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Molarity of KMnO4 is 0.0100 M

Determination of oxalate in Kufe.(0,0...0). Name AB A LOCAL Partner's Name MA R ENDALL Total mass of product synthesized in previous experiment (6) 1720 g Molarity of KMnO. (Record from the label on the container) Trial 1 Trial 2 Trial 2 0.105 1. Mass of sample taken for titration (g) 0.11 2. Initial burette reading (ml) 0 Final burette reading (ml) 20 33.4 22.2 4. Volume of KMnO, used (ml) 2.2 5. Moles of MnOused (mol) 6. Moles of CO2-in sample (mol) 7. Mass of C,0,- in sample (8) mo 8. Mass percent of Co. in sample (%) Average mass percent of C,0in sample (%) Post-Laboratory Question: The coordination compound synthesized could have one of two formulas: KsFe(C,0))3H,0 or [Fe(C,0.)(H,0)). In the first formula, three bidentate oxalate anions form the complex with the iron(III) cation. Three Kions serve to form the neutral compound and the formula has three waters of crystallization. In the second formula, two bidentate oxalate ions and two water molecules act as ligands. The resulting complex has one negative charge that is neutralized by one k'ion. In both formulas, the coordination number of Fe is six, so that there is an octahedral arrangement of ligands around the central atom. Calculate the mass percent of C30/in both formulas, and deduce which formula corresponds to the coordination compound that you synthesized. 2 SECOND TERATION WE ADDED 20.0mL of non inom TO BUPEITE ONCE LIQUID CEACITED

Answer 1

Calculations for Trial 1

Volume of KMnO₄ used for titration=Final buret reading-Initial buret reading

=26.2 mL-0.0 mL

=26.2 mL

Also 1 mol KMnO₄ gives 1 mol K⁺ and 1 mol MnO₄^- in solution

Number of moles of MnO₄^-=Number of moles of KMnO₄

=Molarity x volume (L)

=0.0100 M x 26.2 mL/1000 mL/L (1 L=1000 mL)

=0.000262 mol

Also number of moles of C₂O₄^2- titrated

=number of moles of MnO₄^- x 5 mol C₂O₄^2-/2 mol MnO₄^-

(using formula given in calculations details)

=0.000262 mol MnO₄^- x 5 mol C₂O₄^2-/2 mol MnO₄^-

=0.000655 mol

Relative Molecular weight of C₂O₄^2-=2x relative atomic mass of C+4xrelative atomic mass of O

=2x12 g/mol+4x16 g/mol

=24 g/mol+64 g/mol

=88 g/mol

Mass of C₂O₄^2- in the sample=number of moles of C₂O₄^2- x relative molecular weight (in g/mol)

=0.000655 mol x 88 g/mol

=0.05764 g

Mass % of C₂O₄^2-- in the sample=(Mass of C₂O₄^2-/Mass of sample)x100

=(0.5764 g/0.111 g)x100

=51.93%

Calculations for Trial 2

Volume of KMnO₄ used for titration=Final buret reading-Initial buret reading

=33.4 mL-11.2 mL

=22.2 mL

Also 1 mol KMnO₄ gives 1 mol K⁺ and 1 mol MnO₄^- in solution

Number of moles of MnO₄^-=Number of moles of KMnO₄

=Molarity x volume (L)

=0.0100 M x 22.2 mL/1000 mL/L (1 L=1000 mL)

=0.000222 mol

Also number of moles of C₂O₄^2- titrated

=number of moles of MnO₄^- x 5 mol C₂O₄^2-/2 mol MnO₄^-

=0.000222 mol MnO₄^- x 5 mol C₂O₄^2-/2 mol MnO₄^-

=0.000555 mol

Relative Molecular weight of C₂O₄^2-=2x relative atomic mass of C+4xrelative atomic mass of O

=2x12 g/mol+4x16 g/mol

=24 g/mol+64 g/mol

=88 g/mol

Mass of C₂O₄^2- in the sample=number of moles of C₂O₄^2- x relative molecular weight (in g/mol)

=0.000555 mol x 88 g/mol

=0.04884 g

Mass % of C₂O₄^2-- in the sample=(Mass of C₂O₄^2-/Mass of sample)x100

=(0.04884 g/0.105 g)x100

=46.51%

Average mass percent of C₂O₄^2- in the sample2

=(Mass percent from trial 1+mass percent from trial 2)/2

=(51.93% +46.51%)/2

=98.44%/2

=49.22 %