Some copper 2 sulphate pentahydrate was heated at 120°c with the following result:
Weight of crucible=10.00g
Weight of crucible+CuSo4•5H2O=14.98
Weight of crucible+Residue=13.54g
How many molecule of water of crystalization were lost((H=1,Cu=63.5,O=15,S=32)
First, calculate the weight of CuSO4•5H2O in the original sample:
Weight of CuSO4•5H2O = Weight of crucible+CuSO4•5H2O - Weight of crucible
Weight of CuSO4•5H2O = 14.98g - 10.00g = 4.98g
Next, calculate the weight of residue left after heating:
Weight of residue = Weight of crucible+Residue - Weight of crucible
Weight of residue = 13.54g - 10.00g = 3.54g
The weight difference between the original CuSO4•5H2O and the residue is due to the loss of water molecules during heating:
Weight of lost water = Weight of CuSO4•5H2O - Weight of residue
Weight of lost water = 4.98g - 3.54g = 1.44g
Now, calculate the number of moles of lost water molecules:
Molar mass of H2O = 2(1g/mol) + 15.999g/mol = 18.015g/mol
Number of moles of lost water = Weight of lost water / Molar mass of H2O
Number of moles of lost water = 1.44g / 18.015g/mol = 0.0799 mol
Since the formula of CuSO4•5H2O contains 5 water molecules per one formula unit, the number of lost water molecules is:
Number of lost water molecules = 0.0799 mol * 5 = 0.3995 mol
Convert the number of moles to molecules by multiplying by Avogadro's number (6.022 x 10^23 molecules/mol):
Number of lost water molecules = 0.3995 mol * 6.022 x 10^23 molecules/mol = 2.40 x 10^23 water molecules
Therefore, approximately 2.40 x 10^23 molecules of water of crystallization were lost during the heating process.