Chromatography Forum

Hello guys !

I’m actually trying to determine Toluene Conc in some nails colors. I am actually confused in some quantitative calculation using Headspace GCMS. I have done a calibration curve of three STD points; 1,5,10 ppm Toluene. I weighed 0.05 g of the sample and dissolved in Acetone to final volume of 50 mL. Then, I took only 0.5mL of the sample in Headspace vial even for the three standards and did the running. The result was 8 ppm for my sample in the report.

My question is… should I multiply 8 ppm with the final volume 50ml and divide it by sample weight 0.05 to get the actual concentration ?
Is this called Dilution Factor (Final volume/ sample weight g) or Concentration Factor? What’s the difference?

Hi may i know what is the volume of the standard you have used for head space analysis, also tell me the standard concentration is 1,5,10 in ppm or in µg/mL?

actuallly I prepared three individual standards 1, 5, and 10 ppm . from each std flask, I took only 0.5 ml into the HS Vial which is the same volume I have taken for the samples.

thanks kalidassa

Hi still i am not clear with your dilution, 1,5 & 10 ppm are µg/mL or ppm? That means the concentrations are made with respect to the sample dilution? or with out considering the sample dilution? If u not calculated according to your sample dilution, as per your sample dilution 0.05g (50mg)/50mL your dilution factor is 10.

ppm = µg/mL/ Sample wt(g)*dilution of the sample
For your case if µg/mL is 10 means = 10/0.05*0.5 = 100 ppm.

Sorry Mr.alabsiam Actually your dilution factor is not 10 it is 1000 times.

µg/mL is 10 means = 10/0.05*50 = 10000 ppm. so your dilution factor is 1000.

May i know why you are diluting the sample 0.05g/50 ml, since you are doing head space injection fro GCMS analysis why cant you do 0.05g/0.5ml.

my samples are very concentrated "nails colors" .. when I took it without dilution, I got very high respense and it was higher than the response of my highest std point which is 10 ppm. this is why I tried to take very low sample weight

Again ,,, I prepared 3 different standards 1, 5, 10 ppm in different volumetric flasks which were diluted from stock 1000 ppm

I run these three standards by taking 0.5 ml in the vial and I got a decent and linear curve.

then, I took 0.5 ml sample which was prepared by taking 0.05g from the nails color into volumetric flask 50 ml

after that I measured my sample peak and got a result of 8 ppm in the quantitation report

what's your suggestion ?

alabsiam

You don't understand the questions asked by kalidassa and he will help you if you do so.

Describe step by step and the calculations involved in preparing your standards and the way you measure out the samples you have tested.

Don't repeat what you have stated before, but restate it so it will be understood.

Thanks.

Rod

Hello Rod,

thank you for your valuable reply ... I'm trying to restate and not repet actually ... but it could be due to language difficulty that I have ,,,,,


anyhow... thank you guys


Alabsiam

Hi as per your dilution of the sample which is 0.05g/50mL, you need to inject 0.005 µg/mL of Toluene standard, which is 10 ppm with respect to your sample dilution.


ppm= 0.005/0.05*50/0.5 = 10 ppm


The standard which are injected 10 µg/mL is 2000 ppm with respect to your sample dilution.

If your samples and standards are prepared exactly the same way then you can read the sample concentration directly from the calibration curve.

If I understand your sample/standard preparation corerctly, you dilute 0.05g sample or standard in 50mL solvent then place 0.5mL of this solution in a headspace vial. If the concentration of analyte in the standard prior to dilution was 10ppm, for example, and your sample gives a similar response to this standard then the concentration of analyte in the original/undiluted sample was 10ppm.

Hopefully this example helps you understand the questions on the board.

The primary question is: Is the concentration of the standards you use to define the calibration before or after dilution:

Are the 1, 5, and 10ppm concentrations for the standards before or after dilution (0.05g standard to 50mL solvent)

If the 1ppm standard concentration is before dilution then you can use this curve to read sample concentrations directly since both the standard and sample were diluted the same. Since you want to know the concentration in the undiluted sample this is the simplest way to perform this analysis.

If the 1ppm standard concentration is after dilution then you must multiply the sample result by the dilution factor (50/0.05 =1000) since you want to know the concentration of the analyte in the undiluted sample.

The key issue here is to understand you are measuring the mass of analyte in the headspace vial.

Case 1) If you add 0.5mL of a 1ppm (ug/mL) standard then you have added 0.5ug of analyte to the vial.

Case 2) If you add 0.5mL of a 1ppm standard that was diluted 0.05g/50mL prior to adding to vial then you have added 0.5ng (1000x less = dilution factor) to the vial.

If your diluted sample gives a response simlar to case 1 (0.5ug in vial) then the 0.5mL of the diluted sample contains 0.5ug of analyte (i.e. diluted sample is 1ppm) and therefore the concentration in the undiluted sample is 1000ppm (1000x dilution factor). If your diluted sample gives a response simlar to case 2 then the concentration in the undiluted sample is 1ppm

Dont forget about variations in the weights of sample/standard when diluting, example - exactly 0.05g weighed each time?

Dear carl ,

Thank you so much for your valuable reply . I really appreciate it .

Now its very clear to me. Regarding the standards they were prepared and diluted from stock 1000 ppm toluene and got 1,5,10 ppm then these three standards were injected by taking 0.5 ml into headspace vials . Got calibration

Then for my sample. I weighed 0.05 g of nails color sample into 50 ml solvent After that i took 0.5 ml into headspace vial and was injected

This sample gave a result of 8 ppm .

Now based on your explanation, to get a corrected concentration , i have to multiply 8 by (50/0.05) which comes out as 8 * (1000) = 8000 ppm

Am i right ?

Once again, thank you so much for taking the time and tried to help

Alabsiam

Given the way you prepared the standards the 8ppm result tells you the concentration of the diluted sample solution. So, yes, the concentration in the original/undiluted sample would be 8000ppm.

Good luck!