Higher Chemistry Unit 3 – Chemistry in Society Section 13 – Percentage Yield and Atom Economy

I can carry out percentage yield calculations 3.13 Percentage Yield and Atom Economy Learn how chemists calculate how much reactant is required and how much product this will produce Learning Outcomes/Success Criteria Unit 3 CfE Higher Chemistry I can explain what is meant by percentage yield in terms of reactants converted into products I can carry out percentage yield calculations I can carry out atom economy calculations I can comment on the route for making a chemical based on the percentage yield and atom economy

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. 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.

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

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 the raw material used and energy consumption.

Percentage Yield Measures the proportion of the desired product obtained compared to the theoretical maximum. Gives no indication of the quantity of waste produced.

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

CaCO3 CaO + CO2 PERCENTAGE YIELD RFM: 100 56 44 LIMESTONE (calcium carbonate) is used to make QUICKLIME (calcium oxide) for cement making RAM Ca 40 O 16 C 12 CaCO3 CaO + CO2 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

Old fashioned example: Cement from limestone Limekiln

Very few chemical reactions have a yield of 100% because: Percentage Yield 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 Worked Example The yield in a chemical reaction is the quantity of product obtained. The actual yield can be compared, as a percentage, with the theoretical. 5g of methanol reacts with excess ethanoic acid to produce 9.6g of methyl ethanoate. Calculate the percentage yield. Step 1: determine the theoretical yield (the quantity expected from the balanced equation) CH3OH + CH3COOH  CH3OOCCH3 1 mol 1 mol 32g 74g 5g 5 x 74 32 Theoretical Yield = 11.56g

Step 2: The actual yield is always given in the question Actual Yield = 9.6g Step 3: Percentage Yield = 9.6 x 100 11.56 = 83%

Examples to Try Ammonia is manufactured from hydrogen and nitrogen by the Haber process. 3H2(g) +N2(g) 2NH3(g) If 80kg of ammonia is produced from 60kg of hydrogen, what is the percentage yield? 2. What is the theoretical yield of AlCl3that the reaction can produce when we start with 4.25 g of Cl2our limiting reagent?

Atom Economy Atom economy is a measure of the proportion of reactant atoms which are incorporated into the desired product of a chemical reaction. Calculation of atom economy therefore also gives an indication of the proportion of reactant atoms forming waste products. In developing an atom economical reaction pathway the industrial chemist may well prefer rearrangement and addition reactions over less environmental friendly substitution and elimination reactions.

Example 1: Addition reaction – Halogenation of an alkene Total mass of reactants = 56g + 159.8g = 215.8g (Note: Product mass is also 215.8g)

Mass of desired product (2,3-dibromobutane) = 215.8g % atom economy = 215.8 x 100 = 100% 215.8

Example 2: Elimination Reaction

Total mass of reactants = 161g + 74 = 235g (Note: Total product mass = 235g) Mass of desired product ethylene oxide = 88g % atom economy = 88 x 100 = 37.4% 235 This elimination reaction is therefore only 37.4% atom efficient, with the remaining 62.6% in the form of unwanted waste products (calcium chloride and water).