Answer:
C₂H₄O
Explanation:
What we want to calculate when determining empirical formulas is to obtain the proportion of the coefficients of the atoms in the molecule. This proportion is the same as the ratio in terms of number of moles, and since we have the quantities in grams after combustion , we can calculate the number of moles and thus the empirical formula.
Molar mass CO₂ = 44 gmol⁻¹
Molar mass H₂O = 18 gmol⁻¹
# mol CO₂ = 9.99 g / 44 gmol⁻¹ = 0.227 mol CO₂ ⇒ mol C = 0.227
# mol H₂O =4.09 g H₂O / 18 gmol⁻¹ = 0.227 mol H₂O ⇒ mol H = 0.454
Now to determine the mol of oxygen we would have to use the fact that we have 5.00 g in the combustion, and since we know the mol C and H, it is easy to determine the mass of O. We can not do the relations we did for C and H, since we already have oxygen in the formula.
mass C = 0.227 mol x 12 gmol⁻¹ = 2.72 g C
mass H = 0.454 mol x 1 gmol⁻¹ = 0.454 g H
mass O present in compound = 5.00 g - 2.72 g - 0.454 g = 1.82 g O
mol O = 1.82 g/ 16 gmol⁻¹ = 0.114
Now the composition and proportions are:
mol C = 0.227 / 0.114 = 1.99
mol H = 0.454 / 0.114 = 3.98
mol O = 0.114 / 0.114 = 1
Rounding these numbers we have the empirical formula C₂H₄O
