# What is the length of a unit cell of CuCl assuming that it is fcc?

The density of $\ce{CuCl}$ is given – $x\ \mathrm{g/cm^3}$
The crystal structure is assumed to be fcc.

My teacher is claiming the we can apply the formula

$$\rho=\frac{Z \cdot M}{a^3 \cdot N_\mathrm A}$$ And he took $M = 35.5\ \mathrm{g\ mol^{-1}}+63.5\ \mathrm{g\ mol^{-1}} = 99\ \mathrm{g\ mol^{-1}}$, $Z = 4$ as it is fcc. But my question is how can we take $M = 99\ \mathrm{g\ mol^{-1}}$? It means every particle in the fcc has mass $99\ \mathrm u$ doesn't it?