The aspirin substitute, acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}\right)\), is produced by the following three-step synthesis: I. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{~N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(t) $$ II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(t)+\mathrm{NaCl}(a q) $$ III. \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow\) $$ \mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{~N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l) $$ The first two reactions have percent yields of \(87 \%\) and \(98 \%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}_{3} \mathrm{~N}\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

Firstly, let's determine the overall reaction from the given three-step synthesis: 1. \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N} + 3\mathrm{H}_2 + \mathrm{HCl} \rightarrow \mathrm{C}_6\mathrm{H}_3\mathrm{ONCl} + 2\mathrm{H}_2\mathrm{O}\) 2. \(\mathrm{C}_6\mathrm{H}_8\mathrm{ONCl} + \mathrm{NaOH} \rightarrow \mathrm{C}_6\mathrm{H}_7\mathrm{ON} + \mathrm{H}_2\mathrm{O} + \mathrm{NaCl}\) 3. \(\mathrm{C}_6\mathrm{H}_7\mathrm{ON} + \mathrm{C}_4\mathrm{H}_6\mathrm{O}_{3} \rightarrow \mathrm{C}_8\mathrm{H}_9\mathrm{O}_2\mathrm{~N} + \mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2}\) Overall reaction: \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N} + 3\mathrm{H}_2 + \mathrm{HCl} +\mathrm{NaOH} +\mathrm{C}_4\mathrm{H}_6\mathrm{O}_{3} \rightarrow \mathrm{C}_8\mathrm{H}_9\mathrm{O}_2\mathrm{~N} + 2\mathrm{H}_2\mathrm{O} + \mathrm{NaCl} + \mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2}\) Now, let's use the mole ratios to calculate the number of moles for each species, and then the mass of the reactant and product moles. We are given the overall mole ratio: 3 moles of acetaminophen are produced for every 4 moles of \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N}\) reacted. To simplify the calculation, assume that we initially start with 4 moles of \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N}\): - The mass of \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N}\): \(4 \cdot \left(12 \times 6 + 1 \times 3+ 16 \times 3 + 14\right)= 572\mathrm{g}\) From the stoichiometry of the overall reaction, 3 moles of \(\mathrm{C}_8\mathrm{H}_9\mathrm{O}_2\mathrm{~N}\) would be produced. To find the actual yield in grams: - The mass of \(\mathrm{C}_8\mathrm{H}_9\mathrm{O}_2\mathrm{~N}\) produced: \(3 \cdot \left(12 \times 8 + 1 \times 9 + 16 \times 2 + 14\right) = 471\mathrm{g}\) Now, we can find the overall percent yield by mass by dividing the mass of the product \(\mathrm{C}_{8}\mathrm{H}_{9}\mathrm{O}_{2}\mathrm{~N }\) produced by the mass of the reactant \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N}\) used, and multiplying by 100: Overall percent yield by mass = \(\frac{471}{572} \times 100 \approx 82.3\% \)

In order to find the percent yield by mass of Step III, we need to calculate the yield from Step I and Step II. Let's consider we initially start with the 4 moles of \(\mathrm{C}_6\mathrm{H}_3\mathrm{O}_3\mathrm{~N}\) (mass = 572 g). Step I has a percent yield by mass of 87%. Thus, the actual yield is: Actual yield of Step I: \(0.87 \times 572 \approx 497.64\mathrm{g}\) For Step II, it has a percent yield by mass of 98%. Assuming that all 497.64 g of the product from Step I is used, the actual yield is: Actual yield of Step II: \(0.98 \times 497.64 \approx 487.68\mathrm{g}\) Now, we can use the actual yield from Step II as input for Step III, and use the actual yield from the overall process (471 g of \(\mathrm{C}_8\mathrm{H}_9\mathrm{O}_2\mathrm{~N}\)) to calculate the percent yield by mass of Step III: Percent yield by mass of Step III = \(\frac{471}{487.68} \times 100 \approx 96.6\% \) To summarize the results: a. The overall percent yield by mass = \(82.3\%\) b. The percent yield by mass of Step III = \(96.6\%\)