Re: molecular weight of dry air

[Brian Whatcott <betwys1@SBCGLOBAL.NET>]

I am almost completely innocent of insight in this area.

I checked on the accessible values for molar weight of dry air:
28.9645  g/mol    (Govt.)
28.85264 g/mol     (Rice)
28.9644  g/mol     (Danish)
28.966    g/mol   (Washington)
...which show reasonable agreement.

At 08:34 AM 5/12/2004, you wrote:
> I'm trying to determine the molecular weight of dry air and
> the during the procedure (which I seem to be doing right)
> I've raised a couple of questions that I can't answer.
>
> First, some background: As I understand it, the molecular
> weight of dry air is the average mass per mole of all of
> the various molecules that make up dry air.  The percentage
> of each molecule type is given as:
>
> N2   78.084
> O2   20.9476
> Ar    0.934
> CO2 0.0314
> Ne   0.001818
> He   0.000524
> CH4 0.0002
> Kr    0.000114
> Xe   0.0000087
>
> The average mass per mole of each type is determined by
> using the above percentages with the atomic weight of each
> type.  The atomic weight is determined by averaging the
> mass per mole of each isotope based upon the relative
> abundance of the various naturally-occuring isotopes.
>
> What I want to know is:
>
> 1) Can the isotope distribution of molecules like N2 and O2
> be determined the isotope distribution of the elements N
> and O?  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%?  If so, why?  Couldn't there also be a 33O2?

If atoms combine indifferently with isotopic number
then the proportions of the molecular species would be
16O X 16O = (0.99757)^2
16O X 17O = 2 X (0.99757 X 0.00038)
16O X 18O = 2 X (0.99757 X 0.00205)
17O X 17O = (0.00038)^2
17O X 18O = 2 X (0.00038 X 0.00205)
18O X 18O = (0.00205)^2

Summing:
0.91515 +    (32)
0.00076      (33)
0.00409      (34)
0.0
0.0
0.0
----------------
0.92000

Relative abundance would be
0.91515/0.92  = 0.99473
0.00076/0.92 = 0.00083
0.00409/0.92 = 0.00445

..leading to an average molecular weight of
32 X 0.99473 = 31.83136
33 X 0.00083 =   0.02793
34 X 0.00445 =   0.15130
-------------------------------
                         32.01059

> 2) Is the relative abundance of the isotopes of a given
> element the same for air as for any other material?  For
> example, is the relative abundance of CO2 isotopes the
> same for gaseous CO2 as for solid CO2?  If so, why?
>
> ____________________________________________________
> Robert Cohen; 570-422-3428; www.esu.edu/~bbq
> East Stroudsburg University; E. Stroudsburg, PA 18301


W.F.Libby's 14C dating method supposes that the transmutation
of nitrogen to 14C in the high atmosphere shares this isotopic
proportion in the atmosphere equally with biological material,
 so that the remnant of this isotope remaining after uptake
ceases on death, and decaying with a half life around 5000
years is a measure of the material's age.
On the other hand, heavy water shows some different physical
properties to those of light water - melt and boil points etc.,
so it is unsafe to suppose that atmospheric ratios of all isotopes
 are maintained throughout changes of state.

Further, there is a suggestion that the diffusion methods formerly
used for separating Uranium isotopes may now be replaced by
judiciously chosen chemical reactions which show more sensitivity
to isotopic composition than was formerly known.

Brian Whatcott    Altus OK    Eureka!