Combustion Data
Probably best illustrated by an example:
0.30g of an unknown organic compound X gave 0.733g of carbon dioxide and 0.30g of
water in a combustion analysis. Determine the empirical formula.
How much CO2 and H2O are generated ?
CO2 : 0.733g / 44.009 g/mol = 16.66 mmol. and for H2O: 0.30g / 18.015g/mol = 16.66
mmol.
Remembering that the equation for a combustion reaction tells us that we will get one
molecule of CO2 for each atom of C and one molecule of H2O for every 2 H atoms, then
the ratio of C:H for X is 1: 2 corresponding to 0.30 g of X containing 16.66 mmol of C
and 33.3 mmol of H.
To check the total composition, this should be converted to a %.
% C = (16.66 mmol * 12.011) / 0.30g = 66.6 %
% H = (33.3 mmol * 1.008) / 0.30g = 11.3 %
This does not add up to 100%, therefore, assume that the balance is oxygen, O = 22.1 %
which corresponds to 0.221 * 0.30g = 0.066g or 4.14 mmol.
Now one needs to determine the stoichiometry in X. To do thisI suggest you make a little
table, writing in the contributing atoms and the amount present (mmol) in the sample,
then the empirical ratio can be found by dividing by the smallest.
Atoms
mmol
C
H
16.66 33.3
O
4.14
/ smallest 3.999 7.998 1.000
emp. ratio
4
8
1
THEREFORE, the empirical formula is C4H8O
In order to determine the molecular formula, it is necessary to know the molecular
weight.
(RESIST the temptation to round the numbers by more than a few hundredths, otherwise
you will probably get the wrong answer !)
Elemental Analysis Data
Essentially this is similar to the problem above (but simpler !).
An unknown organic compound X gave the following elemental analysis data : C =
58.8%, H = 9.8%. What is the empirical formula of X ?
First thing to do is check the total %. It is not 100%, therefore, assume that the balance is
oxygen, O = 31.4 %.
To solve the rest of the problem, I suggest you make a little table, writing in the
contributing atoms and the amount present (%) in the sample. If you then imagine that
you are considering a 100g sample then by dividing by the atomic weights you get the
moles of each atom type present. Dividing through by the smallest then gives the
empirical ratio and after rounding out, the empirical formula is obtained.
Atoms
%
C
H
O
58.8 9.8 31.4
/ A.W
4.9
9.8 1.96
/ smallest
2.5
5.0 1.0
emp. ratio
5
10
2
THEREFORE, the empirical formula is C5H10O2
Empirical and molecular formulas for compounds that contain only carbon and
hydrogen (CaHb) or carbon, hydrogen, and oxygen (CaHbOc) can be determined with
a process called combustion analysis. The steps for this procedure are

Weigh a sample of the compound to be analyzed and place it in the apparatus
shown in the image below.

Burn the compound completely. The only products of the combustion of a
compound that contains only carbon and hydrogen (CaHb) or carbon,
hydrogen, and oxygen (CaHbOc) are carbon dioxide and water.

The H2O and CO2 are drawn through two tubes. One tube contains a
substance that absorbs water, and the other contains a substance that
absorbs carbon dioxide. Weigh each of these tubes before and after the
combustion. The increase in mass in the first tube is the mass of H2O that
formed in the combustion, and the increase in mass for the second tube is the
mass of CO2 formed.

Assume that all the carbon in the compound has been converted to CO 2 and
trapped in the second tube. Calculate the mass of carbon in the compound
from the mass of carbon in the measured mass of CO2 formed.

Assume that all of the hydrogen in the compound has been converted to H2O
and trapped in the first tube. Calculate the mass of hydrogen in the compound
from the mass of hydrogen in the measured mass of water.

If the compound contains oxygen as well as carbon and hydrogen, calculate
the mass of the oxygen by subtracting the mass of carbon and hydrogen from
the total mass of the original sample of compound.

Use this data to determine the empirical and molecular formulas in the usual
way.
Apparatus for Combustion Analysis A compound containing carbon and hydrogen (CaHb)
or carbon, hydrogen, and oxygen (CaHbOc) is burned completely to form H2O and CO2. The
products are drawn through two tubes. The first tube absorbs water, and the second tube
absorbs carbon dioxide.
To illustrate how empirical and molecular formulas can be determined from data derived from
combustion analysis, let’s consider a substance called trioxane. Formaldehyde, CH2O, is
unstable as a pure gas, readily forming a mixture of a substance called trioxane and a polymer
called paraformaldehyde. That is why formaldehyde is dissolved in a solvent, such as water,
before it is sold and used. The molecular formula of trioxane, which contains carbon, hydrogen,
and oxygen, can be determined using the data from two different experiments. In the first
experiment, 17.471 g of trioxane is burned in the apparatus shown above, and 10.477 g H 2O
and 25.612 g CO2 are formed. In the second experiment, the molecular mass of trioxane is
found to be 90.079.
We can get the molecular formula of a compound from its empirical formula and its molecular
mass. (See the text for a reminder of how this is done.) To get the empirical formula, we need to
determine the mass in grams of the carbon, hydrogen, and oxygen in 17.471 g of trioxane. Thus
we need to perform these general steps.

First, convert from the data given to grams of carbon, hydrogen, and oxygen.

Second, determine the empirical formula from the grams of carbon, hydrogen, and
oxygen.

Third, determine the molecular formula from the empirical formula and the given
molecular mass.
Because we assume that all the carbon in trioxane has reacted to form in CO 2, we can find the
mass of carbon in 17.471 g trioxane by calculating the mass of carbon in 25.612 g CO 2.
Because we assume that all of the hydrogen in trioxane has reacted to form H2O, we can find
the mass of hydrogen in 17.471 g trioxane by calculating the mass of hydrogen in 10.477 g
H2O.
Because trioxane contains only carbon, hydrogen, and oxygen, we can calculate the mass of
oxygen by subtracting the masses of carbon and hydrogen from the total mass of trioxane.
? g O = 17.471 g trioxane - 6.9899 g C - 1.1724 g H = 9.309 g O
We now calculate the empirical formula, the empirical formula mass, and the molecular formula.
Sample Study Sheet: Calculating Molecular Formulas from Combustion Data
Tip-off – You want to calculate the molecular formula of a compound containing only carbon
and hydrogen (CaHb) or carbon, hydrogen, and oxygen (CaHbOc), and you are given its
molecular mass and the masses of CO2 and H2O formed in the combustion of a given mass of
the compound.
General Procedure

Calculate the number of grams of carbon in the compound by calculating the number of
grams of carbon in the given amount of CO2.

Calculate the number of grams of hydrogen in the compound by calculating the number
of grams of hydrogen in the given amount of H2O.

If the compound contains oxygen, calculate the number of grams of oxygen in it by
subtracting the masses of carbon and hydrogen from the given total mass of compound.
? g O = (given) g total - (calculated) g C - (calculated) g H

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.
EXAMPLE – Obtaining a Molecular Formula from Combustion Data:
Dianabol is one of the anabolic steroids that has been used by some athletes to increase the
size and strength of their muscles. It is similar to the male hormone testosterone. Some studies
indicate that the desired effects of the drug are minimal, and the side effects, which include
sterility and increased risk of liver cancer and heart disease, keep most people from using it.
The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be
determined using the data from two different experiments. In the first experiment, 14.765 g of
Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment,
the molecular mass of Dianabol is found to be 300.44. What is the molecular formula for
Dianabol?
Solution: