Chapter 3, Problem 42E

(a)

Interpretation Introduction

Interpretation: The percentage composition of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is to be determined.

Concept introduction: The percentage composition refers to the relative amounts of element expressed in the form percentages. It is calculated by dividing element’s mass with compound’s molecular mass. The sum of percentage compositions of every element constituting the compound always equals $100\%$ .

Answer to Problem 42E

The percentage composition of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is given below.

  $\begin{array}{c}\%\text{O}=44.44\%\\ \%\text{C}=50.0\%\\ \%\text{H}=5.55\%\end{array}$

Explanation of Solution

The molar mass of all the elements present in ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$ is shown below.

  $\begin{array}{l}\text{C}=12.0\text{g/mol}\\ \text{H}=\text{1}\text{.0}\text{g/mol}\\ \text{O}=\text{16}\text{.00}\text{g/mol}\end{array}$

The molecular formula mass can be calculated by the formula shown below.

  ${M^{\prime }}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{O}}\times M_{\text{O}}$  (1)

Where,

• $n_{\text{C}}$ is the number of atoms of carbon.
• $M_{\text{C}}$ is the molar mass of carbon.
• $n_{\text{H}}$ is the number of atoms of hydrogen.
• $M_{\text{C}}$ is the molar mass of hydrogen.
• $n_{\text{O}}$ is the number of atoms of oxygen.
• $M_{\text{O}}$ is the molar mass of oxygen.
• ${M^{\prime }}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}$ is the molar mass of the substance.

Substitute the known values in equation (1).

  $\begin{array}{c}{M^{\prime }}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}}=3\times 12.00\text{g/mol}+4\times 1.0\text{g/mol}+\text{2}\times \text{16}\text{.00 g/mol}\\ =36.0\text{g/mol}+4.0\text{g/mol}+\text{32 g/mol}\\ =72\text{g/mol}\end{array}$

The mass of each element in the substance is calculated by the formula shown below.

  $\text{Mass}=\text{Number of}\text{atoms}\times \text{Molecular}\text{mass}\text{of}\text{element}$  (2)

Substitute the known values in equation (2). The mass of $\text{O}$ and $\text{C}$ are calculated as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{O}=\text{2}\times \text{16}\text{.00 g/mol}\\ =32.0\text{g/mol}\\ \text{Mass}\text{of}\text{C}=3\times 12.0\text{ g/mol}\\ =36.0\text{g/mol}\end{array}$

Similarly, the mass of $\text{H}$ is calculated by substituting its number of atoms and molar mass in equation (2) as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{H}=4\times 1.0\text{g/mol}\\ =4.0\text{g/mol}\end{array}$

The percentage of each element is calculated by the formula shown below.

  $\text{Mass}\text{percentage}=\frac{\text{Mass}\text{of}\text{element}}{\text{Mass}\text{of}\text{susbstance}}\times 100\%$  (3)

Substitute the values in equation (3). The mass percentage of $\text{O}$ and $\text{C}$ are calculated as shown below.

  $\begin{array}{c}\%\text{O}=\frac{32.0\text{g/mol}}{72.0\text{g/mol}}\times 100\%\\ =44.44\%\\ \%\text{C}=\frac{36.0\text{g/mol}}{72.0\text{g/mol}}\times 100\%\\ =50.0\%\end{array}$

Similarly, mass percentage of $\text{H}$ is calculated by substituting the mass of $\text{H}$ and mass of substance in equation (3) as shown below.

  $\begin{array}{c}\%\text{H}=\frac{4\text{g/mol}}{72\text{g/mol}}\times 100\%\\ =5.55\%\end{array}$

(b)

Interpretation Introduction

Interpretation: The percentage composition of ${\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}$ is to be determined.

Concept introduction: The percentage composition refers to the relative amounts of element expressed in the form percentages. It is calculated by dividing element’s mass with compound’s molecular mass. The sum of percentage compositions of every element constituting the compound always equals $100\%$ .

Answer to Problem 42E

The percentage composition of ${\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}$ is given below.

  $\begin{array}{c}\%\text{O}=37.21\%\\ \%\text{C}=55.81\%\\ \%\text{H}=6.97\%\end{array}$

Explanation of Solution

The molar mass of all the elements present in ${\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}$ is shown below.

  $\begin{array}{l}\text{C}=12.0\text{g/mol}\\ \text{H}=\text{1}\text{.0}\text{g/mol}\\ \text{O}=\text{16}\text{.00}\text{g/mol}\end{array}$

The molecular formula mass can be calculated as by the formula shown below.

  ${M^{\prime }}_{{\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}}=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{O}}\times M_{\text{O}}\cdot \cdot \cdot \cdot \cdot \text{(I)}$

Where,

• $n_{\text{C}}$ is the number of atoms of carbon.
• $M_{\text{C}}$ is the molar mass of carbon.
• $n_{\text{H}}$ is the number of atoms of hydrogen.
• $M_{\text{C}}$ is the molar mass of hydrogen.
• $n_{\text{O}}$ is the number of atoms of oxygen.
• $M_{\text{O}}$ is the molar mass of oxygen.
• ${M^{\prime }}_{{\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}}$ is the molar mass of the substance

Substitute the known values in equation (1).

  $\begin{array}{c}{M^{\prime }}_{{\text{C}}_4{\text{H}}_6{\text{O}}_{\text{2}}}=4\times 12.00\text{g/mol}+6\times 1.0\text{g/mol}+\text{2}\times \text{16}\text{.00 g/mol}\\ =48.0\text{g/mol}+6.0\text{g/mol}+\text{32}\text{.0 g/mol}\\ =86\text{g/mol}\end{array}$

The mass of each element in the substance is calculated by the formula shown below.

