Chapter 13, Problem 91IL

An organic compound contains carbon (71.17%), hydrogen (5.12%) with the remainder nitrogen. Dissolving 0.177 g of the compound in 10.0 g of benzene gives a solution with a vapor pressure of 94.16 mm Hg at 25 °C. (The vapor pressure of pure benzene at this temperature is 95.26 mm Hg.) What is the molecular formula for the compound?

Interpretation Introduction

Interpretation Molecular formula of the compound has to be determined.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

  $P_{solvent}=X_{solvent}{P}^0{}_{solvent}$

  where,

  $P^0{}_{solvent}$ is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 91IL

Molecular formula of the compound is $C_7{H}_6{N}_2$

Explanation of Solution

Molecular formula of the compound is calculated.

Given,

  Molar mass of carbon is $\text{12}\text{.01}\text{g/mol}$

  Molar mass of hydrogen is $\text{1}\text{.01}\text{g/mol}$

  Molar mass of nitrogen is $\text{14}\text{.01}\text{g/mol}$

  Vapour pressure of benzene at ${25}^{0}C$ is $P_{benzene}=94.16mmHg$

  Vapour pressure of pure benzene is $P^0{}_{benzene}=95.26mmHg$

  Mass of the compound is $0.177g$

  Mass of benzene is $10.0g$

  Molar mass of benzene is $\text{78}\text{.11}\text{g/mol}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

  $\begin{array}{c}Numberofmoles\text{of}\text{benzene}\text{=}\frac{10\text{g}}{\text{78}\text{.11}\text{g/mol}}\\ \text{=}\text{0}\text{.128}\text{mol}\end{array}$

Number of moles of benzene is $\text{0}\text{.128}\text{mol}$

Vapour pressure of benzene is calculated by the given equation

  $P_{benzene}=X_{benzene}{P}^0{}_{benzene}$

Mole fraction of benzene is,

  $\begin{array}{c}{X}_{benzene}=\frac{P_{benzene}}{P^0{}_{benzene}}\\ =\frac{94.16mmHg}{95.26mmHg}\\ =0.988\end{array}$

Hence,

$\begin{array}{c}{X}_{benzene}=\frac{{\text{n}}_{benzene}}{{\text{n}}_{\text{benzene}}+{\text{n}}_{compound}}\\ 0.988=\frac{0.128mol}{0.128mol+{\text{n}}_{compound}}\\ 0.12650.128mol+\left(0.988\right){\text{n}}_{compound}=0.128mol\\ {\text{n}}_{compound}=1.518\times {10}^{-3}mol\end{array}$

Number of moles of the compound is $1.518\times {10}^{-3}mol$

Hence,

The molar mass of compound is,

  $\begin{array}{c}\text{Molar}\text{mass}\text{=}\frac{\text{0}\text{.177}\text{g}}{\text{1}\text{.518}\text{×}{\text{10}}^{\text{-3}}\text{mol}}\\ \text{=}\text{116}\text{.6}\text{g/mol}\end{array}$

  $\begin{array}{c}MassofC=\frac{71.17}{100}\times 100g=71.17g\\ MassofH=\frac{5.12}{100}\times 100g=5.12g\\ MassofH=\frac{23.71}{100}\times 100g=23.71g\\ \end{array}$

  $\begin{array}{c}\text{Number of moles}\text{of}\text{C}\text{=}\frac{\text{71}\text{.17}\text{g}}{\text{12}\text{.01}\text{g/mol}}\\ \text{=}\text{5}\text{.9}\text{mol}\end{array}$

Number of moles of carbon is $\text{5}\text{.9}\text{mol}$

  $\begin{array}{c}\text{Number of moles}\text{of}\text{H}\text{=}\frac{\text{5}\text{.12}\text{g}}{\text{1}\text{.01}\text{g/mol}}\\ \text{=}\text{5}\text{.07}\text{mol}\end{array}$

Number of moles of hydrogen is $\text{5}\text{.9}\text{mol}$

  $\begin{array}{c}Numberofmoles\text{of}N\text{=}\frac{\text{23}\text{.71}\text{g}}{\text{14}\text{.01}\text{g/mol}}\\ \text{=}\text{1}\text{.69}\text{mol}\end{array}$

Empirical formula of the compound was determined.

  $\begin{array}{ccccc}C& :& H& :& N\\ \frac{5.9}{1.69}& :& \frac{5.07}{1.69}& :& \frac{1.69}{1.69}\\ 3.5& :& 3& :& 1\\ 7& :& 6& :& 2\end{array}$

Therefore,

The empirical formula of the compound is $C_7{H}_6{N}_2$

Molecular mass of $C_7{H}_6{N}_2$ is $\text{118}\text{.14}\text{g/mol}$

The molar mass of a molecule is a simple multiple of the molar mass of its empirical formula. Hence,

  $\begin{array}{c}\text{n}\text{=}\frac{\text{molar}\text{mass}\text{of}\text{compound}}{\text{molar}\text{mass}\text{of}\text{its}\text{empirical}\text{formula}}\\ \text{=}\frac{\text{116}\text{.6}\text{g/mol}}{\text{118}\text{.14}\text{g/mol}}\\ \text{=0}\text{.98}\approx \text{1}\end{array}$

Therefore,

Molecular formula of the compound is $C_7{H}_6{N}_2$

Conclusion

Molecular formula of the compound is $C_7{H}_6{N}_2$