Chemical Synthesis Lesson 6.

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Presentation transcript:

Chemical Synthesis Lesson 6

Learning objective: To predict theoretical yields and calculate percentage yield. Must: Know the formula for calculating the percentage yield. Grade C Should: Know the steps for predicting the masses of reactants and products. Grade C Could: Be able to calculate the masses of reactants and products used. Grade A Keywords: Reactant, product, ratio, theoretical, percentage and yield. Starter: Calculate the relative formula mass for carbon dioxide (CO2)? (RAM C=12, O=16)

Key skills Numeracy Select the mathematical information to use. Use appropriate mathematical procedures. Find results and solutions. Self managers Work towards goals, showing initiative, commitment and perserverance.

Relative formula mass Relative formula mass

Percentage yield. To calculate the percentage yield use the following formula: Percentage yield = actual yield x 100 theoretical yield Theoretical yield – is the maximum mass of product that could be made. Actual yield – is the mass of product collected.

Calculating percentage yield. In an experiment 32g of CO2 are produced, the theoretical yield was 40g. Calculate the percentage yield. Tip – the answer must always be less than 100%. If it is more than 100% then you have got the calculation the wrong way round.

How did you do? Percentage yield = actual yield x 100 theoretical yield Percentage yield = 32 x 100 40 = 75% Don’t forget the units! Answer question 3 p.225 Additional textbook.

Percentage yield Percentage yield = Actual yield Theoretical yield X 100% Some example questions: The theoretical yield of an experiment to make salt was 10g. If 7g was made what is the percentage yield? Dave is trying to make water. If he predicts to make 15g but only makes 2g what is the percentage yield? Sarah performs an experiment and has a percentage yield of 30%. If she made 50g what was she predicted to make?

Equation calculations The unknown mass of a reactant or product can be worked out by using the known masses of the other reactants and products Equation calculations

Therefore chemical equations must be balanced Chemical equations use the chemical formulae of the reactants and products to describe a reaction The total mass of the product(s) of a chemical reaction must equal the total mass of the reactant(s) Therefore chemical equations must be balanced The total number of atoms of one element should be the same on both sides of the equation Chemical equations

Balancing equations Balancing equations

Calculating the masses of reactants and products. Step 1 – Write a balanced symbol equation. Step 2 – Calculate the RFM for each reactant and product. Step 3 – Work out ratio for each reactant and product.

Ratios. If 1 loaf of bread can make 15 sandwiches. How many sandwiches would; a) 2 loaves make? b) 6 loaves make? c) 1/3 of a loaf make? Practise ratios here.

Calculating the mass of a product E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in air? IGNORE the oxygen in step 2 – the question doesn’t ask for it Step 1: READ the equation: 2Mg + O2 2MgO Step 2: WORK OUT the relative formula masses (Mr): 2Mg = 2 x 24 = 48 2MgO = 2 x (24+16) = 80 Step 3: LEARN and APPLY the following 3 points: 48g of Mg makes 80g of MgO 1g of Mg makes 80/48 = 1.66g of MgO 60g of Mg makes 1.66 x 60 = 100g of MgO

When water is electrolysed it breaks down into hydrogen and oxygen: 2H2O 2H2 + O2 What mass of hydrogen is produced by the electrolysis of 6g of water? Work out Mr: 2H2O = 2 x ((2x1)+16) = 36 2H2 = 2x2 = 4 36g of water produces 4g of hydrogen So 1g of water produces 4/36 = 0.11g of hydrogen 6g of water will produce (4/36) x 6 = 0.66g of hydrogen 2) What mass of calcium oxide is produced when 10g of calcium burns? 2Ca + O2 2CaO Mr: 2Ca = 2x40 = 80 2CaO = 2 x (40+16) = 112 80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO 3) What mass of aluminium is produced from 100g of aluminium oxide? 2Al2O3 4Al + 3O2 Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204 4Al = 4x27 = 108 204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3

Calculations from equations

Calculations from equations

Try some problems. Task: Use Additional textbook p.224-225 answer questions 1 & 2. Remember it is simple ratios. More problems, have a go!