  $\text{Mass}=\text{Number of}\text{atoms}\times \text{Molecular}\text{mass}\text{of}\text{element}\cdot \cdot \cdot \cdot \cdot \text{(II)}$

Substitute the known values in the equation (2). The mass of $\text{O}$ and $\text{C}$ are calculated as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{O}=2\times \text{16}\text{.00 g/mol}\\ =32.0\text{g/mol}\\ \text{Mass}\text{of}\text{C}=4\times 12.0\text{ g/mol}\\ =48\text{g/mol}\end{array}$

Similarly, the mass of $\text{H}$ is calculated by substituting its number of atoms and molar mass in equation (2) as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{H}=6\times 1.0\text{g/mol}\\ =6.0\text{g/mol}\end{array}$

The percentage of each element is calculated by the formula shown below.

  $\text{Mass}\text{percentage}=\frac{\text{Mass}\text{of}\text{element}}{\text{Mass}\text{of}\text{susbstance}}\times 100\%$  (3)

Substitute the values in equation (3). The mass percentage of $\text{O}$ and $\text{C}$ are calculated as shown below.

  $\begin{array}{c}\%\text{O}=\frac{32.0\text{g/mol}}{86.0\text{g/mol}}\times 100\%\\ =37.21\%\\ \%\text{C}=\frac{48.0\text{g/mol}}{86.0\text{g/mol}}\times 100\%\\ =55.81\%\end{array}$

Similarly, the mass percentage of $\text{H}$ is calculated by substituting mass of $\text{H}$ and mass of substance in equation (3) as shown below.

  $\begin{array}{c}\%\text{H}=\frac{6.0\text{g/mol}}{86.0\text{g/mol}}\times 100\%\\ =6.97\%\end{array}$

(c)

Interpretation Introduction

Interpretation: The percentage composition of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is to be determined.

Concept introduction: The percentage composition refers to the relative amounts of element expressed in the form percentages. It is calculated by dividing element’s mass with compound’s molecular mass. The sum of percentage compositions of every element constituting the compound always equals $100\%$ .

Answer to Problem 42E

The percentage composition of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is

  $\begin{array}{c}\%\text{N}=26.41\%\\ \%\text{C}=67.92\%\\ \%\text{H}=5.66\%\end{array}$

Explanation of Solution

The molar mass of all the elements present in ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is shown below.

  $\begin{array}{l}\text{C}=12.0\text{g/mol}\\ \text{H}=\text{1}\text{.0}\text{g/mol}\\ \text{N}=\text{14}\text{.0}\text{g/mol}\end{array}$

The molecular formula mass can be calculated as shown below.

  ${M^{\prime }}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{N}}\times M_{\text{N}}$  (1)

Where,

• $n_{\text{C}}$ is the number of atoms of carbon.
• $M_{\text{C}}$ is the molar mass of carbon.
• $n_{\text{H}}$ is the number of atoms of hydrogen.
• $M_{\text{C}}$ is the molar mass of hydrogen.
• $n_{\text{N}}$ is the number of atoms of nitrogen.
• $M_{\text{N}}$ is the molar mass of nitrogen.
• ${M^{\prime }}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}$ is the molar mass of the substance.

Substitute the known values equation (1).

  $\begin{array}{c}{M^{\prime }}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}=3\times 12.00\text{g/mol}+3\times 1.0\text{g/mol}+\text{1}\times 14.\text{00 g/mol}\\ =36.0\text{g/mol}+3.0\text{g/mol}+\text{14}\text{.0 g/mol}\\ =53\text{g/mol}\end{array}$

The mass of each element in the substance is calculated by the formula shown below.

  $\text{Mass}=\text{Number}\text{of}\text{atoms}\times \text{Molecular}\text{mass}\text{of}\text{element}$  (2)

Substitute the known values in equation (2). The mass of $\text{N}$ and $\text{C}$ are calculated as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{N}=1\times \text{14}\text{.00 g/mol}\\ =14.0\text{g/mol}\\ \text{Mass}\text{of}\text{C}=3\times 12.0\text{ g/mol}\\ =36\text{g/mol}\end{array}$

Similarly, the mass of $\text{H}$ is calculated by substituting its number of atoms and molar mass in equation (2) as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{H}=3\times 1.0\text{g/mol}\\ =3.0\text{g/mol}\end{array}$

The percentage of each element is calculated by the formula shown below.

  $\text{Mass}\text{percentage}=\frac{\text{Mass}\text{of}\text{element}}{\text{Mass}\text{of}\text{susbstance}}\times 100\%$  (3)

Substitute the values in equation (3). The mass percentage of $\text{N}$ and $\text{C}$ are calculated as shown below.

  $\begin{array}{c}\%\text{N}=\frac{14.0\text{g/mol}}{53.0\text{g/mol}}\times 100\%\\ =26.41\%\\ \%\text{C}=\frac{36.0\text{g/mol}}{53.0\text{g/mol}}\times 100\%\\ =67.92\%\end{array}$

Similarly, the mass of $\text{H}$ is calculated by substituting the mass of $\text{H}$ and the mass of substance equation (3) as shown below.

  $\begin{array}{c}\%\text{H}=\frac{3.0\text{g/mol}}{53.0\text{g/mol}}\times 100\%\\ =5.66\%\end{array}$