Aluminum & Copper(II) Sulfate Redox Lab

Introduction:
The purpose of this lab is to determine the number of grams of copper that will be produced from an oxidation reduction reaction when you know the mass of Aluminum that reacted with a known amount of copper 2 sulfate pentahydrate and to compare this to the actual yield of copper.

Materials and Setup:

• measured sample of aluminum foil
• measure amount of copper sulfate pentahydrate
• 80 ml beaker
• Bunsen burner
• Striker
  
• stirring rod
• flask
• funnel
• coffee filter
• water
• safety goggles
• safety apron
• Scale
  

1. react the the sample of aluminum foil and copper sulfate pentahydrate in an aqueous medium sized beaker; stirring frequently. filter, dry and weigh the resulting product.
   

Procedure:

1. obtain a medium sized ( 80 ml) beaker
2. add 80 ml of water to the beaker; set up apparatus to heat your mixture over a Bunsen burner to begin heating.
3. Measure out 10 g of Copper 2 Sulfate pentahydrate ( CuSO4. 5 H2O) and record the mass in the data table. Then slowly add the crystals to the heating water.
4. with a glass stirring rod, stir the solution until the copper 2 sulfate pentahydrate is dissolved.
5. While the copper Sulfate Crystals are dissolving one member of the group can go and get the Aluminum Powder. Carefully weigh out the powder to weight between 0.4-0.7 grams. Record the mass exactly into the data table (thousandths place).
6. Carefully add the powder to the hot solution with continuous stirring until it is all in the beaker.
7. Stirring frequently allow the reaction to occur until you can't see anymore aluminum. This will take 15 to 20 minutes so be patient. Once you can't see anymore Aluminum, heat an additional 3 to 4 minutes, then remove from the heat.
   
8. Write your names around the edge of a filter paper (so you can claim it later), weight and record the mass in the data table.
9. Use the filter paper and your funnel to filter the residue in the beaker, catching the filtrate into the Erlenmeyer flask provided.
10. Rinse out your beaker with a small (amount just covering the bottom of the beaker) of water to be sure you obtained all of the product/residue.
11. Remove the filter paper from the funnel and spread it out on a paper towel to dry overnight.
12. Clean and dry the glassware. Be sure the propane is turned of and Bunsen burner disconnected and put away. Straighten up your area.
13. Upon returning the next day, weigh the filter paper and dry residue and record the mass in the data table. Throw the filter paper and residue away.
14. Construct a Data Table with the following parts:
1. Mass of Copper(II) Sulfate Pentahydrate
2. Mass of Aluminum foil
3. Mass of Coffee Filter
4. Mass of dry residue/product + Filter Paper
   

Analysis:

Mass of Copper(II) Sulfate Pentahydrate	10.01g
Mass of Aluminum foil	0.68g
Mass of Coffee Filter	.90g
Mass of dry residue/product + Filter Paper	3.49g

1. Write the balanced reaction equation.
1. 3CuSO4 + 2Al --> 3Cu + Al2(SO4)3
2. Write the Net Ionic Equation.
1. 3CuSO4(s) + 2Al(s) --> 3Cu + Al2(SO4)3(s)
   
3. What is the reducing agent and what is the oxidizing agent?
1. Reducing: Aluminum; Oxidizing: Copper Sulfate Pentahydrate
   
4. Use the Mass of the Aluminum powder and calculate the mass of the copper you theoretically would form.
   
1. 22.1g Al X 1 mol Al/26.98g Al X 1 mol Cu/1 mol Al X 63.55g Cu/1 mol Cu = 78.1g Cu
   
5. Subtract the filter paper from the dry residue/product in the data table below. This is your actual yield of copper.
1. 3.49g
   
6. Calculate the percent yield of the experiment.
   
1. 78.1g Cu = (actual)/4.3 Al2(SO4)3 = 335.83
   
7. Give three reasons of why the amount of copper that should have formed and the amount of copper that actually formed might be different.
1. Different temperatures
2. Different measurements
3. Different materials
   

A Surprising Twist

• Our lab is a prime example of what NOT to do.
• We performed the lab according to everything we were supposed to do, but we missed one minor detail. Written on the board was the measurement of 0.4-0.7 grams of Aluminum, but we were mistaken for we did not see the decimal point.
• At first the lab seemed fine and everything was going according to what was said would be done. But then our solution bubbled to such an exceeding amount that it overflowed the beaker and spilled all along the counter-top.
• We have learned a valuable lesson from this mistake and we would advise that you learn from our mistake as well and not make it yourself. We have learned to properly read directions and properly weigh the material placed in the concoction.

