Combustion Analysis - 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 combustion reactions. Typically, compounds like hydrocarbons (containing only carbon and hydrogen) or those with carbon, hydrogen, and oxygen react with oxygen gas to produce carbon dioxide (CO₂) and water (H₂O). For example, octane (C₈H₁₈), a hydrocarbon, and glucose (C₆H₁₂O₆), which contains oxygen in addition to carbon and hydrogen, both yield CO₂ and H₂O upon combustion. However, non-hydrocarbons, which may include elements like sulfur, nitrogen, or halogens (fluorine, chlorine, bromine, iodine), can produce additional gaseous products such as sulfur dioxide (SO₂), nitrogen dioxide (NO₂), and diatomic molecules (F₂, Cl₂, Br₂, I₂) during combustion. Understanding the typical products of combustion reactions is crucial for performing accurate combustion analysis and determining the empirical formulas of unknown compounds.

Combustion Analysis

Combustion analysis represents yet another way to determine the empirical formula of a compound. It represents an analytical process that determines the empirical formula of a compound. It does so by utilizing what we call combustion reactions. Now, normally combustion reactions involve a compound composed of just carbon and hydrogen. These we call hydrocarbons because it's hydrogen and carbon together or a compound composed of carbon, hydrogen and oxygen together reacting with O₂ gas.

Now here when we have a hydrocarbon or compound composed of carbon, hydrogen and oxygen, when it reacts by O₂ by combustion reaction, the products form will be CO₂ and water. Now if we take a look here at some common examples of combustion reactions, remember we talked about octane. When we discussed the molecular formula of octane, it was C₈H₁₈. So this is a hydrocarbon. It has only carbons and hydrogens with it. When it reacts with O₂, it creates CO₂ and water.

Another compound that we talked about previously is glucose. Remember, the molecular formula of glucose is C₆H₁₂O₆. This compound is comprised of carbon, hydrogen and oxygen. So when it reacts with O₂, it produces also CO₂ and water. Here we're just concerned with the fact that we have a hydrocarbon and then a compound of carbon, hydrogen and oxygen. When they react with O₂, we produce as a result CO₂ and water.

So that's the basic idea behind a typical combustion reaction. We won't worry about balancing these chemical reactions, we're just concerned with the products that are being formed. So just remember we're going to talk about combustion analysis a little bit more, but it just represents you another way to find empirical formula and it does so to the use of combustion reaction.

Combustion Analysis

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.

An unknown compound is composed of carbon, hydrogen and oxygen. A 4.30 g sample is ignited and creates 8.59 g CO₂ and 3.52 g H₂O. If the molar mass is 176.22 g/mol, what is the molecular formula?

• A. C₃H₅O₃
• B. C₂H₄O₄
• C. C₈H₁₆O₄
• D. C₁₂H₁₆O₄

In the presence of a small amount of oxygen a combustion reaction will not only produce carbon dioxide, but also carbon monoxide. The incomplete combustion of naphthalene, a hydrocarbon used in many dyes, produced 2.80 g CO, 4.40 g CO₂ and 1.44 g H₂O. Determine its empirical formula.

• A. C₃H₃O₃
• B. C₅H₄
• C. C₅H₄O
• D. C₁₀H₈

Combustion Analysis

We've talked about the combustion of hydrocarbons, compounds that possess only hydrogens and carbon, but now we're going to take a look at non hydrocarbons. We're going to say here, a non hydrocarbon represents a compound containing not only carbon and hydrogen, but also sulfur, nitrogen or a halogen.

Now remember, your halogens are the elements from group 7A. So we're talking about fluorine, chlorine, bromine or iodine. We're going to say do combustion analysis. They can create gaseous products such as SO2 which is called sulfur dioxide, NO2, which is called nitrogen dioxide, as well as diatomic molecules.

Remember the halogens in their natural forms exist as F2, Cl2, Br2 and I2. So this is a little bit different from what we're accustomed to seeing with combustion reactions where we typically just see Cl2 and water being formed. But realize Cl2 and water being formed is strictly for hydrocarbons and compounds with carbon hydrogens and oxygens.

