Empirical Formula Lab Report
Steven Leung
9/19/06
Lab Report
The Empirical Formula of a Copper Oxide
Purpose:
To convert an unknown copper oxide to copper (Cu) metal using natural gas to provide a reducing environment as shown below:
Cu O (s) + CH (g) ¨ Cu (s) + Co (g) + H O (g)
From the mass difference between the unknown copper oxide and the Cu metal generated at the completion of the reaction and the molar mass of Cu and oxygen, the empirical formula of the original copper oxide can be calculated.
Materials:
gCopper oxideh (1.0 g)
Bunsen burner
Large test tube
Ring stand
Clamp (with no rubber on the prongs)
Two hole stopper with inlet and outlet tubes
Rubber tubing
Procedures: (in lab notebook)
Raw data: The copper oxide changed colors …show more content…
Test tube with copper oxide 45.904
Test tube with copper after burn 45.740
Mass Measurements (in grams)
Analysis/interpreting data/results: The difference betweent the unknown initial copper oxide and the Cu metal that was produced is .164 grams. This would indicate that the original copper oxide contained .164 grams of oxygen and .833 grams of copper in the original .997 gram copper oxide. .164 grams of oxygen is equivalent to about .010 moles. .833 grams of copper is equivalent to about .013 moles.
Conclusions/comments: The data indicates that the same amount of moles of oxygen and copper were present in the original copper oxide. Based on this conclusion, the empirical formula of this copper oxide would be CuO. This is the formula for Cupric Oxide. The amount of Cupric Oxide used in the experiment was off by a percent error of .3%, as the theoretical value was 1 gram, and the actual amount used was .997 grams. Other possible errors could have occurred in the experiment when the tube began to melt during the heating process. The tube was slightly disfigured but it did not seem to affect the final
9/19/06
Lab Report
The Empirical Formula of a Copper Oxide
Purpose:
To convert an unknown copper oxide to copper (Cu) metal using natural gas to provide a reducing environment as shown below:
Cu O (s) + CH (g) ¨ Cu (s) + Co (g) + H O (g)
From the mass difference between the unknown copper oxide and the Cu metal generated at the completion of the reaction and the molar mass of Cu and oxygen, the empirical formula of the original copper oxide can be calculated.
Materials:
gCopper oxideh (1.0 g)
Bunsen burner
Large test tube
Ring stand
Clamp (with no rubber on the prongs)
Two hole stopper with inlet and outlet tubes
Rubber tubing
Procedures: (in lab notebook)
Raw data: The copper oxide changed colors …show more content…
Test tube with copper oxide 45.904
Test tube with copper after burn 45.740
Mass Measurements (in grams)
Analysis/interpreting data/results: The difference betweent the unknown initial copper oxide and the Cu metal that was produced is .164 grams. This would indicate that the original copper oxide contained .164 grams of oxygen and .833 grams of copper in the original .997 gram copper oxide. .164 grams of oxygen is equivalent to about .010 moles. .833 grams of copper is equivalent to about .013 moles.
Conclusions/comments: The data indicates that the same amount of moles of oxygen and copper were present in the original copper oxide. Based on this conclusion, the empirical formula of this copper oxide would be CuO. This is the formula for Cupric Oxide. The amount of Cupric Oxide used in the experiment was off by a percent error of .3%, as the theoretical value was 1 gram, and the actual amount used was .997 grams. Other possible errors could have occurred in the experiment when the tube began to melt during the heating process. The tube was slightly disfigured but it did not seem to affect the final