Chapter F, Problem H.25E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for the phosphorus oxides 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 H.25E

Empirical formula for phosphorus oxides are ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$ and ${\text{P}}_{\text{2}}{\text{O}}_{\text{3}}$.

Explanation of Solution

Oxide 1:

The mass percentage composition of phosphorus in first oxide is given as $\text{43}\text{.64\%}$.  The mass percentage of phosphorus is calculated as shown below;

$\begin{array}{c}\text{Mass}\text{percentage}\text{of}\text{oxygen}=\text{Total}\text{mass}\text{percentage}-\text{Mass}\text{percentage}\text{of}\text{phosphorus}\\ =(100-43.64)\%\\ =56.36\%\end{array}$

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{43}\text{.64}\text{g}$ of phosphorus, and $\text{56}\text{.36}\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Phosphorus}=\frac{43.64\text{g}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\\ =1.41\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{56.34\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =3.52\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Phosphorus}:\frac{\text{1}\text{.41}\text{mol}}{\text{1}\text{.41}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{3}\text{.52}\text{mol}}{\text{1}\text{.41}\text{mol}}=2.49\end{array}$

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

    $1.00\text{C}:2.50\text{P}$

Thus in compound the atoms are present in the ratio of $\text{P}:\text{O}=1:2.50$.

Therefore, the empirical formula for one of the phosphorus oxide can be given as ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$.

Oxide 2:

The mass percentage composition of phosphorus in second oxide is given as $\text{56}\text{.34\%}$.  The mass percentage of phosphorus is calculated as shown below;

$\begin{array}{c}\text{Mass}\text{percentage}\text{of}\text{oxygen}=\text{Total}\text{mass}\text{percentage}-\text{Mass}\text{percentage}\text{of}\text{phosphorus}\\ =(100-56.34)\%\\ =43.66\%\end{array}$

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{56}\text{.34}\text{g}$ of phosphorus, and $\text{43}\text{.66}\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Phosphorus}=\frac{56.34\text{g}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\\ =1.819\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{43.66\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =2.728\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Phosphorus}:\frac{\text{1}\text{.819}\text{mol}}{\text{1}\text{.819}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{2}\text{.728}\text{mol}}{\text{1}\text{.819}\text{mol}}=1.49\end{array}$

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

    $1.00\text{C}:1.50\text{P}$

Thus in compound the atoms are present in the ratio of $\text{P}:\text{O}=1:1.50$.

Therefore, the empirical formula for another phosphorus oxide can be given as ${\text{P}}_{\text{2}}{\text{O}}_{\text{3}}$.

(b)

Interpretation Introduction

Interpretation:

Molecular formula of the phosphorus oxides 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 H.25E

Molecular formula of the phosphorus oxides are ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ and ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$.  These oxides are named as phosphorus pentoxide and phosphorus trioxide.

Explanation of Solution

Molecular formula for ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$:

Empirical formula of the compound is ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$.  Molar mass of the compound is given as $283.33\text{g}\cdot {\text{mol}}^{-1}$.

Molar mass of the empirical formula is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{P}}_{\text{2}}{\text{O}}_{\text{5}}=2\times 30.97\text{g}\cdot {\text{mol}}^{-1}+5\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =61.94\text{g}\cdot {\text{mol}}^{-1}+80.00\text{g}\cdot {\text{mol}}^{-1}\\ =141.94\text{g}\cdot {\text{mol}}^{-1}\end{array}$

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

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{283.33\text{g}\cdot {\text{mol}}^{-1}}{141.94\text{g}\cdot {\text{mol}}^{-1}}\\ =2.00\end{array}$

The coefficient of empirical formula is multiplied by the factor $2.00$ in order to obtain the molecular formula as shown below;

    $\begin{array}{c}\text{Molecular}\text{formula}\text{of}\text{compound}=2\times ({\text{P}}_{\text{2}}{\text{O}}_{\text{5}})\\ ={\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\end{array}$

Therefore, the molecular formula of one phosphorus oxide is ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$.  This compound is named as phosphorus pentoxide.

Molecular formula for ${\text{P}}_{\text{2}}{\text{O}}_{\text{3}}$:

Empirical formula of the compound is ${\text{P}}_{\text{2}}{\text{O}}_{\text{3}}$.  Molar mass of the compound is given as $219.88\text{g}\cdot {\text{mol}}^{-1}$.

Molar mass of the empirical formula is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{P}}_{\text{2}}{\text{O}}_{\text{3}}=2\times 30.97\text{g}\cdot {\text{mol}}^{-1}+3\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =61.94\text{g}\cdot {\text{mol}}^{-1}+48.00\text{g}\cdot {\text{mol}}^{-1}\\ =109.94\text{g}\cdot {\text{mol}}^{-1}\end{array}$

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

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{219.88\text{g}\cdot {\text{mol}}^{-1}}{109.94\text{g}\cdot {\text{mol}}^{-1}}\\ =2.00\end{array}$

The coefficient of empirical formula is multiplied by the factor $2.00$ in order to obtain the molecular formula as shown below;

    $\begin{array}{c}\text{Molecular}\text{formula}\text{of}\text{compound}=2\times ({\text{P}}_{\text{2}}{\text{O}}_{\text{3}})\\ ={\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\end{array}$

Therefore, the molecular formula of second phosphorus oxide is ${\text{P}}_{\text{4}}{\text{O}}_6$. 

(c)

Interpretation Introduction

Interpretation:

Balanced equation for the formation of two phosphorus oxides has to be written.

Concept Introduction:

Chemical equation is a short form representation of a chemical reaction with all the required conditions necessary for the reaction to take place.  As the atoms are neither created nor destroyed in a chemical reaction, the number of atoms of same element has to be equal on both sides of the equation.  This is known as balanced chemical equation.  Coefficients can be changed but not the subscript.

Explanation of Solution

Balanced equation for the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$:

Phosphorus reacts with oxygen to form phosphorus pentoxide.  The skeletal equation for this reaction can be written as follows;

    $\text{P}+{\text{O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$

Balancing phosphorus atoms:  In the reactant side, there is one phosphorus atom, while on the product side there are four phosphorus atoms.  Adding coefficient $4$ before $\text{P}$ balances the phosphorus atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    $\text{4P}+{\text{O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms, while on the product side there are ten oxygen atoms.  Adding coefficient $5$ before ${\text{O}}_{\text{2}}$ balances the oxygen atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    $\text{4P}+5{\text{O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$

Balanced equation for the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$:

Phosphorus reacts with oxygen to form phosphorus trioxide.  The skeletal equation for this reaction can be written as follows;

    $\text{P}+{\text{O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$

Balancing phosphorus atoms:  In the reactant side, there is one phosphorus atom, while on the product side there are four phosphorus atoms.  Adding coefficient $4$ before $\text{P}$ balances the phosphorus atoms on both sides of the equation.  Therefore, the chemical equation can be written as follows;

    $\text{4P}+{\text{O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$

Balancing oxygen atoms:  In the reactant side, there are two oxygen atoms, while on the product side there are six oxygen atoms.  Adding coefficient $3$ before ${\text{O}}_{\text{2}}$ balances the oxygen atoms on both sides of the equation.  Therefore, the balanced chemical equation can be written as follows;

    $\text{4P}+3{\text{O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$