Now when we introduce these new elements expect to make new types of final products. So just keep that in mind when we're dealing with non hydro, well non hydrocarbons within our question.

Combustion Analysis

Hey, everyone. So in this example question, it says, a solvating agent which has a molar mass of 147 grams per mole that contains carbon, hydrogen, and chlorine is used for spectroscopic processes. Now, determine its molecular formula when a 0.250-gram sample creates 0.451 grams of carbon dioxide and 0.0617 grams of water upon combustion. To solve this question, we're going to employ the following steps. And step 1 is, if present, convert the grams of carbon dioxide to grams of just carbon. Alright. So we're given 0.451 grams of carbon dioxide. The mass of carbon dioxide, one carbon and two oxygens, is 44.01 grams. Again, remember, that comes from adding up one carbon plus two oxygens. We find that from the periodic table for their atomic masses. And that's equal to one mole of carbon dioxide. We're gonna say here, one mole of carbon dioxide has within it, one mole of just carbon. And according to the periodic table, one mole of carbon weighs 12.01 grams. When we work this out, we're gonna get 0.1231 grams of carbon. To avoid rounding errors, make sure you at least use 4 decimal places when doing these types of calculations. Here, I'm giving this as our value for now. Next, convert the grams of water into grams of hydrogen here. So we're gonna have here 0.0617 grams of water. Water is composed of two hydrogens and one oxygen. The combined mass from the periodic table is 18.016 grams, and that's for every one mole of water. Now from the formula, we see that water has in it two hydrogens. So we're gonna have two moles of hydrogen for every one mole of water, and one mole of hydrogen according to the periodic table weighs 1.008 grams. This gives me 0.0069 grams of hydrogen. Now, step 3, if necessary, subtract the grams of steps 1 and 2 from the grams of the sample to determine the third element. So, here we have 0.250 grams of our sample which is made up of carbon, hydrogen, and chlorine. We're gonna subtract out the masses that we just found for carbon and hydrogen. Which would mean that the difference would be the grams of chlorine, which comes out to be 0.1200 grams of chlorine. So we have the grams of all three of our elements. So step 4 says, convert all those masses into moles. Right. So here, we're gonna basically just change each of these into moles by dividing them by their atomic masses. So carbon, one mole of carbon is 12.01 grams. One mole of hydrogen is 1.008 grams. One mole of chlorine, according to the periodic table, is 35.45 grams. All the grams cancel out and we'll have moles of each, which I'm going to write down here. So here for carbon, we'd have 0.01025 moles of carbon. We'd have 0.00685 moles of hydrogen, and we have 0.00339 moles of chlorine. Step 5, it's a long process. Step 5, divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. So the smallest mole answer that we got was 0.00339. So all of them get divided by that number. Now step 6, if you get after you divide them all by the smallest moles, if you get a value of some number point 1 or some number, point 9, then you can round to the nearest whole number. If you can't round, we multiply by a factor to create whole numbers. Luckily, when we divide all these by their lowest mole number, we get whole numbers. We get three carbons here, we get two hydrogens here, and we get one chlorine here. This will give me the empirical formula which would be C3H2Cl. But remember, they're not asking us for the empirical formula, they're asking for the molecular formula. To figure out the molecular formula, we need to figure out what our n value will be. Your n value will equal your molecular mass or molar mass, which was given to us in the very beginning, divided by the empirical mass. So we're told in the very beginning that it weighs 147 grams per mole. We have to find the empirical mass. The empirical mass comes from the empirical formula. We have three carbons, two hydrogens, one chlorine. Multiply them all by their atomic masses from the periodic table. Right? So that'd be 36.03 grams, 2.016 grams, and 35.45 grams. When you add those all up together, we get 73.496 grams per mole. Again, this 73 comes from adding up these numbers in your calculator. So when we do that, we get n equals 2. So now we're going to say that our molecular formula, we're gonna take this number 2 here, and multiply it by the empirical formula. When we do that, we'll get our molecular formula, which comes out to be C 6 H 4 Cl2. So this would be our final answer. That would represent our molecular formula.

A compound composed of carbon, hydrogen and nitrogen undergoes a combustion reaction to produce 264.21 g CO₂, 63.06 g H₂O and 46.4 g NO₂. Determine its empirical formula.

