Determining Compound Formulae

Now we are ready to see how the formulas of some compounds are determined using the

Combustion Technique

All of the percent composition data that we have been using usually (often) comes by using combustion analysis -- particularly for compounds containing C, H, and O. What we see in a combustion reaction is the following if the compound only contains C, H, and O:

Compound + O2 --> CO2 + H2O

Here is then brief analysis of what happens:

1 g C in the compound --> 1 g of C in carbon dioxide 1 mole of C in the compound --> 1 mol of C in 1 mole of CO2 12.01 g of C in the compound --> 12.01 g of c in 44.01 g of CO2 AND 1 g of H in the compound --> 1 g of H in H2O 2 mol of H in the compound --> 2 mol of H in 1 mol of H2O 2.0 g of H in the compound --> 2.0 g of H in 18.02 g of H2O

So if we collect and measure the H2O and CO2 in the products of the reaction we can then determine the amount of H and C that was in the compound.

EXAMPLE

We do an elemental analysis of a compound and determine that it only contains the elements C, H, and O. Then we combust 0.513 g of the compound and then we determine the amount of carbon dioxide and water generated in the reaction to be 0.501 g and 0.103 g respectively. What is the empirical formula of the unknown compound?

We know that 1.000 mol of C in the sample --> 1.000 mole of CO2 so how many moles of carbon dioxide did we get?

0.501 g CO2 * (1 mol CO2/(44.01 g CO2)) = 0.0114 mol CO2

So how many g of C is this 0.0114 mol of CO2?

0.0114 mol CO2 * (1 mol C/(1 mol of CO2) * (12.01 g C/(1 mol C)) = 0.137 g C

So now we know how many grams of C were contained in the original unknown compound.

Now let's get the grams of H in the unknown compound by a similar route:

0.103 g H2O * (1mol H2O/(18.02 g h2O)) = 0.00572 mol H2O

0.00572 mol H2O * (2 mol H/(1 mol H2O)) = 0.0114 mol H

0.0114 mol H * (1.008 g H/(1 mol H)) = 0.0115 g H

What about Oxygen? Well we started with 0.513 g of unknown and now we know that we have 0.137 g of C and 0.0115 g of H in the compound for a total weight of 0.148 grams. So since we know that the compound only contains C, H, and O we can subtract 0.148 from 0.513 and get 0.365 g O.

So now let's find the number of moles of each:

0.137 g C * (1.00 mol C/(12.01 g C)) = 0.0114 mol C

0.0115 g H * (1.0 mol H/(1.01 g H)) = 0.0114 mol H

0.365 g O * (1.00 mol O/(16.00 g O)) = 0.0228 mol O

Now to get the empirical formula we need whole numbers for the number of moles of C, H, and O. We can convert the above into whole numbers by dividing by 0.0114 and this gives us the ratio

C:H:O = 1:1:2 and thus the empirical formula is

CHO2 which has a molar mass of 45.02 g/mol. So how do we get the molecular formula? We need more experimental data, namely the experimental molar mass of the unknown. So we go back to the lab and determine the experimental molar mass to be 90.04 g. So what is the molecular formula? Divide 90.04 by 45.02 and we get 2 so the molecular formula of the unknown compound is C2H2O4

Work homework problems. Also go HERE and choose Chemical Formulas for practice problems.

help Go to this site to practice the combustion problem.

Now take a practice quiz to help you understand if you understand the basic concepts.
You must use your real name when it asks for a name.
The test will only submit when you have answers all of the questions correctly.
If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.

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Web Author: Dr. Leon L. Combs
Copyright 1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED