Determination of Copper Oxide's Percentage Composition and Empirical Formula, Lab Reports of Chemistry

Download Determination of Copper Oxide's Percentage Composition and Empirical Formula and more Lab Reports Chemistry in PDF only on Docsity! Date Performed: 9/18/2007 Name: Yamilette Hernandez Date Submitted: 9/25/2007 Partner: Alex Tobias Instructor: Zelin Liu 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 Mass of empty test tube= 17.57g Mass of test tube plus oxide copper= 19.57g Mass of oxide copper= 2.0g Mass of copper and test tube after heated= 18.762g Mass of copper after heated= 1.192g Sample Calculations Mass Cu oxide: (Mass of test tube w/ copper oxide) – (mass of test tube) (19.57g ± 0.001) – (17.57g ± 0.001g) = = 2.00g ± √0.0012+0.0012 = 2.00g ± 0.001g Mass of Cu: (Mass of test tube w/ Cu) – (Mass of test tube) (18.762g ± 0.001g) – (17.57g ± 0.001g) = = 1.192g ± √0.0012+0.0012 = 1.192g ± 0.001g Mass % of Cu: [(Mass Cu) / (Mass of Cu oxide)*100] [(1.192g ± 0.001g) / (2.00g ± 0.001g)*100] = = 59.6% ± √( 0.0011.192 ) 2 +(0.0012.00 ) 2 =59.6% ± 0.0009 Mass of O: (Mass of Cu Oxide) – (Mass of Cu) (2.0g ± 0.001g) – (1.192g ± 0.001g) = = .808g ± √0.0012+0.0012 = .808g ± 0.001g Mass % of O: [(Mass of Oxygen/Mass of Copper oxide)*100] [(.808g ± 0.001g/2.00g ± 0.001g)*100] = = 40.4% ± √( 0.001.808 ) 2 +(0.0012.00 ) 2 = 40.4% ± 0.001 Moles of Cu: 1.192g*(1mol/63.55g/mol) = 0.0187 mol of Cu Moles of O: 0.808g*(1mol/32.00g/mol) = 0.0253 mol of O Formula: Cu O (59.6g/63.55g/mol)= .938mol (40.4g/32.00g/mol)= 1.26 mol (.938mol/.938mol)=1 (1.26mol/.938mol)= 1.35≈ 1 Results and Conclusion Copper Oxide Copper Oxygen Mass 2.00g ± 0.001g 1.192g ± 0.001g .808g ± 0.001g Mass % 59.6% ± 0.0009 40.4% ± 0.001 Moles 0.0187 mol 0.0253 mol Empirical Formula: CuO During the process of the experiment, the flame was turned off too early leaving some of the carbon oxide not completely turned red. Therefore, the reaction was not completed. The result shows the empirical formula to be CuO, copper (II) oxide. Although, the concluded result could have been copper (I) oxide or Cu2O.