Connect with knowledgeable individuals and get your questions answered on IDNLearn.com. Get comprehensive and trustworthy answers to all your questions from our knowledgeable community members.
Sagot :
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.
### 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.
Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.
