Help identifying solder composition

[high standard 40]

Like others, I have acquired some radiator shop solder drippings. I have already rendered it into clean ingots and have been guessing it to be about 50/50 and have been using it as such. I know it's anybody's guess as to the real composition without a lab test. But today I got to thinking of a way to make an educated guess. I need someone with more knowledge of metallurgy than myself to tell me if I'm on the right track or dead wrong. Here is what I did.

I took one of my molds and cast some boolits of pure lead and set aside a control lot of ten good filled out examples. Then I cleaned my pot and refilled with the solder in question and cast another lot, setting aside 10 more good examples.

The pure lead boolits weighed an average of 92.27 gr each
The solder boolits weighed an average of 77.3 gr each
Tin weighs 64.2% less than an equal volume of lead.
If the solder were in fact 50/50 the solder boolits should weigh 75.75 gr each.
So I figured my radiator shop solder appeared to be a little shy of 50/50, by my calculations, about 45% Tin / 55% Lead.

Then I had to admit that I'm not sure if 50/50 solder is equal weights of lead and tin or equal volumes of lead and tin. This difference would alter the results. I guess for my purposes calling it 50/50 is close enough. Can anybody here confirm or discredit this system?

[Defcon-One]

"Tin/lead solders, also called soft solders, are commercially available with tin concentrations between 5% and 70% by weight. "

I have not checked your math, but you are right on track! I have done this before and was always able to identify the solder very close to one standard or the other. Based on what you have said, yours is likely a mix of two or more types of solder in varying amounts or there's a bit of pure lead in the mix.

********** UPDATE **********

The density of Lead is 0.409 lb. per cubic inch or (2,863 gr. per cubic inch).
The density of Tin is 0.264 lb. per cubic inch or (1,848 gr. per cubic inch).

At 92.27 gr. your bullet is .0322 cubic inches of pure Lead. (92.27/2,863)

So, a bullet of pure Tin from your mould would be 1,848 X .0322 = 59.51gr.

A bullet from your mould of the mystery alloy is 77.3 grains. That works out to about 45.7% Tin and 54.3% Lead per this equation:

(.457 X 59.51 gr.) + (.543 X 92.27 gr.) = 77.298 grains

I’d say that you got it almost exactly right! (You might want to check my math!)

[high standard 40]

Based on what you have said, yours is likely a mix of two or more types of solder in varying amounts or there's a bit of pure lead in the mix.

An understanding of the workings of a radiator shop makes this easy to understand.
I've heard some say that this shop or that shop used only 60/40 solder so my drippings must be 60/40. Maybe, but most likely not. Included in the mix are the solder that they remove from radiators to be repaired and there is no way to tell what that may be. The vast bulk of solder drippings from most radiator shops is not from the solder they use, it comes from the radiators they repair. So the mix can be just about anything. Plus, i'm sure some Tin gets lost to oxidation in the storage and smelting process. I know the drippings I scrounged had more trash (sand, dirt, copper, alum, etc) than solder. It took more heat to smelt than wheel weights because of the insulating qualities of the sand and dirt. I'm pleased with the 45/55 result and I have about 300 lbs of it in ingots now.

[Defcon-One]

Good stuff! With 300 pounds, it will go a long, long way.

The calculations above always assume that there is not a third metal in the mix. If there is a bit of copper or something else, your alloy falls back into the "mystery metal" category. With 300 pounds on hand, it would be worth testing, epecially if it is all from one big homogenious batch.

I'd find someone who could get it to an XRF scanner and have it analyzed and I'd send them a few pounds to test and keep as payment for the service. The results would be exact and you'd know exactly what you have.

Once you know what you have, you could sell some off as needed to finance other materials and equipment purchases!

[badbob454]

you must also be sure the pure lead, is pure. to use this in your calculations ..

[merlin101]

Like others, I have acquired some radiator shop solder drippings. I have already rendered it into clean ingots and have been guessing it to be about 50/50 and have been using it as such. I know it's anybody's guess as to the real composition without a lab test. But today I got to thinking of a way to make an educated guess. I need someone with more knowledge of metallurgy than myself to tell me if I'm on the right track or dead wrong. Here is what I did.

I took one of my molds and cast some boolits of pure lead and set aside a control lot of ten good filled out examples. Then I cleaned my pot and refilled with the solder in question and cast another lot, setting aside 10 more good examples.

