Question

PARTI - Synthesis and purification of [Cu(NH).JSO..H,0 Section: Compound CuSO..5H20 [Cu(NH3).JSO..H20 52.15 Mass (R) 5.07 2.80 PART II - Spectrophotometric analysis of the coordination complex Compound Complex mass (8) Volume of distilled water (ml.) Quantity . NO Absorbance Wavelength (nm) 575 Absorbance 202 Wavelength (nm) 400 425 450 475 500 600 625 650 . 194 .045 675 525 700 133 . 109 550 . 159 PART III - Titrimetric analysis of the coordination complex Concentration of HC/ Solution, mol/L................ Complex Mass (g) Consumed HCl volume (mL) Determination Number Determination 1 Determination 2 4mL 027 ImL

1.

a. determine limiting reagent

b. find the % yield

2.

a. what is the absorption macimum for your complex? how do these relate to the biaibke color of your complex?

b. calculate the molar extinction coefficient

3.

a. mass of the complex for both determinations using the stoichiometric ratio

b. percentage purity of your complex

c. based on the %purity what is the correct range of value of e which you calculated earlier?

Answer 1

GuSO4, 5 H2O + H NHS →[Cu(NH3)4]504. Hao (5.07 g) molao mass of CuSO4.5H2O = 149.5 g/mol I mole of CuSO4'S HOO = 249.9 g/mol : 5.07 g GuSO4.5 H2O = 15.07x11 mol of Cu 504 • ( 249.5 - SHO = 0.0203mol AS per ed". I mole of GuSO4. SHOO Beacts coith 4 mo? of NH3 : 0.0 203 mol of CuSO4.5H20 deacts with (4*0.0203) 0.08199 mole of NH3 Here 4 mole of NH3 is used. Hence Nha is a excess and CuSO4. H20 is a limiting seagent. I mole of CuSO4.5 to produce I mole of product: :: 0.0 203 mol cu 504 .5H2O produce 0.0203 mol of product • 1 mole of [Cu(NH3)44 ] 504.10 = 317.80 g : 0.0203 mole of product = ( 397.80 x 0.0203 ) g Theoretical yield= 6:4519 poachcal Yield = 2.80 g : 1. yield = 2.80g = 43.40 10 6.451 g