1. c) Calculations concerning reactions which involve solutions Learning intention Learn how to calculate quantities of reactant or product for reactions involving solutions, using the concentration.

2. Calculations involving concentration n = C x V 1000 n=number of moles C=concentration mol l -1 V=volume in Litres

3. Standard solutions Standard solutions are solutions of known concentrations. They are made in volumetric glassware and can be made from a solid or from another solution.

4. Making a standard solution What mass of solid should be dissolved in 100cm 3 of water to prepare a 0.1 mol l -1 solution of copper(II) sulphate? n = C x V 1000 n = 0.1 x 100 1000 n= 0.01 moles 1 mole  123.5 g 0.01 moles  =(0.01/1)x123.5 =1.235g

5. Making a standard solution weigh out the solid.

6. Making a standard solution Next, dissolve the solid in a small amount of solvent.

7. Making a standard solution Finally, fill the solvent to the line.

8. Aqueous solutions Most solutions are aqueous i.e. the solvent is water When preparing aqueous solutions deionised water (or distilled water) is added why?

9. Calculations involving concentration 0.01 moles How many moles are in 100cm 3 of sodium hydroxide concentration 0.1 mol l -1 ?

10. Neutralisation Reactions/titrations What volume hydrochloric acid 1.0 mol l -1 is needed to neutralise 50 cm 3 of potassium hydroxide solution concentration 0.25 mol l -1 ? 12.5cm 3

11. Calculations for you to try. 1.Calculate the concentration of potassium hydroxide (KOH ) if 14.8 cm 3 is required to neutralise 20cm 3 of 0.1 mol/l nitric acid (HNO 3 ). 2.Calculate the volume of 0.15 mol/l sulphuric acid(H 2 SO 4 ) if it is neutralised by 25 cm 3 of 0.25mol/l sodium hydroxide (NaOH). 0.135 mol/l 20.83 cm 3

12. g) Percentage Yield and Atom Economy Learning intention Learn how the efficiency of a chemical reaction can be measured in terms of percentage yield and atom economy.

13. Making only as much as we need PERCENTAGE YIELD and ATOM ECONOMY

14. Reactants (raw materials) Products Most of the substances we use every day are made from RAW MATERIALS, often through complex chemical reactions. Chemical reactions

15. Pottington Braunton The chemical industry is a multi billion pound international industry producing millions of products vital to our civilisation and well being. Chemical Engineers play a crucial role and are much in demand – there are many opportunities and high levels of pay! Chemical Engineers are much concerned with: % YIELD and ATOM ECONOMY..

16. What is green chemistry? The sustainable design of chemical products and chemical processes. It minimises the use and generation of chemical substances that are hazardous to human health or the environment.

17. Green chemistry principles Better to prevent waste than to treat it or clean it up. Chemical processes should aim to incorporate all reactants in the final product. Chemical processes should aim to use and generate substances with minimal toxicity to human health and the environment.

18. The green chemical industry Modern chemists design reactions with the highest possible atom economy in order to minimise environmental impact. Chemists achieve this by reducing raw material and energy consumption.

19. Percentage yield Historical method for evaluating reaction efficiency. Measures the proportion of the desired product obtained compared to the theoretical maximum. Gives no indication of the quantity of waste produced.

20. % YIELD is the amount of product you actually make as a % of the amount you should theoretically make + SHOULD make this much ACTUALLY make this much REACTANTS PRODUCT % YIELD ABOUT 75%

21. Old fashioned example: Cement from limestone Limekiln

22. CaCO 3 CaO + CO 2 LIMESTONE (calcium carbonate) is used to make QUICKLIME (calcium oxide) for cement making RFM: 100 56 44 So, THEORETICALLY, 100 tonnes of limestone should produce 56 tonnes of quicklime. BUT the ACTUAL YIELD is only 48 tonnes So..the PERCENTAGE YIELD is only 48 x 100 = 87.5% 56 Why? – next slide PERCENTAGE YIELD 1 RAM Ca 40 O 16 C 12

23. Very few chemical reactions have a yield of 100% because: The raw materials (eg limestone) may not be pure Some of the products may be left behind in the apparatus The reaction may not have completely finished Some reactants may give some unexpected products Some reactions involve an equilibrium Careful planning and design of the equipment and reaction conditions can help keep % yield high PERCENTAGE YIELD 2

24. Atom economy In an ideal reaction, all reactant atoms end up within the useful product molecule. No waste is produced! Inefficient, wasteful reactions have low atom economy. Efficient processes have high atom economy and are important for sustainable development. They conserve natural resources and create less waste.

