Question

2. Methyl salicylate was reacted with sodium hydroxide amounts shown below: give 1.78 grams of salicylic acid using the Millimoles Used Molar Role Chemical name M Mass Amount (/mol)(mmol) reactant Methyl Salicilate 152.154.6 Sodium reactant hydroxide 6 M solution 45 ml product Salicylic acid 138.121.78g A. Draw the reaction below: B. What is the percent yield of this reaction. C, The melting range for the product is 147 to 154 c. what would you do to improve the purity of the product?

Answer 1

A.

B.

Number of millimoles of methyl salicylate $=\frac{\text { 4.6 g}}{\text { 152.15 g/mol}} \times \frac{\text { 1000 mmol}}{\text { 1 mol }}=\text { 30.23 mmol }$

Number of millimoles of salicylic acid $=\frac{\text { 1.78 g}}{\text { 138.12 g/mol}} \times \frac{\text { 1000 mmol}}{\text { 1 mol }}=\text { 12.88 mmol }$

Number of millimoles of NaOH $=\text { 6 mol/L } \times \frac{\text { 45 mL}}{\text { 1000 mL/L}} \times \frac{\text { 1000 mmol}}{\text { 1 mol }}=\text { 270 mmol }$

Percent yield $=\frac{ \text { 12.88 mmol }}{ \text { 30.23 mmol}} \times 100= \text { 42.6 } \%$

C.

Broad melting range indicates that product is impure. To improve the purity of the product, it is recrystallised from suitable solvent.