Question

The aspirin substitute acetaminophen(${\mathbf{\text{C}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{5}}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\text{N}}$) is produced by the following three-stepsynthesis:

I. ${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}{\mathbf{\text{O}}}_{\mathbf{3}}\mathbf{\text{N}}\left(\text{s}\right)\mathbf{+}\mathbf{3}{\mathbf{\text{H}}}_{\mathbf{2}}\left(\text{g}\right)\mathbf{+}\mathbf{\text{HCl}}\left(\text{aq}\right)\mathbf{\to }{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}\mathbf{\text{ONCl}}\left(\text{s}\right)\mathbf{+}\mathbf{2}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\left(l\right)$ role="math" localid="1648648560649" $$\begin{gathered} {\mathbf{\text{II.C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}\mathbf{\text{ONCl}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}\mathbf{+}\mathbf{\text{NaOH}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{7}}\mathbf{\text{ON}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{\text{l}}\mathbf{\right)}\mathbf{+} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{NaCl}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)} \end{gathered}$$role="math" localid="1648648678003" ${\mathbf{\text{III.C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{7}}\mathbf{\text{ON}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}\mathbf{+}{\mathbf{\text{C}}}_{\mathbf{4}}{\mathbf{\text{H}}}_{\mathbf{6}}{\mathbf{\text{O}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{\text{l}}\mathbf{\right)}\mathbf{\to }{\mathbf{\text{C}}}_{\mathbf{8}}{\mathbf{\text{H}}}_{\mathbf{9}}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\text{N}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}\mathbf{+}{\mathbf{\text{HC}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{3}}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{\text{l}}\mathbf{\right)}$

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 ${\mathbf{\text{C}}}_{\mathbf{2}}{\mathbf{\text{H}}}_{\mathbf{5}}{\mathbf{\text{O}}}_{\mathbf{2}}\mathbf{\text{N}}$ reacted.

1. What is the percent yield by mass for the overall process?
2. What is the percent yield by mass of step III?

Short answer

1. The percent yield by mass for the overall process is$\mathbf{75}\mathbf{}\mathbf{\%}$.
2. The percent yield by mass of step III of $\mathbf{\text{Ais88\%}}$.

Step-by-step solution

 Step 1: Definition

The massof any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of overall process

a)

Step 1:

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{C}}_6{\mathrm{H}}_8\mathrm{ONCl}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2:

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{ONCl}\left(\mathrm{s}\right)+NaOH\left(aq\right)\to {\mathrm{C}}_6{\mathrm{H}}_7\mathrm{ON}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+NaCl\left(aq\right)$

Step 3:

${\mathrm{C}}_6{\mathrm{H}}_7\mathrm{ON}\left(\mathrm{s}\right)+{\mathrm{C}}_4{\mathrm{H}}_6{\mathrm{O}}_3\left(l\right)\to {\mathrm{C}}_8{\mathrm{H}}_9{\mathrm{O}}_2\mathrm{N}\left(\mathrm{s}\right)+{\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\left(\mathrm{l}\right)$

For every 4 mol of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}$reacted, 3 moles of acetaminophen product is formed. We can observe from the reactions above that one mole of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}$should yield one mole of product.

So, for every4 moles of a reactant reacted in the first stage, we should receive 4 moles of product.

Determination of percent yield by mass for the overall process

Mass of reactant-${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}$taken role="math" localid="1648645694847" $\mathbf{=}\mathbf{\text{moles×molarmass}}$

$=\text{4moles×139.11}\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$=556.4\text{g}$

Theoretical mass of product- ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}$

$=\text{4moles×151.16}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$=435.5\text{g}$

Percentage yield $=\frac{453.5g}{606.64g}\times 100\%$

$=75\%$

Therefore, the percentage yield is $\mathbf{75}\mathbf{}\mathbf{\%}$.

Determination of mass for the overall process

b)

Iteration 1

Let’s consider that it begins with 100 g of reactant in iteration 1.
Percent yield is 87%
Mass of product $=\left(\frac{87}{100}\right)\times 100g$

$=87\text{g}$

Iteration 2

The product of${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}$ will be the reactant.
Percent yield is 98 %
So the mass of reactant $=87\text{g}$
Mass of product $=\left(\frac{98}{100}\right)\times 87g$

$=85.26\text{g}$

Iteration 3

The product of${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}_3\mathrm{N}$ will be the reactant
Percent yield is 75 %
So the mass of reactant $=85.26\text{g}$
Mass of product $=\left(\frac{75\%}{100}\right)\times 100g$

$=75\text{g}$

Determination of mass of product 

Mass of product$=\left(\frac{A}{100}\right)\times 85.26g$

$=88\%$

Therefore, the massof step III,$\mathbf{\text{Ais88\%}}$