# Why does ammonium carbonate smell?

Why is it that $\ce{(NH4)2CO3}$ smells while other salts with ammonium like $\ce{NH4Cl}$ does not smell? Does it have something to do with the equilibrium reaction? I know it has nothing to do with the coefficient of $\ce{NH4}$ in ammonium carbonate. I am pretty sure the water in our nose causes it to dissolve. I do note that $\ce{CO3^2-}$ is basic along with ammonium is acidic while the chloride ion is neutral. So does it have to do with the fact that $\ce{(NH4)2CO3}$ is more neutral than $\ce{NH4Cl}$?

• Yes it does, though people usually put it in different words, like "Ammonium carbonate, being a salt of a weak base and a weak acid, is prone to..." – Ivan Neretin Sep 30 '15 at 22:53
• @IvanNeretin can you elaborate, i dont quite understand what you are trying to say – TanMath Sep 30 '15 at 22:56
• See, what happens when you heat ammonium salt (any of them)? – Ivan Neretin Sep 30 '15 at 23:06
• @IvanNeretin i see, but why does (nh4)2co3 smell stronger than nh4cl? – TanMath Sep 30 '15 at 23:12
• Basically, because $\ce{H2CO3}$ is a weaker acid than $\ce{HCl}$. – Ivan Neretin Sep 30 '15 at 23:13

1. As mentioned by Ivan Neretin in the comments, carbonate is the conjugate base of a much weaker acid than chloride is the conjugate base of. In other words, carbonic acid is less acidic than hydrochloric acid, and carbonate is more basic than chloride. This means that carbonate solutions will have higher pHs than chloride solutions, in general. At higher pH, the equilibrium between $\ce{NH4^{+}}$ and $\ce{NH3}$ is shifted more towards $\ce{NH3}$, which is volatile and smelly.
2. Carbonate is also potentially volatile. If you leave ammonium carbonate, even in solution, around forever in an open container, it will eventually completely disappear. $$\ce{(NH4)2CO3(s) -> H2O(g) + 2 NH3(g) + 2CO2(g) }$$This also favors more vaporization of ammonia. In contrast, in ammonium chloride, chloride is essentially completely non-volatile. Thus ammonium could not evaporate without a buildup of negative charge in the solution, highly unfavorable.
• Actually, I am curious as to whether there is any appreciable formation of urea via $\ce{(NH4)2CO3 -> CO(NH2)2 + 2H2O}$. Ordinarily I would try to search myself but I'm on mobile now, might get round to it later. – orthocresol Oct 1 '15 at 8:13