1. Quantities in Chemical Reactions

2. the amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction – Law of Conservation of Mass – balancing equations by balancing atoms the study of the numerical relationship between chemical quantities in a chemical reaction is called reaction stoichiometry 2

3. Stoichiometry is…  Greek for “measuring elements” Pronounced “stoy kee ahm uh tree”  Defined as: calculations of the quantities in chemical reactions, based on a balanced equation.  Let's first look at how to interpret a balance equation

4. Making Pancakes the number of pancakes you can make depends on the amount of the ingredients you use 4 this relationship can be expressed mathematically 1 cu flour + 2 eggs + ½ tsp baking powder = 5 pancakes 1 cup Flour + 2 Eggs + ½ tsp Baking Powder → 5 Pancakes

5. Making Pancakes if you want to make more or less than 5 pancakes you can use the number of eggs you have to determine the number of pancakes you can make – assuming you have enough flour and baking powder 5

6. Making Molecules Mole-to-Mole Conversions the balanced equation is the “ recipe ” for a chemical reaction the equation 3 H 2 (g) + N 2 (g) → 2 NH 3 (g) tells us that 3 molecules of H 2 react with exactly 1 molecule of N 2 and make exactly 2 molecules of NH 3 or 3 molecules H 2 + 1 molecule N 2 → 2 molecules NH 3 – in this reaction and since we count molecules by moles 3 moles H 2 + 1 mole N 2 → 2 moles NH 3 6

7. In Terms of Moles  The coefficients tell us how many moles of each substance 2Al 2 O 3 → 4 Al + 3O 2 2Na + 2H 2 O → 2NaOH + H 2  A balanced equation is a Molar Ratio

8. Check for Understanding  What is the molar ratio between hydrogen and ammonia (NH 3 ) when hydrogen gas reacts with nitrogen gas to form ammonia gas?  First write a balanced equation  3H 2(g) + N 2(g) → 2NH 3(g)  3 mol H 2 to 2 mol NH 3

9. Mole to Mole conversions 2Al 2 O 3 → 4 Al + 3O 2 each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are the two possible conversion factors to use in the solution of the problem.

10. Mole-to-Mole Conversions How many moles of NaCl result from the complete reaction of 3.4 mol of Cl 2 in the reaction below? First you need a balanced equation: 2 Na(s) + Cl 2 (g) → 2 NaCl(s) Now, you need to do the math: 10

11. Mole to Mole conversions  How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 2 O 3 → 4 Al + 3O 2 3.34 mol Al 2 O 3 x 2 mol Al 2 O 3 3 mol O 2 = 5.01 mol O 2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

12. Practice: 2C 2 H 2 + 5 O 2 → 4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (9.60 mol) (8.95 mol) (4.94 mol)

13. Making Molecules Mass-to-Mass Conversions we know there is a relationship between the mass and number of moles of a chemical 1 mole = Molar Mass in grams the molar mass of the chemicals in the reaction and the balanced chemical equation allow us to convert from the amount of any chemical in the reaction to the amount of any other 13

14. Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.84 g Al 2 O 3 (6.50 x 1 x 2 x 101.84) ÷ (26.92 x 4 x 1) = 12.3 g Al 2 O 3 are formed

15. Mass-to-Mass Conversions In photosynthesis, plants convert carbon dioxide and water into glucose, (C 6 H 12 O 6 ), according to the following reaction. How many grams of glucose can be synthesized from 58.5 g of CO 2 ? Assume there is more than enough water to react with all the CO 2.

16. Another example:  If 10.1 g of Fe are added to a solution of copper (II) sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4 → Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

17. More Making Pancakes we know that 17 BUT what would happen if we had 3 cups of flour, 10 eggs, and 4 tsp of baking powder? 1 cup Flour + 2 Eggs + ½ tsp Baking Powder  5 Pancakes

18. More Making Pancakes 18

19. More Making Pancakes each ingredient could potentially make a different number of pancakes but all the ingredients have to work together! we only have enough flour to make 15 pancakes, so once we make 15 pancakes the flour runs out, no matter how much of the other ingredients we have 19

20. Making more pancakes The flour limits the amount of pancakes we can make. In chemical reactions we call this the limiting reactant. – also known as limiting reagent The maximum number of pancakes we can make depends on this ingredient. In chemical reactions we call this the theoretical yield. – it also determines the amounts of the other ingredients we will use! More on this to come… 20

21. “Limiting” Reagent  The limiting reagent is the reactant you run out of first.  The excess reagent is the one you have left over.  The limiting reagent determines how much product you can make

22. To Identify the Limiting Reagent  Convert the given masses of the reactants to moles  Divide by the lowest mole value  Divide each by the corresponding coefficient in the balanced equation  The smallest remaining value specifies the limiting reagent  Use the moles of limiting reagent to calculate the amount of product(s)

23. Excess Reagent  Calculate the grams of excess reagent by stoichiometry, using the amount of product produced by the limiting reagent  Subtract this value from the starting mass to get the mass left over (think back to the pancake example!)

24. Example  What mass of iodine is produced when 100.00 g KI is added to a solution containing 12.00 g H 2 O 2 and 50.00 g H 2 SO 4 ? What is the limiting reagent?How much of the excess reagents are left over? (Balance equation 1st!) 2KI + H 2 O 2 + H 2 SO 4  I 2 + K 2 SO 4 + 2H 2 O  Limiting: KI  Amount of I 2 produced: 76.44g  Excess H 2 O 2 : 1.753 g, H 2 SO 4 : 20.46 g

25. Another Example  If 10.3 g of aluminum are reacted with 51.7 g of copper (II) sulfate how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 the CuSO 4 is limiting, so Cu = 20.6 g  How much excess reagent will remain? Excess = 4.5 grams Al

26. The Concept of: A little different type of yield than you see on the road.

27. More Making Pancakes Lets now assume that as we are making pancakes we spill some of the batter, some batter remains on the side of the bowl, drop one on the floor or other uncontrollable events happen so that we only make 11 pancakes. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of our pancake- making by calculating the percentage of the maximum number of pancakes we actually make. In chemical reactions we call this the percent yield. 27

28. What is Yield for chemistry?  Yield is the amount of product made in a chemical reaction.  There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield 3. Percent yield = Actual Theoretical x 100

29. Details on Yield  Percent yield tells us how “efficient” a reaction is.  Percent yield can not be bigger than 100 %.  Theoretical yield should always be larger than actual yield! Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring,etc

30. Theoretical and Actual Yield As we did with the pancakes, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make. The theoretical yield will always be the least possible amount of product. – The theoretical yield will always come from the limiting reactant. Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield. 30

31. Finding Limiting Reactant, Theoretical Yield and Percent Yield When 11.5 g of C are allowed to react with 114.5 g of Cu 2 O in the reaction below, 87.4 g of Cu are obtained. Find the Limiting Reactant, Theoretical Yield and Percent Yield. C (s) + 2Cu 2 O (s)  4Cu (s) + CO 2

32. Example  6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4 → Al 2 (SO 4 ) 3 + 3Cu  What is the actual yield?  What is the theoretical yield?  What is the percent yield? = 6.78 g Cu = 13.9 g Cu = 48.8 %