Chapter 12, Problem 12.105QE

(a)

Interpretation Introduction

Interpretation:

Empirical formula of the compound has to be calculated.

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 12.105QE

Empirical formula of the compound is ${\text{NaCO}}_{\text{2}}$.

Explanation of Solution

Percentage composition of the compound is given as $\text{33}\text{.8}\text{\%}\text{Na}$, $\text{17}\text{.7}\text{\%}\text{C}$, and $\text{47}\text{.0}\text{\%}\text{O}$.  The composition of the compound was $\text{33}\text{.8}\text{g}$ of sodium, $\text{17}\text{.7}\text{g}$ of carbon, $\text{47}\text{.0}\text{g}$ of oxygen.  Moles of the components in the compound can be calculated as shown below;

Moles of sodium:

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{33.8\text{g}}{22.99\text{g}\cdot {\text{mol}}^{-1}}\\ =1.47\text{mol}\end{array}$

Moles of carbon:

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{17.7\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =1.47\text{mol}\end{array}$

Moles of oxygen:

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{47.0\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =2.94\text{mol}\end{array}$

Empirical formula can be obtained by dividing the moles of each element with the least mole.  This is done as follows;

    $\begin{array}{c}\text{Sodium}:\frac{1.47\text{mol}}{1.47\text{mol}}=1.00\\ \text{Carbon}:\frac{1.47\text{mol}}{1.47\text{mol}}=1.00\\ \text{Oxygen}:\frac{2.94\text{mol}}{1.47\text{mol}}=2.00\end{array}$

Therefore, the ratio of the element can be given as $\text{Na}:\text{C}:\text{O}=1:1:2$.

Thus the empirical formula will be ${\text{NaCO}}_{\text{2}}$.

(b)

Interpretation Introduction

Interpretation:

The formula of the white solid has to be given and the equation for the dissociation in water has to be written.

Explanation of Solution

It is given that the solution is made by dissolving $\text{0}\text{.500}\text{g}$ of white solid in $\text{20}\text{.0}\text{g}$ of water.  The freezing point depression of the solution is given as $-1.01°\text{C}$.  The molality of water can be calculated using the freezing point depression equation as follows;

    $\begin{array}{c}\Delta T_f=i\times m\times k_f\\ m=\frac{\Delta T_f}{i\times k_f}\\ =\frac{0.00°\text{C}-(-1.01°C)}{3\times 1.86°\text{C/}m}\\ =0.181m\end{array}$

Thus the molality of water is $0.181m$.  The mass of solvent is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}\text{solvent}=20.0\text{g}\text{water}-0.500\text{g}\text{compound}\\ \text{=19}\text{.5}\text{g}\text{solvent}\end{array}$

Moles of solvent is calculated as follows;

    $\text{19}\text{.5}\text{g}\text{solvent}\times \frac{1\text{kg}}{1000\text{g}}\times \frac{\text{0}\text{.181}\text{mol}}{\text{1}\text{kg}\text{solvent}}=0.0035295\text{mol}$

Molar mass of the compound is calculated using the moles as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\frac{\text{Mass}}{\text{Number}\text{of}\text{moles}}\\ =\frac{0.500\text{g}}{0.0035295\text{mol}}\\ =141.6\text{g/mol}\end{array}$

Molar mass of the empirical formula is calculated as shown below;

    $\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{compound}=1\times (massofNa)+1\times (massofC)+2\times (massof\text{O)}\\ \text{=22}\text{.99}\text{g/mol+12}\text{.01}\text{g/mol+32}\text{.00}\text{g/mol}\\ \text{=67}\text{.0}\text{g/mol}\end{array}$

Molecular formula can be calculated from the molar mass and empirical mass as shown below;

    $\begin{array}{c}n=\frac{\text{molecular}\text{mass}}{\text{empirical}\text{mass}}\\ =\frac{\text{141}\text{.6}\text{g}}{\text{67}\text{.00}\text{g}}\\ =2.11\\ =2\end{array}$

Therefore, the formula of the white solid is given as shown below;

    $\begin{array}{c}\text{Formula}=\text{Empirical}\text{formula}\times n\\ ={\text{NaCO}}_{\text{2}}\times 2.00\\ ={\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\end{array}$

Thus the formula of the white solid is ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$.

Dissolution of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ in water can be given as shown below;

    ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}(s)\to 2N{a}^{+}(aq)+{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)$