Answer:
C₂H₄O
Step-by-step explanation:
In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.
Moles CO₂ = Moles C:
11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =
3.216g C
Moles H₂O = 1/2 moles H:
4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =
0.537 mol H * (1g/mol) = 0.537g H
Mass O to find moles O:
5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O
Ratio of atoms -Dividing in 0.134 moles-:
C = 0.268mol C / 0.134 mol O = 2
H = 0.537mol H / 0.134 mol O = 4
O = 0.134mol O / 0.134 mol O = 1
Empirical formula is:
C₂H₄O