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 Hyper Chem 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 CH 4 (methane). o The bond length from Carbon to all four Hydrogen atom was the same measurement that came to be about 1.
113 Ao 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. 47^0. o The non-bonded distances between all the parts of the molecule are 1. 818 A. o The charge on the Carbon atom is -. 143 o All of the hydrogen atoms have the same charge being.
036 The following data was collected by Hyper Chem Lite for C 2 H 5 OH (ethanol). o All the distances between bonded Carbon and Hydrogen atoms were 1. 115 A. o The distance of the bonded oxygen (3) to hydrogen (9) was. 942 A. o The distance between bonded Carbon (1) to Oxygen (3) was 1.
40 A. o Bond angle 5, 1, 4 was 109. 8^0 o Bond angle 3, 1, 2 was 108. 9^0 o Bond angle 6, 2, 8 was 107.
9^0 o The smallest non-bonded distance was between atoms 9 (hydrogen) and 6 (hydrogen) it was 4. 067 A. o The largest distance between two atoms was from 6 to 8 1. 802 A. o The charge on the Carbon atoms were: Carbon 1 = . 466, and Carbon 2 = -.
076 o Hydrogen charges were as follows: Hydrogen’s 4, 5 = -. 004, 6 = . 021, 7, 8 = . 020, and the charge of hydrogen 9 = . 447 o And lastly the charge on the oxygen atom was -. 891 Calculations: Formulas used in the lab…
? E = hv = hc/? 1 ev = 1. 6022 E-19 Joules? E is the energy difference between the highest occupied and lowest unoccupied molecular orbital (HOMO and LUMP). h = Plank’s constant = 6. 63 E-34 Jsc = speed of light = 3. 00 E 8 m / s formal charge = number of valence electrons – (# of unshared electrons + # of bonds) Volume of an ellipsoid is V = ? /6 (a + 1 A) (b+1 A) (c+1 A) = xA 3 A, b, and c are the skeletal lengths of the molecule. CH 4? E = (-15.
51 ev) – (-24. 49 ev) = 8. 98 ev = 1. 44 E-18 J? = hc/? E = (6. 63 E-34 Js) (3. 00 E 8 m / s ) / (1.
44 E-18 J) = 1. 38 E-7 m = 138 nmThe wavelength of CH 4 is 138 nm 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 These charges (formal) found above are not equal but similar to the charges found by Hyper Chem Lite using Huckel’s theory. Huckel’s numbers are much closer to the actual charge of each atom. Volume of methane = 29 A 3 Cooking gas is one of the main uses of methane. C 2 H 5 OH? E = (-13.
93 ev) – (-33. 94 ev) = 20. 03 ev = 3. 209 E-18 J? = hc/? E = (6. 63 E-34 Js) (3. 00 E 8 m / s ) / (3.
209 E-18 J) = 6. 198 E-8 m = 62 nmThe wavelength of C 2 H 5 OH is 62 nm and resides in the ultra-violet spectrum and is not visible to the naked eye. Formal charge of Carbon atoms = 4 – (0+4) = 0 Formal charge of Oxygen atom = 6 – (4+2) = 0 Formal charge of all 6 Hydrogen atoms = 1 – (0+1) = 0 These charges found above are not equal to the actual Huckel charges but similar to. Volume of ethanol = ? /6 (4. 067 A+1 A) (2. 241 A+1 A) (3.
131 A+1 A) = 35. 52 A 3 Where a is atoms 9 to 6, b is atoms 5 to 9, and c is atoms 4 to 8 Ethanol is the active ingredient in all alcoholic beverages. Conclusions: In this experiment we dissected methane and ethanol and found everything from there bond lengths to there uses, and along the way found there charges, angles, and volumes. This was a good lab because I’m sure being a Pharmacy major that I am I will be using this program in Organic chemistry next year. Hyper Chem Lite was a quick was to find out valuable information about molecules.