Q137CP - Chemistry: Molecular Nature Of Matter And Change

Question

At an air-water interface, fatty acids such as oleic acid lie in a one-molecule-thick layer (monolayer), with the heads in the water and the tails perpendicular in the air. When $\mathbf{2}\mathbf{.50}\mathbf{\text{ mg}}$ of oleic acid is placed on a water surface, it forms a circular monolayer $\mathbf{\text{38.6 cm}}$ in diameter. Find the surface area (in ${\mathbf{cm}}^2$ ) occupied by one molecule ($\mathbf{{\rm M}}$ of oleic acid is $\mathbf{283}\raisebox{1ex}{$\mathbf{\text{ g}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{mol}}$}\right.$ ).

Step-by-Step Solution

Answer

The surface area occupied by one molecule is $2.19\times {10}^{-16}\raisebox{1ex}{${\displaystyle {\text{cm}}^{\text{2}}}$}\!\left/ \!\raisebox{-1ex}{$\text{molecule}$}\right.$.

A concept:

Moles of oleic acid can be calculated by dividing its mass by its molar mass.

The surface area occupied by one molecule by dividing surface area by the moles of oleic acid.

Consider the given data as below.

The mass of oleic acid, $m=2.50\text{ mg}=2.50\times {10}^{-3}\text{ g}$ 

The molar mass of oleic acid, $M=283\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Molecules of oleic acid:

Determine the molecules of oleic acid as below.

$\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}oleic\hspace{0.17em}acid}& =& \frac{\text{mass}}{\text{molar\hspace{0.17em}mass}}\\ & =& \frac{m}{M}\\ & =& \frac{2.50\times {10}^{-3}\text{ g}}{283{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& 8.83\times {10}^{-6}\text{ mol}\end{array}$

 

One mole of substance is $6.022\times {10}^{23}\text{ molecules}$.

 

Hence, the number of oleic acid molecules is,

$(8.83\times {10}^{-6}\text{ mol})\times \frac{6.022\times {10}^{23}\text{ molecules}}{1\text{ mol}}=5.32\times {10}^{18}\text{ molecules}$

Surface area occupied by one molecule of oleic acid:

The diameter of circular monolayer, $d=38.6\text{ cm}$

 The radius is half of diameter, so it becomes,

$\begin{array}{rcl}r& =& \frac{38.6\text{ cm}}{2}\\ & =& 19.3\text{ cm}\end{array}$ 

 

Now, the surface area of molecules can be calculated by using following formula.

$\begin{array}{rcl}A& =& \pi r^2\\ & =& 3.14\times {\left(19.3\text{ cm}\right)}^2\\ & =& 1169.61{\text{ cm}}^2\end{array}$

                                

Thus, it can find out the surface area occupied by one molecule by dividing above calculated surface area by the moles of oleic acid.

$\begin{array}{rcl}\text{surface\hspace{0.17em}area\hspace{0.17em}occupied\hspace{0.17em}by\hspace{0.17em}one\hspace{0.17em}molecule}& =& \frac{1169.61{\text{ cm}}^2}{5.32\times {10}^{18}\text{ molecule}}\\ & =& 2.19\times {10}^{-16}\raisebox{1ex}{${\displaystyle {\text{cm}}^{\text{2}}}$}\!\left/ \!\raisebox{-1ex}{$\text{molecule}$}\right.\end{array}$ 

 

Hence, the surface area occupied by one molecule is  $2.19\times {10}^{-16}\raisebox{1ex}{${\displaystyle {\text{cm}}^{\text{2}}}$}\!\left/ \!\raisebox{-1ex}{$\text{molecule}$}\right.$.