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Sebacoyl Chloride And Hexamethylenediamine Synthesis

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E. Discussion: In order to synthesize the polymer, Nylon 6,10, we had to complete a few steps to create the chemical reaction that combined sebacoyl chloride and hexamethylenediamine. First we measured the mass of the two graduated cylinders when they were empty, and measured it again after they were filled with sebacoyl chloride and hexamethylenediamine. We did this in order to find the measurements of the reactants. When we measured the graduated cylinder when they were emptied, one weighed at 10.99 grams while the other weighed at 10.94 grams. Even though they were the same kind of cylinder, I believe that a systematic error might have caused the second cylinder’s weight to be slightly affected, causing the weight to be lowered by 0.05 grams. After this, we added sebacoyl chloride and hexamethylenediamine into those cylinders and …show more content…

The cylinder with the sebacoyl chloride was 27.14 grams and the cylinder with hexamethylenediamine was 36.14 grams. We then calculated the mass of the reactants, which is found when we found the difference of the weight of the cylinders before and after sebacoyl chloride and hexamethylenediamine were added. The total mass of the reactants was equal to 41.35 grams. After we calculated these results, we started to create the chemical reaction. We put the sebacoyl chloride and the hexamethylenediamine in separate beakers, but then slowly added the hexamethylenediamine to the beaker with the sebacoyl chloride in it. As soon as we did this, a small, clear, gelatin-like clump, formed in the beaker as a result of the chemical reaction between the monomers, sebacoyl chloride and hexamethylenediamine. Next, we kept pulling out pieces of the Nylon strand with the tweezers and wrapping it around the glass rod. It was a little difficult to form strands around the glass rod because when we kept trying to grab the nylon with the tweezers, it kept breaking a part. However, we were eventually able to make the strands after trying

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(0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube

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