Chapter F, Problem F.18E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for the paclitaxel has to be given.

Concept Introduction:

Empirical formula is the one that can be determined from the molar mass of the elements that is present in the compound and the mass percentage of the elements.  The mass percentage of the elements present in the compound is converted into the moles of each element considering the molar mass of each element.  The relative number of moles for each type of atoms is found out finally.

Answer to Problem F.18E

Empirical formula of paclitaxel is ${\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}$.

Explanation of Solution

The mass percentage composition of paclitaxel is given as $\text{66}\text{.11\%}\text{C}$, $\text{6}\text{.02}\text{\%}\text{H}$, and $\text{1}\text{.64}\text{\%}\text{N}$ and the remaining is oxygen.

The mass percentage of oxygen is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{percentage}\text{of}\text{oxygen}=100\%-(66.11+6.02+1.64)\%\\ =(100-73.77)\%\\ =26.23\%\end{array}$

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{66}\text{.11}\text{g}$ of carbon, $\text{6}\text{.02}\text{g}$ of hydrogen, $\text{1}\text{.64}\text{g}$ of nitrogen, and $\text{26}\text{.23}\text{g}$ of oxygen.

Number of moles of each element present in the compound can be calculated using the molar mass and mass of the element as follows;

    $\begin{array}{c}\text{Moles}\text{of}\text{Carbon}=\frac{66.11\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =5.505\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{6.02\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =5.972\text{mol}\\ \text{Moles}\text{of}\text{Nitrogen}=\frac{1.64\text{g}}{14.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.117\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{26.23\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =1.639\text{mol}\end{array}$

Dividing the moles of element obtained using the smallest amount, the ratio can be obtained as shown below;

    $\begin{array}{c}\text{Carbon}:\frac{\text{5}\text{.505}\text{mol}}{\text{0}\text{.117}\text{mol}}=47.05\\ \text{Hydrogen}:\frac{\text{5}\text{.972}\text{mol}}{\text{0}\text{.117}\text{mol}}=51.04\\ \text{Nitrogen}:\frac{\text{0}\text{.117}\text{mol}}{\text{0}\text{.117}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{1}\text{.639}\text{mol}}{\text{0}\text{.117}\text{mol}}=14.00\end{array}$

The ratio of the atoms in the compound is given as follows;

    $47.05\text{C}:51.04\text{H}:1.00\text{N}:\text{14}\text{.00}\text{O=}47.00\text{C}:51.00\text{H}:1.00\text{N}:\text{14}\text{.00}\text{O}$

Thus in compound the atoms are present in the ratio of $\text{C}:\text{H}:\text{N}:\text{O}=47:51:1:14$.

Therefore, the empirical formula for the compound can be given as ${\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}$.

(b)

Interpretation Introduction

Interpretation:

Molecular formula of paclitaxel has to be found out.

Concept Introduction:

Empirical formula is the one that can be determined from the molar mass of the elements that is present in the compound and the mass percentage of the elements.  The mass percentage of the elements present in the compound is converted into the moles of each element considering the molar mass of each element.  The relative number of moles for each type of atoms is found out finally.

Molecular formula of a compound can be found if the empirical formula and molar mass of the compound is known.  The molar mass of compound is divided by the molar mass of the empirical formula in order to obtain the factor which is multiplied with the coefficients of empirical formula in order to obtain the molecular formula.

Answer to Problem F.18E

Molecular formula of paclitaxel is ${\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}$.

Explanation of Solution

Empirical formula of the compound is ${\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}$.  Molar mass of the compound is given as $853.91\text{g}\cdot {\text{mol}}^{-1}$.

Molar mass of the empirical formula is calculated as follows;

$\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}=47\times 12.01\text{g}\cdot {\text{mol}}^{-1}+51\times 1.008\text{g}\cdot {\text{mol}}^{-1}+1\times 14.01\text{g}\cdot {\text{mol}}^{-1}\\ +14\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =564.47\text{g}\cdot {\text{mol}}^{-1}+51.408\text{g}\cdot {\text{mol}}^{-1}+14.01\text{g}\cdot {\text{mol}}^{-1}\\ +224.00\text{g}\cdot {\text{mol}}^{-1}\\ =853.88\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Molar mass of the paclitaxel is divided by the molar mass of empirical formula in order to obtain the factor as shown below;

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{paclitaxel}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{853.91\text{g}\cdot {\text{mol}}^{-1}}{853.88\text{g}\cdot {\text{mol}}^{-1}}\\ =1.00\end{array}$

The coefficient of empirical formula is multiplied by the factor $1.00$ in order to obtain the molecular formula as shown below;

    $\begin{array}{c}\text{Molecular}\text{formula}\text{of}\text{compound}=1\times ({\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}})\\ ={\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}\end{array}$

Therefore, the molecular formula of paclitaxel is ${\text{C}}_{\text{47}}{\text{H}}_{\text{51}}{\text{NO}}_{\text{14}}$.