Combustion Reaction Calculator

Omni's combustion reaction calculator will instantly balance any combustion reaction involving hydrocarbon or C, H, O organic substances. Wondering what's the combustion reaction for methane (CH₄)? Or maybe propane gas (C₃H₈)? Just input these, and the calculator will have your answer! 🔥

Keep reading to learn about:

• Hydrocarbon combustion;
• The combustion reaction equation;
• A simple step-by-step guide on how to balance a combustion reaction equation with an example of hydrocarbon combustion; and
• Examples of combustion reactions.

According to Lavoisier's law of conservation of mass, the total mass of a chemical reaction system doesn't change. That means that the total mass of reactants and products is the same before and after the reaction. With this in mind, when balancing chemical reactions, the goal is to equilibrate the masses of reactants and products of the equation:

$\small \text{Mass of Reactants} = \text{Mass of Products}$

In the case of the combustion of C, H, O organic compounds — substances that contain only carbon (C), hydrogen (H), and oxygen (O) — we'll see that the procedure to balance the reaction is quite simple.

The general formula for the complete combustion of organic fuels is given by:

$\text C_\alpha \text H_\beta \text O_\gamma + a(\text O_2 + 3.76 \text N_2) \longrightarrow b\text C\text O_2 + c\text H_2\text O + d\text N_2$

On the left side, we have:

• $\text C_\alpha \text H_\beta \text O_\gamma$ denotes the generic chemical formula of the fuel.
• The subscripts $\alpha$, $\beta$ and $\gamma$ indicate the respective number of atoms of carbon, hydrogen, and oxygen.
• The term $(\text O_2 + 3.76\text N_2)$ represents the combustion air — assuming that air is composed of 21% oxygen and 79% nitrogen.

On the right side we see:

• The products of complete combustion: carbon dioxide $\text C\text O_2$, water vapor $\text H_2\text O$ and molecular nitrogen $\text N_2$.

Since we're assuming complete theoretical combustion, the nitrogen in the air doesn't react with the fuel, and the reaction is often simplified as:

$\text C_\alpha \text H_\beta \text O_\gamma + a\text O_2 \longrightarrow b\text C\text O_2 + c\text H_2\text O$

This simplification leaves us with an equation that contains only carbon (C), hydrogen (H), and oxygen (O)— and by determining the coefficients $a$, $b$ and $c$, we'll have balanced the reaction.

Let's see how to balance the combustion reaction with a simple procedure that always follows the same order:

1. From the molecular formula of the fuel, identify the number of atoms of each element, $\alpha$, $\beta$ and $\gamma$.

2. Always start by balancing the carbon atoms (C), assigning to the coefficient b the value: $b = \alpha$.

3. Next, balance the hydrogen (H) by assigning water's coefficient $c = \beta/2$.

4. The last element to balance is the oxygen (O): $a = \alpha + \beta/4 - \gamma/2$.

5. Finally, if you got a fraction or decimal value for any of the coefficients, multiply all the coefficients by the quantity that is required to eliminate the fractions.

6. The combustion reaction equation is balanced! 🔥

When the empirical and molecular formulas are not available, these can be obtained via combustion analysis calculation. Once the molecular formula is known, you can proceed to the balancing of the combustion reaction.

Let's implement these steps into an example: What's the balanced combustion reaction of the hydrocarbon hexane C₆H₁₄?

1. First, by inspecting the molecular formula of the fuel C₆H₁₄, identify the values of the subscripts: $\alpha = 6$, $\beta = 14$ and $\gamma = 0$.

2. Once these are determined, continue to assign the values of the coefficients. For balancing carbon the coefficient is $b = 6$.

3. For hydrogen's coefficient, the value corresponds to$c = 14/2 = 7$.

4. The oxygen is balanced with $a = 6 + 14/4 - 0 = 9.5$.

   By substituting theses values into the reaction, the balanced combustion equation is:

   $\text C_6 \text H_{14}+ 9.5\text O_2 \longrightarrow 6\text C\text O_2 + 7\text H_2\text O$

5. Finally, since the oxygen coefficient ($a = 9.5$) is a decimal value, multiply all the coefficients by 2 to express the balance with whole numbers:

   $2\text C_6 \text H_{14}+ 19\text O_2 \longrightarrow 12\text C\text O_2 + 14\text H_2\text O$

That's it! the reaction is now properly balanced 😀.

The combustion reaction calculator will give you the balanced reaction for the combustion of hydrocarbons or C, H, O substances. To use the calculator, enter the molecular formula of your substance:

1. On the first row, Total atoms of carbon C (α), enter the number of carbon atoms of your substance.
2. Then, on the Total atoms of hydrogen H (β) field, input the number of hydrogen atoms.
3. Finally, on the Total atoms of oxygen O (γ) field, enter the total number of oxygen atoms. In the case of hydrocarbons, compounds with only carbon (C) and hydrogen (H) atoms, your input on this field should be zero 0.
4. With these inputs, the calculator will display the balanced combustion reaction equation.