HS-GC methodology question- change in calibration

[JesseG]

Hi,
I have been puzzled by this for weeks now and hope someone can offer some insight.
We run a headspace GC method for residual acetone in a 1 gram product (lactide-collagen polymer). Its a P.E. -HSGC and we use 22 ml vials with 1 gram of material and 5 ml of NaOH to solubilize the product (the headspace method has not changed).

Current we calibrate at a single point of 1000pmm with acetone in MeOH. We use 1 ml of cal standard in 5 ml of NaOH and no product.

We would like to change to a 2-POINT CAL at 1,000 AND 5,000 PPM. Problem is the 5,000 PPM STD spikes the detector. We can fix that by changing the std volume to 0.1 ml from 1.0 ml. We also learned that the material itself interferes with the acetone recovery so we have added 0.9g of "clean" product to the cal sample. When we do this we get a nice cal curve with perfect checks at 100, 500 and 3500 ppm.

The question I have is- What does the change form 1 ml of standard to 0.1ml of standard mean? Is it still relevant to my product? My older data will not correlate and when we reprocess the data through the new cal curve the acetone content increases by 60%. This makes me think that there is something fundamentally wrong with the changes. However if I run a samples from the same lot through both methods (I can't rerun samples since the dissolution is destructive) I get the same ppm value for each sample in each method.

Bigger problem is how do I explain to my QC group that the change from 1 ml to 0.1 is nominal and still represent the actual value?

Am I on the right path or have I totally screwed this up?

[chromatographer1]

Jesse,

I read your post and let me restate it so you can determine if my understanding is correct. I am uncomfortable with your procedure.

You test samples looking for acetone and dissolve the sample in 5 mL of sodium hydroxide (no molarity specified).

You measure acetone recovered from this solution by headspace analysis.

You calibrate your measurement by using a solution of acetone in methanol at 1000 ppm (v/v? or wt%?: which is not specified) added to 5mL of sodium hydroxide.

First let me give you the bad news. I have to say that your original calibration has no basis for accuracy. Using 6mL of solution versus 5mL of solution (which probably increased in volume somewhat when the sample was dissolved in it) will likely give you an inaccurate calibration. It may be linear, it may be precise, but it is not going to be accurate unless you got lucky, but you have no way of knowing based on your procedure. You may not be in error by a large amount but it is almost certain your values are not accurate. (from your post I ascertain that you may have gotten lucky in your reported values but your calibration reponse is 60% incorrect)

You did the right thing when you added 900 mg of clean sample to your calibration standards. You found out that your recovery is linear. This is very good news.

Whatever you got before is history and you should begin anew with the proper calibration. Let me give you the standard addition procedure that I would follow.

Given your difficult matrix I think you should perform a standard addition calibration to determine your acetone content. You should make solutions of NaOH which contain a measured amount of acetone. For example, I would weigh 1000, 3000, and 5000 µg of acetone into three 10mL volumetic flasks each containing 4-6 ml of NaOH. Then I would fill the flask and determine the concentration of the acetone in each flask.

I would then weigh out four 1 gram samples into four HS vials. I would add 1mL of each acetone solution to three vials and 1mL of NaOH to the fourth vial. Then add 4 mL of NaOH to each of the four vials.

Then analyze all four samples and plot the acetone response versus the amount of acetone added to the four vials: 0, 100, 300, and 500 µg. You can of course add different amounts if you choose.

The intercept of the line drawn through the responses will intercept the X-axis at the negative value of the amount of acetone of the unspiked sample. If the line through the responses is linear you will have proof of the linearity of your analysis and its reproducibility, as well as proof that your samples are homogeneous in their acetone content.

You have also performed headspace on 4 identical samples except that they vary in acetone content. Thus you can use Henry's Law to ACCURATELY measure the amount of acetone present in your sample.

If you try to compare the acetone partition from any matrix other than your dissolved sample you are making a dangerous assumption.

Once you ascertain the slope of your calibration you can go ahead and test multiple samples of different lots without doing a std addition to each lot. The one lot should be adequate for that determination. If you do this slope determination over several days and it stays consistent (and it should) then you may decide to only do check samples of determined acetone content with your other samples without doing the standard addition samples if your regulatory department agrees. Two additions are the minimum but I proposed making three in your original determination.

