Q45P - Chemistry: Molecular Nature Of Matter And Change

Question

Acetaminophen (below) is a popular nonaspirin, “over-thecounter” pain reliever. What is the mass % of each element in acetaminophen?

Step-by-Step Solution

Answer

The final mass % each element is:

C = 71.1%

H = 6.7%

N = 10.4%

O = 11.8%

The molecular formula is ${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{9}}\mathbf{N}{\mathbf{0}}_{\mathbf{2}}$

Step 1: Determine the Preliminary formula of the compound

This is the structure of the molecule and has the molecular formula ${\mathrm{C}}_8{\mathrm{H}}_9\mathrm{N}{0}_2$ 

It has 8 C, 9 H, 1N, and 2O, for each element we have to calculate the % of the element.

Step 2: Calculation of the mass % of the C in acetaminophen

The mass % of the element is:

$\mathbf{mass}\mathbf{ }\mathbf{\%}\mathbf{of}\mathbf{ }\mathbf{element}\mathbf{=}\frac{\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{element}\mathbf{ }\mathbf{in}\mathbf{1}\mathbf{ }\mathbf{mole}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{compound}}{\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{1}\mathbf{mole}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{compound}\mathbf{ }}\mathbf{\times }\mathbf{100}\mathbf{\%}$

The mass of 1 mole of compound is:

$$\begin{gathered} \mathbf{1}\mathbf{ }\mathbf{mole}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{compound}\mathbf{ }\mathbf{=}\mathbf{8}\mathbf{\times }\left(12.011g\right)\mathbf{+}\mathbf{9}\mathbf{\times }\left(1.001g\right)\mathbf{+}\mathbf{14}\mathbf{.}\mathbf{007}\mathbf{ }\mathbf{g}\mathbf{+}\mathbf{2}\mathbf{+}\left(15.999g\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{135}\mathbf{.}\mathbf{166}\mathbf{g} \end{gathered}$$

Thus, the mass % of C is calculated as:

$\mathbf{\%}\mathbf{C}\mathbf{=}\frac{\mathbf{8}\mathbf{\times }\left(12.011\right)}{\mathbf{135}\mathbf{.}\mathbf{166}}\mathbf{\times }\mathbf{100}\mathbf{\%}\mathbf{=}\mathbf{71}\mathbf{.}\mathbf{1}\mathbf{\%}$

Step 3: Determine the mass % of the H in acetaminophen

The mass % of H is:

$\mathbf{\%}\mathbf{H}\mathbf{=}\frac{\mathbf{9}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{008}\mathbf{)}}{\mathbf{135}\mathbf{.}\mathbf{166}}\mathbf{\times }\mathbf{100}\mathbf{\%}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{7}\mathbf{\%}$

Step 4: Determine the mass % of the N in acetaminophen

$\mathbf{\%}\mathbf{H}\mathbf{=}\frac{\mathbf{14}\mathbf{.}\mathbf{009}}{\mathbf{135}\mathbf{.}\mathbf{166}}\mathbf{\times }\mathbf{100}\mathbf{\%}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{4}\mathbf{\%}$

Step 5: Determine the mass % of the O in acetaminophen

The mass % of O is:

$\mathbf{\%}\mathbf{O}\mathbf{=}\frac{\mathbf{2}\mathbf{\times }\left(15.999\right)}{\mathbf{135}\mathbf{.}\mathbf{166}}\mathbf{\times }\mathbf{100}\mathbf{\%}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{8}\mathbf{\%}$