Reaction Lab

Introduction:

• Synthesis: ( Direct combustion) two or more substances combine to form a single product. Example~Hydrogen gas and oxygen gas burn to produce water. 2H2+O2 2H2O

• Decomposition: Substance breaks down into two simpler substances. Example~potassium chlorate when heated comes apart into oxygen gas and potassium chloride. 2 KClO3 2 KCl + 3 O2

• Single Displacement (Replacement): one element fakes place of another in a compound. Example~Aluminum and iron trioxide.

• Double displacement (Replacement): positive and negative of two ionic compounds interchange. Example~calcium chloride and silver nitrate are reacted to form insoluble silver chloride.
  

• Combustion: Rapidly combines with oxygen to form oxides and release energy. Example~Hydrocarbon plus oxygen forms carbon dioxide and water.
  

Purpose:

• The purpose of this lab is to be able to see and identify the 5 chemical reactions rather than just being told what they are and what they do.
  

Materials:

• Safety Goggles
• Safety Apron
  
• 4 small test tubes
• Zinc
• CuSO4
• Ba(No3)2
• Magnesium ribbon
• Bunsen burner
• Striker
• Test tube clasp
• H2O2
• MnO2
• test tube rack

Procedure:

1. Put on Safety Gear.
   
2. Obtain 4 small test tubes.
3. In the first test tube, place a piece of zinc and about 1/2 mL of CuSO4 solution. Record observations.
4. In the second test tube add about 1/2 mL Ba(NO3)2 solution to about 1/2 mL of CuSO4 solution. Record Observations.
   
5. In the third test tube place a piece of magnesium ribbon. Add about 1/2 mL of HCl solution. Record Observations.
6. Light a Bunsen burner (burning propane gas, C3H8). Record observations of the flame.
7. In the fourth test tube add about 2 mL H2O2 solution. Lightly heat it. Record Observations.
   
8. Add a pinch of MnO2 (catalyst) to the H2O2 solution. Lightly heat it. Record observations.

Data:

Observations	Reaction Type	Equation	Evidence of Reaction
Metal changed color, started to dissolve, and sentiment at the bottom of the test tube.	Single Displacement	Zn + CuSO4 => Cu + ZnSO4	Formation of Copper
Temperature stayed the same, color combined, and a lot of bubbles.	Double Displacement	Ba(NO3) 2 + CuSO4 => Cu(NO3) 2 + BaSO4	Color/ Precipitate
Sizzling sound, dancing metal, a lot of bubbles, high temperature, and precipitate in the test tube	Single Displacement	Mg + 2HCl => H2 + MgCl2	Hydrogen Gas
Blue at the bottom and orange on the top	Combustion	C3H8 + 5O2 => 3CO2 + 4H2O	Flame
Lots of bubbles, foggy [Picture Unavailable]	Decomposition	2H2O2 => 2H2O + O2	Splint
Lots of bubbles, molecules rising, still bubbling after heating, smell, smoky, light grey. Picture Unavailable]	Decomposition	2H2O2 + MnO2=> 2H2O + MnO2 + O2	Splint

Discussion:
Through this entire experiment we are more capable of understanding the reactions, for we can now connect it with a visual presentation. This caused us to think more logically as we had to come up with our own equations and decided with reaction was being presented.

Conclusion:
Through each individual experiment we saw one of the 5 reactions and evidence of these reactions. We were capable of identifying the reactions from this evidence. The reaction descriptions, type, equation, and evidence of the reaction were placed in a data table for a presentation of the information that is more pleasing to the eye.

