Final answer:
The empirical formula for aluminum oxide, formed from 5.00 g of aluminum and 9.45 g of the final aluminum oxide, is deduced by finding the mass of oxygen, calculating moles of both elements, and then finding the simplest whole-number ratio, which results in Al2O3.
Step-by-step explanation:
To find the empirical formula of aluminum oxide formed when aluminum is burned in oxygen, we must first find the amount of oxygen that has combined with the aluminum. Since the initial mass of aluminum is 5.00 g and the final mass of aluminum oxide is 9.45 g, we can determine the mass of oxygen by subtracting the mass of aluminum from the mass of aluminum oxide:
9.45 g (final mass of Al2O3) - 5.00 g (initial mass of Al) = 4.45 g of Oxygen
Next, we calculate the moles of aluminum and oxygen:
- Moles of Al = 5.00 g / 26.98 g/mol = 0.185 moles
- Moles of O = 4.45 g / 16.00 g/mol = 0.278 moles
To determine the simplest whole-number ratio, we divide the moles by the smallest number of moles:
- Ratio of Al to O = 0.185 moles Al / 0.185 moles = 1
- Ratio of O to Al = 0.278 moles O / 0.185 moles = 1.5
Since we want whole numbers, we multiply each ratio by 2:
- Al ratio becomes 1 * 2 = 2
- O ratio becomes 1.5 * 2 = 3
Therefore, the empirical formula for aluminum oxide is Al2O3.