Combustion Analysis
Ten Examples


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Go to a discussion of empirical and molecular formulas.

Problems 1 - 10


This technique requires that you burn a sample of the unknown substance in a large excess of oxygen gas. The combustion products will be trapped separately from each other and the weight of each combustion product will be determined. From this, you will be able to calculate the empirical formula of the substance. This technique has been most often applied to organic compounds.

A brief discussion (yet quite informative) of the history of this technique can be found here. This technique is also called "elemental analysis"

Some points to make about combustion analysis:

1) The elements making up the unknown substance almost always include carbon and hydrogen. Oxygen is often involved and nitrogen is involved sometimes. Other elements can be involved, but problems with C and H tend to predominate followed by C, H and O and then by C, H, O and N.

2) We must know the mass of the unknown substance before burning it.

3) All the carbon in the sample winds up as CO2 and all the hydrogen in the sample winds up as H2O.

4) If oxygen is part of the unknown compound, then its oxygen winds up incorporated into the oxides. The mass of oxygen in the sample will almost always be determined by subtraction.

5) Often the N is determined via a second experiment and this introduces a bit of complexity to the problem. Nitrogen dioxide is the usual product when nitrogen is involved. Sometimes the nitrogen product is N2, sometimes NH3.

6) Sometimes the problem asks you for the empirical formula and sometimes for the molecular formula (or both). Two points: (a) you have to know the molar mass to get to the molecular formula and (b) you have to calculate the empirical formla first, even if the question doesn't ask for it. A few lines below is a link that goes to a file that discusses how to go from empirical to molecular formulas.


Here is a brief overview of the solution steps before doing some examples:

1) Determine the grams of each element present in the original compound. Carbon is always in CO2 in the ratio (12.011 g / 44.0098 g), hydrogen is always in H2O in the ratio (2.0158 g / 18.0152 g), etc.
2) Convert grams of each element to the number of moles. You do this by dividing the grams by the atomic weight of the element. Many times students will want to use 2.016 for hydrogen, thinking that it is H2. This is wrong, use 1.008 for H.
3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers.

Steps 2 and 3 are the technique for determining the empirical formula. Step one is required because you have all your carbon, for example, in the form of CO2 instead of a simpler problem where it tells you how much carbon is present.

Finally, a common component of this type of problem is to provide the molecular weight of the substance and ask for the molecular formula. For example, the empirical formula of benzene is CH while the molecular formula is C6H6. Several of the problems below include this question and you can go here for a discussion about calculating the molecular formula once you know the empirical formula.


Example #1: A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of H2O. What is the empirical formula of this compound?

Solution:

1) Determine the grams of carbon in 4.40 g CO2 and the grams of hydrogen in 2.70 g H2O.

carbon: 4.40 g x (12.011 g / 44.0098 g) = 1.20083 g

hydrogen: 2.70 g x (2.0158 g / 18.0152 g) = 0.3021482 g

2) Convert grams of C and H to their respective amount of moles.

carbon: 1.20083 g / 12.011 g/mol = 0.09998 mol

hydrogen: 0.3021482 g / 1.0079 g/mol = 0.2998 mol

3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole numbers.

carbon: 0.09998 mol / 0.09998 mol = 1

hydrogen: 0.2998 mol / 0.09998 mol = 2.9986 = 3

4) We have now arrived at the answer:

the empirical formula of the substance is CH3

Note: I did not check for the presence of oxygen. The problem said hydrocarbon, which are compounds with only C and H. Sometimes the problem will be silent about what type of compound it is, give only C and H data, but oxygen will also be in the compound. See comment in Example #3.


Example #2: A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g of H2O. What is the empirical formula of this compound?

Solution:

1) Determine the grams of C in 0.845 g CO2 and the grams of H in 0.173 g H2O.

carbon: 0.845 g x (12.011 g / 44.0098 g) = 0.2306 g

hydrogen: 0.173 g x (2.0158 g / 18.0152 g) = 0.01935 g

2) Convert grams of C and H to their respective amount of moles.

carbon: 0.2306 g / 12.011 g / mol = 0.01920 mol

hydrogen: 0.01935 g / 1.0079 g/mol = 0.01921 mol

3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole numbers.

carbon: 0.01920 mol / 0.01920 mol = 1

hydrogen: 0.01921 mol / 0.01920 mol = 1

4) We have now arrived at the answer:

the empirical formula of the substance is CH.

