A sample of a saturated hydrocarbon is placed in a 1.0L vessel at 107oC. The pressure of the sample at these conditions is .40 atm. A stoichiometrically equivalent amount of oxygen is added to the vessel and the mixture is combusted. Following combustion, the pressure in the vessel is 5.19 atm at 107oC. Assume all of the products are gases. What is the formula of the hydrocarbon?

Take a look at this. If the hydrocarbon is CH4, then others in the homologous series.

CH4 + 2O2 ==> CO2 + 2H2O
C2H6 + 7/2 O2 --> 2CO2 + 3H2O
C3H8 + 5O2 ==> 3CO2 + 4H2O
C4H10 + 13/2 O2 ==> 4CO2 + 5H2O
C5H12 + .....
C6H14.......

So CH4 produces 1+2 = 3 mols gas.
C2H6 produces 2 + 3 = 5 mols gas.
C3H8 produces 7 mols gas
C4H10 produces 9 mols gas
etc.
So substitute into PV = nRT and solve for n initially. Do the same for the second filling and solve for n.
n products/n initial = round to whole number, look at the chart above, and draw conclusions as to the hydrocarbon's formula.