Chromatography Forum

Good noon,

I have one sample in which a gaseous impurity is trapped. I am analyzing it on HPLC with derivatization. I have used one derivatizing agent and derivatized the gaseous impurity in the the sample. Now I need to quantify this gaseous impurity sample. I have used the standard, which is already in derivatized form. Now in external standard calculation I m getting little bit confused, plz help me.

The out line is as follows;

Standard: A, (Already derivatized and weigh as such. The mol weight of derivatized standard is 212.25)

Sample ( for trapped gas) : B (It is a trapped gas and present as impurity in sample. It is derivatized by a derivatizing agent and with this derivatization now it is converted in the same form as standard) The molecular weight of this gaseous compound is 98.92.

Now the calculation-1 is :

Content of gaseous compound B = (Area of compound-B in derivatized form x Weight of already derivatized standard in ppm x Purity of drivatized standard) x Mol wt of derivatized standard (212.25) / (Area of derivatized standard x weight of sample in ppm x Mol weight of gaseous impurity (98.92)

Now the calculation-2 is :

Content of gaseous compound B = (Area of compound-B in derivatized form x Weight of already derivatized standard in ppm x Purity of drivatized standard) x Mol wt of gaseous impurity (98.92) / (Area of derivatized standard x weight of sample in ppm x Mol weight of derivatized standard (212.25)

Now plz help me which calculation is correct.

Best regards
Praveen

Hi Praveen,

if you do the calculation step by step, it should be clear.

The amount (in mg oder whatever unit you prefer to use) of the standard is its weight * purity (e.g. 100 mg * 98% = 98 mg).

This amount divided by the peak area of the standard gives you a response factor (eg. 98 mg / 1000 mAU.min / = 0.098 mg/(mAU.min)

Now you can multiply the area of your sample with the response factor (eg. 1250 mAU.min * 0.098 mg/(mAU.min) =122.5 mg).

So you have 122.5 mg of derivatized analyte in your sample. Now you multiply this by the ratio of the molar masses, like 122.5 mg * 98.92/ 212.25 to get your result of 57.1 mg.

Are you sure a single standard is enough?

Best regards, Hartmut

Thanks Harmut,

for your reply. Yes we have injected 2 runs of standard and then injected sample. From your answer what I am understand that in calculation formula the IInd one is correct:

Content of gaseous compound B = (Area of compound-B in derivatized form x Weight of already derivatized standard in ppm x Purity of drivatized standard) x Mol wt of gaseous impurity (98.92) / (Area of derivatized standard x weight of sample in ppm x Mol weight of derivatized standard (212.25)

Molecular mass of gasesous compound (98.92)/ Molecular mass already derivatized standard (212.25)

Plz experts help me on this