1. MOLE-MOLE AND MASS-MASS CONVERSIONS Stoichiometry 1

2. Using Chemical Equations  Chemical equations are like recipes:  Tells chemists what and how much goes in/comes out of a reaction  Given a balanced chemical equation & the quantity of one substance in the equation, the quantity of any other substance in the same reaction can be calculated  Quantity = the amount of a substance measured in grams, liters, molecules, or moles.

3. Using Chemical Equations  Balanced Chemical Equations(BCEs) show the ratio of chemical compounds in a reaction  Ratio stays the same when referring to:  Single molecules  Molar quantities  The above equation can be read as:  1 molecule of nitrogen gas reacts with 3 molecules of hydrogen to produce 2 molecules of ammonia  1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia **Remember, 1 mole = 6.02x10 23 molecules!** N 2 + 3H 2 → 2NH 3

4. Stoichiometry Stoichiometry – (stoy-key-ometry)The calculation of quantities of substances involved in chemical equations. Balanced chemical equations show molar ratios All quantities must be converted to moles FIRST before using the molar ratio

5. Mole-Mole Stoichiometry Consider the following BCE:  4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) If you only had 1.8 moles of Al, how much product could you make?  Given: 1.8 mol Al  Want: ??? mol Al 2 O 3  Conversion factor: 4 mol Al produce 2 mol Al 2 O 3  1.8 mol Al x 2 mol Al 2 O 3 4 mol Al = 0.90 mol Al 2 O 3

6. Mole-Mole Stoichiometry If you wanted to produce 24 moles of Al 2 O 3, how many moles of each reactant would you need? 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s)  Given: 24 mol Al 2 O 3  Want: ??? mol Al  Want: ??? mol O 2  Conversion factors:  4 mol Al produce 2 mol Al 2 O 3, or  3 mol O 2 produce 2 mol Al 2 O 3 4 mol Al 2 mol Al 2 O 3 3 mol O 2 2 mol Al 2 O 3

7. Mole-Mole Stoichiometry If you wanted to produce 24 moles of Al 2 O 3, how many moles of each reactant would you need? 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s)  24 mol Al 2 O 3 x 4 mol Al 2 mol Al 2 O 3 3 mol O 2 2 mol Al 2 O 3 = 48 mol Al = 36 mol O 2

8. Mole-Mole Stoichiometry You can always do mole-mole conversions if you have a BALANCED chemical equation.  Unbalanced equations are not useful for stoichiometry. moles of substance A moles of substance B BCE

9. Mole-Mole Stoichiometry Practice Octane combusts in the presence of oxygen via the following reaction:  2C 8 H 18 + 25O 2  16CO 2 + 18H 2 O How many moles of oxygen gas can react with 7.50 moles of octane?  7.50 mol C 8 H 18 x If 48 moles of CO 2 are produced during one such reaction, how many moles of H 2 O are also produced?  48 mol CO 2 x 25 mol O 2 2 mol C 8 H 18 18 mol H 2 O 16 mol CO 2 = 93.8 mol O 2 = 54 mol H 2 O

10. Mass-Mass Stoichiometry Moles are not measured directly mass can be measured with a scale Mass (in grams) can be converted to moles and vice-versa  Mass-mass conversions are more common and practical in real life. Given the mass of one reactant or product, the mass of any other reactant or product in the reaction can be calculated  Requires a BCE  the unknown can be either a reactant or a product.

11. Mass-Mass Stoichiometry Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ):  CaC 2 + 2H 2 O  Ca(OH) 2 + C 2 H 2 How many grams of C 2 H 2 are produced by adding water to 5.00 grams of CaC 2 ?  Given: 5.00 g CaC 2  Want: ??? g C 2 H 2  Conversion factors:  1 mol CaC 2 produces 1 mol C 2 H 2  1 mol CaC 2 = 64.099 g CaC 2 (molar mass of CaC 2 )  1 mol C 2 H 2 = 26.038 g C 2 H 2 (molar mass of C 2 H 2 )

12. 26.038 g C 2 H 2 1 mol C 2 H 2 x 1 mol CaC 2 x 64.099 g CaC 2 x Mass-Mass Stoichiometry Make a plan!  Your first step should always be to change to moles!  Next, do a mole-mole conversion using the BCE to guide you.  Finally, convert from moles back to grams. 5.00 g CaC 2  mol CaC 2  mol C 2 H 2  g C 2 H 2  5.00 g CaC 2 = 2.03 g C 2 H 2

13. Mass-Mass Stoichiometry mass A moles A moles B mass B MM A MM B BCE

14. Mass-Mass Stoichiometry Copper metal reacts with concentrated nitric acid to produce copper(II) nitrate, nitrogen dioxide gas, and water:  Cu + 4HNO 3  Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O How many grams of HNO 3 are needed to react completely with 0.450 grams of Cu?  Given: 0.450 g Cu  Want: ??? g HNO 3  Conversion factors:  1 mol Cu reacts with 4 mol HNO 3  1 mol Cu = 63.546 g Cu (molar mass of Cu)  1 mol HNO 3 = 63.012 g HNO 3 (molar mass of HNO 3 )

15. How many grams of HNO 3 are needed to react completely with 0.450 grams of Cu?  Cu + 4HNO 3  Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O Make a plan!  0.450 g Cu  mol Cu  mol HNO 3  g HNO 3  0.450 g Cu 1 mol Cu 63.546 g Cu x 4 mol HNO 3 1 mol Cu x 63.012 g HNO 3 1 mol HNO 3 x Mass-Mass Stoichiometry = 1.78 g HNO 3

16. A Few Reminders Stoichiometry problems can involve reactants, products, or both.  They can also involve moles, grams, liters or milliliters, atoms, molecules, or any other unit that measures the amount of a particular chemical. All stoichiometry problems have a mole-mole conversion at their center.  At the very least, you’ll need a balanced chemical equation in order to solve any stoichiometry problem. It’s a good idea to write not only the unit (g, mol, L, etc) but also the chemical formula when doing conversions.  Ex: write 48.0 g Cu(NO 3 ) 2, not just 48.0 g.