Combustion Analysis Calculator

Welcome to Omni's combustion analysis calculator that will determine the empirical and molecular formulas of C, H, O organic compounds from combustion data 🔥. We invite you to read on and learn about:

• Combustion analysis;
• How to find the empirical formula from combustion analysis; and
• How to find the molecular formula yourself.

In chemistry, combustion analysis is a quantitative analysis used to determine the empirical formula of an unknown organic compound containing carbon (C), hydrogen (H), and oxygen (O).

The unknown substance, initially weighted, undergoes a combustion process on a combustion apparatus that collects the combustion products carbon dioxide (CO₂) and water (H₂O), which are weighed afterwards. Then, the empirical formula and the molar masses of C, H, and O are obtained with this information.

Let's take a look at how to find the empirical formula of a C, H, O organic compound. The process can be divided into three steps:

1. Calculate the masses of each element;
2. Determine each's moles; and
3. Obtain the empirical formula.

Let's see each of these steps in detail 🔎

When calculating the masses, we assume that the organic substance is undergoing complete combustion — that is, the only products of the reaction are carbon dioxide (CO₂) and water vapor (H₂O), as you can see in the combustion reaction equation:

From here, we can tell that all the carbon (C) initially present in the C, H, O compound is now in the dioxide carbon (CO₂) and all the hydrogen (H) is contained in the water vapor (H₂O) molecule. With these assumptions, we can calculate the masses of carbon $m_\text{C}$ and hydrogen $m_\text{H}$ as:

Where:

• $m\_{\text{CO}\_2}$ and $m\_{\text{H}\_2\text{O}}$ are the masses of carbon dioxide and water;
• $M_\text{C}$ and $M_\text{H}$ are the molar masses of carbon and hydrogen; and
• $M\_{\text{CO}\_2}$ and $M_{\text{H}_2\text{O}}$ are the molecular masses of dioxide carbon and water.

The mass of oxygen $m_\text{O}$ is obtained as the difference of carbon and hydrogen masses from the sample mass $m_\text{sample}$:

Once the values of the masses are known, we can calculate the moles of each element. For this, we divide each element's mass by its molar mass:

Finally, to obtain the empirical formula, divide each molar mass by the smallest molar value to get the proportion between the atoms of each element.

Not sure about the difference between molecular weight and molar mass? Check out our molecular weight calculator!

Now that you know how to find the empirical formula of an organic substance, maybe you'd like to know as well how to find its molecular formula. You'll see this is even simpler, all we need is:

• The empirical formula of a given substance; and
• Its molecular mass.

With these known, we can divide the general procedure to get the molecular formula into three steps:

Step 1. From the empirical formula, calculate the empirical molar mass $\text{EFM}$:

Where:

• $\text{mol}\_\text{C}$, $\text{mol}\_\text{H}$ and $\text{mol}\_\text{O}$ are the moles of carbon, hydrogen and oxygen from the empirical formula; and
• $M_\text{C}$, $M_\text{H}$ and $M_\text{O}$ are their molar masses.

Step 2. Determine $n$ as the ratio between the molar mass and the empirical molar mass of the substance:

Step 3. Finally, multiply the moles of each element in the empirical formula by $n$ to get the molecular formula. And that's it! 😀

The combustion analysis calculator will help you find the empirical and molecular formula of C, H, O compound or for a hydrocarbon:

1. Choose the type of substance that you'd like to study.
2. Input the molar mass, sample mass, CO₂ mass, and H₂O mass from the combustion analysis. For hydrocarbons, the sample mass is not required.
3. The calculator will display your substance's empirical formula, empirical mass, and molecular formula.
4. If you'd like to know the masses of C, H, and O of the sample, select Yes from the drop-down menu on the last row.

Let's see how to get the empirical and molecular formulas of a C, H, O compound with a numerical example!

Consider that from a combustion analysis report, we get that after burning a sample of 12.915 g of a C, H, O compound, 18.942 g CO₂ and 7.749 g of H₂O are formed. The molar mass is 90.0779 g/mol. What are the empirical and molecular formulas of the substance? 🤔

To solve the problem, we divide the solution process into two phases. We begin by obtaining the empirical formula, then we obtain the molecular formula.

1. Empirical formula

First, determine the masses of C, H, and O that are present in the sample:

Once we know the values of the masses, next we calculate the number of moles of each element:

Finally, to obtain the empirical formula, we divide the molar masses by the smallest value of them. This way, we can obtain the proportion between the three elements.

In our example, the smallest value of moles corresponds to oxygen:

From here, we get the empirical formula for our unknown substance: $\text C \text H_2 \text O$.

2. Molecular formula

To find the molecular formula, we start by calculating the empirical molar mass $\text{EFM}$:

Next, we calculate the ratio $n$ between the molar masses of the molar and empirical formulas:

Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio n: $(\text C \text H_2 \text O)_3$ or $\text C_3 \text H_6 \text O_3$.

The method to determine the empirical formula of a hydrocarbon by combustion analysis is similar to the one we studied for C, H, O compounds. Again, to make this procedure clear and illustrate the differences between the first one, we’ll check a numerical example.

Suppose that from a combustion analysis, we get the following information: after burning a sample of 12.501 g of a hydrocarbon, we see that 33.057 g CO₂ and 10.816 g of H₂O have formed. The molar mass is 204.35 g/mol. What are the empirical and molecular formulas of the hydrocarbon? 🤔

Again, we'll separate the solution onto two stages:

1. Empirical formula obtention; and
2. Molecular formula calculation.

1. Empirical formula

Following the steps explained before, first we calculate the masses of C and H that are present in the sample compound:

With these values known, next we calculate the number of moles of each element:

Finally, to obtain the empirical formula, we divide each of the amounts of moles by the smallest of them. In this example, the smallest value of moles corresponds to hydrogen:

Note this time we aren't rounding the moles of hydrogen to 2. Doing that will yield an incorrect proportion between the elements. Instead, we express the decimal value 1.5968 into the fraction $\tfrac{8}{5}$ — then the ratio between the moles of carbon and hydrogen is 5 mol C : 8 mol H.

From here, we get the empirical formula for our unknown substance: $\text C_5 \text H_8$.

2. Molecular formula

To get the molecular formula, first we calculate the empirical molar mass $\text{EFM}$:

Next, we calculate the ratio $n$ between the molar masses of the molar and empirical formulas:

Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio $n$ : $(\text C_5 \text H_8)3$ or $\text C_{15} \text H_{24}$.

Combustions are exothermic reactions; this is, heat is released. The amount of heat produced per unit mass of fuel is known as the heat of combustion. Look at the heat of combustion calculator to find out more about this topic!