Answer:
The empirical formula is C3H4O
Explanation:
Step 1: Data given
Mass of the compound = 1.000 grams
The compound contains:
- Carbon
- hydrogen
- oxygen
The combustion of this compound gives:
2.360 grams of CO2
0.640 grams of H2O
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 2.360 grams / 44.01 g/mol
Moles CO2 = 0.05362 moles
In CO2 we have 1 mol
This means for 1 mol CO2 we have 1 mol C
For 0.05362 moles CO2 we have <u>0.05362 moles C</u>
We have 0.05362 moles of C in the compound
Step 3: Calculate mass of C
Mass C = moles C * molar mass C
Mass C = 0.05362 moles * 12.0 g/mol
Mass C = 0.643 grams
Step 4: Calculate moles of H2O
Moles H2O = 0.640 grams / 18.02 g/mol
Moles H2O = 0.0355 moles H2O
For 1 mol H2O we have 2 moles of H
For 0.0355 moles H2O we have 2*0.0355 = <u>0.071 moles H</u>
Step 5: Calculate mass of H
Mass H = moles H * molar mass H
Mass H = 0.071 moles * 1.01 g/mol
Mass H = 0.072 grams
Step 6: Calculate mass of O
Mass of O = Mass of compound - mass of C - mass of H
Mass of O = 1.000 g - 0.643 - 0.072 = 0.285 grams
Step 7: Calculate moles of O
Moles O = 0.285 grams / 16.0 g/mol
<u>Moles O = 0.0178 moles</u>
<u>
</u>
Step 8: Calculate mol ratio
We divide by the smallest amount of moles
C: 0.05362 / 0.0178 = 3
H: 0.071 / 0.0178 = 4
O: 0.0178/0.0178 = 1
The empirical formula is C3H4O
