Chromatography Forum

Hi All

I'm using a std whose potency is expressed as % m/v. When using this std, during HPLC, should I convert the potency from % m/v to % m/m. I feel we should express the std as % m/m by dividing the potency by the density of the std. Some of my colleagues feel we should not use the density in the calculation. Any comments will be appreciated.

Example of Assay calculation
Std prep: 100mg → 100 ml.
Sample prep: Dilute 5 ml to 10 ml

mg/ml=
Ru x mass std (mg) x 10 x Potency of std (% m/v) x 1
Rs x 100 x 5 x 100 x density of std


Many Thanks
Mike

Hi Mike,
Yes, I feel density is not required in calculations, because you are not using the standard in volume. You are using straight away weight in the calculations. You are applying the weight and potency of the standard. That’s enough.

Its early so I may be a bit off but here's my logic:

If the standard potency is expressed in m/v % then how much mass of your active have put in your calibration standard when weighing in 100mg?

If you have an accurate density value for the standard solution then you could use that density value to calculate the mass of active you have added to your standard, i.e. std. conc in mg active/mL soln. If this is a concentrated mixture and a reliable density value for the mixture is not available then a volumetric pipet would allow you to calculate the mass of active in your standard but this could be problematic with viscous solutions.

The key calculation is 100mg std x std purity in m/v (fractional). Obviously the dimensional analysis does not work.

The simplest approach would be to use volumetric pipets and flasks for all standard and sample prep.

You could check the validity of each approach by weighing an accurate volume when making your calibration solution. The mass should be close to what you normally use (100mg) or increase the volume and mass proportionately to allow accurate volume measurement. Calculate results using the standard calculation with mass standard and the alternative calculation using volume to calulate the mass of active added.

Thanks brbabu_1979 and Carls for your response.

I am able to determine the density of the std accurately. The std I'm using is 50% /v Benzalkonium chloride std.

thanks

The std I'm using is 50% /v Benzalkonium chloride std.

So that's 50mg Benzalkonium chloride/100mL solution? Is it dissolved in water? What is the density?

Hi Carls

The 50% m/v std is dissolved in water i.e 50 g of benzalkonium chloride/100ml solution. The density of the std is 0.9740.

So 100mg solution is 102.7uL or 51.3mg added to 100mL = 0.513mg/mL

Alternatively, 100mg x 0.5 = 50mg added to 100mL = 0.5mg/mL

This will give an ~2.5% low bias in your results.

Is someone reinventing the wheel? I hardly understand any of this.

What dont you understand or which way do you see it?

The 50% m/v std is dissolved in water i.e 50 g of benzalkonium chloride/100ml solution. The density of the std is 0.9740.

Did you prepare the standard by weighing out an exact weight of a solid standard or did you pipet an exact weight of a stock solution of known concentration and density?

Thanks Carls for your comments. The density, therefore, needs to be taken into account.
skunked_once - The std is a liquid which a potency of 50% m/v. This std is then weighed off and made up to volume.

thanks

Hi Mike
If you use the below formula finally you will get dimension less value.
mg/ml=
Ru x mass std (mg) x 10 x Potency of std (% m/v) x 1
Rs x 100 x 5 x 100 x density of std
How means
Spl Area X mass std (mg=m) X 10 (ml) X Potency ( %m/v) X 1
Std Area x 100(ml=v) x 5(ml) x 100(m/v) x density of std (m/v)

I hope you will understand the above formula. Finally you will get dimensionless value. If you do not use the standard density in the calculations you will get the value with the dimensions of mg/ml that is you are looking for. I have experience with same analysis. Generally our intension is to prove the claim of the product. We also are calculating the BKC without considering the density of the standard in the calculations. So my suggestion is the density of the standard is not required in the calculation part.

Babu

Molarity and molality, expressed in different units, have been used by generations of chemists. These are concentrations, what is that nonesense "potency"?
Activity expressed in x units have been used if the formular of the substance was unknown.

Hi Mike
If you use the below formula finally you will get dimension less value.
mg/ml=
Ru x mass std (mg) x 10 x Potency of std (% m/v) x 1
Rs x 100 x 5 x 100 x density of std
How means
Spl Area X mass std (mg=m) X 10 (ml) X Potency ( %m/v) X 1
Std Area x 100(ml=v) x 5(ml) x 100(m/v) x density of std (m/v)

Babu - the 100 (m/v) in your formula above is incorrect. The "extra" 100 included in Mike's original formula was used to change the potency in % to the fractional value, i.e. 50%/100 = 0.5 or 50g/100mL = 0.5g/mL

Molarity and molality, expressed in different units, have been used by generations of chemists. These are concentrations, what is that nonesense "potency"?
Activity expressed in x units have been used if the formular of the substance was unknown.

Mueller- please be mindful of personal and cultural preferences. The choice of the word potency over purity or concentration (which is how the "potency" is expressed) does not obscure the intent of the question.

Molarity and molality are not relevant to this discussion.