# Why is the empirical formula of sodium chloride NaCl? [duplicate]

Since we know in sodium chloride crystal sodium ions are surrounded by $6$ chloride ions and each chloride ions are again bonded to $6$ sodium ions. What I want to know is that on what basis empirical formula of sodium chloride was assigned to be NaCl? does that formula mean sodium ion should atleast bonded to one chloride ion?

## marked as duplicate by Jon Custer, paracetamol, Todd Minehardt, Mithoron, Pritt BalagopalJul 5 '17 at 15:22

The empirical formula of, say, $\ce{P4O10}$ is $\ce{P2O5}$ but that doesn't mean that each phosphorus is bonded to only 2.5 oxygens. It's perfectly possible (and in fact it is pretty much a requirement) that if Na has 6 nearest-neighbour Cl and Cl has 6 nearest-neighbour Na, then the overall stoichiometry is NaCl.
Let's say, for example, that there are $n$ sodium atoms in a sample of NaCl. Then, if each $\ce{Na+}$ atom has 6 nearest-neighbour $\ce{Cl-}$ ions, then at first glance we might say there are $6n$ chloride ions.
However, since each $\ce{Cl-}$ ion has 6 nearest-neighbour $\ce{Na+}$ ions, this means that in the earlier calculation we counted each $\ce{Cl-}$ ion 6 times (once per $\ce{Na+}$ neighbour). So in fact there are only $6n/6 = n$ chloride ions, leading to a stoichiometry of $1:1$.
Surely you'd agree that each red tree has two green neighbours, and each green tree has two red neighbours. Yet the "stoichiometry" is obviously $1:1$.
Now if you tile that in three dimensions, you get the $\ce{NaCl}$ structure.