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Answer :
Final answer:
The empirical formula of a compound with 35.9g of aluminum and 64.1g of sulfur is b. Al2S3, as it has a simple mole ratio of 2:3 for aluminum to sulfur.
Explanation:
The student's question pertains to determining the empirical formula of a compound composed of aluminum (Al) and sulfur (S). To find the empirical formula, we calculate moles of each element by dividing their given mass by the respective atomic mass. For aluminum which is 35.9g, the atomic mass is 26.98g/mol, resulting in approximately 1.33 moles of Al. For sulfur, 64.1g divided by its atomic mass of 32.06g/mol gives us approximately 2.00 moles of S. The ratio of Al to S is therefore about 1.33 to 2.00, which simplifies to 2:3 when divided by the smallest number of moles, which is about 1.33. Therefore, the simplest whole number ratio of Al to S is 2:3, indicating that the empirical formula of the compound is Al2S3.
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