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Copper (II) Sulfate pentahydrate, CuSO4 X 5H20 is heated in an open crucible to remove the water. (a) Diagram the lab set up clearly labelling the key chemicals and materials needed. (b) The initial mass of the sample is 8.00 g and after heating the new mass is 6.50 g. Based on this data is the compound completely dried or is additional heating required? Provide evidence including calculations and written reasoning for credit.

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cristisp93

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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