Introduction: The purpose of this lab that we are carrying out is to determine the number of grams of copper that will be produced from the oxidation reduction reaction. Redox (short for reduction-oxidation reaction describes all the chemical reactions in which atoms have their oxidation number(oxidation state) changed. This can be either a simple redox process, such as the oxidation of carbon to yield carbon dioxide or the reduction of carbon by hydrogen to yield methane(NH4), or a complex process such as the oxidation of sugar in the human body through a series of complex electron transfer processes.
The term comes from the two concepts of reduction and oxidation. It can be explained in the the simple terms:
The term comes from the two concepts of reduction and oxidation. It can be explained in the the simple terms:
- Oxidation is the loss of electrons or an increase in oxidation state by a molecule,atom, or ion.
- Reduction is the gain of electrons or a decrease in oxidation state by a molecule,atom, or ion.
We will go through and describe to you every procedure that we take in this lab and their products. We will show you how we found the:
- Actual yield:It is given as the weight in grams or in moles (molar yield)
- Theoretical yield:The quantity of a product obtained from the complete conversion of the limiting reactant in a chemical reaction. Theoretical yield is commonly expressed in the terms of grams or moles.
- Percentage yield: Calculate to be the experimental yield divided by theoretical yield multiplied by 100%.
We will also show you the process of finding out what type of reaction has occurred in this experiment and why it is this type of reaction.
Statement of the Problem: To determine the amount of copper in grams that will be produced from a single displacement reaction when we know the mass of Aluminum that reacted with a known amount of copper II sulfate pentahydrate and to compare this theoretical yield to the actual yield of copper.
Statement of the Problem: To determine the amount of copper in grams that will be produced from a single displacement reaction when we know the mass of Aluminum that reacted with a known amount of copper II sulfate pentahydrate and to compare this theoretical yield to the actual yield of copper.
- Aluminum Powder
- Copper Sulfate Pentahydrate
- Medium Sized Beaker
- Glass Stiring Rod
- Filter Paper
- Scale
- Heating Apparatus
- Bunsen Burner
- Water
- Funnel
- Erlenmeyer Flask
- Apron
- Safety Goggles

Discussion: When we finished our lab we found the balanced equation in order to really understand what reaction had taken place. We found the balanced equation to be 3CuSO4+2Al (arrow) Al2(SO4)3+3Cu. Now we were able to use this balanced equation to figure out our theoretical yield and compare it to our actual yield. Our actual yield was 2.28gCu. We then converted our .757gAl to moles of Al to moles of Cu to grams of Cu in order to find our theoretical yield. After completing this conversion we found our theoretical yield to be 2.68gCu. Once we had both of our yields we were able to find our percent yield by dividing our actual yield by our theoretical yield. This lab allowed us to see how to find percent yield in a real experiment rather than just knowing how to do it on a worksheet.
Conclusion: After we divided our actual yield by our theoretical yield we found that our percent yield was 85.1%. In this lab we were able to watch a reaction take place that helped us to understand what it really means to find a percent yield. We watched as copper was formed from this reaction between Copper (II) Sulfate pentahydrate and Aluminum. After the water changed colors and the aluminum dissolved into the solution were watched the copper form in little clumps. We also were able to learn why the actual yield and the percent yield are different. These yields may not have been equal for a variety of reasons. Some possible explanations are that we did not let the reaction take place completely before we started filtering the solution through, some of the copper pieces may have been small enough to filter through the paper, or the solution may have not have stayed on the heat for long enough after the aluminum pieces had dissolved. In any case this experiment helped us to learn what percent yield really means in a real life situation.