Author Topic: 15% NH3 in EtOH Preparation?  (Read 6727 times)

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  • Guest
15% NH3 in EtOH Preparation?
« on: June 01, 2004, 03:37:00 AM »
At Rhodiums website, the post "Base-catalyzed Hydrolysis of Ergocristine to Lysergic Acid" (

) has a procedure for producing 15% NH3 in EtOH by adding NaOH to absolute Ethanol containing NH4Cl. However no weights of the reagents are given. If one wanted to produce say, 1 liter of 15% Ammoniated Ethanol, how much NH4Cl, NaOH, and CaO would be required?  TIA.


  • Guest
Try this.
« Reply #1 on: June 01, 2004, 04:29:00 AM »
As one litre of ethanol weighs about 790 g, you want 15 percent of those grams to be ammonia, that is 790 x 0,15 = 118,5 g, which is 118,5 / 17 g/molNH3 = 7 moles roughly.

NH4Cl + NaOH -> NH3 + H2O + NaCl

You will need to add 7 moles of NH4Cl and 7 moles of NaOH.

NH4Cl: 7 x 52,5 g/mol = 367 g.
NaOH: 7 x 40,0 g/mol = 280 g.
CaO: Enough to absorb 7 moles of water (CaO + H2O -> Ca(OH)2).

I'm quite positive that a solution of 671,5 g ethanol and 118,5 g of ammonia does not have the volume of 1 litre, so you might have to make a slight excess just to be sure you get enough. Now that you get the idea I'm sure it won't be a problem though...


  • Guest
convienient prep
« Reply #2 on: June 01, 2004, 07:50:00 AM »
one method is to take a 500ml jar of 28% nh4oh and a 500ml jar of c2h5oh and place them together in a 4000ml jar with a lid. after a day or two the ammonia vapors will go into the alcohol in about 15% which is about the saturation point for nh3 in c2h5oh. i got this from the tips in a lancaster catalog for preparing a silicate-free solution of nh3 from regent stored in glass container.


  • Guest
Thanks All!
« Reply #3 on: June 02, 2004, 03:11:00 AM »
Appreciate the math and tips.