Chapter 12, Problem 1DE

Determine the empirical and molecular formulas of each of the following substances:

a. Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% 0 by mass and has a molar mass of 206 g/mol.
b. Cadaverine, a foul-smelling substance produced by the action of bacteria on meat, contains 58.55% C, 13.81% H, and 27.40% N by mass; its molar mass is 102.2 g/mol.
c. Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, contains 59.0% C, 7.1% H, 262% 0, and 7.7% N by mass; its molar mass is about 180 amu.

Interpretation Introduction

Interpretation: The empirical and molecular formula of ibuprofen.

Concept introduction:
A formula that depicts the simplest ratio of the constituent elements in a compound is known as the empirical formula.
The empirical formula mass of a given compound is the total of the atomic masses of the constituent atoms.
The steps to determine the molecular formula from the empirical formula are,

• The empirical formula mass is calculated.
• The gram molecular mass of the compound is divided by the empirical formula mass.
• The subscripts within the empirical formula are multiplied by the number obtained in the previous step.
• The chemical formula obtained having new subscript values is the molecular formula of the compound.

Answer to Problem 1DE

Solution: The empirical formula of ibuprofen is ${\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2$ and its molecular formula is ${\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2$ .

Explanation of Solution

Given that,
Mass percent of carbon is $75.69\%$ .
Mass percent of hydrogen is $8.80\%$ .
Mass percent of oxygen is $15.51\%$ .

The atomic mass of carbon is $12\text{g}$ .
The atomic mass of hydrogen is $1\text{g}$ .
The atomic mass of oxygen is $16\text{g}$ .

The number of moles is calculated using the formula (assuming the sample to be $100\text{g}$ ),
$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{percent}}{\text{Atomic}\text{mass}\text{of}\text{the}\text{substance}}$

Substituting the values of given mass and the molar mass in the above expression,

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{carbon}=\frac{75.69\text{g}}{12\text{g}}\\ =6.3\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}=\frac{\text{8}\text{.80}\text{g}}{1\text{g}}\\ =8.8\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{oxygen}=\frac{\text{15}\text{.51}\text{g}}{16\text{g}}\\ \approx 0.97\text{mole}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent,

For carbon $\left(\text{C}\right)$ ,
  $\text{C}=\frac{6.3\text{mol}}{0.97\text{mol}}=6.5$

For hydrogen $\left(\text{H}\right)$ ,
  $\text{H}=\frac{8.8\text{mol}}{0.97\text{mol}}=9$

For oxygen $\left(\text{O}\right)$ ,
  $\text{O}=\frac{0.97\text{mol}}{0.97\text{mol}}=1$

The calculated value of the number of moles is multiplied by a common multiple $2$ to obtain the simplest whole number values for the same.
Therefore, the empirical formula obtained is ${\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2$ .

Now, to determine the molecular formula of ibuprofen:
Given that,
Empirical formula is ${\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2$ .
Molar mass of the compound is $206\text{g}/\text{mol}$ .

The atomic mass of carbon is $12\text{g}$ .
The atomic mass of hydrogen is $1\text{g}$ .
The atomic mass of oxygen is $16\text{g}$ .

The empirical formula mass of ${\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2$ $\begin{array}{l}=\left(\left(13\times 12\right)+\left(1\times 18\right)+\left(2\times 16\right)\right)\text{g}\\ =\text{206}\text{g}\end{array}$

The given molar mass value is divided by the calculated empirical formula mass value to obtain a whole number multiple,
$\begin{array}{c}\text{Whole}\text{number}\text{multiple}=\frac{206\text{g}/\text{mol}}{206\text{g}/\text{mol}}\\ =1\end{array}$

The subscripts of the empirical formula are multiplied by the whole number multiple obtained to get the molecular formula of the given compound,

  $\begin{array}{c}\text{Molecular formula}={\left({\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2\right)}_1\\ ={\text{C}}_{13}{\text{H}}_{18}{\text{O}}_2\end{array}$

Interpretation Introduction

Interpretation: The empirical and molecular formula of cadeverine.

