Question

-NITRATION OF Methyl benzoate

-addditional information. .( 3.9 g of the product methyl 3-nitrobenzoate for the percentage yield

Materials used: Amount Compound used/ Molecular weight Density (g/mL) Moles produced 3.05g used 136.15 g/mol N/A 0.0224 mol Methyl Benzoate HO-S-OH 6mL used 98.079 g/mol 1.83 g/cm3 0.112 Sulfuric Acid 2mL used 63.01 g/mol 1.51 g/cm3 0.047g o-N-OH Nitric Acid 3.034g of pure product 181.15 g/mol produced N/A 0.0167 mol NO2 Methyl-m- nitrobenzoate

Answer 1

The %yield is just the ratio between the obtained mass of product (experimental) and the theoretical yield of product:

$\%yield=\frac{m_{experimental}}{m_{theoretical}}*100=\frac{3.034g}{4.058g}*100=74.8\%$