1. 1 Stoichiometry Used to predict how much chemical is needed / made by reactions. Requires: –Molecular Weight –Balanced Equation

2. 2 Stoichiometry – Balanced Eqn. A balanced chemical equation has multiple conversion factors in it: 4 Na (s) + O 2 (g)  2 Na 2 O Use these ratios to relate moles of one chemical to moles of another. Always use moles… “mole superhighway”

3. 3 Stoichiometry – Balanced Eqn. C 6 H 12 O 2 + 8 O 2  6 CO 2 + 6 H 2 O How many moles of O 2 do you need to burn 1 mole of glucose? –Write a mole ratio. How much water do you get from each mole of O 2 ? –Write a mole ratio.

4. 4 Stoichiometry Road Map Moles of B Moles of A  Molar Mass Mass of B Gas Laws  Avogadro’s Number  Molarity Volume of Gas B Number of particles of B Volume of Solution B  Molar Mass Mass of A Gas Laws  Avogadro’s Number  Molarity Volume of Gas A Number of particles of A Volume of Solution A Moles B Moles A

5. 5 Mass calculations. Wrap some (mass  mole) around your (mole  mole): mass A  mole A mole A  mole B mole B  mass B

6. 6 Twinkies… Nutrition Facts Serving Size: 1 cake 43g Amount Per Serving Calories 150Calories from Fat 45 % DV** Total Fat 5g 8% Saturated Fat 2g10% Cholesterol 20mg7% Sodium 200mg8% Total Carbohydrate 25g8% Dietary Fiber 0g0% Sugars 14g Protein 1g2% Vitamin A 0%Vitamin C 0% Calcium 0%Iron 4%

7. 7 Mass Calculation - Twinkies How many grams of water do you get after burning 14.0 grams of glucose? Need: –Balanced Equation (Previous Slide) –What’s Chemical A & Chemical B? (on the roadmap) –MW of glucose. –MW of water –Mole ratio

8. 8 Mass Calculation - Twinkies –What’s A, B? (roadmap) A = glucose, C 6 H 12 O 6 B = water, H 2 O –MW of glucose. 180.16 g/mol –MW of water 18.02 g/mol –Mole ratio 1 mol glu / 6 mol H 2 O

9. 9 % Yield – What is it? Yield compares how much product you got in lab to how much you should have gotten. You Got: –Comes from weighing product in lab. Should have gotten: –Comes from Mass  Moles  Moles  Mass (Stoichiometry)

10. 10 % Yield – What can go wrong? What can go wrong in lab? Loss during transfer. Loss during purification Chemistry doesn’t react 100% Make some of the wrong chemical in the kitchen? Lick the cake bowl. Cookie cutter extras. 1 lb of eggs  1 lb of hard-boiled egg (shell) Burned the edges of cake.

11. 11 % Yield – The Formula Here’s the actual formula:

12. 12 % Yield Example 680 kilograms of magnesium oxide react with water to form 805 kilograms of magnesium hydroxide. Steps: 1.Balanced Equation. 2.Stoichiometry Mass 1.Identify A & B 2.Get molecular weights 3.Get mole Ratio 4.Do the calculation – show all work + units! 3.% Yield Calculation

13. 13 % Yield Example - Balanced Equation Mg 2+, O 2-  MgO Mg 2+, OH -  Mg(OH) 2 MgO + H 2 O  Mg(OH) 2  Mg Balanced?   H Balanced?  O Balanced?

14. 14 % Yield Example - Planning A = MgO, B = Mg(OH) 2 (on stoich. map) M.W. MgOM.W. Mg(OH) 2 Mole Ratio: 1 mol MgO 1 mol Mg(OH) 2 Mg1 × 24.31 =24.31 O2 × 16.00 =32.00 H2 × 1.008 =+ 2.016 58.33 g/mol Mg1 × 24.31 =24.31 O1 × 16.00 =+ 16.00 40.31 g/mol

15. 15 % Yield Example – Stoichiometry II Map: (Mass A) ÷ (MW A) × (mol B/mol A) × (MW B)

16. 16 % Yield Example – Finally an Answer

17. 17 You try one with your team: Show all work on a sheet of paper, turn in at end of class. For the reaction: 3 NO 2 + H 2 O  2 HNO 3 + NO 1.Name each chemical. 2.Calculate the theoretical yield and percentage yield of nitrogen oxide if 49.2 grams of nitrogen dioxide is reacted. 3.Calculate the percentage yield if 8.90 grams of nitrogen oxide is made. Final Answer: 83.2%

18. 18 % Yield as a Conversion Factor Percentage is always something per 100 somethings. Mg(OH) 2 yield was 81.3 %, we could use this as the conversion factor: Use this for homework problems #26 & 29

19. 19 Chains of % Yield Modern real-world chemisty methods often take 8 – 10 steps to complete. 90% yield sounds good but what if you need to do 10 steps? What would you get with a 50% yield? (0.098 g or )

20. 20 Taxol In the late 1950s the National Cancer Institute announced a new program aimed at screening plant extracts for chemotherapeutic activity. As a direct result of this program, extracts from the bark of the pacific yew, Taxus brevifolia, were shown to inhibit tumor growth. In 1969 the most active component of the extract was isolated. Its structure was published in 1971. The compound was named taxol. In 1983 the National Cancer Institute began clinical trials of taxol's safety and effectiveness against various types of cancer. In 1992 the Food and Drug Administration approved the use of taxol for treatment-resistant ovarian cancers, and in 1994 the FDA also approved taxol for recurrent breast cancer chemotherapy. Figure 1 presents the structure of taxol.

21. 21 Taxol Taxol seems to be the answer to many cancer patients' prayers, but there are "deep rooted" ethical issues involved that makes its production a controversial issue. Ecologically, the problem with Taxol is that it is produced from a resource that is rapidly being depleted. It is extracted from the bark of the Pacific yew tree, a slow growing tree native of the pacific north west. It takes an average of six, one hundred year old trees to treat each patient. The scarcity of Taxol and the ecological impact of harvesting it have prompted extension searches for alternative sources including semisynthesis, cellular structure production and chemical synthesis. The later has occurred for almost two decades, but these attempts have been thwarted by the magnitude of synthetic challenge.

22. 22 Taxol Taxol’s first complete synthesis was performed independently by two groups in February, 1994. One group was led by K.C. Nicolaou of the Scripps Research Institute, the other by Robert Holton of Florida State University. The complete synthesis of taxol has eluded chemists for the prior twenty years, since its anti-cancer potential was discovered. Both methods of synthesis take more than 30 steps. This is not considered commercially viable. Yields are reported to be about 0.05%.

23. 23 New Methods: Produced in over 50% yield from a chemical 10- deacetylbaccatin III (from Yew tree needles). At least two pharmaceutical companies now uses similar methodology for the commercial production of taxol from 10-deacetylbaccatin III. A number of companies have developed processes for producing taxol from cell cultures of Taxus brevifolia, Taxus cuspidata, and Taxus canadensis. Taxol

24. 24 Outline Limiting Reagents (Chapter 9) Gases – Properties (Chapter 4) Gases – Laws (Chapter 4) Combined Gas Law (Chapter 4)

25. 25 Limiting Reactants – What is it? Example: Cu + S  Cu 2 S We added lotsa Sulfur & reacted it. Then we added extra sulfur & reacted it. We didn’t weigh the sulfur or worry when it burned away. Copper was the “limiting” reagent.

26. 26 Cl 2 + H 2  2 HCl

27. 27 Limiting Reactants