To determine the empirical formula of the compound, we can follow these detailed steps:
Step 1: Determine the moles of carbon in the CO₂ produced.
- We are given the mass of CO₂ produced, which is 1.6004 g.
- The molar mass of CO₂ is 44.01 g/mol.
[tex]\[ \text{Moles of CO₂} = \frac{\text{mass of CO₂}}{\text{molar mass of CO₂}} = \frac{1.6004 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.03636 \, \text{mol} \][/tex]
Since each mole of CO₂ contains 1 mole of carbon:
[tex]\[ \text{Moles of C} = 0.03636 \, \text{mol} \][/tex]
Step 2: Determine the moles of hydrogen in the H₂O produced.
- We are given the mass of H₂O produced, which is 0.6551 g.
- The molar mass of H₂O is 18.02 g/mol.
[tex]\[ \text{Moles of H₂O} = \frac{\text{mass of H₂O}}{\text{molar mass of H₂O}} = \frac{0.6551 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.03635 \, \text{mol} \][/tex]
Each mole of H₂O contains 2 moles of hydrogen:
[tex]\[ \text{Moles of H} = 0.03635 \, \text{mol} \times 2 \approx 0.07271 \, \text{mol} \][/tex]
Step 3: Calculate the masses of carbon and hydrogen.
- The molar mass of carbon (C) is 12.01 g/mol:
[tex]\[ \text{Mass of C} = \text{moles of C} \times \text{molar mass of C} = 0.03636 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 0.43674 \, \text{g} \][/tex]
- The molar mass of hydrogen (H) is 1.008 g/mol:
[tex]\[ \text{Mass of H} = \text{moles of H} \times \text{molar mass of H} = 0.07271 \, \text{mol} \times 1.008 \, \text{g/mol} \approx 0.07329 \, \text{g} \][/tex]
Step 4: Calculate the mass of oxygen in the compound.
The total mass of the compound is the mass of the sample:
[tex]\[ \text{Mass of sample} = 0.8009 \, \text{g} \][/tex]
The mass of oxygen is obtained by subtracting the masses of carbon and hydrogen from the total mass:
[tex]\[ \text{Mass of O} = \text{mass of sample} - (\text{mass of C} + \text{mass of H}) = 0.8009 \, \text{g} - (0.43674 \, \text{g} + 0.07329 \, \text{g}) \approx 0.29087 \, \text{g} \][/tex]
Step 5: Calculate the moles of oxygen in the compound.
- The molar mass of oxygen (O) is 16.00 g/mol:
[tex]\[ \text{Moles of O} = \frac{\text{mass of O}}{\text{molar mass of O}} = \frac{0.29087 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.01818 \, \text{mol} \][/tex]
Step 6: Determine the mole ratio of C, H, and O.
[tex]\[ \text{Mole ratio} = \left( \frac{\text{moles of C}}{\text{smallest number of moles}}, \frac{\text{moles of H}}{\text{smallest number of moles}}, \frac{\text{moles of O}}{\text{smallest number of moles}} \right) \][/tex]
The smallest number of moles is 0.01818 (O).
[tex]\[ \text{Mole ratio} = \left( \frac{0.03636}{0.01818}, \frac{0.07271}{0.01818}, \frac{0.01818}{0.01818} \right) \approx (2.0, 4.0, 1.0) \][/tex]
Step 7: Approximate the ratio to the nearest whole number.
So, the empirical formula ratio is approximately:
[tex]\[ (2, 4, 1) \][/tex]
Thus, the empirical formula of the compound is C₂H₄O.
