3.46P - Chemistry: Molecular Nature Of Matter And Change

Question

Menthol (m = 156.3 g/mol), a strong-smelling substance used in cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g of menthol was subjected to combustion analysis, it produced 0.449 g of CO2 and 0.184 g of H2O. What is menthol’s molecular formula?

Step-by-Step Solution

Answer

The empirical formula mass calculated and the molar mass given are the same. Therefore, the molecular formula of menthol is also ${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{22}}\mathbf{O}$.

So final answer for the molecular formula of menthol is ${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{22}}\mathbf{O}$.

Step1. Preliminary formula.

Given data in the question is as follows:

M_w of menthol = 156.3 g/mol

Mass of menthol = 0.1595 g

Mass of CO₂ produced = 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:

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

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

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

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

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

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

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

Step2. Final calculation.

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

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

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

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

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

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

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