What is the empirical formula of a compound that consists of 40.00% carbon, 6.71% hydrogen l, and the rest is oxygen
1. Convert the percentages to grams:
- 40.00% carbon = 40.00g C
- 6.71% hydrogen = 6.71g H
2. Find the mass of oxygen by subtracting the mass of carbon and hydrogen from the total mass:
- Total mass = 100g
- Mass of oxygen = 100g - 40.00g - 6.71g = 53.29g O
3. Convert the mass of each element to moles by dividing by their respective atomic masses:
- Moles of C = 40.00g / 12.01g/mol = 3.33 mol
- Moles of H = 6.71g / 1.01g/mol = 6.65 mol
- Moles of O = 53.29g / 16.00g/mol = 3.33 mol
4. Divide each mole value by the smallest mole value to get the simplest whole number ratio:
- C:H:O = 3.33 mol : 6.65 mol : 3.33 mol
- Simplified ratio = 1 : 2 : 1
Therefore, the empirical formula of the compound is CH2O.