The pure lead boolits weighed an average of 92.27 gr each
The solder boolits weighed an average of 77.3 gr each
Tin weighs 64.2% less than an equal volume of lead.
If the solder were in fact 50/50 the solder boolits should weigh 75.75 gr each.
So I figured my radiator shop solder appeared to be a little shy of 50/50, by my calculations, about 45% Tin / 55% Lead.

Then I had to admit that I'm not sure if 50/50 solder is equal weights of lead and tin or equal volumes of lead and tin. This difference would alter the results. I guess for my purposes calling it 50/50 is close enough. Can anybody here confirm or discredit this system?

You made my head hurt! Math isn't my best subject so I look for other ways. If you used your test batch of 10 lead boolits and made more control batches of other known alloy's and then just compared them to the solder. Would you be close enough?
This brain work's to hard I'm going to go shovel the driveway!

[Defcon-One]

you must also be sure the pure lead, is pure. to use this in your calculations ..

Excellent POINT!

Funny how you can be so careful and still miss something critical. If the pure was not pure, then it could really be 50/50, couldn't it?

Something to think about!

[high standard 40]

Excellent POINT!

Funny how you can be so careful and still miss something critical. If the pure was not pure, then it could really be 50/50, couldn't it?

Something to think about!

True, but then the lead I used and considered "Pure" resulted in a boolit weighing 92.27 grains from a mold that will usually produce a bollit weighing 90 grains from wheel weight alloy. The reality of this conversation is that you can never tell with absolute certainty what an alloy is unless you have it tested. My attempt was to get a realistic idea of unknown solder composition using tools available to the average person.
And let's be honest, for most of our purposes, close really is close enough when blending alloy as long as the blend is reasonably easy to reproduce and works for the intended purpose.

[Defcon-One]

Hey, I'm happy if your happy! I know that we are very close with 45.7/55.3 using the math.

You are 100% right, if you really need to know, then you need to get it tested and with 300 pounds I still think it might be worthwhile.

For me testing is not practical. The cost of lab testing is too high and I do not have an XRF tester available anywhere near me!

[a.squibload]

If the junkyard thinks you might sell it
they would x-ray test it for free.
Seems they like to show off that test gun anyway...

[10 ga]

The "pure" lead does not fill out the mold as well as the high tin solder alloy. I've measured enough bullets of different alloys to know the soft "pure" lead ones come out the smallest but heaviest, the high tin alloy ones come out bigger but lighter. They aint the same size! The volumes of the 2 different bullets, alloy vs pure, are not the same. I'm thinking the alloy really is 1/1 ah err 50/50. What say the physics majors out there? LOL! 10 ga

[high standard 40]

That's a good point 10ga. The diameter of the pure lead bullet will be smaller, giving it less volume. For my purposes , I use it like it was 50/50. My reason for testing it was to to get an approximate idea of whether it was closest to 40/60, 50/50, 60/40, or something else.

[rsrocket1]

You could cast about 20 bullets each.
Fill up a jar of water to the brim with water and put the jar in a pot.
Put the 20 bullets in the jar. Let the water spill out into the pot, measure the water that was displaced.

Now you know the volume of the 20 bullets.

Best wishes,
Archimedes

[ckcadavona]

Thanks to Defcon1 and Highstandard40 for posting this information. Without it I wouldn't have been able to confirm what type of solder I had. That's some nasty stuff to smelt. gag

[Lloyd Smale]

most radiator shops use 60/40

[Defcon-One]

Happy to help!

I was just reading this again and I thought of an easier way to figure this out, and it will be dead on accurate! No worries about Pure being PURE or size/volume differences due to alloy.

1.) Cast a few good bullets from the mistery alloy. Weigh them.

2.) Using the same mold, same temp., cast a few good bullets from known solder 40/60, 50/50, 60/40. Then weight them all and see which is closest to the weight of the mystery metal bullets.

Closest one is what you have!

3.) If you have a bit of each alloy and a mold you can save the samples or record their weights for another test on different metal later. Just cast the new mystery metal in the same mold, same temp. and see what you have.

Why didn't I think of this before? It saves all the math and hassles and it is easy to do!

DC-1

[smokemjoe]

John Deere made all there own radiators, I had to work next to it where they soldered them togther. It was 60/40 from jonhson Iowa company.

