Chapter 13, Problem 72QRT

Interpretation Introduction

Interpretation:

The molar mass and molecular formula of the compound has to be calculated.

Concept Introduction:

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$. That is water changes from liquid phase to gas phase at temperature ${\text{100}}^{\text{o}}\text{C}$.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}{\text{- T}}_{\text{b}}^{\text{°}}$.

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point.

  ${\text{T}}_{\text{b}}-$ Boiling point of the solution.

  ${\text{T}}_{\text{b}}^{\text{°}}-$ Boiling point of pure solvent.

TheBoiling point elevation${\text{(ΔT}}_{\text{b}}\text{)}$ is indicated as ${\text{ΔT}}_{\text{b}}{\text{= K}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point.

  ${\text{K}}_{\text{b}}-$ Molal boiling point constant.

  $m-$ Molality of the solution.

Molecular formula of a compound is derived from empirical formula mass and molar mass of the compound as,

    $\frac{molarmass}{empiricalmolarmass}.$

Answer to Problem 72QRT

Molecular mass of the compound is $3.6\times 10^{2}g/mol$ andmolecular formula is ${\left(C_{10}{H}_{8}Fe\right)}_2.$

Explanation of Solution

Change in boiling point is calculated as shown below,

Given information as follows, ${\text{T}}_{\text{b}}\left(\text{solution}\right)\text{=}\text{80}{\text{.26}}^{\text{o}}\text{C}\text{;}{\text{T}}_{\text{b}}^{\text{°}}\left(\text{benzene, solvent}\right)\text{=}\text{80}{\text{.10}}^{\text{o}}\text{C}$

By substituting the known values, change in boiling point is,

$\begin{array}{c}{\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}\left(\text{solution}\right){\text{- T}}_{\text{b}}^{\text{°}}\left(\text{solvent}\right)\\ \\ {\text{ΔT}}_{\text{b }}\text{=}\text{80}{\text{.26}}^{\text{o}}\text{C}\text{-}\text{80}{\text{.10}}^{\text{o}}\text{C}\text{=}\text{0}{\text{.16}}^{\text{o}}\text{C}\text{.}\end{array}$

Calculate molality of the unknown compound:

Substituting the values of ${\text{ΔT}}_{\text{b }}$ and $K_b={2.53}^{\circ }Ckgmo{l}^{-1}$, the molality of unknown compound was calculated as shown below,

  $\begin{array}{c}{\text{ΔT}}_{\text{b }}\text{= }\text{0}{\text{.16}}^{\text{o}}\text{C}\\ \\ {\text{ΔT}}_{\text{b }}{\text{= K}}_{\text{b}}{\text{m}}_{{\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}}\\ \\ {\text{m}}_{{\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}}\text{= }\frac{{\text{ΔT}}_{\text{b }}}{{\text{K}}_{\text{b}}}\text{=}\frac{\text{0}{\text{.16}}^{\text{o}}\text{C}}{\text{2}{\text{.53}}^{\text{o}}\text{C}\text{kg}{\text{mol}}^{\text{-1}}}\text{=}\text{0}\text{.063mol/kg}\text{.}\end{array}$

Calculate moles of the unknown compound:

Mass of benzene $\text{=}\text{11}\text{.12}\text{g}$, molality $\text{=1}\text{.52 mol/kg}\text{.}$

  $\begin{array}{c}{\text{molality, m}}_{{\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}}\text{ =}\frac{\text{no}\text{.of mol}}{\text{mass of benzene in kg}}\\ \\ \text{no}\text{.of mol}\text{= 11}\text{.12}\text{g}\text{benzene}\text{×}\frac{\text{1}\text{kg}\text{benzene}}{\text{1000}\text{g}\text{benzene}}\text{×}\frac{\text{1}{\text{.52 mol C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}}{\text{1}\text{kg}\text{benzene}}\\ \\ \text{=}\text{0}\text{.00070}\text{mol }{\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}\text{.}\end{array}$

Molar mass is calculated as shown below;

Given mass of compound is $\text{0}\text{.255}\text{g}$ and number of moles is $\text{0}\text{.00070}\text{mol}\text{.}$

  $\begin{array}{c}\text{molar mass =}\frac{\text{mass}}{\text{no}\text{.of moles}}\\ \\ \text{no}\text{.of mol}\text{=}\frac{\text{0}\text{.255}{\text{g C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}}{\text{0}\text{.00070}\text{mol }{\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}}\\ \\ =3.6\times 10^{2}g/mol.\end{array}$

Thus, molecular mass of the compound is $3.6\times 10^{2}g/mol.$

As known, the molecular formula is a multiple of the empirical formula: ${\left({\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}\right)}_n.$

The molar mass of the empirical formula ${\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\text{Fe}$ is $\text{184}\text{.01}\text{g/mol}\text{.}$

Hence, $\text{n}\text{=}\frac{\text{3}{\text{.6×10}}^{\text{2}}\text{g/mol}}{\text{184}\text{.01}\text{g/mol}}\text{=}\text{2}\text{.0}\text{=}\text{2}\text{.}$

Therefore, the molecular formula is ${\left(C_{10}{H}_{8}Fe\right)}_2.$