Discover new information and insights with the help of IDNLearn.com. Join our interactive Q&A platform to receive prompt and accurate responses from experienced professionals in various fields.
Sagot :
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]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.