Question

Nitration of Methyl Benzoate Calculations - please fill out every box and expected and percent yield.

- eqt: methyl benzoate + HNO₃ ----->(H₂SO₄ above arrow) Methyl-m-nitrobenzoate + H₂O

- end product= 2.105 g

-2ml of nitric acid and 2ml of sulfuric acid were used

Calculation of Theoretical Yield Balanced Equation: RAH OH M.W.*: 18.00 g/mol 136.15 g mos 63.0lgimo 4.026g 31.505g 181.15 Horg grams*: moles*: * Include these values for all compounds, underneath that compound, in the balanced equation. Limiting Reagent Exp't yield (g) Calc. of % yield % yield

Answer 1

Given,

mass of ethyl benzoate = 4.026 g

mass of nitric acid = 31.505 g

So, moles of methyl benzoate = 4.026 / 136.15 = 0.0296

moles of nitric acid = 31.505 / 63.01 = 0.5

Remember that H2SO4 is used so one H+ ion is coming from that to form water molecule.

1 mol of methyl benzoate reacts with 1 mol of nitric acid in presence of sulfuric acid to produce 1 mol methyl-m-nitrobenzene and 1 mol of water.

Since we only have 0.0296 mol methyl benzoate so it will be limiting reagent.

Theoretical amount of methyl-m-nitrobenzene and water formed will be 0.0296 mol each.

Theoretical yield of methyl-m-nitrobenzene and water will be 0.0296(181.15 + 18.00) = 5.895 g

But actual yield is 2.105 g of methyl-m-nitrobenzene and water.

% yield = (actual yield / theoretical yield)*100

= 35.71 %

Expected yield or theoretical yield for this reaction is 5.895 g and percent yield is 35.71 %.