A LITTLE REMINDER!
You have learned previously that during combustion reactions some type
of fuel is burned and water and carbon dioxide are the products.
Sometimes the fuel is a hydrocarbon (contains only carbon and
hydrogen) and sometimes there are other components to the fuel.
Ex. Methane combusts in the presence of oxygen
CH4 +
2O2 
CO2 + 2H2O
Methanol combusts in the presence of oxygen
2CH3OH +
3O2 
2CO2 + 4H2O
FINDING THE EMPIRICAL FORMULA OF A FUEL
If given the components of the fuel, mass of original substance, mass of
products (water and carbon dioxide) we can determine the empirical
formula. If given the molecular mass of the original substance, we
can also determine the molecular formula. However, this only works
if combustion is complete – meaning that all of the fuel is consumed
in the reaction.
LET’S DO AN EXAMPLE!
Combustion of 0.255 grams of isopropyl (a fuel composed of C,H, & O)
produces 0.561 g CO2, 0.306 g H2O
Step 1: Find the grams of carbon
Step 2: Find the grams of hydrogen
Subtract the mass of the carbon & hydrogen from the original mass.
That gives you the mass of the oxygen
Now that you have the masses for all three components, you can
convert grams to moles, divide all by the small, etc.
LET’S TRY ANOTHER! 
What makes dirty socks so stinky? Caproic acid – composed of carbon, hydrogen
and oxygen. A 0.225 g sample has 0.512 g CO2 and 0.209 g H2O
Find the mass of the carbon.
Find the mass of the hydrogen.
Find the mass of the oxygen.
MASS TO MOLES: DIVIDE ALL BY SMALL TO GET THE
EMPIRICAL FORMULA
IF THE MOLECULAR MASS OF “STINKY SOCKS” IS 116.01
GRAMS PER MOLE, WHAT IS THE MOLECULAR FORMULA?
YOU TRY THIS ONE ON YOUR OWN!
A 0.548 g sample of a fuel composed of carbon, hydrogen and nitrogen
is completely combusted. 0.312 grams of water and 1.525 grams of
carbon dioxide are produced. The molecular formula mass is 79.01
grams per mole. What are the empirical and molecular formulas?