Answer:
The empirical formula is C10H20O
Explanation:
Step 1: Data given
Mass of the sample = 0.2010 grams
Mass of CO2 = 0.5658 grams
Molar mass CO2 = 44.01 g/mol
Mass of H2O = 0.2318 grams
Molar mass H2O = 18.02 g/mol
Atomic mass C = 12.01 g/mol
Atomic mass O = 16.0 g/mol
Atomic mass H = 1.01 g/mol
Step 2: calculate moles CO2
Moles CO2 = 0.5658 gram / 44.01 g/mol
Moles CO2 = 0.01286 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.01286 moles CO2 we have 0.01286 moles C
Step 4: Calculate mass C
MAss C = 0.01286 moles * 12.01 g/mol
Mass C =0.1544 grams
Step 5: Calculate moles H2O
Moles H2O =0.2318 grams / 18.02 g/mol
Moles H2O = 0.01286 moles
Step 6: Calculate moles H
For 1 mol H2O we have 2 moles H
For 0.01286 moles H2O we have 2*0.01286 = 0.02572 moles H
Step 7: Calculate mass H
Mass H = 0.02572 moles * 1.01 g/mol
Mass H = 0.0260 grams
Step8: Calculate mass O
Mass O = 0.2010 - 0.1544 - 0.0260
Mass O = 0.0206 grams
Step 9: Calculate moles O
Moles O = 0.0206 grams / 16.0 g/mol
Moles O = 0.00129 moles
Step 10: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.01286 moles / 0.00129 moles = 10
H: 0.02572 moles / 0.00129 moles = 20
O: 0.00129 moles / 0.00129 moles = 1
The empirical formula is C10H20O
