Useful Concepts

Combustion Data

This is probably best illustrated by working through some examples. The problems are each slightly different, but the basic strategy is similar, you need to work out how much of each element is present and the simplest ratio between them.

A 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 many moles of 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.

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 !)

B An unknown organic compound Z gave 8.8 g of carbon dioxide and 4.5 g of water in a combustion analysis for C and H. Z was found to have a molecular weight = 58.124 g/mol. Determine the molecular formula.

How many moles of CO₂ and H₂O are generated ?
CO₂ : 8.8 g / 44.009 g/mol = 0.20 mol. and for H₂O: 4.50g / 18.015g/mol = 0.25 mol.

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 Z is 0.20 : 0.50 indicating an empirical ratio C₂H₅.

Alternatively, to do this, you can 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.

This empirical unit has mass: = (2 * 12.011) + (5 * 1.008) = 29.062

How many of these units are in the molecule ? 58.124 / 29.062 = 2

(RESIST the temptation to round the numbers by more than a few hundredths, otherwise you will probably get the wrong answer !)

C An unknown organic compound Y gave 8.8 g of carbon dioxide and 4.5 g of water in a combustion analysis for C and H. Z was found to have a molecular weight = 74.123 g/mol. Determine the molecular formula.

How many moles of CO₂ and H₂O are generated ?
CO₂ : 8.8 g / 44.009 g/mol = 0.20 mol. and for H₂O: 4.50g / 18.015g/mol = 0.25 mol.

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 Y is 0.20 : 0.50 indicating an empirical ratio C₂H₅.

Alternatively, to do this, you can 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.

This empirical unit has mass: = (2 * 12.011) + (5 * 1.008) = 29.062

How many of these units are in the molecule ? 74.239 / 29.062 = 2.55 (!)

Since this is not an integer, another element must be present....

Need some trial and error now.... the minimum number of those units that can be present is 2...... 2 x 29.062 = 58.124
Which implies we are missing 74.123 - 58.124 = 15.999
Now check the periodic table to see what is most likely to be missing.... in this case it suggests one oxygen atom...
(Note this can be tricky if you need to consider multiple atoms etc. However, if this is normally the case, then other information would have to be provided)

(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 problem A above (but simpler !). Three examples are presented.

A. An unknown organic compound X gave the following elemental analysis data : C = 88.16%, H = 11.84%. What is the empirical formula of X ?

First thing to do is check the total %. In this case it is 100%, therefore, we have a hydrocarbon.

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 scaling to give the simplest integer ratios, the empirical formula is obtained.

(note.... the scaling is often the tough part, here we have an initial ratio of 1 : 1.60, so you need to find the first multiple of 1.6 that is a whole number (or very close to it). Avoid the temptation to round the first two decimal places.

B. An unknown organic compound Y gave the following elemental analysis data : C = 58.8%, H = 9.8%. What is the empirical formula of Y?

First thing to do is check the total %. It is not 100%, therefore, since we are not given any other information, we should assume that the balance is oxygen, O = 31.4 % (see footnote).
To solve the rest of the problem, we 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 scaling to give the simplest integer ratios, the empirical formula is obtained.
 

© Dr. Ian Hunt, Department of Chemistry