Questions: 1. Complete the following table: Compound MW equiv. mmol density mL ------------------------------------------------------------------- methyl benzoate 136.15 g / m 31 0.01674 0.930 sulfuric acid (18M) 0.4507 1.088 0.414 nitric acid 0.01674 0.310 (15.9M) 5.088 0.320 2. What is the theoretical yield for this reaction? 4. The textbook shows the nitration of benzene to occur with heating (50-55°C), why is an ice bath employed in the nitration of methyl benzoate?

Solution Steps

The table provided contains information about the compounds involved in a reaction, including methyl benzoate, sulfuric acid, and nitric acid. We need to fill in the missing values for molecular weight (MW), equivalent weight, and other parameters.

1. Molecular Weight (MW) and Equivalent Weight:

   • For sulfuric acid (H₂SO₄), the molecular weight is approximately 98.08 g/mol. Since it is a diprotic acid, the equivalent weight is half of the molecular weight, which is 49.04 g/equiv.
   • For nitric acid (HNO₃), the molecular weight is approximately 63.01 g/mol. It is a monoprotic acid, so the equivalent weight is the same as the molecular weight, 63.01 g/equiv.

2. Density and Volume:

   • The density of sulfuric acid is given as 1.088 g/mL, and the volume is 0.414 mL.
   • The density of nitric acid is not provided, but the volume is 0.310 mL.

1. Sulfuric Acid:

   • Given: \( \text{mmol} = 0.4507 \)
   • Using the formula: \[ \text{mass} = \text{mmol} \times \text{MW} = 0.4507 \times 98.08 = 44.22 \, \text{mg} \]
   • Volume is given as 0.414 mL, so density can be verified: \[ \text{density} = \frac{\text{mass}}{\text{volume}} = \frac{44.22}{0.414} = 1.068 \, \text{g/mL} \]
   • This is close to the given density of 1.088 g/mL, confirming the calculation.

2. Nitric Acid:

   • Given: \( \text{mmol} = 0.01674 \)
   • Using the formula: \[ \text{mass} = \text{mmol} \times \text{MW} = 0.01674 \times 63.01 = 1.055 \, \text{mg} \]
   • Volume is given as 0.310 mL, so density can be calculated: \[ \text{density} = \frac{\text{mass}}{\text{volume}} = \frac{1.055}{0.310} = 3.403 \, \text{g/mL} \]

Final Answer