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To find empirical formulas from mass data such as are presented in this question, first convert the masses given to gram atoms by dividing the stated mass for each element by the gram Atomic Mass of the element. For aluminum, 3.704/26.9815 is about 0.137279; for oxygen, 3.295/15.9994 is about 0.205945. Then divide the larger of these numbers of gram atoms by the smaller to produce a quotient of 1.500. The smallest integers that approximate this ratio are 3 and 2. Therefore, the empirical formula is Al2O3.

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Q: What is the empirical formula of a substance that consists of 3.704g aluminum and 3.295g oxygen?
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We need to know the elements contained in this molecule and the percentages.


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No. The empirical formula of a substance is the formula in which each atomic symbol has the lowest possible subscript that gives the correct ratio between atoms for the compound as a whole. For C6H12, the empirical formula is CH2, but for C6H14, the empirical formula is C3H7.


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Because unlike the empirical formula, the molecular formula does not have to be the simplest ratio.If by chance you are given the percent composition of the elements in a substance, you could calculate the empirical formula and then the empirical formula's mass. However, the molecular formula equation is molecular formula= (empirical formula)n, where n is the mass of the molecular formula divided by the mass of the empirical formula. You would, therefore, need to know the mass belonging to the molecular formula, which you are not given.


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