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Re: molecular weight of dry air



I do not understand Brian Whatcott's assertion (under the
Dumb... thread) that I used a different set of proportions
other than the binomial proportions Denker listed.

Using p and m16 for the isotopic abundance and mass of 16O
Using q and m17 for the isotopic abundance and mass of 17O
Using r and m18 for the isotopic abundance and mass of 18O...

The formula I used (for average mass of natural O2) was...

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)

Using data for p,q,r and m16, m17, m18 from the CRC
Handbook of Chemistry and Physics this calculation
yields 31.99881 as the molecular mass of O2.

This is the same number you get by doubling the average
atomic mass of O from the periodic table. This is also the
same number you find if you look up the molecular mass
of O2 in the CRC handbook.

What other number is there? Are some people doing
(p*m16 +q*m17)*2 rather than p*q*(m16 + m17)*2?

John... what is not quite right about this? Are you talking about
isotope effects, or something else?


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