25. Atom economy A measure of the proportion of reactant included in the final useful product. A reaction may have a high percentage yield but a low percentage atom economy, or vice versa.

26. High atom economy All reactant atoms included in the desired product.

27. Low atom economy Some reactant atoms not included in the desired product.

28. CaCO 3 → CaO + CO 2 2Mg + O 2 → 2MgO USEFUL PRODUCT (antacids, fire resistant coatings, electrical insulators) USEFUL PRODUCT (cement, glass, agriculture etc) Atom Economy Compare these two industrial reactions What do you notice about each one? Think raw materials, useful products, waste products

29. REACTANTS + PRODUCTS ATOM ECONOMY is the mass of the product you want as a % of the mass of all the products you make Stuff you want Stuff you also get but don’t want ATOM ECONOMY about 50%

30. CALCULATING ATOM ECONOMY Often, chemical reactions produce unwanted products along with the product you want. CaCO 3 → CaO + CO 2 RFM: 100 56 44 Useful product Waste product ATOM ECONOMY mass useful product mass of all reactants =X 100% Atom Economy = 56 / (56 + 44) = 56 / 100 = 56 % ATOM ECONOMY is the mass of product you want as a % of the mass of all the reactants

31. Fe 2 O 3 + 3CO → 2Fe + 3CO 2 160 84 112 132 Atom Economy = 80 / 80 x 100% = 100 % (obviously) ATOM ECONOMY is the mass of product you want as a % of the mass of all the reactants 2Mg + O 2 → 2MgO RFM: 48 32 80 RAM Mg 24 Fe 56 C 12 O 16 Atom Economy = 112 / 244 x 100% = 45.9 %

32. Find the atom economy for these 2 methods of extracting copper: 1. Heat copper oxide with carbon 2. Heat copper sulphide with oxygen 2CuO + C → 2Cu + CO 2 CuS + O 2 → Cu + SO 2 RAM Cu 64, O 16, C 12, S 32 2(80) 12 128 44 96 32 64 64 Mass of all reactants 160 + 12 = 172 Mass of all reactants = 96 + 32 = 128 ATOM ECONOMY 128 172 X 100 = ATOM ECONOMY 64 128 X 100 = = 74.4 %= 50 % Mass of copper = 128 Mass of copper = 64

33. Real example: Paracetamol The non-prescription analgesic market (paracetamol, aspirin, ibuprofen) is worth about £21 billion annually. Maximising % yield and atom economy in the reactions at left is vital to save money and conserve energy and resources

34. ATOM ECONOMY is the mass of the product you want as a % of the mass of all the products you make Stuff you want Stuff you also get but don’t want ATOM ECONOMY about 50% % YIELD is the amount of product you actually make as a % of the amount you should theoretically make SHOULD make this much ACTUALLY make this much % YIELD ABOUT 75% SUMMARY

35. Choosing the most EFFICIENT REACTION to make the product to… Using the most EFFICIENT REACTION CONDITIONS & APPARATUS to… Raw materials are scarce and expensive and so must be carefully conserved. Also, chemical processes need to produce as little waste as possible, minimise costs, energy use and pollution. CHEMICAL ENGINEERS must plan to maximise: PERCENTAGE YIELD by… ATOM ECONOMY by… Reduce energy use, costs and conserve raw materials Reduce waste and pollution

36. Example 1 What is the percentage atom economy for the following reaction for making hydrogen by reacting coal with steam? C (s) + 2H 2 O (g) → CO 2 (g) + 2H 2 (g) 12 g 2(2 + 16) g [12 + (2 × 16)] g 2(2 × 1) g 12 g 36 g 44 g 4 g Total mass of reactants Mass of desired product = 12 + 36 = 48 g = 4 g

37. Example 1 (contd) % atom economy = mass of desired product × 100 total mass of reactants = 4 × 100 48 = 8.3% This reaction route has a very low atom economy and is an inefficient method of producing hydrogen.

38. Example 2 Calculate the percentage atom economy for the reaction below. C 6 H 12 C 6 H 12 Total mass of reactants Mass of desired product = [(6 × 12) + (12 × 1)] = 84 g

39. Example 2 (contd) % atom economy = mass of desired product × 100 total mass of reactants = 84 × 100 84 = 100% This reaction route has a very high atom economy as all reactant atoms are incorporated into the desired product.

