Relying on a standard curve operating through a set of concentrations, a pre-prepared standard compound

With this device the curve can be made using concentrations against the area under the curve of the standard compound

In this way, when injecting a known volume of the extract containing this compound, it is possible through an area under the curve to determine the concentration using the equation of the straight line

best regsrds sabheljoe R Balqees Al-Musawi

Shafq Al-azzawi

You have to check if the standard is the same compound you isolate from plant. After you are sure of the identity of compound I suggest you to prepare a calibration curve for the standard (at least three solutions: 100; 500 and 800 ppm). If correlation coefficient of the calibration curve is good (at least 0.95) then you can calculate the concentration of your sample by interpolation of the response of detector.

Thank you @Daniele

Yes the standard is authenticated from the same plant.

I already made calibration curve for the standard ( 3 different concentrations) 0.5 , 1 and 1.5 mg/0.5ml and produced the calibration curve with r2 was 0.992258

but I am confused how can I calculated the concentration ?

any other comments and suggestions

You extrapolate from the graph the concenteration of your extract from the peak area u get for ur compound

thank you @Wangkheirakpam actually i did that and used the equation mentioned up

but i am confused how can the concentration is 49.3 mg/ml in the case that my original fraction concentration was 15mg/20ml !!!! any comments

Hello sabheljoe R : I saw your attached file and the equation could be work . but there is other way ,you can make preparative method by collect the area of your sample by fractional collector to get only single component which lead to determination the concentration.

Thank @Alaa actually it is difficult to do the preparative method as we do not have fractional collector and our HPLC is analytical HPLC

So I have to you what is available !!

I am still worry about the calculations and the results

I tried to follow all the procedures but the results that I got made me in doubt !! as how can the concentration is 49.3 mg/ml in the case that my original fraction concentration was 15mg/20ml !!!! any comments

You are plotting peak area and concentration. So simply use the peak area of the peak you got from the extract and find the corresponding concentration .i think you can see from the size of the peak the extract might be smaller . But your answer might be right as in 15mg/20ml is 300mg/ml so your answer 49.3mg/ml is quite lesser than your 15mg/20ml. So if you have done correctly i feel your answer is right.

You dont needto use the equation as you have drawn calibration curve

Thank you so much @M. Farooq your answer gave me clear explain.

but I have some questions if you can clear please

I understand that mass Y is the mg that in the injection volume which is in this case 0.01 mg but not the original 5 mg which are in the original stock ? am I correct ?

so for example if the injection volume was 1 ul the mass would be 0.0005 correct ?

another thing can you explain how can I calculate the mass of the compound present in 20 ml ?

thank you a lot for your response

@Wangkheirakpam thank you for your valuable addition

@Wangkheirakpam can you please let me know how did you calculate that 15mg/20ml is 300mg/ml ?

another calculation from the HPLC software >> it made an automatic calibration curve and automatic calculation and gave me an answer but still not sure if I can consider the machine answer or not

the result of the software answer was 284.9 mg/ml !!! which is very different on my calculation results !!!!

Peak intensity in the term of area percent will change according to quantity used

your calculation is right

@M. Farooq and @Shiv Yadav and all who tried to help me thank you so much and

thanks in advance to any body who would like to give me more comments and suggestions as I will keep the window opened to learn more and more from the experienced people

sorry i wrote in hurry i will be careful even i work in thisfield so i was interested . I think you have to see the injection volume and your loop capacity . But for confirmation please repeat ur expt and keep the conditions of extract and sample same

Sorry but something looks wrong for me here, assuming your extract it's so good that correspond to a 100% pure standard. 15 mg in 20 ml the max concentration you can get is 0.75 mg/mL (you are not diluting the sample on the way, true?. So that indicate for me two choices:1- I did'n understood how was your sample preparation (15 mg extract in a volumetric to 20 mL solvent?) or you have something else coming at the same RT.

@Roberto that's why I am confusing from what I got from the calculation !! by that equation I've mentioned in the question and that published in the attached article !!!

in fact I noted that when I run the samples in triplicate I find that the Peak areas changed differently from run to run, and this happens to the standard peaks too.

