Residual solvents calculation headspace

[chemist23]

Hello all,

I am trying to quantify a sample using a 5000 ug/ml Class 3 residual solvent standard using GC headspace technique.

The procedure for the sample preparation indicates to add 5uL of sample to 20 mg of blank matrix to a head space vial.

There are no instructions of how to prepare the standard.

If I add 5uL of 5000 ug/ml standard to a headspace vial, how would I calculate the concentration of compounds in the sample in ug/g?

I would like to build a curve using the 5000 ug/ml class 3 standard.

[pumpedupchemist]

Hi,
The way I would calculate it would be..
standard = 5000ug/mL amount taken = 5uL
5000/1000uL divide both amounts by 200 to get your 5uL needed
so in the 5uL solvent the concentration will be 25ug.

If you add this to 20mg of blank matrix = 25ug/20mg but you need 1g.
20mg into 1g = x 50.
25ug x 50 = 1250ug. 20mg x 50 = 1g.
final concentration in the vial = 1250ug/g.

This may not be 100% accurate but that is how I would calculate it..

[rb6banjo]

Is it possible for you clarify exactly what you're trying to measure and in what matrix? If it's a secret then I understand if you can't share but if it's "I want to measure residual methylene chloride in a polyol matrix" then I don't know why you couldn't share it.

I believe that pumpedupchemist is correct but it might be more clearly stated as:

5,000 µg/mL x 0.005 mL = 25 µg analyte added to the vial
25 µg/0.020 g = 1,250 ppm in the matrix

That of course presupposes that the added analyte in the vial partitions uniformly into the matrix prior to the analysis.

If the "blank matrix" does mimic your sample matrix accurately then add 20 mg of the matrix to a series of vials, add different amounts of the standard to the vials that bracket your expected sample concentrations and run them like samples. I can't stress enough that you must ensure the added analyte partitions well into the matrix after you add it. This can be challenging if the matrix is a solid.

[chemist23]

I am trying to measure solvent standards (USP class 3 standards) in a base oil matrix.

My sample must be prepared by adding 5 uL of unknown sample to 0.020 g of matrix.

I have a 5000 ug/mL standard to use.

How can I prepare the headspace vials using the 5000 ug/ml standard to create a curve from 100 ug/g to 10000 ug/g

[rb6banjo]

It's good that it's an oil matrix because diffusion in liquids is faster than diffusion in solids.

To make 100 µg/g from a 5,000 ppm stock:

5,000 x V/0.020 = 100

V = 0.0004 mL (0.4 µL)

5,000 x V/0.020 = 10,000

V = 0.04 mL (40 µL)

To decrease the error in making the lower-concentrations standards, you may want to take more of the matrix to make the standard. For 1.00 g of your matrix:

5,000 x V/1.00 = 100

V = 0.020 mL (20 µL)

Then, just analyze 20 mg of the spike as a sample.

It's easier to add 20 µL accurately than it is to add 0.4 µL. You can then choose some other standards between the low and high to construct your calibration curve (determine the response factor).

[chemist23]

Then, just analyze 20 mg of the spike as a sample.

I also need to add 5 uL of sample (spike) to 20 mg of base matrix. How do I account for the "5 uL" once I have the curve built in order to perform an accurate calculation?

[LALman]

Hello all,

I am trying to quantify a sample using a 5000 ug/ml Class 3 residual solvent standard using GC headspace technique.

The procedure for the sample preparation indicates to add 5uL of sample to 20 mg of blank matrix to a head space vial.

There are no instructions of how to prepare the standard.

If I add 5uL of 5000 ug/ml standard to a headspace vial, how would I calculate the concentration of compounds in the sample in ug/g?

I would like to build a curve using the 5000 ug/ml class 3 standard.

I'm not running headspace but Purge and Trap. My nominal volume/mass of sample is 5mL/5g. So if I were adding 5uL of 5000 ug/mL standard to 5mL just do the math to cancel the units as follows. [5uL*5000ug/mL] * [1mL/1000uL] = 25 ug. That is how much standard you are adding. Now in my case that is going into 5mL or 5g nominal sample size. So.... 25ug/5mL * 5ug/mL = 5 ppm.

P.S. Are you sure you have a sample size of 20mg? I thought headspace vials held 20-40mL of sample. So, do you mean 20mL or 20g rather than 20mg?

Well say it is 20mg then thats 25ug/20mg which is 1.25 ug/mg. But recast that to ug/g for ppm and you get 1.25ug/mg * 1000mg/g = 1250 ug/g = 1250 ppm.

[chemist23]

Hello all,



I would like to build a curve using the 5000 ug/ml class 3 standard.

