Preparation Of Bis (Acetylacetonato)Copper (Ii) Essay

1374 words - 6 pages

Title : Preparation of bis (acetylacetonato)copper (II)
Objective : I ) To synthesize the bis (acetylacetonato)copper (II) complex
II) To determine the percentage yield of bis (acetylacetonato)copper (II) complex
Background of study :
Metal acetylacetonates, formed by a metal and multiple acetylacetonate anions, are prime examples of coordination complexes. In this experiment, the metal use is copper (II) ion which is crystal blue in colour when it is copper(II)nitrate. All metal ions in solution can react well with water. The water molecules can also be weakly bonded or more strongly as a ligand to form a complex ion, and these can also present in solid ‘hydrated’ salts of ...view middle of the document...

The solution of 1:1 Ammonia was added slowly with stirring until the precipitate first formed has just redissolved.
3. The solution was continuing stirring then 6cm3 of acetyl acetone was added drop by drop into the solution.
4. The crude product was filtered off by using suction filtration.
5. It was washed well with water and it was drained thoroughly.
6. The percentage(%) yield was calculated.

Result :
Mass of copper (II) nitrate | 6.0065g |
Mass of watch glass | 36.1443g |
Mass of filter paper | 0.3288g |
Mass of Bis(Acetylacetonato)copper(II) + Mass of watch glass + Mass of filter paper | 38.4764g |
Mass of Bis(Acetylacetonato) copper (II) | 2.0033g |
Table 1
Chemical equation :
Equation 1 : (Cu(NO3)2∙3H2O) → Cu(H2O)62+ + 2NO3-
Equation 2 :
Equation 3 : [Cu(NH3)4(H2O)6]2+ + C5H8O2 → C10H16CuO4 + 4NH3 + 4OH-

Calculation :
Molar mass of (Cu(NO3)2∙3H2O) = 241g/mol
Molar mass of C10H16CuO4 = 263.78 g/mol
Number of mole of (Cu(NO3)2∙3H2O) = mass of copper IInitratemolar mass of copperIInitrate
= 6.0065g241g/mol
= 0.02492mol
According to the equation above from 1 until 3, 1 mol of (Cu(NO3)2∙3H2O) produce 1 mol of C10H16CuO4.
Thus , number of mol of C10H16CuO4 = 0.02492mol
Theoretical yield of C10H16CuO4 = number of mol of C10H16CuO4 x molar mass of C10H16CuO4\
= 0.02492mol x 263.78g/mol
= 6.5734g
Percentage yield of C10H16CuO4 = actual yield Theoretical yield x 100 %
= 2.0033g6.5734g x 100%
= 31%

Discussion :
In this experiment, the objective is to synthesize the bis (acetylacetonato)copper (II) complex. The objective was determined and the percentage yield of the product is 31%. The product is in dark blue powder form.
In the first step of the experiment, 6g of copper(II) nitrate (Cu(NO3)2∙3H2O) was dissolved in 60cm3 of water. The purpose of this action is dissolve the blue crystal to get the blue solution which contain of copper(II) hexaaqua complexes which is Cu(H2O)62+.The equation is shown below :
(Cu(NO3)2∙3H2O) → Cu(H2O)62+ + 2NO3-
Next, the solution of 1:1 Ammonia was added slowly with stirring until the precipitate first formed has just redissolved. The purpose of this step is to replace the water molecule by ammonia. Ammonia solution can react with hexaaqua metal ions in two quite distinct ways because it can act as a base as well as a ligand. When we added a small amount of ammonia solution, we get milky solution which is precipates of the copper(II) hydroxide – the ammonia is acting as a base. When we added more ammonia to the solution, the precipates redissolved to give dark blue solutions in which a ligand exchange reaction has occurred.
There is a slightly untypical case beause only four of the six water molecules get...

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