molecular structure

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Molecular Structure Lab

Objective:
For this experiment we took two different molecule and virtually dissected them finding everything about them including: bond length, bond angles, the charge on each atom, the non bonded distances between atoms and the energy difference between the highest and lowest molecular orbital.
Procedure:
The procedure is to use HyperChem Lite to get the information needed from each molecule. But explained in full on page 16 and 17 in the lab manual.
Data:
The following data was found about CH4 (methane).
• The bond length from Carbon to all four Hydrogen atom was the same measurement that came to be about 1.113Å
• Not only are the lengths between all the Hydrogen atoms to the Carbon atoms the same so are the angles between the hydrogen atoms, all coming out to be about 109.47°.
• The non-bonded distances between all the parts of the molecule are 1.818Å.
• The charge on the Carbon atom is -.143
• All of the hydrogen atoms have the same charge being .036
The following data was collected by HyperChem Lite for C2H5OH (ethanol).
• All the distances between bonded Carbon and Hydrogen atoms were 1.115Å.
• The distance of the bonded oxygen (3) to hydrogen (9) was .942Å.
• The distance between bonded Carbon (1) to Oxygen (3) was 1.40Å.
• Bond angle 5,1,4 was 109.8°
• Bond angle 3,1,2 was 108.9°
• Bond angle 6,2,8 was 107.9°
• The smallest non-bonded distance was between atoms 9 (hydrogen) and 6 (hydrogen) it was 4.067Å.
• The largest distance between two atoms was from 6 to 8 1.802Å.
• The charge on the Carbon atoms were: Carbon 1=.466, and Carbon 2=-.076
• Hydrogen charges were as follows: Hydrogen’s 4,5=-.004, 6=.021, 7,8=.020, and the charge of hydrogen 9=.447
• And lastly the charge on the oxygen atom was -.891
Calculations:
Formulas used in the lab…
E = hv = hc/
1 ev = 1.6022E-19 Joules
E is the energy difference between the highest occupied and lowest unoccupied molecular orbital (HOMO and LUMO). h = Plank’s constant = 6.63E-34 Js c = speed of light = 3.00E8 m/s formal charge = number of valence electrons – (# of unshared electrons + # of bonds)
Volume of an ellipsoid is V = /6 (a + 1Å)(b+1Å)(c+1Å) = xÅ3
A,b,and c are the skeletal lengths of the molecule.
CH4
E = (-15.51ev)-(-24.49ev) = 8.98ev = 1.44E-18 J
 = hc/E = (6.63E-34 Js)(3.00E8 m/s)/(1.44E-18 J) = 1.38E-7 m = 138nm
The wavelength of CH4 is 138nm and this resides in the ultra-violet spectrum and is not visible to the naked eye.
Formal charge of Carbon = 4-(0+4) = 0
Formal Charge of Hydrogen’s = 1-(0+1) = 0

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