A sample of an ionic compound contains 2.82 g Na, 4.35 g CI, and 7.83 g O. Theempirical formula of this compound is

Explanation

To determine the empirical formula of the ionic compound, we need to calculate the number of moles of each element in the compound.Let's assume that we have 100 grams of the compound. Then, the mass of sodium (Na) is 2.82 grams, the mass of chlorine (Cl) is 4.35 grams, and the mass of oxygen (O) is 7.83 grams.The number of moles of each element can be calculated as follows:Moles of Na = 2.82 g / 23.0 g/mol = 0.1226 molMoles of Cl = 4.35 g / 35.45 g/mol = 0.1227 molMoles of O = 7.83 g / 16.0 g/mol = 0.4894 molThe mole ratios of the elements can be determined by dividing each number of moles by the smallest number of moles. In this case, the smallest number of moles is 0.1226, so we divide each number of moles by 0.1226:Moles of Na = 1Moles of Cl = 1.001Moles of O = 3.992Rounding these values to the nearest integer, we get the empirical formula of the compound: NaClO4.Therefore, the answer to your question is d. NaCIO4.