Question

in an experiment, wintergreen oil (methyl salicylate) and water were made from salicylic acid.

calculate the theoretical yield of methyl salicylate when you start with 50 grams of salicylic acid, 40mL of concentrated sulfuric acid (H2SO4) and 250 mL of methanol (CH3OH).

Answer 1

Step 1

Given:

Mass of Salicylic Acid =50g

Volume of H₂SO₄ = 40mL

Volume of Methanol = 250 mL

To Calculate: the theoretical yield of methyl salicylate

Step 2

Preparation of methyl salicylate is shown below:

Step 3

Determine the Limiting Reagent:

Moles of Salicylic Acid

Molar mass of salicylic Acid =138.121 g mol^-1

Moles of Salicylic Acid = Mass/ Molar Mass

=50 g/ 138.121 g mol^-1

=0.3617 mol

Moles of Salicylic Acid =0.3617 mol

Moles of H₂SO₄

Density of H₂SO₄ = 1.83 g/mL

Mass of H₂SO₄ = Density * Volume

=1.83 g/mL * 40 mL

=73.2 g

Moles of H₂SO₄ = Mass/ Molar mass

=73.2 g /98g mol^-1

=0.746 mol

Moles of H₂SO₄ =0.746 mol

Moles of Methanol

Density of Methanol =0.792 g/mL

Mass of Methanol = Density * Volume

= 0.792 g/mL * 250 mL

=198 g

Molesof Methanol = Mass/ Molar Mass

= 198g / 32.04 g mol^-1

=6.179 mol

Moles of Methanol = 6.179 mol

Moles of Salicylic Acid =0.3617 mol

Moles of H₂SO₄ =0.746 mol

Moles of Methanol = 6.179 mol

Since moles of Salicylic Acid is less, So Salicylic acid will be the Limiting Reactant.

Step 4

According to the reaction

1 mol Salicylic acid gives 1 mol methyl salicylate

So, 0.3617 mol Salicylic acid will give 0.3617 mol methyl salicylate.

Moles of methyl salicylate = 0.3617 mol

Molar Mass of methyl salicylate =152.14 g mol^-1

Mass of methyl salicylate = Moles * Molar Mass

= 0.3617 mol * 152.14 g mol^-1

= 55.03 g

Theoretical Yield of methyl salicylate = 55.03g

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