What does the Combustion Analysis Calculator do?

The combustion analysis calculator determines the empirical and molecular formulas of C, H, O, and hydrocarbon compounds. Use the calculator above for instant results in your browser.

Is the Combustion Analysis Calculator free to use?

Yes. All Try To Calculator tools are free and do not require an account.

Are my inputs stored or sent to a server?

No. Calculations run locally in your browser. We do not collect the numbers you enter or the results shown.

Can I use the Combustion Analysis Calculator for professional decisions?

This tool is for education and quick estimates. For medical, legal, tax, or financial decisions, verify results with a qualified professional.

Where can I find related calculators?

Browse more Chemistry tools on Try To Calculator at /chemistry, or use the related calculators section on this page.

Combustion Analysis Calculator

Last updated: February 11, 202643 people find this calculator helpful

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Welcome to 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:

Calculate the masses of each element;
Determine each's moles; and
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:

C_{α} H_{β} O_{γ} + a O_{2} ⟶ b C O_{2} + c H_{2} O

CαHβOγ+aO2⟶bCO2+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 $m_{C}$ mC and hydrogen $m_{H}$ mH as:

m_{C} m_{H} = m_{CO_{2}} \cdot \frac{M _{C}}{M _{CO_{2}}} = m_{H_{2} O} \cdot \frac{2 M _{H}}{M _{H_{2} O}}

mCmH=mCO2⋅MCO2MC=mH2O⋅MH2O2MH

Where:

$m_CO_2$ m_CO_2 and $m_H_2 O$ m_H_2O are the masses of carbon dioxide and water;
$M_{C}$ MC and $M_{H}$ MH are the molar masses of carbon and hydrogen; and
$M_CO_2$ M_CO_2 and $M_{H_{2} O}$ MH2O are the molecular masses of dioxide carbon and water.

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

m_{O} = m_{sample} - m_{C} - m_{H}

mO=msample−mC−mH

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:

mol_{C} = \frac{m _{C}}{M _{C}} mol_{H} = \frac{m _{H}}{M _{H}} mol_{O} = \frac{m _{O}}{M _{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 $EFM$ EFM:

EFM = mol_{C} \cdot M_{C} + mol_{H} \cdot M_{H} + mol_{O} \cdot M_{O}

EFM=molC⋅MC+molH⋅MH+molO⋅MO

Where:

$mol_C$ mol_C, $mol_H$ mol_H and $mol_O$ mol_O are the moles of carbon, hydrogen and oxygen from the empirical formula; and
$M_{C}$ MC, $M_{H}$ MH and $M_{O}$ 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:

n = \frac{Molar mass}{Empirical formula mass}

n=Empirical formula massMolar mass

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:

Choose the type of substance that you'd like to study.
Input the molar mass, sample mass, CO₂ mass, and H₂O mass from the combustion analysis. For hydrocarbons, the sample mass is not required.
The calculator will display your substance's empirical formula, empirical mass, and molecular formula.
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:

m_{C} m_{H} m_{O} = 18.942 g CO_{2} \cdot \frac{12.011 \frac{g}{mol} C}{44.010 \frac{g}{mol} C O _{2}} = 5.1694 g C = 7.749 g H_{2} O \cdot \frac{2 \cdot 1.00797 \frac{g}{mol} H}{18.0153 \frac{g}{mol} H _{2} O} = 0.8669 g H = 12.915 g - 5.1694 g C - 0.8669 g H = 6.879 g O

mCmHmO=18.942gCO2⋅44.010molgCO212.011molg C=5.1694g C=7.749gH2O⋅18.0153molgH2O2⋅1.00797molg H=0.8669g H=12.915g−5.1694g C−0.8669g H=6.879g O

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

mol_{C} mol_{H} mol_{O} = \frac{5.1694 g C}{12.011 \frac{g}{mol} C} = 0.43039 mol C = \frac{0.8669 g H}{1.00797 \frac{g}{mol} H} = 0.8600 mol H = \frac{6.879 g O}{15.9994 \frac{g}{mol} O} = 0.42995 mol O

molCmolHmolO=12.011molgC5.1694gC=0.43039molC=1.00797molgH0.8669gH=0.8600molH=15.9994molgO6.879gO=0.42995mol O

