Chapter 3, Problem 49E

(a)

Interpretation Introduction

Interpretation: The composition of formaldehyde $\left({\text{CH}}_2\text{O}\right)$ as the mass percentage of its elements is to be determined.

Concept introduction: The compound’s composition explains what percent of the compound is formed by a particular element. The percentage composition is obtained by various experimental methods like combustion analysis. This percentage composition plays significant role in reaching the compound’s empirical formula.

Answer to Problem 49E

The percentage composition of ${\text{CH}}_2\text{O}$ is given below.

  $\begin{array}{c}\%\text{O}=53.33\%\\ \%\text{C}=40.00\%\\ \%\text{H}=6.66\%\end{array}$

Explanation of Solution

The molar mass of all the elements present in ${\text{CH}}_2\text{O}$ 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 shown below.

  $M^{\prime }=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{O}}\times M_{\text{O}}$  (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.
• $M^{\prime }$ is the molar mass of the substance.

Substitute the known values in equation (1).

  $\begin{array}{c}{M^{\prime }}_{{\text{CH}}_2\text{O}}=1\times 12.00\text{g/mol}+2\times 1.0\text{g/mol}+\text{1}\times \text{16}\text{.00 g/mol}\\ =36.0\text{g/mol}+4.0\text{g/mol}+\text{32 g/mol}\\ =30\text{g/mol}\end{array}$

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 $\text{O}$ and $\text{C}$ is calculated as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{O}=1\times \text{16}\text{.00 g/mol}\\ =16.0\text{g/mol}\\ \text{Mass}\text{of}\text{C}=1\times 12.0\text{ g/mol}\\ =12.0\text{g/mol}\end{array}$

The mass of $\text{H}$ is calculated by substituting its molar mass and number of atoms in equation (2).

  $\begin{array}{c}\text{Mass}\text{of}\text{H}=2\times 1.0\text{g/mol}\\ =2.0\text{g/mol}\end{array}$

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

  $\begin{array}{c}\%\text{O}=\frac{16.0\text{g/mol}}{30.0\text{g/mol}}\times 100\%\\ =53.33\%\\ \%\text{C}=\frac{12.0\text{g/mol}}{30.0\text{g/mol}}\times 100\%\\ =40.0\%\end{array}$

The mass percentage of $\text{H}$ are calculated by substituting its mass and mass of ${\text{CH}}_{\text{2}}\text{O}$ in equation (3) as shown below.

  $\begin{array}{c}\%\text{H}=\frac{2.0\text{g/mol}}{30.0\text{g/mol}}\times 100\%\\ =6.66\%\end{array}$

(b)

Interpretation Introduction

Interpretation: The composition of glucose $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ as the mass percentage of its elements is to be determined.

Concept introduction:The compound’s composition explains what percent of the compound is formed by a particular element. The percentage composition is obtained by various experimental methods like combustion analysis. This percentage composition plays significant role in reaching the compound’s empirical formula.

Answer to Problem 49E

The percentage composition of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is

  $\begin{array}{c}\%\text{O}=53.33\%\\ \%\text{C}=40.0\%\\ \%\text{H}=6.66\%\end{array}$

Explanation of Solution

The molar mass of all the elements present in ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is shown below.

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

Substitute the known values in equation (1) to find the molar mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ .

  $\begin{array}{c}{M^{\prime }}_{{\text{C}}_6{\text{H}}_{12}{\text{O}}_6}=6\times 12.00\text{g/mol}+12\times 1.0\text{g/mol}+\text{6}\times \text{16}\text{.00 g/mol}\\ =72.0\text{g/mol}+12.0\text{g/mol}+\text{96}\text{.0 g/mol}\\ =180\text{g/mol}\end{array}$

Substitute the number of atoms and molar mass in equation (2) to calculate the mass of each element. The mass of $\text{O}$ and $\text{C}$ is calculated as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{O}=6\times \text{16}\text{.00 g/mol}\\ =96.0\text{g/mol}\\ \text{Mass}\text{of}\text{C}=6\times 12.0\text{ g/mol}\\ =72\text{g/mol}\end{array}$

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

  $\begin{array}{c}\text{Mass}\text{of}\text{H}=12\times 1.0\text{g/mol}\\ =12.0\text{g/mol}\end{array}$

Substitute the mass of element and the mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ in equation (3) to calculate mass percentage. The mass percentages of $\text{O}$ and $\text{C}$ is calculated as shown below.

