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3. Lesson: Solutions Objectives: Understand the relationship between concentration, volume and moles Pose and solve problems involving solutions (of the chemical kind not the answers kind)

4. Solutions Basics Aqueous copper sulfate solution: + SOLUTE SOLVENT SOLUTION

5. Concentration This is the strength of a solution. Most ConcentratedLeast Concentrated

6. Molarity The number of moles of a substance dissolved in one litre of a solution. Units: mol dm -3 Pronounced: moles per decimetre cubed Units often abbreviated to ‘M’ (do not do this in an exam!) Volume must be in litres (dm 3 ) not ml or cm 3 This is the most useful measure of concentration but there are others such as grams per litre, % by weight, % by volume and molality. moles concentration volume x

7. Example 1: 25.0 cm 3 of a solution of hydrochloric acid contains 0.100 mol HCl. What is it’s concentration? Answer: Concentration = moles / volume = 0.100 / 0.0250 = 4.00 mol dm -3 Note: the volume was first divided by 1000 to convert to dm 3

8. Example It is found by titration that 25.0 cm 3 of an unknown solution of sulfuric acid is just neutralised by adding 11.3 cm 3 of 1.00 mol dm -3 sodium hydroxide. What is the concentration of sulfuric acid in the sample. H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O Use: (C 1 x 25.0)/1 = (1.00 x11.3)/2 C 1 = ((1.00 x 11.3) / 2)/25.0) = 0.226 mol dm -3 Where: n = coefficient in balanced equation C = concentration V = volume ‘1’ refers to H 2 SO 4 ‘2’ refers to NaOH

9. Questions 1. You have 75.0 cm 3 of a 0.150 mol dm -3 solution of zinc sulphate (ZnSO 4 ). What mass of zinc sulphate crystals will be left behind on evaporation of the water? 2. What volume of water should be added to 3.23g of copper (II) chloride (CuCl 2 ) to form a 0.100 mol dm -3 solution? 3. A 10.0 cm 3 sample is removed from a vessel containing 1.50 dm 3 of a reaction mixture. By titration, the sample is found to contain 0.00530 mol H +. What is the concentration of H + in the main reaction vessel? 4. In a titration, 50.0 cm 3 of an unknown solution of barium hydroxide was fully neutralised by the addition of 12.2 cm 3 of 0.200 mol dm -3 hydrochloric acid solution. What concentration is the barium hydroxide solution? Ba(OH) 2 + 2 HCl  BaCl 2 + 2 H 2 O Answers: 1) 1.82g, 2) 3.23g 240 cm 3, 3)0.530 mol dm -3, 4) 0.0244 mol dm -3,

10. Questions 1. You have 75.0 mL of a 0.150 M solution of zinc sulphate (ZnSO 4 ). What mass of zinc sulphate crystals will be left behind on evaporation of the water? 2. What volume of water should be added to 3.23g of copper (II) chloride (CuCl 2 ) to form a 0.100M solution? 3. A 10.0 mL sample is removed from a vessel containing 1.50 L of a reaction mixture. By titration, the sample is found to contain 0.00530 mol H +. What is the concentration of H + in the main reaction vessel? 4. In a titration, 50.0 mL of an unknown solution of barium hydroxide was fully neutralised by the addition of 12.2 mL of 0.200 M hydrochloric acid solution. What concentration is the barium hydroxide solution? Ba(OH) 2 + 2 HCl  BaCl 2 + 2 H 2 O Answers: 1) 1.82g, 2) 3.23g 240 cm 3, 3)0.530 mol dm -3, 4) 0.0244 mol dm -3,

11. Mole Ratios and Theoretical Yields

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13. Lesson: Mole Ratios and Theoretical Yields Objectives: Know how to construct and balance symbol equations Apply the concept of the mole ratio to determine the amounts of species involved in chemical reactions Discuss the idea of the ‘limiting reagent’

14. Mole Ratios This is the ratio of one compound to another in a balanced equation. For example, in the previous equation 2 H 2 + O 2  2 H 2 O Hydrogen, oxygen and water are present in 2:1:2 ratio. This ratio is fixed and means for example: 0.2 mol of H 2 reacts with 0.1 mol of O 2 to make 0.2 mol H 2 O 5 mol of H 2 reacts with 2.5 mol of O 2 to make 5 mol of H 2 O To make 4 mol of H 2 O you need 4 mol of H 2 and 2 mol of O 2

15. Mole Ratios in Calculations You will often have questions that ask you how many moles of X can be made from a amount of Y Or various similar questions Use the following: Where: wanted = the substance you want to find out more about given = the substance you are given the full info for n(wanted) = the number of moles you are trying to find out n(given) = the number of moles of you are given in the question wanteds = the number of wants in the balanced equation givens = the number of givens in the balanced equation The mole ratio!

