Answer :
After calculating the moles for aluminum and sulfur from the given masses and atomic weights, the resulting molar ratio simplifies to 1:1.5. By doubling these numbers to get whole integers, the empirical formula is determined to be Al2S3.
The question pertains to determining the empirical formula of a compound made up of aluminum and sulfur. Given the masses of aluminum and sulfur in the compound, one must first calculate the number of moles of each element. With 35.9 g of aluminum and 64.1 g of sulfur, and using the atomic weights Al = 26.98 g/mol and S = 32.07 g/mol, we find:
- Number of moles of Al = 35.9 g / 26.98 g/mol ≈ 1.33 mol
- Number of moles of S = 64.1 g / 32.07 g/mol ≈ 2.00 mol
To find the empirical formula, we divide the number of moles of each element by the smallest number of moles calculated. This gives:
- Mole ratio of Al to S ≈ 1.33 / 1.33 : 2.00 / 1.33
- Mole ratio of Al to S = 1:1.5
Since we cannot have half an atom in an empirical formula, we multiply by 2 to get whole numbers, resulting in a mole ratio of 2:3. Therefore, the empirical formula for this compound is Al2S3.