If your peaks areas changed differently from run to run you should not take these numbers in consideration. I advice you to check and clean your HPLC system (injector, pre-column, column....). After that at least you should get consistent values and then you can calculate your sample conc.

If you still get an over estimation of what you expect to be the real conc of your sample, then run using different conditions, maybe you have two chemicals coming at the same RT.

@Roberto I am not sure if the system needs to be cleaned because it is brand new and I am the 2nd person using it after my friend who used it only for 10 samples.

the RT is stable actually and RT of the standard and the sample is matched

I would rather that there are Two chemicals at the same RT , but How can come and overlay over each other ? and no any sign for that ? !!!

is there any possibility to say that there may be an isomer of the original compound comes at the same RT ?

the formula is correct. you can try to run samples in duplicate and at different concentrations in order to obtain more accurate results. Maybe if it is the first time that you use the instrument, the manufacturer could help you with the software for the construction of the calibration curve.

Cheers

Try to make the serial dilutions of your standard and inject into the HPLC system, measure the peak areas of all, plot the calibration plot (Peak area VS concentrations) then you get the regression equation (y= mx+ c). From which calculate the unknown sample. Other wise include the response factor in your equation.

RESPONSE FACTOR= SLOPE OF THE REGRESSION LINE.

yes, you should go for the panikumar anumolu comments. Moreover, to reduce sample preparation error you can use internal std. the method issame just like drawing calibration curve. Just add same amount of IS to all dilution and test too.

Take ratio of std/IS peak area ratio and draw calibration graph vs concentration of std. and extra plot to get unknown concentration. Validate your method first.

Hope this will work.

Cheers

Peak intensity in the term of area percent will change according to quantity used

your calculation is right

@Nerijus Karlonas @Shiv Yadav Can you please explain what is the meaning of "Peak intensity in the term of area percent will change according to quantity used" ???

thanks in advance

Dear all who suggested and helped thank you ,

please see this curve that I already did according to the results and I already did the calculation according to the equation, However I got very high concentrations !! I run the same sample twice and I calculated the concentration according to the peak areas the first run I got 198 the other one got 150 !!!

and it is more different than the one that i got from the calculation that I mentioned in the question up !!!?

which detector are you using UV?RI?

Peak intensity is peak height .i think something is wrong with your calculation or while taking the peak area. How did u integrate the peak area that also will change your answer.

@ Roberto using UV !

@Wangkheirakpam I am still confused about the areas too

but I run the samples many time it gives me different area and when I make the calculations it gives me the same confusion concentrations !!!

my next step is I will collect fractions by prepHPLC and then will measure how much I will get ?!! do you think it makes sense ?

many factors contribute the mobile phase if impurities are there then the way you integrate i also have that knowledge only but Practical yield is important that is what you get practically and weigh it parctically so i think thats better.@sabheljoe R

Wangkheirakpam Sujata, thanks for your contributions and all others. It is an interesting topic although am very late.Your explanation is well understood, but would you explain how 15mg/20ml could be equivalent to 300mg/ml? and further how  49.3mg/ml is quite lesser than the initial concentration of 15mg/20ml used by @sabheljoe R ? I thought the highest concentration that can be achieved here is 0.75mg/ml and even less if we consider the injection loop and not the initial concentration as you and others have rightly explained above? In my view, @sabheljoe R's fears are justified because the concentration is actually high.However, when the loop injection is considered, then calculated value should be smaller.

There is no universal detector for HPLC, and till now MS is most reliable, so best approach is cross examine standard sample using MS detector and compare with your result.

good luck

Is there a formula to convert a peak area of HPLC to concentration?

Peak area or peak height can be used to calculate the concentration or mole fraction

can You please give the reference?

Relying on a standard curve operating through a set of concentrations, a pre-prepared standard compound

With this device the curve can be made using concentrations against the area under the curve of the standard compound

In this way, when injecting a known volume of the extract containing this compound, it is possible through an area under the curve to determine the concentration using the equation of the straight line

best regsrds sabheljoe R Balqees Al-Musawi

Shafq Al-azzawi

0.5 of std ist mg/ml?

You have to make dilution to the extract until you obtain a peak area close to std. Solution to an accurate comparison