Polarity and Molecule Shape Lab

Statement of the Problem:

• Construct Models of Molecules
• Determine Molecule Shapes
• Predict Polarity of Molecules

1. Background Information:
• The most common type of bond between two atoms is a covalent bond. A covalent bond is formed when two atoms share a pair of electrons. If both atoms have the same electronegativity or tendency to attract electrons the bond is nonpolar covalent. When atoms have different electronegativity the electrons are attracted to the atom with the higher electronegativity. The bond that forms is a polar covalent
• Molecules made up of covalently bonded atoms may themselves be polar or nonpolar. If the polar bonds are symmetrical around the central atom, the bonds offset each other and the molecule is nonpolar. If the polar bonds are not symmetrical, the electrons will be pulled to one end of the molecules and the molecule will be polar. (If the polar end is "sticking out" it will be a polar molecule)
• Many physical properties of matter are the result of the shape and polarity of molecules. Water, for example, has unusual properties that can be explained by the shape of its molecule and the distribution of charge on the molecule.

Hypothesis:

• The polarity of molecules that we build will show nonpolar/polar characteristics depending on electronegativity.

Materials:

• Molecule Model Kit

Procedure:

1. Build a model for each of the molecules listed on the date table on the back of the page given us.
2. Draw the 3D structure of each molecule in table 1. Use solid lines to represent bonds in the plane of the paper, dashed lines for bonds that point back from the plane on the paper, and wedged lines for bonds that point out from the plane of the paper toward the viewer.
3. Note the shape of each molecule in 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.

Results:

Lewis Structure- -Model- -Shape- -Bond Angle- -Polarity- -Resonance-

CH4

-[Model Unavailable]- -Tetrahedral- 109.5 degrees- -No- -No-

BF3

-Pyramidal- -90-109.5 degrees- -Yes- -No-

C3H8

-Tetrahedral- -109.5 degrees- -No- -No-

H20

-Angular- -90-109.5 degrees- -Yes- -No-

Si2H6

-Tetrahedral- -109.5 degrees- -No- -No-

HF

-Linear- -180 degrees- -Yes- -No-

CH3NH2

-Terahedral/pyramidal- -109.5/90-109.5 degrees- -No/Yes- -No/No

H202

-Angular- -90/109.5 degrees- -Yes- -No-

N2

-Linear- -180 degrees- -No- -No-

SeF4

-Tetrahedral- -109.5 degrees- -Yes- -No-

C2H4

-Pyramidal- -90/109.5 degrees- -No- -No-

SiH20

-Trigonal Planar- -120 degrees- -Yes- -No-

IF3

-Pyramidal- -90/109.5 degrees- -Yes- -No-

SF6

-Square Pyramidal- -90 degrees- -Yes- -No-

CO2

-Linear- -180 degrees- -Yes- -No-

SO3 [-2]

-SeeSaw- -120/90 degrees- -Yes- -Yes-


Conclusion:

• Each atom followed the octet rule and the shapes of the adjacent molecules corresponded with this action.
  

Questions and Answers:

Q- Explain how water's Shape causes it to be polar.
A- The negative and the positive elements are on the opposite ends of the molecule thus causing it to be polar.

Q- Describe how water's properties would be different if the molecules were linear instead of bent.
A- It would "just float away" as Mr. Howell said, respectively.

Q- Based on the results of this experiment, list the molecules that would be water-soluble.
A- H20, HF, C2H4

Chromatography

Introduction:
-Paper Chromatography is an important separation technique that depends on difficulties in both absorption and solubility. A small amount of the mixture to be separated is placed close to the edge of a piece of chromatography paper. The edge of the paper is then immersed in a developing solution. As the developing solution ascends up the paper by capillary action, the components of the sample are carried along at different rates. Each component of the mixture will move a definite distance on the paper in proportion to the distance that the solvents move. Retention factors values are dependent upon the paper, developing solution, the sample size.

