Step 1. Procedure Obtain a weighted amount of the Copper (II) Sulfate hydrate (s

Question

Step 1. Procedure Obtain a weighted amount of the Copper (II) Sulfate hydrate (say 30 g), place in an 100 ml beaker Step 2. Heat compound with bunsen burner until all of the water is driven-off (you can verify this be weighing the beaker while heating it, when the weight stops declining you’ve removed all the water. Step 3. Weigh final compound.

write your observations

Observations:

Initial Weight: 30 grams

Final Weight: 19.1773 grams

Weight of Water driven off (Initial weight - final weight) : 30g - 19.1773g = 10.8227grams (water)

moles of water loss: Weight of water / GMW H2O (18.016 )
moles of CuSO4: final weight/ GMW CuSO4 (159.61)

X = moles of water loss/ moles of CuSO4

Given that hydrate is: CuSO4 (XH2O)

Formula of hydrate:

Explanation / Answer

Copper (II) Sulfate is penta-hydrated with chemical formulae CuSO4.5H2O

When CuSO4.5H2O is heated it loses all the five H2O molecule to form anhydrated Copper (II) Sulfate, CuSO4

CuSO4.5H2O -- > CuSO4 + 5H2O

1 mol 1 mol 5 mol

Let's consider 30 g of CuSO4.5H2O.

Molecular mass of CuSO4.5H2O = 249.68 g/mol

Moles of CuSO4.5H2O = mass / mlecular mass = 30 g / (249.68 g/mol) = 0.120 mol

From the above balanced reaction, 1 mol of CuSO4.5H2O forms 1 mole of CuSO4.

Hence 0.120 mol of CuSO4.5H2O will also form 0.120 mol of CuSO4

Molecular mass of CuSO4 = 159.6 g/mol

Hence mass of 0.120 mole of CuSO4 = 0.120 mol x (159.6 g/mol) = 19.15 g

Hence the weight of the final compound after complete removal of water will be around 19.15 g

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