Concept introduction:
A formula that depicts the simplest ratio of the constituent elements in a compound is known as the empirical formula.
The empirical formula mass of a given compound is the total of the atomic masses of the constituent atoms.
The steps to determine the molecular formula from the empirical formula are,

• The empirical formula mass is calculated.
• The gram molecular mass of the compound is divided by the empirical formula mass.
• The subscripts within the empirical formula are multiplied by the number obtained in the previous step.
• The chemical formula obtained having new subscript values is the molecular formula of the compound.

Answer to Problem 1DE

Solution: The empirical formula of cadeverine is ${\text{C}}_5{\text{H}}_{14}{\text{N}}_2$ and its molecular formula is ${\text{C}}_5{\text{H}}_{14}{\text{N}}_2$ .

Explanation of Solution

Given that,
Mass percent of carbon is $58.55\%$ .
Mass percent of hydrogen is $13.81\%$ .
Mass percent of nitrogen is $27.40\%$ .

The atomic mass of carbon is $12\text{g}$ .
The atomic mass of hydrogen is $1\text{g}$ .
The atomic mass of nitrogen is $14\text{g}$ .

The number of moles is calculated using the formula (assuming the sample to be $100\text{g}$ ),
$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{percent}}{\text{Atomic}\text{mass}\text{of}\text{the}\text{substance}}$

Substituting the values of given mass and the molar mass in the above expression,

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{carbon}=\frac{58.55\text{g}}{12\text{g}}\\ =4.87\text{moles}\\ \approx 4.9\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}=\frac{13.81\text{g}}{1\text{g}}\\ =13.81\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}=\frac{27.40\text{g}}{14\text{g}}\\ =1.95\text{moles}\\ \approx 2\text{moles}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent,

For carbon $\left(\text{C}\right)$ ,
  $\text{C}=\frac{4.9\text{mol}}{2\text{mol}}\approx 2.5$

For hydrogen $\left(\text{H}\right)$ ,
  $\text{H}=\frac{13.81\text{mol}}{2\text{mol}}\approx 7$

For nitrogen $\left(\text{N}\right)$ ,
  $\text{N}=\frac{2\text{mol}}{2\text{mol}}=1$

The calculated value of the number of moles is multiplied by a common multiple $2$ to obtain the simplest whole number values for the same.
Therefore, the empirical formula obtained is ${\text{C}}_5{\text{H}}_{14}{\text{N}}_2$ .

Now, to determine the molecular formula of cadaverine:

Given that,
Empirical formula is ${\text{C}}_5{\text{H}}_{14}{\text{N}}_2$ .
Molar mass of the compound is $102.2\text{g}/\text{mol}$ .

The atomic mass of carbon is $12\text{g}$ .
The atomic mass of hydrogen is $1\text{g}$ .
The atomic mass of nitrogen is $14\text{g}$ .

The empirical formula mass of ${\text{C}}_5{\text{H}}_{14}{\text{N}}_2$ $\begin{array}{l}=\left(\left(5\times 12\right)+\left(14\times 1\right)+\left(2\times 14\right)\right)\text{g}\\ \text{=102}\text{g}\end{array}$

The given molar mass value is divided by the calculate empirical formula mass value to obtain a whole number multiple,

  $\begin{array}{c}\text{Whole}\text{number}\text{multiple}=\frac{102.2\text{g}/\text{mol}}{102\text{g}/\text{mol}}\\ \approx 1\end{array}$

The subscripts of the empirical formula are multiplied by the whole number multiple obtained to get the molecular formula of the given compound,

  $\begin{array}{c}\text{Molecular formula}={\left({\text{C}}_5{\text{H}}_{14}{\text{N}}_2\right)}_1\\ ={\text{C}}_5{\text{H}}_{14}{\text{N}}_2\end{array}$

Interpretation Introduction

Interpretation: The empirical and molecular formula of epinephrine.

