42.70g of aluminium is combined with 10.65g of chlorineWhat is the empirical formula of the new substance?
0.300396 mol Cl Divide by the smaller number of moles: (0.10007 mol Al) / 0.10007 mol 1.00 (0.300396 mol Cl) / 0.10007 mol 3.00 Round to the nearest whole numbers to find the empirical formula: AlCl3
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