QUESTION about calculation in ( purge and trap GC-MS)

[eight2015]

Hi, I am new to this technique. I searched the forum but I did not find my request.

Now I know how to do external/internal calibration using direct GCMS. However I am lost when using purge and trap.

I have vegetable samples, I spiked them with 10 mg/L compound A, B and C LETS say D is an internal standard. then how can I measure recovery using purge and trap?

what if I have water or liquid solution, say I spiked 10 mg/L compound A,B AND C ( D is an internal standard ). I feel I take 5 ml from the sample then just divide the area of sample by standard and get recovery.

I would be grateful if could anyone explain in it in a very basic matter ( training level).


My suggection for part 1: take 1 g sample, spike by 10 mg/L then put in purge and trap and the peak area that come will determined from my calibration curve, then this will be in 1g,, i am worry here from dilution etc etc

I read many EPA method but don't understand I feel it is quite advance

Please some time to teach me here and if you send me any paper or books I will read and buy them all as I really want to learn by taking my self from primary level into advance

Thanks

[eight2015]

Hi, kindly spare some time to help me here.

[Bigbear]

Purge and trap is just another injection technique. Your GC software should allow you to build an internal standard method. All you need do then is prepare your calibration standards and purge them. You will need to add the same amount of internal standard to everything you purge. Most of the newer P&T systems will spike the IS for you.

[Steve Reimer]

If you take 1 gram of vegetable matter and place it in your purge vessel then your sample amount is 1 gram. Don't worry about how much water is in it.
If you extract 1 gram of vegetable matter in methanol or water and then take a portion of that into your purge vessel then your sample will be less than 1 gram and you may need to account for the water in the vegetation. It will dilute your extract volume.

[tangaloomaflyer]

Are you extracting or just purging the vegetable matter directly?

[eight2015]

Thanks all

Tanga, i am looking to use both methods so my question is rely on calculations/quantitation I really wish to see worked calculation in excel or step by step clear papers

Thank my email ( eight2015@hotmail.com)

[James_Ball]

Thanks all

Tanga, i am looking to use both methods so my question is rely on calculations/quantitation I really wish to see worked calculation in excel or step by step clear papers

Thank my email ( eight2015@hotmail.com)

The simple way is to not do any calculations at all actually.

What we normally use is a 5ml or 5g sample size.

For Water samples:
For all samples, blanks, and calibrations standards we use 5ml sample volume. Each calibration is made in a 100ml volumetric flask to a known concentration. Lets say we do a 5ug/l , 10ug/l, 25ug/l, 50ug/l, 100ug/l and 200ug/l made into 6 volumetric flasks. Take 5ml from each and add 10ul of your internal standard solution to make a known concentration in the 5 ml. If you use a 25ppm solution of your internal standard in methanol, then inject 10ul of that solution into every 5ml sample it gives a concentration of 50ug/l.

Since you are preparing each water sample the exact same way, you just enter your calibration curve into your calculations as 5ug/l=area of peak in standard 1, or 10ug/l= area of peak in standard 2, ect and all internal standards will use 50ug/l=area of internal standard peak in standard 1, 50ug/l = area of internal standard peak in standard 2 ect.

From this just treat it as a normal internal standard calculation, ignoring volumes purged, since they are all the same they cancel out. You end up with a response factor of a water with 5ug/l, 10ug/l, ect to plot into your curve. As long as you always purge 5ml samples you don't have to calculate anything but a simple concentration directly from the peak areas.

Solid samples:

Here is the really simple part. If you have the calibration done for the waters, you can do it the exact same way for solids. Simple assumption is that 5ml water = 5g water = 5g solid sample. If you always purge 5 grams of your solid sample then you already have a calibration that will work. It is better if you have a clean matrix that you can weigh 5g of into each calibration standard, but that will only correct for any adsorption the matrix may have on the analytes of interest. You can always run a spiked sample to see what the matrix effects are and use that to correct the data if you wish. Plus finding a truly clean matrix of your sample may be difficult. For soils we normally use sea sand that has been baked in an oven overnight after washing with methanol.

Any deviation from 5g of sample can be corrected for by a simple ratio correction factor. If you use 5.5g sample then multiply your result by 5/5.5 to correct for weight.

Other considerations:

Most systems will have a good linear range for 5ml samples in the concentration range of 5ug/l to 200ug/l. You can extend it by increasing your split ratio at the expense of low end sensitivity. I have managed to calibrate some of mine from 0.2 ug/l up to 200ug/l but is is not always possible to get that much linearity.

If you expect high ppm range concentrations then you should probably dilute your samples prior to analysis or do direct injection.

[eight2015]

Thanks James , this explanation is extremely helpfull. I reallly understood every point from you. Thanks indeed indeed sir, god bless you.

[James_Ball]

Thanks James , this explanation is extremely helpfull. I reallly understood every point from you. Thanks indeed indeed sir, god bless you.

Glad I could help, and have a blessed day also.