In this example problem, we will find the empirical formula of an organic compound containing carbon, hydrogen, and oxygen using combustion analysis. Our problem reads combustion analysis of a twenty-eight-point-four gram sample of an organic compound containing carbon, hydrogen, and oxygen. It produced fifty-four-point-three grams CO2 and fifteen-point-six grams H2O. Find the empirical formula for the compound. Our first step will be to calculate the moles of carbon, hydrogen, and oxygen in this problem. We assume that all of the carbon in our organic compound is incorporated into the fifty-four-point-three grams of carbon dioxide. So we begin our dimensional analysis with fifty-four-point-three grams of carbon dioxide and convert it into moles of carbon dioxide using the molar mass. Then, we convert it into moles of carbon using the ratio from the chemical formula, and we'll get 1.23 moles of carbon. We also make the assumption that all of the hydrogen in the organic compound gets incorporated into the H2O. So beginning with fifteen-point-six grams of H2O, we can convert it into moles of H2O using its molar mass, and then two moles of hydrogen using the ratio from the chemical formula. We'll get 1.73 moles of hydrogen in our sample. To calculate the moles of oxygen in our sample, we first have to determine the mass of oxygen. We know that the total mass of our sample is composed of the mass of carbon, the mass of hydrogen, and the mass of oxygen. So we can solve for the mass of oxygen by taking the mass of the sample and subtracting the mass of carbon and the mass of hydrogen. We're given the mass of the sample in the problem, which is 28.4 grams. We calculated that we have 1.23 moles of carbon in our sample, and we can convert this moles of carbon in our sample into grams of carbon using its molar mass. We also calculated that we have 1.73 moles of hydrogen in our sample, and we convert it into grams of hydrogen using its molar mass. Performing this calculation, we should get 11.9 grams of oxygen. Then we just need to convert our 11.9 grams of oxygen into moles of oxygen using oxygen's molar mass, and we'll get 0.743 moles of oxygen. Now we have the molar ratio of carbon, hydrogen, and oxygen in our sample, and we just need to convert this molar ratio into the smallest whole number ratio of elements in our compound, which is the empirical formula. In the strategy we will use, we'll divide first all of our numbers by the smallest number so that all of our numbers are 1 or greater. And then, if there are any fractions in our molar ratio, we'll multiply by a factor to eliminate those fractions so we have the smallest whole number ratio. And anything we do to one number, we will do to all of our numbers to maintain our current ratio. So first, we'll divide all of our numbers by 0.743, and we'll get numbers 1 or greater. When we look at our molar ratios, there are fractions of 2/3 (0.66) and 1/3 (0.33). So to get rid of these fractions, we'll multiply all of our numbers by 3, and we'll end up with 5, 7, and 3. Therefore, our empirical formula is C5H7O3.