Sciencemadness Discussion Board

Doing the math with Ammonium Hydroxide

Organicsynth - 4-9-2020 at 04:53

I want to use 2mmol of NH4OH in a synthesis, but I'm having problems with the math, I may be even doing it right but at this point I'm with such a brain fog attack that I prefer to ask for help than screwing it up.

Ammonium Hydroxide 28% 1L (NH4OH | NH3 + H2O)
Ammonium Molar Mass: 17,03 g/mol
Ammonium Hydroxide Molar Mass: 35,04 g/mol
Density @ 28% (20ºC) = 0,898 g/cm3
Concentration: 280mL x 0,898 = 251,44 g/L

n = m / M.W.
n = 251,44/17,03 g (should I use ammonium hydroxide MM instead?)
n = 14,77 M/L

I want to use 2 mmol so:
14,77 x X = 2 mmol
X = 0.135 mL

So my question is: 2 mmol of Ammonium Hydroxide 28%/1L is 0,135mL?

njl - 4-9-2020 at 07:05

Get the concentration in Moles/liter (molarity) (you're correct ~14.8 M) and then rearrange to solve for volume:

M (14.8) = Moles / volume solvent

14.8 x volume = .002 moles

.002 moles / 14.8 = volume = .0001428 liters = .1428 mL

So yes close enough. If you're doing sensitive work you might want to control for temperature and titrate your ammonia. If you're just using it for neutralization or some thing like that it might be ok to add an excess in which case it would probably be far easier to just measure out 1 mL.

unionised - 4-9-2020 at 09:22

If the density is 0.898 then I doubt that the 28% figure refers to (virtually non existent) ammonium hydroxide.

https://airgasspecialtyproducts.com/wp-content/uploads/2016/...

It's 28% NH3 w/w
That's 280 g/kg
251 g/litre

or about 14.8M

You have made two errors that cancel out
Quote: Originally posted by Organicsynth  


... (should I use ammonium hydroxide MM instead?)
...


You should- if the concentration of ammonium hydroxide was 28% but it isn't.

Organicsynth - 4-9-2020 at 09:34

Quote: Originally posted by unionised  
If the density is 0.898 then I doubt that the 28% figure refers to (virtually non existent) ammonium hydroxide.

https://airgasspecialtyproducts.com/wp-content/uploads/2016/...

It's 28% NH3 w/w
That's 280 g/kg
251 g/litre

or about 14.8M

You have made two errors that cancel out
Quote: Originally posted by Organicsynth  


... (should I use ammonium hydroxide MM instead?)
...


You should- if the concentration of ammonium hydroxide was 28% but it isn't.

I think that I didn't understood what you've said.

What were my errors? I've calculated 251,44 g/L, it matches what you said (251 g/litre).
I have a 1 litre bottle of Ammonium Hydroxide with 28% concentration, why you say it isn't?

Ahh this is so confusing.

Organicsynth - 4-9-2020 at 15:24

Ammonium Molar Mass: 17,03 g/mol
Water Molar Mass: 18,01 g/mol
Ammonium Hydroxide Molar Mass: 35,04 g/mol

17,03 + 18,01 = 35,04

I've now understood this, but still not sure if 2mmol of Ammonium Hydroxide 28% solution in 1L of water is equal to 0,135mL.

unionised - 5-9-2020 at 04:50

The solution in the bottle contains 28% NH3 by weight.
It contains practically no ammonium hydroxide which (virtually) does not exist.

Organicsynth - 5-9-2020 at 10:03

I think this means I have 280mL of NH3 (ammonia) diluted in 720mL of H2O (distilled water).

So the synthesis requires 2mmol of NH4OH, I should use the necessary solution amount to contain 2mmol of NH3.

BUT, I don't know how to calculate the volume I need to use from my solution, to get 2mmol of NH3. |:<

Tsjerk - 5-9-2020 at 13:00

Forget the volumes, the 28% is weight/weight, not volume/volume. The 28% means 28% of the mass is NH3.

One liter weighs 898 grams, of that 898 grams 28% is NH3, so 0.28 x 898 = 251 gram. One mol weighs 17.03 grams so every liter contains 14.76 moles. So 28% ammonia is 14.76 M

You want a solution of 2 mM, that is 14.76/0.002 times less, so 7380 times diluted. If you want 1 liter of 2 mM you have to dilute 7380 times.

1000 ml/7380 = 0.135 ml

Edit: and what unionised means is that ammonium hydroxide doesn't really exist in a solution of ammonia. It is better to just use NH3 instead of NH4OH. No calculation uses it and I'm pretty sure your synthesis doesn't either.

You could also calculate how much ammonia you need the other way around. 2 millimoles weigh 0.002 x 17.03 = 0.03406 gram. Divided by 28 times 100 (the ammonia is only 28%,not 100) = 0.121 grams divided by 0.898 (density of ammonia) = 0.135 ml.

[Edited on 5-9-2020 by Tsjerk]

Organicsynth - 5-9-2020 at 15:30

Thanks for the explanation, really nicely done.

Now I understand how to properly do this, thanks!

fusso - 3-10-2020 at 23:41

Quote: Originally posted by unionised  
https://airgasspecialtyproducts.com/wp-content/uploads/2016/...
What does "20/4" (shown as fraction in pdf) means?

B(a)P - 4-10-2020 at 00:15

It is meant to be 20 degrees C. Maybe a typo? The densities definitely match those for 20 degrees C.

beta4 - 4-10-2020 at 00:49

Tsjerk's solution is correct, but for the lazy there are also tables that map weight percent to molarity for various common reagents.

I usually refer to thisk booklet, which used to be found on merck's website but now I could only find through this link: http://pro.unibz.it/staff2/sbenini/documents/laboratory_and_safety/lab_tables.pdf

The table for ammonia is at page 68, with that you map 28% to 14.76M and continue from there. I find such tables useful to speed up computation when planning a reaction.

unionised - 4-10-2020 at 02:20

Quote: Originally posted by fusso  
Quote: Originally posted by unionised  
https://airgasspecialtyproducts.com/wp-content/uploads/2016/...
What does "20/4" (shown as fraction in pdf) means?

It's the density measured at 20C (Room temp) compared to the density of water at 4C (The maximum density)

[Edited on 4-10-20 by unionised]

Tsjerk - 4-10-2020 at 03:22

Quote: Originally posted by beta4  
Tsjerk's solution is correct, but for the lazy there are also tables that map weight percent to molarity for various common reagents.

I usually refer to thisk booklet, which used to be found on merck's website but now I could only find through this link: http://pro.unibz.it/staff2/sbenini/documents/laboratory_and_safety/lab_tables.pdf

The table for ammonia is at page 68, with that you map 28% to 14.76M and continue from there. I find such tables useful to speed up computation when planning a reaction.



If you practice these calculations often enough they become so easy looking up takes more time compared to calculating the value.
Besides, everyone should be able to do this from the top of the head. How stupid do you look if you ever have to quickly calculate something when there is no chart available, and you just can't do it?

beta4 - 4-10-2020 at 03:50

Yes, it's important to also be able to do the computation yourself, but you still have to look up the density somewhere, and if you find the entire table, why not read the answer directly? Maybe I'm just lazy.