Chemistry

Chemistry
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HS Chromatography Lab

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Thursday, May 20, 2010

Copper Lab




Copper Lab


Introduction
In this lab we will be mixing aluminum powder with a mixture of copper sulfate and water. The purpose of this lab is to measure out different quantities of chemicals and along with there molar mass incorporate those numbers into equations that will give us info such as the limiting reactant, percent yield, and theoretical yield. With this lab you are also given the opportunity to observe multiple chemical reactions.

Supplies


Copper Sulfate Pentahydrate (CuSO4)
Aluminum Powder (Al)
Medium Sized Beaker
Glass Stirring Rod
Periodic Table
Paper Towels
Asbestos Pad
Paper Filter
Appron
Goggles
Stand
Scale

Flask

Safety
For this lab there are should be more safety standards in place than previous labs. The chemicals when mixed together can and are very reactive when heated. Due to this you need to wear aprons, goggles, and it is highly recommended to not wear open toed shoes due to the high possibility of an overflow of the hot contents. Keep the chemicals under the hood vent at all times, follow other instructions, and don't be stupid.


Procedure
We began our lab by aquiring all of our equipment and dressing out in the proper gear. We then measured out 73 grams of water in our beaker along with 10 grams of CuSO4. We accidentally measured out 4.12 grams of aluminum so we will later alalyze how this effects our final results. We heated our water until it reached a slow boil then slowly added the CuSO4 while stirring with our glass rod. Afterwards we turned off the bunsen burner so as to keep from reaching a rapped boil and added the aluminum still stirring thoroughly. We then waited and observed the reactions taking place including spontaneous heat, copper chunks rising to the surface, and the change in color of the liquid. After it was done we seperated the copper chunks from the liquid through a paper funnel. The liquid that ended up in the bottom of the beaker after the seperation was perfectly clear due to the large amount of aluminum used in the experiment.




Equations
Balanced Equation
2Al + 3CuSO4 -> 2Al3(SO4) + 3Cu

Limiting Reactant
4.12g. Al X 1mol Al/ 26.98 g. Al= .153 mol Al
10g. CuSO4 X 1mol CuSO4/ 111.66 g. CuSO4= .09mol CuSO4
.09mol CuSO4X 2mol Al/3mol CuSO4= .06mol Al needed
Aluminum is the limiting reactant...

Grams of Copper
4.12g Al X 1 mol Al/26.98g Al X 1 mol Cu/1 mol Al X 63.55g Cu/1 mol Cu=9.7g. Cu

Theoretical Yield4.12 g. Al X 1mol Al/ 26.98g. Al X 3mol Cu/ 2mol Al X 63.6g. Cu/ 1mol Cu= 14.57 g. Cu

Percent Yield9.7g. actual/ 14.57g. theoretical X 100= 67%

Mole Ration
Al 2:3 Cu
Molar MassAl= 26.98gCu= 63.46gCuSO4=111.66g





Observations
- White Foam - Spontaneous Heat - Formation of a Solid (Copper) - Constant Bubbles - Change in Color of Water



Conclusion
From our equations and observations during the lab we came to some interesting conclusions. We found our limiting reactant to be Aluminum along with our percent yield being 67%. We used a large amount of Aluminum which did change the out come of our experiment. Our theoretical yield came out to be 14.57 g. of copper. There were also a large number of chemical reactions taking place. After the bunson burner was turned off the liquid continued to heat up, a thin foam formed at the top of the liquid, there was a formation of a solid(Cu), and there was constant bubbling from the copper located at the bottom of the beaker.






Thursday, March 18, 2010

Chemical Reactions



Introduction/ Background


The purpose of this lab is to observe the different reaction types and the macroscopic events they caused. These reactions include decomposition, synthesis, single replacement, double replacement, and combustion. Synthesis is where two or more substances combine to form a single product, in a decomposition reaction a compound breaks down to form two or more simpler substances, in single replacement one element takes the place of another compound metal takes place of a metal, in double replacement the positive and negative portions of two ionic compounds interchange two metals are replaced by two metals, and in a combustion reaction a substance rapidly reacts with oxygen to form one or more oxides and releases energy. Through this experiment we were able to observe three of these reactions which we will identify further on.












We began by obtaining 3 test tubes, a test tube rack, a test tube holder, a bunsen burner, and other needed the materials. We then put on our safety equipment and began the experiment!


We added approximately 1/2 ml. of CuSO4 to one of the beakers. A small amount of Zinc was added to the solution and was closely observed. Bubbles were almost immediately visible and that the zinc was changing a dark rust color and later on seemed to begin breaking down. We tested for hydrogen gas and were assured none was being released. This leads us to assuming that this reaction represents the equation Zn+CUSO4 -> ZNSO4+CU and is a single displacement reaction type. In this it is obvious that the SO4 stays in place from start to finish with the only change being the switch between the copper and zinc ions and referring back to the top a single displacement reaction is when one element takes the place of another which is what we see in this reaction.

For the next mixture we will be using approximately 1/2 ml. Ba(NO3)2 and 1/2 mL of CuSO4. After mixing them together we noticed immediately the formation of a white foam separating itself from the darker blue solution. Our hypothesis was that the equation of this reaction is CuSO4 + Ba(NO3)2 --> BaSO4 + Cu(NO3)2 and the reaction type is the double displacement reaction type. As you can see the Cu and Ba ions replace each other on the SO4 and (NO3)2 ions. Double displacement is two ionic compounds interchanging two metals thus our reasoning.



In the third experiment we placed some magnesium in about 1/2 mL of HCL solution. This proved to be a very interesting mixture and immediately bubbles were noticeable. We tested for hydrogen gas and
found that there was indeed hydrogen gas being released from the still active bubbles. The most fascinating part of this experiment was the heat produced by the mixture and was the major source of our hypothesis. The equation for this is Mg2+HCl= Mg+HCl2 and is a combustion reaction due to the energy produced creating the heat we observed.

The last experiment we conducted used H2O2 mixed with MnO2. As soon as the two were mixed the otherwise clear H2O2 turned to gray then to a
black color within a few seconds. The bubbles were very intense and if heated to quickly it would bubble out of control as some individuals discovered. The chemical equation of this is 4H2O2+MnO2 → 2O2 + 4H2O+MnO2 and we know this to be a Decomposition reaction. As you can see on the right side of the equation H2O2 has lost one oxygen creating H2O and producing O2. Just as the decomposition reaction is defined as a compound breaking down to form two or more simple compounds so does H2O2 break down to form the simpler compounds O2 and H2O.



Conclusion

As you can see each of the five reaction types has specific characteristics in what happens when one mixes two compounds together. This is what allowed us to determine what type of reactions each one was and what its chemical equation turned out to be.



Please click for clearer image of the data table.



Monday, March 1, 2010

Molecular Shapes Lab

Molecular Shapes and Polarity

Problem Statement:

Determine Molecule Shapes

Construct Models of Molecules

Predict Polarity of Molecules

1. Background Information:

    • The most common type of bond between two atoms is a covalent bond which is formed when two atoms share a pair of electrons. This bond will be nonpolar covalent if both atoms have the same electronegativity.On the other hand if these atoms have different electronegativities the bond between them is considered to be a polar covalent bond.
    • Molecules made up of covalently bonded atoms may be polar or nonpolar. If the polar bonds are symmetrical around the central atom, the bonds are offest by each other thus making the molecule nonpolar. If the polar bonds are not symmetrical, the electrons will be pulled to one end of the molecule making it polar.
    • 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

This experiment will produce results showing nonpolar or polar makeups depending on each individual atom’s electronegativity.

Materials

Molecule Model Kit

Lab Procedures

Draw the 3D structure of each molecule in table

Obtain needed materials.

Build a model for each of the molecules listed in the lab table

Take note of each model and list its shape, bond angle, Polarity, and resonance.



C3H8
-Tetrahedral- -109.5 degrees- -No- -No-

H20
-Angular- -90-109.5 degrees- -Yes- -No-



HF
-Linear- -180 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-

The shapes of the adjacent molecules corresponded with the octet rule which is fulfilled in each molecular shape.

Conclusion


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

A- C2H4, H20, HF

Q- Explain how water's Shape causes it to be polar.
A- Its polarity is caused by the negative and positive being on different ends of the molecule.

Q- Describe how water's properties would be different if the molecules were linear instead of bent.
A- It would simply float away like a gas.