Final answer:
The empirical formula of a compound consisting of 53.70% iron and 46.30% sulfur is Fe2S3. This is determined by converting the mass percentages to moles and then finding the simplest whole-number mole ratio.
Step-by-step explanation:
To find the empirical formula of a compound containing 53.70% iron and 46.30% sulfur, you first convert these percentages to grams, assuming a 100g sample. Next, you convert the masses of iron (Fe) and sulfur (S) to moles by using their molar masses (Fe = 55.85 g/mol, S = 32.07 g/mol). After calculating the moles of each element, you find the simplest mole ratio of Fe to S to determine the empirical formula.
For 53.70g of Fe, the moles of Fe are 53.70 g / 55.85 g/mol = 0.961 mol of Fe. For 46.30g of S, the moles of S are 46.30 g / 32.07 g/mol = 1.443 mol of S. To find the ratio, you divide both by the smallest number of moles. Thus, 0.961 mol of Fe and 1.443 mol of S become 1 mol Fe and 1.5 mol S, which simplifies to the whole number ratio of 2:3. Hence, the empirical formula of the compound is Fe2S3.