                             MolarMas

(To view this file in Windows, use Write with conversion.  Notepad
shows funny characters.)

This free program determines the molar mass and percent composition
from a formula.  Features:

        Uncertainties in values are displayed.  These are
         calculated from a "propagation of errors" analysis
         using the published uncertainties in the masses of the
         elements.
      
        Few keystrokes are needed.  Symbols can be in uppercase, 
         lowercase, or both.  The actual keystrokes used to enter 
         sulfuric acid are "h2s ENTER o4 ENTER".  Symbols not separated 
         by numbers are separated by pressing the ENTER key.  
         NOTE:  The final ENTER must be pressed, or else the last 
         symbol or number entered will only be displayed, but not 
         used in the calculation.
      
        Handles enormous molecules.  Over 100,000,000 atoms of each 
         of 30 different elements, amino acids, and
         functional groups can be present.  Common abbreviations
         that are recognized include Me (CH3), Et (C2H5),
         Ph (C6H5), cp (C5H5), and Gly (glycine, -C(O)CH2NH-).
      
        Results can be printed.

        Other features:  A mouse is supported.  Press Escape to replace
         a field with the previous entry.  Control/Insert and Shift/Insert
         copy and paste selected text to a Windows-like clipboard.


OTHER FILES

MOLARMAS.ICO is an icon for Windows.  To add MolarMas to Windows, 
     Drag molarmas.exe from file manager onto program manager.
        It is given a default icon.  
     To change the icon, from program manager select File, Properties..., 
        Change Icon..., OK, Browse, and select MOLARMAS.ICO from the 
        appropriate directory.
        
MMSOURCE.ZIP contains the source code, written using visual basic for
DOS.  Unless you are interested in modifying the code and have VB, you 
might as well delete the file.


DISTINGUISHED FLUORINE

So you enter F2 and get these results:
    Element  Number  % Composition   Atomic Mass
    _______  ______  _____________   ___________
      F      2          100           18.9984032(9)

                      Molar Mass:     37.996806(2)

Fine, no problem.

Now you enter F1F1 and get these results:
    Element  Number  % Composition   Atomic Mass
    _______  ______  _____________   ___________
      F      1        50.000000(3)    18.9984032(9)
      F      1        50.000000(3)    18.9984032(9)

                      Molar Mass:     37.996806(1)

How come the uncertainty in the molar mass is .000001 for F1F1,
and .000002 for F2?  The reason is that the fluorine atoms in
F1F1 are distinguishable, and indistinguishable in F2.  That is,
you can tell one F from the other in F1F1.  The bottom line is
that the uncertainty in the molar mass is less if you introduce
the fiction that you can tell one F from another.  Or one carbon
from another.  As when you enter toluene as Ph1Me1 instead of
C7H8.  In short, any time abbreviations are combined that contain
the same atoms, such as Ala2Gly1, the calculated uncertainties
will be low.  

Sometimes atoms are distinguishable.  For example, to determine 
just the C bonded to O in CH3CO2H, you might decompose the 
compound and collecting CO2.  To get the uncertainty in percent 
C as CO2 in the compound, enter the formula as C1C1H4O2 instead 
of C2H4O2.


THE MATH

Molar mass is of the form w = ax + by + cz + ...  For CH4, this is
                   16.04276 = 1 * 12.011 + 4 * 1.00794
Standard deviation of molar mass, s, depends on standard deviation 
of x, s(x), and y, s(y) as follows:
          s = square root [as(x) + bs(y)]
That is, s equals the square root of a-squared times s(x) squared
plus b-squared times s(y) squared.  Uncertainties for C and H are
.001 and .00001, giving an uncertainty in CH4 of
          s = square root [1*1*.001*.001 + 4*4*.00001*.00001]
            = .0010007
So the mass of CH4 is given as 16.043(1).

Standard deviation of percent composition, s(%), is given by:
         s(%) = percent * square root[(s(x)/a) + (s/molar mass)]
The uncertainty in percent carbon in methane, then, is given by:
         s(%) = 74.868663 * square root[(.001/12.001) + (.0010007/16.043)]
              = .00779
So the % C in CH4 is reported as 74.869(8)

     
Holler if you find a bug, or have a suggestion.

     Christopher King
     Department of Chemistry
     Eastern Oregon State College
     La Grande, OR 97850-2899
     cking@eosc.osshe.edu
