Density Calculation Still Raises Questions

Folks,

It is hard to believe, but I have been blogging now for over 8 years. In all of that time, the most popular topic by far has been the calculation of alloy densities. Many people are troubled by the required equation. At first blush it doesn’t seem logical. I have derived it, but here is an effort to try and make it more intuitive.

A reader wrote the following, which inspired my explanation that follows.

Craig writes:

Why do we calculate the theoretical density of an alloy by using the formula of:
1/Da = x/D1 + y/D2 (binary) or
1/Da = x/D1 + y/D2 + z/D3 (ternary) as opposed to multiplying the individual metals’ densities by their percentages in the alloy and adding them together ? I thought the density of an alloy would be analogous to a weighted average of the densities of the metals in the alloy. Obviously, that is wrong, but I don’t understand why?

The reason you have to use the non-obvious formula is that density is given as mass/volume. So there are two quantities to be concerned with, and one is inverse (i.e. volume is 1/volume in the density formula). So you have to use the equations above, or you will be in error. I derived these equations in a previous post. However, many people do not find the result intuitive.

Here is, hopefully, a more intuitive example. Let’s say, for some strange reason, you were interested in inverse height. So let’s say John is 5 feet at his shoulders, or 1/5 inverse feet. His girlfriend Kathy is 5 feet tall, also 1/5 inverse feet. Kathy stands on John’s shoulders. How many inverse feet is their total height? The tendency would be to say they are 1/5 +1/5 = 2/5 inverse feet. But we know that when Kathy stands on John’s shoulders they span 10 feet, so they must be 1/10 inverse feet. So, to calculate their height in inverse feet, we need to use:

1/IFT = 1/John IF + 1/Kathy IF = 1/(1/5) +1/(1/5) = 10

IFT = 1/10

Note: (IFT = Inverse Feet Total)

The same is true for density, since density = mass/volume (inverse volume is like inverse feet).

Another engineering example is electrical resistivity and conductivity. If the resistance of one wire is 2 ohms, and another is 1 ohm, and we connect them in series, the total resistance is 3 ohms. However, if we consider conductivity, one wire has a conductivity of 0.5 mho and the other 1 mho. Again, to get the total conductivity we don’t add the conductivities, we must use:

1/Ctotal = 1/C1+1/C2 = 1/0.5 + 1/1 = 2+1 = 3. So Ctotal = 1/3 mho.

I hope this helps.

Cheers,

Dr. Ron

Wet Gold II: Measuring Gold Content in an Ore With Only a Scale

Folks,

In my last post we saw how you could measure density with only a scale.  In this post, we will expand on that technique and learn how to measure metal content in gold/quartz ore.  In principle, this technique could be used for other ore, but the ores can only be two part (e.g. gold and quartz) systems. Gold is a “natural” for this analysis as it is typically pure gold with quartz.

Gold is often found “veined” in quartz. I was certain that this was the origin of the “Golden Fleece,” the fleece being the white quartz with the gold on top. However, a little research did not clarify this belief.

Anyway, let’s assume you take a few weeks off from work. Leaving the world of solder paste, TIMS, ITO, wave solder flux and solder preforms behind, you set out for the west in search of some large gold nuggets.  Fate was with you in that, in a short time, you find a gold/quartz specimen as shown below.   The images, and the new “wet gold” weighing technique I will discuss, are from Bill and Linda Prospecting.

You are so excited you are shaking.  The only tools you brought are a scale, some string and a beaker.  To determine that gold content, you need to measure the weight of the gold in air and under water.  But you only have the scale as shown below.  What can you do?

After measuring the weight of the ore in air, fill the beaker part way with water, place it on the scale and zero the weight.  Then insert the ore on a string as shown below.  The scale will now read the weight of the volume of water that the ore displaces.  Let’s call this weight of the water displaced WD .  The wet weight of gold (weight of gold under water) will be the weight in air minus WD.  So we now have the weight in air and the weight in water.

The derivation of the equation that tells us how much gold is in the ore is at the end of this post.  The final equation we need is WAu = 3.07WW – 1.91WAir.  For our ore sample WAir = 25.1 pennyweight (pw). A pennyweight is 1/20th of a troy oz.  WD as shown in the photo above is 8.3 pw.  So WW = WAir – WD = 25.1-8.3 = 16.8 pw.  So WAu = 3.05*16.8 – 1.91*25.1 = 3.635 pw.  Subsequent analysis showed that the gold content was actually 3.9 pw and error less than 7%.  Not too bad for a simple field measurement.  At \$1600/oz our ore sample contained. a little over \$300 dollars of gold.

This technique could be used to measure the density of an alloy as in the last post.

Cheers,

Dr.Ron

The Derivation of the Equation

Measuring Alloy Density

Folks,

In the category of interesting requests, Ron, a gold worker, from Guyana, sent me the following note:

Dr. Ron,

My colleagues use a “wet” gold technique to measure gold alloy density.  Is this valid?  Where does the formula come from?

Sincerely,

Ron

Well, to tell the truth, I had never heard of it and was skeptical. How can you measure density (mass/volume) by only measuring weight? So, I investigated. The technique claims that one can measure density with only a scale, by measuring the alloy’s weight in air and in water.

I could find no derivation, so I thought about it and derived it on my own. As far as measurements go, as stated, you only have to measure the weight in air and water. If you don’t have a scale that can handle being immersed in water, you can use a hanging scale (think weighing a fish). So, after weighing the alloy in air, you immerse it in water. It will weigh the amount of water it displaces less.  The derivation is below:

As an example, let’s say you have a gold alloy ingot that weighs 1,000 grams (OK, I know grams is mass, but we are all sloppy and use it as weight, too) in air.  You weigh it in water and it weighs 930 grams. From the formula below, the alloys density is:

r = 1000/(1000-930) = 14.29g/cc

Since the density of gold is 19.3g/cc, the alloy is not pure gold.  If you knew the alloying element, say copper, you could use a Solder Alloy Density Calculator to determine that the alloy was 69.8% gold, 30.2% copper. If there are multiple alloying elements, since most of the common elements have a density of about 9 g/cc, you can even estimate the fineness of the gold.

Could this technique be used to measure the alloy density of say a handful of solder preforms? Sure, you could put them in a woven bag of non-hygroscopic material and weigh them in air and water. Admittedly, measuring the density of solder paste, with this technique, would be a challenge.

Next posting, I will show how this technique is used to measure the quantity of gold in gold/quartz ore.

Cheers,

Dr. Ron