Experiment Melting Point Experiment

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When performing the experiment in test tubes A, B, and C different components of various chemicals were placed in each test tube and one was placed in the light and one in the dark. Test tube A containing DCM, dimethyl maleate, bromine, and being placed in the light. Test tube B contained DCM and dimethyl maleate which was the control substance and what test tube A and C were compared to. Test tube C contained DCM, dimethyl maleate, bromine, and was placed in the dark. After the completion of the experiment, test tube A contained a solid and the solvent was a light-yellow tint. Test tube B contained no solid and was a clear liquid. Finally, test tube C did not contain a solid and the solvent was a deep red-orange color. Therefore, when the…show more content…
The melting point obtained by doing the experiment was 82 degrees Celsius. The melting point is the temperature at which the solid melts and turns into a liquid. The melting point helps determine the compound if given a list of values to compare it to and it also helps determine if a compound is pure or not. Impure solids melt at lower temperatures while pure solids melt at higher temperatures. A possible source of error is possibly having impurities that remained in the crystalline solid after filtering. This could have been due to the fact that it was not washed enough with DCM to get rid of all of the impurities. Another potential source of error was not having enough crystalline solid to test for in the capillary tubes which gave an inaccurate…show more content…
This was calculated by taking 0.5 mL of bromine multiplied by the density of dimethyl maleate which is 1.15 g/mL to give 0.575 g, the actual value. The ratio to dimethyl maleate to dimethyl fumarate is 1:1. Therefore, it is 0.575 g for both of them. To calculate the percent yield, the grams of precipitate, 0.014 g, was divided by 0.575 g and then multiplied by 100. The percent yield is the actual amount compared to the experimental amount and it explains how much of the product was recovered through the filtering process. The percent yield obtained from performing the experiment was not good compared to the value of 60%, a good percent yield. The source of loss of the product was probably due to not having an accurate ratio between the reactants of bromine and the dimethyl maleate which would have affected how much crystalline solid would have formed. Another source of error could have been some of the crystalline solid being washed away and escaping through the

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