Could someone advise me of a good method of testing gold and silver purity.
I'd like to know how the various methods work.
Miles
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Gold and silver testing methods
Hi Miles,
I just came across this on some old notes " A drop of a dilute solution of silver nitrate applied to an item of silver will leave no mark, but if the item is a base metal it will form a dark patch".
Years ago I used to use a testing kit with five bottles of different acids, this would determine the purity of gold and silver but if an item were heavily plated you have to deep file to expose a possible base metal underneath thus scaring the object.
The best method I find is that good old gut instinct when something is right or wrong.
Trev.
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I just came across this on some old notes " A drop of a dilute solution of silver nitrate applied to an item of silver will leave no mark, but if the item is a base metal it will form a dark patch".
Years ago I used to use a testing kit with five bottles of different acids, this would determine the purity of gold and silver but if an item were heavily plated you have to deep file to expose a possible base metal underneath thus scaring the object.
The best method I find is that good old gut instinct when something is right or wrong.
Trev.
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I usually do a specific gravity test on all silver and gold objects [that don't have voids]. If accurately done on numerous items, the data can be fairly reliable in differentiating .800 from .925 silver (but not .800 from .835 or .900 from .925). Gold is a bit more difficult due to variations in alloy composition, but 14K can be distinguished from 18K or 10K.
The virtue of this test is that it is non-destructive and only requires a scale with good precision.
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The virtue of this test is that it is non-destructive and only requires a scale with good precision.
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I have tried testing silver. A product called Probiersäure für Silber from Ballerstaedt & co tel +49(0)7222 95150. You put one drop at the item. If it is silver it turns red if not it remains brown. It is true that silverplated also turns red, I scrach a bit at the edge of the item. You dont realy destroy the item. Afterwards it is almost invisible. You have to know where it is. You just add asid to the scratched part.
In theory you should be able to determin 800 fra 925 silver. The more silver the more red it should be. Take a picture for reference. But mind picture should be taken in same light and same amount of seconds after you have aded the asid.
I have not been able to do this - i soppose try and error - try and error should be performed a number of times.
When I tested 688, 750, 830, 925 and silverplate. the 11L (688) silver came out whit the most red colour. I think that I just need to pratice more bu havent had the time.
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In theory you should be able to determin 800 fra 925 silver. The more silver the more red it should be. Take a picture for reference. But mind picture should be taken in same light and same amount of seconds after you have aded the asid.
I have not been able to do this - i soppose try and error - try and error should be performed a number of times.
When I tested 688, 750, 830, 925 and silverplate. the 11L (688) silver came out whit the most red colour. I think that I just need to pratice more bu havent had the time.
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SilverSurfer
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Besides the precision scale, how does one get an equally precise determination of the silver/gold piece volume? Or, if performing the differential weight method (weigh container of water and then container + water + silver/gold piece), how does one get the volumes exactly equal? Since this method involves subtracting nearly equal values, small errors are greatly magnified. TIA for any advice on how to get the required precision.I usually do a specific gravity test on all silver and gold objects [that don't have voids]. ... The virtue of this test is that it is non-destructive and only requires a scale with good precision.
SS
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SilverSurfer
- contributor
- Posts: 267
- Joined: Tue May 31, 2005 12:42 am
Thanks, but I still have the same questions. Using a digital platform scale good to +/- one gram, I weighed my six mystery salt spoons at 60 grams. I then weighed a cup filled with water, with and without the mystery spoons. The mass of all three was 350 grams, and that of the cup full of water and no spoons was 296 grams. So, letting M = mass, D = density, and V = volume, I have for the case of all three
Mcup + Mwater + Mspoons = 350 g
Then, without the spoons, I have the mass of the cup plus the mass of the original water plus the mass of the additional water occupying the space that was previously occupied by the spoons (this mass is equal to the density of water times the volume of the spoons), or
Mcup + Mwater + (Dwater x Vspoons) = 296 g
Subtracting the second line from the first, I get
Mspoons - (Dwater x Vspoons) = 54 g,
and since Mspoons = 60 g and Dwater = 1 g/ml,
Vspoons = (60 g - 54 g)/(1 g/ml) = 6 ml, so
Dspoons = Mspoons/Vspoons = 60 g/6 ml = 10 g/ml
The density of pure silver is 10.49 g/ml, that of sterling around 10.4 g/ml, and that of copper, nickel and other likely base metals is under 9 g/cc. So it would appear that the spoons are most likely sterling. BUT ... an error of just 1 gram (the limit of my scale) in the subtraction calculation (60 g - 54 g) is more than enough to throw the answer the other way. ALSO ... it is very difficult to determine when the cup is precisely "filled". I used an eye dropper and stopped as I just began to see a reverse meniscus form as I sighted across the plane of the rim. So the likely errors don't give me a sense of assurance. Is there a technique that can more accurately determine the density, using a simple platform scale? TIA!
