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Author Topic: Prepare antibiotic stock concentration according to potency and hydratation  (Read 135 times)

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CLG

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Question:
I want to prepare an antibiotic at 100mg of cefotaxime in 10mL of water.

I have the cefotaxime sodium salt powder from sigma and here is the information I have:
Molecular formula:  C16H16N5O7S2Na
Molecular weight: 477.5
Molecular weight of cefotaxime: 455.47
Moisture content < 6%
Assay (Anhydrous Basis) ug Cefotaxime/mg:  916 - 964 (0,916 -0,964 mg/mg)

Iam really confuse with all this info... How much antibiotic should I weight to have 100mg of cefotaxime pure...Could someone help me with the calculation?  ???
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Arkcon

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So... you want 100 mg of cefotaxime in 10 ml of water.  So just mass out 100 mg, and dissolve it in 10 mL total volume.  Simple, right?

But wait, you know from the label, its not pure, so you know you need some more.  Just a little bit more.  But how much?

Can you begin to use the data you provided to solve this problem for yourself?  Can you at least exclude some information?

Sorry to answer questions with more questions, but this is what we do here.  Check out the Forum Rules{click},
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That all depends on how reasonable we're all willing to be.  I just want my friends back, except for Cartman, you can keep him.

CLG

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Thanks a lot for your reply   ;D

No problem, I understand the rules! I agree that it is the best way to learn! ;) Here is my first guess:

1) Using the molecular weight, I found that the weight of the antibiotic should be 95.39mg instead of 100mg ((100mg*455.47)/477.5). As the molecular weight of the cefotaxime is smaller, the weight will be smaller.. Is it right?

2) My second guess is that the purity of the product is about 95% ((455.47/477.5) *100). So, if we want a purity of 100%, I should mass out 104,84mg ((100mg * 100%)/95.38%).

Also, I don't understand what does the potency (assay mg/mg) is? Does the moisture content changes the final number?

« Last Edit: Today at 04:02:34 AM by CLG »
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Arkcon

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I don't think we need to use the molecular weight for anything, unless we need to determine molarity, as an example.

Moisture is a dominant impurity, the assay tells you how much added weight the sample has, so you should use proportionally more.  Warning: on this planet, moisture is everywhere.  The value may change.  You may be called to determine moisture content, by Karl-Fisher titration, just before using this reagent.  You will find such regulations in The Merck Index, or whatever pharmacopeia you're using.

You might also want to try to figure out what the assay results mean, and if they're significant for your application.

You can also just prepare a 1 mg/Ml solution, then, when you have results, scale them according to your purity.  You will probably have to do this anyway.  Consider:  How would you prepare 1 mg/mL of a substance you knew was 100.0000% pure?  Would you weigh out 1.00000 mg?  What if you try, and get 1.00010 mg?  Are you going to remove and discard 0.0001 mg? 
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That all depends on how reasonable we're all willing to be.  I just want my friends back, except for Cartman, you can keep him.

CLG

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Thanks so much for your answer!  :)

So,

1. (If I don't check the water content before use:) The moisture content is between 0-6% (purity 94%-100%), the weight will be 100mg-106.38mg ((100mg*100%)/94%).
2. As I don't want the weight of the sodium salt, I will consider the assay (0.916-0.964mg/mg) and so, the weight will be between 91.6 mg-102.6mg...

What value should I consider...Is it ok if I weigh 100mg in 10 ml of water total?
« Last Edit: Today at 03:31:54 PM by CLG »
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