Q143P - Chemistry: Molecular Nature Of Matter And Change

Question

Tartaric acid occurs in crystalline residues found in wine vats. It is used in baking powders and as an additive in foods. It contains $\mathbf{32}\mathbf{.3}\mathit{\%}$ by mass carbon and $\mathbf{3}\mathbf{.}\mathbf{97}\mathbf{\%}$ by mass hydrogen; the balance is oxygen. When $\mathbf{0}\mathbf{.981}\mathbf{\text{ g}}$ of tartaric acid is dissolved in $\mathbf{\text{11.23 g}}$ of water, the solution freezes at $\mathbf{-}\mathbf{126}\mathbf{°}\mathbf{\text{C}}$ . Find the empirical and molecular formulas of tartaric acid?

Step-by-Step Solution

Answer

The empirical formula and the molecular formula is  ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{3}}$ and ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$ respectively.

A concept:

A formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

Consider the given data below.

The mass of tartaric acid is  $0.981\text{ g}$.

Assume $100\text{ g}$  of tartaric acid. So,  $\text{32.3 g}$ of  $C$,  $\text{3.97 g}$ of $H$ and the remaining amount of oxygen is,

$100\text{ g}-(32.3+3.97)\text{ g}=63.73\text{ g}$ 

 

Now, calculate the moles of  $C,H,$ and $O$.

$\begin{array}{rcl}\text{Moles of carbon}& =& 32.3\text{ g of }C\times \frac{1\text{ mol of }C}{12.01\text{ g of\hspace{0.17em}}C}\\ & =& 2.69\text{ mol}\end{array}$

$\begin{array}{rcl}\text{Moles of hydrogen}& =& 3.97\text{ g of\hspace{0.17em}}H\times \frac{1\text{ mol of }H}{1\text{\hspace{0.17em}g of\hspace{0.17em}}H}\\ & =& 3.97\text{\hspace{0.17em}mol}\end{array}$

$\begin{array}{rcl}\text{Moles of hydrogen}& =& 3.97\text{ g of\hspace{0.17em}}H\times \frac{1\text{ mol of }H}{1\text{\hspace{0.17em}g of\hspace{0.17em}}H}\\ & =& 3.97\text{\hspace{0.17em}mol}\end{array}$ 

$\begin{array}{rcl}\text{Moles of oxygen}& =& 63.73\text{ g of\hspace{0.17em}}O\times \frac{1\text{ mol of \hspace{0.17em}}O}{16\text{ g of }O}\\ & =& 3.98\text{\hspace{0.17em}mol}\end{array}$ 

Empirical formula:

Empirical formula can be calculated as follow.

Now, the smallest amount is of  $C$. Hence, divided moles of  $C,H,$ and $O$ by moles of $\begin{array}{rcl}& & C\end{array}$.

$\begin{array}{rcl}C& =& \frac{2.69}{2.69}\\ & =& 1\end{array}$

$\begin{array}{rcl}H& =& \frac{3.97}{2.69}\\ & =& 1.47\\ & & ~1.5\end{array}$

$\begin{array}{rcl}O& =& \frac{3.98}{3.33}\\ & =& 1.47\\ & & ~1.5\end{array}$

Hence, the empirical formula is ${\text{CH}}_{\text{1.5}}{\text{O}}_{\text{1.5}}$. 

Empirical formula is in fraction, so we can multiply it with $2$  to get the empirical formula in integer. It becomes  $\left({\text{CH}}_{\text{1.5}}{\text{O}}_{\text{1.5}}\right)\times 2={\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{3}}$.

 

Hence, the empirical formula is ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{3}}$.

Molecular formula:

Given freezing point of tartaric acid is $-126°\text{C}$ and freezing point of water is $0°\text{C}$.

Hence, the freezing point depression is,

$\Delta T_f=0°\text{C}-(-126°\text{C})$ 

Molality of solution can be calculated as follow.

$\text{Molality}=\frac{\Delta T_f}{K_b}$ 

Here, $K_b$ for water is $1.86\raisebox{1ex}{${\displaystyle °\text{C}}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.$.

 

Hence, 

$\begin{array}{rcl}\text{Molality}& =& \frac{0°\text{C}-(-126°\text{C})}{1.86{\displaystyle \raisebox{1ex}{$°\text{C}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.}}\\ & =& 0.677\text{ m}\end{array}$

 

Now, it has to calculate the moles of tartaric acid. Given the mass of water is $11.23\text{ g}$.  Hence you can calculate the moles of tartaric acid by using following formula of molality.

$\begin{array}{rcl}\text{molality}& =& \frac{\text{moles\hspace{0.17em}of\hspace{0.17em}tartaric\hspace{0.17em}acid}}{\text{mass\hspace{0.17em}of\hspace{0.17em}water}}\end{array}$ 

 $\begin{array}{rcl}\text{moles\hspace{0.17em}of\hspace{0.17em}tartaric\hspace{0.17em}acid}& =& \text{molality}\times \text{mass\hspace{0.17em}of\hspace{0.17em}water}\left(\text{kg}\right)\\ & =& 0.677\raisebox{1ex}{${\displaystyle \text{mol}}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.\times 11.23\text{ g}\times \frac{1\text{ kg}}{1000\text{ g}}\\ & =& 0.76\text{ moles}\end{array}$

Molar mass of tartaric acid can be calculated by dividing given mass of tartaric acid by moles of tartaric acid.

$\begin{array}{rcl}\text{molar\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}tartaric\hspace{0.17em}acid}& =& \frac{\text{mass\hspace{0.17em}of\hspace{0.17em}tartaric\hspace{0.17em}acid}}{\text{moles\hspace{0.17em}of\hspace{0.17em}tartaric\hspace{0.17em}acid}}\\ & =& \frac{0.981\text{ g}}{0.76\text{ mol}}\\ & =& 1.29\raisebox{1ex}{${\displaystyle \text{g}}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\end{array}$

This is the calculated molar mass of tartaric acid but actual molar mass of tartaric acid is $150\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$. 

Molar mass of empirical formula  $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{3}}\right)$ is $\text{75 }\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

 

Molecular formula of tartaric acid can be calculated as follows.

$\begin{array}{rcl}\text{molecular\hspace{0.17em}formula\hspace{0.17em}}& =& \text{empirical\hspace{0.17em}formula}\times \frac{\text{molar\hspace{0.17em}mass}}{\text{empirical\hspace{0.17em}formula\hspace{0.17em}mass}}\\ & =& {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{3}}\text{×}\frac{\text{150\hspace{0.17em}}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}{\text{75\hspace{0.17em}}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}\\ & =& {\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{3}}\times 2\\ & =& {\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}\end{array}$ 

 

Hence, the molecular formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}{\text{O}}_{\text{6}}$.