Question

A hydrated form of copper sulfate (CuSO4·x H2O) is heated to drive off all the water. If we start with 8.79 g of hydrated salt and have 6.57 g of anhydrous CuSO4 after heating, find the number of water molecules associated with each CuSO4 formula unit.

Answer 1

Answer:

3 water molecules

Explanation:

The molar ratio between CuSO₄ and H₂O needs to be calculated.

The mass of water that was removed by heating is calculated, then converted to moles. The molecular weight of water is 18.02g/mol.

(8.97 g - 6.57 g) = 2.22 g H₂O

(2.22 g)/(18.02g/mol) = 0.1232 mol H₂O

The moles of anhydrous copper sulfate are calculated. The molecular weight is 159.609 g/mol

(6.57 g)/(159.609 g/mol) = 0.04116 mol CuSO₄

Now the molar ratio between CuSO₄ and H₂O can be calculated:

0.1232 mol H₂O ÷ 0.04116 mol CuSO₄ = 3 H₂O / CuSO₄

Answer 2

The molar ratios determine how many moles of product are created from a given amount of reactant, as well as how many moles of one reactant are required to thoroughly react with another reactant.

• It is necessary to compute the molar ratio of $CuSO_{4}$ and $H_{2} O$.
• The mass of water removed by heating is determined, and the moles are transformed. Water has a molecular weight of 18.02g/mol.

$(8.97 g - 6.57 g) = 2.22 g H_{2} O(2.22 g)/(18.02g/mol) = 0.1232 mol H_{2} O$

• Calculate the moles of anhydrous copper sulfate. 159.609 g/mol is the molecular weight.

$(6.57 g)/(159.609 g/mol) = 0.04116 mol CuSO_{4}$

• The molar ratio of $CuSO_{4}$ to $H_{2} O$ can now be calculated:

$0.1232 mol H_{2} O / 0.04116 mol CuSO_{4} = 3 H_{2} O / CuSO_{4}$

Thus, the molar ratio is 3:1 meaning 3 water molecules are associated.

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