Concept introduction:
A formula that depicts the simplest ratio of the constituent elements in a compound is known as the empirical formula.
The empirical formula mass of a given compound is the total of the atomic masses of the constituent atoms.
The steps to determine the molecular formula from the empirical formula are,

• The empirical formula mass is calculated.
• The gram molecular mass of the compound is divided by the empirical formula mass.
• The subscripts within the empirical formula are multiplied by the number obtained in the previous step.
• The chemical formula obtained having new subscript values is the molecular formula of the compound.

Answer to Problem 1DE

Solution: The empirical formula of epinephrine is ${\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}$ and its molecular formula is ${\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}$ .

Explanation of Solution

Given that,
Mass percent of carbon is $59.0\%$ .
Mass percent of hydrogen is $7.1\%$ .
Mass percent of nitrogen is $7.7\%$ .
Mass percent of oxygen is $26.2\%$ .

The atomic mass of carbon is $12\text{g}$ .
The atomic mass of hydrogen is $1\text{g}$ .
The atomic mass of nitrogen is $14\text{g}$ .
The atomic mass of oxygen is $16\text{g}$ .

The number of moles is calculated using the formula (assuming the sample to be $100\text{g}$ ),

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{percent}}{\text{Atomic}\text{mass}\text{of}\text{the}\text{substance}}$

Substituting the values of given mass and the molar mass in the above expression,

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{carbon}=\frac{59.0\text{g}}{12\text{g}}\\ =4.91\text{moles}\\ \approx 5\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{hydrogen}=\frac{7.1\text{g}}{1\text{g}}\\ =7.1\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{nitrogen}=\frac{7.7\text{g}}{14\text{g}}\\ =0.55\text{moles}\end{array}$

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{oxygen}=\frac{26.2\text{g}}{16\text{g}}\\ =1.64\text{moles}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent,

For carbon $\left(\text{C}\right)$ ,
$\text{C}=\frac{5\text{mol}}{0.55\text{mol}}\approx 9$

For hydrogen $\left(\text{H}\right)$ ,
  $\text{H}=\frac{7.1\text{mol}}{0.55\text{mol}}\approx 13$

For nitrogen $\left(\text{N}\right)$ ,
  $\text{N}=\frac{0.55\text{mol}}{0.55\text{mol}}=1$

For oxygen $\left(\text{O}\right)$ ,
  $\text{O}=\frac{1.64\text{mol}}{0.55\text{mol}}\approx 3$

Therefore, the empirical formula obtained is ${\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}$ .

Now, to determine the molecular formula of epinephrine:
Given that,
Empirical formula is ${\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}$ .
Molar mass of the compound is $180\text{amu}$ .

The atomic mass of carbon is $12\text{amu}$ .
The atomic mass of hydrogen is $1\text{amu}$ .
The atomic mass of nitrogen is $14\text{amu}$ .
The atomic mass of oxygen is $16\text{amu}$ .

The empirical formula mass of ${\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}$ $\begin{array}{l}=\left(\left(9\times 12\right)+\left(13\times 1\right)+\left(3\times 16\right)+14\right)\text{amu}\\ =183\text{amu}\end{array}$

The given molar mass value is divided by the calculate empirical formula mass value to obtain a whole number multiple,

$\begin{array}{c}\text{Whole}\text{number}\text{multiple}=\frac{180\text{g}/\text{mol}}{183\text{g}/\text{mol}}\\ \approx 1\end{array}$

The subscripts of the empirical formula are multiplied by the whole number multiple obtained to get the molecular formula of the given compound,

  $\begin{array}{c}\text{Molecular formula}={\left({\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}\right)}_1\\ ={\text{C}}_9{\text{H}}_{13}{\text{O}}_3\text{N}\end{array}$