Calculate the percentage yield based on the quantity of benzoic acid used?

Methanol used = 8.00cm^3

Density of methanol = 0.79

Relative molecular weight of methanol = 32g



An ester was made using benzoic acid and methanol.



5.005g of benzoic acid was used and the final weight of the ester was 3.001g.



how do i calculate the percentage yield?Calculate the percentage yield based on the quantity of benzoic acid used?
C6H5COOH + CH3OH --%26gt; C6H5COOCH3 + H2O

5.005 g benzoic acid / 122.1 g/mol = 0.04098 mol

8.00 cc x 0.79 g/cc = 6.32 g methanol

6.32 g methanols / 32.0 g/mol = 0.1972 mol

You have excess methanol

0.04098 mol benzoic acid will yiel 0.04098 mol methyl benzoate

C6H5COOCH3 molar mass = 136.15 g/mol

0.04098 mol x 136.15 g/mol = 5.579 g

3.001 g / 5.579 g = 0.538 (53.8%)