2lab6 - JessicaGaboriault 2/6/08 LabPartnerJackieLerea...

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Jessica Gaboriault 2/6/08 Lab Partner- Jackie Lerea Section A Tuesday @ 2 pm Hutch 115 Experiment 6: Infrared Spectroscopy and the covalent bond week 1 Purpose :  The purpose of lab for this week is demonstrating Hooke’s law by measuring how far 3  different springs extend when various masses are suspended from each spring.   We will then  find the force constant using Hooke’s law. Also, we will be measuring the strength of the bonding  forces between atoms and evaluate the structure of the complex molecules. We want to then find  the stretch of the C-D bond and the C-D bend of CDCl 3 . Procedure : see pages 51-59 in the lab manual. Lab data prior to lab: I. Hooke’s law   3 springs with varying stiffness a. Loose – 20-200 g b. Medium – up to 250 g c. Stiff – 100-300 g d. Use reasonable increments so you can graph force vs. extension II. IR a. With single bonds (CHCl 3 , CDCl 3 ) b. With multiple bonds (m-hexane, 1-hexene, 1-hexyne) c. Effects of resonance on stretching frequencies d. CH 3 COOCH 3 , CH 3 COONa, CH 3 CONHCH 3 e. 8 photocopies for 8 compounds (liquids, solids)
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Data and observations : Loose Spring weight (g) Extensio n Started at 49.0 cm difference (cm) 20 45.5 3.5 40 41.5 7.5 60 38 11 80 34.5 14.5 100 30.5 18.5 120 27 22 140 23 26 160 19.5 29.5 180 16 33 200 12 37 Medium Spring weight(g ) extensio n started at 49.5 cm difference (cm) 25 49 0.5 50 46 3.5 75 43 6.5 100 40.25 9.25 125 37.5 12 150 34.5 15 175 31.5 18 200 28.5 21 225 26 23.5 250 23 26.5 Stiff Spring weight(g ) extensio n started at 49.5 cm difference (cm) 100 49.2 0.3 120 49 0.5
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140 48.7 0.8 160 48.5 1 180 48.2 1.3 200 47.9 1.6 220 47.7 1.8 240 47.4 2.1 260 47.2 2.3 280 46.9 2.6 300 46.6 2.9 Calculations : PART A Sample for the calculation of difference from orginal: Original loose spring (49.0 cm) – distance of new weight = difference for loose spring 49.0 cm – 45.5 cm = 3.5 cm 3.5 cm/100 cm/m =  0.035 m Sample for the calculation of constant k F = ma = kx Loose spring (0.02 kg)(9.8 m/s 2 ) = k(0.035) K = 5.6 LOOSE SPRING weight  (kg) Extensio n difference  (cm) cm diff/100 =  m applied force= constant  k mass(kg)*9.8m/s ^2 0 49 0 0 0 0 0.02 45.5 3.5 0.035 0.196 5.6 0.04 41.5 7.5 0.075 0.392 5.227 0.06 38 11 0.11 0.588 5.345 0.08 34.5 14.5 0.145 0.784 5.407
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0.1 30.5 18.5 0.185 0.98 5.297 0.12 27 22 0.22 1.176 5.345 0.14 23 26 0.26 1.372 5.287 0.16 19.5 29.5 0.295 1.568 5.315 0.18 16 33 0.33 1.862 5.642 0.2 12 37 0.37 1.96 5.297 MEDIUM SPRING applied force = weight(kg ) extensio n difference (cm) cm diff/100 = m mass(kg)* 9.8 m/
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2lab6 - JessicaGaboriault 2/6/08 LabPartnerJackieLerea...

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