Question

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers.It contains only C, H, and O. Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO₂and 6.45 mg H₂O. If 0.250 mole of terephthalic acid has a mass of 41.5 g, determine themolecular formula of terephthalic acid.

Short answer

The molecular and empirical formulas are$C_4{H}_3{O}_2\&C_8{H}_6{O}_4$.

Step-by-step solution

Reaction of Terephthalic acid combustion

The combustion reaction of terephthalic acid combustion can be,

$a{C}_x{H}_y{O}_z+b{O}_2\to cC{O}_2+d{H}_2$

To calculate the mass of H and C can be denoted as,

$\begin{array}{rcl}\mathrm{m}\left(\mathrm{C}\right)& =& 0.04198\mathrm{g}{\mathrm{CO}}_2·\frac{12.011\mathrm{g}\mathrm{c}}{44.009\mathrm{g}{\mathrm{CO}}_2}\\ & =& 0.011\mathrm{gC}\\ \mathrm{m}\left(\mathrm{H}\right)& =& 0.000645\mathrm{g}{\mathrm{H}}_2\mathrm{O}·\frac{2.016\mathrm{g}\mathrm{H}}{18.015\mathrm{g}{\mathrm{H}}_2\mathrm{O}}\\ & =& 0.00072\mathrm{gH}\end{array}$

Calculating C and H percent

On calculating C and H percent in acid,

$\begin{array}{rcl}\mathrm{Percent}\mathrm{C}& =& \frac{\mathrm{m}\left(\mathrm{C}\right)}{\mathrm{m}\left(\mathrm{acid}\right)}·100\mathrm{Percent}\\ & =& \frac{0.011\mathrm{g}}{0.01981\mathrm{g}}·100\mathrm{Percent}\\ & =& 55.53\mathrm{Percent}\\ \mathrm{Percent}\mathrm{H}& =& \frac{0.00072\mathrm{g}}{0.01981\mathrm{g}}·100\mathrm{Percent}\\ & =& 3.64\mathrm{percent}\\ \mathrm{Percent}\mathrm{O}& =& 100\mathrm{Percent}-55.53\mathrm{Percent}-3.64\mathrm{Percent}\\ & =& 40.826\mathrm{percent}\end{array}$

Calculating empirical formula

On finding empirical formula,

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{C}\right)& =& \frac{55.53\mathrm{g}}{12.001{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 4.623\mathrm{mol}\\ \mathrm{n}\left(\mathrm{H}\right)& =& \frac{3.64\mathrm{g}}{1.008{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 3.611\mathrm{mol}\\ \mathrm{n}\left(\mathrm{O}\right)& =& \frac{40.826\mathrm{g}}{15.999{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 2.55\mathrm{mol}\\ \mathrm{n}\left(\mathrm{C}\right):\mathrm{n}\left(\mathrm{H}\right):\mathrm{n}\left(\mathrm{O}\right)& =& 4:3:2\end{array}$

The empirical formula is$C_4{H}_3{O}_2$

Calculating molecular formula

On calculating molecular formula,

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{formula}& =& \left(\mathrm{Theem}\mathrm{pirical}\mathrm{formula}\right)\\ \mathrm{n}& =& \frac{\mathrm{m}}{\mathrm{M}}\\ \mathrm{M}& =& \frac{41.5\mathrm{g}}{0.25\mathrm{mol}}\\ \mathrm{M}& =& 166\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

The mass of empirical formula is. Then,$4.12+3.1+2.16=83g$

$\begin{array}{rcl}\mathrm{x}& =& \frac{166\mathrm{g}}{83\mathrm{g}}\\ \mathrm{x}& =& 2\end{array}$

Molecular formula is$C_4{H}_3{O}_2\&C_8{H}_6{O}_4$.