Everyone, in this video, we're going to take a look at the combustion of carbon compounds. Here in this example, it says 23 Dihydroxytuxinic acid is responsible for the acidulous flavor of some candies and is composed of carbon, hydrogen, and oxygen. If the combustion of 12.01 grams of this acid creates 14.08 grams of CO₂ and 4.32 grams of water, what is its empirical formula? Alright.

So, for a question like this, the first thing we're going to do is convert the grams of Carbon Dioxide into grams of just carbon. We have 14.08 grams of CO₂, and we're going to convert grams of CO₂ into moles of CO₂. One mole of CO₂, the combined weight of the one Carbon and the two Oxygens, is 44.01 grams. Grams cancel out. Then we're going to say that for every one mole of this compound of CO₂, there's one carbon within the formula, so there's one mole of carbon. Moles of CO₂ cancel out. Now that we have moles of carbon, we can convert those to grams of carbon and say for every one mole of carbon, its atomic mass from the periodic table is 12.01 grams. When we work all this out, this gives me 3.8423 grams of carbon.

Next, step 2, we're going to convert the grams of water into just grams of hydrogen. We're focusing on finding carbon and hydrogen from these two compounds because that's the only place they're located. Carbon dioxide is the only place for carbon. Water is the only place for hydrogen. We have 4.32 grams of water. For every one mole of water, which has 2 hydrogens and one oxygen, their combined mass together is 18.016 grams. Grams cancel out. Here for every one mole of H₂O, we can see from the formula that there are two hydrogens. So there are two moles of H. For every one mole of H, its atomic mass from the periodic table is 1.008 grams. So this is 0.4834 grams of H.

Now that we've found the grams of carbon and the grams of H, we go to step 3. If necessary, subtract the grams we found from the grams of the sample to determine the third element, oxygen, in this case. Our substance weighs 12.01 grams, and that's the carbon, hydrogen, and oxygen altogether. If we subtract what we just found for carbon and for hydrogen, the difference of what's left would be the grams of oxygen. So, 7.6843 grams of oxygen.

Now we move to step 4: convert all the masses into moles. Since we already have all their grams here, let's change them to moles. One mole of carbon is 12.01 grams, one mole of hydrogen is 1.008 grams, and one mole of oxygen is 16 grams. This gives us 0.3199 moles of carbon, 0.4796 moles of hydrogen, and 0.4803 moles of oxygen.

The next step is to divide each mole answer by the smallest mole value to obtain whole numbers for each element. So, we have 0.3199 moles of C, 0.4796 moles of H, and 0.4803 moles of O. The smallest mole answer from all of these was 0.3199. This gives us 1 carbon, approximately 1.5 hydrogen, and 1.5 oxygen. As we can only divide to the nearest whole number if we get a 0.1 or a 0.9, here we have 0.5s. We need to multiply in order to create whole numbers. Multiplying by 2 gives us 2 carbons, 3 hydrogens, and 3 oxygens. Thus, the empirical formula, based on this given question, becomes C₂H₃O₃.