Give The Empirical Formula Of The Following Compound If A Sample Contains 57.8 Percent C, 3.6 Percent H, And 38.6 Percent O By Mass?


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1) If we had 100g of this substance, find the number of moles for each element
-dividing 57.8g/12.01g/mol gives 4.81mol C, doing the same gives 1.35mol H (1.008g/mol) and 2.41mol O (16.00g/mol)
2) We can see that 4.81 is twice 2.41 so if we set the ratios of the elements to 2.41, diving through by that number we get C2-H0.56-O1
3) We need to multiply through to get the lowest integer value (can't have 0.56) so the easiest thing would be to multiply by 100 (make 0.56 into 56) giving us C200-H56-O100
4) Reduce as much as possible to get the empirical formula (the actual formula of the compound might be a multiple of these numbers) -- we get, dividing by 4:  C50-H14-O25

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