A compound has 40 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen. I don't know the steps to turn these percentages into an empirical formula.
Treat the percentages as grams in a 100 g sample. So you have 40 g C, 6.7 g H and 53.3 g O. Next, divide each mass by its atomic mass to get moles: carbon 40 / 12 = 3.33, hydrogen 6.7 / 1 = 6.7, oxygen 53.3 / 16 = 3.33. Now divide every value by the smallest of these (3.33): carbon 3.33 / 3.33 = 1, hydrogen 6.7 / 3.33 = 2, oxygen 3.33 / 3.33 = 1. These give the simplest whole-number ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH2O. If the ratios had come out as decimals like 1.5, you would multiply all of them by a small number to clear the fraction. This particular formula matches formaldehyde or, scaled up, glucose.
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