Chromatography Forum

Trying to work this out with a colleague of mine. If you are supplied a free base salt and want to describe a solution in terms of molarity, do you use the molecular weight of the salt or the molecular weight of the free base?

Please note the values below are made up but illustrate the idea of the problem:

I have a method that calls for me to make 200 mL of a 100 uM solution of compound Y. The salt form of compound Y is 350 g/mol and the free base form is 325 g/mol. The material was supplied as a salt.

My colleague thinks the correct way to make this solution is to use 0.0065 g the material in 200 mL (0.0065g/(325g/mol)).

I think the amount to use should be calculated by (0.200 L*(100 umol free base/1L)*(1 umol salt/1 umol free base)*(350 ug salt/1 umol salt) = 0.0070 g of the material in 200 mL.

I think my colleague's way only works if you ALSO apply a correction factor (350/325).

Please let me know what you think. thanks

In answer to your question, both of your suggestions are correct. When expressing concentration you need to indicate whether it is a base or salt (and type of salt). It generally seems to be assumed that it is a base if a salt isnt indicated.

For example with thebaine bitartrate, to make a 1 molar solution (of thebaine bitartrate) you would weigh out 467 grams of the bitartrate salt per litre of diluent, but if you wanted a 1 molar solution of thebaine base you would weigh out 311 grams of the base per litre. The actual thebaine concntration is 1 molar in both solutions. Its easy when working in moles.

It's a little different when working in w/v concentrations, for example to get a 1 g/L solution of thebaine, one would weigh 1 g of thebaine base, or 1.50 grams of the bitartrate salt (467/311 x 1) per litre to acheive the same concentration. BUT if the intended concentration was 1 g/L thebaine bitartrate, you would weigh 1 g of the bitartrate salt per litre, or 0.67 grams of thebaine base per litre.

Hope this helps

Can you clarify this for me:

"For example with thebaine bitartrate, to make a 1 molar solution (of thebaine bitartrate) you would weigh out 467 grams of the bitartrate salt per litre of diluent, but if you wanted a 1 molar solution of thebaine base you would weigh out 311 grams of the base per litre. The actual thebaine concntration is 1 molar in both solutions. Its easy when working in moles."

Does that mean if you have the salt you would weight out 467 grams, but if you were supplied the free base you would weight out 311 grams?

This is the crux of the question. In my colleague's example she would be weighing out the salt, but using only the molecular weight of the free base to calculate the molarity of the solution. To mean that doesn't make sense, but she seems to think this is how such concentrations are commonly described when talking about salts.

"Moles is moles". If you want to weigh out 1 mole of something, you have to know how big a molecule is attached and use that MW. (i.e.. the weight of salt vs. free base used will be different).