Chromatography Forum

Hi guys!

I've got a problem with the ppm calculation. You probably say "This is very ignorant guy". It's correct. I must learn more things!

I can't determinate the concentration of my standard. For example:

My sample is weigh 500 mg. The limit of impurity is 20 ppm, how can I determinate the final concentration, after dilution, of my impurity standard?

20 ppm are 20 mg/kg, then 0,02 mg/g, then 0,01 mg/g for 500 mg sample. It is correct or no?

Can you teach me the right formula?

Your calculation is correct.

Ok. Perfcet! Thanks Tom!

Then if my Standard concentration is 0,01 mg/g if I prepare a solution, in water solvent, the final concentration of my standard solution is 0,00001 mg/ml. It' correct?

What volume of solvent do you use?

What volume of solvent do you use?

Why you ask me this thing? Sorry but I can't understand... You want know the diluition that I must do?

The density of water is approximately 1g/mL. That means that 0.01 mg/g is the same as 0.01 mg/mL of water.

Your standart concentration is 0.01 mg/g. Then you make solution in water, is that correct?
Calculation way in this case is:
ppm = Mass of standart sample * 0.01 / volume of water * 1000

Ok. Perfcet! Thanks Tom!

Then if my Standard concentration is 0,01 mg/g if I prepare a solution, in water solvent, the final concentration of my standard solution is 0,00001 mg/ml. It' correct?

"The density of water is approximately 1g/mL. That means that 0.01 mg/g is the same as 0.01 mg/mL of water." (Tom)

So you have a 1000 times dilution in going from 0.01 mg/g to 0,00001 mg/ml, which is probably why Axar was asking about volumes.

Are you sure that you are not confusing liters with milliliters ?

Peter

Ok. Perfcet! Thanks Tom!

Then if my Standard concentration is 0,01 mg/g if I prepare a solution, in water solvent, the final concentration of my standard solution is 0,00001 mg/ml. It' correct?

That is why I guessed, that Alemaggot is going to dissolve his standart sample in some volume of water.
English is not my first language, so I can miss some valueble information.

Ok. Perfcet!

well, not quite "perfect"...

Ok. Perfcet!

well, not quite "perfect"...

Sorry

Am I missing something here? You said:

"The limit of impurity is 20 ppm"

That means the most your impurity can be is 20 ppm, not that its concentration in the API actually IS 20 ppm, right? So, I don't see how you can calculate actual concentration in a dilution--you can only calculate the upper limit of the concentration.

Guys I've got new question.

I must analyze one thing in Head space vial. I must quantificate it, then I must prepare my standard solution.

I put in my head space vial the powder, without solution. For example 100mg. I must reserch a residual solvent in this powder. The limit of solvent is 3000ppm, then I must prepare a STD solution with 3000ppm of my impurity.

Well. Let's go. 3000 ppm are 3 mg/g. Then 3 mg for 1 g of powder. For my standard solution (liquid) 3000 ppm are 3ml/l. Then 0,003 ml of impurity in 1 ml of solution. It's not the same thing.
How can I compare grams with liters? I must use density? 1 ml of water is 1 g, then I've got 0,3 ml/g. But I don't think that is the right way.

I can't understand this calculation. Can you help me?

You are going to compare impurities in different substancies (powder against water). It's not right.
Try to add some internal standart to your powder. Do not use standart solution.

Mmm ok...

But If I add a few microliters of solvent to my powder?

For example 200 ul to 100 mg of powder. In this mode I can use standar solution. But my sample concentration is varied, or not? Before I've got 3mg/g then 0,3 mg for 100mg of sample. If I add 200 ul of solvent the impurity concentration is go from 0,3 mg to 0,075 mg?