Combustion Apparatus - Online Tutor, Practice Problems & Exam Prep

Combustion analysis is a method used to determine the empirical formula of a compound by analyzing the products of its combustion. In this process, a sample is vaporized in a combustion apparatus consisting of multiple chambers. Oxygen gas is introduced as a reactant, essential for combustion. As the sample transitions to a gaseous state, any hydrogen present is converted to water, while nonmetals form gases such as:

• CO₂ for carbon
• NO₂ for nitrogen
• SO₂ for sulfur
• X₂ for halogens

Excess oxygen may also be present post-reaction. Understanding the function of each chamber in the apparatus is crucial for interpreting results and solving related analytical problems in chemistry.

Combustion Apparatus

Now remember that combustion analysis is just an analytical process that helps us to determine the empirical formula of a compound. Combustion analysis is accomplished through the use of a combustion apparatus. Now this apparatus is where the sample is vaporized and goes through combustion by traveling between chambers.

So here we have a basic type of combustion apparatus. Here it has chambers A through D. What happens first is that we have to place our sample within chamber A and we allow oxygen gas to enter this. Remember a key component of a combustion reaction is the presence of O₂ as a reactant. Now chamber A, we've placed our sample within it and this is where the sample is vaporized so it's changed into a gaseous state.

It then travels into chamber B and chamber B is where hydrogen is converted into water. So if this sample possesses hydrogen, it gets converted into water and the non-metal portions of this sample get converted into some type of gas. Now if that gas, if that nonmetal happens to be carbon, it's normal for it to become CO₂. We talked about non-hydrocarbons, so if it's nitrogen it would become NO₂. If it's sulfur it becomes SO₂. If it's one of the halogens it becomes a diatomic molecule.

Now traveling out of chamber 2. This is where we have the splitting of water and our gas in chamber C. This is where water is being trapped and again based on the identity of the non-metal we could make different types of gases. CO₂ if the non-metal is carbon, NO₂ if it's nitrogen, SO₂ if it's sulfur, and then one of the diatomic halogens if it happens to be a halogen.

Now what comes out after all of this is any excess oxygen that remains, and this is a possibility. Sometimes a combustion reaction can happen in an environment that's rich in oxygen. The reaction takes place, but there's still oxygen all over the place. So just remember, combustion analysis requires the use of a combustion apparatus. And now that we've seen this apparatus, let's take a look at some different types of combustion analysis questions that kind of go hand in hand with understanding this apparatus.

Combustion Apparatus Example 1

The compound that makes up an experimental jet contains carbon, hydrogen, and oxygen. Suppose that in one experiment, the combustion of 11.5g of the liquid gave the following results. Here we're given the masses of carbon dioxide and water before combustion and after combustion. Remember, when it comes to our combustion apparatus, we have our chambers, which basically collect the amount of carbon dioxide and water that's produced during a combustion reaction. Within these chambers, there had already been some carbon dioxide and some water, and what we're seeing is after combustion, how much more water and carbon dioxide was created.

From this information, we need to determine the empirical formula of the unknown liquid. Alright, so if we take a look at the steps, so step zero, we're going to subtract the grams after combustion by the grams before combustion to find the grams of CO₂ and water. All right, so we're going to say we have 32.815 grams of CO₂ after combustion, -10.815 grams of CO₂ before combustion. That tells us how much CO₂ was created. When we do that, we're going to get 22.0 grams of CO₂ and then we're going to do 23.610 grams of water minus 10.110 grams of water. So that gives me 13.5 grams of water. So this is how much CO₂ and water that was produced.

So at this point it then continues into a regular combustion analysis type of question. We're going to do step one, where we're going to convert the grams of CO₂ into moles or into grams, actually grams of carbon. So we're going to have 22.0 grams of CO₂. What we want to do first is we want to cancel out. It's right over here. We have 22 grams of CO₂. We want to change those grams of CO₂ into moles of CO₂, so one mole of CO₂ on top. At this point we've seen carbon dioxide numerous times, so the one carbon and the two oxygens weigh 44.01 grams. Grams cancel out. Now convert the moles of CO₂ into just moles of carbon. Within this formula, we have only one carbon, so that's one mole of C and then finally for every one mole of C we're told according to the periodic table, its atomic mass is 12.01 grams. When we do that, we're going to get the mass of C, which comes out to 6.0036 grams of C.

Next we're going to convert grams of water into grams of hydrogen, so we have 13.5 grams of water. We're going to say here that one mole of water which possesses 2 hydrogens and one oxygen is 18.016 grams. Then we're going to convert moles of water into just moles of H. And remember, for every one mole of water, we see that there's two hydrogens within the formula. And then convert the one mole of H into grams of H. So when we do that, we're going to get 1.5107 grams of H.

Step three if necessary. Subtract the grams of step one and two from the grams of the sample to determine the third element. What we're told within the question that our third element that makes up this jet fuel is oxygen. Our total amount of the liquid, which contains carbon, hydrogen and oxygen, is 11.5, so we're going to do 11.5 grams, which contains all three elements. Subtract out the grams of carbon plus the grams of hydrogen gives us 3.9857 grams oxygen.

Now that we have the grams of all three elements that comprise my compound, we move on to step 4. In step 4, you convert all the masses into moles. We're going to bring down these masses here, so 6.0036 grams carbon, so one mole of C and on the bottom the mass of C. Then we have 1.5107 grams H. For every one mole H it's 1.008 grams H and then we have 3.9857 grams O, and so for every one mole of O, it's 16 grams. Remember, these masses are coming from the periodic table. When we do that, we're going to get the moles of each of these elements. This comes out to be 0.4999 moles of C. This equals 1.4987 moles of H and this equals 0.2491 moles of O.

At this point, step five, you divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. So our smallest moles that we obtained were the moles of the oxygen. So everyone gets divided by 0.2491. So when we do that, that's going to give us ratios of each of these elements. So that gives me two carbons, 6 hydrogens and one oxygen. Now, since we got whole numbers, we don't have to worry about rounding, but if we did, we'd look at step 6. If you get a value of 0.1 or 0.9, then you can round. But again, at this point we've determined what our empirical formula is. We've determined that it's C₂H₆O in terms of our answer.

The following results are obtained from burning 5 g of an organic molecule (MW = 233.07 g/mol), which contains C, H, and O, in a combustion chamber:

Determine the molecular formula of linolenic acid.

• A. C₉H₁₅O
• B. C₃H₅O₁₂
• C. C₆H₅O
• D. C₁₈H₄₅O₃