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sesame ☆ 2010-11-01 20:14 (5718 d 20:40 ago) Posting: # 6100 Views: 11,419 |
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Basic question regarding potency. I don't know where posting this question. I have to weight equal to 44.4mg of atorvastatin from atorvastatin calcium trihydrated working standard. 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. Thank you very much. Edit: Category changed. [Helmut] |
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Helmut ★★★   Vienna, Austria, 2010-11-01 21:45 (5718 d 19:10 ago) @ sesame Posting: # 6101 Views: 10,328 |
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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. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! ![[image]](https://static.bebac.at/pics/Blue_and_yellow_ribbon_UA.png) Helmut Schütz ![[image]](https://static.bebac.at/img/CC by.png) The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
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ElMaestro ★★★ Denmark, 2010-11-01 22:07 (5718 d 18:48 ago) @ sesame Posting: # 6102 Views: 10,153 |
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Hi sesame, ❝ I have to weight equal to 44.4mg of atorvastatin from atorvastatin calcium trihydrated working standard. ❝ 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. Need more info. My best guess is that you can use 44.4 mg x 0.946 to get your A-weight. Seems to me the trihydrate dried up a little bit. Therefore, my guess is that your assayed (apparent) content of the trihydrate exceeds to 100-mark slightly. What does it say on your CoA?  Best regards EM. |
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Helmut ★★★ Vienna, Austria, 2010-11-01 22:20 (5718 d 18:35 ago) @ ElMaestro Posting: # 6103 Views: 10,181 |
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Hi ElMeastro! ❝ My best guess is that you can use 44.4 mg x 0.946 to get your A-weight. No way. If the potency is 94.6%, more should be weighed in in order to compensate to 100%, not less: 44.4 mg / 0.946 = 46.9 mg. But this would only be true if the (free acid of) atorvastatin is available. Maybe I messed up my calculation entirely (rusty brain syndrome), but still one mole of the salt is equivalent to two moles of the acid (which we don't have). Therefore it's a little bit tricky. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
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ElMaestro ★★★ Denmark, 2010-11-01 22:50 (5718 d 18:04 ago) @ Helmut Posting: # 6104 Views: 10,114 |
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Dear HS, ❝ No way. The way I understand the given info is this: Sesame has 44.4 mg of a powder which is assayed to contain what corresponds to 94.6% A. Way? EM. |
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Helmut ★★★ Vienna, Austria, 2010-11-01 23:04 (5718 d 17:51 ago) @ ElMaestro Posting: # 6105 Views: 10,114 |
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Hi ElMaestro! ❝ ❝ No way. ❝ You sound like my girlfriend.  ❝ The way I understand the given info is this: Sesame has 44.4 mg of a powder which is assayed to contain what corresponds to 94.6% A. Well, the original statement was: ❝ I have to weight equal to 44.4mg of atorvastatin from atorvastatin calcium trihydrated working standard. Sounds like a math exercise to me. Given: Salt · 3 H2O. Wanted: Amount of salt which equals a target amount of acid (44.4mg 100% pure). First complication: Mass (mg) vs. amount of substance (mole). Second complication: Salt is not pure. I wouldn't bother too much about the second complication (~5% error), but on getting the stoichiometry right (first complication) - error >50%. — Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
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moblak ☆ 2010-11-03 10:00 (5717 d 06:54 ago) @ Helmut Posting: # 6108 Views: 10,169 |
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Hi Sesame, It's 15 years since my days at university, so I hope my memory still serves me well Given the information you have provided, the calculation below would be correct (94.6% is the potency for atorvastatin AS IS)....46.9 mg of your standard for 44.4 mg of atorvastatin ❝ No way. If the potency is 94.6%, more should be weighed in in order to compensate to 100%, not less: 44.4 mg / 0.946 = 46.9 mg. However, according to our experince with atorvastatin calcium only (not trihydrate), the atorvastatin standard is HIGHLY hygroscopic and therefore, we measure the water content by Karl-Fischer weekly (and water content has increased dramatically in some cases...a few %). Therefore, the potency of 94.6% declared on your CoA might not be completely correct. Cheers Marko |
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Helmut ★★★ Vienna, Austria, 2010-11-06 19:18 (5713 d 21:37 ago) @ moblak Posting: # 6110 Views: 10,112 |
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Dear Marko, Sesame, ElMaestro & all! OK, I was completely wrong! After consulting my old textbook* and with three chemists they agreed in the following: The formula to get a potency corrected weighted sample is radical (mg desired) molecular mass (radical)or based on atomic masses from NIST in full precision 44.4 mg 558.6398032 g/molWe can also think about “molar content” of atorvastatin in the salts. For the Ca-salt it’s 96.71% (2×558.64/1115.34, “rest” is Ca) and for the trihydrate 92.38% (2×558.64/1209.39, “rest” is Ca and water). One of my friends noted that he could not follow the discrepancy between potencies of the trihydrate given as atorvastatin (94.6%) and atorvastatin Ca (98.7%). So do I – please check with the vendor.
— Dif-tor heh smusma 🖖🏼 Довге життя Україна! Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes |
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sesame ☆ 2011-01-13 20:11 (5645 d 20:43 ago) @ sesame Posting: # 6412 Views: 9,734 |
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Dear all, I was wrong about potency amount of atorvastatin an atorvastatin ca..I confused with other working standard...please forgive me.. potency as atorvastatin is 91.5% potency as atorvastatin calcium is 94,8% I found somme reply from difrent forum.. 44.4 mg atorvastatin from atorvastatin ca mean 44.4/91.5%=48.53mg to get 100% of desired weight please excuse me for this stupid mistake. thank you for all. |
Ing. Helmut Schütz