High School

We appreciate your visit to 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. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

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.

Thanks for taking the time to read 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. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada