Higher Chemistry The Chemistry of Making Money - Challenge NEW LEARNING Calculating the volume of gases. Excess reactants. Percentage Yield and Atom Economy. REVISION Moles, mass and concentration. Calculations from equations.

Starter Questions What mass would 5 moles of carbon dioxide have? S3 Revision

Starter Questions How many moles are in 150g of calcium carbonate, CaCO3? S3 Revision

Lesson 1: Chemical Industry – The Challenge Today we will learn to Use chemistry to predict how feasible a profit is in a chemical process. We will do this by Learning about the chemical industry in the UK and considering the factors affecting profitability. We will have succeeded if We can explain the relevance of basic chemical calculations to profitability.

Chemical Industry in Scotland Basic chemicals: The basic chemicals sector refers to the production of basic organics/inorganics, fertilisers, industrial gases, plastic, rubber and man-made fibres. Speciality chemicals: The speciality chemicals sector refers to the manufacture of products such as specialised organics & inorganics; paints, varnishes, inks, coatings & mastics and soaps & detergents.  

Chemical Industry in Scotland Pharmaceuticals: The pharmaceuticals sector refers to the manufacture of basic pharmaceutical products and pharmaceutical preparations. Animal health products and cell cultures and reagents are also manufactured in Scotland. Consumer Chemicals: The consumer chemicals sector refers to the manufacture of products such as perfumes and toiletries.  

The Challenge in Unit 3 ¨ What is your product and what will it be used for? ¨ What are the reactants? ¨ How much product is produced per kilogram or litre of reactant? ¨ What is your predicted % yield and how green is your atom economy? ¨ How will you ensure a fast rate of reaction and a high yield? ¨ How much energy is required for your reaction, and will your process be energy efficient? ¨ How will you ensure your product is pure and free from contaminants? ¨ How can we be sure of this? ¨ What factors have influenced your choice of industrial process for your product?   The challenge is for your team to make a pitch to the “Den”, this can be in the form of a PowerPoint or other type of presentation. Your pitch is to try and raise funds for the starting up of a chemical plant. You will be given lots of examples of chemical reactions that produce useful products in this booklet, you can choose any of these, or you can do some research and choose one of your own.  

Raw Materials and Feedstocks A feedstock is a chemical from which other chemicals are manufactured. Feedstocks are made from raw materials; the basic resources that the earth supplies to us. They are: Fossil fuels–coal, oil and natural gas metallic ores–e.g. aluminium extracted from bauxite(Al2O3) minerals–chlorine from sodium chloride water and air – water in hydration of ethene to ethanol and nitrogen in the Haber Process, oxygen in the catalytic oxidation of ammonia organic materials–of plant and animal or vegetable oils and starch Crude oil is a raw material from which naphtha is obtained by fractional distillation. Naphtha is a feedstock that can be cracked to produce ethene.

Batch and Continuous Processes In a batch process the chemicals are loaded into the reaction vessel. The reaction is monitored and at the end of the reaction the product is separated and the reaction vessel cleaned out ready for the next batch. In a continuous process the reactants are continuously loaded at one end of the reaction vessel and the products are removed at the other end. Each process has advantages and disadvantages.

Batch and Continuous Processes

How Much Product? We know that each atom of a different element has a different mass; in order to compare substances in chemistry we cannot just compare masses. This is because 12g of helium will contain three times the number of atoms as 12g of carbon, as helium atoms are three times lighter than carbon atoms. Its a bit like comparing ping pong balls with bowling balls, we know that the same number of ping pong balls and bowling balls would weigh completely different masses. We therefore need to use the number of atoms or molecules rather than their mass. However as atoms are so small we use the number of moles, where one mole of a substance is 602,000,000,000,000,000,000,000 or 6.02 x 1023 particles of that substance, this number is fixed, just like a dozen always equals 12. It is known as Avogadro’s constant or number and often given the symbol L.