Example #3: A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound?

Solution:

1a) Determine the grams of carbon in 0.3664 g CO2 and the grams of hydrogen in 0.1500 g H2O.

carbon: 0.3664 g x (12.011 g / 44.0098 g) = 0.1000 g

hydrogen: 0.1500 g x (2.0158 g / 18.0152 g) = 0.01678 g

1b) Determine the grams of oxygen in the sample by subtraction.

0.2500 − (0.1000 g + 0.01678) = 0.1332 g

Notice that the subtraction is the mass of the sample minus the sum of the carbon and hydrogen in the sample. Also, it is quite typical of these problems to specify that only C, H and O are involved.

A warning: sometimes the problem will give the CO2 and H2O values, but FAIL to say that C, H, and O are involved. Make sure you add the C and H values (or sometimes the C, H, and N values) and check against the mass of the sample. Any difference would be an amount of oxygen present (or it might be a mistake!! Keep double checking your work as you do each calculation.)

2) Convert grams of C, H and O to their respective amount of moles.

carbon: 0.1000 g / 12.011 g / mol = 0.008325 mol

hydrogen: 0.01678 g / 1.0079 g/mol = 0.01665 mol

oxygen: 0.1332 g / 15.9994 g/mol = 0.008327 mol

3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers.

carbon: 0.008325 mol / 0.008325 mol = 1

hydrogen: 0.01665 mol / 0.008325 mol = 2

oxygen: 0.008327 mol / 0.008325 mol = 1

4) We have now arrived at the answer:

the empirical formula of the substance is CH2O

Example #4: Quinone, which is used in the dye industry and in photography, is an organic compound containing only C, H, and O. What is the empirical formula of the compound if you find that 0.105 g of the compound gives 0.257 g of CO2 and 0.0350 g of H2O when burned completely? Given a molecular weight of approximately 108 g/mol, what is its molecular formula?

Comment: as a reminder, the following link goes to a discussion of how to calculate the molecular formula once you get the empirical formula.

Solution:

1) mass of each element:

carbon ⇒ 0.257 g x (12.011 / 44.0098) = 0.07014 g
hydrogen ⇒ 0.0350 g x (2.016 / 18.015) = 0.00391674 g
oxygen ⇒ 0.105 g − (0.07014 + 0.00391674) = 0.03094326 g

2) moles of each element:

carbon ⇒ 0.07014 g / 12.011 g/mol = 0.005840 mol
hydrogen ⇒ 0.00391674 g / 1.008 g/mol = 0.0038856 mol
oxygen ⇒ 0.03094326 g / 16.00 g/mol = 0.0019340 mol

3) Look for smallest whole-number ratio:

carbon ⇒ 0.005840 / 0.0019340 = 3
hydrogen ⇒ 0.0038856 / 0.0019340 = 2
oxygen ⇒ 0.0019340 / 0.0019340 = 1

4) Empirical formula:

C3H2O

5) Molecular formula:

the weight of C3H2O is 54

108 / 54 = 2

C6H4O2


Example #5: A 1.000 g sample of a compound is combusted in excess oxygen and the products are 2.492 g of CO2 and 0.6495 g of H2O.

a) Determine the empirical formula of the compound.
b) Given that its molar mass is 388.46 g/mol, determine the compound's molecular formula.

Solution:

1) mass of each element:

carbon ⇒ 2.492 g x (12.011 / 44.0098) = 0.68011 g
hydrogen ⇒ 0.6495 g x (2.016 / 18.015) = 0.07268343 g
oxygen ⇒ 1.000 − (0.68011 + 0.07268343) = 0.24720657 g

Notice that there was oxygen in the compound and that the problem did not tell you that.

2) moles of each element:

carbon ⇒ 0.68011 g / 12.011 g/mol = 0.0566240 mol
hydrogen ⇒ 0.07268343 g / 1.008 g/mol = 0.0721066 mol
oxygen ⇒ 0.24720657 g / 16.00 g/mol = 0.01545 mol

3) Look for smallest whole-number ratio:

carbon ⇒ 0.0566240 / 0.01545 = 3.665
hydrogen ⇒ 0.0721066 / 0.01545 = 4.667
oxygen ⇒ 0.01545 / 0.01545 = 1

Do NOT round these off. You should only round off with numbers very close to a whole number. How close? Something like 2.995 goes to 3.