Statement of problem:
-Which solvent will move quicker than the other solvents
-Which inks are a combination of different colors.
background info:
-Paper chromatography is an important separation technique that depends on differences in both absorption and solubility. Each component of the mixture will move definite distance on the paper on proportion to the distance that the solvent moves. this ratio: (retention factor)=distance component moves/ distance solution moves, can be calculated for each component, to aid in identification. retentions factor values are dependent upon paper, developing solution, and sample size.

Hypothesis:
-the solute with less elements will reach the distance the components move faster and will separate the pigments in the ink more efficiently.

Materials:
-H2O, CH3OH, C3H7OH, C6H14
-mixtures to separate: water-soluble overhead pens ( part 1-black), (part 2-red, green, purple, blue)
-Chromatography paper strips(10)-1cm*8cm
-well plate

Procedure:
-To start off this lab we obtained our safety equipment, which included safety goggles and safety aprons. We then gained all of our materials for part 1 listed above, filled the well plate with the different solutes and folded and marked the chromatography paper and marked 3 black dots. We then placed the 4 strips of paper marked accordingly to solutes names and placed them in the correct well plates, then placing the well plate under the fume hood we waited. We recorded the results from the following procedure.
-Then in part 2 we picked out the four colors: red, green, blue and purple and picked our solution as being H2O, we then repeated part one but with the different colors in the place of the black. We allowed the solvent to wick up the paper for approx. 1/2 hour and noted the information.

Results:
-We did the first part and our data we collected were as follows: After some time the solute containing H2O reached its mark and began to separate the pigments from the black ink, the other solutes soon followed in consecutive order as followed: H2O, CH3OH, C6H14,C3H7OH.

H2O	CH3OH	C3H7OH	C6H14
Reached dots approx. 5min	Reached dot approx. 7 min	Did not reach dots in time allotted	Reached dot approx.15 min

- In part 2 we observed the following: The blue stayed blue and ran up the paper, the red went from orange to dark red as it went up the paper strip, the green went from dark green to light blue as it progressed. Green as it progressed further then changed colors again to yellow to green.

Conclusion:

-We accept our hypothesis; it was accurate.
-Our hypothesis stated the H2O would move up the chromatography paper the fastest because there are less elements in it. This was supported by our data showing the H2O moving up the paper in the fastest amount of time.
- The fastest solute to the slowest are as follows: H2O, Ch3OH, C3H7Oh,C6H14
- In this lab we learned that the water-soluble ink in the pens are made with many different colors and that H2O or water is the best solvent. We can use this information in every day situation such as when you are doing laundry or trying to mix certain colors to get your desired color.
- Errors that could occur are as follows: The Hexane evaporates at a fast rate and this lab takes a lot of patience and time.

Question & Answers:

Q: What is the mobile phase in paper chromatography investigation?
A: The Solute

Q: What is the stationary phase in a paper chromatography investigation?
A: The ink in the Water-soluble pens

Q: What is meant by the tern retention time:
A: The amount of time it takes for the solute to completely separate all the colors of the ink.

Q: Would a polar molecule or a non polar molecule be a better mobile phase for a chromatography experiment? Why?
A: non-polar because the atoms are equal and there fore will move quicker on paper and are not being dragged by the larger end of a polar molecule.

Q: Rank the solvents as to which produced the best separation to your ink.
A: H2O, CH3OH, C3H7OH, C6H14

Q: Explain in terms of polarity why some solvents worked better than other did on your ink?
A: The polarity of some molecules was larger then the polarity in other molecules thus the speed were different.

Q: Is the black overhead pen a pure substance or a mixture of polar molecules? give evidence
A: it is a mixture because it did not stay black instead it separated into many different colors.

Q: Why wouldn't C6H14 be an appropriate solvent for part 2?
A: We would not get efficient data in the time allotted it moves to slowly.

Q: Which color inks would you classify as pure substances and which were mixtures?
A: The colors that stayed the same color but only turned lighter were pure substances, while the other ones that changed their color all together are mixtures.

Q: What is Chromatography?
A: A technique we use for separating components of a mixture by placing the mixture in a mobile phase that is passed as a station phase.