Equivalence

What was the volume of the submitted piece? From there it's just arithmetic.

Tornado,

I do not know the volume.

You mean the displacement of water after the peace of metal is introduced in a liter?

If so, I do not know.

Thank you for answering.

H. L.

Hubert, it looks like the analysis was done by AAS (Atomic Absorption Spectroscopy). These numbers should convert directly to ppm. This means that there is 170ppm copper (0.17%), 65ppm silver (0.065%), and 65.9ppm gold (0.0659%). I would assume that if you are in the mining industry that there would be tin and nickle as the balance, but that is just a guess.

If these numbers look way off, then you need to call up Polytechnique and find out how much of your sample they dissolved, and how much liquid they used in the analysis. Next, you divide the mass (in mg) by the volume (in L) to get your total ppm.

To change this to percent, divide the metal analyzed value by the total value. Lets use some numbers to make this a little more clear.

Sample dissolved: 300 mg (this value needs to be in mg)

Final total volume sample is dissolved in: 0.250 L (this value needs to be in L)

300 / .250 = 1200 mg/L or ppm

65.9 mg/L gold / 1200 mg/L total = 5.5% gold in your sample.

Same guest as above.

The numbers that you received should be already standardized to the amount of sample used. This means that the number should be correct as it stands.

Thank you GUEST,

Yes, the analysis was done by Atomic Absortion.

I spent 25 years in precious metal recycling and this is the first time I received a result in mg/L. I will call up the Université de Montréal monday.

You may have a good answer but I still think there is more gold because I melted it myself.

But we never know...

Hubert

You are correct Tad, I guess that will teach me for writing a technical response while I am in a hurry.

And to the person modding everything as Off Topic, SOMETHING IS NOT OFF TOPIC IF YOU DO NOT AGREE WITH IT!

-the guest from above

I am surprised that the lab expressed their report in mg/L units, which are more suitable for substances in water. Until we know some densities, we don't know what a liter of metal weighs. Conversely, we don't know from 94 grams what the volume of the sample was. If the sample was irregularly shaped, putting it in a graduate of water and comparing the difference in readings would give the volume.

Converting mg/L into a fractional equivalent such as ppm or percent requires dividing by the density involved.

well, they need to answer in better terms than that. This is not a proper assay report. How much did they charge? or was this a freeby student assay?

if this was 3 ounces of gold = over $3000 worth if pure.

What is the purpose behind this assay?

Why not use an assay company.

there are many of them, use your country, and look for mining +assayers

Since they found only copper, silver and gold, we must assume that is nearly 100% of the metal. Ignoring any trace elements present, the ratio of Cu to Ag to Au is 170:65.8:65.9. It doesn't matter how much sample was taken or how much solution was made from it, only the ratios of the concentrations are relevant. The composition of your metal is therefore:

56.35 % copper

21.81 % silver

21.84 % gold

Is this about what you expected?

The percentages of Cu, Ag, and Au are much less than that. Enough clues are here on the thread to figure out why.

I'd like to assume that apart from Copper, Silver and Gold that piece of metal contained other stuff that is not considered interesting. Also, that the whole lump dissolved completely in whatever quantity of solvent that was used to dissolve it (and/or that the lump is very homogenous and/or the sample taken from it is very representative).

If so, the lump contains: 0.18085% Copper, 0.07011% Silver, 0.0700% Gold and 99.67904% other stuff. All percentages being by weight.

Or am I wrong too?

A typical lab would weigh the sample, and if it was a cast sample and showed no stratification would dissolve a small representative sample in aqua regia.

If you look here you will find test procedures in the links.

http://www.alsglobal.com/mineralDownloads.aspx

google assay labs, there are many more.

There is nothing that tells us anything about the concentrations of elements in the original sample, only that 94 g. of the metal sample was submitted. Some, or all, would have been dissolved in aqua regia and analysed by AA (ICP is also a possibilty). Whether the resulting solution was diluted to 10 litres or 100 litres is irrelevant, the AA analysis merely gives us the ratio of weights between the 3 elements found in the solution. If there were other elements present in large quantities then that should have been reported - otherwise the assay is useless.

Other elements don't matter for AA. it only tests for the specific element that you tell it to, and gives back the concentration in the liquid tested. Cheaper AA can only test for one element at a time, but the more expensive ones can test for 3 or more during each run.

Thank you Guest.

The mystery should be resolved tomorrow.

Hubert

Hello all,

The technician at Polytechnique is on vacation until June 29.

I will have a ICP analysis next week and I let everyone hnow a correct answer and at the end of June, I hope I will have the comments from the technician who did the test.

Thank you to all of you .

Hubert

This is an obvious error by the lab. No lab would report concentrations of a solid in mg/l. You will have to call them to get the correct answer.

The problem probably occurred by having one technician prepare the sample (dissolve and dilute it) while another technician analyzed it by AA. Someone should have taken the concentration, and calculated the actual concentration in the original sample. Instead, it looks like you got just the concentration from the tech. that analyzed the diluted sample.

You can not even assume that the ratios of concentrations in the original sample match the ratios of concentrations given. The sample may have been split, and different dilutions used for different analysis.

Ja. Right, Tad. All correct !

Ha, ha.

Ha, ha.

Your analysts gave you raw experimental data which can be interpreted as follow:

Cu, Ag and Au exist in your sample at ratio 170:65.8:65.9. or 2.57:1:1.

So, in your 94g of metal,

You have Cu as Cu=170/(170+65.8+65.9)x94=52.97grams

Ag will be 65.8/(170+65.8+65.9)x94=20.50grams

Au will be 65.9/(170+65.8+65.9)=20.53grams

Check: 52.97+20.50+20.53=94grams

Erm - This is merely a rehash of my entry.

"The technician at Polytechnique is on vacation until June 29.

I will have a ICP analysis next week and I let everyone hnow a correct answer and at the end of June, I hope I will have the comments from the technician who did the test.

Thank you to all of you ." ......... Hubert Lefebvre

Well June 29th has come and gone, but still no results from you. When are you going to put us out of our agony?

Her answer will be on this site within 2 days.

Thank you "Energyconversion" for still remembering.

Hubert