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Lab Report 2

# Lab Report 2 - Date Created Submission Date Instructor Bing...

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Date Created: 10-4-2009 Name: Tyler Campos Submission Date: 10-6-2009 Partner: Conor McKenna Instructor: Bing Yan Determination of a Chemical Formula Objective Determine the percentage composition and the simplest formula for an oxide of copper by its reduction with methane gas at approximately 500°C. Experimental Data Mass of test tube =17.28 ± 0.001 g Mass of test tube and Copper Oxide =19.28 ± 0.001 g Mass of test tube and Copper =18.93 ± 0.001 g Sample Calculations Mass of Copper Oxide = Mass of test tube and Copper Oxide (g) – Mass of test tube (g) = (19.28 ± 0.01 g) – (17.28 ± 0.01 g) = 2.00 ± √(0.01 2 ) + (0.01 2 ) g = 2.00 ± 0.001 g Mass of Copper = Mass of test tube and Copper (g) – Mass of test tube (g) = (18.93 ± 0.01 g) – (17.28 ± 0.01 g) = 1.65 ± √(0.01 2 ) + (0.01 2 ) g = 1.65 ± 0.001 g Mass of Oxygen = Mass of Copper Oxide (g) – Mass of Copper (g) = (2.00 ± 0.01 g) – (1.65 ± 0.01 g) = 0.350 ± √(0.01 2 ) + (0.01 2 ) g = 0.350 ± 0.01 g Mass percentage of Copper = Mass of Copper (g) / Mass of Copper Oxide (g) x 100 = (1.65 ± 0.01 g /2.00 ± 0.01 g) x 100 (S r ) 2 /(0.825) 2 = (0.01) 2 /(1.65) 2 + (0.01) 2 /(2.00) 2 S r = .0006 = 82.5 ± 0.01 % Mass percentage of Oxygen = Mass of Oxygen (g) / Mass of Copper Oxide (g) x 100 = (0.350 ± 0.01 g / 2.00 ± 0.01 g) x 100

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