Sciencemadness Discussion Board

Substituting nitric acid with ammonium nitrate

ManyInterests - 6-12-2024 at 06:39

I'm setting up to make nitric acid for several synthesis experiments. Specifically WFNA for PETN and RDX. While PETN can be made with nitric acid of lower concentration (in the presence of sulfuric acid), RDX requires WFNA as the intermediaries are destroyed by sulfuric acid and thus needs very dry and highly concentrated nitric acid. I've noticed that most other energetics can be made with just nitrate salts, like nitroglycerin, EGDN ( and DEGDN I assume), and even TNT. The last part actually surprised me, BTW.

I've also seen a lot of other things just made with either azeotropic nitric acid or ammonium nitrate (superior to other nitrate salts because it keeps the nitrating mixture much more liquid and easy to stir), and I was told there was a formula to discover how much ammonium nitrate to use vs 68% nitric acid. I don't have my lab notes with me right now, but I would like to know how to approach the issue.

For example when I made ETN with nitric acid I used 4ml of nitric acid per gram of erythritol. I also did make methyl nitrate with 1.22ml of nitric acid to 1 ml of methanol (I used RFNA for that one). How should I approach substituting the nitric acid with ammonium nitrate. Also would the ratio of sulfuric acid need to be adjusted or kept the same?

I wish to produce the material in a variety of ways because that's where the fun is.

[Edited on 6-12-2024 by ManyInterests]

UndermineBriarEverglade - 6-12-2024 at 10:56

I would compare molar ratios. For instance, last time I made ETN I used 6.5 moles of nitrate per mole of erythritol. Theoretically that made 6.5 mol nitric acid in situ:

2(KNO3) + H2SO4 -> 2(HNO3) + K2SO4

And 6.5 mol HNO3 per 1 mol erythritol = 3.35g nitric acid per gram erythritol. Divided by 68%, divided by 1.41 g/cm3, that comes out to 3.55mL per g erythritol. There's only 4 nitronium ions in erythritol tetranitrate but I think any method will need to make an excess available. I've seen 5-8 recommended and by my math you were using 7.4.

So I think that for nitration reactions that are OK with a lower concentration of nitric acid and the presence of sulfuric acid, each mol of nitric acid in the original reaction can be substituted with 1 mol of nitrate salt and 0.5 mol of sulfuric acid. Doing the math, that would be 1.22g of ammonium nitrate and .75g sulfuric acid per mL of azeotropic nitric acid. That's in addition to however much sulfuric acid the original reaction may have called for.

greenlight - 6-12-2024 at 11:10

The examples you listed are three different types of nitration:

RDX and HMX is N-nitration (NO2 bonded to a nitrogen) or nitrolysis and requires anhydrous nitric acid for best results and no sulfuric acid present as well as other reagents like acetic anhydride or ammonium nitrate. It has been found that the cation of the nitrate salt is the factor responsible for yield obtained in a study done with varying nitrate salts used in HMX synthesis from DPT and the ammonium cation presence resulted in the highest yield with some salts like copper nitrate being responsible for yields as low as 30%.

PETN and NG are O-nitration with the NO2 bonding to am oxygen usually of a sugar. These reactions can easily be performed with good yield using 60% nitric acid and concentrated sulfuric acid or nitrate salt and sulfuric acid.

The third is C-nitration as in TNB and TNT on a benzene ring with the NO2 bonded to you guessed it, a carbon. These require a mixed acid solution with increasing temperatures and acid strength to add further stubborn nitro groups and can be a pain in the ass. In fact, TNB outperforms TNT and probably would have been very widely used and we would be singing a different ACDCs song if it wasn't for the fact that you have to really, really bully that last nitro on to the ring for the complete trinitro package deal making it more expensive and time consuming to produce.

I've always been highly in favour of using azeptropic nitric acid for all the nitrate esters as you have everything in solution and just know your going to get quality product in good yield if you run it carefully with sufficient bath time.

I would not try and distill nitric acid from a mixture of ammonium nitrate and sulfuric acid as that sounds like a possible bad time at high temps. But you can use it as far as I know fine as a nitration mixture as the reaction supplies the water to dissolve the NO2 to form your nitric acid in situ.

