what is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen
To determine the empirical formula, we need to find the ratio of zinc to oxygen atoms in the compound.
1. Suppose we have 100 grams of the compound.
- 67.1% of 100 grams is 67.1 grams of zinc.
- The remaining mass (100 grams - 67.1 grams) is 32.9 grams of oxygen.
2. Convert the masses to moles.
- The molar mass of zinc (Zn) is 65.38 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
- Moles of zinc: 67.1 g / 65.38 g/mol ≈ 1.025 mol
- Moles of oxygen: 32.9 g / 16.00 g/mol ≈ 2.056 mol
3. Divide the number of moles by the smallest value to find the simplest whole-number ratio.
- Zinc: 1.025 mol / 1.025 mol ≈ 1 mol
- Oxygen: 2.056 mol / 1.025 mol ≈ 2 mol
Thus, the empirical formula for the compound is ZnO.