Topic 1 Stoichiometric Relationships 1.3 - Solutions 1

Refresh A student reacted some salicylic acid with excess ethanoic anhydride. Impure solid aspirin was obtained by filtering the reaction mixture. Pure aspirin was obtained by recrystallisation. The following data was recorded by the student. Mass of salicylic acid used 3.15 ± 0.02 g Mass of pure aspirin obtained 2.50 ± 0.02 g Determine the amount, in mol, of salicylic acid, C6H4(OH)COOH, used. 2

Lesson 5: Solutions Objectives: Understand the relationship between concentration, volume and moles Prepare a standard solution of silver nitrate. Pose and solve problems involving solutions (of the chemical kind not the answers kind) 3

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

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

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 (dm3) not ml or cm3 This is the most useful measure of concentration but there are others such as grams per litre, % by weight, % by volume and molality. moles x volume concentration 6

Example 1: 25.0 cm3 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 dm3 7

Example 2: Water is added to 4.00 g NaOH to produce a 2.00 mol dm-3 solution. What volume should the solution be in cm3? Calculate quantity of NaOH: n(NaoH) = mass / molar mass = 4.00 g / 40.0 g mol ̶ 1 = 0.100 mol Calculate volume of solution: Volume = moles / concentration = 0.100 mol / 2.00 mol dm3 = 0.0500 dm3 = 50.0 cm3 8

C1 = ((1.00 mol dm ̶3 x 11.3 cm3) / 2)/25.0 cm3) = 0.226 mol dm-3 Example 3: It is found by titration that 25.0 cm3 of an unknown solution of sulfuric acid is just neutralised by adding 11.3 cm3 of 1.00 mol dm ̶3 sodium hydroxide. What is the concentration of sulfuric acid in the sample. H2SO4 + 2 NaOH  Na2SO4 + 2 H2O Use: (C1 x 25.0 cm3)/1 = (1.00 mol dm ̶3 x 11.3 cm3) / 2 C1 = ((1.00 mol dm ̶3 x 11.3 cm3) / 2)/25.0 cm3) = 0.226 mol dm-3 Where: n = coefficient in balanced equation C = concentration V = volume ‘1’ refers to H2SO4 ‘2’ refers to NaOH   9

MnO4- + 5 Fe2+ + 8 H+  Mn2+ + 5 Fe3+ + 4 H2O Questions You have 75.0 cm3 of a 0.150 mol dm-3 solution of zinc sulphate (ZnSO4). What mass of zinc sulphate crystals will be left behind on evaporation of the water? What volume of water should be added to 3.23g of copper (II) chloride (CuCl2) to form a 0.100 mol dm-3 solution? A 10.0 cm3 sample is removed from a vessel containing 1.50 dm3 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? In a titration, 50.0 cm3 of an unknown solution of barium hydroxide was fully neutralised by the addition of 12.2 cm3 of 0.200 mol dm-3 hydrochloric acid solution. What concentration is the barium hydroxide solution? Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O In a redox titration, 25.0 cm3 of an unknown solution of Fe2+ is found to react with 5.6 cm3 of a 0.100 mol dm-3 solution of manganate ions (MnO4-). What is the concentration of Fe2+ ions? What mass of iron was present in the solution? MnO4- + 5 Fe2+ + 8 H+  Mn2+ + 5 Fe3+ + 4 H2O 10 Answers: 1) 1.82g, 2) 3.23g 240 cm3, 3)0.530 mol dm-3, 4) 0.0244 mol dm-3, 5) 0.112 mol dm-3, 0.156g

Preparing a Standard Solution A standard solution is one whose concentration is well known And usually a round number like 1.00 or 0.250 mol dm-3 Prepare standard solutions sodium hydroxide with concentrations of 0.25-0.50 mol dm-3 You will use these in a future lesson so make them well! 11

Problem Time Write three stoichiometry problems using any of the ideas from the unit so far (make sure you know the answer). In ten minutes time you will need to give your problems to a classmate to solve. 12

Key Points 13

Homework! Practice Questions Problems 6, 12, 14 14

Topic 1 Stoichiometric Relationships 1.3 - Titrations 15

Lesson 6 Titrations

Refresh Which statement about solutions is correct? When vitamin D dissolves in fat, vitamin D is the solvent and fat is the solute. In a solution of NaCl in water, NaCl is the solute and water is the solvent. An aqueous solution consists of water dissolved in a solute. The concentration of a solution is the amount of solvent dissolved in 1 dm3 of solution A student added 27.20 cm3 of 0.200 mol dm–3 HCl to 0.188 g of eggshell. Calculate the amount, in mol, of HCl added.