[40Super]

Now I've got some numbers to be checked by someone.

A bullet that is cast out of pure lead(known pure) weighs 208gr(207.3-209),

The same bullet cast out of a solder alloy(unknown) weighs 178gr(177.8-178.9).

If my calculations are correct that would make this 40% tin/60%lead, am I correct?
I want to verify this just to get a good % to mix and be able to repeat down the road. Thanks.

[Defcon-One]

Same math!

The density of Lead is 0.409 lb. per cubic inch or (2,863 gr. per cubic inch).
The density of Tin is 0.264 lb. per cubic inch or (1,848 gr. per cubic inch).

At 208 gr. your bullet is .0727 cubic inches of pure Lead. (208/2,863)

So, a bullet of pure Tin from your mould would be 1,848 X .0727 = 134.35 gr.

A bullet from your mould of the mystery alloy is 178 grains. That works out to about 40.7% Tin and 59.3% Lead per this equation:

(.407 X 134.35 gr.) + (.593 X 208 gr.) = 178.02 grains

I’d say that you are exactly right at 40% Tin/60% Lead! It is always good to know what you have!

[Tom Myers]

Another approach to determine the percentage of tin in a Tin / Lead alloy is to:

STEP 1. Determine the specific gravity of the alloy using a balance beam scale and then:

STEP 2. Use a formula to determine the percentage of tin in the alloy as related to the specific gravity.


STEP 1:
If you have a balance beam scale, it is fairly easy to determine the specific gravity of a chunk of alloy.

First, arrange a balance beam scale so that the sample may be suspended from the pan by a fine thread in such a manner that a container of distilled water may be brought up under the sample to the point where the sample is just completely immersed in the water.

Next, place the sample in the scale pan, without the thread, and carefully weigh the sample then record that weight and label the value as SW (Sample Weight).

Next suspend the sample from the pan with the thread and record the combined weight of the sample and thread or wire and label that value as DW (Dry Weight).

Now, bring a container of distilled water up under the sample until the sample is just immersed in the water and balance the beam out and record that weight and label the value as WW (Wet Weight).

We can determine the specific gravity of the sample with this formula.

Specific Gravity = SW / ( DW - WW )

Simply stated. Subtract the Wet Weight from the Dry Weight and divide that difference into the Sample weight.

For instance a Lyman 40 Caliber Snover Bullet Cast from a 1/20 Tin/Lead alloy weighs 408.6 grains.

Suspended from the scale pan with some fine thread it weights 409.3 grains. Immersed in distilled water, the scale shows 372.3 grains.

409.3 - 372.3 = 37 grains

408.6 / 37 = 11.04324324 = The estimated specific gravity of the alloy.

The actual calculated specific gravity of 1/20 Tin/Lead alloy is 11.0433658. (within 0.05%)


STEP 2: This is the formula to determine the percentage of tin in a Tin / Lead Alloy using the specific gravity of that alloy.

P = (20.7680302 / SG - 1.83058882) * 100
Where P = The percentage of Tin in a Tin / Lead Alloy

Plug in the Snover Bullet Specific gravity and we have

20.7680302 / 11.04324324 = 1.880634298
1.880634298 - 1.83058882 = 0.050045478
0.050045478 * 100 = 5.005% of tin in the alloy

For those interested in how we arrive at that formula:

Let SG = The Specific Gravity of the alloy
Let T = The Percentage of Tin / 100
Let 1- T = The Percentage of Lead / 100
7.337 = The Specific Gravity of Tin
11.345 = The Specific Gravity of Lead

1 / SG = T / 7.337 + (1 - T) / 11.345

Then multiply both sides of the equation by 7.337 and
Then multiply both sides of the equation by 11.345
Then
83.238265 / SG = 11.345 * T + 7.337 - 7.337 * T

Then subtract 7.337 from both sides of the equation
Then combine T
Then
83.238265 / SG - 7.337 = 4.008 * T

Then divide both sides of the equation by 4.008

Now we have the simplified formula to find the percentage of tin divided by 100.

20.7680302 / SG - 1.83058882 = T = the ratio of tin in the alloy

Finally we multiply T by 100 to arrive at the percentage of tin in the Alloy
100 * T = Percentage of Tin in the Alloy