# Combustion Analysis Calculator (2023)

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.

## What is combustion analysis?

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.

## How to find the empirical formula from combustion analysis?

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:

$\small \text C_\alpha \text H_\beta \text O_\gamma + a \text O_2 \longrightarrow b \text C \text O_2 + c \text H_2 \text O$CαHβOγ+aO2bCO2+cH2O

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 $m_\text{C}$mC and hydrogen $m_\text{H}$mH as:

\begin{align*}\footnotesize m_\text{C} & \footnotesize = m_{\text{CO}_2}\cdot \dfrac{M_\text{C}}{M_{\text{CO}_2}} \\[1em]\footnotesize m_\text{H} & \footnotesize = m_{\text{H}_2\text{O}}\cdot \dfrac{2 M_\text{H}}{M_{\text{H}_2\text{O}}}\end{align*}mCmH=mCO2MCO2MC=mH2OMH2O2MH

(Video) Introduction to Combustion Analysis, Empirical Formula & Molecular Formula Problems

Where:

• $m\_{\text{CO}\_2}$m_CO_2 and $m\_{\text{H}\_2\text{O}}$m_H_2O are the masses of carbon dioxide and water;
• $M_\text{C}$MC and $M_\text{H}$MH are the molar masses of carbon and hydrogen; and
• $M\_{\text{CO}\_2}$M_CO_2 and $M_{\text{H}_2\text{O}}$MH2O are the molecular masses of dioxide carbon and water.

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

$\footnotesize m_\text{O} = m_\text{sample} - m_\text{C} - m_\text{H}$mO=msamplemCmH

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:

$\footnotesize \text{mol}_\text{C}=\dfrac {m_\text{C}}{M_\text{C}} \\[1em]\footnotesize \text{mol}_\text{H}=\dfrac {m_\text{H}}{M_\text{H}} \\[1em]\footnotesize \text{mol}_\text{O}=\dfrac{m_\text{O}}{M_\text{O}}$molC=MCmCmolH=MHmHmolO=MOmO

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!

💡 Did you know that the air-fuel ratio or AFR represents the ratio between the mass of air and fuel needed for the complete combustion of the fuel? You can learn more about this with our AFR calculator.

## How to find the molecular formula?

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 $\text{EFM}$EFM:

\begin{align*}\footnotesize \text{EFM} & \footnotesize = \text{mol}_\text{C} \cdot M_\text{C} + \text{mol}_\text{H} \cdot M_\text{H} \\& \footnotesize + \text{mol}_\text{O}\cdot M_\text{O}\end{align*}EFM=molCMC+molHMH+molOMO

Where:

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

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

(Video) Combustion Analysis Sample

$\footnotesize n = \dfrac {\text{Molar mass}}{\text{Empirical formula mass}}$n=EmpiricalformulamassMolarmass

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

## How to use the combustion analysis calculator

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.

💡 If you don't need the molecular formula, it's not necessary to input the substance's molar mass. The combustion analysis calculator will still give you the empirical formula.

## Empirical and molecular formula of C, H, O compounds - An example

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:

\begin{align*}\footnotesize m_\text{C} & \footnotesize = 18.942\: \text{g} \: \text{CO}_2 \cdot\dfrac{12.011\: \tfrac{\text{g}}{\text{mol}} \ \text C}{44.010 \: \tfrac{\text{g}}{\text{mol}} \: \text C \text O_2} \\ & \footnotesize = 5.1694 \: \text{g} \ \text C \\\footnotesize m_\text{H} & \footnotesize = 7.749 \: \text{g} \: \text H_2 \text O \cdot\dfrac{2 \cdot 1.00797 \: \tfrac{\text{g}}{\text{mol}} \ \text H}{18.0153 \: \tfrac{\text{g}}{\text{mol}} \: \text H_2 \text O} \\ & \footnotesize = 0.8669 \: \text{g} \ \text H \\\footnotesize m_\text{O} & \footnotesize = 12.915 \: \text{g} - 5.1694 \: \text{g}\ \text C - 0.8669 \: \text{g}\ \text H \\ & \footnotesize = 6.879 \: \text{g} \ \text O\end{align*}mCmHmO=18.942gCO244.010molgCO212.011molgC=5.1694gC=7.749gH2O18.0153molgH2O21.00797molgH=0.8669gH=12.915g5.1694gC0.8669gH=6.879gO

