Chapter F, Problem J.16E

(a)

Interpretation Introduction

Interpretation:

Molecular formula for the compound A 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.

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 J.16E

Molecular formula for the compound A is ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$.

Explanation of Solution

The mass percentage composition of compound A is given as $\text{43}\text{.64\%}\text{P}$.  The remaining mass percentage is said to be oxygen.  The mass percentage of oxygen in the compound is calculated as follows;

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

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{43}\text{.64}\text{g}$ of phosphorus, and $56.36\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Phosphorus}=\frac{43.64\text{g}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\\ =1.41\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{56.36\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =3.52\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Phosphorus}:\frac{\text{1}\text{.41}\text{mol}}{\text{1}\text{.41}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{3}\text{.52}\text{mol}}{\text{1}\text{.41}\text{mol}}=2.50\end{array}$

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

    $1.00\text{P}:2.50\text{O}$

Thus in compound the atoms are present in the ratio of $\text{P}:\text{O}=1:2.5$.  Rounding it of two whole number, the empirical formula can be given as ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$.

Empirical formula of compound A is ${\text{P}}_{\text{2}}{\text{O}}_{\text{5}}$.  Molar mass of compound A is given as $283.9\text{g}\cdot {\text{mol}}^{-1}$.

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

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{A}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{283.9\text{g}\cdot {\text{mol}}^{-1}}{141.94\text{g}\cdot {\text{mol}}^{-1}}\\ =2.00\end{array}$

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

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

Therefore, the molecular formula of compound A is ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$.

(b)

Interpretation Introduction

Interpretation:

Molecular formula for the compound B has to be given.

Concept Introduction:

Refer part (a).

Answer to Problem J.16E

Molecular formula for the compound B is ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$.

Explanation of Solution

The mass percentage composition of compound B is given as $\text{31}\text{.60\%}\text{P}$, and $3.087\%\text{H}$.  The remaining mass percentage is said to be oxygen.  The mass percentage of oxygen in the compound is calculated as follows;

    $\begin{array}{c}\text{Mass}\text{\%}\text{of}\text{O}=\text{Total}\text{mass}\text{\%}-(\text{Mass}\text{\%}\text{P+Mass}\text{\%}\text{H})\\ =(100-(31.60+3.087))\%\\ =(100-34.687)\%\\ =65.313\%\end{array}$

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{31}\text{.60}\text{g}$ of phosphorus, $\text{3}\text{.087}\text{g}$ of hydrogen, and $65.313\text{g}$ 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;

    $\begin{array}{c}\text{Moles}\text{of}\text{Phosphorus}=\frac{31.60\text{g}}{30.97\text{g}\cdot {\text{mol}}^{-1}}\\ =1.02\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{65.313\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =4.08\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{3.087\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =3.06\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Phosphorus}:\frac{\text{1}\text{.02}\text{mol}}{\text{1}\text{.02}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{4}\text{.08}\text{mol}}{\text{1}\text{.02}\text{mol}}=4.00\\ \text{Hydrogen}:\frac{\text{3}\text{.06}\text{mol}}{\text{1}\text{.02}\text{mol}}=3.00\end{array}$

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

    $1.00\text{P}:4.00\text{O:3}\text{.00}\text{H}$

Thus in compound the atoms are present in the ratio of $\text{P}:\text{O}:\text{H}=1:4:3$.

Empirical formula of compound B is ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$.  Molar mass of compound B is given as $97.99\text{g}\cdot {\text{mol}}^{-1}$.

Molar mass of the empirical formula is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}=3\times 1.008\text{g}\cdot {\text{mol}}^{-1}+1\times 30.97\text{g}\cdot {\text{mol}}^{-1}+4\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =3.024\text{g}\cdot {\text{mol}}^{-1}+30.97\text{g}\cdot {\text{mol}}^{-1}+64.00\text{g}\cdot {\text{mol}}^{-1}\\ =97.99\text{g}\cdot {\text{mol}}^{-1}\end{array}$

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

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{B}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{97.99\text{g}\cdot {\text{mol}}^{-1}}{97.99\text{g}\cdot {\text{mol}}^{-1}}\\ =1.00\end{array}$

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

    $\begin{array}{c}\text{Molecular}\text{formula}\text{of}\text{B}=1\times ({\text{H}}_{\text{3}}{\text{PO}}_{\text{4}})\\ ={\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\end{array}$

Therefore, the molecular formula of compound B is ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$.

(c)

Interpretation Introduction

Interpretation:

Balanced chemical equation for the preparation of compound A, compound B, and compound C has to be given.

Explanation of Solution

Balanced equation for compound A:

Compound A is said to have formed from phosphorus burning in air.  Therefore, the reaction between phosphorus and oxygen is given as shown below;

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

Balancing oxygen atoms:  In the above chemical equation, there are two oxygen atoms on the left side of the equation, while in the product side, there are ten oxygen atoms.  Adding coefficient $5$ before ${\text{O}}_{\text{2}}$ in the reactant side balances the oxygen atoms on both sides of the equation.  The balanced chemical equation is given as shown below;

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

Balanced equation for compound B:

Compound B is said to have formed from compound A by reacting with water.  Therefore, the reaction between compound A and water is given as shown below;

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Balancing phosphorus atom:  In the above chemical equation, there are four phosphorus atoms on the left side of the equation, while in the product side, there is one phosphorus atom.  Adding coefficient $4$ before ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ in the product side balances the phosphorus atoms on both sides of the equation.  The chemical equation is given as shown below;

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+H}}_{\text{2}}\text{O}\to 4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Balancing oxygen atoms:  In the above chemical equation, there are eleven oxygen atoms on the left side of the equation, while in the product side, there are sixteen oxygen atoms.  Adding coefficient $6$ before ${\text{H}}_{\text{2}}\text{O}$ in the reactant side balances the oxygen atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{+6H}}_{\text{2}}\text{O}\to 4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Balanced equation for compound C:

Compound B reacts with calcium hydroxide to form compound C as a white precipitate.  Compound B is an acid while calcium hydroxide is a strong base.  The product formed will be salt and water.  Therefore, the reaction between compound B and calcium hydroxide is given as shown below;

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+Ca(OH)}}_{\text{2}}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}})}_2+{\text{H}}_{\text{2}}\text{O}$

Balancing calcium atom:  In the above chemical equation, there is one calcium atom on the left side of the equation, while in the product side, there are three calcium atoms.  Adding coefficient $3$ before ${\text{Ca(OH)}}_{\text{2}}$ in the reactant side balances the calcium atoms on both sides of the equation.  The chemical equation is given as shown below;

    ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+3Ca(OH)}}_{\text{2}}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}})}_2+{\text{H}}_{\text{2}}\text{O}$

Balancing phosphate group:  In the above chemical equation, there is one phosphate ion on the left side of the equation, while in the product side, there are two phosphate ions.  Adding coefficient $2$ before ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ in the reactant side balances the phosphate ions on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+3Ca(OH)}}_{\text{2}}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}})}_2+{\text{H}}_{\text{2}}\text{O}$

Balancing hydrogen atom:  In the above chemical equation, there are twelve hydrogen atoms on the left side of the equation, while in the product side, there are two hydrogen atoms.  Adding coefficient $6$ before ${\text{H}}_{\text{2}}\text{O}$ in the product side balances the hydrogen atoms on both sides of the equation.  This step balances the other atoms also.  The balanced chemical equation is given as shown below;

    ${\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{+3Ca(OH)}}_{\text{2}}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}})}_2+6{\text{H}}_{\text{2}}\text{O}$