3. After completing this lesson you should be able to : For a particular set of reaction conditions, the percentage yield provides a measure of the degree to which the limiting reagent is converted into the desired product. It is possible to calculate the percentage yield using the equation below. Percentage yield equation: The ‘actual yield’ is the quantity of the desired product formed under the prevailing reaction conditions whilst the ‘theoretical yield’ is the quantity of desired product which would be obtained, assuming full conversion of the limiting reagent, as calculated from the balanced equation. By considering a balanced equation, the limiting reactant and the reactant(s) in excess can be identified by calculation. Given costs for the reactants, a percentage yield can be used to calculate the feedstock’s cost for producing a given mass of product.

4. 4 To calculate the percent yield, the actual yield and theoretical yield are needed. You prepared cookie dough to make 60 cookies. The phone rings and you answer. While talking, a tray of 12 cookies burn and you have to throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies Calculating Percent Yield

5. 5 Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product calculated using the balanced equation. Actual yield The amount of product obtained when the reaction takes place. Percent yield The ratio of actual yield to theoretical yield. percent yield = actual yield (g) x 100 theoretical yield (g) COPY

6. Percentage yield CH 3 COOH + CH 3 CH 2 CH 2 OH CH 3 COOCH 2 CH 2 CH 3 + H 2 O Q1) 4.3 g of propyl ethanoate was produced when 6 g of ethanoic acid was reacted with propan-1-ol (in excess). What is the percentage yield of the ester? COPY Ethanoic acidPropan-1-olPropyl ethanoate Step 1 – Write down mole ratio. 1111 44g 102g 6g Xg Step 2 – Convert to gram formula mass Step 3 – Calculate theoretical yield ((6 ÷44) x 102 = 13.9g) Step 4 – Use the actual yield with the theoretical yield to calculate percentage yield ((4.3g ÷13.9g) x 100 = 30.93 %

7. Percentage yield is calculated using the formula:- percentage yield = x 100 The actual yield is given in the question. The theoretical yield is calculated from the balanced equation. actual yield theoretical yield Worked example. In a reaction 320 kg of sulphur dioxide reacted with oxygen to form 360kg of sulphur trioxide. Calculate the percentage yield. Find the theoretical yield using the balanced equation:- 2SO 2 + O 2  2SO 3 2 mol  2 mol 128.2 g  160.2 g 128.2 kg  160.2 kg 320 kg  399.9 kg percentage yield = x 100 = 90.0% 360 399.9 COPY

8. Q2) 18.72g of benzene, C 6 H 6, enters a reaction chamber with excess nitric acid. After a time, 22.14g of nitrobenzene C 6 H 5 NO 2 is obtained. C 6 H 6 + HNO 3 -> C 6 H 5 NO 2 + H 2 O Calculate the percentage yield. COPY Answer 78g 123g 18.72gxg theoretical X g = 18.72 x 123 78 = 29.52g % yield = actual x 100 = 22.14 x 100 = 75% theoretical 29.52

9. Ammonia is produced in industry by the Haber Process. N 2 (g) + 3H 2 (g) --> 2NH 3 (g) Under certain conditions, 500 kg of nitrogen reacts with excess hydrogen to produce 405 kg of ammonia. Calculate the percentage yield of ammonia under these conditions. Show your working clearly. COPY

10. Calculation of percentage yield. Calculations for you to try. 1.0.5 tonnes of hydrogen reacts with excess nitrogen to form 0.66 tonnes of ammonia. N 2 + 3H 2  2NH 3 Calculate the percentage yield of ammonia. N 2 + 3H 2  2NH 3 3 mol  2 mol 6 g  34 g 6 tonnes  34 tonnes 0.5 tonnes  2.83 tonnes percentage yield = x 100 = 23.32% 0.66 2.83 COPY

11. Calculation of percentage yield. Calculations for you to try. 2. Ethene reacts with steam to ethanol C 2 H 4 + H 2 O  C 2 H 5 OH Calculate the percentage yield of ethanol when 140g of ethene reacts to form 170 g of ethanol. C 2 H 4 + H 2 O  C 2 H 5 OH 1 mol  1 mol 28 g  46 g 140 g  140 / 28 x 46 g  230 g percentage yield = x 100 = 73.9% 170 230 COPY

12. Calculation of reactant in excess/limiting reagent As soon as one of the reactant in a chemical reaction is used up the reaction stops. Any other reactant which is left over is said to be ‘in excess’. The reactant which is used up determines the mass of product formed and is called the limiting reactant/reagent. The limiting reagent (or limiting reactant) in a chemical reaction is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. Worked example. Which reactant is in excess when 10g of calcium carbonate reacts with 100cm 3 of 1 mol l -1 hydrochloric acid? Write the balanced equation for the reaction and show mole ratio:- CaCO 3 + 2HCl  CaCl 2 + CO 2 + H 2 O 1 mol 2 mol Calculate the number of moles of each reactant:- Number of moles in 10g of CaCO 3 = 10 / 100 = 0.1 Number of moles of HCl = 1 x 100 / 1000 = 0.1 mass n gfm n C V (l) From equation 0.1 mol of CaCO 3 needs 0.2 mol of HCl and as we only have 0.1 mol of HCl the CaCO 3 is in excess. COPY

13. Calculation of excess Calculations for you to try. 1.What mass of calcium oxide is formed when 0.4 g of calcium reacts with 0.05 mole of oxygen? 2Ca + O 2  2CaO 2 mol 1 mol Number of moles of Ca in 0.4 g = 0.4 / 40 = 0.01 From equation 2 mol of Ca reacts with 1 mol of O 2. So 0.01 mol of Ca reacts with 0.005 mol of O 2. As we have 0.05 mol of O 2 it is in excess. All 0.01 mol of Ca is used up From equation 0.01 mol of Ca will produce 0.01 mol of CaO 1 mole of CaO (1 x 40) + (1 x 16) = 56g So 0.01 moles weighs 0.01 x 56 = 0.56 g COPY

14. Calculation of excess Calculations for you to try. 2.What mass of hydrogen is formed when 3.27g of zinc is reacted with 25cm 3 of 2 mol l -1 hydrochloric acid? Zn + 2HCl  ZnCl 2 + H 2 1 mol 2 mol 1 mol Number of moles of Zn in 3.27 g = 3.27 / 65.4 = 0.05 Number of moles of HCl = 2 x 25 / 1000 = 0.05 From equation 1 mol of Zn reacts with 2 mol of HCl. So 0.05 mol of Zn reacts with 0.1 mol of HCl. As we have only 0.05 mol of HCl it is the zinc that is in excess. All 0.05 mol of HCl is used up From equation 0.05 mol of HCl will produce 0.5 x 0.05 mol H 2 0.025 mol of H 2 weighs 0.025 x 2 = 0.05 g COPY

15. CfE Higher Chemistry – Specimen paper COPY

16. CfE Higher Chemistry Exemplar paper COPY

17. CfE Higher Chemistry past paper (2015) COPY

18. Answer Scheme