A compound is found to contain 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound?

To find the empirical formula of a compound, follow these steps:

1. Assume that you have 100g of the compound. This means that the percentages can be read as grams. So, you have 23.3g of Magnesium (Mg), 30.7g of Sulfur (S), and 46.0g of Oxygen (O).

2. Convert the mass of each element to moles. The atomic mass of Mg is approximately 24.31g/mol, S is approximately 32.07g/mol, and O is approximately 16.00g/mol.

   Moles of Mg = 23.3g / 24.31g/mol = 0.958 moles Moles of S = 30.7g / 32.07g/mol = 0.957 moles Moles of O = 46.0g / 16.00g/mol = 2.875 moles

3. Divide each mole quantity by the smallest number of moles calculated. In this case, the smallest number is 0.957.

   Ratio of Mg = 0.958 / 0.957 = 1.001 ≈ 1 Ratio of S = 0.957 / 0.957 = 1 Ratio of O = 2.875 / 0.957 = 3.002 ≈ 3

4. The empirical formula is the elements with their respective ratios. So, the empirical formula of the compound is MgSO3.