In the lab Nitration of Methyl Benzoate calculate the theore

In the lab Nitration of Methyl Benzoate, calculate the theoretical yield and the actual yield of methyl m-nitrobenzoate, to get percent yield.

Starting methyl benzoate: 0.13g

Final product after isolating methyl m-nitrobenzoate: 0.033g

Please show all calculations.

1. Prepare a sulfuric acid/nitric acid mixture. Add 0.75 mL of concentrated sulfuric acid to a small clean dry test tube. Cool this test tube in an ice-water bath in the hood. Add 0.25 mL of concentrated nitric acid to the sulfuric acid slowly with swirling. Let this mixture sit in the ice bath. 2. Add approximately 0.25 mL of methyl benzoate to a clean, dry conical vial of known mass. Weigh again. Top the vial with an air condenser and clamp the condenser so that the vial is immersed in the ice-water bath in the hood. Wait a few minutes to insure that the contents of both containers are below 15 degrees Celsius. 3. Use a Pasteur pipette to add the acid mixture to the methyl benzoate DROP-WISE, WITH SWIRLING, OVER A 15-MINUTE PERIOD through the top of the air condenser. 4. Warm the reaction mixture to room temperature by replacing the ice water bath with warm water baths. Allow the reaction mixture to sit at room temperature for fifteen minutes untouched. Poor yields result when the mixture is not warmed sufficiently. 5. Transfer the reaction mixture onto 2.0 g of crushed ice in a small beaker. Stir. Once the ice has melted, follow Pavia\'s procedure for isolating the methyl m-nitrobenzoate by vacuum filtration and washing the product.

Solution

The word equation for the formation of methyl m-nitrobenzoate from methyl benzoate by using the nitrating mixture (HNO₃ + H₂SO₄) can be written as

Methyl Benzoate ------------> Methyl m-nitrobenzoate

As per the stoichiometric equation,

1 mole methyl benzoate = 1 mole methyl m-nitrobenzoate

Molar mass of methyl benzoate, C₈H₈O₂ = (8*12.011 + 8*1.008 + 2*15.999) g/mol = 136.15 g/mol.

We started with 0.13 g methyl benzoate; therefore, mole(s) of methyl benzoate corresponding to 0.13 g = (0.13 g)/(136.15 g/mol) = 9.5483*10^-4 mole.

As per the balanced equation, mole(s) of methyl m-nitrobenzoate produced = mole(s) of methyl benzoate taken = 9.5483*10^-4 mole.

Molar mass of methyl m-nitrobenzoate, C₈H₇NO₃ = (8*12.011 + 7*1.008 + 1*14.007 + 3*15.999) g/mol = 165.148 g/mol.

Theoretical yield of methyl m-nitrobenzoate = (mole of methyl m-nitrobenzoate)*(molar mass of methyl m-nitrobenzoate) = (9.5483*10^-4 mole)*(165.148 g/mol) = 0.1577 g (ans).

However, only 0.033 g methyl m-nitrobenzoate was produced; therefore, percent yield of methyl m-nitrobenzoate = (0.033 g)/(0.1577 g)*100 = 20.9258% 20.92% (ans).