/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 172 The aspirin substitute. acetamin... [FREE SOLUTION] | 91Ó°ÊÓ

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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. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCl}(a q)\) 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 \(C_{6} H_{5} O_{3} N\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

Short Answer

Expert verified
a: The overall percent yield by mass for the process is 85.3%. b: The percent yield by mass of Step III is 98.0%.

Step by step solution

01

Calculate theoretical mass of products from the given reaction yield

We know that 3 moles of acetaminophen (C8H9O2N) are produced for every 4 moles of C6H5O3N reacted. We can use this information to determine the theoretical mass of products from the overall reaction. First, find the molar masses of C6H5O3N and C8H9O2N: - C6H5O3N: (6 × 12.01) + (5 × 1.01) + (3 × 16.00) + (1 × 14.01) ≈ 137.12 g/mol - C8H9O2N: (8 × 12.01) + (9 × 1.01) + (2 × 16.00) + (1 × 14.01) ≈ 151.17 g/mol From the given molar ratio, calculate the theoretical mass of products for each mole of C6H5O3N reacted: (3 moles of C8H9O2N / 4 moles of C6H5O3N) × 151.17 g/mol = 113.38 g/mol
02

Calculate the mass of products after Steps I and II

Now, consider the percent yields of Steps I (87%) and II (98%). Multiply the theoretical mass of products from Step 1 by the percent yields to find the mass of products after Steps I and II: - 113.38 g/mol × 0.87 (Step I yield) = 98.64 g/mol - 98.64 g/mol × 0.98 (Step II yield) = 96.67 g/mol
03

Calculate the overall percent yield by mass

The overall percent yield by mass is the mass of products after all steps divided by the theoretical mass from Step 1, multiplied by 100: Overall percent yield by mass = (96.67 g/mol / 113.38 g/mol) × 100 ≈ 85.3%
04

Calculate the percent yield by mass of Step III

We can now determine the percent yield by mass of Step III by dividing the mass of products after Step II (96.67 g/mol) by the mass of products after Step I (98.64 g/mol) and then multiplying by 100: Percent yield by mass of Step III = (96.67 g/mol / 98.64 g/mol) × 100 ≈ 98.0% #Answer#a: The overall percent yield by mass for the process is 85.3%. b: The percent yield by mass of Step III is 98.0%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Synthesis
Chemical synthesis is a sequence of chemical reactions designed to transform simple substances into complex molecules. In the context of the provided exercise, acetaminophen, a common pain reliever, is synthesized through a multi-step process involving various starting materials.

Each step of a synthesis can vary in efficiency, which leads to diferent yields. The efficiency is affected by reaction conditions, purity of reagents, and mechanisms occurring at the molecular level. In the case of acetaminophen, it involves an amide bond formation in the final step. Understanding the intricacies of each reaction step is crucial for both conducting the experiment in a lab and for calculating the theoretical and actual yields in exercises.

In real-world lab scenarios, chemists aim to optimize synthesis methods to improve yield, reduce costs, and minimize the environmental impact, hence understanding chemical synthesis is foundational for any student embarking on a career in chemistry.
Stoichiometry
Stoichiometry is all about the quantitative relationships between the reactants and products in a chemical reaction. It is a central concept in chemistry that allows for predictions about the amounts of substances consumed and produced in a reaction.

Theoretical Yield and Percent Yield

In the given exercise, stoichiometry aids in the calculation of the theoretical yield of acetaminophen. The theoretical yield is the amount of product expected based on stoichiometric ratios, assuming complete conversion of reactants to products. However, in practice, reactions rarely proceed perfectly, which is where the concept of percent yield comes into play. It is defined as the ratio of the actual yield (the amount actually produced) to the theoretical yield, times 100.

Through stoichiometry, we can determine that three moles of acetaminophen should theoretically be produced for every four moles of the starting material, under perfect conditions. This exercise uses stoichiometry to connect the theoretical concepts with practical outcomes, such as determining yields at different stages of the synthesis.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is determined by adding up the atomic masses of all the atoms in the molecule, as found on the periodic table. In the context of the exercise, calculating the molar mass of reactants and products is an essential step for solving stoichiometric problems.

For instance, the molar mass of a molecule of acetaminophen, \(C_8H_9O_2N\), is calculated by multiplying the number of each type of atom by its respective atomic mass, then summing those values. The molar mass calculations are vital for converting moles to grams and vice versa, which is a fundamental aspect of determining the theoretical yield and the percent yield in synthesis reactions. By mastering molar mass calculations, students are equipped to handle various problems in chemistry, including those related to stoichiometry and synthesis.

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