[alt-photo] Re: Stoichiometry for the nonscientist

Loris Medici mail at loris.medici.name
Wed Aug 4 17:04:59 GMT 2010

18% Potassium Oxalate: K2C2O4, 166.216g/mol, 180g (in final volume of 1000ml) / 166.216  = 1.083M solution.

1. 1.083 mol K2CO3 (mol. mass = 138.206g) = 1.083M x 138.206g/mol ~= 150g

2.a. 1.083 mol "anhydrous" C2O4H2 (mol. mass = 90.035g) = 1.083M x 90.035g/mol = 97.5g
2.b. 1.083 mol C2O4H2.2H2O  (that's "dihydrate" - mol. mass = 126.066g) = 1.083M x 126.066g/mol ~= 136.5g

for a total / final solution volume of 1000ml.

Yes, your figures are right.

BTW, you'll have to add some more oxalic acid to set the pH to something around 5... (You add in small amnts, test with a strip of pH test paper.) Potassium oxalate alone will give a pH around 8 (if I'm not mistaking), and you don't want that!


On 04.Ağu.2010, at 18:50, David Ashcraft wrote:

> My past post:
>> 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?
> And now after some tricky shenanigans I arrive at: 149.64 g of K2CO3 and 97.59 g of C2H2O4
> Would anybody agree?

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