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Re: molecular weight of dry air



Robert Cohen wrote:
1) Can the isotope distribution of molecules like N2 and O2
be determined the isotope distribution of the elements N
and O?

Sure. It can easily be calculated to a good approximation.

> For example, if 16O, 17O and 18O have relative
abundance of 99.757%, 0.038% and 0.205%, would 32O2, 34O2
and 36O2 also have an abundance of 99.757%, 0.038% and
0.205%?

No.

> Couldn't there also be a 33O2?

There most certainly is 33O2 ... and 35O2 (although
purists might cringe at the notation).

The key assumption/approximation is that the chemistry
depends only on Z (atomic number), not A (atomic weight)
... which is not always exactly true, but good enough
for present purposes.

Calculating the probabilities is easy if you organize
it the right way. Given the atomic probabilities, we
get the molecular probabilities the same way we would
calculate any binomial. If there are two atomic
isotopes with probabilities p and q, then the diatomic
probabilities are
p*p p*q
q*p q*q
where I have grouped the four contributions into
columns according to their mass. Note that the
middle column (the odd-mass molecule) is twice as
likely as you might have guessed. Buzzwords to
throw around at this point include "multiplicity"
which is related to "entropy".

Note we are using a classical "ball-and-stick"
model of the molecules, i.e. we are not worrying
about quantum-mechanical identical-particle effects.
This is appropriate given the masses and temperatures
of interest. The molecular quantum physics that
Feynman describes in Volume III chapter 9 (the
ammonia maser) is true for ammonia, but for almost
anything heavier than ammonia the ball-and-stick
model is better. Why this is so, and where the
boundary is, is a verrry deep question that we can
discuss later.

For three isotopes with probabilities p, q, and r, the
calculation can be extended in the obvious way:
p*p p*q p*r
q*p q*q q*r
r*p r*q r*r

So do the indicated multiplications, then add up the
columns.