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Lab 8 (Reduction of CuO)

Lab 8 (Reduction of CuO). Using data to determine Percent Composition Empirical Formula. Pre-Lab Questions:. (Theoretical) % composition of Cu. Cu 2 O. 63.55 g. 16.00 g. x 2. x 1. = 143.10 g. +. 127.10 g. 16.00 g. 127.10 g. =. %Cu. X 100 =. 88.8 % Cu.

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Lab 8 (Reduction of CuO)

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  1. Lab 8 (Reduction of CuO) Using data to determine Percent Composition Empirical Formula

  2. Pre-Lab Questions: (Theoretical) % composition of Cu Cu2O 63.55 g 16.00 g x 2 x 1 = 143.10 g + 127.10 g 16.00 g 127.10 g = %Cu X 100 = 88.8 % Cu 143.10 g

  3. Pre-Lab Questions: (Theoretical) % composition of Cu CuO 63.55 g + 16.00 g = 79.55 g 63.55 g = %Cu X 100 = 79.9 % Cu 79.55 g

  4. Lab 8 (Reduction of CuO) 33.09 Mass of empty test tube ___________ grams Data and Observations (Sample Data) Mass of test tube + CuO 37.93 ___________ grams Mass of test tube + Copper 36.94 ___________ grams Black Reddish-Brown Color of copper oxide ___________ Color of product left after heating ________________

  5. 33.09 Mass of empty test tube ___________ grams 37.93 Mass of test tube + CuO ___________ grams Mass of test tube + Copper 36.94 ___________ grams Calculations and Graphs (SHOW WORK) 1. Original mass of CuO before heating (Test tube + CuO) – mass of empty test tube - = 37.93 g 33.09 g 4.84 g CuO Data and Observations (Sample Data) 4.84 Copper (II) oxide = _____________g Final mass of Copper product (Test tube + Cu) – mass of empty test tube - = 36.94 g 33.09 g 3.85 g Cu 3.85 Copper = _____________g Mass of oxygen in original CuO sample (Mass of CuO) – (mass of Cu product) - = 4.84 g 3.85 g 0.99 g O 0.99 Oxygen = _____________g

  6. Calculations and Graphs (SHOW WORK) 1. Original mass of CuO before heating (Test tube + CuO) – mass of empty test tube - = 37.93 g 33.09 g 4.84 g CuO 4.84 Copper (II) oxide = _____________g Final mass of Copper product (Test tube + Cu) – mass of empty test tube - = 36.94 g 33.09 g 3.85 g Cu 3.85 Copper = _____________g Mass of oxygen in original CuO sample (Mass of CuO) – (mass of Cu product) - = 4.84 g 3.85 g 0.99 g O 0.99 Oxygen = _____________g 2. Calculate the experimental percent composition of copper 3.85 g 79.5 % Cu x 100 = % Cu = 4.84 g

  7. Calculations and Graphs (SHOW WORK) 2. Calculate the experimental percent composition of copper 3.85 g 79.5 % Cu x 100 = % Cu = 4.84 g 3. Compare your answer from question 2 to the pre-lab calculations. Which formula most closely matches your results? Pre-Lab results: (Theoretical %) Copper (I) oxide, Cu2O = 88.8 % Cu Copper (II) oxide, CuO = 79.9 % Cu 79.5 % Cu(experimental result) matches Copper (II) oxide, CuO, 79.9 % Cu

  8. Calculations and Graphs (SHOW WORK) 3. Compare your answer from question 2 to the pre-lab calculations. Which formula most closely matches your results? 79.5 % Cu(experimental result) = CuO, 79.9 % Cu (theoretical result) 4. Determine the percent error of your lab data. Theoretical % - Experimental % % error = x 100 Theoretical % - 79.9 79.5 % error = x 100 = 0.50 % error 79.9

  9. Calculations and Graphs (SHOW WORK) 1. Original mass of CuO before heating - = 37.93 g 33.09 g 4.84 g CuO 4.84 Copper (II) oxide = _____________g Final mass of Copper product - 3.85 = 36.94 g 33.09 g 3.85 g Cu Copper = _____________g Mass of oxygen in original CuO sample - 0.99 = 4.84 g 3.85 g 0.99 g O Oxygen = _____________g 5. Calculate the number of moles of Cu product. 29 Cu 63.55 1 mol Cu 3.85 g Cu 0.0606 mol Cu x = 63.55 g Cu 6. Calculate the number of moles of oxygen lost. 8 O 16.00 1 mol O 0.99 g O 0.0619 mol O x = 16.00 g O

  10. Calculations and Graphs (SHOW WORK) 5. Calculate the number of moles of Cu product. 29 Cu 63.55 1 mol Cu 3.85 g Cu 0.0606 mol Cu x = 63.55 g Cu 6. Calculate the number of moles of oxygen lost. 8 O 16.00 1 mol O 0.99 g O 0.0619 mol O x = 16.00 g O 7. Show the ratio of moles of Copper to moles of Oxygen (Predict empirical formula) mol Cu 0.0606 mol Cu 0.98 1 = = = mol O 0.0619 mol O 1 1 Cu O Empirical formula

  11. Conclusion Questions: 1. Identify all the physical properties that confirmed the product was copper metal. • Color: The color matched that of copper (reddish-brown). • Luster: When the sample was rubbed on the table it became shiny. • Conductivity: The sample conducted electricity easily.

  12. Conclusion Questions: • Methane gas was used as a reducing agent in this experimient. The gas was allowed to pass over the CuO inside the test tube as the reactant was heated. • How did the methane help reduce the CuO to copper metal? • The methane helped remove the oxygen from the CuO so it would not recombine with the hot copper left over. • Equation: CH4 (g)+ O2 (g) CO2 (g)+ H2O(g) 2 2

  13. Conclusion Questions: 3. Why was the methane gas ignited at the top of the test tube? • The methane was ignited to prevent unburned methane being able to escape the test tube and blow up the classroom. 4. Why was the methane kept burning for 5 minutes after the Bunsen burner was shut off below the test tube? • The methane was kept burning to prevent oxygen from the air getting into the test tube and reacting with the hot copper product.

  14. Conclusion Questions: 5. Would not heating the copper oxide sample long enough make the final ratio of moles of copper to moles of oxygen higher, lower, or have no effect? Explain. mol Cu = mol O • The mole ratio would be higher. • If you did not heat the sample long enough, the final mass of the sample would be a mixture of CuO and Cu instead of pure Cu. • Since CuO has more mass than Cu, the mass would be higher and as a consequence our calculation of Moles Cu would also be higher.

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