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Lesson 3: Solutions Objectives: Conduct an experiment involving the technique of titration Analyse experimental data to determine the concentration of two unknown solutions

Titration Titration involves using a solution whose concentration is known, to find the concentration of another which isn’t known. An exact volume of one solution is in a conical flask, a second solution is added to it from a burette. It doesn’t matter which solution is in the burette, and which is in the conical flask. When the reaction reaches its ‘endpoint’, we record how much was added There is always some kind of indicator which changes colour to tell us when we have reached the end. This is one of the most useful tools in the experimental chemist’s toolbox. Determine the concentration of acids/bases Determine concentrations of other reactants Following the rate of a reaction Determining equilibrium constants

The mathematics of titrations Use this equation (met last lesson…check example here) Where: n = coefficient in balanced equation C = concentration V = volume ‘1’ refers to H2SO4 ‘2’ refers to NaOH

Your turn… Complete the experiment here. If you made a standard solution in the last lesson, make sure you use it here. Manage your time so that you are able to complete the experiment and the calculations by the end of the lesson.

Key Points:

Lesson 7 Mole Ratios and Theoretical Yields

Refresh What volume, in cm3, of 0.200 mol dm–3 HCl(aq) is required to neutralize 25.0 cm3 of 0.200 mol dm–3 Ba(OH)2(aq)? 12.5 25.0 50.0 75.0

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Lesson 7: 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 Meet the idea of the ‘limiting reagent’

Mole Ratios This is the ratio of one compound to another in a balanced equation. For example, in the previous equation 2 H2 + O2  2 H2O Hydrogen, oxygen and water are present in 2:1:2 ratio. This ratio is fixed and means for example: 0.2 mol of H2 reacts with 0.1 mol of O2 to make 0.2 mol H2O 5 mol of H2 reacts with 2.5 mol of O2 to make 5 mol of H2O To make 4 mol of H2O you need 4 mol of H2 and 2 mol of O2

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!

Example 1… What quantity of Al(OH)3 in moles is required to produce 5.00 mol of H2O? 2 Al(OH)3 + 3 H2SO4  Al2(SO4)3 + 3 H2O H2O is given, Al(OH)3 is wanted. n(Al(OH)3) = 5.00 x (2/3) n(Al(OH)3) = 3.33 mol Check for balanced equation Assign ‘wanted’ and ‘given’ State the equation Sub in your numbers Evaluate the sum

Example 2…you try What quantity of O2 in moles is required to fully react with 0.215 mol of butane (C4H10) to produce water and carbon dioxide? 2 C4H10 + 13 O2  8 CO2 + 10 H2O C4H10 is given, O2 is wanted. n(O2) = 0.215 x (13/2) n(O2) = 1.40 mol Check for balanced equation Assign ‘wanted’ and ‘given’ State the equation Sub in your numbers Evaluate the sum

Your Turn… Try questions 1-5 on the sheet here. If you finish early, set your own problems for others to try.

