Question

A solid sample weighing 0.2376 g contained only malonic acid and aniline hydrochloride. It required 34.02 mLof 0.8771 M NaOH to neutralize the sample. Find the weight percent of each component in the solid mixture. The reactions are.

Short answer

The percentages of the weight of each species in the given mixture are

$$\begin{gathered} \%\mathrm{of}\mathrm{ani}\mathrm{linium}\mathrm{chloride}=57.90\% \\ \%\mathrm{of}\mathrm{malonic}\mathrm{acid}=42.09\% \end{gathered}$$

Step-by-step solution

Define the molarity

The total number of moles of solute per litre of solution is known as a solution's molarity. Because, unlike mass, the volume of a system fluctuates with changes in physical conditions, the molality of a solution is affected by changes in physical parameters of the system such as pressure and temperature. Molarity is represented by the letter M, which stands for a molar.

The molarity of a solution is defined as one gramme of solute dissolved in one litre of solution.

Calculate the percentage of the weight of each species in the given mixture

Malonic acid makes up 42.09 percent of the weight in a particular composition.

The weight proportion of anilinium chloride in a given combination is 57.90%.

The reaction

$$\begin{gathered} {\mathrm{CH}}_2({\mathrm{CO}}_2^2+2{\mathrm{OH}}^{-}\to {\mathrm{CH}}_2({\mathrm{CO}}_2)+2{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{NH}}_3+{\mathrm{Cl}}^{-}+{\mathrm{OH}}^{-}\to {\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{NH}}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}^{-} \end{gathered}$$

Number of moles of$\mathrm{NaOH}=34.02\mathrm{mL}\times 0.08771\mathrm{M}$

$=2.9839\mathrm{mmolNaOH}$

The percentage of anilinium is given by

$2(\mathrm{mass}\mathrm{of}\mathrm{malonic}\mathrm{acid})+(\mathrm{mass}\mathrm{of}\mathrm{ani}\mathrm{linium}\mathrm{chloride})=0.0029839\mathrm{mol}$

$2\left(\frac{\mathrm{mass}\mathrm{of}\mathrm{malonic}\mathrm{acid}}{129.59\mathrm{g}/\mathrm{mol}}\right)+\left(\frac{\mathrm{mass}\mathrm{of}\mathrm{ani}\mathrm{linium}\mathrm{chloride}}{104.06\mathrm{g}/\mathrm{mol}}\right)=0.0029839\mathrm{mol}$

$y=0.2376-x,thenx=0.10001g$

$\%ofaniliniumchloride=\frac{0.13759g}{0.2376g}\times 100=57.90\%$

$\%ofmalonicacid=\frac{0.10001g}{0.2376g}\times 100=42.09\%$

Thus, the percentage of weight of each species in the given mixture are

$$\begin{gathered} \%\mathrm{of}\mathrm{ani}\mathrm{linium}\mathrm{chloride}=57.90\% \\ \%\mathrm{of}\mathrm{malonic}\mathrm{acid}=42.09\% \end{gathered}$$