Do I need to include lattice energy when calculating the energy required to vaporize NaCl?
Lattice energy


  Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol.


First, we have to understand the term lattice energy.
Here is the textbook explanation (Fleming: Physical Chemistry): The lattice energy is the energy required to separate the ions in an ionic lattice so that they are at infinite distance (but still ions). This would be difficult to do experimentally, but the value may be determined by using a Born Haber cycle. As the diagram below shows, we know the enthalpy of reaction for:

$$\ce{Na(s) + 1/2 Cl2(g) -&gt; NaCl(s)}$$

We can get the same energy in a thought experiment (path 2 in the diagram), first turning all species into gas, then into atoms, then transferring the electron and - finally - forming the lattice from the individual anions and cations. All these processes are experimentally accessible except for the last, so you can determine the lattice energy this way.


Source: diagram adapted from the cited textbook.

Enthalpy of vaporization


  But do we need to add this energy again when we are trying to vaporize an ionic compound?


You problem is not well-defined. What are the initial and final temperatures, what is the pressure? What does it mean to vaporize solid NaCl or any ionic compound, i.e. what is the product? If we boil sodium chloride at 1 atm pressure, it produces mostly NaCl monomers in the gas phase, see https://chemistry.stackexchange.com/a/14560. If you are interested in this process, you could write it as:

$$\ce{NaCl(s) -&gt; NaCl(g)}$$

So to get the required energy, you could first separate all ions (lattice energy) and then calculate the energy of forming individual pairs of sodium and chloride ions (assuming the interaction is purely ionic).

$$\ce{NaCl(s) -&gt; Na+(g) + Cl-(g)}\tag{1}$$

$$\ce{Na+(g) + Cl-(g) -&gt; NaCl(g)}\tag{2}$$

It should be obvious that process (1) cost much more energy than what you get out of process (2) because you are going from 6 nearest neighbors to just one. If you look up the atomic distance of $NaCl(g)$, you could estimate the enthalpy of (2) pretty accurately and get a numeric answer.


  [Comment from https://chemistry.stackexchange.com/a/14177]: Roughly, somewhere around half of the lattice enthalpy of a salt comes solely from the binding of the smallest electrically neutral agglomerate; that is, it takes about as much energy to break a macroscopic solid NaCl crystal into a gas of ion pairs as it does to break all the ion pairs and create a true plasma. Thus, it is rather unlikely that, at reasonable temperatures, an ionic gas will break down any further than the smallest possible electrically neutral aggregates. – Nicolau Saker Neto Jul 8 '14 at 16:39 


According to the comment above, the heat of vaporization should be about half of the lattice energy (with opposite sign).


  I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.


Because enthalpy is a state function, it does not matter if you choose the actual (realistic) path or take a different path. You are right that if you step through what actually happens, you need to know heats of fusion and vaporization as well as heat capacities of all phases for the relevant temperature ranges.