Chapter F, Problem L.38E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for the acid 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 L.38E

Empirical formula for the acid is ${\text{CHO}}_{\text{2}}$.

Explanation of Solution

Given sample of acid is said to contain $\text{0}\text{.224}\text{g}$ of hydrogen, $\text{2}\text{.67}\text{g}$ of carbon, and the remaining is oxygen.  The mass of oxygen in $\text{10}\text{.0}\text{g}$ of acid is calculated as follows;

    $\begin{array}{c}\text{Mass}\text{of}\text{oxygen}=\text{Total}\text{mass}-(\text{Mass}\text{of}\text{hydrogen}+\text{Mass}\text{of}\text{carbon})\\ =10.0g-(0.224g+2.67\text{g)}\\ \text{=10}\text{.0}\text{g}-2.894\text{g}\\ \text{=7}\text{.106}\text{g}\end{array}$

Therefore, the mass of oxygen present in $\text{10}\text{.0}\text{g}$ of acid is $\text{7}\text{.106}\text{g}$.

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

    $\begin{array}{c}\text{Moles}\text{of}\text{Carbon}=\frac{2.67\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =0.222\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{7.106\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =0.444\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{0.224\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =0.222\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Carbon}:\frac{\text{0}\text{.222}\text{mol}}{\text{0}\text{.222}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{0}\text{.444}\text{mol}}{\text{0}\text{.222}\text{mol}}=2.00\\ \text{Hydrogen}:\frac{\text{0}\text{.222}\text{mol}}{\text{0}\text{.222}\text{mol}}=1.00\end{array}$

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

    $1.00\text{C}:2.00\text{O}:\text{1}\text{.00}\text{H}$

Thus in compound the atoms are present in the ratio of $\text{C}:\text{H}:\text{O}=1:1:2$.

Therefore, the empirical formula for the compound can be given as ${\text{CHO}}_{\text{2}}$.

(b)

Interpretation Introduction

Interpretation:

Molecular formula of the unknown acid 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 L.38E

Molecular formula of the unknown acid is ${\text{C}}_2{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$.

Explanation of Solution

Empirical formula of the acid is ${\text{CHO}}_{\text{2}}$.  The acid is said to be diprotic.  Consider the unknown acid as ${\text{H}}_{\text{2}}\text{A}$.  Therefore, the reaction of unknown acid with sodium hydroxide is given as follows;

    ${\text{H}}_{\text{2}}\text{A(aq)}+2\text{NaOH(aq)}\to {\text{Na}}_{\text{2}}\text{A}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$

From the above equation, it is found that one mole of unknown acid reacts with two moles of sodium hydroxide.

Volume of sodium hydroxide is given as $\text{50}\text{.0}\text{mL}$ and the concentration is given as $\text{0}\text{.040}\text{M}$.  Therefore, the number of moles of sodium hydroxide is calculated as follows;

    $\begin{array}{c}{n}_{\text{NaOH}}=50.0\text{mL}\times \frac{0.040\text{mol}}{1\text{L}}\\ =0.050\text{L}\times \frac{0.040\text{mol}}{1\text{L}}\\ =0.002\text{mol}\end{array}$

From the stoichiometric relation, the moles of ${\text{H}}_{\text{2}}\text{A}$ can be calculated as follows;

    $\begin{array}{c}{n}_{{\text{H}}_{\text{2}}\text{A}}=\text{0}\text{.002}\text{mol}\text{NaOH}\times \frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{A}}{\text{2}\text{mol}\text{NaOH}}\\ =0.001\text{mol}{\text{H}}_{\text{2}}\text{A}\\ =1.0\times {10}^{-3}\text{mol}\end{array}$

Mass of the unknown acid taken is given as $\text{0}\text{.0900}\text{g}$.  Considering the moles of unknown acid and mass of unknown acid, the molar mass of acid can be calculated as follows;

    $\begin{array}{c}M=\frac{m_{{\text{H}}_{\text{2}}\text{A}}}{n_{{\text{H}}_{\text{2}}\text{A}}}\\ =\frac{0.0900\text{g}}{1.0\times {10}^{-3}\text{mol}}\\ =0.09\times {10}^3\text{g}\cdot {\text{mol}}^{-1}\\ =90\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Molar mass of the empirical formula of unknown acid is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{CHO}}_{\text{2}}=1\times 12.01\text{g}\cdot {\text{mol}}^{-1}+1\times 1.008\text{g}\cdot {\text{mol}}^{-1}+2\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =12.01\text{g}\cdot {\text{mol}}^{-1}+1.008\text{g}\cdot {\text{mol}}^{-1}+32.00\text{g}\cdot {\text{mol}}^{-1}\\ =45.018\text{g}\cdot {\text{mol}}^{-1}\end{array}$

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

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{unknown acid}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{90\text{g}\cdot {\text{mol}}^{-1}}{45.018\text{g}\cdot {\text{mol}}^{-1}}\\ =2.00\end{array}$

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

    $\begin{array}{c}\text{Molecular}\text{formula}\text{of}\text{unknown acid}=2\times ({\text{CHO}}_{\text{2}})\\ ={\text{C}}_2{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\end{array}$

Therefore, the molecular formula of unknown acid is ${\text{C}}_2{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$.

(c)

Interpretation Introduction

Interpretation:

Balanced chemical equation has to be written for the neutralization of the unknown acid with sodium hydroxide.

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.

Answer to Problem L.38E

Balanced chemical equation is ${\text{C}}_2{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq)}+2\text{NaOH(aq)}\to {\text{Na}}_{\text{2}}{\text{C}}_2{\text{O}}_{\text{4}}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$.

Explanation of Solution

The unknown acid is said to be diprotic.  Consider the unknown acid as ${\text{H}}_{\text{2}}\text{A}$.  Therefore, the reaction of unknown acid with sodium hydroxide is given as follows;

    ${\text{H}}_{\text{2}}\text{A(aq)}+2\text{NaOH(aq)}\to {\text{Na}}_{\text{2}}\text{A}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$

The unknown acid has a molecular formula of ${\text{C}}_2{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$.  Therefore, the balanced chemical equation for the neutralization of the unknown acid by sodium hydroxide is given as follows;

    ${\text{C}}_2{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}\text{(aq)}+2\text{NaOH(aq)}\to {\text{Na}}_{\text{2}}{\text{C}}_2{\text{O}}_{\text{4}}(aq)+2{\text{H}}_{\text{2}}\text{O}(l)$