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 H2 + O2  2 H2O For example, if you have 2.0 mol H2 and 2.0 mol O2 H2 is the limiting reactant – it will run out O2 is present in excess – there is more than enough To determine this, divide the quantity of each reactant by its coefficient in the equation. The smallest number is the limiting reactant: H2: 2.0 / 2 = 1.0 – smallest therefore limiting O2: 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

Example 1: What quantity, in moles, of MgCl2 can be produced by reacting 10.5 g magnesium with 100 cm3 of 2.50 mol dm-3 hydrochloric acid solution? Check for balanced equation Mg + 2HCl  MgCl2 + 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 MgCl2 n(MgCl2) = (0.100 x 2.50) x (1 / 2) n(MgCl2) = 0.125 mol Determine limiting reagent Assign ‘wanted’ and ‘given’ State the equation Sub in your numbers Evaluate the sum

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? C3H8 + 5O2  3CO2 + 4H2O Determining limiting reagent: C3H8: 2.00 / 1 = 2.00 (smallest therefore is L.R.) O2: 12.0 / 5 = 2.40 Given is C3H8, wanted is CO2 n(CO2) = 2.00 x (3 / 1) = 6.00 mol N(O2) remaining = n(O2) at start – n(O2) used = 12.00 – (2.00 x (5 / 1) ) = 2.00 mol Check for balanced equation Determine limiting reagent Assign ‘wanted’ and ‘given’ State the equation Sub in your numbers Evaluate the sum

Your Turn… Try questions 6-10 on the sheet here. If you finish early, set your own problems for others to try.

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.

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? AgNO3(aq) + NaCl(aq)  AgCl(aq) + NaNO3(aq) Determine theoretical yield: AgCl is wanted, AgNO3 is given n(AgCl) = 0.150 x (1/1) = 0.150 mol % Yield = 0.125 / 0.150 x 100 = 83.3% Check for balanced equation Calculate theoretical yield using previous maths Determine % yield

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? CaCO3  CaO + CO2 theoretical yield = (0.437 / 77.4) x 100 = 0.565 mol Check for balanced equation Rearrange yield equation   Sub-in the numbers

Your Turn… Try questions 11-12 on the sheet here. If you finish early, set your own problems for others to try.

Key Points Balance equations by playing with coefficients Use mole ratios to work quantities of chemicals involved in reactions Divide the quantity of each reactant by its coefficient to determine the limiting reactant

Lesson 8 Stoichiometry in Practice

Refresh Equal masses of the metals Na, Mg, Ca and Ag are added to separate samples of excess HCl(aq). Which metal produces the greatest total volume of H2(g)? Na Mg Ca Ag Chloroethene, C2H3Cl, reacts with oxygen according to the equation below. 2C2H3Cl(g) + 5O2(g) → 4CO2(g) + 2H2O(g) + 2HCl(g) What is the amount, in mol, of H2O produced when 10.0 mol of C2H3Cl and 10.0 mol of O2 are mixed together, and the above reaction goes to completion? 4.00 8.00 10.0 20.0

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Lesson 8: Stoichiometry in Practice Objectives: Apply your knowledge of stoichiometry to determine the proportions of a mixture Practice stoichiometry past-paper questions

Preparation of Silver Chromate In this experiment you will prepare silver chromate The challenge is that you will start with an unknown mixture of the reactants and will have to use your knowledge of stoichiometry to work out its composition

2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4 (s) + 2 KNO3(aq) The Reaction The reaction is as follows: 2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4 (s) + 2 KNO3(aq) The challenge is that you won’t know the individual amounts of each reactant you start with, just their combined mass. To determine the proportions of the mixture, you will need to piece together the information you gain like a puzzle. A table like this could help you. You will find out what is in the yellow box first, and should be able to use that to help you fill in all the other boxes AgNO3(aq) K2CrO4(aq) Ag2CrO4 (s) KNO3(aq) Amount (mol) Mass (g)

To do: Complete the experiment here: Preparation of Silver Chromate.docx During any wait times you should be: Working out how to do your calculations Working through the past-paper question pack

Key Points Calculate moles of all reactants Determine limiting reactant Use mole ratios to calculate moles of product expected Convert moles of product to mass/volume/solution etc.

Lesson 9 Molar Volumes of Gases

2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l) Refresh 0.600 mol of aluminium hydroxide is mixed with 0.600 mol of sulfuric acid, and the following reaction occurs: 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l) Determine the limiting reactant. Calculate the mass of Al2(SO4)3 produced.

