2. Stoichiometry is… Greek for “measuring elements” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There are several ways to interpret a balanced chemical equation

3. #1. In terms of Particles Element= made of atoms Molecular compound (made of only nonmetals) = molecules Ionic Compounds (made of a metal and nonmetal parts) = formula units (ions) 2H 2 + O 2   2H 2 O We’d say two molecules of hydrogen and one molecule of oxygen form two molecules of water.

4. #2. In terms of Moles The coefficients tell us how many moles of each substance A balanced equation is a Molar Ratio 2H 2 + O 2   2H 2 O We’d say two moles of hydrogen and one mole of oxygen form two moles of water.

5. A balanced equation is can be used to make mole to mole ratios! Example:

6. Another example:

7. Calculating Stoichiometric Problems (mole to mole) 1. Balance the equation. 2. Set up mole ratios. 3. Use mole ratios to calculate moles of desired chemical.

8. Example: How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2 Al 2 O 3  Al + 3O 2 X mol O 2 = 3.34.mol Al 2 O 3 x 2 mol Al 2 O 3 3 mol O 2 =5.01 mol O 2

9. Calculating Stoichiometric Problems (moles to grams, grams to moles, grams to grams) 1. Balance the equation. 2. Convert mass in grams to moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into grams, if necessary.

10. Example: If 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2  2Al 2 O 3 = 6.50 g Al x 1 mol Al x 26.98 g Al 4 mol Al 2 mol Al 2 O 3 x 1 mol Al 2 O 3 101.96 g Al 2 O 3 12.3 g Al 2 O 3 x g Al 2 O 3 =

11. How do you get good at this?

12. Try it: Page 115 # 4, 5, 6 Page 121 # 11, 12, 13