1. Balancing Chemical Equations

2. Chemical Equation A representation of a chemical reaction. For example, burning sugar: C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O

3. Chemical Equation C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O ReactantsProducts

4. Chemical Equation C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O This equation is not balanced. There are not the same number of each TYPE of atom as a reactant and as a product.

5. Chemical Equation C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O Reactants: –6 carbon atoms –12 hydrogen atoms –6 oxygen atoms in sugar –2 oxygen atoms alone

6. Chemical Equation C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O Products –1 carbon atom –2 oxygen atoms in carbon dioxide & –1 oxygen atom in water –2 hydrogen atoms in water

7. Chemical Equation C 6 H 12 O 6 + O 2 ---> Reactants: –6 carbon atoms –12 hydrogen atoms –6 oxygen atoms in sugar –2 oxygen atoms alone CO 2 + H 2 O Products –1 carbon atom –2 oxygen atoms in carbon dioxide & –1 oxygen atom in water –2 hydrogen atoms in water

8. Chemical Equation C 6 H 12 O 6 + O 2 ---> Reactants: –The reactants need to have the same number of each type of atom as the products, CO 2 + H 2 O Products –So we multiply each type of atom in a molecule by a number to change how many atoms of that type we have.

9. Chemical Equation C 6 H 12 O 6 + O 2 ---> Reactants: –6 carbon atoms –12 hydrogen atoms –6 oxygen atoms in sugar –2 oxygen atoms alone 6 CO 2 + H 2 O Products –Now if we multiply the CO 2 by six (showing we have six molecules of CO 2 instead of only one), we would have 6 carbon atoms 12 oxygen atoms in CO 2 The number placed in front of a formula is called a coefficient. It is multiplied by the number each type of atom.

10. Chemical Equation C 6 H 12 O 6 + O 2 ---> Reactants: –6 carbon atoms –12 hydrogen atoms –6 oxygen atoms in sugar –2 oxygen atoms alone 6 CO 2 + 6 H 2 O Products –If we multiply the H 2 O by six (showing we have six molecules of H 2 O instead of only one), we would have 12 hydrogen atoms 6 oxygen atoms in water

11. Chemical Equation C 6 H 12 O 6 + O 2 ---> Reactants: –Beginning count: 6 carbon atoms 12 hydrogen atoms 6 oxygen atoms in sugar 2 oxygen atoms alone 6 CO 2 + 6 H 2 O Products –Totals are now : 6 carbon atoms 12 oxygen atoms in CO 2 12 hydrogen atoms 6 oxygen atoms in water There are not enough oxygen atoms at the beginning for this to happen.

12. Chemical Equation C 6 H 12 O 6 + 6 O 2 ---> Reactants: –If we multiply the O 2 by six, we will have the needed 12 additional oxygen atoms to complete the reaction. 6 CO 2 + 6 H 2 O Products –Totals are now : 6 carbon atoms 12 oxygen atoms in CO 2 12 hydrogen atoms 6 oxygen atoms in water

13. Chemical Equation C 6 H 12 O 6 + 6 O 2 ---> Reactants: –Totals are now: 6 carbon atoms 12 hydrogen atoms 6 oxygen atoms in sugar 12 oxygen atoms alone 6 CO 2 + 6 H 2 O Products –Totals are now : 6 carbon atoms 12 oxygen atoms in CO 2 12 hydrogen atoms 6 oxygen atoms in water We now have a balanced equation! There are equal numbers of each type of atom as a reactant and as a product.

14. Chemical Equation C 6 H 12 O 6 + 6 O 2 --> 6 CO 2 + 6 H 2 O What does a balanced chemical equation tell us? –It tells us what reactants are –It tells us what the products are –It tells us the proportions of each of the reactants to each product.

15. Let’s look at another example.

16. H 2 + O 2 --> H 2 0 The above equation does not have equal number of each type of atom as a reactant and a product. This equation needs to be balanced.

17. H 2 + O 2 --> H 2 0 How do we do it? Count how many of each type of atom there are as reactants- also as products.

18. H 2 + O 2 --> H 2 0 As reactants: 2-H 2-O As products: 2-H 1-O

19. H 2 + O 2 --> 2 H 2 0 It is obvious we don’t have enough oxygen as products. Atoms are neither created nor destroyed on chemical reactions, so it must mean that another water (H 2 O) molecule was formed. This is represented by placing a 2 in the equation in front of the formula.

20. H 2 + O 2 --> 2 H 2 0 In making this change, we also added the number of hydrogen atoms as products. We must now change the number of hydrogen atoms as reactants. Recall that the coefficient multiplies everything in the formula; there are now four hydrogen atoms as products.

21. 2 H 2 + O 2 --> 2 H 2 0 We need to start with four hydrogen atoms to end with four hydrogen atoms. We must use a coefficient in front of the H 2 that will make it so we end up with four hydrogen atoms. That number is two.

22. 2 H 2 + O 2 --> 2 H 2 0 We now have a balanced equation.

23. Mg 3 (PO 4 ) 2 + Na 2 CO 3 --> MgCO 3 + Na 3 PO 4 Mg: 3 P: 2 O: 8O: 3 Na: 2 C: 1 Mg: 1 C: 1 O: 3O: 4 Na: 3 P: 1 ReactantsProducts

24. Mg 3 (PO 4 ) 2 + Na 2 CO 3 --> 3 MgCO 3 + Na 3 PO 4 Mg: 3x1=3 P: 2x1=2 O: 8x1=8O: 3 Na: 2 C: 1 Mg: 1 C: 1 O: 3O: 4 Na: 3 P: 1 ReactantsProducts Assume that we have one of the most complicated molecule. 1

25. Mg 3 (PO 4 ) 2 + Na 2 CO 3 --> 3 MgCO 3 + Na 3 PO 4 Mg: 3x1=3 P: 2x1=2 O: 8x1=8O: 3 Na: 2 C: 1 Mg: 1x3=3 C: 1x3=3 O: 3x3=9O: 4 Na: 3 P: 1 ReactantsProducts We need to have 3 Mg atoms if we started with 3 Mg atoms.

26. Mg 3 (PO 4 ) 2 + 3 Na 2 CO 3 --> 3 MgCO 3 + Na 3 PO 4 Mg: 3x1=3 P: 2x1=2 O: 8x1=8O: 3x3=9 Na: 2x3=6 C: 1x3=3 Mg: 1x3=3 C: 1x3=3 O: 3x3=9O: 4 Na: 3 P: 1 ReactantsProducts We now finished with 3 C atoms; we must begin with 3 C atoms.

27. Mg 3 (PO 4 ) 2 + 3 Na 2 CO 3 --> 3 MgCO 3 + Na 3 PO 4 Mg: 3x1=3 P: 2x1=2 O: 8x1=8O: 3x3=9 Na: 2x3=6 C: 1x3=3 Mg: 1x3=3 C: 1x3=3 O: 3x3=9O: 4x2=8 Na: 3x2=6 P: 1x2=2 ReactantsProducts We now finished with 3 C atoms; we must begin with 3 C atoms.

28. The end