Chapter F, Problem F.12E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for talc 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.12E

Empirical formula for talc is ${\text{Mg}}_{\text{3}}{\text{Si}}_{\text{4}}{\text{O}}_{\text{10}}{\text{H}}_{\text{34}}$.

Explanation of Solution

The compound given is talc.  The mass percentage composition of talc is given as $\text{19}\text{.2\%}\text{Mg}$, $\text{29}\text{.6}\text{\%}\text{Si}$, $42.2\%\text{O}$, and $\text{9}\text{.0}\text{\%}\text{H}$.  Considering $\text{100}\text{g}$ of talc, it is understood that talc contains $\text{19}\text{.2}\text{g}$ of magnesium, $\text{29}\text{.6}\text{g}$ of silicon, $\text{42}\text{.2}\text{g}$ of oxygen, and $\text{9}\text{.0}\text{g}$ of hydrogen.

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

    $\begin{array}{c}\text{Moles}\text{of}\text{Magnesium}=\frac{19.2\text{g}}{24.305\text{g}\cdot {\text{mol}}^{-1}}\\ =0.7899\text{mol}\\ \text{Moles}\text{of}\text{Silicon}=\frac{29.6\text{g}}{28.08\text{g}\cdot {\text{mol}}^{-1}}\\ =1.05\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{42.2\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =2.63\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{9.0\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =8.928\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Magnesium}:\frac{0.7899\text{mol}}{0.7899\text{mol}}=1.00\\ \text{Silicon}:\frac{1.05\text{mol}}{0.7899\text{mol}}=1.329\\ \text{Oxygen}:\frac{\text{2}\text{.63}\text{mol}}{0.7899\text{mol}}=3.329\\ \text{Hydrogen}:\frac{\text{8}\text{.928}\text{mol}}{0.7899\text{mol}}=11.302\end{array}$

The atoms are present in the compound always in whole numbers.  Hence, multiplying by $3$, the ratio of each element is obtained as shown below;

    $\text{Mg}:\text{Si}:\text{O}:\text{H}=3:4:10:34$

Therefore, the empirical formula for talc can be given as ${\text{Mg}}_{\text{3}}{\text{Si}}_{\text{4}}{\text{O}}_{\text{10}}{\text{H}}_{\text{34}}$.

(b)

Interpretation Introduction

Interpretation:

Empirical formula for saccharin has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem F.12E

Empirical formula for saccharin is ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{NO}}_{\text{3}}\text{S}$.

Explanation of Solution

The mass percentage composition of saccharin is given as $\text{45}\text{.89\%}\text{C}$, $\text{2}\text{.75}\text{\%}\text{H}$, $\text{7}\text{.65}\text{\%}\text{N}$, $\text{26}\text{.20}\text{\%}\text{O}$, and $\text{17}\text{.50}\text{\%}\text{S}$.  Considering $\text{100}\text{g}$ of saccahrin, it is understood that the compound contains $\text{45}\text{.89}\text{g}$ of carbon, $\text{2}\text{.75}\text{g}$ of hydrogen, $\text{7}\text{.65}\text{g}$ of nitrogen, $\text{26}\text{.20}\text{g}$ of oxygen, and $\text{17}\text{.50}\text{g}$ of sulfur.

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

    $\begin{array}{c}\text{Moles}\text{of}\text{Carbon}=\frac{45.89\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =3.8209\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{2.75\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =2.728\text{mol}\\ \text{Moles}\text{of}\text{Nitrogen}=\frac{7.65\text{g}}{14.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.546\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{26.20\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =1.637\text{mol}\\ \text{Moles}\text{of}\text{Sulfur}=\frac{17.50\text{g}}{32.06\text{g}\cdot {\text{mol}}^{-1}}\\ =0.546\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Carbon}:\frac{\text{3}\text{.8209}\text{mol}}{0.546\text{mol}}=7.00\\ \text{Hydrogen}:\frac{\text{2}\text{.728}\text{mol}}{\text{0}\text{.546}\text{mol}}=5.00\\ \text{Nitrogen}:\frac{\text{0}\text{.546}\text{mol}}{\text{0}\text{.546}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{1}\text{.637}\text{mol}}{\text{0}\text{.546}\text{mol}}=3.00\\ \text{Sulfur}:\frac{\text{0}\text{.546}\text{mol}}{\text{0}\text{.546}\text{mol}}=1.00\end{array}$

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

    $7.00\text{C}:5.00\text{H:}\text{1}\text{.00}\text{N}:3.00\text{O}:1.00\text{S}$

Thus in saccharin, the atoms are present in the ratio of $\text{C}:\text{H}:\text{N}:\text{O}:\text{S}=7:5:1:3:1$.

Therefore, the empirical formula for saccharin can be given as ${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{NO}}_{\text{3}}\text{S}$.

(c)

Interpretation Introduction

Interpretation:

Empirical formula for salicylic acid has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem F.12E

Empirical formula for salicylic acid is ${\text{C}}_{\text{7}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$.

Explanation of Solution

The mass percentage composition of salicylic acid is given as $\text{60}\text{.87\%}\text{C}$, $\text{4}\text{.38}\text{\%}\text{H}$, and $\text{34}\text{.75}\text{\%}\text{O}$.  Considering $\text{100}\text{g}$ of salicylic acid, it is understood that the compound contains $\text{60}\text{.87}\text{g}$ of carbon, $\text{4}\text{.38}\text{g}$ of hydrogen, and $\text{34}\text{.75}\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Carbon}=\frac{60.87\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =5.068\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{4.38\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =4.34\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{34.75\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =2.17\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Carbon}:\frac{\text{5}\text{.068}\text{mol}}{\text{2}\text{.17}\text{mol}}=2.33\\ \text{Hydrogen}:\frac{\text{4}\text{.34}\text{mol}}{\text{2}\text{.17}\text{mol}}=2.00\\ \text{Oxygen}:\frac{\text{2}\text{.17}\text{mol}}{\text{2}\text{.17}\text{mol}}=1.00\end{array}$

The ratio of the atoms in salicylic acid is given as follows;

    $2.33\text{C}:2.00\text{H:}1.00\text{O}$

Multiplying by three in order to make all as whole numbers, in salicylic acid, the atoms are present in the ratio of $\text{C}:\text{H}:\text{O}=7:6:3$.

Therefore, the empirical formula for salicylic acid can be given as ${\text{C}}_{\text{7}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$.