The value of A.‘s number is arbitrary in that it is tied to the units of mass we choose to use. If we used, say, pounds, we’d have to calculate how many atoms of a given element are in the number of pounds corresponding to the atomic weight. If we used a unit of mass of pithyfeathers (I don’t know what they are either!), we’d calculate the number of molecules or atoms in a atomic-number’s mass worth of pithyfeathers.

Atomic weights are essentially counting numbers, they measure the number of (roughly) proton-sized masses in an atom or molecule. Logical thought leads to the idea that one can “unitize” the counting number by making it grams, kilograms, pounds, slugs, or pithyfeathers. So instead of “atomic mass 16” we end up with “16 kilograms”. It’s also clear that in atomic (or proton) mass units, it takes exactly one atom/molecule to have “enough” mass to add up to it’s atomic weight. When we decide to count in kilograms or pithyfeathers, we need a number of atoms/molecules identical to the ratio of 1 AU expressed in the weight unit to 1 weight unit (in our case, 1 AU = how many kilograms, or lbs, or pithyfeathers).

To put it simply: to have exactly 1 atomic weight of a molecule, you need only one molecule. To have enough molecules to make exactly one unit of mass in another set of units (say, 1 kg), the number you need is equal to the ratio between the weight of one atomic-mass *in kilograms*, and the arbitrary unit in question (say, a kilogram).

So (1 AU) is to (1 Kg) as (1 atom) is to (n atoms) where we solve for “n”.

1 AU / 1 Kg = (1.660538782(83) × 10−27 kg) / 1kg

If it takes 1 atom to make 1 AU, how many atoms to make one kg?

1AU (in kg) / 1 kg = 1 atom / n atoms

n = 1 / 1AU(in kg) = 1 / 1.660538782(83) × 10−24 kg = 6.023x10+23 Notice that A.‘s number is just = 1 / (1AU mass expressed in kg)

Similarly, if I said a single AU was 1/100th of a pithyfeather (notice how completely arbitrary this is, just like the kilogram), I’d need 100 1AU atoms to make one pithyfeather of mass. Analogously, if I had molecules each weighing 16AU, 100 of them would equal 16 pithyfeathers. If I had huge protien molecules, each weighing 205 AU, then 100 of them would weigh 205 pithyfeathers. In any case, the “magic” number either 100 or Avogodro’s comes from understanding the mass of a single AU expressed in my arbitrary unit system.

As an interesting aside, here’s a question for you: Does Avogadro’s number HAVE to be an integer? If you can think about and answer this question clearly, you’ve almost certainly got a great grasp of the “magic” number.

-Scott