and if you trust in the potency given as atorvastatin) in order to get 44.4 mg acid you should weigh 20.51 / 94.6% = 21.68 mg (well, I guess your balance could measure only to a tenth of a milligram: 21.7).

Warning: Check with a chemist who didn't finish his/her studies 30 years ago like I do.