Chapter 13, Problem 91IL

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 $C_7{H}_6{N}_2$