Water of Hydration

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Waterof Hydration

EXPERIMENT8

PrelabQuestions

  1. The final product will weigh less than the starting material because of the loss of water of hydration.

  2. Water of hydration has been lost from the hydrated compound to make it anhydrous.

  3. The process is endothermic because it takes in heat.

  4. Anhydrous salt does not contain any water.

  5. The final product is anhydrous.

  6. Percentage of water

BaCl2.2H2O

Totalmolecular mass = 137+(35.5*2) +2( (2*1)+(16))

= 137 + 71 + 36 = 244

*100 = 14.75%

ZnSO4.7H2O

Totalmolecular mass = 65+ 32 + (16*4) +7( (2*1)+(16))

= 65 + 32+ 64 + 126 = 287

*100 = 43.90%

MgSO4.7H2O

Totalmolecular mass = 24+ 32 + (16*4) +7((2*1)+(16))

= 24 + 32+ 64 + 126 = 246

*100 = 51.22%

  1. FeCl3.XH20

FeCl3=56 + (35.5*3) = 162.5

nFeCl3= 0.0185

nH20= 0.1111

Molesof FeCl3= 1

Molesof H20= 6

Empiricalformula FeCl3.6H20

  1. Na2CO3.XH20

Na2CO3=46 + 12 + (16*3) = 106

H20= (2*1) + 16 = 18

nNa2CO3= 0.0714

nH20= 0.0711

Molesof Na2CO3= 1

Molesof H20= 1

Empiricalformula Na2CO3.H20

PostlabQuestions

  1. Percentage of water

CaCl2.2H2O

Totalmolecular mass = 40+ (35.5*2) + 2( (2*1)+(16))

= 40 + 71 + 36 = 147

*100 = 24.49%

  1. CuSO4.XH20

CuSO4=63.5 + 32 + (16*4) = 159.5

H20= (2*1) + 16 = 18

nCuSO4= 0.01

nH20= 0.05

Molesof CuSO4= 1

Molesof H20= 5

Empiricalformula CuSO4.5H2

  1. MgSO4.XH20

MgSO4=24 + 32 + (16*4) = 120

H20= (2*1) + 16 = 18

nMgSO4= 0.4067

nH20= 2.8444

Molesof MgSO4= 1

Molesof H20= 7

Empiricalformula MgSO4.7H20

  1. MgSO4.XH20

MgSO4=24 + 32 + (16*4) = 120

H20= (2*1) + 16 = 18

nMgSO4= 0.055

nH20= 0.3844

Molesof MgSO4= 1

Molesof H20= 7

Empiricalformula MgSO4.7H20

  1. NiCl2.XH20

NiCl2=58 + (35.5*2) = 129

H20= (2*1) + 16 = 18

nNiCl2= 0.0211

nH20= 0.1267

Molesof NiCl2= 1

Molesof H20= 6

Empiricalformula NiCl2.6H20

  1. K2CO3.XH20

K2CO3=(39*2) + 12 + (16*3) = 138

H20= (2*1) + 16 = 18

nK2CO3= 0.01

nH20= 0.02

Molesof K2CO3= 1

Molesof H20= 2

Empiricalformula K2CO3.2H20

  1. The percentage of water recorded in the hydrated copper II sulphate could be less than 36.1% because of two reasons. One is that the water vapor which had escaped from the salt and condensed would have been absorbed back into the salt recording less water lost. The other reason could be that not all water was lost due to inadequate heating.

  2. Overheating causes a higher percentage than 36.1%, this is because anhydrous copper II sulphate is further broken down to copper II sulphide which weighs less than copper II sulphate and thus the loss in weight is assumed to be water of hydration.