Aluminum and Copper II Sulfate Redox Lab

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:

• 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.

Materials:

• 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

Procedure: First we obtained a medium sized beaker and filled it with 75 - 100 ml of water. Once our beaker was filled to the proper amount of water we set up our heating apparatus and began heating the water. Next, we measured out 8-10g Copper (II) sulfate pentahydrate, wrote down the exact measurement in our table, and slowly added the crystals to the heating water. With our glass stirring rod we stirred the solution until the Copper II sulfate pentahydrate was dissolved. While the solution was being stirred one member of the group went and measured out 0.7-1.0g Aluminum and recorded the exact measurement in our table. Once the Copper II sulfate pentahydrate was dissolved we added the aluminum to the hot solution and continued to stir until the reaction had taken place. After we could no longer see any amount of aluminum we heated our substance for 3 to 4 minutes, and then removed it from the heat. After we took our substance off of the heat we used our filter paper and our funnel to filter the residue in the beaker. We caught the filtrate in the Erlenmeyer flask. Then we rinsed out our beaker with a small amount of water to be sure that we got all of the residue at the bottom. We removed our filter paper from the funnel and layed it flat on a paper towel to dry overnight. We also cleaned and dried all of our also tools and glassware. We made sure to turn the propane off and disconnect the Bunsen burner and put that and all of our other equipment away. The next day we weighed the filter paper and dry residue and recorded that mass in our data table. Finally, we threw the filter paper and residue away. The next day, after letting the residue dry all night, we weighed the dry residue and the filter paper together and recorded the mass in our table. Next, we subtracted the mass of the filter paper from the mass of the dry residue and the filter paper. The product of this was our actual yield. Then we calculated our theoretical yield and compared the two together.

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.

Reactions Lab. [:

Introduction:
      There are five types of reactions: synthesis, single displacement, double displacement, decomposition, and combustion. The first reaction, synthesis, is a chemical reaction in which atoms or simple molecule combine to form one single compound. In a single displacement reaction, one element replaces another element in a compound that is in a solution. A double displacement reaction is when ions from two compounds interact in a solution to form a product. Combustion is a violently exothermic reaction with oxygen to form oxides. Decomposition is when a single compound is broken down to produce two or more simpler substances. Lastly, a combustion reaction is an exothermic reaction, usually with oxygen, that forms oxides. It also forms CO2 and H2O.


Materials:

• 3 test tubes
• test tube rack
• CuSO4 solution
• Ba(NO3)2 solution
• HCL solution
• H2O2 solution
• MnO2
• Bunsen burner
• Zinc
• Magnesium ribbon

Procedure:
      Since we were once again working with potentially dangerous chemicals we took safety precautions and put on goggles and lab aprons before we started any of our work. We then took our first test tube placed a piece of zinc in the bottom and added about 1/2 mL CuSO4 solution. Next, we took our second test tube and added about 1/2 mL Ba(NO3)2 solution to about 1/2 mL of CuSO4 solution. In our third test tube we placed a piece of magnesium ribbon and added about 1/2 mL of HCL solution. During each of these reactions we recorded our observations in our journals. After this we lit a bunsen burner (burning propane gas) and recorded our observations of the flame. Next, we rinsed out the first test tube and added about 1/2 mL of H2O2 solution and lightly heated it over the flame. We recorded our observations of this reaction as well. Then we added a pinch of MnO2 (catalyst) to the H2O2 and lightly heated it to watch the reaction again. Then we recorded our observations one final time. Once we were done cleaning up our equipment and putting it away we made up a data table to show our results. We also identified what reactions took place and wrote out the balanced equations of each.

Discussion:     

How do we know that the reactions that occurred were chemical reactions?

Reaction 1: In this reaction between CuSo4 and Zn, the zinc changed color from

Silver into a reddish, copper color, an outcome of chemical reactions. This reaction

Is an example of a single displacement reaction as shown in the balanced formula-

Zn+CuSo4>Cu+ZnSo4.


Reaction 2: In this second reaction between Ba(No3)2 solution and CuSo4, the
Mixture bubbled, a precipitate formed on the bottom of the tube, and it also changed
Color a sign of a chemical reaction. This reaction is an example of a double displacement- Ba(No3)2+CuSo4>BaSo4+Cu(No3)2


Reaction 3: In this reaction between Magnesium Ribbon and HCl solution, the
Mixture began to fizz and turned into a gray color. This reaction is an example of
A single displacement reaction-Mg+2HCl>H2+MgCl2.

Reaction 4: In this reaction with lighting the Bunsen burner, and the form

of a flame, it gave off heat and was a blue and orange color. This reaction

Is an example of a combustion reaction-C3HS+5O2>3Co2+4H2O.


