Experiment no 1                                                26/08/2013

Preparation of Tetra amine Copper(II) Sulphate and [Cu(NH₃)₄]SO₄ also calculate the percentage yield_.

Apparatus:

                   Following apparatus was used in the preparation of the complex salts.

·        Test tubes

·        Test tube holder

·        Test tube stand

·        Copper Sulphate

·        Ammonia Solution

·        Ethyl Alcohol.

Reaction:

CuSO4.5H2O + 4NH3 → [Cu(NH₃)₄(H₂O)₅]SO₄

Procedure:

                  First of all 3 gm of CuSO4.5H2O was dissolved in distilled water in the 250 ml beaker to make a concentrated solution. Than ammonia solution was added in it drop vise with a constant stirring. Blue precipitates were formed and were dissolved in excess of ammonia. After that 25 ml of ethyl alcohol was added in this solution and that solution was allowed to stand for 10-15 minutes. Blue crystals of complex Tetra amine Copper (II) sulphate were formed and these were separated by filtration. These crystals were allowed to dry and percentage yield was calculated and other reactions were performed.

Observations and Calculations:

Theoretical yield:

249.5 gm of CuSO.5H2O produces Tetra  aminecopper (II) Sulphate=317.5gm

1 gm of CuSO.5H2O produces  Tetra amine  copper (II) Sulphate=317.5gm/249.5gm

3gm of CuSO.5H2O produces  Tetra amine  copper (II) Sulphate=317.5gm/249.5*3gm

= 3.8

Actual Yield:

Weight of Petri dish = 43.51gm

Weight of petridish + complex = 45.31 gm

Wight of complex = 45.31-43.51

= 1.80

Percentage yield:

 percentage yield = Actual yield / Theoretical yield * 100

% yield = 1.8/ 3.8 *100

% yield = 47%

water molecules in it:

Weight of crucible = M1= 16.18 gm

Wt.of Compound+ crucible = M2 = 16.81 gm

Wt. of hydrated complex = M3 = M2-M1= 16.81-16.18 = 0.63gm

Constant wt.of crucible + salt = M4 = 16.73 gn

Mass of removed water = M5 = M2-M4 = 16.81-16.73 =  0.08 gm

Mass of anhydrous salt = M4 – M1 = 16.73 – 16.18 =  0.55gm

0.55 gm of [Cu(NH3)4]SO4 have water = 0.08 gm

1 gm of [Cu(NH3)4]SO4 have water = 0.08/ 0.55 gm

2276.76 gm of [Cu(NH3)4]SO4 have water = 0.08/0.55*227.76=33.12gm

Moles of water = 33.12/18 =  1.84 m

 Test for Sulphate ion: 

Solution of [Cu(NH3)4] SO4.5H2O + pb(CH3COO)2 → 2CH3COO + white PPT of pbSO4

Test for CU:

Solution of [Cu(NH3)4] SO4.5H2O + BaCl2 → White PPT   

Results:

The percentage yield for that experiment was 47%.

Sulphate ions when react with the lead acetate they form white ppt..

Cu reacted with the BaCl2 and form the White ppt.