16. Example 1… What quantity of Al(OH) 3 in moles is required to produce 5.00 mol of H 2 O? 2 Al(OH) 3 + 3 H 2 SO 4  Al 2 (SO 4 ) 3 + 3 H 2 O H 2 O is given, Al(OH) 3 is wanted. n(Al(OH) 3 ) = 5.00 x (2/3) n(Al(OH) 3 ) = 3.33 mol Assign ‘wanted’ and ‘given’ Sub in your numbers State the equation Evaluate the sum Check for balanced equation

17. Example 2…you try What quantity of O 2 in moles is required to fully react with 0.215 mol of butane (C 4 H 10 ) to produce water and carbon dioxide? 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O C 4 H 10 is given, O 2 is wanted. n(O 2 ) = 0.215 x (13/2) n(O 2 ) = 1.40 mol Assign ‘wanted’ and ‘given’ Sub in your numbers State the equation Evaluate the sum Check for balanced equation

18. The Limiting Reagent In a reaction, we can describe reactants as being ‘limiting’ or in ‘excess’ Limiting – this is the reactant that runs out Excess – the reaction will not run out of this 2 H 2 + O 2  2 H 2 O For example, if you have 2.0 mol H 2 and 2.0 mol O 2 H 2 is the limiting reactant – it will run out O 2 is present in excess – there is more than enough

19. To determine this, divide the quantity of each reactant by its coefficient in the equation. The smallest number is the limiting reactant: H 2 : 2.0 / 2 = 1.0 – smallest therefore limiting O 2 : 2.0 / 1 = 2.0 The limiting reactant will be your ‘given’ in all further calculations: Determining amounts of products formed Determining amounts of other reactants used

20. Example 1: What quantity, in moles, of MgCl2 can be produced by reacting 10.5 g magnesium with 100 cm 3 of 2.50 mol dm -3 hydrochloric acid solution? Mg + 2HCl  MgCl 2 + H Determining limiting reagent: Mg: (10.5 / 24.31)/1 = 0.432 HCl: (0.100 x 2.50)/2 = 0.125 ( smallest therefore is L.R.) Given is HCl, wanted is MgCl 2 n(MgCl 2 ) = (0.100 x 2.50) x (1 / 2) n(MgCl 2 ) = 0.125 mol Assign ‘wanted’ and ‘given’ Sub in your numbers State the equation Evaluate the sum Determine limiting reagent Check for balanced equation

21. Example 2 (you try): What quantity, in moles, of carbon dioxide would be formed from the reaction of 12.0 mol oxygen with 2.00 mol propane, and how much of which reactant would remain? C 3 H 8 + 5O 2  3CO 2 + 4H 2 O Determining limiting reagent: C 3 H 8 : 2.00 / 1 = 2.00 ( smallest therefore is L.R.) O 2 : 12.0 / 5 = 2.40 Given is C 3 H 8, wanted is CO 2 n(CO 2 ) = 2.00 x (3 / 1) = 6.00 mol N(O 2 ) remaining = n(O2) at start – n(O2) used = 12.00 – (2.00 x (5 / 1) ) = 2.00 mol Assign ‘wanted’ and ‘given’ Sub in your numbers State the equation Evaluate the sum Determine limiting reagent Check for balanced equation

22. Theoretical, actual and percentage yield Theoretical yield is the maximum amount of product you would make if the limiting reactant was fully converted to product. Use the limiting reactants maths to work this out Actual yield is the actual amount of product collected in after a reaction It is always less than the theoretical yield Percentage yield reflects how close you got to achieving the theoretical yield: Your actual and theoretical yields can be in either moles or grams, so long as they are both the same units.

23. Example: 0.150 mol of silver nitrate was reacted with excess sodium chloride. After filtration, 0.125 mol of silver chloride was collected. What was the % yield? AgNO 3 (aq) + NaCl(aq)  AgCl(aq) + NaNO 3 (aq) Determine theoretical yield: AgCl is wanted, AgNO 3 is given n(AgCl) = 0.150 x (1/1) = 0.150 mol % Yield = 0.125 / 0.150 x 100 = 83.3% Determine % yield Calculate theoretical yield using previous maths Check for balanced equation

24. Example: After the thermal decomposition of some calcium carbonate, I collected 0.437 mol of calcium oxide, which was a 77.4% yield. How much calcium carbonate did I start with? CaCO 3  CaO + CO 2 theoretical yield = (0.437 / 77.4) x 100 = 0.565 mol Sub-in the numbers Rearrange yield equation Check for balanced equation

25. Exit Ticket In a redox titration, 25.0 cm 3 of an unknown solution of Fe 2+ is found to react with 5.6 cm 3 of a 0.100 mol dm -3 solution of manganate ions (MnO 4 - ). What is the concentration of Fe 2+ ions? What mass of iron was present in the solution? MnO 4 - + 5 Fe 2+ + 8 H +  Mn 2+ + 5 Fe 3+ + 4 H 2 O 0.112 mol dm -3, 0.156g