The percent yield by mass of Step III is approximately 18.36%.
The overall reaction states that 4 moles of C₆H₅O₃N yield 3 moles of C₈H₉O₂N. Hence, 3.41x moles of C₈H₉O₂N correspond to 3 moles of C₈H₉O₂N.
Solving for x gives: x = 3 / 3.41
= 0.88.
molar mass of C₈H₉O₂N (151.17 g/mol) to find the actual mass of acetaminophen: 0.88 * 151.17 g/mol = 133.04 g.
4 moles of C₆H₅O₃N (molar mass 181.14 g/mol),
the theoretical mass of the product would be:
4 * 181.14 g/mol = 724.56 g.
We then divide the actual mass of acetaminophen by the theoretical mass and multiply by 100%:
(133.04 g / 724.56 g) * 100%
= 18.36%.