Question

An organic compound has a mass composition of 93.46% C and 6.54% H. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C.\({\bf{Kf}}\)for camphor is 37.7 °C/m. What is the molecular formula of the solute? Show your calculations.

Short answer

The molecular formula of the solute is \({{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{10}}}}\).

Step-by-step solution

Molecular Formula

The molecular formula may be defined as the formula which tells about the proportion and constituent present in the molecules.

Calculate the molecular mass of the substance

\({\rm{\Delta Tf = }}{{\rm{K}}_{\rm{f}}}{\rm{ \times m = }}{{\rm{K}}_{\rm{f}}}{\rm{ \times }}\dfrac{{\left( {{\rm{Number of Moles}}} \right)}}{{{\rm{Mass of Solvent}}}}{\rm{ }}\)

\(Number\:{\rm{ }}of\:{\rm{ }}Moles{\rm{ }} = {\rm{ }}\dfrac{{Mass}}{{Molar{\rm{ }}Mass}}{\rm{ }}\)

\(\begin{aligned}{\rm{Molar}}\;{\rm{Mass}}\; &= \;{{\rm{K}}_{\rm{f}}}\dfrac{{\left( {{\rm{Mass}}\;{\rm{of}}\;{\rm{Solute}}} \right)}}{{{\rm{\Delta }}{{\rm{T}}_{\rm{f}}}\left( {{\rm{Mass}}\;{\rm{of}}\;{\rm{Solvent}}} \right)}}\\{\rm{Molar}}\;{\rm{Mass}} &= \;\dfrac{{\left( {{\rm{37}}{\rm{.7^\circ C}}} \right)\left( {{\rm{0}}{\rm{.045}}\;{\rm{kg}}} \right)}}{{\left( {{\rm{20}}{\rm{.0^\circ C}}} \right)\left( {{\rm{0}}{\rm{.000550}}\;{\rm{kg}}} \right)}}\\{\rm{Molar}}\;{\rm{Mass}} &= \;{\rm{154}}{\rm{.2}}\;{\rm{g}}\;{\rm{mol}}{{\rm{e}}^{{\rm{ - 1}}}}\end{aligned}\)

Calculate the empirical formula of the substance

Empirical formula:

\(\begin{aligned}{\rm{C}} & = \;\dfrac{{{\rm{93}}{\rm{.46}}\;{\rm{g}}}}{{{\rm{12}}{\rm{.011}}\;{\rm{g}}\;{\rm{mol}}{{\rm{e}}^{{\rm{ - 1}}}}}}\\{\rm{C}} &= \;{\rm{7}}{\rm{.781}}\;{\rm{mole}}\end{aligned}\)

By dividing with the smallest mole to get the empirical formula

\(\begin{aligned}{\rm{C}} &= \;\dfrac{{{\rm{7}}{\rm{.781}}}}{{{\rm{6}}{\rm{.4885}}}}\\{\rm{C}} &= \;{\rm{1}}{\rm{.2}}\end{aligned}\)

\(\begin{aligned}{\rm{H}} &= \;\dfrac{{{\rm{6}}{\rm{.5}}\;{\rm{g}}}}{{{\rm{1}}{\rm{.0794}}\;{\rm{g}}\;{\rm{mol}}{{\rm{e}}^{{\rm{ - 1}}}}}}\\{\rm{H}} &= \;{\rm{6}}{\rm{.4885}}\;{\rm{mole}}\\{\rm{H}} &= \;\dfrac{{{\rm{6}}{\rm{.4885}}}}{{{\rm{6}}{\rm{.4885}}}}\\{\rm{H}} &= \;{\rm{1}}\end{aligned}\)

Ratio : \({{\rm{C}}_{{\rm{1}}{\rm{.2}}}}{\rm{H}}\;{\rm{or}}\;{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{10}}}}\)

The formula mass of \({{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{10}}}}\) agrees with the calculated formula mass.