[alt-photo] Re: Stoichiometry for the nonscientist

[etienne garbaux]

David wrote:

>OK - What if I needed to mix a formula like: 180 gms potassium oxalate
>K2C204.H2O mixed with water to make 1 liter.  I only have potassium
>carbonate and oxalic acid on hand.
>
>I know that I can make this formula of potassium oxalate by mixing
>potassium carbonate and oxalic acid; PC is K2CO3 and OC is C2H204.
>Now how would I know how much of each of these to make up my original
>formula?

As Eric already said, you have to figure out what happens to 
everything you put into the reaction.  You also need to know the 
thermodynamics of the reaction you are contemplating -- whether it is 
exothermic or endothermic, and by how much.  Then, it is a matter of 
figuring how much of each reactant you need.  That is where mols come 
in (which, as I recall, is what started this thread).  "180 grams" 
doesn't tell you how many molecules of potassium oxalate you hope to 
end up with and, consequently, how many molecules of each reactant 
you need.  No getting around it, you need to figure this out using 
the molecular weights of the compounds involved.  You can do this 
every time you have a chemical weight (like "180 grams of potassium 
oxalate," or you can do it beforehand by making up solutions that 
have a known number of molecules per milliliter and then just 
measuring a volume of each solution.

That's what the concept of "molar concentration" is.  Instead of 
counting molecules, we find it easier to count by bunches of 
molecules -- 6.022 x 10^23 molecules, to be precise.  That is handy 
because of one Sig. Amedeo Avogadro and one M. Jean Perrin, who 
brought us the concept of "gram molecular weight" ("GMW" or 
mol).  GMW is simply the realization that 6.022 x 10^23 atoms or 
molecules weigh, in grams, the atomic or molecular weight of the atom 
or compound.  So, 58.443 grams of table salt (NaCL), which has a 
molecular weight of 58.443, will have 6.022 x 10^23 molecules (give 
or take one or two).  If you then dissolve this salt in water and add 
water to make one liter, you have a "one molar" solution -- it 
contains one GMW of salt in every liter.  If you dissolve 2 x 58.443 
= 116.89 g and make up one liter, you have a "two molar" solution, 
and so forth (limited, of course, by the solubility of the compound 
in question).

By inspection, you can see that 1 mol of potassium carbonate reacts 
with 1 mol of oxalic acid to make 1 mol of potassium oxalate, with 
one mol of H2 and one mol of CO3 left over.  The H2 and one oxygen 
atom react to make one mol of water (H2O), leaving only carbon 
dioxide (CO2) in excess.  So, we expect this reaction to evolve one 
mol of CO2 for every mol of potassium carbonate and oxalic acid we react.

The molecular weight of anhydrous potassium oxalate is 166; the 
monohydrate is 184.  Thus, the solution you postulated is very close 
to 1 molar (0.978 molar).  If you react 489 ml of 2 molar potassium 
carbonate 489 ml of 2 molar oxalic acid, then add water to make 1 
liter, you would have what you want.

Note that if you started with potassium hydroxide (KOH) instead of 
potassium carbonate, you would require two mols for each mol of 
oxalic acid and the only byproduct would be water.

Best regards,

etienne