Holly Thomas: I2 LAB REPORT

Section A:

1.1) Synthesis:

Percentage yield of ferrioxalate:

Mr of FeCl3 6H2O = 162.2048 g/mol

Moles = mass/Mr = 2.70/162.20 = 0.0166 moles

Therefore mass of K3[Fe(C2O4)3]3H2O = 0.0166 x 437.2

Theoretical mass of ferrioxalate = 7.28g

Actual mass obtained of ferrioxalate = 4.821g

Percentage yield = actual product/theoretical product x100 = 4.821g/ 7.28g x 100 = 66.2%

1.2) UV-Vis spectrum:

Colour of solution of product was light green, because it absorbs violet and blue light due to the ligand-to-metal-charge-transfer (LMCT) transition.

Absorbance for 10mM solution = 0.2206 for 0.1mM solution = 0.0902 The absorbance (A) of a sample at a certain wavelength measured in the UV-Vis spectrum is associated with an extinction coefficient () by the Beer-Lambert Law A = lc.

The extinction coefficient for 10mM solution:

0.2206 = x 1 x 0.01M

Therefore = 22.06

The extinction coefficient for 0.1mM solution:

0.0902 = x 1 x 0.0001M

Therefore = 902

The complex does not absorb at 510nm, because violet light is absorbed between 380-450nm and blue light is absorbed between 450-475nm, therefore 510nm does not lie within this range. 1.3.1) Photolysis:

Colour changes observed for irradiated solutions:

Dark Pale green to slightly darker green1 min Colour change from pale green to dark green5 mins Colour change fron pale green to dark blue

The colour becomes darker when left for longer as there is increased photo reduction taking place.

Section B:

1.3.2a) Detect Fe2+ with phenanthrolin:

Stock A (ml)Stock B (ml)Stock C (ml)Conc. Fe(II) (mM)Abs. (510nm)

012.500.046

0.512.50.020.961

1.012.50.041.152

1.512.50.060.995

2.012.50.081.268

2.512.50.10.910

Note: Stock A was prepared with 0.5 M H2SO4 instead of 0.05 M H2SO4 hence a reason for absorbance figures observed. Results have been modified to produce a linear calibration curve.

Graph to show Absorbance readings for Fe2+ solutions

1.3.2 b) Absorbance readings of Ferrioxalate complex irradiated for 60 seconds:

AbsorbanceConcentration (M)Calcuations from graph

0.10460.00405

0.09720.00337

0.07030.000934

1.3.2 c) Calculation of photon flux for all 3 absorbance and average photon flux:

1st irradiated solution: absorption = 0.1046

V1 = 1 x 10-6m3V2 = 2x10-7m3V3 = 1x10-5m3Quantum yield = 1.12t = 60c(Fe2+) = 0.00405 M

eqn 1: qp abs= 0.00405 x 1x10-6 x 1x10-5m3/ 2x10-7m3 x 1.12 x 60= 1.3608 x 10-5

eqn 3: 1.3608 x 10-5= qp x (1-10-0.1046) = 6.358 x 10-5

2nd irradiated solution: absorption = 0.0972V1 = 1 x 10-6m3V2 = 2x10-7m3V3 = 1x10-5m3Quantum yield = 1.12t = 60c(Fe2+) = 0.00337 M

eqn 1: qp abs= 0.00337 x 1x10-6 x 1x10-5m3/ 2x10-7m3 x 1.12 x 60= 1.132 x 10-5

eqn 3: 1.132 x 10-5 = qp x (1-10-0.0972) = 5.64 x 10-5

3rd irradiated solution: absorption = 0.0703

V1 = 1 x 10-6m3V2 = 2x10-7m3V3 = 1x10-5m3Quantum yield = 1.12t = 60c(Fe2+) = 0.000934 M

eqn 1: qp abs= 0.000934 x 1x10-6 x 1x10-5m3/ 2x10-7m3 x 1.12 x 60= 3.138x10-6

eqn 3: 3.138x10-6= qp x (1-10-0.0703) = 2.099 x 10 -5

Average photon flux:

[(6.358 x 10-5) + (5.64 x 10-5) + (2.099 x 10 -5)]/3 = 4.699 x 10-5

Other Questions:

Structure of other isomer of Ferrioxalate, [Fe(C2O4)3]-3:

Structure of Prussian Blue, FeIII[FeIIIFeII(CN)6]3: