Corondum, the 9th hardestmineral, contains 52.94%Al and 47.06% O. Basedon this composition, whatis the empirical formula ofthe compound?​

Here's how to determine the empirical formula of corundum: 1. Assume a 100g Sample - This makes the calculations easier. We'll have 52.94 g of Al and 47.06 g of O. 2. Convert Grams to Moles - Divide the mass of each element by its molar mass:- Al: 52.94 g / 26.98 g/mol = 1.96 mol- O: 47.06 g / 16.00 g/mol = 2.94 mol 3. Find the Simplest Mole Ratio - Divide each mole value by the smallest mole value (1.96 mol in this case):- Al: 1.96 mol / 1.96 mol = 1- O: 2.94 mol / 1.96 mol ≈ 1.5 4. Adjust to Whole Numbers - Since we can't have half an atom, multiply both subscripts by 2:- Al: 1 x 2 = 2- O: 1.5 x 2 = 3 Empirical Formula The empirical formula of corundum is Al₂O₃.

Answered by Anonymous | 2024-09-16