The water molecules are not completely removed so additional heating is required.
Explanation:
We have the copper (II) sulfate pentehydrate with the chemical formula CuSO₄ · 5H₂O.
molar mass of CuSO₄ · 5H₂O = 159.6 + 5 × 18 = 249.6 g/mole
Knowing this, we devise the following reasoning:
if in 249.6 g of CuSO₄ · 5H₂O there are 90 g of H₂O
then in 8 g of CuSO₄ · 5H₂O there are Y g of H₂O
Y = (8 × 90) / 249.6 = 2.88 g of water
mass of dried CuSO₄ = mass of CuSO₄ · 5H₂O - mass of H₂O
mass of dried CuSO₄ = 8 - 2.88 = 5.12 g
5.12 g is less that the weighted mass of 6.50 g. We deduce from this that the sample needs additional heating in order to remove all the water (H₂O) molecules.
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