Q132CP - Chemistry: Molecular Nature Of Matter And Change

Question

A pharmaceutical preparation made with ethanol $\mathbf{\left(}{\mathbf{\text{C}}}_{\mathbf{\text{2}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{OH}}\mathbf{\right)}$ is contaminated with methanol $\mathbf{\left(}{\mathbf{\text{CH}}}_{\mathbf{\text{3}}}\mathbf{\text{OH}}\mathbf{\right)}$. A sample of vapor above the liquid mixture contains a $\raisebox{1ex}{$\mathbf{97}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{1}$}\right.$ mass ratio of ${\mathbf{\text{C}}}_{\mathbf{\text{2}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{OH}}$ to ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}\mathbf{\text{OH}}$. What is the mass ratio of these alcohols in the liquid? At the temperature of the liquid, the vapor pressures of ${\mathbf{\text{C}}}_{\mathbf{\text{2}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{OH}}$ and ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}\mathbf{\text{OH}}$ are $\mathbf{\text{60.5 torr}}$ and $\mathbf{\text{126.0 torr}}$, respectively.

Step-by-Step Solution

Answer

The mass ratio of ethanol and methanol in liquid is $\raisebox{1ex}{$204$}\!\left/ \!\raisebox{-1ex}{$1$}\right.$.

A concept:

Mole ratio of ethanol and methanol in liquid can be calculated by dividing moles of ethanol in vapor and partial vapor pressure of ethanol and moles of methanol divided by partial vapor pressure of methanol.

Moles of ethanol and methanol in vapor:

The mixture contains a $\raisebox{1ex}{$97$}\!\left/ \!\raisebox{-1ex}{$1$}\right.$ mass ratio of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ and ${\text{CH}}_{\text{3}}\text{OH}$.

Hence,

The mass of ethanol is $97\text{ g}$.

The mass of methanol is $\text{1 g}$.

 

Calculate the mass of ethanol as follow.

$\begin{array}{rcl}\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}ethanol }\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)& =& 2\times 12+1\times 6+16\\ & =& 46\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$ 

 

Calculate the mass of methanol as follow.

$\begin{array}{rcl}\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}methanol }\left({\text{CH}}_{\text{3}}\text{OH}\right)& =& 12+1\times 4+16\\ & =& 32\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$  

 

Moles of ethanol can be calculated as follow.

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}methanol }\left({\text{CH}}_{\text{3}}\text{OH}\right)& =& \text{1\hspace{0.17em}g}\times \frac{\text{1\hspace{0.17em}mol\hspace{0.17em}methanol}}{\text{32\hspace{0.17em}}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 0.031\text{ mol}\end{array}$ 

 

Moles of methanol can be calculated as follow.

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}ethanol }\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)& =& \text{97g}\times \frac{\text{1\hspace{0.17em}mol\hspace{0.17em}methanol}}{\text{46\hspace{0.17em}}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 2.11\text{ mol}\end{array}$

Mole ratio of ethanol and methanol in liquid:

Given the partial vapor pressure of ethanol and methanol is $60.5\text{ torr}$ and  $\text{126 torr}$ respectively.

 

Calculate the mole ratio of ethanol and methanol in liquid as below. 

$\begin{array}{rcl}\frac{\text{moles\hspace{0.17em}of\hspace{0.17em}ethanol\hspace{0.17em}in\hspace{0.17em}liquid}}{\text{moles\hspace{0.17em}of\hspace{0.17em}methanol\hspace{0.17em}in\hspace{0.17em}liquid}}& =& \frac{{\displaystyle \left(\frac{\text{moles\hspace{0.17em}of\hspace{0.17em}ethanol\hspace{0.17em}in\hspace{0.17em}vapor}}{\text{partial\hspace{0.17em}vapor\hspace{0.17em}pressure\hspace{0.17em}of\hspace{0.17em}ethanol}}\right)}}{{\displaystyle \left(\frac{\text{moles\hspace{0.17em}of\hspace{0.17em}methanol\hspace{0.17em}in\hspace{0.17em}vapor}}{\text{partial\hspace{0.17em}vapor\hspace{0.17em}pressure\hspace{0.17em}of\hspace{0.17em}methanol}}\right)}}\\ & =& \frac{{\displaystyle \left(\frac{\text{2.11}}{\text{60.5 torr}}\right)}}{{\displaystyle \left(\frac{\text{0.031}}{\text{126 torr}}\right)}}\\ & =& \frac{142}{1}\end{array}$

Mass ration of ethanol and methanol in liquid:

It knows the molar mass of ethanol and methanol are $46.7\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mo}$}\right.$ and $32.04\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$ respectively.

Since it can calculate the mass of any substance by multiplying its moles and molar mass.

 

Hence, you can calculate the mass ratio of ethanol and methanol by multiplying their moles with their molar mass as follow.

$\begin{array}{rcl}\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}ethanol\hspace{0.17em}in\hspace{0.17em}liquid}}{\text{mass\hspace{0.17em}of\hspace{0.17em}ethanol\hspace{0.17em}in\hspace{0.17em}liquid}}& =& \frac{142\text{\hspace{0.17em}mol}\times 46.07{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}{1\text{ mol}\times 32.04{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& \frac{204}{1}\end{array}$  

 

Hence, the mass ratio of ethanol and methanol in liquid is $\raisebox{1ex}{${\displaystyle 204}$}\!\left/ \!\raisebox{-1ex}{$1$}\right.$.