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

(Video) Combustion Analysis Calculations: Chemistry Sample Problem

\begin{align*}\footnotesize \text{mol}_\text{C} & \footnotesize =\dfrac {5.1694 \: \text{g} \: \text C }{12.011 \: \tfrac{\text{g}}{\text{mol}} \: \text C}= 0.43039 \: \text{mol} \: \text C \\ \footnotesize \text{mol}_\text{H} & \footnotesize =\dfrac {0.8669 \: \text{g}\: \text H }{1.00797 \: \tfrac{\text{g}}{\text{mol}} \: \text H} = 0.8600 \: \text{mol} \: \text H \\\footnotesize \text{mol}_\text{O} & \footnotesize =\dfrac {6.879 \: \text{g}\: \text O }{15.9994 \: \tfrac{\text{g}}{\text{mol}} \: \text O} = 0.42995 \: \text{mol}\ \text O\end{align*}molCmolHmolO=12.011molgC5.1694gC=0.43039molC=1.00797molgH0.8669gH=0.8600molH=15.9994molgO6.879gO=0.42995molO

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:

\begin{align*}\footnotesize \text{mol}_\text{C} & \footnotesize= \dfrac {0.43039 \: \text{mol} \ \text C}{0.42995} = 1.0010 \approx 1 \ \text{mol}\ \text C \\\footnotesize \text{mol}_\text{H} & \footnotesize= \dfrac {0.8600 \: \text{mol} \ \text H }{0.42995} = 2.0002 \approx 2 \ \text{mol}\ \text H \\\footnotesize \text{mol}_\text{O} & \footnotesize= \dfrac {0.42995 \: \text{mol} \ \text O}{0.42995} = 1\: \text{mol}\ \text O\end{align*}molCmolHmolO=0.429950.43039molC=1.00101molC=0.429950.8600molH=2.00022molH=0.429950.42995molO=1molO

From here, we get the empirical formula for our unknown substance: $\text C \text H_2 \text O$CH2O.

💡 Notice that we approximate the number of moles to the closest integer when calculating the proportion between the elements.

#### 2. Molecular formula

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

\begin{align*}\footnotesize \text{EFM} & \footnotesize = \bigg(\dfrac{1 \: \text{mol}\ \text C}{\text{mol} \: \text{substance}} \cdot \dfrac{12.011 \: \text{g}}{\text{mol} \ \text C}\bigg) \\ & \footnotesize +\bigg(\dfrac{2 \: \text{mol}\ \text{H}}{\text{mol} \: \text{substance}}\cdot \dfrac{1.00797 \: \text{g}}{\text{mol}\ \text H}\bigg) \\ & \footnotesize+\bigg(\dfrac{1 \: \text{mol}\ \text O} {\text{mol} \: \text{substance}}\cdot \dfrac {15.9994 \: \text{g}}{\text{mol}\ \text O}\bigg) \\& \footnotesize = 30.031 \: \dfrac{\text{g}}{\text{mol}}\end{align*}EFM=(molsubstance1molCmolC12.011g)+(molsubstance2molHmolH1.00797g)+(molsubstance1molOmolO15.9994g)=30.031molg

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

$\footnotesize n = \dfrac {90.0779 \ \tfrac{\text{g}}{\text{mol}}}{30.031 \ \tfrac{\text{g}}{\text{mol}}}\approx 3$n=30.031molg90.0779molg3

Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio n: $(\text C \text H_2 \text O)_3$(CH2O)3 or $\text C_3 \text H_6 \text O_3$C3H6O3.

## Empirical and molecular formula of hydrocarbons — an example

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? 🤔

(Video) ALEKS: Combustion analysis

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:

\begin{align*}\footnotesize m_\text{C} &\footnotesize= 33.057 \: \text{g} \: \text C\text O_2 \cdot \dfrac{12.011\: \tfrac{\text{g}}{\text{mol}}\: \text C}{44.010\: \tfrac{\text{g}}{\text{mol}}\: \text C\text O_2} \\& \footnotesize = 9.0218\: \text{g}\: \text C \\\footnotesize m_\text{H} &\footnotesize= 10.816\: \text{g}\: \text H_2 \text O \cdot \dfrac{2 \cdot 1.00797 \: \tfrac{\text{g}}{\text{mol}} \: \text H}{18.0153\: \tfrac{\text{g}}{\text{mol}}\: \text H_2\text O} \\ & \footnotesize = 1.2103\: \text{g}\: \text H\end{align*}mCmH=33.057gCO244.010molgCO212.011molgC=9.0218gC=10.816gH2O18.0153molgH2O21.00797molgH=1.2103gH

💡 Notice this time, we didn't use the sample mass value in our calculation! This amount is used to find the mass of oxygen in the case of a C, H, O substance.

