91Ó°ÊÓ

Given data in the question is as follows:

M_wof menthol = 156.3 g/mol

Mass of menthol = 0.1595 g

Mass of CO₂produced = 0.449 g

Mass of H₂O produced = 0.184 g

Now let’s assume complete combustion so that all of the carbon presents in the sample is found in the carbon of CO₂ produced, and all of the hydrogen present in the sample is found in the hydrogen of H₂O produced. Whereas the molar mass of CO₂ and H₂O are 44.01 and 18.02 g/mol, respectively.

The moles of carbon and hydrogen can be calculated as below:

$\mathrm{Moles}\left(\mathrm{C}\right)=0.449\mathrm{g}\left({\mathrm{CO}}_2\right)\left(\frac{{\mathrm{molCO}}_2}{44.01{\mathrm{gCO}}_2}\right)\left(\frac{1\mathrm{molC}}{1{\mathrm{molCO}}_2}\right)=1.0202×{10}^{-2}\mathrm{molC}$

$\mathrm{Moles}\left(\mathrm{H}\right)=0.184\mathrm{g}\left({\mathrm{H}}_2\mathrm{O}\right)\left(\frac{{\mathrm{molH}}_2\mathrm{O}}{18.02{\mathrm{gH}}_2\mathrm{O}}\right)\left(\frac{2\mathrm{molH}}{1{\mathrm{molH}}_2\mathrm{O}}\right)=2.0422×{10}^{-2}\mathrm{molH}$

Determination of the masses of C and H in the sample by multiplying the calculated number of moles by their molar masses.

$\mathrm{Mass}\mathrm{of}\mathrm{C}=1.0202×{10}^{-2}\mathrm{molC}\left(\frac{12.01\mathrm{gC}}{\mathrm{molC}}\right)=0.1225\mathrm{gC}$

$\mathrm{Mass}\mathrm{of}\mathrm{H}=2.0422×{10}^{-2}\mathrm{molH}\left(\frac{1.01\mathrm{gH}}{\mathrm{molH}}\right)=0.0206\mathrm{gH}$

Subtracting the masses of C and H from the given mass of menthol to get the mass of O.

$\begin{array}{rcl}\mathrm{Moles}\mathrm{of}\mathrm{O}& =& \mathrm{Mass}\mathrm{of}\mathrm{menthol}-\mathrm{Mass}\mathrm{of}\mathrm{C}-\mathrm{Mass}\mathrm{of}\mathrm{H}\\ \mathrm{Moles}\mathrm{of}\mathrm{O}& =& 0.1595\mathrm{g}-0.1225\mathrm{g}-0.0206\mathrm{g}\\ & =& 0.0164\mathrm{g}\end{array}$

Multiplying the calculated mass of O by the reciprocal of its molar mass to get the number of moles of O.

$\mathrm{Moles}\mathrm{of}\mathrm{O}=0.0164\mathrm{g}\mathrm{O}\left(\frac{\mathrm{molO}}{16.00\mathrm{gO}}\right)=1.0250×{10}^{-3}\mathrm{molO}$

The preliminary formula for the compound is ${\mathbf{C}}_{\mathbf{1}\mathbf{.}\mathbf{0202}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}{\mathbf{H}}_{\mathbf{2}\mathbf{.}\mathbf{0422}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}{\mathbf{O}}_{\mathbf{1}\mathbf{.}\mathbf{0250}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}$

Dividing all the above subscripts by the subscript with the smallest value which is 1.0250´10^-3

$$\begin{gathered} {\mathrm{C}}_{\frac{1.0202×{10}^{-2}}{1.0250×{10}^{-3}}}{\mathrm{H}}_{\frac{2.0422×{10}^{-2}}{1.0250×{10}^{-3}}}{\mathrm{O}}_{\frac{1.0250×{10}^{-3}}{1.0250×{10}^{-3}}} \\ {\mathrm{C}}_{10}{\mathrm{H}}_{20}{\mathrm{O}}_1 \end{gathered}$$

The empirical formula is${\mathrm{C}}_{10}{\mathrm{H}}_{20}{\mathrm{O}}_1$and its empirical formula mass is calculated below:

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{empirical}\mathrm{formula}& =& 10×\left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{C}\right)+20\left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{H}\right)+\left({\mathrm{M}}_{\mathrm{w}}\mathrm{of}\mathrm{O}\right)\\ & =& 10\left(12.01\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\right)+20×\left(1.01\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\right)+\left(16.00\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\right)\\ & =& 156.3\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

The empirical formula mass calculated and the molar mass given are the same. Therefore, the molecular formula of menthol is also${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{20}}\mathbf{O}$

So final answer for the molecular formula of menthol is${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{20}}\mathbf{O}$