Let's determine the number x of water molecules in $CuSO_4\cdot x H_2O$ Reminder: Mass of the empty vessel = 78.584 g Mass of the vessel + hydrated copper sulfate = 94.815 g Masse of the vessel + anhydrous copper sulfate = 89.476 g
Mass of anhydrous copper sulfate = 89.476-78.584 = 10.892 g Number of moles of anhydrous copper sulfate = $\frac{m}{M}$ = $\frac{10.892}{63.54+32+4\cdot 16}$ = 6.782·10-2 mol Mass of hydration water = mass of hydrated copper sulfate - mass of anhydrous copper sulfate = 94.815-89.476= 5.339 g Number of moles of hydration water = $\frac{m}{M}$ = $\frac{5,339}{16+2\cdot 1}$ = 2.97· 10-1 mol Number of moles of hydration water/Number of moles of anhydrous copper sulfate $\approx 4.4$ So x is between $4$ and $5$. The real value is $x=5$