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

\begin{align*}\footnotesize \text{mol}_\text{C} &\footnotesize=\dfrac {9.0218 \ \text{g} \ \text C }{12.011 \ \tfrac{\text{g}}{\text{mol}} \ \text C}\\& \footnotesize = 0.75112 \ \text{mol}\ \text C \\[1em]\footnotesize \text{mol}_\text{H}&\footnotesize=\dfrac {1.2103 \ \text{g} \ \text H }{1.00797 \ \tfrac{\text{g}}{\text{mol}} \ \text H}\\ & \footnotesize= 1.20076 \ \text{mol} \ \text H\end{align*}molCmolH=12.011molgC9.0218gC=0.75112molC=1.00797molgH1.2103gH=1.20076molH

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:

\begin{align*}\footnotesize \text{mol}_\text{C} &\footnotesize= \dfrac {0.75112\: \text{mol}\: \text C} {0.75112} \\&\footnotesize= 1\: \text{mol}\: \text C \\[1em]\footnotesize \text{mol}_\text{H} &\footnotesize= \dfrac {1.20076 \: \text{mol}\: \text H }{0.75112} \\ &\footnotesize= 1.5968\: \text{mol}\: \text H\end{align*}molCmolH=0.751120.75112molC=1molC=0.751121.20076molH=1.5968molH

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}$58 — 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: $\text C_5 \text H_8$C5H8.

#### 2. Molecular formula

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

\begin{align*}\footnotesize \text{EFM} &\footnotesize= \bigg(\dfrac{5 \ \text{mol}\ \text C}{\text{mol substance}}\cdot \dfrac{12.011 \ \text{g}}{ \text{mol}\ \text C}\bigg) \\&\footnotesize + \bigg(\dfrac{8 \ \text{mol}\ \text H}{\text{mol substance}}\cdot \dfrac{1.00797 \ \text{g}}{ \text{mol}\ \text H}\bigg) \\&\footnotesize = 68.119\: \dfrac{\text{g}}{\text{mol}}\end{align*}EFM=(molsubstance5molCmolC12.011g)+(molsubstance8molHmolH1.00797g)=68.119molg

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

$\footnotesize n = \dfrac {204.35\: \tfrac{\text{g}}{\text{mol}}}{68.119 \: \tfrac{\text{g}}{\text{mol}}}\approx 3$n=68.119molg204.35molg3

(Video) Empirical Formula from Combustion Analysis (Example)

Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio $n$n : $(\text C_5 \text H_8)3$(C5H8)3 or $\text C_{15} \text H_{24}$C15H24.

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!

## FAQs

### How do you calculate combustion analysis? ›

Calculate the empirical formula of the compound from the grams of carbon, hydrogen, and oxygen. Calculate the formula mass for the empirical formula and divide the given molecular mass by the empirical formula mass to get n. Multiply each of the subscripts in the empirical formula by n to get the molecular formula.

What is the combustion equation formula? ›

Meaning of Combustion Reaction

The usual equation for a complete combustion reaction is CH4 + O2 → CO2 + H2O.

Which compound can not be analyzed by combustion analysis? ›

So oxygen cannot be determined directly in combustion analysis.

What is produced when a 1.50 g sample of hydrocarbon undergoes complete combustion? ›

1.5 g of hydrocarbon undergoes complete combustion to give 4.4 g of CO2 and 2.7 g of H2O.

How do you calculate combustion air requirements? ›

The International Fuel Gas Code requires the following combustion air openings for a boiler room:
1. Vertical opening. One-inch free area for each 4,000 Btu/hr. ...
2. Horizontal duct opening. One-inch free area for each 2,000 Btu/hr. ...
3. Mechanical fan. One CFM of air for each 2,400 Btu/hr. ...
4. Indoor air. 50 cu.
Sep 28, 2016

What are the 3 types of combustion? ›

They are:
• Rapid Combustion,
• Spontaneous Combustion, and.
• Explosive Combustion.
Mar 10, 2022

What are the 5 types of combustion? ›

• Suggested Videos. Combustion and flames. ...
• Browse more Topics under Combustion And Flame. Flame and Structure of a Flame. ...
• 1] Complete Combustion. One of the types of combustion is Complete Combustion. ...
• 2] Incomplete Combustion. ...
• 3] Rapid Combustion. ...
• 4] Spontaneous Combustion. ...
• 5] Explosive Combustion.

What are the 4 stages of combustion? ›

The cycle includes four distinct processes: intake, compression, combustion and power stroke, and exhaust.

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