Final answer:
The empirical formula of acetic acid is CH2O, and the molecular formula is C2H4O2 or CH3COOH. These formulas are determined by converting the percentage composition to moles, finding the simplest mole ratio, and then relating the molar mass of the empirical formula to the experimental molar mass of acetic acid.
Step-by-step explanation:
To calculate the empirical formula of acetic acid based on its percentage composition, we start by assuming we have 100 grams of the compound. This gives us 39.9 grams of carbon, 6.7 grams of hydrogen, and 53.4 grams of oxygen. We then convert these masses to moles by dividing by the respective atomic masses of these elements (C: 12.01 g/mol, H: 1.008 g/mol, and O: 16.00 g/mol).
- C: 39.9 g ÷ 12.01 g/mol = 3.324 moles
- H: 6.7 g ÷ 1.008 g/mol = 6.647 moles
- O: 53.4 g ÷ 16.00 g/mol = 3.338 moles
Next, we divide each mole value by the smallest number of moles obtained, in this case, the moles of carbon. The ratios determine the proportion of atoms in the empirical formula.
- C: 3.324 / 3.324 = 1
- H: 6.647 / 3.324 = 2
- O: 3.338 / 3.324 = 1
Therefore, the empirical formula of acetic acid is CH2O.
To find the molecular formula, we use the given molecular mass of 60 amu (which is close to the molar mass in grams per mole). The mass of the empirical formula (CH2O) is 12.01 + (2 × 1.008) + 16.00 = 30.026 g/mol. The ratio of the molecular mass to the empirical formula mass is approximately 2 (60.06 / 30.026 ≈ 2). This means the molecular formula contains twice as many of each atom as the empirical formula.
Thus, the molecular formula of acetic acid is C2H4O2 or written as CH3COOH.