To determine the empirical formula of the compound, we will first convert the mass percentages to moles.1. Assume we have a 100 g sample of the compound. This means we have: - 44.44 g of magnesium Mg - 22.22 g of sulfur S - 33.33 g of oxygen O 2. Convert the masses to moles using the molar masses of the elements: - Molar mass of Mg = 24.31 g/mol - Molar mass of S = 32.07 g/mol - Molar mass of O = 16.00 g/mol Moles of Mg = 44.44 g / 24.31 g/mol = 1.827 moles Moles of S = 22.22 g / 32.07 g/mol = 0.692 moles Moles of O = 33.33 g / 16.00 g/mol = 2.083 moles3. Divide the moles of each element by the smallest number of moles to get the mole ratio: Mole ratio of Mg = 1.827 / 0.692 = 2.64 3 Mole ratio of S = 0.692 / 0.692 = 1 Mole ratio of O = 2.083 / 0.692 = 3.01 34. The empirical formula is Mg3S1O3, or simply Mg3SO3.Now, let's determine the molecular formula using the molar mass of the compound:1. Calculate the molar mass of the empirical formula: Molar mass of Mg3SO3 = 3 24.31 + 32.07 + 3 16.00 = 72.93 + 32.07 + 48.00 = 153.00 g/mol2. Divide the molar mass of the compound by the molar mass of the empirical formula to find the ratio: Ratio = 120 g/mol / 153.00 g/mol = 0.784Since the ratio is not close to a whole number, it is likely that there was an error in the given molar mass of the compound. The empirical formula Mg3SO3 is the best answer we can provide based on the given information.