Re: molecular weight of dry air

[Michael Edmiston <edmiston@BLUFFTON.EDU>]

In case you are wondering about the algebra,
but haven't wanted to tackle it.  I did, and it
works out, as it must that

p*p*(m16 + m16) +
p*q*(m16 + m17)*2 +
p*r*(m16 + m18)*2 +
q*r*(m17 + m18)*2 +
q*q*(m17 + m17) +
r*r*(m18 + m18)

is the same thing as

2*(abundance-weighted-atomic mass)

which is

2*(p*m16 + q*m17 + r*m18)

If you want a few of the details,
here they are.

Expand the first summation as

2*p*p*m16 +
2*p*q*m16 + 2*p*q*m17 +
2*p*r*m16 + 2*p*r*m18 +
2*q*r*m17 + 2*q*r*m18 +
2*q*q*m17 +
2*r*r*m18

In the second line, replace the first q with (1-p-r)
In the second line, replace the second p with (1-q-r)
In the third line, replace the second p with (1-q-r)

Next, expand everything and add it all up.  Most
terms drop out, leaving us with

2*p*m16 + 2*q*m17 + 2*r*m18

which is the desired result.

As John Denker pointed out (and so did I in an earlier post)
isotope effects during reactions can make this invalid because
isotope effects mean that the probabilities are not
exactly as stated in the top summation.

However, isotope effects are small, especially as we move
toward higher-mass elements.

Deliberate "isotope effects" as John described for lithium,
or another example would be depleted uranium, are an
entirely different mater.

Michael D. Edmiston, Ph.D.
Professor of Physics and Chemistry
Bluffton College
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu