Thursday, March 18, 2010

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.

Friday, February 19, 2010

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

Thursday, February 4, 2010

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.