Chapter F, Problem F.11E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for cryolite 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.11E

Empirical formula for cryolite is ${\text{Na}}_{\text{3}}{\text{AlF}}_{\text{6}}$.

Explanation of Solution

The compound given is cryolite.  The mass percentage composition of cryolite is given as $\text{32}\text{.79\%}\text{Na}$, $\text{13}\text{.02}\text{\%}\text{Al}$, and $\text{54}\text{.19}\text{\%}\text{F}$.  Considering $\text{100}\text{g}$ of cryolite, it is understood that cryolite contains $\text{32}\text{.79}\text{g}$ of sodium, $\text{13}\text{.02}\text{g}$ of aluminium, and $\text{54}\text{.19}\text{g}$ of fluorine.

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

    $\begin{array}{c}\text{Moles}\text{of}\text{Sodium}=\frac{32.79\text{g}}{22.99\text{g}\cdot {\text{mol}}^{-1}}\\ =1.43\text{mol}\\ \text{Moles}\text{of}\text{Aluminium}=\frac{13.02\text{g}}{26.98\text{g}\cdot {\text{mol}}^{-1}}\\ =0.483\text{mol}\\ \text{Moles}\text{of}\text{Fluorine}=\frac{54.19\text{g}}{18.99\text{g}\cdot {\text{mol}}^{-1}}\\ =2.853\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Sodium}:\frac{\text{1}\text{.43}\text{mol}}{0.483\text{mol}}=2.96\\ \text{Aluminium}:\frac{0.483\text{mol}}{0.483\text{mol}}=1.00\\ \text{Fluorine}:\frac{\text{2}\text{.853}\text{mol}}{0.483\text{mol}}=5.906\end{array}$

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

    $2.96\text{Na}:1.00\text{Al:}\text{5}\text{.906}\text{F}=3.00\text{Na}:1.00\text{Al}:6.00\text{F}$

Thus in cryolite, the atoms are present in the ratio of $\text{Na}:\text{Al}:\text{F}=3:1:6$.

Therefore, the empirical formula for cryolite can be given as ${\text{Na}}_{\text{3}}{\text{AlF}}_{\text{6}}$.

(b)

Interpretation Introduction

Interpretation:

Empirical formula for used to generate oxygen has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem F.11E

Empirical formula for compound is ${\text{KClO}}_{\text{3}}$.

Explanation of Solution

The mass percentage composition of compound is given as $\text{31}\text{.91\%}\text{K}$, $\text{28}\text{.93}\text{\%}\text{Cl}$, and the remaining is oxygen.  Therefore, the percentage of oxygen is calculated as follows;

    $\begin{array}{c}\text{Mass}\text{percentage}\text{of}\text{oxygen}=(100-(31.91+28.93))\%\\ =(100-60.84)\%\\ =39.16\%\end{array}$

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{31}\text{.91}\text{g}$ of potassium, $\text{28}\text{.93}\text{g}$ of chlorine, and $\text{39}\text{.16}\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Potassium}=\frac{31.91\text{g}}{39.09\text{g}\cdot {\text{mol}}^{-1}}\\ =0.816\text{mol}\\ \text{Moles}\text{of}\text{Chlorine}=\frac{28.93\text{g}}{35.45\text{g}\cdot {\text{mol}}^{-1}}\\ =0.816\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{39.16\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =2.447\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Potassium}:\frac{\text{0}\text{.816}\text{mol}}{\text{0}\text{.816}\text{mol}}=1.00\\ \text{Chlorine}:\frac{\text{0}\text{.816}\text{mol}}{\text{0}\text{.816}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{2}\text{.447}\text{mol}}{\text{0}\text{.816}\text{mol}}=2.99\end{array}$

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

    $1.00\text{K}:1.00\text{Cl:}\text{3}\text{.00}\text{O}$

Thus in compound, the atoms are present in the ratio of $\text{K}:\text{Cl}:\text{O}=1:1:3$.

Therefore, the empirical formula for compound can be given as ${\text{KClO}}_{\text{3}}$.

(c)

Interpretation Introduction

Interpretation:

Empirical formula for the fertilizer has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem F.11E

Empirical formula for fertilizer is ${\text{NH}}_{\text{6}}{\text{PO}}_{\text{4}}$.

Explanation of Solution

The mass percentage composition of compound is given as $\text{12}\text{.2\%}\text{N}$, $\text{5}\text{.26}\text{\%}\text{H}$, $\text{26}\text{.9}\text{\%}\text{P}$, and $\text{55}\text{.6}\text{\%}\text{O}$.  Considering $\text{100}\text{g}$ of fertilizer, it is understood that the compound contains $\text{12}\text{.2}\text{g}$ of nitrogen, $\text{5}\text{.26}\text{g}$ of hydrogen, $\text{26}\text{.9}\text{g}$ of phosphorus, and $\text{55}\text{.6}\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Nitrogen}=\frac{12.2\text{g}}{14.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.871\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{5.26\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =5.218\text{mol}\\ \text{Moles}\text{of}\text{Phosphorus}=\frac{26.9\text{g}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\\ =0.8685\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{55.6\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =3.475\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Nitrogen}:\frac{\text{0}\text{.871}\text{mol}}{\text{0}\text{.8685}\text{mol}}=1.00\\ \text{Hydrogen}:\frac{\text{5}\text{.218}\text{mol}}{\text{0}\text{.8685}\text{mol}}=6.00\\ \text{Phosphorus}:\frac{\text{0}\text{.8685}\text{mol}}{\text{0}\text{.8685}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{3}\text{.475}\text{mol}}{\text{0}\text{.8685}\text{mol}}=4.00\end{array}$

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

    $1.00\text{N}:6.00\text{H:}\text{1}\text{.00}\text{P:}\text{4}\text{.00}\text{O}$

Thus in fertilizer, the atoms are present in the ratio of $\text{N}:\text{H}:\text{P}:\text{O}=1:6:1:4$.

Therefore, the empirical formula for fertilizer can be given as ${\text{NH}}_{\text{6}}{\text{PO}}_{\text{4}}$.