1. Plan for Wed, 1 Oct 08 Lecture –Limiting Reactants and Percent Yield (3.10) –Molecular vs. Ionic compounds. What happens when they dissolve in water? (4.1-2) –…undergo a phase change? (1.9) –Composition of solutions (4.3) Also read –4.5-10

2. 2 Al (s) + 3 CuCl 2(aq)  3 Cu (s) + 2 AlCl 3(aq) Stoichiometry Concept Quiz In which case will all the Al AND all the CuCl 2 be used up? A. 1:1 mass ratio (1 gram Al and 1 gram CuCl 2 ) B. 1:1 mole ratio (1 mol Al and 1 mol CuCl 2 ) C. 2:3 mass ratio (2 grams Al and 3 grams CuCl 2 ) D. 2:3 mole ratio (2 mol Al and 3 mol CuCl 2 )

3. Limiting Reactants Given a balanced chemical equation, the amount of reactants available to react determines the amount of products we can produce. 1 ¾ cup cake flour 3 tsp baking powder 2 oz. baking chocolate 1 ½ cup sugar ½ cup butter 4 eggs ½ cup milk 1 ¾ cup cake flour 3 tsp baking powder 2 oz. baking chocolate 1 ½ cup sugar ½ cup butter 2 eggs ½ cup milk one whole chocolate cake one half of a chocolate cake Having only 2 eggs available limits the amount of cake we can make, even though we have plenty of the other ingredients.

4. Limiting Reactants (cont) The same kind of thing happens in chemical reactions… …because we have to count particles by weighing, it is rare to measure the exact number of particles of each reactant required in a chemical equation. More often than not one reactant is completely used up before the other reactant(s) are.

5. Limiting Reactants (cont) Consider the combination of solid carbon with oxygen: C(s) + O 2 (g)  CO 2 (g)

6. LRs in Action CO(g) + 2H 2 (g)  CH 3 OH( l ) Initial 1 mol2 mol0 mol Change - 1 mol- 2 mol+ 1 mol End 0 mol 1 mol Both reactants are completely used up…neither CO or H 2 limit the production of CH 3 OH.

7. LRs in Action (cont) CO(g) + 2H 2 (g)  CH 3 OH( l ) Initial 0.5 mol2 mol0 mol Change - 0.5 mol- 1 mol+ 0.5 mol End 0 mol1 mol0.5 mol CO is used up before H 2 … CO limits the production of CH 3 OH.

8. LRs in Action (cont) CO(g) + 2H 2 (g)  CH 3 OH( l ) Initial 1 mol 0 mol Change - 0.5 mol- 1 mol+ 0.5 mol End 0.5 mol0 mol0.5 mol H 2 is used up before CO… H 2 limits the production of CH 3 OH.

9. LRs in Action (cont) Chemistry happens at the particulate level… …so the limiting reactant in a chemical equation must be determined by comparing numbers of moles. In the last example, we were given moles of reactants, so we could determine the LR by inspection. In the lab, you will measure reactant amounts in terms of g, mg, kg, etc. These measurements will need to be converted to moles.

10. Example Suppose 25.0 kg of nitrogen gas and 5.00 kg of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion. N 2 (g) + H 2 (g)  NH 3 (g) 3 2 25.0 kg 5.00 kg ? kg

11. Use mole ratios to determine LR Example (cont) N 2 (g) + 3H 2 (g)  2NH 3 (g) mol N 2 MM mol H 2 MM mol LRmol NH 3 mol LR MM 25.0 kg 5.00 kg ? kg

12. N 2 (g) + 3H 2 (g)  2NH 3 (g) 25.0 kg 5.00 kg ? mol N 2 = 25.0 kg N 2 1 kg N 2 1000 g N 2 28.01 g N 2 1 mol N 2 = 892.54 mol N 2 ? mol H 2 = 5.00 kg H 2 1 kg H 2 1000 g H 2 2.016 g H 2 1 mol H 2 = 2480.16 mol H 2 To determine which of this reactants is limiting, ask the following question: “How many moles of N 2 would we need to react completely with 2480.16 mol of H 2 ??” ? kg

13. N 2 (g) + 3H 2 (g)  2NH 3 (g) 25.0 kg 892.5 mol 5.00 kg 2480.2 mol ? mol N 2 = 2480.2 mol H 2 3 mol H 2 1 mol N 2 = 826.73 mol N 2 “How many moles of N 2 would we need to react completely with 2480.16 mol of H 2 ??” We have 892 mol N 2, but we only need 827 mol N 2 to react completely with 2480 mol of H 2. H 2 IS THE LIMITING REAGENT. ? kg

