Question

PART I - Synthesis and purification of [Cu(NH3)4]SO4.H2O

Compound	Mass (g)
CuSO4.H2O	5.090 g
[Cu(NH3)4]SO4.H2O	1.880 g

1. limiting reagent determination (use the stoichiometric ratio to determine the limiting reactant)

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

2. percentage yield of crude product (remember to include the 0.025 g of compound you take for determining the absorption spectrum).

PART II - Spectrophotometric analysis of the coordination complex

Compound	Quantity
Complex mass (g)	0.026 g
Volume of distilled water (mL)	20 mL

Wavelength (nm)	Absorbance	Wavelength (nm)	Absorbance
400	0.740	575	0.825
425	0.845	600	0.865
450	0.930	625	0.875
475	0.942	650	0.740
500	0.920	675	0.750
525	0.784	700	0.685
550	0.875

3. What is/are the absorption maximum/maxima (λmax) for your complex? How do these relate to the visible color of your complex? Use the color wheel in explaining the visible color.

4. Using your absorbance data, calculate Ɛ, the molar extinction coefficient for your complex (A = Ɛlc)

Answer 1

Part I:

1. In the reaction, as CuSO4.5H2O:NH3 = 1:4, so limiting reagent is CuSO4.5H2O.

2. Mass of CuSO4.5H2O = 5.090 g

Molar mass of CuSO4.5H2O = 249.5 g/mol

moles of CuSO4.5H2O = mass/molar mass = 5.090/249.5 = 0.020 mol

Mass of [Cu(NH3)4]SO4.H2O = 1.880 g + 0.025 g = 1.905 g

Molar mass of [Cu(NH3)4]SO4.H2O = 245.5 g/mol

moles of [Cu(NH3)4]SO4.H2O = mass/molar mass = 1.905/245.5 = 0.008 mol

% yield of crude product = (0.008/0.020)*100 = 40 %

Part II:

3. We can see in data table, the maximum value of absorbance is 0.942 and the wavelength corresponding to that value is 475 nm. So, λmax for the complex is 475 nm.

It lies within the visible range. The visible range is between 390 to 700 nm.

The complex absorbs blue color and transmits orange color. So, the complex will be orange in color.

4. Absorbance = A = 0.942

Molar extinction coefficient = Ɛ

length = l = 1 cm

concentration = c = moles of complex/volume

moles of complex = mass/molar mass = 0.026/245.5 = 0.0001 mol

volume = 20 ml = 0.020 L

concentration = c = moles of complex/volume = 0.0001/0.020 = 0.005 M

Molar extinction coefficient = Ɛ = A/lc = 0.942/(1*0.005) = 188.4 M^-1 cm^-1