1. Topic 1 Stoichiometric Relationships
1.3 - Solutions 1

2. 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 ± 0.02 g
Mass of pure aspirin obtained ± 0.02 g
Determine the amount, in mol, of salicylic acid, C6H4(OH)COOH, used. 2

3. 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

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

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

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 (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

7. Example 1:
25.0 cm3 of a solution of hydrochloric acid contains mol HCl. What is it’s concentration?
Answer:
Concentration = moles / volume
= /
= 4.00 mol dm-3
Note: the volume was first divided by 1000 to convert to dm3 7

8. 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
= mol
Calculate volume of solution:
Volume = moles / concentration
= mol / 2.00 mol dm3
= dm3
= 50.0 cm3 8

9. 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 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) = mol dm-3
Where:
n = coefficient in balanced equation
C = concentration
V = volume
‘1’ refers to H2SO4
‘2’ refers to NaOH 9

10. MnO4- + 5 Fe2+ + 8 H+  Mn2+ + 5 Fe3+ + 4 H2O
Questions
You have 75.0 cm3 of a 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 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 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 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 mol dm-3 solution of manganate ions (MnO4-). What is the concentration of Fe2+ ions? What mass of iron was present in the solution?
MnO Fe H+  Mn Fe H2O 10
Answers: 1) 1.82g, 2) 3.23g 240 cm3, 3)0.530 mol dm-3, 4) mol dm-3, 5) mol dm-3, 0.156g

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

12. 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

13. Key Points 13

14. Homework!
Practice Questions
Problems 6, 12, 14 14

15. Topic 1 Stoichiometric Relationships
1.3 - Titrations 15

16. Lesson 6
Titrations

17. 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 cm3 of mol dm–3 HCl to g of eggshell. Calculate the amount, in mol, of HCl added.

18. We Are Here

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

20. 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

21. 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

22. 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.

23. Key Points:

24. Lesson 7
Mole Ratios and Theoretical Yields

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

26. We Are Here

27. 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’

28. 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

29. 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!

30. 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

31. Example 2…you try
What quantity of O2 in moles is required to fully react with mol of butane (C4H10) to produce water and carbon dioxide?
2 C4H O2  8 CO H2O
C4H10 is given, O2 is wanted.
n(O2) = 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

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

33. 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

34. 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 = (smallest therefore is L.R.)
Given is HCl, wanted is MgCl2
n(MgCl2) = (0.100 x 2.50) x (1 / 2)
n(MgCl2) = mol
Determine limiting reagent
Assign ‘wanted’ and ‘given’
State the equation
Sub in your numbers
Evaluate the sum

35. 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
= – (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

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

37. 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.

38. Example: mol of silver nitrate was reacted with excess sodium chloride. After filtration, 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) = x (1/1) = mol
% Yield = / x 100 = 83.3%
Check for balanced equation
Calculate theoretical yield using previous maths
Determine % yield

39. Example: After the thermal decomposition of some calcium carbonate, I collected 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
= mol
Check for balanced equation
Rearrange yield equation
Sub-in the numbers

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

41. 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

42. Lesson 8
Stoichiometry in Practice

43. 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 mol of O2 are mixed together, and the above reaction goes to completion?
4.00
8.00
10.0
20.0

44. We Are Here

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

46. 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

47. 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)

48. 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

49. 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.

50. Lesson 9
Molar Volumes of Gases

51. 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 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.

52. We Are Here

53. 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.

54. 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

55. 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.

56. In Calculations….
What volume of H2 gas is produced when 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) = x (1 / 2) = 0.025
V(H2) = x 22.4 = 0.56 dm3
Check for balanced equation
Determine moles of product
Calculate volume

57. 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 to get back to volumes…so why bother?
Volume CO2 produced:
V(CO2) = 6.00 x 4 = 24.0 cm3

58. 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

59. 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!)

60. 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

61. Lesson 10
Ideal Gases

62. 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

63. We Are Here

64. 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

65. 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

66. 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

67. 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

68. 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

69. Example 1: 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

70. 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

71. 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 / = 6.00 / T2
T2 = (6.00 x / 3.00) = 653 K
Where doe this come from?
54.0 converted to Kelvin by adding 273

72. Time to practice some sums
Complete the questions found here

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

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

75. Lesson 11
Determining The Molar Mass of A Gas

76. 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?

77. We Are Here

78. 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

79. 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.

80. Key Points

81. Lesson 12
Test

82. Good Luck
You have 80 minutes!

83. Lesson 13
Test Debrief

84. 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.

85. 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

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

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