The Spectrochemical Series Lab Report Example | Topics and Well Written Essays - 1000 words

The Spectrochemical Series - Lab Report Example Also ammonium chloride (NH4Cl), 16.00 g was dissolved in concentrated aqueous ammonia (NH4OH), 16 m ,there was formation of white suspension as a result of this. The dissolving process was done in separate beakers. The 2 solutions were then mixed in a casserole. These led to the formation of an orange solution. Following this, there was small increment addition of 4 ml of 30% hydrogen peroxide (H2O2) into this solution; this resulted to the formation of black solution with fizzing of oxygen gas. Lastly, the solution changed to maroon. The solution was heated on a steam bath until thick slurry was obtained; it was then put in an ice bath to cool. The rinsing of the solution was carried using 3 M hydrochloric acid (65 ml) in a 200 ml beaker and heated 10 minutes at 60?C. The solution was placed on an ice bath to cool and it was later filtered under a vacuum and the resulting precipitate was washed with acetone and cold water (iced). As a result of this there was formation of a pink pre cipitate. Week 2: There was addition of 2 M aqueous ammonia (NH4OH) (200 ml) in a beaker with the solid then heated to 50-60?C. this resulted to the formation of a cherry red solution. Then later, there was addition of concentrated hydrochloric acid (65 ml) to the solution. This resulted to the formation of white fumes. This solution was then heated for 10 min. ... Then later the precipitate was put at 110 ?C in an oven for 1 hour. Lastly, the solid mass was obtained using an electronic balance. Results and Discussion: Complex Formula of complex Color of solution Absorbance peak (nm) Splitting energy (J) A [Co(NH3)5Cl]2+ Pink 530.76 374519E-19 B [Co(NH3)6]3+ Orange 474.86 4.18228E-19 C [Co(NH3)5F]2+ (Not Given) 500-515 3.9756E-19 - 3.85981E-19 D [Co(NH3)5Br]2+ Purple 548.24 3.62578E-19 E [Co(NH3)5H2O]3+ Light red/orange 497.04 3.9994E-19 Table 1: Shows formulae of the complex, Solution Color, peak absorbance, and the splitting energy Of the 5 cobalt complexes employed in the experiment. Calculations (sample B): ?E = hc/? ? E = (6.626x10-34Js)(3x108m/s)/(474.86x10-9m) ? E = 4.18228E-19J Overall Reaction: 2CoCl2.6H2O +10NH4OH +2NH4Cl +H2O2 +3Cl- Â  ? 2[Co(NH3)5Cl]Cl2 + 24H2O + NH3Â   Re-crystallized solid Mass = solid mass + vial mass) – vial mass =15.76 – 13.25 = 2.51 g Reactions: CoCl2Â ·6H2O ? [Co(H2O)6]2+ + 2 Cl- [Co(H2O)6]2+ (aq) + 2NH3(aq) ? [Co(OH)2(H2O)4](s) + 2NH4+(aq) [Co(OH)2(H2O)4](s) + 5NH3(aq)? [Co(NH3)5]2+(aq) + 4H2O(l) + 2OH?(aq) [Co(NH3)5]2+(aq)+ ? H2O2? [Co(NH3)5]3+ + OH- (aq) [Co(NH3)5]3+ + 3Cl- ?[Co(NH3)5Cl]2+ + 2Cl- ?[Co(NH3)5Cl]Cl2 Mol of CoCl2.6H2O = mass/molar mass = 5.00/238 = 0.021008 mol CoCl2.6H2O : [Co(NH3)5Cl]Cl2 = 2 : 2 = 1 : 1 Mol of [Co(NH3)5Cl]Cl2 = 0.021008 mol Mass of [Co(NH3)5Cl]Cl2 = 0.021008 X 250.5 = 5.26 g % yield = (actual yield / theoretical yield) X 100 = (2.51 / 5.26) X 100 = 47.71 % This experiment was able to determine the identity of each of these five cobalt complexes. Basing on the table information provided, the Cobalt complexes can be determined easily by arranging them according to wavelength from smallest to the largest: D < A < C < E < B This arrangement corresponds