Solder Alloy Density Equation: Why What Most People Think is Right is Wrong

Folks,

It’s hard to believe but I have been blogging for over 10 years. In all of this time, with the hundreds of posts I have made, the most popular topic by far has been calculating density in a metal alloy. One of the reasons for this popularity has been the belief that the density of an alloy is determined by the equation

Eq. 1     eq1

Where x is the mass fraction of metal 1, y the mass fraction of metal two, ? (rho) the respective densities and ?t the total or alloy density. I have shown in the past that the correct formula to calculate the alloy’s density is:

Eq. 2    eq2

This formula is derived below again.

People continue to ask why equation number 1 is not correct, so I have posted an explanation that has been modestly helpful.  I have thought of an example that shows that Eq. 1 cannot be correct and have now derived an equation in the form of Eq. 1 that uses volume fractions instead of mass fractions.  This derivation is also below and the equation is:

Eq. 3       eq3

However, Eq. 3 is not very useful as the volume fraction of each metal is not as readily available as the mass fraction, which is easily measured with a scale.

Now, to give an example that shows that Eq. 1 is unreasonable, let’s consider a thought experiment that will help us conclude that Eq. 1 can be way off. Consider a cubic meter of air in a container 1 meter on a side at room temperature. The cubic meter of air will weigh 1.225 kg. (The fact air weighs this much surprises many people.) Inside the container is 1.225 kg of a fine gold powder. We blow the gold powder into the air and it covers all of the inside with an equal concentration. The powder is so fine that it will remain suspended for a short time. So we will consider this an alloy of gold and air.  The mass fractions x and y are equal at 0.50.  So if Eq. 1 were to hold true the density of the “alloy” would be:

Eq. 4    eq4

Figure 1. The gold dust and air density experiment.

Figure 1. The gold dust and air density experiment.

The weight of the 1 cubic meter container would now be 9650.6 kg/m3 * 1 meter3 = 9650.6 kg!  Whereas we know it to be 1.225 kg + 1.225 kg = 2.45 kg. Eq. 2 or 3 will provide the correct answer.

The correct derivations are below:

 

Eq. 5 6

Cheers,

Dr. Ron

 

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About Dr. Ron

Materials expert Dr. Ron Lasky is a professor of engineering and senior lecturer at Dartmouth, and senior technologist at Indium Corp. He has a Ph.D. in materials science from Cornell University, and is a prolific author and lecturer, having published more than 40 papers. He received the SMTA Founders Award in 2003.