Chapter 3, Problem 60E

Interpretation Introduction

Interpretation: The molecular formula of the terephthalic acid is to be predicted.

Concept introduction: The empirical formula is the simplified version of molecular formula. It gives the relative ratio of elements forming the compound.Two distinct compounds may possess same empirical formula however this does not apply to molecular formula. This property makes molecular formula important for a compound’s identity. Both molecular and empirical formulas are related as

  $\text{Molecular}\text{formula}={\left(\text{Empirical}\text{formula}\right)}_n$

Answer to Problem 60E

The molecular formula of terephthalic acid is ${\text{C}}_8{\text{H}}_6{\text{O}}_4$ .

Explanation of Solution

The reaction for the combustion of general hydrocarbon is shown below.

  ${\text{C}}_x{\text{H}}_y{\text{O}}_z+{\text{O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

The moles of $\text{C}$ is same as that of ${\text{CO}}_{\text{2}}$ . Therefore, first, the number of moles of ${\text{CO}}_{\text{2}}$ is to be calculated.

The number of moles of ${\text{CO}}_{\text{2}}$ is calculated by using the formula given below.

  $n=\frac{m}{M}$  (1)

Where,

• $n$ is the number of moles.
• $m$ is the given mass.
• $M$ is the molecular mass.

The molar mass of ${\text{CO}}_{\text{2}}$ is $\text{44}\text{.01 g/mol}$ .

The conversion of mg into g is shown below.

  $\begin{array}{l}1\text{mg}={10}^{-3}\text{g}\\ 41.98\text{mg}=41.98\times {10}^{-3}\text{g}\end{array}$

The number of moles of ${\text{CO}}_{\text{2}}$ is calculated by substituting its mass and molar mass in equation (1) as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{CO}}_{\text{2}}\left(n_{{\text{CO}}_{\text{2}}}\right)=\frac{41.98\times {10}^{-3}\text{g}}{44.01\text{g/mol}}\\ =0.9538\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of $\text{C}$ is $0.9538\times {10}^{-3}\text{mol}$ .

The mass of $\text{C}$ is calculated using the formula shown below.

  $\text{Mass}=\text{}\text{Moles}\times \text{Molar}\text{mass}$  (2)

The molar mass of $\text{C}$ is $\text{12}\text{.0}\text{g/mol}$ .

Substitute the number of moles and molar mass of carbon in equation (2).

  $\begin{array}{c}\text{Mass of}\text{C}=\text{}0.9538\times {10}^{-3}\text{mol}\times 12.\text{0}\text{g/mol}\\ =11.445\times {10}^{-3}\text{g}\\ =11.445\text{mg}\end{array}$

The number of moles of $\text{H}$ is twice as that of ${\text{H}}_{\text{2}}\text{O}$ . Therefore, the number of moles of ${\text{H}}_{\text{2}}\text{O}$ is to be calculated.

The molar mass of ${\text{H}}_{\text{2}}\text{O}$ is $\text{18}\text{.015 g/mol}$ .

The number of moles of ${\text{H}}_{\text{2}}\text{O}$ can be calculated by substituting the mass and molar mass in equation (1).

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\left(n_{{\text{H}}_{\text{2}}\text{O}}\right)=\frac{6.45\times {10}^{-3}\text{g}}{18.015\text{g/mol}}\\ =0.358\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of $\text{H}$ is $2\times 0.358\times {10}^{-3}\text{mol}=0.716\times {10}^{-3}\text{mol}$ .

The molar mass of $\text{H}$ is $1.\text{0}\text{g/mol}$ .

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

  $\begin{array}{c}\text{Mass}\text{of H}=0.716\times {10}^{-3}\text{mol}\times 1.\text{0}\text{g/mol}\\ =0.716\times {10}^{-3}\text{g}\\ =0.716\text{mg}\end{array}$

The mass of oxygen is calculated as shown below.

  $\text{Mass}\text{of O}=\text{Mass}\text{of compound}-\left(\text{Mass}\text{of C}+\text{Mass}\text{of H}\right)$  (3)

Substitute the mass of hydrogen and carbon in equation (3).

