The percentage composition of acetic acid is found to be 39.9% C, 6.7% H, and
53.4% O and its molecular mass was determined by experiment to be 60 amu
Calculate the empirical and molecular formula of acetic acid.
To find the empirical formula of acetic acid, we need to determine the simplest whole number ratio of the atoms present in the compound.
1. Assume we have 100g of acetic acid.
2. Calculate the number of moles for each element using their respective percentages:
C: (39.9 g/100g) * (1 mol/12.01 g) ≈ 3.323 mol
H: (6.7 g/100g) * (1 mol/1.008 g) ≈ 6.65 mol
O: (53.4 g/100g) * (1 mol/16.00 g) ≈ 3.34 mol
3. Divide the number of moles by the smallest number of moles calculated to find the simplest ratio:
C: 3.323 mol / 3.323 mol ≈ 1
H: 6.65 mol / 3.323 mol ≈ 2
O: 3.34 mol / 3.323 mol ≈ 1
So, the empirical formula of acetic acid is CH2O.
To find the molecular formula, we need to determine the actual number of each type of atom in the molecule. To do this, we need to know the molar mass of acetic acid.
The molecular mass of acetic acid was determined to be 60 amu. The empirical formula mass (CH2O) can be calculated as follows:
C: 1 (atomic mass of C) * 12.01 g/mol = 12.01 g/mol
H: 2 (atomic mass of H) * 1.008 g/mol = 2.016 g/mol
O: 1 (atomic mass of O) * 16.00 g/mol = 16.00 g/mol
Summing these atomic masses:
12.01 g/mol + 2.016 g/mol + 16.00 g/mol ≈ 30.03 g/mol
To find how many empirical formula units are needed to make up the molecular formula, divide the molar mass of the compound by the empirical formula mass:
60 g/mol / 30.03 g/mol ≈ 2
Therefore, the molecular formula of acetic acid is C2H4O2.