# Stoichiometry: Salicylic Acid and Ethanol Reaction ## (a) Balanced Equation with Molecular Structures **Esterification Reaction:** Salicylic acid (contains a carboxylic acid and a phenol group) reacts with ethanol (an alcohol) to produce ethyl salicylate (an ester) and water. **Molecular structures:** ``` HO | C6H4—COOH + HOCH2CH3 → C6H4—COOCH2CH3 + H2O (salicylic (ethanol) (ethyl salicylate) (water) acid) ``` - **Salicylic acid:** Benzene ring with an -OH and -COOH group attached ortho to each other. - **Ethanol:** Two-carbon alcohol (CH3CH2OH). - **Ethyl salicylate:** Benzene ring with an -OH and an -COOCH2CH3 group. - **Water:** H2O. ## (b) Identify Limiting Reagent ### Step 1: Molar Masses - **Salicylic acid (C7H6O3):** \( (7 \times 12.01) + (6 \times 1.01) + (3 \times 16.00) = 84.07 + 6.06 + 48.00 = \textbf{138.13 g/mol} \) - **Ethanol (C2H6O):** \( (2 \times 12.01) + (6 \times 1.01) + (1 \times 16.00) = 24.02 + 6.06 + 16.00 = \textbf{46.08 g/mol} \) ### Step 2: Moles of Each Reactant #### Salicylic Acid - Given: **800 mg = .800 g** - Moles: \( n = \frac{.800\, \text{g}}{138.13\, \text{g/mol}} = .00579\, \text{mol} \) #### Ethanol - Given: **1.2 mL**, density = **.789 g/mL** - Mass: \( 1.2\, \text{mL} \times .789\, \text{g/mL} = .947\, \text{g} \) - Moles: \( n = \frac{.947\, \text{g}}{46.08\, \text{g/mol}} = .0206\, \text{mol} \) ### Step 3: Limiting Reagent **Balanced reaction:** 1 mol salicylic acid : 1 mol ethanol - Salicylic acid: .00579 mol - Ethanol: .0206 mol **Limiting reagent:** Salicylic acid (fewer moles) --- ## (c) Predict Grams of Products (Assume 100% Yield) ### Step 1: Moles of Products Formed - Both **ethyl salicylate** and **water** are formed in a 1:1 molar ratio with the limiting reagent. - Thus, both will be formed in **.00579 mol**. ### Step 2: Molar Masses of Products - **Ethyl salicylate (C9H10O3):** - \( (9 \times 12.01) + (10 \times 1.01) + (3 \times 16.00) = 108.09 + 10.10 + 48.00 = \textbf{166.19 g/mol} \) - **Water (H2O):** - \( (2 \times 1.01) + 16.00 = 2.02 + 16.00 = \textbf{18.02 g/mol} \) ### Step 3: Mass of Products #### Ethyl Salicylate \[ .00579\, \text{mol} \times 166.19\, \text{g/mol} = .963\, \text{g} \] #### Water \[ .00579\, \text{mol} \times 18.02\, \text{g/mol} = .104\, \text{g} \] --- ## **Final Answers** 1. **Balanced equation (using structures):** - See part (a) above. 2. **Limiting reagent:** - **Salicylic acid** (.00579 mol, fewer than ethanol) 3. **Predicted product masses:** - **Ethyl salicylate:** **.963 g** - **Water:** **.104 g**