This will give you confidence in the accuracy of your test and will give any reviewer of your work confidence as well.

best wishes,

Rod

[JesseG]

Thanks for the response Rod!

I can add that the NaOH is 14N, the acetone in MeOH std is prepared ug/ml (we order them from an outside supplier).

It is reassuring to hear that I am on the right track (I was pretty sure the old method was junk).

I will need to re-read your suggestion for checking my calibration in the morning (it has been a long day) but we have one other issue.
As I stated, we have an "interference" between our sample and the acetone recovery. So much so that we actually have to inject the the standard into our sample (its porous) to get proper recover.
If you just inject the standard in the NaOH and drop a sample in we don't get the same values as if you inject the standard into the "blank" product. Since the acetone in processing is actually sprayed onto the product we feel this is the best way for our cal standards to duplicate our samples.
Could this be an additional methodology issue since we are completely dissolving the test matrix anyway?

[chromatographer1]

I proposed putting the acetone into the NaOH before dissolving the sample in it. Does this cause a change in the acetone measured? It should not. If you inject the acetone into the same and then dissolve it you should get the same value if you add the acetone to the sample and then dissolve the sample.

You must have the sample matrix and the same solvent volume in order to compare the partition of a measured headspace sample.

To do anything else is to compare apples to oranges. And sometimes the difference is not important but there will be a difference. Of course in analytical chemistry accuracy is job 1, is it not?

Feel free to mail me directly if you have questions about details that you don't want to share with the internet.

best wishes,

Rodney George
Rodney.George@sial.com

[chromatographer1]

Good morning,

Let me expand on the procedure I proposed to you.

You make 4 vials of samples containing 1g of sample and 5mL of 14N NaOH. Each vial contains the residual acetone found in the sample plus an added amount of acetone in levels of 0, 100, 300, and 500 µg amounts.

If you have a consistent level of 1000ppm of acetone in the samples (1mg) then your vials contain 1mg, 1.1mg, 1.3mg, and 1.5mg of acetone.

If you plot the headspace analysis results you will get a straight line (hopefully) when plotting the acetone area versus the added amount of acetone in the vials. This line will intercept the x-axis where x = -1000 given the points plotted were x=0, 100, 300, and 500 and the y values were the areas of the acetone measured by HS.

I hope this is clear. Once you know the slope of the line you can calculate any sample's acetone level. Any area of Y divided by the slope will give you X, the amount of acetone in the sample.

You can of course add an internal standard like MIBK or MEK to the sample as well and determine acetone levels as well. But the standard addition method gives you proof that your analysis is consistent in hardware and in chemistry. IT PROVES your measured value is correct and that your samples are consistent in their acetone content. No one can question your results as long as your recovery is linear.

best wishes,

Rodney George
Senior Research and Development Scientist
Gas Separations Research
Supelco
595 North Harrison Road
Bellefonte, PA 16823

814-359-5737 voice
814-359-5459 fax
rodney.george@sial.com

[Willy]

Hi there,

Just want to know something about your calibration thing. Do you need to weight 1g of sample? Is lower quantity can be used? (like 100 mg). Acetone is a very volatile compound. Probably some people will not agree with me but do you really need to solubilize the sample with NaOH to be able to recover the acetone in the samples? You often only need a suspension of your matrix in your vial to be able to perform your analysis. Samples heated and shake in headspace with proper time can extract easily Acetone.
If you spike Acetone in samples to make few calibrations points and you see that you recover 100% from each of them there is not need to add that amount of sample and NaOH solution.
One thing i would recommend is weigh for example 100mg of sample and add 1 mL total (standard-solvent) too match the target concentration.


The more you add in the vials, more scrap you inject in the system.



I just want to make it more simple...


Any toughts on this?


Willy the GC

[chromatographer1]

I developed methods for residual solvents that only used 500ng.

Yes, 0.5mg. My LOD was 1ppm of residual solvents (each of 18 possible). See AC 1997 June issue. My usual sample was 5-20 mg.

Residual solvents can be trapped in crystal matrices. Dissolution is the best policy for quanitative recovery.

You should test samples and standards with the same volume and the same ending matrix in HS analysis.

best wishes,

Rod