Lec_06 - Today 1. Report for Experiment 4 is due (40 pts)...

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Unformatted text preview: Today 1. Report for Experiment 4 is due (40 pts) 2. Post-lab for Experiment 4 is due (10 pts) 3. Unknown Summary Sheet for Experiment 4 is due (20 pts) 4. Pre-lab for Experiment 5 is due (5 pts) 5. Experiment 5. Synthesis and Analysis of a Complex Iron Compound. Part 1 have a write-up in your notebook Next week 1. Report for Experiment 5 is due (40 pts) 2. Post-lab for Experiment 5 is due (10 pts) 3. Pre-lab for Experiment 6 is due (5 pts) 4. Experiment 6. Synthesis and Analysis of a Complex Iron Compound. Part 2 have a write-up in your notebook Spring 2007 Experiment 5. Synthesis and Analysis of a Complex Iron Compound. Part 1 PURPOSE AND LEARNING OBJECTIVES To synthesize potassium oxalatoferrate coordination compound To use redox titration to determine the oxalate content of a prepared sample To use visible spectrophotometry to determine the iron content of a prepared sample 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 Water content Oxalate content x? y? z? Iron content Oxalate anion C2O42 Parent acid (oxalic acid) H2C2O4 Spring 2007 Another Look at Potassium Oxalatoferrate Ionic salt Kx[Fey(C2O4)x] zH2O K+ cation [Fey(C2O4)x]x anion Waters of hydration Waters of hydration water molecules that are part of the crystalline structure of the compound and are present in a fixed molar ratio compared to the number of moles of the compound itself Spring 2007 Coordination Compounds Coordination (or complex) compounds contain coordinate covalent bonds Coordinate covalent bond formed when the shared pair of electrons is provided by a single atom Most d-transition metal ions have vacant d-orbitals that can accept electron pairs Example: Hemoglobin is a protein that is responsible for the oxygen transport in human bodies Fe is the oxygen binding site Spring 2007 Synthesis Step 1. Synthesis of an iron (II) oxalate dihydrate, FeC2O42H2O Starting materials: Fe(NH4)2(SO4)26H2O ammonium iron (II) sulfate hexahydrate H2C2O4 oxalic acid (1.0 M solution) The reaction: Fe(NH4)2(SO4)26H2O + H2C2O4 FeC2O42H2O (s) + (NH4)2SO4 + 4 H2O + H2SO4 Net ionic equation: Fe2+ + C2O42 + 2 H2O FeC2O42H2O (s) intermediate product Spring 2007 Synthesis 2. Step 2 Synthesis of a complex iron compound, Kx[Fey(C2O4)x]zH2O Starting materials: FeC2O42H2O iron (II) oxalate dihydrate (from Step 1) H2O2 hydrogen peroxide (3% solution) added to oxidize Fe2+ to Fe3+ K2C2O4 potassium oxalate (2.0 M solution) H2C2O4 oxalic acid (1.0 M solution) added to to provide excess of oxalate, C2O4 2, anions and K+ counter cations The reaction: a FeC2O42H2O (s) + b H2C2O4 + c H2O2 + d K2C2O4 e Kx[Fey(C2O4)x]zH2O Spring 2007 Synthesis (cont). Step 2 (cont) Synthesis of a complex iron compound, Kx[Fey(C2O4)x]zH2O After the reaction is complete: Aqueous solution of Kx[Fey(C2O4)x]zH2O Add ethanol (ethyl alcohol), CH3CH2OH, and grow crystals of your product Kx[Fey(C2O4)x]zH2O soluble in H2O insoluble in ice-cold ethanol Isolate (by vacuum filtering) and dry the crystals Determine the actual yield of your product (weight it) Next week Spring 2007 Percent Yield Calculation Percent yield = Actual yield (g) 100% Theoretical yield (g) Actual yield the amount of product that is actually obtained in a given procedure Theoretical yield maximum amount of product that could be obtained in the reaction based on the stoichiometry of its balanced chemical equation assuming that the reaction goes to completion Theoretical yield is always calculated from the limiting reactant Limiting reactant a substance that stoichiometrically limits the amount of product(s) that can be formed Spring 2007 Theoretical Yield Calculation Example: 100.0 g of Al react with 800.0 g of BaO. Determine theoretical yield of Al2O3. 2 Al (s) + 3 BaO (s) Al2O3 (s) + 3 Ba (s) 1. Determine the limiting reactant Moles Al = 100 . 0 g = 3 . 706 mole 26 . 9815 g / mole 800 . 0 g = 5.218 mole 1 53.326 g / mole HAVE Moles B aO = To completely react 3.706 moles of Al we will NEED 5.559 moles of BaO: Moles B aO = 3 3 . 706 mole = 5.559 mole 2 NEED (5.559 moles) more than we HAVE (5.218 moles) BaO is the limiting reactant Spring 2007 Theoretical Yield Calculation (cont.) 2 Al (s) + 3 BaO (s) Al2O3 (s) + 3 Ba (s) 2. Determine theoretical yield of Al2O3 from the limiting reactant (BaO) Moles A l 2 O 3 = 1 1 Moles BaO = 5 . 218 mole = 1 . 739 mole 3 3 G rams A l 2 O 3 = 1 . 739 mole 101 . 9612 g/mole = 177.3 g 1. Determine the limiting reactant 2. Starting Material (limiting reactant): convert g into moles 3. Relate moles of starting material to the moles of final product using the chemical equation stoichiometry 4. Final Product: convert moles into g Spring 2007 Miscellaneous Tips Add H2O2 SLOWLY drop by drop while controlling the solution temperature at about 40 C If you see a lot of yellow precipitate in the beaker after addition of H2O2 and oxalic acid, go back to the H2O2 addition step Make the distinction between the K2C2O4 and H2C2O4 solutions Final solution of your synthesized compound will be of a bright green color Make sure you have some precipitate before leaving for the day Spring 2007 ...
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