Question

Calculate the percent yield of your product (methyl nitrobenzoate).

Experimental Results Questions: Product 1 Methyl nitrobenzoale 3. Calculate the percent yield of your product. Assume the methyl benzoate is the limiting reagent. Show your work. (1 pt) 0.2931 g methyl benzoate 0.0932 g of product Theo .2931 x 1 mol x INOX 181.5 = .38979. 136.15 no Twok. Experimental: 23. yield -13°C 78-80 94-ano

Answer 1

Reaction -

Methylbenzoate

Page No. # Reaction scheme - nitration of methyl benzoate. T outh Nitrating mixure. Tooth methyl benzoate (Reactant) molar mals = 136.15 g/mol methyl citrobenzoate product Molar mass= 181.04 g/mol

# Mass of methylbenzoate = 0.2931 g

# molar mass of methyl benzoate = 136.15 g/mol

#mass of methyl nitrobenzoate = 0.0932 g

# molar mass of methyl nitrobenzoate = 181.04 g/mol

We have to calculate number of moles of reactant and product .

We know that,

We are going to calculate number of moles of methyl benzoate

Number of moles (n) = (mass,  m) / Molar mass (M)

= (0.2931 g /136.15 g/mol)

= 0.002153 mol

Now moles of product ,methyl nitrobenzoate-

Number of moles of methyl nitrobezoate (n) = (mass ,m) /molar mass (M)

= (0.0932 g /181.04 g/mol)

=0.0005148 mol of Methyl nitrobenzoate

Now,

Theoretical yield of Methyl nitrobenzoate = moles of Methyl benzoate x Molar mass of Methyl nitrobenzoate

= (0.002153 mol x 181.04 g/mol)

=0.390 g of Methyl nitrobenzoate (theoretical yield )

But you have isolated mass of methyl nitrobenzoate =0.0932 g

So

Percentage yield =( practical yield /theoretical yield ) x 100

= (0.0932 /0.390 ) x100

=23.9 %