Chapter 12, Problem 81A

(a)

Interpretation Introduction

Interpretation: The empirical formula of pentyl acetate is to be calculated.

Concept Introduction: The empirical formula for a compound represents the simplest atomic ratio between its constituent elements.

Answer to Problem 81A

The empirical formula of pentyl acetate is ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ .

Explanation of Solution

The mass of pentyl acetate is $7.44\text{ g}$ .

The mass of ${\text{CO}}_2$ is $17.6\text{ g}$ .

The mass of ${\text{H}}_2\text{O}$ is $7.21\text{ g}$ .

The molar mass of $\text{C}$ is $12.01{\text{ g mol}}^{-1}$ .

The molar mass of $\text{H}$ is $1.008{\text{ g mol}}^{-1}$ .

The molar mass of $\text{O}$ is $16.00{\text{ g mol}}^{-1}$ .

The calculation for the mass of carbon is as follows:

  $17.6\bcancel{{\text{g CO}}_2}\times \frac{\bcancel{1{\text{ mol CO}}_2}}{44.0\bcancel{{\text{g CO}}_2}}\times \frac{\bcancel{1\text{ mol C}}}{\bcancel{1{\text{ mol CO}}_2}}\times \frac{12.01\text{ g C}}{\bcancel{1\text{ mol C}}}=4.804\text{ g C}$

Therefore, the mass of carbon is $4.804\text{ g}$ .

The calculation for the mass of hydrogen is as follows:

  $7.21\bcancel{{\text{g H}}_2\text{O}}\times \frac{\bcancel{1{\text{ mol H}}_2\text{O}}}{18.0\bcancel{{\text{g H}}_2\text{O}}}\times \frac{2\bcancel{\text{mol H}}}{\bcancel{1{\text{ mol H}}_2\text{O}}}\times \frac{1.008\text{ g H}}{1\bcancel{\text{mol H}}}=0.8075\text{ g H}$

Therefore, the mass of hydrogen is $0.8075\text{ g}$ .

The calculation for the mass of oxygen is as follows:

  $7.44\text{ g}-4.804\text{ g}-0.8075\text{ g}=1.8285\text{ g}$

Therefore, the mass of oxygen is $1.8285\text{ g}$ .

The calculation for the number of moles of carbon is as follows:

  $4.804\bcancel{\text{g C}}\times \frac{1\text{ mol C}}{12.01\bcancel{\text{g C}}}=0.4\text{ mol C}$

Therefore, the number of moles of carbon is $0.4\text{ mol}$ .

The calculation for the number of moles of hydrogen is as follows:

  $0.8075\bcancel{\text{g H}}\times \frac{1\text{ mol H}}{1.008\bcancel{\text{g H}}}=0.8011\text{ mol H}$

Therefore, the number of moles of hydrogen is $0.8011\text{ mol}$ .

Calculate the number of moles of oxygen as follows:

  $1.8285\bcancel{\text{g O}}\times \frac{1\text{ mol O}}{16.00\bcancel{\text{g O}}}=0.1143\text{ mol O}$

Therefore, the number of moles of oxygen is $0.1143\text{ mol}$ .

Divide the number with the smallest number, $0.1143$ as follows:

  $\begin{array}{l}\text{C:H:O}=\frac{0.4}{0.1143}:\frac{0.8011}{0.1143}:\frac{0.1143}{0.1143}\\ =3.4996:7.01489:1\\ \approx 3.5:7:1\end{array}$

The empirical formula is the whole number as follows:

The ratio of atoms is as follows:

  $\left(3.5:7:1\right)\times 2=7:14:2$

The ratio of $\text{C:H:O}$ is $7:14:2$ .

Therefore, the empirical formula of pentyl acetate is ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ .

(b)

Interpretation Introduction

Interpretation: The molecular formula of pentyl acetate is to be determined.

Concept Introduction: The molecular formula is an expression that describes the number of atoms of each element in a molecule. It indicates the exact number of atoms in a molecule.

Answer to Problem 81A

The molecular formula of pentyl acetate is ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ .

Explanation of Solution

The given molar mass of ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ is $130.0{\text{ g mol}}^{-1}$ .

The empirical formula of pentyl acetate is ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ .

The molar mass of ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molar mass C}}_7{\text{H}}_{14}{\text{O}}_2=7\times \text{Molar mass C+14}\times \text{Molar mass H+2}\times \text{Molar mass O}\\ =7\times 12.01{\text{ g mol}}^{-1}+\text{14}\times 1.008{\text{ g mol}}^{-1}+2\times 16.00{\text{ g mol}}^{-1}\\ =84.07{\text{ g mol}}^{-1}+14.112{\text{ g mol}}^{-1}+32{\text{ g mol}}^{-1}\\ =130.182{\text{ g mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ is $130.182{\text{ g mol}}^{-1}$ .

The ratio of molecular molar mass to the ratio of the empirical molar mass is calculated as follows:

  $\begin{array}{c}\frac{130.0\bcancel{{\text{g mol}}^{-1}}}{130.182\bcancel{{\text{g mol}}^{-1}}}=0.9986\\ \approx 1\end{array}$

The molecular formula of pentyl acetate is calculated as follows:

  $\begin{array}{c}\text{Molecular formula}=1\times {\text{C}}_7{\text{H}}_{14}{\text{O}}_2\\ ={\text{C}}_7{\text{H}}_{14}{\text{O}}_2\end{array}$

Therefore, the molecular formula of pentyl acetate is ${\text{C}}_7{\text{H}}_{14}{\text{O}}_2$ .

