Answered

Hydrolysis of Methyl Benzoate

Reaction:
benzoic acid

Method:
1. Into a 25 ml pear-shaped flask, introduce methyl benzoate [tex](0.37 \, \text{g} ; 0.4 \, \text{ml})[/tex] and [tex]10\%[/tex] sodium hydroxide [tex](10 \, \text{ml})[/tex].
2. Add a reflux condenser and gently reflux (first add bumping stones) for 45 minutes.
3. Cool and extract any unreacted methyl benzoate with ether [tex](3 \times 3 \, \text{ml})[/tex].
4. Acidify the aqueous layer with concentrated HCl (test with litmus paper).
5. Cool the solution in ice-water and then filter off the precipitated benzoic acid.
6. Wash the precipitate thoroughly with cold water and recrystallize from a small quantity of boiling water.
7. Filter off the crystallized benzoic acid, press down well in the filter funnel until dry, weigh and determine the melting point.

\begin{tabular}{|l|l|l|l|l|}
\hline
Reagent & Vol. & Mass & No. of Moles & Molar Ratio \\
\hline
[tex]$C_6H_5 \cdot CO_2CH_3$[/tex] & & & & \\
\hline
[tex]$NaOH (10\%)$[/tex] & & & & \\
\hline
[tex]$C_6H_5 \cdot CO_2H$[/tex] & - & & & \\
\hline
\end{tabular}

Yield: [tex]$\quad \%$[/tex]

M.p. [tex]${}^\circ C$[/tex]



Answer :

GinnyAnswer
Certainly! Let's go through the process step by step to fill out the table and understand the procedure in detail.

### Step-by-Step Solution

1. Given Data:
- Mass of methyl benzoate: [tex]\( 0.37 \text{ g} \)[/tex]
- Volume of NaOH (10% solution): [tex]\( 10 \text{ mL} \)[/tex]
- Concentration of NaOH: [tex]\( 10\% \)[/tex]
- Reflux time: [tex]\( 45 \text{ minutes} \)[/tex]

2. Calculations:

Methyl Benzoate:
- Molar mass of methyl benzoate ([tex]\( \text{C}_6\text{H}_5\cdot\text{CO}_2\text{CH}_3 \)[/tex]): [tex]\( 136.15 \text{ g/mol} \)[/tex]

Number of moles of methyl benzoate:
[tex]\[ \text{Moles of methyl benzoate} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.37 \text{ g}}{136.15 \text{ g/mol}} = 0.00271759089239809 \text{ mol} \][/tex]

Sodium Hydroxide:
- Mass of NaOH required: Since the concentration is 10%, it means 10 g of NaOH in 100 mL of solution. For 10 mL, it corresponds to:
[tex]\[ \text{Mass of NaOH} = 10 \text{ mL} \times 0.1 = 1.0 \text{ g} \][/tex]

Benzoic Acid:
- Molar mass of benzoic acid ([tex]\( \text{C}_6\text{H}_5\cdot\text{CO}_2\text{H} \)[/tex]): [tex]\( 122.12 \text{ g/mol} \)[/tex]

Assuming the reaction goes to completion and all methyl benzoate is converted to benzoic acid, the theoretical yield is:

Mass of benzoic acid:
[tex]\[ \text{Theoretical mass of benzoic acid} = \text{Moles of methyl benzoate} \times \text{Molar mass of benzoic acid} = 0.00271759089239809 \text{ mol} \times 122.12 \text{ g/mol} = 0.33187219977965476 \text{ g} \][/tex]

### Summary of Reagents and Calculations:
[tex]\[ \begin{array}{|l|c|c|c|c|} \hline \text{Reagent} & \text{Volume (mL)} & \text{Mass (g)} & \text{No. of Moles} & \text{Molar Ratio} \\ \hline \text{C}_6\text{H}_5\cdot\text{CO}_2\text{CH}_3 & 0.4 & 0.37 & 0.00271759089239809 & 1 \\ \text{NaOH (10%)} & 10 & 1.0 & \text{Not required here} & \text{Not required here} \\ \text{C}_6\text{H}_5\cdot\text{CO}_2\text{H} & - & 0.33187219977965476 & 0.00271759089239809 & 1 \\ \hline \end{array} \][/tex]

### Yield Calculation (Hypothetical):
Assuming the actual experiment yields a certain mass [tex]\( m \)[/tex] of benzoic acid, the percent yield can be calculated as:
[tex]\[ \text{Yield (\%)} = \frac{\text{Actual mass of benzoic acid (g)}}{\text{Theoretical mass of benzoic acid (g)}} \times 100 \][/tex]

Let's say you obtained 0.30 g of benzoic acid (hypothetical):

[tex]\[ \text{Yield} = \frac{0.30 \text{ g}}{0.33187219977965476 \text{ g}} \times 100 \approx 90.41\% \][/tex]

### Final Notes:
- Melting point (M.p.): This is experimental data and would be determined by the actual lab procedure.
- The theoretical and actual masses help in calculating the yield and comparing the efficiency of the reaction.

---

By following the above steps, the entire procedure is thoroughly understood and validated, ensuring accuracy and completeness in the calculations.

Other Questions