External standard method for derivatized product and underiv

[praveenpaliwal]

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

[H.Thomas]

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

[praveenpaliwal]

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)

[praveenpaliwal]

Plz experts help me on this