[RogueRose]

If this is TLDR/TLCR then jump to the part under the ***********.  The top explains the process and gives data, amounts, etc, the bottom is what I did.

I'm trying to find out the proper amounts to make Sodium Acetate Trihydrate using 5% vinegar and baking soda. 

I weighed the vinegar in 10, 100 and 250 amounts with a graduated cylinder for a total of 360ml the weights were 10g, 100.0g and 250.65g respectively for a total of 360.75.  360.75 / 360 = 1.002083

I weighed 64fl oz and the weight was 1942g for what should be 1892.5 ml.    Now if the vinegar is 5% then 1942g / 20 = 97.1g of 100% Acetic Acid

A mole of NaHCO3, Acetic Acid and Sodium Acetate are 84.0066g-mol , 60.05g-mol amd 82.03g-mol and the densities are 1.1g/cc-1.3g/cc (Na bi carb powder), 1.049g/ml (Acetic Acid) and 1.45g/cc or 1.528g/cc (NaOAc Trihydrate and Anhydrous - respecively)
   
Sodium Acetate Melting point 136F and 615F for Trihydrate/Anhydrous respectively.  BP is 252F (decomposition point) and 1,615 for the trihydrate and anhydrous respectively)

Acetic acid has a melting point of 61-62F and a BP of 244 to 246F. 

The thing is that Sodium Diacetate is formed the same way but 2x the acetic acid is used so a half neutralization is preformed. 

   
The equation is NaHCO3 + CH3COOH = NaCH3COOH + CO2 + H2O


I've been told that the best way to make the sodium acetate is to add more acetic acid (vinegar) than what is called for to ensure full conversion of the bicarb. 

I also found that when boiling off the water, the boiling point is raised considerably and found that it was boiling at 225 the first time I checked.  the temp continued to rise until I got to 241-242 and removed it from heat.  I had placed about an extra 2-4% vinegar than what was needed and IDK if that could have messed thing up. 

****************************************

After looking back at my calculations I think I made some mistakes.  I took the total ml of 1892.5 / 20 = 94.625ml which would be 100% AA.  I then divided the 94.625 by 60.05g (mole of AA) to get 1.5757 moles.  From there I calculated that 1.5757 x 84.007 (mole of NaHCO3) = 132.37g   so for the 1/2 gallon I added the 132g until completely mixed and then added about another 25ml of vinegar. 

I think my mistake was calculating the moles of acetic acid. if there were 94.625ml then I needed to convert that to grams (1.049g/ml) so 94.625 * 1/049 = 99.26g.    99.26g/60.05g-mole = 1.653 moles.  1.653moles * 84.007g-mol = 138.86g NaHCO3 so a difference of almost 5% (the amount difference between weight of H2O and AA per ml). 

The thing is that I'm not sure exactly how the weight vs volume is calculated when using a 5% solution of a substance and water.  If anyone could help me out I would really appreciate it!  Thank your!

Also, I need to figure out some kind of initiator/activator to start the nucleation of the acetate in a sealed pouch (no dropping hardened crystal into it).  In the heating packs they have a little disk that you "snap" and it starts the reaction.  I need to come up with something similar or learn how to make these (not out of the question, just need to know what metal and some other things).  Any suggestions are greatly welcomed!

[Borek]

Perhaps I am missing something, but how are you removing excess water?

[AWK]

Density of 5 % acetic acid is close to 1.005. For most calculations you may safely use density of water instead.
Then 1 L of vinegar contains ~50 g of acetic acid which is 5/6 of mole. Hence for 1 L of acid, according to a balanced reaction,  you need 5/6 x 84 = 70 g of NaHCO₃. From this mixture you should evaporate about 0.9 L of water before crystallization.

[RogueRose]

Perhaps I am missing something, but how are you removing excess water?

Boiling after adding the bicarb.

[AWK]

How you remove 0.9 L of water? Evaporation at RT will need a few weeks.

[Borek]

How you remove 0.9 L of water? Evaporation at RT will need a few weeks.

Boiling after adding the bicarb.

Answered just before you asked.