Weight of methyl 3-nitrobenzoate = 1.8 g m.p. of methyl 3-nitrobenzoate = 65.3 °C

Weight of methyl 3-nitrobenzoate = 1.8 g m.p. of methyl 3-nitrobenzoate = 65.3 °C
Weight of 3-nitrobenzoic acid = 1.5 g m.p. of 3-nitrobenzoic acid = 133.2 °C
Calculate the percent yield for the formation of methyl 3-nitrobenzoate & 3-nitrobenzoic acid Based on the data provided above

The Correct Answer and Explanation is :

To calculate the percent yield of methyl 3-nitrobenzoate and 3-nitrobenzoic acid, we need to follow these steps:

Step 1: Calculate Molar Masses

• Molar mass of methyl 3-nitrobenzoate (C8H7NO4):
• Carbon (C): 8 × 12.01 = 96.08 g/mol
• Hydrogen (H): 7 × 1.008 = 7.056 g/mol
• Nitrogen (N): 1 × 14.01 = 14.01 g/mol
• Oxygen (O): 4 × 16.00 = 64.00 g/mol
• Total = 96.08 + 7.056 + 14.01 + 64.00 = 181.146 g/mol
• Molar mass of 3-nitrobenzoic acid (C7H6NO4):
• Carbon (C): 7 × 12.01 = 84.07 g/mol
• Hydrogen (H): 6 × 1.008 = 6.048 g/mol
• Nitrogen (N): 1 × 14.01 = 14.01 g/mol
• Oxygen (O): 4 × 16.00 = 64.00 g/mol
• Total = 84.07 + 6.048 + 14.01 + 64.00 = 168.128 g/mol

Step 2: Convert the Weights of Products to Moles

• Moles of methyl 3-nitrobenzoate (1.8 g):
  [
  \text{Moles of methyl 3-nitrobenzoate} = \frac{1.8 \, \text{g}}{181.146 \, \text{g/mol}} = 0.00994 \, \text{mol}
  ]
• Moles of 3-nitrobenzoic acid (1.5 g):
  [
  \text{Moles of 3-nitrobenzoic acid} = \frac{1.5 \, \text{g}}{168.128 \, \text{g/mol}} = 0.00892 \, \text{mol}
  ]

Step 3: Determine the Theoretical Yield

Assuming a 1:1 molar conversion between methyl 3-nitrobenzoate and 3-nitrobenzoic acid (in a simple ester hydrolysis or similar reaction), the theoretical yield is based on the limiting reagent.

From the calculations above:

• Moles of methyl 3-nitrobenzoate (0.00994 mol) and 3-nitrobenzoic acid (0.00892 mol) are the starting amounts.

Thus, 3-nitrobenzoic acid is the limiting reagent, so theoretical yield is 0.00892 mol of 3-nitrobenzoic acid.

Step 4: Calculate the Percent Yield

The percent yield is calculated using the formula:
[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
]
Using 1.5 g as the actual yield of 3-nitrobenzoic acid and the molar mass of 3-nitrobenzoic acid:

• Theoretical yield in grams = ( 0.00892 \, \text{mol} \times 168.128 \, \text{g/mol} = 1.5 \, \text{g} )

Thus, percent yield of 3-nitrobenzoic acid = 100% (no loss or gain).

Conclusion

• Percent yield for methyl 3-nitrobenzoate = 100% (assuming no other data, it’s implied the product was recovered at full efficiency).
• Percent yield for 3-nitrobenzoic acid = 100% as well.

Let me generate the corresponding diagram for the calculations.

Here is the diagram illustrating the chemical yield calculations for methyl 3-nitrobenzoate and 3-nitrobenzoic acid. It highlights the steps from the molar masses, grams to moles conversion, limiting reagent, to the final percent yield calculation. Let me know if you need further clarification or additional information!