Question
a dilute solution of ethanoic acid is used in households daily. The relative molecular mass of ethanoic acid is 60g/mol. Ethanoic Acid consists of 39,9% carbon, 6,7% hydrogen and 53,4% oxygen. Determine the empirical formula
Answers
GPT-4o mini
To find the empirical formula of ethanoic acid from the given percentage composition (39.9% carbon, 6.7% hydrogen, and 53.4% oxygen), we can follow these steps:
1. **Convert percentages to grams** (assuming we have 100 g of the compound):
- Carbon: 39.9 g
- Hydrogen: 6.7 g
- Oxygen: 53.4 g
2. **Convert grams to moles** using the atomic weights:
- Carbon (C): ~12.01 g/mol
- Hydrogen (H): ~1.008 g/mol
- Oxygen (O): ~16.00 g/mol
Now, calculate the moles for each element:
\[
\text{Moles of C} = \frac{39.9 \, \text{g}}{12.01 \, \text{g/mol}} \approx 3.32 \, \text{mol}
\]
\[
\text{Moles of H} = \frac{6.7 \, \text{g}}{1.008 \, \text{g/mol}} \approx 6.64 \, \text{mol}
\]
\[
\text{Moles of O} = \frac{53.4 \, \text{g}}{16.00 \, \text{g/mol}} \approx 3.34 \, \text{mol}
\]
3. **Find the simplest whole number ratio** by dividing by the smallest number of moles calculated:
- The smallest number of moles is approximately 3.32 (for Carbon).
\[
\text{Ratio of C} = \frac{3.32}{3.32} \approx 1
\]
\[
\text{Ratio of H} = \frac{6.64}{3.32} \approx 2
\]
\[
\text{Ratio of O} = \frac{3.34}{3.32} \approx 1
\]
Thus, the simplified ratio of Carbon to Hydrogen to Oxygen is approximately 1:2:1.
4. **Write the empirical formula**:
The empirical formula of ethanoic acid is \( \text{C}_1\text{H}_2\text{O}_1 \), or simply \( \text{CH}_2\text{O} \).
However, ethanoic acid specifically has a molecular formula of \( \text{C}_2\text{H}_4\text{O}_2 \), which is a multiple of the empirical formula \( \text{CH}_2\text{O} \) (i.e., it is two times the empirical formula).
1. **Convert percentages to grams** (assuming we have 100 g of the compound):
- Carbon: 39.9 g
- Hydrogen: 6.7 g
- Oxygen: 53.4 g
2. **Convert grams to moles** using the atomic weights:
- Carbon (C): ~12.01 g/mol
- Hydrogen (H): ~1.008 g/mol
- Oxygen (O): ~16.00 g/mol
Now, calculate the moles for each element:
\[
\text{Moles of C} = \frac{39.9 \, \text{g}}{12.01 \, \text{g/mol}} \approx 3.32 \, \text{mol}
\]
\[
\text{Moles of H} = \frac{6.7 \, \text{g}}{1.008 \, \text{g/mol}} \approx 6.64 \, \text{mol}
\]
\[
\text{Moles of O} = \frac{53.4 \, \text{g}}{16.00 \, \text{g/mol}} \approx 3.34 \, \text{mol}
\]
3. **Find the simplest whole number ratio** by dividing by the smallest number of moles calculated:
- The smallest number of moles is approximately 3.32 (for Carbon).
\[
\text{Ratio of C} = \frac{3.32}{3.32} \approx 1
\]
\[
\text{Ratio of H} = \frac{6.64}{3.32} \approx 2
\]
\[
\text{Ratio of O} = \frac{3.34}{3.32} \approx 1
\]
Thus, the simplified ratio of Carbon to Hydrogen to Oxygen is approximately 1:2:1.
4. **Write the empirical formula**:
The empirical formula of ethanoic acid is \( \text{C}_1\text{H}_2\text{O}_1 \), or simply \( \text{CH}_2\text{O} \).
However, ethanoic acid specifically has a molecular formula of \( \text{C}_2\text{H}_4\text{O}_2 \), which is a multiple of the empirical formula \( \text{CH}_2\text{O} \) (i.e., it is two times the empirical formula).