Answer:
The empirical formula is C3H6O
Explanation:
Step 1: Data given
Mass of the sample =2.088 grams
The mass contains carbon, hydrogen, and oxygen
Mass of CO2 produced = 4.746 grams
Mass of H2O produced = 1.943 grams
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 4.746 grams/ 44.01 g/mol
Moles CO2 = 0.1078 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.1078 moles CO2 we'll have 0.1078 moles C
Step 4: Calculate mass C
Mass C: moles C * atomic mass C
Mass C: 0.1078 moles * 12.01 g/mol
Mass C= 1.295 grams
Step 5: Calculate moles H2O
Moles H2O = 1.943 grams / 18.02 g/mol
Moles H2O = 0.1078 moles
Step 6: Calculate moles H
For 1 mol H2O we'll have 2 moles H
For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H
Step 7: Calculate mass H
Mass H = 0.2046 moles * 1.01 g/mol
Mass H = 0.218 grams
Step 8: Calculate mass O
Mass O = 2.088 grams - 1.295 grams - 0.218 grams
Mass O = 0.575 grams
Step 9: Calculate moles O
Moles O = 0.575 grams / 16.0 g/mol
Moles O = 0.0359 moles
Step 10: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.1078 moles / 0.0359 moles = 3
H: 0.2156 moles / 0.0359 moles = 6
O: 0.0359 moles / 0.0359 moles =1
The empirical formula is C3H6O
