Cheapest way to get accurate lead quantity tests on suppressor residue?

[Odd Job]

I want to find out what percentage of the residues (by weight) is lead in each of the six baffles on my A-TEC CMM4(6). I will do this at the end of my testing, so it is a one-off procedure for this can which already has 10,000 rounds through it (.22LR)

Getting the residues into 6 little containers is the easy part. But I need these residues professionally tested. One private company I called, quoted me in excess of £600 to do it here in the UK.

Does anyone here have any suggestions on how I can get this done cheaper? Anyone done anything like this before?

[doubloon]

Maybe dissolving it in the dip then precipitating the lead yourself will be accurate enough for the variance between one 10,000 round sample and another or interweb postings.

Without reclaiming the lead from multiple 10,000 round samples "accuracy" is dubious at best anyway.

But most people just weigh the can with the buildup and subtract the original weight.

[Bendersquint]

Only way to get an accurate measurement is to have the lead separated from the carbon and other buildups.

This is not a cheap process and on top of that you are needing it done on 6 different samples.

600 pounds for that is not that bad a price considering.

[Fulmen]

Check with a university (or lab trade school). XRF should handle this sample without any special workup or treatment, AAS or ICP would require the sample to be digested in nitric acid.

[mbogo]

I assume that you are in the UK, where sound moderators are not as restricted as in the USA.


According to the A-Tec web site, that item weighs 220grs. new. Subtract that weight from the weight of your A-TEC CMM4(6), and the difference will be the lead residue (plus some powder if you haven't flushed it yet).

If you don't think the manufacturer's specification is accurate enough, bring an accurate digital scale to your sound moderator dealer and weigh a new A-TEC CMM4(6). Then, do your arithmetic.

mbogo