Amounts wise, just use the stoichiometric reaction equation and convert your grams of reagents to moles for the size you are performing. I believe the molar ratio is something like 1:2 sulfuric acid to ammonium nitrate. The ability to write a stoichiometric balanced reaction equation is vital for chemistry as it holds all the secrets to the quantities of reagent and expected yield.



greenlight - 6-12-2024 at 11:13

I see @underminebrianeverglades beaten me to it and also been kind enough to crunch the numbers for you :)

Sir_Gawain - 6-12-2024 at 11:57

Quote: Originally posted by greenlight  

I would not try and distill nitric acid from a mixture of ammonium nitrate and sulfuric acid as that sounds like a possible bad time at high temps.

I’ve done it multiple times before without incident. I think there is some decomposition of ammonium nitrate because I haven’t been able to get over 85% acid from it. Using the same sulfuric acid, and sodium nitrate, 98% nitric was obtained.
If your ammonium nitrate contained some organic material, I could see it being a problem.

greenlight - 7-12-2024 at 08:06

Quote: Originally posted by Sir_Gawain  
Quote: Originally posted by greenlight  

I would not try and distill nitric acid from a mixture of ammonium nitrate and sulfuric acid as that sounds like a possible bad time at high temps.

I’ve done it multiple times before without incident. I think there is some decomposition of ammonium nitrate because I haven’t been able to get over 85% acid from it. Using the same sulfuric acid, and sodium nitrate, 98% nitric was obtained.
If your ammonium nitrate contained some organic material, I could see it being a problem.


I have not tried manufacture of nitric acid from any nitrate salts before, only ever distilled azeotropic nitric with sulfuric acid for higher strength.

The lower strength you obtain from using ammonium salt must be like you stated due to decomposition of the ammonium nitrate or the only other standout variable would be the actual ammonium cation causing unfavorable side reactions.

I would be interested to know if increasing the molar ratio of ammonium nitrate to sulfuric acid results in a higher purity nitric acid. If so, the evidence would definitely point towards decomposition of the ammonium nitrate under heating.

ManyInterests - 8-12-2024 at 13:04

Quote: Originally posted by Sir_Gawain  
Quote: Originally posted by greenlight  

I would not try and distill nitric acid from a mixture of ammonium nitrate and sulfuric acid as that sounds like a possible bad time at high temps.

I’ve done it multiple times before without incident. I think there is some decomposition of ammonium nitrate because I haven’t been able to get over 85% acid from it. Using the same sulfuric acid, and sodium nitrate, 98% nitric was obtained.
If your ammonium nitrate contained some organic material, I could see it being a problem.


Yes. I have heard that ammonium nitrate is not a good reagent to use in making nitric acid as part of the ammonium nitrate decomposes into plain ammonia which neutralizes part of the end product. This is probably why you were not able to get to the WFNA levels of concentration with it.

Sodium nitrate, I find, is much much more effective. While it is more hygroscopic, I have never had an issue with that since I always dry that part in the oven for a few hours (and letting it cool down in the the sealed flask for a few hours more in the fridge) before I add the other reagents and commence the distillation. I will be using sodium bisulfate going forward in very large part due to the safety. Adding sulfuric acid to the sodium/potassium nitrate immediately released toxic NO2 gasses which are always a problem even with a gas mask on and other precautions taken. With the sodium bisulfate method they only start to release those gasses once heating is underway, and by that point all the safety precautions are in place and I have no danger of inhalation.
Quote: Originally posted by UndermineBriarEverglade  
I would compare molar ratios. For instance, last time I made ETN I used 6.5 moles of nitrate per mole of erythritol. Theoretically that made 6.5 mol nitric acid in situ:

2(KNO3) + H2SO4 -> 2(HNO3) + K2SO4

And 6.5 mol HNO3 per 1 mol erythritol = 3.35g nitric acid per gram erythritol. Divided by 68%, divided by 1.41 g/cm3, that comes out to 3.55mL per g erythritol. There's only 4 nitronium ions in erythritol tetranitrate but I think any method will need to make an excess available. I've seen 5-8 recommended and by my math you were using 7.4.

So I think that for nitration reactions that are OK with a lower concentration of nitric acid and the presence of sulfuric acid, each mol of nitric acid in the original reaction can be substituted with 1 mol of nitrate salt and 0.5 mol of sulfuric acid. Doing the math, that would be 1.22g of ammonium nitrate and .75g sulfuric acid per mL of azeotropic nitric acid. That's in addition to however much sulfuric acid the original reaction may have called for.


