1. Balancing Chemical equations. For each skeleton equation to become a complete (final form) chemical equation, coefficients must be inserted into the skeleton equation. Coefficients are whole numbers which indicate the relative amounts of each substance. A chemical equation is said to be “balanced” when, the number of atoms of each element found on the reactant side of the equation is equal to the number of atoms of that same element on the product side of the equation.

2. In a balanced equation, each side of the equation will have the same numbers of atoms of each element. To balance an equation, a coefficient is placed before each substance in the equation. The coefficient is a “multiplier”. It doubles, triples, quadruples (etc.) the amount of the substance it is placed in front of. The coefficients used for an equation should be the lowest whole numbers possible to achieve balancing of the atoms.

3. Strategies or tips for balancing chemical equations. Balance one element at a time; don’t try to balance the whole equation at once. First pick an element that appears in only one reactant and one product.

4. Determine the LCM for the element you have picked. Ex (below) – let’s pick fluorine. The subscripts are 2 and 3; the least common multiple of 2 and 3 is 6. “count by 2” and “count by 3”… when you find the first # that is the same, you have identified the LCM

5. Example: Place a coefficient of 3 on the left and 2 on the right, so as to now have 6 atoms of F both on the left and on the right. _____Al + ___ F 2  ___ AlF 3 Think momentarily only about the LCM of “F”. 3 F 2  2 F 3 (LCM of 2, 3 = 6) _____Al + _ 3 F 2  __2_ AlF 3 (Remember, coefficients are multipliers

6. After balancing one element, pick a second element to work with that appears in the same compound as the already- balanced element. Again, determine the LCM Think momentarily only about the LCM of “Al”, in __ Al  __2 AlF 3 You need, at the moment, the LCM of 1, 2 which is 2

7. After balancing one element, pick a second element to work with that appears in the ___________ compound as the already-balanced element. Again, determine the LCM. 2 Al + 3 F 2  2 AlF 3 is correct because 2 Al + 6 F  2 Al + 6 F

8. Start an atom inventory. The atom inventory for the above example is this: Only when the atom inventory is equal on both sides, is the balancing correctly done. 2 Al 6 F

9. Balance these two atoms last: hydrogen and oxygen The reason is, many times H and O occur in 2 substances to the left of the arrow and/or 2 substances to the right of the arrow. When this occurs, you must add as well as multiply to solve the problem. The example shown below will illustrate this point.

10. __ CH 4 + __ O 2  __ CO 2 + __H 2 O Balance the “C” first: 1 CH 4 + __ O 2  _1_ CO 2 + __H 2 O atom inventory: 1C 4H 2 Ox 1C 2 Ox 2H 1 Ox 1C 2H 3 Ox

11. Balance the “H” second LCM of 4, 2 = 4 this means, aim for 4 H’s. ? X 2 = 4 ? = 2 Insert “2” here _1_ CH 4 + __ O 2  _1_ CO 2 + _2_H 2 O

12. atom inventory atom inventory Now work with oxygen Notice, to balance the “O”, you must now add the number of oxygen atoms from two compounds (CO 2, H 2 O): CO 2 has 2 oxygen atoms; 2 H 2 O also has 2 oxygen atoms. 2 +2 = 4 1C 4H 2 Ox 1C 2 Ox 4H 2 Ox 1C 4H 4 Ox

13. _1_ CH 4 + __ O 2  _1_ CO 2 + _2_H 2 O Continue working with oxygen- LCM 2,4 is 4 aim for 4 oxy’s on the left; ? X 2 = 4 ? = 2 _1_ CH 4 + _2_ O 2  _1_ CO 2 + _2_H 2 O Balance the ___________________ atoms second. Save the _______________ atoms till last because you will have to ___________ together the oxygen atoms from CO 2 and H 2 O. a.Practice: Predict the products and balance: C 3 H 8 + O 2 

14. _1_ CH 4 + _2_ O 2  _1_ CO 2 + _2_H 2 O This equation is correctly balanced as shown by the atom inventory. 1C 4H 4 Ox

15. If you are having difficulty with a combustion problem (or any problem), double a coefficient and re-work the problem based on the assumption that the new double coefficient is correct. This corrects problems with needing a “half” coefficient, which might occur during combustion reactions in which there are an even number of carbon atoms. Ex: 2 ½ can’t be used as a coefficient; but, 5 can be used as a coefficient.

16. As long as you have a polyatomic ion both to the left and to the right of the arrow, work with it as a whole unit (don’t work with it as individual atoms). Ex: in Al(NO 3 ) 3, there are 3 (NO 3 ) Re-write H 2 O as HOH whenever (OH) is present on the opposite side of the arrow from H 2 O.