1 / 14

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.

barbie
Download Presentation

Lab 8 (Reduction of CuO)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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.

More Related