Humboldt State University ® Department of Chemistry

Richard A. Paselk

	Science 331
Fall 2004	Lecture/Activity	Office: SA560a
Notes: 15 September	return to lecture	Phone: x 5719 Home: 822-1116
		e-mail: rap1

Propane gas (C3H8) gas plus oxygen gas gives carbon dioxide plus water  C3H8 + O2 -> CO2  +  H2O  Now we need to have the same number of each kind of atoms on each side. Let's count the atoms: Carbons  1x3=3C 1x1=1C Hydrogens  1x8=4H            1x2=2H Oxygens      1x2=2O    1x2=2O  +  1x1=1O Again, the best strategy for balancing is to start with the molecule containing atoms other than oxygen, in this case, propane. Comparing the two sides we can first balance Carbon: C3H8  +  O2  -> 3CO2  +   H2O Carbons  1x3=3C      =  3x1=3C     Hydrogens  1x8=8H             1x2=2H Oxygens       1x2=2O     3x2=6O     1x1=2O Comparing the two sides again, we now have the same number of carbons. Now let's balance the hydrogens by adding 3 waters to the right side:   C3H8 +    O2  ->  3CO2 +   4 H2O  Carbons  1x3=3C      =  3x1=3C      Hydrogens  1x8=8H      =       4x2=8H  Oxygens       1x2=2O     3x2=6O     4x1=4O Carbon and hydrogen are now balanced, while oxygen is not, we have 2 oxygen atoms on the left, but we now have 10 on the right. We can balance oxygen by adding 8 oxygen atoms, or four molecules to the left:    C3H8 +    5 O2 ->   3CO2 +   4 H2O Carbons  1x3=3C   =  3x1=3C    Hydrogens  1x8=8H     =        4x2=8H Oxygens     5x2=10O =   3x2=6O     4x1=4O voila! We have now balanced the entire equation. We have different substances on the two sides, but the same number of atoms of each type - elements and mass are conserved!

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© R A Paselk

Last modified 15 September 2004