A morphism of rings $R\to S$ is called essentially smooth if it is formally smooth and essentially finitely presented, where essentially finitely presented means that $S$ is the localization of some finitely presented $R$-algebra $T$ at some multiplicative system $A \subset T$, that is, $S=A^{-1}T$. 

Can we factor an essentially smooth map into a smooth map and a localization?  
Certainly the converse is true, i.e. the localization of any smooth $R$-algebra is essentially smooth, since localizations are formally étale (except of course when the multiplicative system contains zero.)