Chapter F, Problem J.15E

(a)

Interpretation Introduction

Interpretation:

Empirical formula for the compound X 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 J.15E

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

Explanation of Solution

The mass percentage composition of compound X is given as $\text{26}\text{.68\%}\text{C}$, and $\text{2}\text{.239}\text{\%}\text{H}$.  The remaining mass percentage is said to be oxygen.  The mass percentage of oxygen in the compound is calculated as follows;

    $\begin{array}{c}\text{Mass}\text{\%}\text{of}\text{O}=\text{Total}\text{mass}\text{\%}-(\text{Mass}\text{\%}\text{C+Mass}\text{\%}\text{H})\\ =(100-(26.68+2.239))\%\\ =(100-28.919)\%\\ =71.081\%\end{array}$

Considering $\text{100}\text{g}$ of compound, it is understood that the compound contains $\text{26}\text{.68}\text{g}$ of carbon, $71.081\text{g}$ of oxygen, and $\text{2}\text{.239}\text{g}$ of hydrogen.

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

    $\begin{array}{c}\text{Moles}\text{of}\text{Carbon}=\frac{26.68\text{g}}{12.01\text{g}\cdot {\text{mol}}^{-1}}\\ =2.22\text{mol}\\ \text{Moles}\text{of}\text{Oxygen}=\frac{71.081\text{g}}{16.00\text{g}\cdot {\text{mol}}^{-1}}\\ =4.44\text{mol}\\ \text{Moles}\text{of}\text{Hydrogen}=\frac{2.239\text{g}}{1.008\text{g}\cdot {\text{mol}}^{-1}}\\ =2.22\text{mol}\end{array}$

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

    $\begin{array}{c}\text{Carbon}:\frac{\text{2}\text{.22}\text{mol}}{\text{2}\text{.22}\text{mol}}=1.00\\ \text{Oxygen}:\frac{\text{4}\text{.44}\text{mol}}{\text{2}\text{.22}\text{mol}}=2.00\\ \text{Hydrogen}:\frac{\text{2}\text{.22}\text{mol}}{\text{2}\text{.22}\text{mol}}=1.00\end{array}$

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 $\text{C}:\text{O}:\text{H}=1:2:1$.

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

(b)

Interpretation Introduction

Interpretation:

Molecular formula of the compound X 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 J.15E

Molecular formula of the compound X is ${\text{C}}_{\text{12}}{\text{O}}_{\text{12}}{\text{Os}}_{\text{3}}$.

Explanation of Solution

Empirical formula of compound X is ${\text{CHO}}_{\text{2}}$.  Molar mass of compound X is given as $90.0\text{g}\cdot {\text{mol}}^{-1}$.

Molar mass of the empirical formula is calculated as follows;

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

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

    $\begin{array}{c}\frac{\text{Molar}\text{mass}\text{of}\text{X}}{\text{Molar}\text{mass}\text{of}\text{empirical}\text{formula}}=\frac{90.0\text{g}\cdot {\text{mol}}^{-1}}{45.018\text{g}\cdot {\text{mol}}^{-1}}\\ =1.99\\ \approx 2.00\end{array}$

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

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

Therefore, the molecular formula of compound X is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{O}}_{\text{4}}$.

(c)

Interpretation Introduction

Interpretation:

Balanced chemical equation and net ionic equation for reaction of compound X with sodium hydroxide has to be written.

Concept Introduction:

Complete ionic equation is the one that shows all the species that is present in the chemical reaction.  The dissolved ionic compounds are alone showed as ions.  Precipitates are not showed as ions.

Net ionic equation is the one that is obtained from the complete ionic equation by cancelling out the spectator ions.

Explanation of Solution

Compound X is said to have two acidic hydrogen atoms.  Therefore, the reaction between compound X and sodium hydroxide is given as shown below;

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

Balancing sodium atoms: In the reactant side, there is one sodium atom while on the product side, there are two potassium atoms.  Adding coefficient $2$ before $\text{NaOH}$ in the reactant side balances the sodium atoms on both sides of the equation.  The chemical equation obtained is given as follows;

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

Balancing oxygen atoms:  In the above chemical equation, there are six oxygen atoms on the left side of the equation, while in the product side, there are five oxygen atoms.  Adding coefficient $2$ 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;

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

Complete ionic equation:

The complete ionic equation can be written considering the ionic compounds in aqueous medium to be written into respective ions.  Therefore, the complete ionic equation can be given as follows;

${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{+2Na}}^{\text{+}}(aq)+2{\text{OH}}^{-}(aq)\to 2{\text{Na}}^{\text{+}}(aq)+{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)+2{\text{H}}_{\text{2}}\text{O}(aq)$

Net ionic equation:

The net ionic equation can be obtained from the complete ionic equation by cancelling out the spectator ions on both sides of the equation.

${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{+}\cancel{{\text{2Na}}^{\text{+}}(aq)}+2{\text{OH}}^{-}(aq)\to \cancel{2{\text{Na}}^{\text{+}}(aq)}+{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)+2{\text{H}}_{\text{2}}\text{O}(aq)$

Thus, the net ionic equation can be given as shown below;

  ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{+}2{\text{OH}}^{-}(aq)\to {\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{2-}(aq)+2{\text{H}}_{\text{2}}\text{O}(aq)$