SS
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Mcup + Mwater + Mspoons = 350 g
Then, without the spoons, I have the mass of the cup plus the mass of the original water plus the mass of the additional water occupying the space that was previously occupied by the spoons (this mass is equal to the density of water times the volume of the spoons), or
Mcup + Mwater + (Dwater x Vspoons) = 296 g
Subtracting the second line from the first, I get
Mspoons - (Dwater x Vspoons) = 54 g,
and since Mspoons = 60 g and Dwater = 1 g/ml,
Vspoons = (60 g - 54 g)/(1 g/ml) = 6 ml, so
Dspoons = Mspoons/Vspoons = 60 g/6 ml = 10 g/ml
The density of pure silver is 10.49 g/ml, that of sterling around 10.4 g/ml, and that of copper, nickel and other likely base metals is under 9 g/cc. So it would appear that the spoons are most likely sterling. BUT ... an error of just 1 gram (the limit of my scale) in the subtraction calculation (60 g - 54 g) is more than enough to throw the answer the other way. ALSO ... it is very difficult to determine when the cup is precisely "filled". I used an eye dropper and stopped as I just began to see a reverse meniscus form as I sighted across the plane of the rim. So the likely errors don't give me a sense of assurance. Is there a technique that can more accurately determine the density, using a simple platform scale? TIA!
SS
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Hi,
According to Galileo, Archimedes must have faced the same problem. Without a well calibrated container that will be almost impossible, hence Galileo thought Archimedes used a balance instead of a bucket of water.
Using a pycnometer is a more precise way to do what you are trying to do.
I would use hydrostatic weighing though, but since specific gravity is calculated to 2 decimals, you will need a more precise scale. Or build one yourself with a long, thin, end of a fishing rod.
See http://gemologyproject.com/wiki/index.p ... ic_Balance" onclick="window.open(this.href);return false; .
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According to Galileo, Archimedes must have faced the same problem. Without a well calibrated container that will be almost impossible, hence Galileo thought Archimedes used a balance instead of a bucket of water.
Using a pycnometer is a more precise way to do what you are trying to do.
I would use hydrostatic weighing though, but since specific gravity is calculated to 2 decimals, you will need a more precise scale. Or build one yourself with a long, thin, end of a fishing rod.
See http://gemologyproject.com/wiki/index.p ... ic_Balance" onclick="window.open(this.href);return false; .
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Frankly, you need a more precise scale. For small objects, I use an Ohaus Cent-o-gram - calibrated to .01 grams. For objects over 10 oz., I use a balance beam scale calibrated to .1 grams. Also, by testing the same object multiple times and testing a variety of objects of known composition, baseline values and accuracy can be determined.
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SilverSurfer
- contributor
- Posts: 267
- Joined: Tue May 31, 2005 12:42 am
Thank you, Doos, for the link, and I apologize for being dense, I finally got it. Suspending the mystery piece in a beaker of water placed on a platform scale will, of course, lighten its apparent weight by the amount of displaced water, this weight discrepancy being then hydrostatically transmitted to the beaker bottom and hence the scale (it was this latter piece of wizardry that somehow escaped me, doh!). And so one need not accurately measure the beaker water volume. With this technique, I came up with 7 ml of volume, yielding a density of about 8.6 g/ml for the 60 gram dry weight of the spoons, so now looking more like base metal than sterling. But known coin silver spoons of 66 grams dry weight came up with the same 7 ml of volume, for about 9.4 g/ml, about one g/ml short of the true value. Yes, salmoned, I need a more accurate scale, but am afraid to look to see what a 500 gram scale accurate to +/- 0.1 gram might cost. May be better just to build my own beam balance and use inexpensive calibrated brass weights. Thanks again to both for your feedback.
SS
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SS
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