Chemistry: A Molecular Approach (3rd Edition)

Its formula is $IrBr_{3}.4H_{2}O$
The cation of iridium(III) bromide tetrahydrate has a charge of (III), i.e., it is $Ir^{3+}$. The anion is bromide, i.e $Br^{-}$. To balance the charges, 3 $Br^{-}$ combines with 1 $Ir^{3+}$. Then it also has 4 (tetra) water molecules of hydration. So, its formula is $IrBr_{3}.4H_{2}O$