A 48.3 g sample of an aluminum iodine compound contains 3.2 g of aluminum. What is the empirical formula for the compound?

• A. AlI3
• B. Al2I3
• C. Al3I2
• D. AlI

Answer: (A)

Verified step by step guidance

1. Step 1: Determine the mass of iodine in the compound by subtracting the mass of aluminum from the total mass of the compound. This is calculated as: 48.3 g (total mass) - 3.2 g (mass of aluminum) = 45.1 g (mass of iodine).
2. Step 2: Convert the mass of aluminum to moles using its molar mass. The molar mass of aluminum (Al) is approximately 26.98 g/mol. Use the formula: \( \text{moles of Al} = \frac{3.2 \text{ g}}{26.98 \text{ g/mol}} \).
3. Step 3: Convert the mass of iodine to moles using its molar mass. The molar mass of iodine (I) is approximately 126.90 g/mol. Use the formula: \( \text{moles of I} = \frac{45.1 \text{ g}}{126.90 \text{ g/mol}} \).
4. Step 4: Determine the simplest whole number ratio of moles of aluminum to moles of iodine. Divide the moles of each element by the smallest number of moles calculated in the previous steps.
5. Step 5: Use the ratio from Step 4 to write the empirical formula of the compound. The empirical formula is based on the simplest whole number ratio of atoms in the compound.