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:

mol_{C} mol_{H} mol_{O} = \frac{0.43039 mol C}{0.42995} = 1.0010 \approx 1 mol C = \frac{0.8600 mol H}{0.42995} = 2.0002 \approx 2 mol H = \frac{0.42995 mol O}{0.42995} = 1 mol O

molCmolHmolO=0.429950.43039mol C=1.0010≈1 mol C=0.429950.8600mol H=2.0002≈2 mol H=0.429950.42995mol O=1mol O

From here, we get the empirical formula for our unknown substance: $C H_{2} O$ 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 $EFM$ EFM:

EFM = (\frac{1 mol C}{mol substance} \cdot \frac{12.011 g}{mol C}) + (\frac{2 mol H}{mol substance} \cdot \frac{1.00797 g}{mol H}) + (\frac{1 mol O}{mol substance} \cdot \frac{15.9994 g}{mol O}) = 30.031 \frac{g}{mol}

EFM=(molsubstance1mol C⋅mol C12.011g)+(molsubstance2mol H⋅mol H1.00797g)+(molsubstance1mol O⋅mol O15.9994g)=30.031molg

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

n = \frac{90.0779 \frac{g}{mol}}{30.031 \frac{g}{mol}} \approx 3

n=30.031 molg90.0779 molg≈3

Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio n: $(C H_{2} O)_{3}$ (CH2O)3 or $C_{3} H_{6} O_{3}$ 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? 🤔

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

Empirical formula obtention; and
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:

m_{C} m_{H} = 33.057 g C O_{2} \cdot \frac{12.011 \frac{g}{mol} C}{44.010 \frac{g}{mol} C O _{2}} = 9.0218 g C = 10.816 g H_{2} O \cdot \frac{2 \cdot 1.00797 \frac{g}{mol} H}{18.0153 \frac{g}{mol} H _{2} O} = 1.2103 g H

mCmH=33.057gCO2⋅44.010molgCO212.011molgC=9.0218gC=10.816gH2O⋅18.0153molgH2O2⋅1.00797molgH=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:

mol_{C} mol_{H} = \frac{9.0218 g C}{12.011 \frac{g}{mol} C} = 0.75112 mol C = \frac{1.2103 g H}{1.00797 \frac{g}{mol} H} = 1.20076 mol H

molCmolH=12.011 molg C9.0218 g C=0.75112 mol C=1.00797 molg H1.2103 g H=1.20076 mol H

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:

mol_{C} mol_{H} = \frac{0.75112 mol C}{0.75112} = 1 mol C = \frac{1.20076 mol H}{0.75112} = 1.5968 mol H

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 $\frac{8}{5}$ 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: $C_{5} H_{8}$ C5H8.

2. Molecular formula

To get the molecular formula, first we calculate the empirical molar mass $EFM$ EFM:

EFM = (\frac{5 mol C}{mol substance} \cdot \frac{12.011 g}{mol C}) + (\frac{8 mol H}{mol substance} \cdot \frac{1.00797 g}{mol H}) = 68.119 \frac{g}{mol}

EFM=(mol substance5 mol C⋅mol C12.011 g)+(mol substance8 mol H⋅mol H1.00797 g)=68.119molg

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

n = \frac{204.35 \frac{g}{mol}}{68.119 \frac{g}{mol}} \approx 3

n=68.119molg204.35molg≈3

Finally, to go from the empirical formula to the molecular formula, multiply the former by the ratio $n$ n : $(C_{5} H_{8}) 3$ (C5H8)3 or $C_{15} H_{24}$ 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

What does the Combustion Analysis Calculator do?: The combustion analysis calculator determines the empirical and molecular formulas of C, H, O, and hydrocarbon compounds. Use the calculator above for instant results in your browser.
Is the Combustion Analysis Calculator free to use?: Yes. All Try To Calculator tools are free and do not require an account.
Are my inputs stored or sent to a server?: No. Calculations run locally in your browser. We do not collect the numbers you enter or the results shown.
Can I use the Combustion Analysis Calculator for professional decisions?: This tool is for education and quick estimates. For medical, legal, tax, or financial decisions, verify results with a qualified professional.
Where can I find related calculators?: Browse more Chemistry tools on Try To Calculator at /chemistry, or use the related calculators section on this page.

Combustion Analysis Calculator

What is combustion analysis?

How to find the empirical formula from combustion analysis?

How to find the molecular formula?

How to use the combustion analysis calculator

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

1. Empirical formula

2. Molecular formula

Empirical and molecular formula of hydrocarbons — an example

1. Empirical formula

2. Molecular formula

FAQs

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