Miniscale nitration of methyl benzoate was performed in the lab: Procedure below
ID: 483003 • Letter: M
Question
Miniscale nitration of methyl benzoate was performed in the lab:
Procedure below:
1. Prepare and ice bath in a 100mL beaker on a magnetic stir plate.
2. To a clean, dry 25 mL Erlenmeyer flask, add 1.0 mL of methyl benzoate via pipet.
3. Calculate the mass of methyl benzoate.
4. To the methyl benzoate, add 2.25 mL of concentrated (18M) sulfuric acid and a spin bar.
5. Carefully clamp the flask so that it is immersed in the ice bath and begin stirring.
6. Prepare the nitrating solution by adding 0.75mL of concentrated (18M) sulfuric acid and 0.75mL of concentrated (16M) nitric acid to a clean, dry vial.
7. Cool the mixture in a second ice bath.
8. With a Pasteur pipet, slowly add the cooled sulfuric acid-nitric acid solution to the stirred solution of methyl benzoate. *Addition should take approximately 15 minutes. Adding the nitrating agent more rapidly will increase the amount of by-products and decrease the yield
9. After all of the nitrating agent has been added, carefully remove the ice bath.
10. Continue stirring until the reaction has come to room temperature.
11. Then let the reaction flask stand undisturbed for 15 minutes.
12. Place 10 g of ice in a 100mL beaker.
13. With a Pasteur pipet, carefully transfer the reaction mixture to the beaker.
14. Rinse the flask with 5mL of cold water.
15. When the ice has melted, vacuum filter the crude crystals using a Buchner funnel.
16. Wash the crystals with 2 5mL portions of cold water and finally with 2 1.5mL of cold methanol.
17. The crude product can be recrystallized from methanol.
18. Record the mass of the dry product and calculate the yield.
QUESTION: Calculate percent yield of the product.
data:
mass of methyl benzoate: 1.111 grams
mass of dry product: 0.885 grams
Explanation / Answer
In this first we have to balance the equation
CH3COOC6H5 + Con H2SO4+ Con HNO3------------------>CH3COOC6H4NO2 + HSO4- + H3O+
Next we have to find the limiting agent
Moles of H2SO4 = 0.75 L x 18Mol/L = 13.5 mol
Moles of HNO3 = 0.75 L x 16mol/L = 12 mol
Moles of CH3COOC6H5 ( methyl benzoate) = 1.111g / molar mass(136g/mol)
= 0.0082 mol
Since methyl benzoate is the limiting reagent
Next we calculate the theoretical yield
1mole of methylbenzoate + 1mole of NO2 (HNO3) .gives 1 mol of methyl 3-nitrobenzoate
molar mass methyl benzoate = 136 g
molar mass of methyl 3-nitrobenzoate = 181g
so theoretical yield will be no of moles limiting reactant x molar mass of the product = 0.0082 x 181
= 1.4842g
therefore
% yield = actual amount of product were obtained / theoretical yield x 100 = 0.885 /1.4842 x 100
= 59.62 %