14. N 2 (g) + 3H 2 (g)  2NH 3 (g) 25.0 kg 892.5 mol 5.00 kg 2480.2 mol ? mol H 2 = 892.5 mol N 2 1 mol N 2 3 mol H 2 = 2677.5 mol H 2 What if instead we asked: “How many moles of H 2 would we need to react completely with 892.5 mol of N 2 ??” We need 2680 mol H 2 to react completely with 892 mol of N 2, but we only have 2480 mol H 2. H 2 IS THE LIMITING REAGENT. ? kg

15. N 2 (g) + 3H 2 (g)  2NH 3 (g) 25.0 kg 892.5 mol 5.00 kg 2480.2 mol ? kg ? kg NH 3 = 2480.2 mol H 2 3 mol H 2 2 mol NH 3 1 mol NH 3 17.03 g NH 3 = 28.1585 kg NH 3 LIMITING REAGENT 1000 g NH 3 1 kg NH 3 = 28.2 kg NH 3

16. Making the LR work for YOU. To determine which reactant is the LR, we must compare the following quantities… –what we have (based on the problem set up) –what we need (based on the balanced chemical equation) The mole ratios in a chemical equation are crucial for determining the LR… The reactant for which we have fewer moles is not necessarily the LR!!

17. Theoretical vs. Actual Yield The theoretical yield of a reaction is the amount of product that would be formed under ideal reaction conditions in which starting materials are completely consumed (up to the LR). This is a calculated number. The actual yield is the amount of product that is actually produced in real life (in the lab). The actual yield is always less than the theoretical yield because… –starting materials may not be completely consumed –side reactions may occur –the reverse reaction may occur –there may be loss of product during purification steps

18. Percent Yield A comparison of… how much product we actually produced to how much product we could theoretically produce Gives us the percent yield of a chemical reaction. % yield = actual yield theoretical yield 100

19. N 2 (g) + 3H 2 (g)  2NH 3 (g) 25.0 kg 892.5 mol 5.00 kg 2480.2 mol 28.2 kg LIMITING REAGENT Theoretical Yield In a certain experiment, the actual yield of this reaction was only 26.7 kg NH 3. What was the percent yield? % yield = 26.7 kg NH 3 28.2 kg NH 3 100 = 94.68% Percent yield should never be greater than 100%. If it is, there is something wrong with your theoretical yield calculation, or with the actual yield you obtained experimentally. = 94.7%

20. Tips for Stoichiometric Success Make sure your chemical equation is balanced. Always compare moles, not mass. Use mole ratios from your balanced chemical equation to determine the limiting reactant. Remember… –theoretical yield is a calculated quantity. –actual yield is a measured quantity. If your percent yield is greater than 100%... –something is wrong with the theoretical yield you calculated, or… –something is wrong with the actual yield you measured in your experiment.

21. QUESTION If two compounds are about to react, which statement about the reaction states an accurate observation? 1.If temperature conditions can be kept optimal, both compounds will fully react with no excess. 2.The reactant present with the fewest grams will be the limiting reactant. 3.The reactant present with the fewest moles will be the limiting reactant. 4.None of the statements are accurate. A limiting reactant can be determined only when both the moles of the two compounds AND the mole ratio by which the compounds react are known.

22. Ch 4 – Chemical Rxns and Solution Stoichiometry –Molecular vs. Ionic compounds. What happens when they dissolve in water? (4.1-2) –Composition of solutions (4.3) Also read… –4.5-10

23. Ionic vs. Molecular Compounds Let’s compare sodium chloride and water. Solid sodium chloride Solid water What are the microscopic particles in the samples of these two compounds?

24. Molecular Solid: H 2 O Covalent O-H bond (need ~100-1000 kJ/mol to break) Intermolecular attraction (need ~10-40 kJ/mol to break) Molecular solids consist of discrete molecules or atoms held together by intermolecular forces.

25. Ionic Solid: NaCl Formula Unit: NaCl Ionic solids consist of arrays of anions and cations held together in a net of mutual electrostatic attraction. Ionic bond: ~100-1000 kJ/mol needed to break bond

26. What about ionic compounds with polyatomic ions? Copper sulfate, CuSO 4 O S O O O S O O – O O – Cu 2+

27. Ionic compounds in solution Polar bond

28. Molecular compounds in solution Some molecular compounds have polar bonds like water does. These compounds will be somewhat soluble in water.

29. Fig. 16-8, p. 464 Anti-freeze is soluble in water.Motor oil is not soluble in water. No polar bonds to attract water

30. Electrical Conductivity of Aqueous Solutions Charged species in solution conduct electricity…these are called electrolytes. Electrolytes are either strong or weak –Strong: soluble salts (NaCl, KNO 3, etc), strong acids (HCl, H 2 SO 4, etc) –Weak: insoluble salts (BaSO 4, Mg(OH) 2, etc), weak acids (CH 3 COOH, HF, etc) Solutions of non- electrolytes (like sugar in water) will not conduct electricity StrongWeakNon