Answer :
To determine the empirical formula of a compound, we follow these steps:
1. Convert the percent composition to masses.
Assuming we have a 100 g sample of the compound:
- Carbon: [tex]\( \frac{52.1}{100} \times 100\, \text{g} = 52.1\, \text{g} \)[/tex]
- Hydrogen: [tex]\( \frac{13.2}{100} \times 100\, \text{g} = 13.2\, \text{g} \)[/tex]
- Oxygen: [tex]\( \frac{34.7}{100} \times 100\, \text{g} = 34.7\, \text{g} \)[/tex]
2. Convert the masses to moles.
Using the atomic masses:
- Carbon (C): [tex]\( \text{atomic mass} = 12.01 \, \text{g/mol} \)[/tex]
- Hydrogen (H): [tex]\( \text{atomic mass} = 1.008\, \text{g/mol} \)[/tex]
- Oxygen (O): [tex]\( \text{atomic mass} = 16.00\, \text{g/mol} \)[/tex]
Moles of each element:
- Moles of Carbon: [tex]\( \frac{52.1\, \text{g}}{12.01\, \text{g/mol}} = 4.338 \, \text{mol} \)[/tex]
- Moles of Hydrogen: [tex]\( \frac{13.2\, \text{g}}{1.008\, \text{g/mol}} = 13.095 \, \text{mol} \)[/tex]
- Moles of Oxygen: [tex]\( \frac{34.7\, \text{g}}{16.00\, \text{g/mol}} = 2.169 \, \text{mol} \)[/tex]
3. Determine the mole ratio by dividing each element's mole count by the smallest mole count.
- Smallest mole count is for Oxygen: [tex]\( 2.169 \, \text{mol} \)[/tex]
Ratios:
- Carbon: [tex]\( \frac{4.338 \, \text{mol}}{2.169 \, \text{mol}} = 2.000 \)[/tex]
- Hydrogen: [tex]\( \frac{13.095 \, \text{mol}}{2.169 \, \text{mol}} = 6.038 \)[/tex]
- Oxygen: [tex]\( \frac{2.169 \, \text{mol}}{2.169 \, \text{mol}} = 1.000 \)[/tex]
4. Convert these ratios to the smallest whole numbers.
The ratios suggest approximately:
- Carbon to Hydrogen to Oxygen is [tex]\( 2:6:1 \)[/tex]
Thus, the empirical formula coefficients are:
- [tex]\( C = 2 \)[/tex]
- [tex]\( H = 6 \)[/tex]
- [tex]\( O = 1 \)[/tex]
Therefore, the empirical formula of the compound is [tex]\( \mathbf{C_2 H_6 O} \)[/tex]. The correct answer from the given choices is:
[tex]\[ \boxed{C_2 H_6 O} \][/tex]
1. Convert the percent composition to masses.
Assuming we have a 100 g sample of the compound:
- Carbon: [tex]\( \frac{52.1}{100} \times 100\, \text{g} = 52.1\, \text{g} \)[/tex]
- Hydrogen: [tex]\( \frac{13.2}{100} \times 100\, \text{g} = 13.2\, \text{g} \)[/tex]
- Oxygen: [tex]\( \frac{34.7}{100} \times 100\, \text{g} = 34.7\, \text{g} \)[/tex]
2. Convert the masses to moles.
Using the atomic masses:
- Carbon (C): [tex]\( \text{atomic mass} = 12.01 \, \text{g/mol} \)[/tex]
- Hydrogen (H): [tex]\( \text{atomic mass} = 1.008\, \text{g/mol} \)[/tex]
- Oxygen (O): [tex]\( \text{atomic mass} = 16.00\, \text{g/mol} \)[/tex]
Moles of each element:
- Moles of Carbon: [tex]\( \frac{52.1\, \text{g}}{12.01\, \text{g/mol}} = 4.338 \, \text{mol} \)[/tex]
- Moles of Hydrogen: [tex]\( \frac{13.2\, \text{g}}{1.008\, \text{g/mol}} = 13.095 \, \text{mol} \)[/tex]
- Moles of Oxygen: [tex]\( \frac{34.7\, \text{g}}{16.00\, \text{g/mol}} = 2.169 \, \text{mol} \)[/tex]
3. Determine the mole ratio by dividing each element's mole count by the smallest mole count.
- Smallest mole count is for Oxygen: [tex]\( 2.169 \, \text{mol} \)[/tex]
Ratios:
- Carbon: [tex]\( \frac{4.338 \, \text{mol}}{2.169 \, \text{mol}} = 2.000 \)[/tex]
- Hydrogen: [tex]\( \frac{13.095 \, \text{mol}}{2.169 \, \text{mol}} = 6.038 \)[/tex]
- Oxygen: [tex]\( \frac{2.169 \, \text{mol}}{2.169 \, \text{mol}} = 1.000 \)[/tex]
4. Convert these ratios to the smallest whole numbers.
The ratios suggest approximately:
- Carbon to Hydrogen to Oxygen is [tex]\( 2:6:1 \)[/tex]
Thus, the empirical formula coefficients are:
- [tex]\( C = 2 \)[/tex]
- [tex]\( H = 6 \)[/tex]
- [tex]\( O = 1 \)[/tex]
Therefore, the empirical formula of the compound is [tex]\( \mathbf{C_2 H_6 O} \)[/tex]. The correct answer from the given choices is:
[tex]\[ \boxed{C_2 H_6 O} \][/tex]
