Determination of Copper Oxide's Chemical Formula: An Experiment in Stoichiometry - Prof. P, Lab Reports of Chemistry

Download Determination of Copper Oxide's Chemical Formula: An Experiment in Stoichiometry - Prof. P and more Lab Reports Chemistry in PDF only on Docsity! Date Performed: 9/30/09 Name: Ben Maddock Dater Submitted: 10/7/09 Partner: Alec Jones Instructor: Brian Harding Determination of a Chemical Formula Objective: To determine the percentage composition and simplest formula for an oxide of copper by its reduction with methane gas (CH4) at approximately 500 OC. This experiment is a study of stoichiometry. Experimental Data: Correction Factor: Graduated Cylinder = ± 0.05 mL Balance = ± 0.001 g Atomic Mass: Copper = Cu = 63.55 ± 0.01 g/mol Oxygen = O = 16.00 ± 0.01 g/mol Masses: Large Test Tube = 17.462 g Large Test Tube + Copper Oxide = 19.468 g Large Test Tube + Copper Oxide After Heating = 19.173 g Possible Reduction Reactions: 500 oc 4Cu2O + CH4  CO2 + 2H2O + 8Cu 500 oc 4CuO + CH4  CO2 + 2H2O + 4Cu Sample Calculations: (mass of test tube + oxide) – (mass of test tube) = mass of oxide (19.468 ± 0.001 g) - (17.462 ± 0.001 g) = 2.006 ± 0.001 g Copper Oxide (mass of test tube + oxide after heat) – (mass of test tube) = mass of copper (19.173 ± 0.001 g) – (17.462 ± 0.001 g) = 1.711 ± 0.001 g Cu (mass of copper)/(mass of oxide) x 100 = mass % of Cu (1.711 ± 0.001 g)/(2.006 ± 0.001) = 85.29 % Cu (mass of copper) x (1 mol Cu/ 63.55 ± 0.01 g) = moles of copper (1.711 ± 0.001 g) x (1 mol Cu/ 63.55 ± 0.01 g) = 0.0269 ± 0.01 mol Cu (mass of test tube + oxide) – (mass of test tube + oxide after heat) = mass of oxygen (19.468 ± 0.001 g) - (19.173 ± 0.001 g) = 0.295 ± 0.001 g O (mass of oxygen)/(mass of oxide) x 100 = mass % of O (0.295 ± 0.001 g)/(2.006 ± 0.001 g) x 100 = 14.71 %O (mass of oxygen) x (1 mol O/16.00 ± 0.01 g) = moles of oxygen (0.295 ± 0.001 g) x (1 mol O/16.00 ± 0.01 g) = 0.0184 ± 0.01 mol O Empirical Formula of The Oxide: (mol O)/(mol O) = 1 (0.0184 ± 0.01)/(0.0184 ± 0.01) = 1 ± 0.7