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Lesson 9: The Molar Volume of Gases Objectives: Understand that a fixed quantity (in moles) of gas, always occupies the same volume at room temperature Perform calculations using the molar volume of a perfect gas Apply the above concept to design the best possible bottle rocket.

The Molar Volume of an Ideal Gas At standard temperature and pressure (STP, T = 273K, P = 1.01x105 Pa): Molar Volume of Gas = 2.24x10-2 m3 mol-1 = 22.4 dm3 mol-1 At room temperature and pressure (RTP, T = 298K, P = 1.01x105 Pa): Molar Volume of Gas = 2.45x10-2 m3 mol-1 = 24.5 dm3 mol-1 Note: Although you may have learned the volume at RTP prior to IB, it is not something you will use at IB, and is only mentioned to help you link with what you know

Avogadro’s Law We can generalise from the previous slide to say that: "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules". Alternatively: At a given temperature and pressure, the volume occupied by one mole of a gas is the same, regardless of the gas. This is true (ish) no matter what the gas!...we will look at why next lesson This can really help us to simplify mole calculations involving only gaseous reactants.

In Calculations…. What volume of H2 gas is produced when 0.0500 mol Mg reacts with excess acid at S.T.P.? 2 Li(s) + 2 HCl(aq)  2 LiCl(aq) + H2(g) n(H2) = n(Li) x (1 / 2) = 0.050 x (1 / 2) = 0.025 V(H2) = 0.025 x 22.4 = 0.56 dm3 Check for balanced equation Determine moles of product Calculate volume

More calculations At STP, 30.0 cm3 ethane reacts with 60.0 cm3 oxygen. Which reactant is present in excess, how much remains after the reaction and what volume of CO2 is produced? 2 C2H6(g) + 10 O2(g)  6 H2O(l) + 4 CO2(g) Limiting reactant: C2H6: 30.0/2 = 15.0 O2: 60.0/10 = 6.00 therefore O2 limiting, ‘6.00’ will be the number used in all further calculations as there is enough O2 for ‘6’ of the reaction C2H6 remaining: V(C2H6 used) = 6.00 x 2 = 12.0 cm3 V(C2H6 remaining) = 30.0 – 12.0 = 18.0 cm3 Note: there is no need to convert to moles as they are all in a ratio of moles as you would divide by 22.4 to get to moles, do your sums and then multiply by 22.4 to get back to volumes…so why bother? Volume CO2 produced: V(CO2) = 6.00 x 4 = 24.0 cm3

Time to practice What is the minimum volume of H2 gas required to fully reduce 10.0 g copper (II) oxide to copper (assume STP…poor assumption but will do for the time being!)? CuO(s) + H2(g)  Cu(s) + H2O(l) In a car airbag, sodium azide (NaN3) decomposes explosively to make N2 gas. What is the minimum mass of sodium azide required to fully inflate a 60.0 dm3 airbag, assuming STP? 2 NaN3(s)  2 Na(s) + 3 N2(g) 500 cm3 methane reacts with 600 cm3 oxygen to produce carbon dioxide and water. What are the final volumes of each of the three gases on completion of the reaction? CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) Answers: 1) 2.82 dm3, 2) 116 g, 3) CH4: 200 cm3, O2: 0 cm3, CO2: 300 cm3

Bottle Rockets Hydrogen reacts explosively with oxygen Bad if you hate fun! Excellent if you like to make stuff go bang or whoosh! You will need to design and build rockets (from standard 500 ml drinks bottles). The best will win an awesome prize. You need to consider: A suitable reaction to generate hydrogen Allow about 250 cm3 to bubble away first before you start collecting, to ensure you have pushed any air out of the conical flask The stoichiometry of the reaction The oxygen content of the air Suitable ways to decide the winner Making it look cool (if you finish early!)

