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A 100.0-g sample of a compound is made up of 35.9 g of aluminum and 64.1 g of sulfur. The empirical formula of the compound is:

A. AlS
B. Al₂S₃
C. Al₃S₂
D. Al₄S₆

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.

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Rewritten by : Jeany

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