(c)

Interpretation Introduction

Interpretation: The balance equation of the complete combustion of pentyl acetate is to be written.

Concept Introduction: The components of a chemical equation are reactants, products, and a directional arrow. For a chemical reaction to be considered "balanced," both sides of the equation must have the same number of atoms.

Answer to Problem 81A

The balance equation of the complete combustion of pentyl acetate is as follows:

  ${\text{2C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}\left(l\right)+19{\text{O}}_2\left(g\right)\to 14{\text{CO}}_2\left(g\right)+14{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

During the combustion of pentyl acetate, carbon dioxide and water are produced.

The chemical equation is as follows:

  ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}\left(l\right)+{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

On the left side $7$

  $\text{C}$ atoms, $14$

  $\text{H}$ atoms, and $4$

  $\text{O}$ atoms. On the right side, $1$

  $\text{C}$ atoms, $2$

  $\text{H}$ atoms, and $3$

  $\text{O}$ atoms are present in this equation. To balance this equation, put the coefficient of $2$ in front of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ , $19$ in front of ${\text{O}}_{\text{2}}$ , and $14$ in front of ${\text{CO}}_2$ , and ${\text{H}}_{\text{2}}\text{O}$ .

The balanced chemical equation is as follows:

  ${\text{2C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}\left(l\right)+19{\text{O}}_2\left(g\right)\to 14{\text{CO}}_2\left(g\right)+14{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation: The mass of carbon dioxide and water is to be calculated.

Concept Introduction: To get the mass of a compound, multiply the molar mass of each component by the number of atoms of that component.

Answer to Problem 81A

  $17.6270\text{ g}$ of ${\text{CO}}_2$ and $7.2111\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ are produced during the complete combustion of $7.44\text{ g}$

  ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ .

Explanation of Solution

During the combustion of pentyl acetate, carbon dioxide and water are produced.

The chemical equation is as follows:

  ${\text{2C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}\left(l\right)+19{\text{O}}_2\left(g\right)\to 14{\text{CO}}_2\left(g\right)+14{\text{H}}_{\text{2}}\text{O}\left(l\right)$

  $\text{2}$ moles of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ react with $19$ moles of ${\text{O}}_2$ to give $14$ moles of ${\text{CO}}_2$ and $14$ moles of ${\text{H}}_{\text{2}}\text{O}$ .

The mass of $\text{2}$ moles ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ is as follows:

  $\text{2 }\bcancel{{\text{mol C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}\times \frac{130.0{\text{ g C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}{1\bcancel{{\text{mol C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}}=260.0{\text{ g C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$

Therefore, the mass of $\text{2}$ moles of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ is $260.0\text{ g}$ .

The mass of $19$ moles ${\text{O}}_2$ is as follows:

  $\text{19 }\bcancel{{\text{mol O}}_2}\times \frac{\text{16}{\text{.00 g O}}_2}{1\bcancel{{\text{mol O}}_2}}=304.0{\text{ g O}}_2$

Therefore, the mass of $19$ moles of ${\text{O}}_2$ is $304.0\text{ g}$ .

The mass of $14$ moles ${\text{CO}}_2$ is as follows:

  $\text{14 }\bcancel{{\text{mol CO}}_2}\times \frac{44{\text{ g CO}}_2}{1\bcancel{{\text{mol CO}}_2}}=616{\text{ g CO}}_2$

Therefore, the mass of $14$ moles of ${\text{CO}}_2$ is $\text{616 g}$ .

The mass of $14$ moles ${\text{H}}_{\text{2}}\text{O}$ is as follows:

  $\text{14 }\bcancel{{\text{mol H}}_{\text{2}}\text{O}}\times \frac{{\text{18 g H}}_{\text{2}}\text{O}}{1\bcancel{{\text{mol H}}_{\text{2}}\text{O}}}=252{\text{ g H}}_{\text{2}}\text{O}$

Therefore, the mass of $14$ moles of ${\text{H}}_{\text{2}}\text{O}$ is $252\text{ g}$ .

  $260.0\text{ g}$ of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ reacts with $304.0\text{ g}$ of ${\text{O}}_2$ to give $\text{616 g}$ of ${\text{CO}}_2$ and $252\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ .

The mass of ${\text{CO}}_2$ produces by $7.44\text{ g}$ of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ is as follows:

  $7.44\bcancel{{\text{g C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}\times \frac{{\text{616 g CO}}_{\text{2}}}{260.0\bcancel{{\text{g C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}}=17.6270{\text{ g CO}}_{\text{2}}$

Therefore, $17.6270\text{ g}$ of ${\text{CO}}_2$ is produced by $7.44\text{ g}$ of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ .

The mass of ${\text{H}}_{\text{2}}\text{O}$ produces by $7.44\text{ g}$ of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ is as follows:

  $7.44\bcancel{{\text{g C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}\times \frac{252{\text{ g H}}_{\text{2}}\text{O}}{260.0\bcancel{{\text{g C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}}}=7.2111{\text{ g H}}_{\text{2}}\text{O}$

Therefore, $7.2111\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ is produced by $7.44\text{ g}$ of ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}{\text{O}}_{\text{2}}$ .