Specific Gravity Question

[Lmarc1]

I looked through the search function, but could not find an answer to my question.

I have done lots of assay work, sampling and melting, but I've never done specific gravity tests. I have a friend who has some very good gold-bearing quartz which I would like to test by specific gravity to determine at least a close approximation of it's gold content, but I'm having some difficulty getting results that I can believe.

I did a test on a silver coin of known content but I may be doing something wrong. The coin is a Honduran 1 Lempira coin of 0.900 silver content. I trust the coins stated content because it was minted in the United States at Philadelphia. I used an AND scale, weighing the coin in both Oz/T as well as grams, with a three decimal place resolution. The results were:

Dry Weight - 0.400 Oz/Troy and 12.441 grams
Displacement Weight - .040 Oz/T and 1.244 grams
Specific Gravity - 10.00 exactly (this seems a little strange to me, since the Displacement Weight is exactly one tenth the dry weight)

Using 10.53 as the specific gravity of pure silver, I get around 95% silver in the coin, though I know it to be actually 90%.

I used a glass container with purified drinking water and a thin thread to do the displacement weighing, and held by hand. Is a five percent error normal in such a case? This seems excessive to me. That could be a major error if testing gold.

 

[g_axelsson]

I think you have forgotten to include the weight of the other alloying metal. When I just divide with the silver density I get 94.97% silver but that is an alloy with air.

What we want to do is solve for x in the following formula.
x*(density of silver=10.53) + (1-x)*(density of copper=8.93) = density of object
Which gives...
x*((density of silver=10.53)-(density of copper=8.93)) = (density of object=10.0) - (density of copper=8.93)
Which gives...
x=((density of object=10.0) - (density of copper=8.93))/((density of silver=10.53)-(density of copper=8.93)) = 0.67 or 67% silver

The coin should have a density of 10.37 (0,9*10,53+0,1*8,93), plugging 10.37 into the formula we get... 0.9 which is spot on.

I'm a bit curious about the weight of the coin, according to http://www.ngccoin.com/price-guide/world/honduras-lempira-km-75-1931-1937-cuid-12195-duid-41931 it should weigh 12.500 g exactly and not 0.400 Oz/T exactly. I get a feeling that your scale is showing too fine and rounded numbers even though it is showing three decimal places.

So there could be three different errors here
- Your weights are off thanks to measuring errors / scale broken / scale not calibrated / worn coin / surface tension / water evaporating
- The coin is fake or there is an error in the literature.
- The coin is not a silver / copper alloy but contains something else.

Although a silver - copper alloy isn't linearly in density versus composition the error is a lot less than the above numbers gives.
http://link.springer.com/article/10.1007%2FBF03216580#page-1

Göran

 

[Lmarc1]

Thank you very much for the reply and the formula. As far as the three possible errors:

The coin isn't of sufficient value to warrant faking, so that is not likely.

I trust the US Mint to alloy the coin properly, using copper or copper/nickel for the additional ten percent, though so far I haven't found an exact composition for this coin.

The scale is a very good possibility for error. I did not calibrate it before weighing. I will also try to increase it's resolution.

Thanks again! I didn't realize that SG could be that complicated.

 

[Grelko]

You didn't specify which year the coin was. Maybe that's the difference?

In 1931, coins were introduced in denominations of 5, 20 & 50 centavos & 1 lempira. 1, 2 & 10 centavos coins were added in 1935, 1939 & 1932 respectively. The silver 1 lempira coins ceased production in 1937, with the other silver coins (20 & 50 centavos) replaced by cupro-nickel in 1967.

https://en.wikipedia.org/wiki/Honduran_lempira

I may have been mistaken.

900/1000 Silver .3617 oz. ASW. 100/1000 copper
http://colnect.com/en/coins/list/year/1931/currency/130-L_-_Honduran_lempira

 

[Lmarc1]

The coin is 1937. I tried again today with a different scale and got the same results.

I'm going to try again tomorrow with a newly calibrated scale. I'm only off by a bit more than 0.06 grams, and that may be due to the coin being a little worn, or maybe the thread I'm using for suspension.

 

[MarcoP]

If the original coin weight is 12.500g but today's weight is 12.446g, could it simply be warn out? After all it's only missing 54mg eighty years later. I have no clue whatsoever about coins so apologies in advance.

Marco

 

[Grelko]

could it simply be worn out?
Marco

Definately, just imagine how many times that coin was traded around, banged against other coins, in the wash, rubbed for luck, etc.

Edit -

I'm only off by a bit more than 0.06 grams, and that may be due to the coin being a little worn, or maybe the thread I'm using for suspension.

My apologies from my other post, I mis-read and thought you said that the coin was 1.224 grams low. Even still, enough wear and tear could drop the weight that much after all those years.

As for only being off by "a bit more than 0.06 grams", I bet it still looks almost brand new.

 

[Lmarc1]

could it simply be worn out?
Marco

Definately, just imagine how many times that coin was traded around, banged against other coins, in the wash, rubbed for luck, etc.
...
As for only being off by "a bit more than 0.06 grams", I bet it still looks almost brand new.

Yeah, these coins were struck in fairly high relief, especially the rims, to protect the design against wear. This one shows some wear on the face and hair of Chief Lempira, and the back shows some wear also. I think that could account for the weight difference.

For accurate specific gravity measurements I think I need a more sensitive scale. The resolution of this AND is 0.1 for grams, 0.005 for Oz/T, and 0.0005 for Lb. It will measure individual grains, which may be more accurate than the former, but I'm not crazy about doing weight conversions.

 

[g_axelsson]

0.005 Oz/T resolution is a lot when measuring the volume. It means you can be off by 0.0025g at least and at displacement Weight - .040 Oz/T it means 0.0025/0.040 = 6,25% error in volume. Add the error in weight also and I think we know why the result is so bad.

Try this with ten coins at one time, then you would lower the systematic error to a tenth of your original test. This is possible to do if you have large samples enough (a lot of similar coins, large gold specimens and so on...) A small sample gives large errors from the systematic errors in any measuring device.

When I googled the AND brand I found precision scales so I never suspected that it was a systematic error that plagued your measurement. I should have known when the weight came out with so even numbers.

Göran