Friday, February 19, 2010

Polarity and Molecular Shape Lab

Statement of the problem:

We have been talking in class about how to arrange atoms in Lewis Structures. Many people just disregard the electron pairs because they think they don't amount to anything. I think this is bogus. But with that in mind we searched for the real answer, How are molecular shapes determined?

Hypothesis:

We predict that unbonded pairs of electrons determine the shape of the molecule.

Materials:

  • Molecular Model kit
  • Polarity and Molecular Shape LAB Worksheet
  • 6-3 worksheet
Procedure:

  1. We built a model out of the molecules listed on the data table on the back of the Polarity and Molecular Shape LAB Worksheet. (CH4, BF3, C3H8, H2O, Si2H6, HF, CH3NH2, H2O2, N2, SeF4, C2H4, SiH2O, IF3, SF6, CO2, and SO3(-2))
  2. Next, we drew all the lewis structures of the molecules.
  3. Then, after we wrote down the structure the molecule was.
  4. Next, we stated the bonds and if it was polar or non-polar
ANALYSIS:
  1. Explain how water's shape causes it to be polar.
  2. Describe how water's properties would be different if the molecules were linear instead of bent.
  3. Based on the results of this experiment.
List the molecules from the experiment that would be water- soluble.
1. CH4 Tetrahedron 109 degrees non polar No resonance structure.

2. BF3 Triangular Planar 130.7 degrees Polar No resonance structure.

3. C3H8 Octahedral unknown degree non polar No resonance structure.

4. H2O Linear 104.5 degrees non polar No resonance structure.

5. Si2H6 Octahedral 130 degrees non polar With a resonance structure.

6. HF Linear 180 degrees Polar and No resonance structure.

7. CH3NH2 Triangular bi pyramid 150 degrees non polar With a resonance structure.

8. H2O2 Angular bent unknown degree Polar and With a resonance structure.

9. N2 Linear 180 degrees non polar With a resonance structure.

10. SeF4 Tetrahedron 90 degrees non polar with No resonance structure.

11. C2H4 120 degrees non polar With resonance structure.

12. SiH2O Triangular Planar Polar With resonance structure.

13. IF3 Triangular Planar 86 degree non polar with No resonance structure.

14. SF6 Octahedral 90 degree non polar with No resonance structure.

15. CO2 Linear 180 degree non polar with No resonance structure.

16. SO3 Triangular Planar 90 degree Polar with No resonance structure.

1. CH4-

2. BF3-

3.C3H8-

4.H2O-

5.Si2H6-



6.HF-

7.CH3NH2-

8.H2O2-

9.N2-

10.SeF4-

11.C2H4-

12.SiH2O-


13.IF3-


14.SF6-


15.CO2-



16.SO3-2-

Thursday, February 4, 2010

Chromatography Lab


EXTREME COLOR SEPARATION


Introduction:

The Chromatography Lab is a way in which we can examine how different solvents separate a mixture into their pure components. The mixture that will be used in the lab are markers of different color that will be marked onto chromatography paper. The solvents will then separate the markers into different colors onto the chromatography paper. After the solvents separate the mixtures into their pure components, we can then examine and analyze our results and come up with a conclusion of which solvent is the best at separating mixtures.

Statement of the Problem:

What is the best solvent in separating a mixture into its pure components?
We will be using these solvents:
Water (H2O), Methanol (CH3OH), Proponol (C3H7OH), Hexane (C6H14)


Hypothesis:

We believe that water (H2O) is the best solvent for separating mixtures into their pure components.

Materials:

  • Goggles
  • Aprons
  • 24 well plate
  • Chromatography paper
  • Water (H2O),
  • Methanol (CH3OH)
  • Proponol (C3H7OH)
  • Hexane (C6H14)
  • Markers (Black, red, blue, green, yellow)

Keep goggles on at all times while working with solvents and keep solvents under fume hood.

Procedure:

First we got 4 strips of chromatography paper and labeled each one according to the solvent it was in. We then bent each strip 1.5 cm. at one end of the paper. We marked the creases of the paper with a pencil and then we put 2 dots of black marker ink. We filled 4 of the wells in our 24 well plate with these solvents:Water (H2O), Methanol (CH3OH), Proponol (C3H7OH), and Hexane (C6H14). We placed a strip of chromatography paper in each of the filled wells and examined them as they separated the black marker ink into its pure components. We took notes and wrote down our results. We repeated this procedure with the colors: blue, green, yellow, and red. Also we only used the solvent water in each of the 4 wells for it had the greatest ability to separate a mixture into its pure components. After examining the results we answered the questions on our Chromatography lab papers.

Results (Data):

The results of the lab in order from greatest ability to separate a mixture into its pure components to the least are: Water (H2O), Methanol (CH3OH), Proponol (C3H7OH), Hexane (C6H14). After our experiment we can clearly conclude that Water (H2O) is the best solvent for separating a mixture into their pure components. As you can see on the left side of the photo below, water went all the way across the chromatography paper separating the mixture very well.








Conclusions:

After examining our results we can conclude that our hypothesis was correct and Water (H2O) has the greatest ability to separate a mixture into its pure components. The results of the lab in order from greatest ability to separate a mixture into its pue components to the least are: Water (H2O), Methanol (CH3OH), Proponol (C3H7OH), Hexane (C6H14). I have learned that water has a great ability for separating mixtures into their pure components and Hexane (C6H14) has a poor ability for separating mixtures into their pure components.

1. The solvents that produced the best separation of ink to the least are: Water (H2O), Methanol (CH3OH), Proponol (C3H7OH), Hexane (C6H14)

2. Some solvents worked better than others did on our ink because they are less dense.

3. The ink in the black overhead pen is a mixture of polar molecules, because as the time goes by the different colors appear down the chromatography paper.

4. Hexane (C6H14) would not be an appropriate solvent choice, because it does not separate pigments very well.

5. All colors should be classified as mixtures, because the all had colors in them.

6. Chromatography is a technique for seperating components of a mixture by placing the mixture in a mobile phase that is passed over a stationary phase.