How Much Product? 1mole  18g 2moles  (18 x 2) ÷ 1 = 36g For example the formula mass of H2O is 18a.m.u. So the mass of one mole will be 18g. If we had 36g of water we would have 2 moles of water molecules. 1mole  18g 2moles  (18 x 2) ÷ 1 = 36g This leads to the following expression: number of moles = mass given GFM n = m

ConsolidationTask Complete Quick Test 1 using one of the methods on the previous slide. You will have 15 minutes total. S3 Revision

Starter Task Complete Quick Test 1 using one of the methods on the previous slide. You will have 15 minutes total. S3 Revision

Quick Test Answers Quick Test 1 1a. 0.2 b. 0.01 c. 0.1 d. 0.009 2a. 413g b. 85g c. 2.22g d. 0.6g S3 Revision

Lesson 2: Calculations from Balanced Equations Today we will learn to Use balanced equations to predict the quantity of product available. We will do this by Revising Nat 5 calculations and practicing! We will have succeeded if We can use a worded question to produce a balanced equation and calculate a mass of product feasible.

Balanced Equations. p7 In order to work out how much product you can produce in your reaction it is important that you can write out a balanced chemical equation for it. This gives you the proportions in terms of moles of reactants to products. If you know how much reactant you start with, you can use the balanced equation’s proportions to work out how much product you can produce. Example 1. Industrial ethanol, used as a solvent in the pharmaceutical industry, can be made by the catalytic hydration of ethene. If we started with 56g of ethene, how much ethanol could we make?

Balanced Equations. p8 Example 2 Ammonia (nitrogen hydride) is used to make fertilisers and explosives, it is made by reacting nitrogen with hydrogen using an iron catalyst. How much ammonia is made when 120 tonnes of hydrogen react with the correct amount of nitrogen?

ConsolidationTask Complete Quick Test 2 using the pattern of the worked examples you have been given. We will do one at a time using the whiteboards. S3 Revision

Quick Test Answers Quick Test 2 1. 88g 72g 817kg 140kg 8.6x106kg S3 Revision

Helpful Tips Write 5 top tips or golden rules about the topic for students taking the lesson next year.

Starter Task S3 Revision

Starter Task S3 Revision

Lesson 3: Reactions Involving Gases Today we will learn to Use MOLAR VOLUME to simplify calculations. We will do this by Doing an experiment to measure the molar volume of a gas and practicing calculations. We will have succeeded if We can define and use the molar volume of a gas in chemical calculations.

Molar Volume Molar Volume The molar volume is the volume which one mole of a gas occupies. Worked Example 1 The volume of 8g of oxygen is 5.5 litres. Calculate the volume of 3 mol of oxygen. gfm O2 = 2 x 16 = 32g 8g  5.5 96g  5.5 x 96 8 = 66 litres

Molar Volume

Molar Volume 23 litres  1mole 46 litres  1 x 46 = 2moles 23

Molar Volume Example 1 Sulfur dioxide is an important gas in the manufacture of sulfuric acid. It is made by reacting sulfur with oxygen. If 64g of sulfur is reacted with an excess of oxygen what volume of sulfur dioxide would be produced, assuming the molar volume is 23 litres. S(s) + O2(g) à SO2(g) 1 mole à 1 mole 32g à 23L 1g à 23 32 64g à 23x64 = 460 litres

Molar Volume Try the questions (12 and 13) on p135/136 of the textbook.

ConsolidationTask Complete Quick Test 3 using the pattern of the examples you have been given. S3 Revision

Quick Test Answers Quick Test 3 1. 0.28g 2. 3.4L 3. 1.2L S3 Revision

Reactions with JUST gases If a reaction just involves gases then it is possible to just compare the reaction ratios and volumes without calculating the actual numbers of moles. Example 1: What is the volume and composition of the resultant gas mixture when 10 litres of sulfur dioxide reacts with 50 litres of oxygen gas to form sulfur trioxide? 2SO2(g) + O2(g)  2SO3(g) Reaction ratio: 2 : 1 : 2 Start 10 50 0 Used/made 10 5 10 (use ratio to work it out) End 0 45 10 (what you end up with) Resultant gas composition and volume = 45 litres of unreacted O2(g) and 10 litres of SO3(g)

Reactions with JUST gases

ExitTask Complete Quick Test 4 using the pattern of the examples you have been given. S3 Revision

Quick Test Answers Quick Test 4 450cm3 O2 & 400 cm3 50 cm3 3. x = 3, y = 6 S3 Revision

Starter Task How would you prepare a 500cm3 solution of sodium chloride with a concentration of 1 mol l-1? S3 Revision

Starter Task  1000 cm3 of NaCl solution would need to contain 1 mole So 500cm3 would need to contain? How many grams is that? S3 Revision Once I’ve weighed out my mass of NaCl, what do I do, and how do I do it? 