Key Points The volume of gas depends on the temperature, pressure and number of moles, NOT THE TYPE OF GAS Molar Volume at STP = 22.4 dm3

Lesson 10 Ideal Gases

Refresh What volume of sulfur trioxide, in cm3, can be prepared using 40 cm3 sulfur dioxide and 20 cm3 oxygen gas by the following reaction? Assume all volumes are measured at the same temperature and pressure. 2SO2(g) + O2(g) → 2SO3(g) 20 40 60 80

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Lesson 10: Ideal Gas Equation Objectives: Understand the ideal gas equation Complete a circus of short experiments to explore the ideal gas equation Perform calculations using the ideal gas equation

Molar Volume of a Perfect Gas We learnt about the molar volume of gases last lesson….how can they be the same? The distance between particles is much bigger than the size of the particles….so particle size makes very little difference: The blue particle is twice the size of the red particle, but the blue particles are not taking up twice the amount of space. In reality, the relative distance between the molecules is much much greater than this. 10 units

The Ideal Gas Equation The volume a gas takes up is determined by: Pressure Temperature Moles of gas This combines to form the ideal gas equation PV = nRT Where: P = pressure in Pa V = volume in m3 n = moles of gas R = gas constant, 8.31 J K-1 mol-1, this appears in many places in chem T = temperature in K

Ideal Gas Assumptions Particles occupy no volume Particles have zero intermolecular forces These are not always valid, particularly at: Low temperature – when intermolecular forces become significant High pressure – when particle volume becomes significant

Study the equation and predict the lines you would expect on these graphs (assuming the third factor is fixed):   Temperature Volume Temperature Pressure Volume Pressure

Example 1: 1.048 g of unknown gas A, occupies 846 cm3 at 500K and standard pressure. What is it’s molecular mass? State ideal gas equation Rearrange for chosen subject Sub in numbers with unit conversion and evaluate   Complete further calculations

Example 2 (your turn): A fire extinguisher containing 45 Example 2 (your turn): A fire extinguisher containing 45.4 mol CO2 has a volume of 3000 cm3. What is the pressure inside at 50OC? State ideal gas equation Rearrange for chosen subject Sub in numbers with unit conversion and evaluate

Where doe this come from? One last example: The volume of an ideal gas at 54.0 °C is increased from 3.00 dm3 to 6.00 dm3. At what temperature, in °C, will the gas have the original pressure? Use a modified version of the ideal gas equation: Since original and final pressure should be the same, we can remove this from the equation as they cancel out: 3.00 / 327.0 = 6.00 / T2 T2 = (6.00 x 327.0 / 3.00) = 653 K Where doe this come from? 54.0 converted to Kelvin by adding 273

Time to practice some sums Complete the questions found here

Exploring Ideal Gases The best way to explore the behaviour of gases is to have a little play. Complete these four short experiments

Key Points The Ideal Gas equation: PV = nRT Also: Provided that: Molecules have zero volume Molecules experience no attraction to each other

Lesson 11 Determining The Molar Mass of A Gas

Refresh What volume of carbon dioxide, in dm3 under standard conditions, is formed when 7.00 g of ethene (C2H4, Mr = 28.1) undergoes complete combustion?

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Lesson 11: Determining the Molar Mass of a Gas Objectives: Apply the ideal gas equation to determine the molar mass of a gas from experimental data. Complete some independent practice of stoichiometry calculations

Determining the Molar Mass of A Gas Two important things you already know how to do: Calculate the number of moles of gas given the volume, temperature and pressure. Calculate the molar mass of a substance given a mass and a number of moles. In this experiment you will combine these two things to determine the molar mass of the gas used in cigarette lighters. Once you have finished and performed your calculations, you should do practice questions from the stoichiometry question-pack, and prepare for the coming test.

Key Points  

Lesson 12 Test

Good Luck You have 80 minutes!

Lesson 13 Test Debrief

Personal Reflection Spend 15 minutes looking through your test: Make a list of the things you did well Use your notes and text book to make corrections to anything you struggled with.

Group Reflection Spend 10 minutes working with your classmates: Help classmates them with corrections they were unable to do alone Ask classmates for support on questions you were unable to correct

Go Through The Paper Stop me when I reach a question you still have difficulty with.

Targeted Lesson PREPARE AFTER MARKING THE TEST SHORT LESSON ON SPECIFIC AREAS OF DIFFICULTY