Answer:
The empirical formula is C3H3O
Explanation:
Step 1: Data given
Suppose the mass of the molecule = 100 grams
The molecule contains:
65.5 % Carbon = 65.5 grams
5.5 % Hydrogen = 5.5 grams
29.0% Oxygen = 29.0 grams
Molar mass of C = 12 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles C = 65.5 grams / 12 g/mol = 5.46 moles
Moles H = 5.5 / 1.01 g/mol = 5.45 moles
Moles O = 29.0 grams / 16 g/mol = 1.8125 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
C: 5.46 / 1.8125 = 3
H = 5.45 / 1.8125 = 3
O = 1.8125/1.8125 = 1
The empirical formula is C3H3O
