Determining silver content of coins with varying fineness?

So, I've gotten into world silver coins as a way to build the 'junk' silver stash and have a basic question.

Is there a method to calculate the silver content of a group of silver coins without having to know/lookup the individual fineness/weight of each coin?

For example: Let's say I have a 1957 German 5 Mark, a 1942 Great Britain Half Crown, a 1950 Belgium 20 Frank and a 1918 France 50 centimes.

If I didn't know the fineness and weight of each coin, can you weigh them all together and closely know the total amount of silver in the group of coins?

I searched for such as answer elsewhere on the internets, but couldn't seem to come up with something.

Oh wise coin collectors and bullion gurus, share your wisdom.

 

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No.

 

Are you sure. Your answer seems to leave some space for debate.

 

Please explain how you think it could be done.

 

No. Go do your homework!

 

Years of experience. You look at a random handful, see how you feel about 75% as an average silver content, mentally tweak it a bit one way or another as your instinct dictates and use that as your figure, you may settle on 65% or 80%, depending on the content of your sample or even the general colour of the pile.

It's as hard to explain as telling someone how to catch a ball. Easy to do, impossible to explain. Most purchases of quantities of stuff involve taking a sample and extrapolating a value for the lot.

 

It would depend on how accurate you wish to be.
If you took the total weight times 30% time spot silver price you would find lots of buyers.
Better to put them in known percentages (homework) then do the math.
More time consuming, more math, better accuracy, better profit.
Good luck.

 

Sorry. I meant a .

Just having some fun.

Thanks all. I figured there really wasn't a 'simple' way. I mainly ask because I bought some random world silver coins at a coin shop recently and he was selling them for $20/oz. I assumed he was weighing the silver content of the coins I chose (fineness varied - .50, .625, .835), but I guess he was actually just measuring the full weight of the group. But he did give me a $10 discount (I paid $15), because he was Polish and I have alot of Polish-American ancestry in my and my wife's immediate extended family.

I like you guesstimate method, afantiques. I'll likely just do that from now on and/or search for overlooked silver coins in common world junk bins.

 

Sure you can! Put it all on a scale, get the weight, divide by the number of coins you weighted and multiply by .5 you will come as close at humanly possible to the accurate weight of nothing important. Smiles everyone, smiles.

 

(?) Room for what? Did he mean NO for No?

After a while, you'll keep a lot of the "fineness" of silver in commonly seen junk coins. Then you can separate the coins by fineness keeping the .850, .900, etc. apart. Then you may be able to weigh the batch and get an estimate of the amount of silver. I don't recall the multipliers to use for each grade of purity anymore as I haven't played with bullion in a long, long time.

 

The problem is that some of the coins might be 10% silver, some 50% silver, 90% silver or 92.5% (sterling). I habitually buy coins from a junk bin that I am sure are silver and 90% of the time they are Cu-Ni.

 

That's why you need to separate them by fineness (as best you can - without using sp gr. Experience saves time.)

 

I think @afantiques said it all.
Thats how I do it, anyway. I typically use .800, unless there is a lot of later British 50% stuff in there.

 

The only way you could be accurate without having to look up each one, is to have put in the time and experience to have learned the fineness of each series. There is no shortcut or magic scale that will calculate that for you.

I'm sure that dealer you bought those coins from at $20.00 an ounce was not charging for the weight of the coin, but rather the silver content of that coin. He more than likely had the fineness written on a 2x2 along with the ASW of the coin. That way he would not have to look up the information later. He would only have to do the simple math at the time of purchase.

 

Well, the 5 DM coin mentioned in the initial post is a .625 piece ...

 

I agree that knowing the fineness of many different series and their gradual debasements is a much more definitive way, but the dealer I picked from had .50, .625, .835, .925 etc fineness coins of varying sizes/denoms loose in a bin.

From hearing what everyone is saying here, I presume he was giving a bulk price in relation to the absolute weight of the coins I chose and he weighed them in front of me. Or he knew the various finenesses and weights of each coin by sight and would calculate the 'value' in his head.

Basically, I suppose I was over-thinking the method he used to determine the price and simply averaged the value when pricing the bulk loose coins.

 

Try the Specific Gravity test on each one!

 

There really is no simple substitute for knowing the material. But in the case of older coins, the melt value is likely not all that important unless the coins are significantly worn.

 

Here's a more mathematical approach...if anyone cares. Assuming the four coins are only comprised of silver and copper, I came up with the following system of equations:



Where the coins are numbered 1 through 4 (i.e. - the subscripts that contain 2's all apply to coin 2 for example), the X's are the percentages of each metal for each coin, and the M's are the masses of each coin, or the mass of silver/copper in each coin. The first four equations basically say that the fineness of silver times the mass of the silver in the coin plus the fineness of the copper times the mass of the copper in the coin equals the total mass of the coin. The 5th equation says that all four coin's masses add up to get the total mass. The last four equations say that the finenesses of the silver and copper have to add up to 1 (100%).

Since we only would know the total mass (underlined), we would be left with 20 unknowns and only 9 equations. We need an equal number of equations and unknowns to solve, so you'd need to find 11 of the unknowns (i.e. - finenesses, coin masses, mass of the silver in the coins, etc.) to solve.

TL;DR: I wasted a bunch of time confirming what everyone said because I have no better way to spend Friday nights.