A reaction was performed in which 3.7 g of benzoic acid was reacted with excess methanol to make 2.1 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.

The theoretical yield is 4.13 grams methyl benzoate

The percent yield is 50.8 %

Explanation:

Step 1: Data given

Mass of benzoic acid = 3.7 grams

Mass of methyl benzoate = 2.1 grams

Molar mass of benzoic acid = 122.12 g/mol

Molar mass of methyl benzoate = 136.15 g/mol

Step 2: The balanced equation

C7H6O2 + CH3OH → C8H8O2 + H2O

Step 3: Calculate moles benzoic acid

Moles benzoic acid = mass benzoic acid / molar mass

Moles benzoic acid = 3.7 grams / 122.12 g/mol

Moles benzoic acid = 0.0303 moles

Step 4: Calculate moles methy benzoate

For 1 mol benzoic acid we'll have 1 mol methyl benzoate

For 0.0303 moles benzoic acid we'll have 0.0303 moles methyl benzoate

Step 5: Calculate mass methyl benzoate

Mass methyl benzoate = moles methyl benzoate * molar mass

Mass methyl benzoate = 0.0303 moles * 136.15 g/mol

Mass methyl benzoate = 4.13 grams

Step 6: Calculate % yield

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

% yield = (2.1 grams / 4.13 grams) * 100%

% yield = 50.8 %

The theoretical yield is 4.13 grams methyl benzoate

The percent yield is 50.8 %