# Exercise 2 | The periodic table and Chemical Periodicity

Calculate the mass of 3.12 x 1022 formula units of aluminum chloride.

nAlCl3 = $\frac{{\mathrm{m}}_{\mathrm{AlCl}3}}{{\mathrm{M}}_{\mathrm{AlCl}3}}$ = $\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}$

⇒ mAlCl3 =  x MAlCl3 = 6.91 g