  $\begin{array}{c}\text{Mass}\text{of O}=19.81\text{mg}-\left(11.445\text{mg}+0.716\text{mg}\right)\\ =7.649\text{mg}\end{array}$

The molar mass of $\text{O}$ is $\text{16}\text{.00}\text{g/mol}$ .

The number of moles of oxygen is calculated by substituting its mass and molar mass in equation (1) as shown below.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{O}\left(n_{\text{O}}\right)=\frac{7.649\times {10}^{-3}\text{g}}{\text{16}\text{.00 g/mol}}\\ =0.478\times {10}^{-3}\text{mol}\end{array}$

The ratio of carbon to hydrogen to oxygen is calculated by dividing their moles with the lowest number of moles.

  $\begin{array}{c}\text{C:H:O}=\frac{0.9538\times {10}^{-3}\text{mol}}{0.478\times {10}^{-3}\text{mol}}:\frac{0.716\times {10}^{-3}\text{mol}}{0.478\times {10}^{-3}\text{mol}}:\frac{0.478\times {10}^{-3}\text{mol}}{0.478\times {10}^{-3}\text{mol}}\\ \text{C:H:O}=2:1.5:1\end{array}$

Multiply the above ratio by $2$ to get whole numbers.

  $\text{C:H:O}=4:3:2$

Therefore, the empirical formula of the given compound is ${\text{C}}_4{\text{H}}_3{\text{O}}_2$ .

The molecular formula of a compound is related to the empirical formula as shown below.

  $\text{Molecular}\text{formula}={\left(\text{Empirical}\text{formula}\right)}_n$  (4)

Where,

• $n$ is the number of empirical formula units.

Empirical formula units can be calculated as shown below.

  $n=\frac{\text{Molar}\text{mass}\left(M_{{\text{C}}_4{\text{H}}_3{\text{O}}_2}\right)}{\text{Empirical}\text{formula}\text{mass}\left({M^{\prime }}_{{\text{C}}_4{\text{H}}_3{\text{O}}_2}\right)}$  (5)

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

  ${M^{\prime }}_{{\text{C}}_4{\text{H}}_3{\text{O}}_2}=n_{\text{C}}\times M_{\text{C}}+n_{\text{H}}\times M_{\text{H}}+n_{\text{O}}\times M_{\text{O}}$  (6)

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.
• ${M^{\prime }}_{{\text{C}}_4{\text{H}}_3{\text{O}}_2}$ is the empirical formula mass of the substance.

Substitute the known values in equation (6).

  $\begin{array}{c}{M^{\prime }}_{{\text{C}}_4{\text{H}}_3{\text{O}}_2}=4\times 12.0\text{g/mol}+3\times 1.00\text{g/mol}+\text{2}\times 16.00\text{g/mol}\\ =83\text{g/mol}\end{array}$

The molar mass of terephthalic acid is calculated by the formula shown below.

  $\text{Molecular}\text{mass}\left(M\right)=\frac{\text{Mass}\left(m\right)}{\text{Number}\text{of}\text{moles}\left(n\right)}$  (7)

Substitute the known values in equation (7).

  $\begin{array}{c}\text{Molecular}\text{mass}\left(M\right)=\frac{\text{41}\text{.5}\text{g}}{\text{0}\text{.25}\text{mol}}\\ =166\text{g/mol}\end{array}$

Substitute the molecular mass and empirical formula mass in equation (5).

  $\begin{array}{c}n=\frac{166\text{g/mol}}{83\text{g/mol}}\\ =2\end{array}$

Substitute the value of empirical formula units in equation (4) to find the molecular formula.

  $\begin{array}{c}\text{Molecular}\text{formula}={\left({\text{C}}_4{\text{H}}_3{\text{O}}_2\right)}_2\\ ={\text{C}}_8{\text{H}}_6{\text{O}}_4\end{array}$

Therefore, the molecular formula of the compound is ${\text{C}}_6{\text{H}}_8{\text{O}}_4$ .

Conclusion

The molecular formula of terephthalic acid is ${\text{C}}_8{\text{H}}_6{\text{O}}_4$ .