Answer:
Empirical formula = CH3O
Molecular formula = C2H6O2
Explanation:
Step 1: Data given
Mass of the sample = 23.46 grams
Mass of H2O = 20.42 grams
Molar mass of H2O = 18.02 g/mol
Mass of CO2 = 33.27 grams
Molar mass of CO2 = 44.01 G:mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Atomic mass of H = 1.01 g/mol
Molar mass of the compound = 62.0 g/mol
Step 2: Calculate moles of H2O
Moles H2O = 20.42 grams / 18.02 g/mol
Moles H2O = 1.133 moles
Step 3: Calculate moles H
For 1 mol H2O we have 2 moles H
For 1.133 moles H2O we have 2* 1.133 = 2.266 moles H
Step 4: Calculate mass H
Mass H = 2.266 moles * 1.01 g/mol
Mass H = 2.29 grams
Step 5: Calculate moles CO2
Moles CO2 = 33.27 grams / 44.01 g/mol
Moles CO2 = 0.7560 moles
Step 6: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.7560 moles CO2 we have 0.7560 moles C
Step 7: Calculate mass C
Mass C = 0.7560 moles * 12.01 g/mol
Mass C = 9.08 grams
Step 8: Calculate mass O
Mass O = 23.46 grams - 9.08 grams - 2.29 grams
Mass O =12.09 grams
Step 9: Calculate moles O
Moles O = 12.09 grams / 16.0 g/moles
Moles O = 0.7556
Step 10: Calculate mol ratio
We divide by the smallest amount of moles
C: 0.7560 moles / 0.7556 moles =1
H: 2.266 moles / 0.7556 moles =3
O; 0.7556 / 0.7560 moles = 1
This means for 1 mol C we have 3 moles H and 1 mol O
The empirical formula is CH3O
Step 11: Calculate the molecular formula
The molar mass of the empirical formula is 31 g/mol
Step 11: Calculate molecular formula
We have to multiply the empirical formula by n
n = 62.0 g/mol / 31g/mol = 2
Molecular formula = 2*(CH3O)
Molecular formula = C2H6O2
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