The Law of Conservation of Mass  Chemist Antoine Lavoisier’s work in the 1700s resulted in the Law of Conservation of Mass. It states that: In a chemical reaction, the total mass of the products is always the same as the total mass of the reactants.

 For example: The electrolysis of water  H 2 O  H 2 + O 2 is NOT a balanced chemical equation.  Why? Because the number of H and O atoms in the reactants does not equal the number of H and O atoms in the products.  How can we make a balanced chemical equation?

Example of Balancing Chemical Equations  The electrolysis of water yields oxygen gas and hydrogen gas.  a) Write the word equation. water   b) Write the skeleton equation. H 2 O (l)   c) Write the balanced chemical equation. 2H 2 O (l)  O 2(g) + 2H 2(g) oxygen +hydrogen O 2(g) + H 2(g)

we will work through a second example Rules for Balancing Chemical Equations iron +oxygen  iron(III) oxide Word Equation Skeleton Equation Fe (s) +O 2(g)  Fe 2 O 3(s) the rusting of iron

Rules for Balancing Chemical Equations 1. Write the number of each element present on each side of the equation eg. Fe (s) + O 2(g)  Fe 2 O 3(s) Fe O Fe O We will remove the ‘states’ to neaten things up eg. Fe + O 2  Fe 2 O 3

Fe O Fe O 1 2. Place a co-efficient in front of one element on one side and then recount the atoms Hint: if you have an odd number (not 1) of atoms on one side and even on the other, multiplying the odd number by an even number will be necessary eg. Fe + O 2  2Fe 2 O 3 Fe O Fe O x 2 = 4 2 x 3 = 6

eg. Fe + O 2  2Fe 2 O 3 Fe O Fe O a. Repeat this process until all elements are balanced eg. 4Fe + O 2  2Fe 2 O 3 Fe O Fe O 1 x 2 = eg. 4Fe + 3O 2  2Fe 2 O 3 Fe O Fe O x 3 = 6 3b. Repeat this process until all elements are balanced We are done!

4. Hints: a. Leave elements by themselves to the end b. Leave oxygen and hydrogen to the end Another example: butane lighter burning butane

butane = C 4 H 10 butane +oxygen  carbon dioxide Word Equation Skeleton Equation + water C 4 H 10(g) +O 2(g)  CO 2(g) +H 2 O (l)

Another example: C 4 H 10 + O 2  CO 2 + H 2 O CHOCHO CHOCHO Place a co-efficient in front of one element on one side and then recount the atoms C 4 H 10 + O 2  4CO 2 + H 2 O CHOCHO CHOCHO x 4 = 4 2 (2 x 4) + 1 = 9

Place a co-efficient in front of one element on one side and then recount the atoms C 4 H 10 + O 2  4CO 2 + 5H 2 O CHOCHO CHOCHO x 5 = 10 (2 x 4) + (1 x 5) = 13 We now have a problem  odd number of oxygen on one side and even number on the other side!! C 4 H 10 + O 2  4CO 2 + 5H 2 O CHOCHO CHOCHO x 6 ½ = x 5 = 10 (2 x 4) + (1 x 5) = 13 We can start again OR we can take a short cut 6 ½

C 4 H ½ O 2  4CO 2 + 5H 2 O We seem to be done  BUT we can not have fractions so… 2C 4 H O 2  8CO H 2 O CHOCHO CHOCHO 4 x 2 = 8 10 x 2 = 20 2 x 13 = 26 1 x 8 = 8 2 x 10 = 20 2 x 8 = 16 1 x 10 = = 26 CHOCHO CHOCHO x 6½ = x 5 = 10 (2 x 4) + (1 x 5) = 13 Now we are done! We multiply by 2

Now let’s do some practice