Lesson 4: Calculations with Solutions Today we will learn to Use mass, moles and concentration combined in a calculation. We will do this by Trying some examples and practicing our problem solving techniques individually. We will have succeeded if We can solve chemistry problems involving more than one proportion/calculation.

Calculations With Solutions Reactions involving solutions: If a reaction involves solutions then it is possible to calculate the number of moles of solute in the solution and use this to calculate the amount of product produced. Number of moles of solute = concentration x volume in litres n = cv

Calculations With Solutions Example 1: What mass of carbon dioxide gas is produced when 250cm3 of a 0.5moll-1 solution of hydrochloric acid is completely neutralised by calcium carbonate powder? 2HCl(aq) + CaCO3(s) à CaCl2(aq) + CO2g) + H2O(l) 2 1 n = cv = 0.5 x 0.25 (note volume is in litres) = 0.125 n = ½x 0.125 = 0.0625 m = n x GFM = 0.0625 x 44 = 2.75g

Calculations With Solutions Example 2: What volume of hydrogen gas is produced when 100 litres of a 2moll-1 solution of sulfuric acid is completely neutralised by magnesium powder? Assume the molar volume is 23 litres.   H2SO4(aq) + Mg(s) àMgSO4(aq) + H2g) 1 1 n = cv = 2 x 100 (note volume is in litres) = 200 n = 200   volume= n x mol vol = 200 x 23 = 4600 litres

Calculations With Solutions Example 2: What volume of hydrogen gas is produced when 100 litres of a 2moll-1 solution of sulfuric acid is completely neutralised by magnesium powder? Assume the molar volume is 23 litres.   H2SO4(aq) + Mg(s) àMgSO4(aq) + H2g) 1 1 n = cv = 2 x 100 (note volume is in litres) = 200 n = 200   volume= n x mol vol = 200 x 23 = 4600 litres

ConsolidationTask Complete Quick Test 5 using the pattern of the examples you have been given. S3 Revision

Quick Test Answers Quick Test 5 0.5moll-1 5cm3 0.83moll-1 S3 Revision

Starter Task S3 Revision

Starter Task S3 Revision

Lesson 5: % Yield and Atom Economy Today we will learn to Calculate how efficient a process is using different methods. We will do this by Learning about the differences between % yield and atom economy. We will have succeeded if We can solve chemistry problems involving more than one proportion/calculation.

% yield and Atom Economy S3 Revision

% yield and Atom Economy S3 Revision

Atom Economy S3 Revision

Atom Economy S3 Revision

Judging Efficiency 1. Read through pages 16 and 17 of the pupil booklet, then use the whiteboards to write a sentence explaining the difference between % yield and atom economy. 2. Explain why atom economy is a better measure of efficiency that % yield alone. 3. Use your two sentences to write a note on judging the efficiency of a chemical process. S3 Revision

StarterTask Complete Quick Test 6 using the pattern of the examples you have been given. S3 Revision

Quick Test Answers S3 Revision

Lesson 6: Reactants in Excess Today we will learn to Calculate how much reactant is in excess. We will do this by Doing (yet more!) calculations from equations. We will have succeeded if We can work out which reactant is in excess, and by how much.

Reactants in Excess S3 Revision

Reactants in Excess S3 Revision

ConsolidationTask Complete Quick Test 7 using the pattern of the examples you have been given. We will complete Q 1 together, then try 2 and 3 individually. S3 Revision

Quick Test Answers Quick Test 7 1a. 77% b. 92.5% 2ai. 78% aii. 22.7% b. All the reactant atoms are converted into products so there is no waste and a 100% atom economy. 3a. 83% b. 52.6% S3 Revision

Objective Traffic Lights How do you feel about the lesson objectives? Red = don’t think I have grasped this Amber = feeling OK about this, have just about got there Green = Confident I have achieved this If there are no coloured cards available (e.g. in planners), Mark corners in the room for each colour and ask pupils to move to indicate confidence.