Lec_09 - Today 1. Post-lab for Experiment 7 is due (10 pts)...

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Unformatted text preview: Today 1. Post-lab for Experiment 7 is due (10 pts) Next week 1. Report for Experiment 7 is due (40 pts) 2. Unknown Summary Sheet for Experiments 6-7 is due (20 pts) 3. Pre-lab for Experiment 8 is due (5 pts) 4. Experiment 8. Thermochemistry. have a write-up in your notebook Spring 2007 Experiment 7. Synthesis and Analysis of a Complex Iron Compound. Part 3 (cont.) PURPOSE AND LEARNING OBJECTIVES To synthesize and isolate potassium oxalatoferrate coordination compound To use redox titration to determine the oxalate content of a prepared sample To determine the iron content of a prepared sample using visible spectrophotometry To determine molecular stoichiometry of a synthesized sample using redox titration data in combination with the spectrophotometric data Kx[Fey(C2O4)x] zH2O Potassium content x? y? h Iron content Oxalate content h Water content z? Spring 2007 Experimental Procedure Part 1. Sample preparation Part 2. Last week Part 3. Last week Part 4. Sample preparation (cont.); determination of the iron molar content in the sample Determination of the potassium oxalatoferrate complex formula, Kx[Fey(C2O4)x]zH2O Determination of the percent yield of the product x, y, z Last week: determined molar absorptivity of Fe(phen)32+ Spring 2007 Sample preparation Objective: Kx[Fey(C2O4)x]zH2O Fe(phen)32+ Dissolve sample in water: Kx[Fey(C2O4)x]zH2O (s) x K+ (aq) + [Fey(C2O4)x]x (aq) + z H2O (l) Free up the iron from the [Fey(C2O4)x]x anion by reacting it with KMnO4: a [Fey(C2O4)x]x + b MnO4 + 2c H+ b Mn2+ + d Fe3+ + c H2O + e CO2 (g) Compare to your redox titration reaction (Lab 6): 5 C2O42 + 2 MnO4 + 16 H+ 2 Mn2+ + 8 H2O + 10 CO2 (g) [Fey(C2O4)x]x Spring 2007 Sample preparation (cont.) Objective: Kx[Fey(C2O4)x]zH2O Fe(phen)32+ Add 10% hydroxylamine, NH2OH, to reduce Fe3+ to Fe2+: 4 Fe3+ + 2 NH2OH 4 Fe2+ + N2O (g) + 4H+ + H2O Add excess of 0.3% solution of 1,10-phenanthroline: Fe2+ + 3 phen Fe(phen)32+ Add 1 M sodium acetate, NaCH3COO, to buffer the solution Let the solution sit for at least 30 minutes to insure the reaction goes to completion Spring 2007 Sample preparation. Dilutions Green solution of Kx[Fey(C2O4)x]zH2O Dilutions: Mass (g) 25 ml solution (M ?) 5 ml 25 ml solution Light purple solution of Fe3+ (aq) Red solution of Fe(phen)32+ 5 ml 25 ml solution last dilution (will vary) 5 5 X Dilute red solution of Fe(phen)32+ Total Dilution Factor = 5 5 X Final absorbance measurement has to be within the range of 0.1 1.0 Spring 2007 Sample Calculation Beer's Law: A = l c Asample = 0.3500 Dilution factor = 5 5 8 = 200 = 5,200 M1cm1 l = 1.00 cm 0.3500 A c= = = 6.731 10 -5 M l 5,200 M -1cm -1 1 cm # moles Fe3+ = # moles Fe(phen)32+ Moriginal (Fe3+) = 6.731105 M 200 = 0.0135 M In 25 ml of solution: msample = 0.1565 g [Fe3+] = ? (mole/g) Moles Fe3+ = 0.0135 M 0.0250 L = 3.36 104 mole 3.36 10- 4 mole = 0.00215 mole/g Moles Fe /g of sample = 0.1565 g 3+ Spring 2007 Determination of Kx[Fey(C2O4)x] zH2O moles Fe3+ = 0.002150 mole/g moles C2O42 = 0.004632 mole/g moles K+ = 0.004632 mole/g msample = 1.0000 g y=1 moles C 2 O 4 2 - 0.004632 = = 2.154 2 3+ 0.002150 moles Fe x = 2, y = 1 C2O42 : Fe3+ = x : y = 2 : 1 K2[Fe(C2O4)2] zH2O Spring 2007 Determination of Kx[Fey(C2O4)x] zH2O (cont.) moles Fe3+ = 0.002150 mole/g moles C2O42 = 0.004632 mole/g moles K+ = 0.004632 mole/g msample = 1.000 g y = 1, x = 2 g Fe3+ /g sample = 0.002150 mole Fe3+ 55.847 g/mole = 0.1201 g Fe3+ /g sample g K + /g sample = 0.004632 mole K + 39.0983 g/mole = 0.1811 g K + /g sample g C 2O 2- /g sample = 0.004632 mole C 2O 2- MW (C 2O 2- ) = ? g C 2O 2- /g sample 4 4 4 4 g H2O = 1.000 g 0.1201 g 0.1811 g ? g = ? g H2O mole H 2 O = g H 2O = ? mole H 2 O/g sample MW (H 2 O) moles H 2 O moles H 2O = =z 3+ 0.002150 moles Fe Spring 2007 Theoretical Yield Calculation limiting reactant Fe(NH4)2(SO4)26H2O + H2C2O4 FeC2O42H2O + (NH4)2SO4 + 4 H2O + H2SO4 1 mole 1 mole product a FeC2O42H2O + b H2C2O4 +c H2O2 + d K2C2O4 e Kx[Fey (C2O4)x]zH2O final a mole e mole Need to determine the molar ratio between the starting limiting reactant and the final product Starting Limiting Reactant: convert g into moles Calculate moles of final product from the moles of starting limiting reactant Final Product: convert moles into g NOTE: If you cannot balance the equation above, use 1 : 1 ratio of FeC2O42H2O : Kx[Fey(C2O4)x]zH2O in theoretical yield calculations a = e = 1 Spring 2007 Miscellaneous Tips Most common source of error DILUTION ERRORS In the very beginning (Step 1), dissolve crystals in a small amount of water in a beaker, then transfer quantitatively into a volumetric flask. In Part 4, use pale purple (not green) solution from Part 1 Report writing: do not skip through the calculations that are outlined in your procedure in Part 4 AFTER the data table (steps 49). Show detailed calculations in your report (under Sample Calculations) Spring 2007 ...
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This note was uploaded on 11/10/2008 for the course CH 204 taught by Professor Leytner during the Spring '08 term at University of Texas.

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