Stoichiometry

Combustion reactions of organic molecules (molecules containing only Carbon, Hydrogen, and Oxygen) are sometimes used to determine the elemental constituency of an unknown. We examined a simple combustion reaction before: 
CH4 + 2 O2 <--> CO2 + 2 H2O                       (1)
This reaction can be performed using an excess of oxygen for the combustion so the amount of products that are generated is limited by the amount of Carbon and Hydrogen in the organic reagent. The following shows an apparatus that is used for the combustion of such an organic unknown. The CuO chamber is there to ensure the complete reaction of all CO to CO2 and the other two chambers are used to separate and collect the two products. The water is usually absorbed by Mg(ClO4)2 and the CO2 is usually collected by NaOH on asbestos.

A combustion is run in the following way: The water and carbon dioxide collection chambers are pre-weighed (this is the so called tare weight). A known amount (mass) of organic sample is admitted into the furnace, burned completely in excess oxygen, and the collection chambers are reweighed. The mass increase in each collector is simply the mass of the corresponding reaction product.  Again, the amount of each product is limited by the amount of carbon and hydrogen in the sample (oxygen is in excess and is provided by the experimenter)  The combustion reaction stoichiometry is then a useful way of identifying at least the constituency of an unknown.  For a sample that consists of only C, H and O, the combustion reaction that would take place in the above procedure would be: 
CxHyOz + (X + Y/4 - Z/2)O2 = (X) CO2 + (Y/2) H2O        (2)
Check for yourself that this is a balanced chemical equation.

With a simple understanding of reaction stoichiometry, we can predict the outcome of chemical transformations.
How much CO2 and H2O will be produced by the combustion of 2.45 g of Methane?  We start with a balanced chemical equation (1).  Equation (1) says that for every 1.00 moles of methane, 2.00 moles of water and 1.00 moles of carbon dioxide will be produced.  How many moles of methane is 2.45 grams?  We need to know how much a molecule of methane weighs (on average).  The molecular weight of methane is: 
MW(methane) = (12.011 g/mol) + 4*(1.00794 g/mol) = 16.043 g/mol
Therefore a 2.45 g sample of methane will contain 
(2.45 g) / (16.043 g/mol) = 0.1527 moles of methane
So how many moles of carbon dioxide will be collected?  0.1527 moles!  The mass of carbon dioxide can be simple obtained from the molecular weight of carbon dioxide.
  
Mass{CO2} = (0.1527 mol) * (44.0098 g/mol) =  6.7210 g CO2
                                    'Rounded' 6.72 g CO2
What about the water? 
Mass{H2O} = 2*(0.1527 mol) * (18.0153 g/mol) =  5.5024 g H2O
                                      'Rounded' 5.50 g H2O
A Note on the Calculation:  Since the stochiometric ratio of 2 waters per 1 methane molecule is EXACT, the number 2 should have an infinite number of significant figures, but that is silly so it is just written as an integer.  The least number of significant figures in the problem is the mass of the sample (3 sig figs) so the final answers (masses of products) should be quoted with the same precision.  All the intermediate results and atomic weights obtained from the periodic table should be carried with GREATER precision than the final answer so no 'round off' error is incurred.   Please consult your textbook for proceedures involving Significant Figures in chemical calculations.

But that's no fun, we knew what we would get before we ran the experiment.  Can we use the combustion experiment to help us learn about something we don't know?  Of Course!
Combustion analysis is routinely performed to determine the empirical formula of organic compounds.    Here is an example.
A sample of an unknown organic compound (which contains only C, H, and O) is known from Osmotic pressure measurements to have a molecular weight of approximately 150. g/mol.  Combustion of 1.00 grams of this material in excess oxygen in the apparatus above yields 0.8397 g of H2O and 1.762 g of CO2.  What is the molecular formula of the unknown organic compound?
First and foremost, remember that reaction stoichiometry relates the number of moles(molecules) of reactants and products, and not the relative masses (except by calculation).  Therefore, lets figure out how many moles of everything we have. 

Moles(Sample) = (1.00 g) /(150. g/mol) = 6.667 x 10-3
Moles(Carbon Dioxide) = (1.762 g)/(44.098 g/mol) = 4.002 x 10-2
Moles(Water) = (0.8397 g)/(18.015 g/mol) =  4.661 x 10-2


How many moles of carbon atoms are there per mole of sample? 

moles carbon/mole sample = 4.002 x 10-2 / 6.667 x 10-3
                         = 6.00


How many moles of Hydrogen atoms per mole of sample? 

moles hydrogen/mole sample = 2*(4.661 x 10-2) / 6.667 x 10-3 
                         = 14.0


Now, whatever is left in the total mass (1.00 g) of the sample must be oxygen!
  

Mass oxygen = 1.00g
             - (4.002 x 10-2)*(12.011 g /mol)
             - (9.322 x 10-2)*(1.00797 g/mol)
            =  4.254 x 10-1 g
Moles oxygen/mole sample = ((4.254 x 10-1 g)/(15.9994 g/mol))/6.667 x 10-3
                         = 3.99
So, it looks like the molecular formula of our unknown is C6H14O4
Can you write a balanced chemical equation for the combustion of this substance? answer
Syllabus || Staff || Operations || TOP
PJ Brucat || University of Florida