Raising the Fineness of Silver Alloys

Hello, I have been trying to research how to raise the millesimal fineness of a few ounces of 835 silver I have to 925 silver. I have found previous discussions on here and the formulars don’t seem right like one says to add 40 grams of fine silver to 10 grams of 625 sliver to get 925 silver. just seem to much.
In my research I found to get 925 from fine silver (lets work with 1 toz, 31.1 grams) you divide 31.1 by .925 which gives you 33.6 so then subtracting the 31.1 gives you 2.5 grams of alloy to add to get 925.
So I thought, to get 835 silver you would do the same sum using .835 which would give you 37.2 then subtracting the 31.1 giving you 6.1 of alloy to add.
I then subtracted 2.5 off the 6.1 giving you 3.6 so in my thinking, I have to add 3.6 grams of fine silver to 31.1 grams of 835 silver to make 925 sliver. Is this right, if not, please explain why not.
Also, I am only using one decimal point here but when I go ahead, I will be using three decimal points.
Any feedback would be great,

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Please correct my math where I’m wrong!

For every gram of .925 silver there is .925 g of silver and .075 g copper.

For every gram of .835 silver there is .835 g of silver and .165 g of copper.

The math is to get .165/x = .075/1 I think.

For brevity, here’s the math with wolfram alpha :

So the total weight of the material after adding pure silver would be 2.2 g. Therefore you need to add 1.2 g of pure silver to every 1 g of .835 silver to reach .925 Sterling.

Following is a link to a calculator that might help answer your question…Rob

Hi Rob, thank you for the link but it only tells you how much alloy to add to fine silver to get 925, it will come in handy when I alloy the 8 ounces of fine silver I’ve got, thanks again.

Hi Brennan,
Thanks for the input and link. I didn’t think of it that way (.075 per gram) I found that Wolfram Alpha a bit confusing, I’m not sure how you come up with 1.2 grams to every gram .835 but I find it not right. I had a look at the calculator at United Precious Metals Refining that Rob posted and found that to get .925 from 31.1 grams of fine silver, you only have to add 2.521 grams of alloy which give you 33.621 grams of .925. .835 is only .09 lower than .925, I just have to keep working on it. Thanks again

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I wouldn’t think about it as 31 g right now. I think that makes it more complicated. If you think about it as 1 g then you only have to think about it as how to convert 1 g of 835 to 925.

Using that as a simplification, 835 has more than double the amount of copper as 925. Because of that you need to more than double the amount of silver to get the same quotients.

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More thoughts:

We know we have to add pure silver to get the purity to .925. Because of that, we know the copper amount is not going to increase. It’s a fixed amount.

For each g of .835 silver we have .165 g of copper. That’s 2.2 times the copper amount in .925 silver. That means, if we keep the copper amount fixed at .165 g, the total amount of .925 silver will 2.2 g.

2.2 g of .925 silver has .165 g of copper in it. So all we have to do is add pure silver until we have 2.2 g of total silver.

Multiply the amount if .835 you have by 1.2 and add that much pure silver.

Here are two pages from Dr. Brepohl’s book that explain raising the karet of gold. Since he is dealing with parts-per-thousand using this info with silver shouldn’t be a problem.

From “The Theory and Practice of Metalsmithing” by Erhard Brepohl, posted with permission from Brynmorgen Press.


Thanks again Brennan for your input, still a little confusing but getting a handle on it (maybe I should of listen more in maths class instead of being the class clown)
A big thanks to alonzo and Brynmorgen Press for allowing you to post those pages. I’ve got my head around Example 2, still working on Example 1 but I will get there.
I liked those pages that much, I jumped on Amazon and bought a copy of the book and look forward to reading it.
Thanks again guys for all your help.

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I just thought of another way to wrap our heads around it because it is definitely not intuitive.

This is an abstraction so it won’t be perfectly accurate but take rice and beans.

Take 8 beans and 2 grains of rice. That signifies 80% silver and 20% copper (a rounded version of .835 silver.

Now add beans until it is 90% silver. I think what we discover is that we have to double the beans.

Anyways. It’s a fun math problem but I also found it unintuitive.


in order to get a true alloy, the atomic weights of silver and copper have to be considered. This makes it confusing. Atom percent gives you the desired alloy. Weight percent neglects the difference in the weight of pure silver and pure copper. The atomic weight of silver is 108. That of copper is 64… due to the presence of isotopes, these numbers are not 100% accurate, silver has an atomic weight of 107.868, which is close enough to 108, copper has an atomic weight of 63.546…using 64 or 63,5 makes it closer. Using atom percent, gives you the most accurate proportions of silver and copper to mix. 108 divided by 63.5 is 1.7…this ratio is the amount of copper by weight to add to 108 grams of pure silver to get any alloy based on atomic weight. Conversely the amount of pure silver that has to be added to copper to get any alloy based on atomic weight is 63.5/108… less silver than copper than by simple weight percent. If you have an alloy of 90% silver and want to bring it up to sterling, you would have to add pure silver by atomic weight. These calculations can be done for any alloy, including gold which is far heavier at 197. These ratios are used by the calculator as shown above.
The basis of atomic weight is the weight of one mole of a pure element… a mole contains 6.023 X 10 to the 23rd power of atoms… one mole of gold weighs 197 grams. a mole of silver weighs 107.87 grams and a mole of copper weighs 63.546 grams. The decimal place number are isotopes of each metal with their respective abundance. Gold has only one isotope.