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UT CH 204 - Cheat Sheet for Experiment 7 - Visible Spectrophotometry.

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CH204 Fall 2008 Cheat Sheet for Experiment 7. Visible Spectrophotometry. Molecular Formula Determination 1 Part 3. Standard Iron Solution Calculations Dilution calculations: M1V1 = M2V2, where M1 and M2 are the molarities of the original concentrated and the final dilute solutions and V1 and V2 are the volumes of the original concentrated and the final dilute solutions Calculations: 1. Convert g/L into a mole/L concentration for the standard Fe2+ solution 2. Calculate the molarity of the stock Fe(phen)32+ solution prepared from the standard Fe2+ solution using the stoichiometry of the following equation (note that 1,10-phenanthroline is used in excess in this reaction): Fe2+ (aq) + 3 phen (aq) → Fe(phen)32+ (aq) 3. Calculate the molarities of the five standard Fe(phen)32+ solutions prepared from the stock Fe(phen)32+ solution (dilution calculations) Part 4. Iron Content Determination by Visible Spectrophotometry Beer’s law: A = ε l c Known: Absorbance of the Fe(phen)32+ sample, A Molar absorptivity for Fe(phen)32+ at 510 nm, ε (M–1cm–1) Dilution factors Pathlength, l = 1 cm Mass of the complex iron sample analyzed (g) To be determined: Moles of Fe3+ per gram of sample (mole/g) Calculations: 1. εlA=+)(Fe(phen) M23 (mole/L) 2. Total dilution factor = 5 × 5 × X, where X is your final dilution 3. Moriginal (Fe3+) = M ))phen(Fe(23+ × Total dilution factor 4. In 25 ml of solution: moles Fe3+ = L 025.0)(Fe M3original×+ 5. (g) sample of masseF olesmsample /geF Moles33++= KxFey(C2O4)x·zH2O Molecular Formula Determination Known: Moles of C2O42– per gram of sample (mole/g sample) Moles of K+ per gram of sample (mole/g sample) Moles of Fe3+ per gram of sample (mole/g sample) y = 1CH204 Fall 2008 Cheat Sheet for Experiment 7. Visible Spectrophotometry. Molecular Formula Determination 2 To be determined: x and z Calculations: 1. sample g/Fe of molessample g/OC of moles 3242+−= yx ⇒ x – ? 2. g Fe3+/g sample = moles Fe3+/g sample × MW (Fe) 3. g C2O42–/g sample = moles C2O42–/g sample × MW (C2O42–) 4. g K+/g sample = moles K+/g sample × MW (K) 5. g H2O/g sample = 1.000 g – g Fe3+/g sample – g C2O42–/g sample – g K+/g sample 6. O)(HMW sample O/gH g sample O/gH mole222= 7. sample g/Fe of molessample g/OH of moles32+= yz ⇒ z – ? Determination of the Theoretical Yield and Percent Yield of KxFey(C2O4)x·zH2O Chemistry involved: Fe(NH4)2(SO4)2·6H2O + H2C2O4 → FeC2O4·2H2O + (NH4)2SO4 + H2SO4 + 4 H2O (1) a FeC2O4·2H2O + b H2C2O4 + c H2O2 + d K2C2O4 → e Kx[Fey(C2O4)x]·zH2O (2) Note: Equation (2) has to be balanced before you start working on these calculations. Known: Actual yield of the product (g) Mass of the starting material, Fe(NH4)2(SO4)2·6H2O (g) Moles of FeC2O4·2H2O, produced in reaction (1) and used as a starting material in reaction (2). See Post-lab 5, question 5 for reference Assume FeC2O4·2H2O to be the limiting reactant in reaction (2) To be determined: Theoretical yield (g) and Percent yield (%) Calculations: 1. Determine moles of Kx[Fey(C2O4)x]·zH2O from moles of FeC2O4·2H2O using the stoichiometry of equation (2)* 2. Theoretical yield (g) = Moles Kx[Fey(C2O4)x]·zH2O × MW (Kx[Fey(C2O4)x]·zH2O) 3. %100(g) yield ltheoretica(g) yield actual yieldPercent ×= *If you are unable to balance equation (2) due to the erroneous molecular formula determination, leave equation (2) unbalanced and use a 1-to-1 molar ratio of FeC2O4·2H2O to Kx[Fey(C2O4)x]·zH2O at this


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UT CH 204 - Cheat Sheet for Experiment 7 - Visible Spectrophotometry.

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