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best value for percent purity and how did you arrive at the best value

best value for percent purity and how did you arrive at the best value

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I want the best value for percent purity and how did you arrive at the best value 

Part III-Commercial Aspirin Trial 1 Trial 2 .371g .381g Mass of aspirin tablet (g) 2.06 x 10 ^-3 2.11 x 10 ^3 mol asp Theoretical moles in tablet Show calculations for theoretical moles: 1:.371 g asp x 1.00 mol / 180.16 g asp = 2.06 x 10 ^-3 mol asp 2:.381 g asp x 1.00 mol / 180.16 g asp = 2.11 x 10 ^-3 mol Molarity of NaOH (mol/L) 216 Volume of NaOH used (mL) 1:5.31 mL 2:7.31 mL Color of Solution 1: very light pink, faded away 2: better color than 1, light pink color Moles of NaOH used 1:115moles 2:.158 moles Show calculations for moles NaOH: 1:216 mol NaOH/1.00 L x .531 NaOH used = .115 moles 2: .216 mol NaOH / 1.00 Lx.731 L NaOH used =.158 moles Percent purity of aspirin (%) 65.4% Example Calculation: 1:.246 g asp/.371 x 100 66.2 % 2:246 g asp/.381 g asp = 64.6 % avg: .246 g asp/.376 g asp x 100 = 65.4%
Transcribed Image Text:Part III-Commercial Aspirin Trial 1 Trial 2 .371g .381g Mass of aspirin tablet (g) 2.06 x 10 ^-3 2.11 x 10 ^3 mol asp Theoretical moles in tablet Show calculations for theoretical moles: 1:.371 g asp x 1.00 mol / 180.16 g asp = 2.06 x 10 ^-3 mol asp 2:.381 g asp x 1.00 mol / 180.16 g asp = 2.11 x 10 ^-3 mol Molarity of NaOH (mol/L) 216 Volume of NaOH used (mL) 1:5.31 mL 2:7.31 mL Color of Solution 1: very light pink, faded away 2: better color than 1, light pink color Moles of NaOH used 1:115moles 2:.158 moles Show calculations for moles NaOH: 1:216 mol NaOH/1.00 L x .531 NaOH used = .115 moles 2: .216 mol NaOH / 1.00 Lx.731 L NaOH used =.158 moles Percent purity of aspirin (%) 65.4% Example Calculation: 1:.246 g asp/.371 x 100 66.2 % 2:246 g asp/.381 g asp = 64.6 % avg: .246 g asp/.376 g asp x 100 = 65.4%
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