Ethyl butanoate is a major component of the taste and smell of pineapple. A laboratory analysis of benzyl acetate reveals the following mass percent composition: C \( 62.04 \% \) H \( 10.41 \% \) O \( 27.55 \% \) Find the empirical formula of ethyl butanoate. View Available Hint(s)

To find the empirical formula of ethyl butanoate based on its mass percent composition, we can follow these steps: 1. **Convert the mass percentages to grams**: Assume we have 100 grams of the compound. This means we have: - Carbon (C): 62.04 g - Hydrogen (H): 10.41 g - Oxygen (O): 27.55 g 2. **Convert grams to moles**: Use the molar mass of each element to convert grams to moles. - Molar mass of C = 12.01 g/mol - Molar mass of H = 1.008 g/mol - Molar mass of O = 16.00 g/mol Now, calculate the moles of each element: \[ \text{Moles of C} = \frac{62.04 \, \text{g}}{12.01 \, \text{g/mol}} \approx 5.17 \, \text{mol} \] \[ \text{Moles of H} = \frac{10.41 \, \text{g}}{1.008 \, \text{g/mol}} \approx 10.33 \, \text{mol} \] \[ \text{Moles of O} = \frac{27.55 \, \text{g}}{16.00 \, \text{g/mol}} \approx 1.72 \, \text{mol} \] 3. **Find the simplest mole ratio**: Divide the number of moles of each element by the smallest number of moles calculated. - Smallest number of moles is for O (1.72 mol): \[ \text{C ratio} = \frac{5.17}{1.72} \approx 3.01 \quad \text{(approximately 3)} \] \[ \text{H ratio} = \frac{10.33}{1.72} \approx 6.01 \quad \text{(approximately 6)} \] \[ \text{O ratio} = \frac{1.72}{1.72} = 1 \] 4. **Write the empirical formula**: Based on the ratios, we can write the empirical formula as: \[ \text{C}_3\text{H}_6\text{O} \] Thus, the empirical formula of ethyl butanoate is \( \text{C}_3\text{H}_6\text{O} \).