Question

Acetaminophen is the generic name of the pain reliever in Tylenol and some other headache remedies. The compound has the molecular formula ${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{9}}{\mathbf{NO}}_{\mathbf{2}}$. Compute, to four significant figures, the mass percentage of each of the four elements in acetaminophen.

Short answer

The mass percentages of carbon, hydrogen, nitrogen, and oxygen are 63.5%, 5.95%, 9.266%, and 21.1%, respectively.

Step-by-step solution

Calculation of molar mass of acetaminophen.

The molar mass of a compound can be calculated by adding the molar masses of each atom in the compound.

The molar mass of carbon, ${\text{m}}_1=12.001{\text{gmol}}^{-1}$

The molar mass of hydrogen, ${\text{m}}_2=1.0079{\text{gmol}}^{-1}$

The molar mass of nitrogen,${\text{m}}_3=14.007{\text{gmol}}^{-1}$

The molar mass of oxygen, ${\text{m}}_4=15.999{\text{gmol}}^{-1}$

The molar mass of acetaminophen;

$\begin{array}{rcl}{\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{9}}{\mathbf{NO}}_{\mathbf{2}}& =& \left(8\times 12.011{\text{gmol}}^{-1}\right)+\left(9\times 1.0079{\text{gmol}}^{-1}\right)+\left(1\times 14.007{\text{gmol}}^{-1}\right)+\left(2\times 15.999{\text{gmol}}^{-1}\right)\\ & =& 151.164{\text{gmol}}^{-1}\end{array}$

Step 2: Mass percentage of each element in the compound

The mass percentage of carbon,

${\text{m}}_1\%=\frac{{\text{m}}_1}{\text{M}}\times 100$

Here role="math" localid="1663405779574" ${\text{m}}_1$is the molar mass of chlorine, and M is the molar mass of acetaminophen.

Substituting the values;

$\begin{array}{rcl}{\text{m}}_1\%& =& \frac{8\times 12.001}{\text{151.164}}\times 100\\ & =& 63.5\end{array}$

Similarly, the mass percentage of hydrogen,

${\text{m}}_{\text{2}}\%=\frac{{\text{m}}_2}{\text{M}}\times 100$

Substituting the values;

$\begin{array}{rcl}{\text{m}}_{\text{2}}\%& =& \frac{9\times 1.0079}{\text{151.164}}\times 100\\ & =& 5.95\end{array}$

Mass percentage of nitrogen,

$\begin{array}{rcl}{\text{m}}_3\%& =& \frac{{\text{m}}_3}{\text{M}}\times 100\\ & =& \frac{14.007}{151.164}\times 100\\ & =& 9.266\end{array}$

Mass percentage of oxygen,

$\begin{array}{rcl}{\text{m}}_4\%& =& \frac{{\text{m}}_4}{\text{M}}\times 100\\ & =& \frac{2\times 15.999}{151.164}\times 100\\ & =& 21.1\end{array}$