# The theoretical yield value was used to determine the

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The theoretical yield value was used to determine the percent yield by substituting it into Equation 3. This was completed as follows: Percent Yield = Weight of aspirin x 100 Theoretical Yield of Aspirin Percent Yield = 0.5895 g x 100 3.927 g Percent Yield = 0.1788 x 100 = 17.88% After the percent yield was determined, a melting point was needed to establish the level of impurities in the synthesized aspirin. The literature melting point, with no impurities, was 138- 140°C. The measured melting point of the synthesized aspirin in the experiment was slightly 4
lower at 132-135°C. Also, as another way to acknowledge impurities in the product, 1% FeCl 3 was added to salicylic acid and the synthesized aspirin. When the 1% FeCl 3 was added to the salicylic acid, the product turned a dark purple. When the 1% FeCl 3 was added to the synthesized aspirin, it turned yellow-orange in color. Both of these tests, the melting point determination and the addition of 1% FeCl 3 , served to identify impurities in the synthesized aspirin product. The melting point uncovered impurities in the synthesized aspirin because it was slightly lower than the literature melting point value. The second test, the addition of 1% FeCl 3 , was a way of testing if phenol was present in the aspirin. The iron (III) chloride was added to the salicylic acid as a control; it showed a dark purple color because there was a high presence of phenols. When the iron (III) chloride was added to the synthesized aspirin, it did not turn a dark shade of purple because there were impurities present. Also, there would be less phenol present in the aspirin than in the salicylic acid itself. The concentration of the standard iron solution was determined next. To do so, Equation 5 would be needed. The calculation proceeded as follows: Mass of FeCl 3 · 6H 2 O = 5.4 grams Concentration of iron solution = M 1 V 1 = M 2 V 2 Concentration of iron solution = 5.4 g x 1 mol / 264.25 g/mol 1L Concentration of iron solution = 0.0204 M This value was used to determine the concentrations of Solutions A through E, so that the Beer’s Law plot could be made. Equation 5 was also used to find the concentration of these solutions. Each of these calculations ensued as follows: Solution A: M 1 V 1 = M 2 V 2 (0.0204 M) x (.005 L) = M (1 L) Concentration (M) = .00102 M 5
Solution B: M 1 V 1 = M 2 V 2 (0.0204 M) x (.004 L) = M (1 L) Concentration (M) = 8.16E-4 M Solution C: M 1 V 1 = M 2 V 2 (0.0204 M) x (.003 L) = M (1 L) Concentration (M) = 6.12E-4 M Solution D: M 1 V 1 = M 2 V 2 (0.0204 M) x (.002 L) = M (1 L) Concentration (M) = 4.08E-4 M Solution E: M 1 V 1 = M 2 V 2 (0.0204 M) x (.001 L) = M (1 L) Concentration (M) = 2.04E-4 M The absorbance readings of these solutions were obtained and placed into Table 1 with the concentration values. These values were used to create the Beer’s Law plot that is also attached. From this graph, the equation for the best fit line is determined. Equation 6 is this best fit line, or y = 9.152x – 0.0989. The slope of this line is also equal to the molar absorptivity at the maximum wavelength; this value is 915.2 M -1 cm -1 . The absorbance of solution G was found to be 0.346, which is equivalent to that of solution C. The concentration of this solution was determined by substituting the absorbance in for y of Equation 6. This occurred as follows: y = 9.152x – 0.0989 0.346 = 9.152x – 0.0989 X (concentration in M) = 4.86E-4 M
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