1. Mathematics of Chemical Equations By using “mole to mole” conversions and balanced equations, we can calculate the exact amounts of substances that will be used up or produced during a reaction.

2. What you’ll learn: Write mole ratios from a balanced chemical equation Calculate the number of moles and the mass of a reactant or product when given the number of moles or the mass another reactant. Identify the limiting reactant in a chemical reaction. Determine the percent yield of a chemical reaction.

3. Basic Example N 2 + 3H 2  2NH 3 From this balanced equation, we know that for every 1 mole of N 2 we need ___ moles of H 2 to react in order for ___ moles NH 3 to be produced. From this balanced equation, we know that for every 1 mole of N 2 we need ___ moles of H 2 to react in order for ___ moles NH 3 to be produced. How many moles of N 2 would we need to react with 6 moles of H 2 ? How many moles of N 2 would we need to react with 6 moles of H 2 ? 6 moles H 2 moles H 2 3 moles N 2 1 = 2 moles of N 2 x 3 2

4. N 2 + 3H 2  2NH 3 How many moles of Ammonia (NH 3 ) are produced if you have an excess of Nitrogen gas, and a sample of 0.43 moles of Hydrogen? How many moles of Ammonia (NH 3 ) are produced if you have an excess of Nitrogen gas, and a sample of 0.43 moles of Hydrogen? 0.43 moles H 2 moles H 2 3 moles NH 3 2 = 0.287 moles of NH 3 x

5. Mass – Mass Conversions You MUST BALANCE the equation first! You MUST BALANCE the equation first! NaClO 3  NaCl + O 2 How many grams of O 2 would be produced if the final mass of NaCl produced was 375.6 g? How many grams of O 2 would be produced if the final mass of NaCl produced was 375.6 g? 375.6 g NaCl 223 x 58.5 g NaCl mole NaCl1 xx mole NaCl mole O 2 2 3 mole O 2 1 32 Grams O 2 = 308.18g O 2

6. Stoichiometric Conversions A Three Step Process Stoichiometric Conversions A Three Step Process Step 1 Convert the value of the given substance to moles of that substance. Step 1 Convert the value of the given substance to moles of that substance. Step 2 Convert from moles of the given substance to moles of the unknown substance (use the mole ratio from the balanced equation). Step 2 Convert from moles of the given substance to moles of the unknown substance (use the mole ratio from the balanced equation). Step 3 Convert from moles of the unknown to proper unit. Step 3 Convert from moles of the unknown to proper unit.

7. Limiting Reactants When calculating your theoretical products, it is important to factor in which reactant you will run out of first! When calculating your theoretical products, it is important to factor in which reactant you will run out of first! This is called the “Limiting Reactant”. The other reactant(s) is called the “Excess Reactant”. This is called the “Limiting Reactant”. The other reactant(s) is called the “Excess Reactant”.

8. Ex: If you have 10grams of O 2 and 10 Liters of H 2, how many molecules of water can you create? 2H 2 + O 2  2H 2 O Step 1: Convert both to moles. Step 1: Convert both to moles. 10gO 2 x 1 mole O 2 = 0.31 moles O 2 1 32g O 2 1 32g O 2 10L H 2 x 1 mole H 2 = 0.45 moles H 2 10L H 2 x 1 mole H 2 = 0.45 moles H 2 1 22.4L H 2 1 22.4L H 2

9. Step 2: Complete a mole to mole conversion, from one reactant to the other reactant to find the theoretical need. Step 2: Complete a mole to mole conversion, from one reactant to the other reactant to find the theoretical need. 2H 2 + O 2  2H 2 O 2H 2 + O 2  2H 2 O 0.31 moles O 2 x 2 moles H 2 = 0.62 moles H 2 1 mole O 2 NEEDED 1 mole O 2 NEEDED YOU KNOW that you have 0.45 moles H 2 from your given values. So, that means you do not have enough H 2. It is therefore the Limiting Reactant.

10. IF YOU COMPLETED STEP 2 FINDING MOLES OF O 2 : IF YOU COMPLETED STEP 2 FINDING MOLES OF O 2 : 2H 2 + O 2  2H 2 O 0.45 moles H 2 x 1 moles O 2 = 0.225 moles O 2 2 mole H 2 NEEDED 2 mole H 2 NEEDED You would come to the same conclusion, because this shows that you have more than enough O 2. (You have.31 moles)

11. Step 3: Decide from Step 2 which reactant is the L.R. and use its moles to calculate your answer. (MUST have a balanced Chemical Equation!) Step 3: Decide from Step 2 which reactant is the L.R. and use its moles to calculate your answer. (MUST have a balanced Chemical Equation!) 2H 2 + O 2  2H 2 O 0.45mol H 2 x 2 mol H 2 O x 6.02x10 23 molecules H 2 O = 0.45mol H 2 x 2 mol H 2 O x 6.02x10 23 molecules H 2 O = 1 2 mol H 2 1 mole H 2 O 2.71x10 23 molecules H 2 O 2.71x10 23 molecules H 2 O

12. Percent Yield Percent Yield Most chemical reactions never succeed in producing the predicted amount of product. Most chemical reactions never succeed in producing the predicted amount of product. So, the actual amount of product is less than expected, due to experimental error So, the actual amount of product is less than expected, due to experimental error This is generally due to experimental error such as evaporation, product left on filter paper etc. This is generally due to experimental error such as evaporation, product left on filter paper etc.

13. Percent Yield Percent Yield Percent yield is the ratio of actual yield to the theoretical yield expressed as a percent. Percent yield is the ratio of actual yield to the theoretical yield expressed as a percent.