When I read that I looked in some of my old notes that I made with discussions previously with other chemists. Your explanation answered some questions of mine like why do some nitrate salt synthesis have such an excess of sulfuric acid. I am still scratching my head when trying to translate the synthesis to the other stuff... so I am going to try to apply your formula for the methyl nitrate synthesis (which I got from the Improvised Munitions Handbook).

The IMH calls for 120ml of sulfuric acid, 82.5ml of 90% nitric acid, and 67.5ml of methanol (they use teaspoons technically...). So 82.5ml of 90% acid at 20C is a SG of 1.4911, meaning 123 grams in total. We can substract the 10% which is most likely water and maybe a touch of NO2 contamination, leaving us with 112g of pure nitric acid, this leaving us with a 1.777 mols of HNO3. so for a 1:1 molar rate, that would mean 142.27g of ammonium nitrate + an additional 86.8 grams of sulfuric acid to the mixture.

Am I on the right track here?

ManyInterests - 22-12-2024 at 20:12

Bumping the thread up since I am still curious if I was on the right track!

greenlight - 23-12-2024 at 00:11

The reaction equation for the production of nitric acid from ammonium nitrate and 98% sulfuric acid is:

H2SO4 + 2NH4NO3 = 2HNO3 + (NH4)2SO4

* Molar mass 98% Sulfuric acid = 98.079g/mol
* Molar mass nitric acid =63.01g/mol
* Molar mass ammonium nitrate = 80.043g/mol

Using your numbers, you require 82.50ml of 90% nitric acid. Specific gravity of 90% nitric acid is approximately 1.48 g/ml. The mass required is therefore (82.50ml)(1.48g/ml) = 122.10 g.
Excluding the 10% water you have (0.90)(122.10g) = 109.89 g nitric acid round up to 110g for simpler numbers.

Converting this mass of nitric acid to moles gives: 110g / 63.01g/mol = 1.745 moles nitric acid.
The sulfuric acid is in a 1:2 ratio with the nitric acid product so the one half the molar mass is required: 1.745mol / 2 = 0.8725 mol sulfuric acid required. So,: (0.8725mol)(98.079g/mol) = 85.57g sulfuric acid
Factoring in the density of the sulfuric acid (1.84g/ml) gives: 85.57g / 1.84g/ml = 46.50ml sulfuric acid.

The ammonium nitrate has a molar ratio of 1:1 with the product nitric acid, so: (1.745mol)(80.043g/mol) = 139.67g ammonium nitrate.

By the reaction equation, you would need 46.50ml 98% sulfuric acid and 139.67 g ammonium nitrate plus the additional 86.8 g for the nitration. This is calculated using the stoichiometric reaction equation for production of nitric acid with the reagents you are using, but to be honest, nothing runs exactly stoichiometrically perfect, so a little more sulfuric acid in excess can only help hold more water away from interfering.

ManyInterests - 24-12-2024 at 17:37

Quote:
By the reaction equation, you would need 46.50ml 98% sulfuric acid and 139.67 g ammonium nitrate plus the additional 86.8 g for the nitration. This is calculated using the stoichiometric reaction equation for production of nitric acid with the reagents you are using, but to be honest, nothing runs exactly stoichiometrically perfect, so a little more sulfuric acid in excess can only help hold more water away from interfering.


Quote:
The IMH calls for 120ml of sulfuric acid, 82.5ml of 90% nitric acid, and 67.5ml of methanol (they use teaspoons technically...). So 82.5ml of 90% acid at 20C is a SG of 1.4911, meaning 123 grams in total. We can substract the 10% which is most likely water and maybe a touch of NO2 contamination, leaving us with 112g of pure nitric acid, this leaving us with a 1.777 mols of HNO3. so for a 1:1 molar rate, that would mean 142.27g of ammonium nitrate + an additional 86.8 grams of sulfuric acid to the mixture.


So 139.67 grams of ammonium nitrate plus an additional 86.8 grams of nitrate? That sounds like an awful lot of nitrate for that amount of sulfuric acid. I have a feeling I wasn't that off in my own calculation if it is something else.