  $\begin{array}{c}\%\text{O}=\frac{96.0\text{g/mol}}{180.0\text{g/mol}}\times 100\%\\ =53.33\%\\ \%\text{C}=\frac{72.0\text{g/mol}}{180.0\text{g/mol}}\times 100\%\\ =40.0\%\end{array}$

The mass percentage of $\text{H}$ is calculated by substituting its mass and mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ in equation (3) as shown below.

  $\begin{array}{c}\%\text{H}=\frac{12.0\text{g/mol}}{180.0\text{g/mol}}\times 100\%\\ =6.66\%\end{array}$

(c)

Interpretation Introduction

Interpretation: The percentage composition of acetic acid $\left({\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}\right)$ is to be determined and the type of formula obtained from elemental analysis that gives mass percentage composition is to be stated.

Concept introduction:The compound’s composition explains what percent of the compound is formed by a particular element. The percentage composition is obtained by various experimental methods like combustion analysis. This percentage composition plays significant role in reaching the compound’s empirical formula.

Answer to Problem 49E

The percentage composition of ${\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}$ is given below.

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

The type of formula obtained from elemental analysis that gives mass percentage composition is empirical formula.

Explanation of Solution

The molar mass of all the elements present in ${\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}$ is shown below.

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

Substitute the known values in equation (1) to find the molar mass of ${\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}$ .

  $\begin{array}{c}{M^{\prime }}_{{\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}}=2\times 12.00\text{g/mol}+4\times 1.0\text{g/mol}+2\times \text{16}\text{.00 g/mol}\\ =24.0\text{g/mol}+4.0\text{g/mol}+32.\text{0 g/mol}\\ =60\text{g/mol}\end{array}$

Substitute the number of atoms and molar mass in equation (2) to calculate the mass of each element. The mass of $\text{O}$ and $\text{C}$ is calculated as shown below.

  $\begin{array}{c}\text{Mass}\text{of}\text{O}=2\times \text{16}\text{.00 g/mol}\\ =32.0\text{g/mol}\\ \text{Mass}\text{of}\text{C}=2\times 12.0\text{ g/mol}\\ =24.0\text{g/mol}\end{array}$

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

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

Substitute the mass of elements and the mass of ${\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}$ in equation (3) to calculate mass percentage. The mass percentage of $\text{O}$ and $\text{C}$ is calculated as shown below.

  $\begin{array}{c}\%\text{O}=\frac{32.0\text{g/mol}}{60.0\text{g/mol}}\times 100\%\\ =53.33\%\\ \%\text{C}=\frac{24.0\text{g/mol}}{60.0\text{g/mol}}\times 100\%\\ =40.0\%\end{array}$

The mass percentage of $\text{H}$ calculated by substituting its mass and mass of ${\text{HC}}_2{\text{H}}_3{\text{O}}_{\text{2}}$ in equation (3) as shown below.

  $\begin{array}{c}\%\text{H}=\frac{4.0\text{g/mol}}{60.0\text{g/mol}}\times 100\%\\ =6.66\%\end{array}$

As seen, the mass percentage of elements in formaldehyde, glucose and acetic acid is same.

This is due to the fact that they all have same empirical formula that is ${\text{CH}}_{\text{2}}\text{O}$ . However, the molecular formula is different for all the three compounds.

This observation proves that, the formula obtained from the elemental analysis is not molecular formula but empirical formula.