A reaction was performed in which 2.60 mL of methyl benzoate was reacted with a

Question

A reaction was performed in which 2.60 mL of methyl benzoate was reacted with a mixture of concentrated nitric and sulfuric acids to make 2.64 g of methyl m-nitrobenzoate. Calculate the theoretical yield and percent yield for this reaction.

Consider the nitration by electrophilic aromatic substitution of methyl benzoate to methyl m-nitrobenzoate conc. HNO.3 conc. H2SO4 O2 methyl m-nitrobenzoate methyl benzoate d = 1.09 g/mL A reaction was performed in which 2.60 mL of methyl benzoate was reacted with a mixture of concentrated nitric and sulfuric acids to make 2.64 g of methyl m-nitrobenzoate. Calculate the theoretical yield and percent yield for this reaction Theoretical yield Percent yield Number Number

Explanation / Answer

1 mole methylbenzoate   ----> 1 mole methyl m-nitrobenzoate.

no of moles of methyl benzoate = w/mwt

w = v*d = 2.6*1.09 =   2.834 grams

mwt of methyl benzoate = 136.14792 g/mol

= 2.834/136.14792 = 0.0208156 mole.

no of moles of methyl m-nitrobenzoate produced = 0.0208156 mole.

theoretical yield of m-nitrobenzoate produced = 0.0208156 * mwt

mwt of m-nitrobenzoate = 181.14548 g/mol

= 0.0208156 *181.14548 = 3.77 grams.

percent yield = practical yield / theoretical yield *100

    = 2.64 / 3.77*100 = 70.026%

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