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Chem 112 Lab 2 Report

# Chem 112 Lab 2 Report - 4 =.021 0.0043 x.021 = 9e-5(9e-5(5...

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Lab 2 Report Data: C 2 O4 2- Analysis Sample 1 Sample 2 1)Molarity of KMnO 4 .021 .021 2)Weight of Sample .217 .220 3)Final Buret Reading 50.00 33.20 4)Initial Buret Reading 20.00 0.00 5)Volume of KMnO 4 dispensed 30.00 33.20 6)Moles of KMnO 4 .700 .6325 7)Moles of C 2 O4 2- .280 .253 8)Final Buret Reading (2 nd ) 34.30 44.50 9)Initial Buret Reading (2 nd ) 30.00 40.00 10)Volume 4.30 4.50 Calculations: 1) Calculate the moles of C 2 O4 2- present from the molarity and volume of KMnO 4 solution Sample 1: Volume = 30ml, Molarity of KMnO 4 = .021 0.021 x .03L = 6.3e-4 mol/L (6.3e-4) ( 5/2) = 0.001575 moles Sample 2: Volume = 33.2ml, Molarity of KMnO 4 = .021 0.021 x .0332 = .000697 moles/L (6.97e-4)(5/2) = 0.001743 moles 2) Calculate the moles of Fe 3+ present Sample 1: Volume = 4.30ml, Molarity of KMnO

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Unformatted text preview: 4 = .021 0.0043 x .021 = 9e-5 (9e-5)(5) = 4.52e-4 Sample 2: Volume 4.50ml, Molarity of KMnO 4 = .021 .0045 x .021 = 9.5e-5 (9.5e-5)(5) = 4.73e-4 3) Find: Ratio of C 2 O4 2- / Fe 3+ Sample 1: .001575 mol / 4.52e-4 mol= 3.48 Sample 2: .001743 mol /4.73e-4 mol = 3.68 Discussion: We obtained a 3.58 average ratio of C 2 O4 2- to Fe 3+ ions. This is consistent with the expected 3:1 ration in the original formula of K 3 [Fe(C 2 O 4 ) 3 ] * 3 H 2 O where the ratio is 3 to 1. The inconsistency might have arose from the incomplete reaction with zinc, because the zinc was still reacting when I took it off the burner....
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