How would you calculate the empirical formula to form an aluminum oxide if your are given 18 g of aluminum and 16 g of molecular oxygen (O2)?
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Supposing the reaction is complete: (18 g Al) / (26.98154 g Al/mol) 0.6671 mol Al (16 g O2) / (31.99886 g O2/mol) 0.5000 mol O2 Divide by the smaller number of moles: (0.6671 mol Al) / 0.5000 mol 1.33 (0.5000 mol O2) / 0.5000 mol 1.00 To achieve integer coefficients, multiply by 3, then round to the nearest whole number to find the empirical formula: Al4O3 (I've never heard of this oxide of Al, but that's what the data here say.)