I'm not running headspace but Purge and Trap. My nominal volume/mass of sample is 5mL/5g. So if I were adding 5uL of 5000 ug/mL standard to 5mL just do the math to cancel the units as follows. [5uL*5000ug/mL] * [1mL/1000uL] = 25 ug. That is how much standard you are adding. Now in my case that is going into 5mL or 5g nominal sample size. So.... 25ug/5mL * 5ug/mL = 5 ppm.

P.S. Are you sure you have a sample size of 20mg? I thought headspace vials held 20-40mL of sample. So, do you mean 20mL or 20g rather than 20mg?

Yes I am using a sample size of 20 mg. So if I am using a nominal sample size of 20 mg, I am adding lets say 25 ug of standard to 20 mg sample --> 1250 ug / g.

But now for my sample 'unknown' I am adding 5 uL to 20 mg. Let's say it lands on my calibration curve at 1000 ug/g. How do I account for the 5 uL in the calculation?

[rb6banjo]

Let's say your stock standard is in water (just for convenience).

0.005 mL x 1.0 g/mL = 0.005 g added

That means your sample size is really 0.025 g (a dilution by 20%, 0.020/0.025 = 0.80). Usually, you want the volume/mass of your overspike to not affect the overall volume/mass of your sample. In your case, it does affect it.

Here you go. So, 0.4 µL of 5,000 ppm standard to 0.020 g of oil matrix gives a total mass of 0.0204 g of sample analyzed.

5,000x0.0004/(0.020+0.0004) = 98 ppm (not the 100 you want but not too bad)

If you have to add 40 µL of the stock standard to 0.020 g of the blank matrix then:

5,000x0.04/(0.020+0.040) = 3,333 ppm (not the 10,000 ppm you want)

You really need to make your standards in your blank matrix but use a large mass of blank matrix relative to the amount of standard you add. This way, you don't dilute it dramatically when you add your standard.

[chemist23]

Let's say your stock standard is in water (just for convenience).

0.005 mL x 1.0 g/mL = 0.005 g added

That means your sample size is really 0.025 g (a dilution by 20%, 0.020/0.025 = 0.80). Usually, you want the volume/mass of your overspike to not affect the overall volume/mass of your sample. In your case, it does affect it.

Here you go. So, 0.4 µL of 5,000 ppm standard to 0.020 g of oil matrix gives a total mass of 0.0204 g of sample analyzed.

5,000x0.0004/(0.020+0.0004) = 98 ppm (not the 100 you want but not too bad)

If you have to add 40 µL of the stock standard to 0.020 g of the blank matrix then:

5,000x0.04/(0.020+0.040) = 3,333 ppm (not the 10,000 ppm you want)

You really need to make your standards in your blank matrix but use a large mass of blank matrix relative to the amount of standard you add. This way, you don't dilute it dramatically when you add your standard.

Thanks, this makes a lot of sense.

My next question though, is how do I calculate the true result of my sample?

Lets say it lands in between the 98 ppm and 3,333 pm at 2000 ppm on the calibration curve.

Since I prepared the sample by adding 5 uL of sample to 0.020 g of blank matrix would it still be 2000 ppm? My result needs to be in ug/g

[rb6banjo]

If you calculate the concentrations of your standards correctly (as I described above) when you generate your curve, you just need to multiply the result you obtained by the dilution factor (0.025/0.005 = 5).

So, 1,000 ppm from your curve means 5,000 ppm in your sample.

Again, let me emphasize, all of this is assuming there is not a substantial change in the partitioning behavior of your analyte when you dilute it by this much with the Class 3 standard (Restek's has analytes dissolved in DMF). My gut tells me this is a big problem. Even with the lower concentration standards. 20 mg of material just isn't a lot. Barely adding any DMF will change your "matrix" quite a lot.

[cody841]

I'm working on this right now. I decided to build my calibration curve based on µg of standard in the HS vial.
So if you put 5 µL of 5000 µg/mL standard in a vial you have a total of 25 µg standard in that vial like everyone says. The matrix does not matter.

If you build your curve in total µg (instead of ppm/concentration) it's easy to get concentration from sample weight in µg/g.

Say measured amount in sample is 16 µg and the weight of your sample was 20 mg or 0.020 g, then 16µg / 0.020 g = 800 µg/g.

Using amount instead of concentration also allows you to test more sample types.

[rb6banjo]

If the matrix doesn't matter, what's the point of adding your standard to the "blank matrix"? It'd be a whole lot easier - not to mention straightforward - to just add the standard to the vial.

[cody841]

If using weight the matrix doesn't matter. If using concentration for the curve it does.

You need matrix in there for matrix effects to match sample, but when making the curve with total weight in headspace vial, it doesn't count.