• A. C₂H₁₁N₃
• B. C₆H₇N
• C. C₆H₁₂N₄
• D. CHN

The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO₂ and 2.52 g H₂O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?

The combustion of 4.16 grams of a compound, which contains only carbon, hydrogen, oxygen, and fluorine, yields 7.7 grams of carbon dioxide and 2.52 grams of water. Another sample of the compound with a mass of 3.63 grams is found to contain 0.58 grams of fluorine. What is the empirical formula of the compound? Alright. So our compound itself contains carbon, hydrogen, oxygen, and fluorine. From the amount of carbon dioxide and water, we know that we can determine the grams of hydrogen and the grams of carbon. So we're going to have 7.7 grams of CO2, and we have here 2.52 grams of water. We're going to change those into grams of C and grams of H, respectively. So we've seen this before with other combustion analysis types of questions. We're going to say 1 mole of CO2 weighs 44.01 grams. So that's the weight of carbon and the 2 oxygens together. Then we're going to convert moles of CO2 into just moles of C. And if we know the moles of C, then we can determine the grams of C. One mole of C weighs 12.01 grams. So when we do that, we're going to get 2.1012 grams of C. Now let's do it for water. One mole of water weighs 18.016 grams. When we take into account, it's made up of 2 hydrogens and 1 oxygen. One mole of H2O has within it 2 moles of H, because we can see that there's 2 hydrogens from the subscript here. Then we're going to say that 1 mole of H weighs 1.008 grams according to our periodic table. So we've determined the grams of water as 0.2820 grams of hydrogen. At this point, what we normally do is we say that the total mass of our sample, which in this case is made up of carbon, hydrogen, oxygen, and fluorine, we'd subtract out the amount of carbon and the amount of hydrogen, and that would help us determine the grams of oxygen remaining. In this case, we can't do it because we also have fluorine involved. So before we can determine the amount of oxygen remaining, we have to also determine the grams of fluorine present. So how do we do that? Well, in the second sentence, it tells us another sample of the same compound when it has a mass of 3.63 grams. It contains 0.58 grams of fluorine. What's going to help us get to the grams of fluorine is something called the law of definite proportions. Now the law of definite proportions basically says that if you have the same sample or compound, it doesn't matter if you have it in 1 sample or in multiple samples. If they're the same compound, they'll have the same mass ratio. So what I'm saying here is, we know that when we have 3.63 grams of our compound, there are 0.58 grams of fluorine involved. And if we're dealing with the law of definite proportions, when we have this 4.16 gram sample of the same compound, we can determine the grams of fluorine by setting up a proportion and making them equal to one another. Okay? So the law of definite proportions is the key to helping us determine the amount of fluorine. Those two proportions have to equal each other, the 2 ratios, because they both represent the same compound. So the ratio of fluorine to the compound's weight will remain constant. Alright. So we're going to cross multiply, so 3.63 times x and then 0.58 times 4.16. When we do that, we're going to have 3.63x equals 0.58 times 4.16. Divide both sides here by 3.63, and when we do that we'll find x. X here will represent the mass of fluorine. So when we do that, we're going to get 0.6647 grams of fluorine. So with that number, we can add it here. So we can subtract out from our total mass of 4.16 the grams of carbon, hydrogen, and fluorine, and what's left would have to represent the grams of oxygen. So here, oxygen's mass will come out to 1.1121 grams of oxygen. So at this point, we have the grams of each of the elements. So then it becomes a simple, few steps in order to find the empirical formula of the compound. We're going to change each one of these grams into moles. So let's write them all down here and then we're going to have oxygen, and I'm writing in the order that they're stated within this question. C H O F, so that's the order I'm writing them down here. So we're going to convert each of these grams into moles and we do that by dividing them all by their atomic masses. So one mole of C here, 1 mole of H here, 1 mole of oxygen, and 1 mole of fluorine. Plug in their atomic masses from the periodic table and that'll give us the moles of each. Oxygen weighs 16 and then fluorine on the periodic table is 19 grams. So here, grams cancel out and I'll have moles of each one of these elements at the end. So I'll have here 0.1750 moles of C, I'll have here 0.2798 moles of H, I have here, let's see, 0.0695 moles of O, and then finally I'm going to have 0.0350 moles of F. Remember the steps for determining the empirical formula. Now that I have the moles of each, I divide them all by the smallest mole answer, and that's going to give me their ratios. So all of them get divided by 0.0350. So when we have 0.0350, that's going to give me 1 fluorine, 2 oxygens, it's going to give me 8 hydrogens, and 5 carbons. All of them are whole numbers, so using this information, I can find my empirical formula, which comes out as C5H8O2F. So this one was a little bit different. We had to employ the law of definite proportions to just help us find fluorine. But other than that, it's pretty similar to other types of combustion analysis questions. You find the amounts of the elements in grams, you deduct it from the sample, which usually helps you find a missing element. Once you have the grams of all of them, convert them into moles, divide all the moles by the smallest moles, and the whole number ratios that you get will be part of the empirical formula.

