Question

An unknown compound contains 13.04% of hydrogen,34.78% of oxygen, and 52.17% of carbon by mass. Determine the empirical formula of the unknown compound. (Atomic massof O-15.999 g/mol; H-1.008 g/mol; C-12.011 g/mol)


CH2OH


C2H6O


C2H6OH


C6H12O6

Answer 1

To find the empirical formula, we need to determine the simplest whole number ratio of atoms in the compound.

1. Convert the mass percentage of each element to mass:
- Hydrogen: 13.04% of the total mass = 13.04 g (since we can assume a 100g sample)
- Oxygen: 34.78% of the total mass = 34.78 g
- Carbon: 52.17% of the total mass = 52.17 g

2. Convert the mass of each element to moles using their atomic mass (in g/mol):
- Hydrogen: 13.04 g ÷ 1.008 g/mol = 12.94 mol
- Oxygen: 34.78 g ÷ 15.999 g/mol = 2.17 mol
- Carbon: 52.17 g ÷ 12.011 g/mol = 4.34 mol

3. Find the simplest whole number ratio of moles by dividing each value by the smallest mole value:
- Hydrogen: 12.94 mol ÷ 2.17 mol = 5.96 ≈ 6
- Oxygen: 2.17 mol ÷ 2.17 mol = 1
- Carbon: 4.34 mol ÷ 2.17 mol = 2

4. Write the empirical formula using the whole number ratio of atoms:
C2H6O

Therefore, the empirical formula of the unknown compound is C2H6O.