- How's Your Chemistry? -
This was the question that my friend Paul Morris asked me one
time in an email in year 2009.
It is an important question because in order to get beyond the
guesswork stage of these wax experiments, some knowledge of chemistry
is required.
Here is the most brief synopsis of the situation as it existed then:
The old recipe calls for 1 ounce of aluminum oxide.
Think of this as a word problem in a high school chemistry book:
How much aluminum metal would be needed to obtain the same amount
by weight of aluminum atoms as are found in 1 ounce of aluminum oxide?
Solving this problem is fun, and here are a few hints:
Aluminum Oxide is AL2O3.  That means that each molecule of AL2O3
contains 2 atoms of aluminum and 3 atoms of oxygen.
Next comes the concept of Atomic Mass Units (AMU).
AL = 27 AMU
O  = 16 AMU
The total AMUs contained in one AL2O3 molecule is therefore 102.
102 is said to be the Formula Mass of AL2O3.
Out of that 102, only 54 is pure aluminum, the rest is oxygen.
So, the ratio of the weight of plain pure AL metal to the total
weight of AL2O3 is: 54/102.
Conversely: inverting that number and looking at it as the ratio
of AL2O3 to AL it becomes 1.89
Example:  27 grams of AL is equivalent to 51 grams of AL2O3.