Here’s what students ask on this topic:

What is combustion analysis?

Combustion analysis is a laboratory procedure used to determine the elemental composition of a compound, particularly organic compounds containing elements like carbon and hydrogen. In this process, a sample of the substance is burned in the presence of excess oxygen to ensure complete combustion. The combustion products typically include carbon dioxide and water, and sometimes nitrogen oxides if nitrogen is present in the original compound.

By measuring the masses of these combustion products, you can calculate the ratios of the elements within the original compound. For example, the mass of carbon dioxide produced can be used to determine the amount of carbon in the sample, while the mass of water can be used to find the amount of hydrogen. This technique is particularly useful for determining the empirical formulas of unknown organic compounds. It's a fundamental method in chemical analysis and has applications in fields like environmental science, pharmaceuticals, and materials science.

How to do combustion analysis?

Combustion analysis is a method used to determine the elemental composition of a compound, particularly hydrocarbons. To perform a combustion analysis, follow these steps:

1. Weigh the Sample: Accurately weigh a known mass of the compound being analyzed.
2. Burn the Sample: The compound is combusted in a stream of oxygen, which converts the carbon in the compound to carbon dioxide (CO2) and the hydrogen to water (H2O).
3. Absorb the Products: The combustion gases are passed through substances that absorb the CO2 and H2O. Anhydrous calcium chloride can be used to absorb water, and sodium hydroxide can be used to absorb carbon dioxide.
4. Weigh the Absorbents: After the combustion process, the increase in mass of the absorbents is measured. This increase corresponds to the amount of water and carbon dioxide produced.
5. Calculate the Composition: Use the masses of CO2 and H2O to calculate the number of moles of carbon and hydrogen in the original sample. From the moles, you can determine the percentage composition of carbon and hydrogen.
6. Determine Other Elements: If there are elements other than carbon and hydrogen, such as sulfur or nitrogen, additional steps are required to measure the products of their combustion (e.g. sulfur dioxide).

How do you calculate the empirical formula from combustion analysis?

To calculate the empirical formula from combustion analysis, follow these steps:

1. Combustion Data: Obtain the mass of carbon dioxide (CO₂) and water (H₂O) produced during the combustion.
2. Moles of Carbon and Hydrogen: Convert these masses to moles by dividing by their respective molar masses (44.01 g/mol for CO₂ and 18.02 g/mol for H₂O).
3. Elemental Moles: Calculate the moles of carbon and hydrogen. Each mole of CO₂ contains one mole of carbon, and each mole of H₂O contains two moles of hydrogen.
4. Oxygen Moles: If the mass of the original compound is known, subtract the mass of carbon and hydrogen from the total mass to find the mass of oxygen. Then convert this mass to moles by dividing by the molar mass of oxygen (16.00 g/mol).
5. Simplest Ratio: Determine the simplest whole number ratio of moles of each element by dividing all the moles by the smallest number of moles calculated.
6. Empirical Formula: Use this ratio to write the empirical formula, with each element symbol followed by its corresponding subscript indicating the number of atoms in the simplest ratio.

Remember, the empirical formula represents the simplest whole number ratio of elements in a compound.