Reaction 5: In this reaction we put a tube of H2O over a flame, it fizzed and oxygen
Bubbles rose to the surface. This reaction is an example of a decomposition reaction-
2H2O2>2H2O+O2

Conclusions:

       We found that though there are five reactions only four occurred in our experiment. In our five reactions synthesis never took place because none of the reactions consisted of two elements/compounds reacting together and forming one compound. Our first and third reaction were both single displacement reactions. As we know from the introduction single displacement means that one element replaced another element in a compound as we can see in the equations in the data table. The second reaction in our experiment was double displacement because elements from two compounds switched places. In this reaction there were bubbles and a precipitate was formed. You can see this in our data table as well. The fourth reaction was combustion because it was an exothermic reaction that produced oxides. We knew that this reaction was exothermic because the test tube was hot while the reaction was taking place and there were sparks. The last reaction was decomposition because one compound broke down into two simpler compounds. Some of our observations from this reaction was steam and bubbles.

polarity and molecular shape lab

Introduction: In this lab we learned to build models of molecules in order to determine the shape and polarity of each molecule. To find the shape of a molecule you need to look at the number of bonding pairs and the number of un-bonded pairs. You can determine these pairs by looking at the lewis structure of the molecule. From the pairs you can determine the shape of the molecule. To find the polarity of a molecule you need to make a ball/stick model. If the ball/stick model contains exposed positive or negative ends you will know that the molecule is polar. We constructed models of molecules to determine the molecular shape and predict the polarity of each molecule.


Materials: Molecular model kit.
Pencil and paper.
Camera.
Brains.

Procedure: We built a model for each of the molecules listed on the data table on the back of our lab paper. Next we drew the three-dimensional structure of each molecule in Table 1, using solid lines to represent bonds in the plane of the paper, dashed lines for bonds that point back from the plane of the paper, and wedged lines for bonds that point out from the plane of the paper toward the viewer. Then we noted the shape of each molecule in the third column of Table 1, the bond angles in column 4, whether or not they will be polar in column 5, and whether or not they exhibit resonance structure in column 6.

The Ball and Stick Models of Each Molecule:

C3H8:

CO2:

SF6:

IF3:

C2H4:

H2O:

H2O2:

N2:

CH4:

SiH2O:

HF:

SeF4:

SO3-2

 

BF3:

CH3NH2:

Conclusion: This experiment has taught us not only how to build a ball/stick model but also how to identify the shapes and the polarity of molecules. When you draw the lewis structure by itself it is more difficult to determine whether there are exposed positive or negative ends. Also, when you only draw the molecule on paper it is only in 2-D which makes it more difficult to see what the shape of the molecule truly is. Making the ball/stick model allows you to see the shape in 3-D which allows you to see the actual shape rather than try to imagine what it might look like in 3-D. This lab has been successful in teaching us about polarity and the shape of molecules. 

chromatography lab

Introduction:
Chromatography is a technique for separating mixtures. In this lab we compared different solvents' polarity and ability to separate a mixture into its pure components. Before we started this lab we had to know what solvents are used for. In our book we found that solvents are the substances in which the solute dissolves to make the solution. In this lab our solvents consisted of water, methanol, isopropyl alcohol, and hexane. We then took these solvents and used a piece of filter paper and markers to find out which solvent worked the best. Yahoooo!!


Statement of the problem:§ To determine which solvent will work the best to separate the pigments of ink from overhead pens.

Hypothesis:
§ We believe water would be the best solvent because of its strong bond and stickiness caused by its high polarity.

Materials:§ One color of overhead pen to test with different solvents.
§ Four strips of filter paper approximately 1cm. x 8cm.
§ Solvents: water (H2O), methanol (CH3OH), isopropyl alcohol (C3H7OH) and hexane (C6H14).
§ Well plate.

Procedure:Since we used potentially dangerous chemicals, we wore lab aprons and safety goggles at all times. We kept the potent chemicals under the fume hood as much as possible and disposed of them in the sink under the fume hood as well. We began our experiment by taking four of the eight strips of filter paper and made 90 degree creases about 1 ½ cm. away from one end. Then we put multiple black dots across the creases on each strip. We filled four of the wells in the 24 well plate, each with a different solvent. Next we placed one strip with the short end down into each different solvent making sure to keep the ink out of the solvent. After half an hour and keeping close observations, we carefully took the strips out and rinsed out the 24 well plate.
Then, for the second part of the experiment, we took the remaining four strips of filter paper and made creases just like we did in the first part of the experiment. This time, we marked one with green pen, one with red pen, one with orange and one with yellow pen. We put water in all four wells in the 24 well plate and put one strip in each of the wells, waited and made careful observations again for another half an hour. Then we recorded our observations for the first part of and the second part of the experiment.

Results:We found that water is the best solvent due to the fact that it carried the ink along the filter paper much better than any of the other solvents. On our second experiment, we observed that red and yellow are pure substances while green and orange are mixtures because when separated along the paper, it consisted of many different pigments. Bam!

Conclusion:

First of all, our hypothesis was correct. After analyzing our observations, we came to the conclusion that water was the best solvent and then methanol then the isopropyl alcohol and lastly hexane. We learned that water worked the best because it was polar and hexane did not work due to its non-polarity. Yes.