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Part 1: Calculate the empirical formula for a substance with the following composition: H 6.71%; C 40.00%; O 53.29%

C = 12 g/molO = 16 g/molH = 1 g/mol
Ratio of C = 40/12 = 3.33Ratio of O = 53.29/16 = 3.33Ratio of H = 6.71/ 1 = 6.71Divide largest value by smallest value = 6.71/3.33 = 2Empirical formula = CH2O

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Part 1: Calculate the empirical formula for a substance with the following composition: H 6.71%; C 40.00%; O 53.29%

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