40. Example 3 Hydrazine (N 2 H 4 ) is used for rocket fuel. Calculate the atom economy for hydrazine production. Total mass of reactants Mass of desired product = 34 + 74.5 = 108.5 g = 32 g NH 3 2 mol 34 g NaOCl 1 mol 74.5 g N 2 H 4 1 mol 32 g NaCl 1 mol 58 g H 2 O 1 mol 18 g

41. Example 3 (contd) % atom economy = mass of desired product × 100 total mass of reactants = 32 × 100 108.5 = 30% This reaction route has an atom economy of 30%. The remaining 70% is waste product (NaCl and H 2 O).

42. Catalysts Have a crucial role in improving atom economy. Allow the development of new reactions requiring fewer starting materials and producing fewer waste products. Can be recovered and re-used. Allow reactions to run at lower temperatures, cutting energy requirements.

43. Methanol and atom economy

44. Methanol Methanol is a potential fuel of the future It can be made from methane which in turn could be made by anaerobic fermentation of waste organic material Laptop with methanol fuel cell green technology

45. Methanol It is added to other fuels, it is the fuel for drag car racing and it is being widely used in fuel cells to power vehicles as it has not been easy to set up the infrastructure needed for hydrogen based fuel cells

46. Carbon neutral? If methane is combusted in a limited supply of oxygen, carbon monoxide forms CH 4 + ½ O 2 → CO + 2H 2 The products are then passed over a catalyst at 50-100 atmospheres and 250 o C CO + 2H 2 → CH 3 OH

47. Carbon neutral? CO + 2H 2 → CH 3 OH 28 4 32 Atom economy = mass of desired products x 100 Total mass of reactants = 32 x 100 = 100% 16 + 16

48. Carbon neutral? All the carbon in the methane is converted into carbon in the methanol – it is said to be carbon neutral The carbon dioxide released when methanol burns contains the same amount of carbon that was reclaimed from the CO The amount of carbon dioxide released when the methanol burns is the same as the amount the methane would have released if it was burned completely. No additional carbon dioxide results from the process

49. Carbon neutral? What were the conditions needed for step 1? How were they achieved? What were the conditions needed for step 2? How would these have been achieved? In your opinion, how true is the claim that the process is carbon neutral? Limited supply of oxygen High pressure and temperature – ie energy costs (where will this energy come from?

50. Green Chemistry Look at the 12 principles of Green Chemistry How many of them apply to this process? The catalysts are zinc oxide, aluminium oxide or copper.

51. % yield

52. Percentage yield = actual yield/theoretical yield x 100%

53. Preparation of zinc sulphate Zinc sulphate is used as a paste, mixed with zinc oxide to treat acne. Aim to produce zinc sulphate and calculate the % yield

54. Preparation of zinc sulphate It is an astringent – it closes the pores of the skin to keep out bacteria This product contains peroxide, zinc sulphate, tea tree oil and sage

55. Preparation of zinc sulphate It can also be used to prevent sunburn!

56. What to do Add a weighing boat to the balance and zero it Add between 2.2 – 2.5 g zinc oxide Record the accurate mass Measure out 25cm 3 1.0 mol l -1 H 2 SO 4 Heat the acid to about 50 o C Add the zinc oxide, stirring constantly

57. What to do next Cool the reaction mixture Weigh an evaporating dish and record the mass accurately Weigh a filter paper, fold the filter paper, place in a funnel and filter the zinc sulphate solution into the dish Heat to nearly dry, then leave to fully evaporate and cool. Reweigh the dish and product

58. Calculation % Yield Write the balanced equation Which reactant is in excess and which one limits the amount of product formed? How much product is expected (using the limiting reactant in the calculation) Calculate the yield = actual mass x 100 expected mass

59. Preparation of sodium citrate % yield Sodium citrate is a food additive (E331) e.g. in pepsi

60. What to do Weigh a beaker accurately Add approx 2g citric acid to the beaker and reweigh Add 25cm 3 1.0 mol l -1 NaOH and stir until all the crystals have dissolved Evaporate the beaker almost to dryness then leave to dry completely Reweigh the beaker

61. For the calculation 3 mol NaOH reacts with 1 mol citric acid Na + 3NaOH ↔ + 3H 2 O

62. The calculation 1. 3NaOH + C 6 H 8 O 6 → CH 5 O 7 Na 3 + 3H 2 O 2. Work out the theoretical yield. 3. Use the actual yield from your results and the theoretical yield to calculate the percentage yield.