Titration of Iron by Permanganate Lab

Titration of Iron by Permanganate Lab - Stephanie Warner...

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Stephanie Warner Chem 153 Shidong Chu 27 September 2005 Titration of Iron by Permanganate Purpose: The purpose of this lab is to determine the percent weight of Fe in a solid by using a classical redox titration. Calculations: Standardization: O.7699g Fe (NH 4 ) 2 (SO 4 ) 2 ∙ 6H 2 O/ 392.14 g/mol = 0.001963 mol Fe 2+ 0.001963 mol Fe 2+ x (1 mol MnO 4 - / 5 mol Fe 2+ ) = 3.926 x 10 -4 mol MnO 4 - 38.07 mL x (1 L/ 1000 mL) = 0.03807 L 3.926 x 10 -4 mol MnO 4 - / 0.03807 L = 0.01031 M MnO 4 - Q Test for Standardization: (0.01035-0.01031)/ (0.01035-0.009956) = 0.10 (0.009956-0.01004)/ (0.01035-0.009956) = 0.21 This means there are no outliers Average for Standardization: 0.009956 + 0.01004 + 0.01031 + 0.01035 = 0.04067/4 = 0.01016 average M MnO 4 - Unknown: 27.42 Ml x (1 L/ 1000 mL) = 0.02742 L MnO 4 - (0.02742 L MnO 4 -) (0.01016 M MnO 4 - ) = 2.786 x 10 -4 mol MnO 4 - 2.786 x 10 -4 mol MnO 4 - x (5 mol Fe 2+ /1 mol MnO 4 - ) = 0.001393 mol Fe 2+ {[(0.001393 mol Fe 2+ )(55.85 g/mol Fe)]/ 0.9167 g Unknown }x 100 = 8.49% w.t. Fe Q Tests for Unknown: (8.69-8.49)/(8.69-7.60) = 0.18 (7.60-7.74)/(8.69-7.60) = 0.13 This means that there are no outliers. Average for Unknown: 8.49 + 7.84 + 8.69 + 7.74 + 7.60 = 40.36/ 5 = 8.07% average w.t. Fe Standard Deviation: d 1 = (x mean value – x measured value ) = (8.07-8.49) = -0.42 d 2 = (8.07-7.84) = 0.23
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d 3 = (8.07-8.69) = -0.61 d 4 = (8.07-7.74) = 0.33 d 5 = (8.07-7.60) = 0.47 S = [(d 1 2 + d 2 2 + d 3 2 + d 4 2 + d 5 2 ) / (number of measurements – 1)] ½ = [((-0.42)
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