4) I want to change the numbers to improper fractions

carbon ⇒ 3.665 = 11/3
hydrogen ⇒ 4.667 = 14/3
oxygen ⇒ 1 = 3/3

Sometimes, a textbook will "magically" tell you what factor to multiply by and you will wonder why that particular factor was selected. Notice that 3.665 is three-and-two-thirds, 4.667 is four-and-two-thirds. I changed everything to fractions to try and highlight why three is used.

5) Multiply by three to get the whole-number ratio:

11 : 14 : 3

empirical formula = C11H14O3

6) The weight of the empirical formula is 194:

388 / 194 = 2

the molecular formula is C22H28O6


Example #6: A carbohydrate is a compound composed solely of carbon, hydrogen and oxygen. When 10.7695 g of an unknown carbohydrate (MW = 128.2080 g/mol) was subjected to combustion analysis with excess oxygen, it produced 29.5747 g CO2 and 12.1068 g H2O. What is its molecular formula?

Solution:

1) Carbon:

mass of CO2 = 29.5747 g ⇒ 0.6722 moles of CO2 ⇒ 0.6722 moles of C ⇒ 8.066 g of C

2) Hydrogen:

mass of H2O = 12.107 g ⇒ 0.67260 moles of H2O ⇒ 1.3452 moles of H ⇒ 1.3558 g of H

3) Oxygen:

mass of compound burnt = 10.770 g

mass of C + H = 9.422 g

10.770 g − 9.422 g = 1.348 g of O ⇒ 0.08424 mol of O

4) Determine empirical and molecular formula:

molar ratio of C : H : O ⇒ 0.6722 : 1.3452 : 0.08424

after dividing by the smallest

molar ratio of C : H : O ⇒ 7.98 : 15.97 : 1.00 <--- the first two are close enough so as to be rounded off

empirical formula is C8H16O

"empirical formula weight" = 96 + 16 + 16 = 128 which is the molecular weight so the molecular formula is also

C8H16O

Note: I use "EFW" for the term "empirical formula weight." Neither is in standard usage in the world of chemistry.


Example #7: Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine produces 1.813 mg CO2, 0.4639 mg H2O, and 0.2885 mg N2. Estimate the molar mass of caffeine, which lies between 150 and 200 g/mol.

Solution to Example #7

Example #8: A 1.50 gram sample of cadaverine gives 3.23 g of CO2, 1.58 g of N2O5, and 1.865 g of H2O. Its molar mass is 102.2 g/mol. Determine the empirical and molecular formulas.

Solution to Example #8

Example #9: Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 2.175 g of lysine was combusted to produce 3.94 g of CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine was burned to produce 0.436 g of NH3. The molar mass of lysine is approximately 150 g/mol. Determine the empirical and molecular formula of lysine.

Solution to Example #9


Example #10: A 3.09 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This produced 3.34 g CO2 and 1.82 g H2O. A second sample of this compound with a mass of 4.50 g produced 2.95 g SO3. The third sample of this compound with a mass of 8.33 g produced 4.30 g of HNO3. Determine the empirical formula of the compound.

Solution:

Key point: there were three different masses of the compound. As we get the masses of each element, we will need to determine their percentages in the compound. We cannot uses the masses we obtain just below in the calculations to get moles of each element

1) Determine the mass of each element in the original compound (units that cancel are not written:

mass of carbon ---> (3.34 g) (12.011 / 44.009) = 0.91156 g
mass of hydrogen ---> (1.82 g) (2.016 / 18.015) = 0.20367 g
mass of oxygen ---> cannot do this one yet
mass of sulfur ---> (2.95 g) (32.065 / 80.062) = 1.18148 g
mass of nitrogen ---> (4.30 g) (14.007 / 63.0119) = 0.955853 g

2) Determine the percent composition:

C ---> 0.91156 / 3.09 = 29.500%
H ---> 0.20367 / 3.09 = 6.5913%
O ---> cannot do this one yet
S ---> 1.18148 / 4.50 = 26.255%
N ---> 0.955853 / 8.33 = 11.4748%

3) Since we know the percent composition adds up to 100%, we can now do oxygen:

100 − (29.500 + 6.5913 + 26.255 + 11.4748) = 26.1789%

4) Now we can assume 100 g of the compound is present. This means the percentages become masses. We then calculate moles of each (unit left off):

C ---> 29.500 / 12.011 = 2.456
H ---> 6.5913 / 1.008 = 6.539
O ---> 26.1789 / 16.00 = 1.63618125
S ---> 26.255 / 32.065 = 0.8188
N ---> 11.4748 / 14.007 = 0.8192

5) Divide through by smallest:

C ---> 2.456 / 0.8188 = 3
H ---> 6.539 / 0.8188 = 8
O ---> 1.6362 / 0.8188 = 2
S ---> 0.8188 / 0.8188 = 1
N ---> 0.8192 / 0.8188 = 1

6) Empirical formula:

C3H8O2SN

Video Example: Compound A contains 5.2% by mass of nitrogen as well as C, H and O. Combustion of 0.0850 g of compound A gave 0.224 g of CO2 and 0.0372 g of H2O. Calculate the empirical formula of A.

Determine the Empirical Formula Using Combustion Analysis


Bonus Example #1: A 2.00 g sample of a compound was found to produce 3.99 g of CO2 and 1.63 g of H2O upon combustion. In addition to C and H, the compound also contains sulfur. The molar mass of the compound was found to be 88.17 g/mol. What is the molecular formula?

Solution:

1) Mass of each element present in sample:

carbon ---> (3.99 g) (12.011 g / 44.009 g) = 1.089 g
hydrogen ---> (1.63 g) (2.016 g / 18.015 g) = 0.1824 g
sulfur ---> 2.00 − (1.089 + 0.1824) = 0.7286 g

2) Moles of each element present in sample:

carbon ---> 1.089 g / 12.011 g/mol = 0.09067 mol
hydrogen ---> 0.1824 g / 1.008 g/mol = 0.18095 mol
sulfur ---> 0.7286 g / 32.065 g/mol = 0.0227226 mol

3) Divide through by smallest:

carbon ---> 0.09067 mol / 0.0227226 mol = 3.99
hydrogen ---> 0.18095 mol / 0.0227226 mol = 7.96
sulfur ---> 0.0227226 mol / 0.0227226 = 1

4) The empirical formula is:

C4H8S

The molecular formula is:

The empirical formula weighs 88.17. The molecular formula is known to weigh 88.17

The molecular formula is C4H8S, same as the empirical formula.


Bonus Example #2: An unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 10.00 mg of this compound is burned, 21.38 mg of CO2 and 2.50 mg of H2O are produced. The freezing point of camphor is lowered by 18.4 °C when 1.326 g of the compound is dissolved in 10.00 g of camphor (Kf = 40.0 °C kg/mol).

Solution:

1) Determine milligrams of C, H, and O

carbon ---> (21.38 mg) (12.011 g/mol / 44.009 g/mol) = 5.835 mg
hydrogen ---> (2.50 mg) (2.016 g/mol / 18.015 g/mol) = 0.280 mg
oxygen ---> 10.00 mg − (5.835 + 0.280) = 3.885 mg

2) Determine millimoles:

C ---> 5.835 mg / 12.011 mg/mmol = 0.4858 mmol
H ---> 0.280 mg / 1.008 mg/mmol = 0.2778 mmol
O ---> 3.885 mg / 15.999 mg/mmol = 0.2428 mmol

3) Divide through by smallest to get whole-number ratio:

C ---> 0.4858 mmol / 0.2428 mmol = 2
H ---> 0.2778 mmol / 0.2428 mmol = 1.14 <--- close enough to round
O ---> 0.2428 mmol / 0.2428 mmol = 1

4) State the empirical formula:

C2HO

5) Freezing point lowering:

ΔT = i Kf m

18.4 = (1) (40.0 °C kg/mol) (x / 0.01000 kg)

x = 0.00460 mol

6) Determine molecular weight:

1.326 g / 0.00460 mol = 288 g/mol

7) Determine molecular formula:

C2HO weighs 41

288 / 41 = 7

C2HO times 7 = C14H7O7

This is an unusual formula. I found some mentions of it from the 1840's and 1850's plus a 1921 mention. It's in the upper right area of the page.

I also found mention of Alizrian Red S, which isn't the exct same formula as in the problem, but I name counded cool. The formula is C14H7O7SNa


Problems 1 - 10

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