1. Stoichiometry
Mr Field

2. Using this slide show
The slide show is here to provide structure to the lessons, but not to limit them….go off-piste when you need to!
Slide shows should be shared with students (preferable electronic to save paper) and they should add their own notes as they go along.
A good tip for students to improve understanding of the calculations is to get them to highlight numbers in the question and through the maths in different colours so they can see where numbers are coming from and going to.
The slide show is designed for my teaching style, and contains only the bare minimum of explanation, which I will elaborate on as I present it. Please adapt it to your teaching style, and add any notes that you feel necessary.

3. Menu: Lesson 1 – The Mole Lesson 2 – Moles and Molar Mass
Lesson 3 - Solutions
Lesson 4 - Formulas
Lesson 5 - Equations
Lesson 6 – Theoretical Yields
Lesson 7 – Molar Volume of Gases
Lesson 8 – Ideal Gases
Lesson 9 - TEST
Lesson 10 – Test Debrief

4. Lesson 1
The Mole and the Avogadro Constant

5. Intro to Unit
Brainstorm everything you know about quantitative chemistry (equations, moles and all that stuff!)

6. Overview
Copy this onto an A4 page. You should add to it as a regular review throughout the unit.

7. Assessment
Test at the end of the unit (100%)
Approx lesson 9

8. We Are Here

9. Lesson 1: Meet the Mole Objectives:
Understand the phrase ‘a mole’ as just a large number
Define ‘a mole’ according to Avogadro’s number
Complete an experiment to determine Avogadro’s number
Calculate numbers of particles using Avogadro’s number

10. What do the following have in common?
A dozen A grand A score A pair A trio

11. A Mole A mole is a large number: A mole is:
6.02 x 1023
602,000,000,000,000,000,000,000
Six hundred and two thousand quadrillion
Given the symbol, L
This number is called Avogadro’s number after the Italian scientist Amedeo Avogadro who first proposed it
Picture – all the grains of sand, on all the beaches and in all the deserts in the world; that is about a tenth of a mole!

12. Some Calculations
We can use this equation to calculate a number of moles from a number of particles
Where:
n = quantity in moles
N = number of particles
L = Avogadro’s Constant (6.02x1023)
Example 1: You have 3.01x1022 atoms of carbon. How many moles is this?
n = N / L
n(C) = 3.01x1022 / 6.02x1023
n(C) = mol
Example 2: You have 6.02x1024 molecules of water. How many moles of hydrogen atoms are present?
n(H) = 2 x n(H2O)
n(H) = 2 x 6.02x1024 / 6.02x1023
n(H) = 20.0
Example 3: How many atoms of hydrogen are there in 2.5 moles of methane (CH4)?
N(H) = 4 x N(CH4)
N(H) = 4 x n(CH4) x L
N(H) = 4 x 2.5 x 6.02x1023
N(H) = 6.02x1024

13. Some Calculations
How many moles of ethene (C2H4) are there in 1.5x1022 molecules?
How many moles of oxygen atoms in 1.20x1024 molecules of sulphuric acid (H2SO4)? How many moles of oxygen molecules could they make?
How many molecules in 3.0 moles of nitrogen gas?
How many chloride ions are released on dissolving moles of calcium chloride (CaCl2)?
BONUS QUESTION: How many moles of people are there on the planet?

14. How do you determine Avogadro’s number?
There are a number of ways including:
Electrolysis
Measurement of electron mass
X-ray crystal density
Molecular monolayers
You will be determining L by measuring the area of a molecular monolayer
Not the most accurate but feasible in our lab!

15. Key Points

16. Lesson 2
Moles and Molar Mass

17. Refresh
How many molecules are present in a drop of ethanol, C2H5OH, of mass 2.3 × 10–3 g? (L = 6.0 × 1023 mol–1)
A. 3.0 × 10 19
B. 3.0 × 1020
C. 6.0 × 1020
D. 6.0 × 1026
How many oxygen atoms are there in 0.20 mol of ethanoic acid, CH3COOH?
A. 1.2 × 1023
B. 2.4 × 1023
C. 3.0 × 1024
D. 6.0 × 1024

18. We Are Here

19. Lesson 2: Moles and Molar Mass
Objectives:
Calculate the relative mass and molar mass of substances
Relate the mass of a substance to a quantity in moles
Conduct an experiment to determine the number of moles of water of crystallisation

20. Atomic and Formula MassAr
is the relative atomic mass of an element
found in the periodic table Mr
is the relative molecular mass of a compound
To calculate Mr, you add the Ar of all the atoms in a compound
What does the term ‘relative’ mean?
What determines the mass of an atom and how can Ar be a decimal number?

21. Calculating Mr HCl C2H4 H2SO4 Mg(OH)2 Ar(H) = 1.01 Ar(H) = 1.01
Ar (Cl) = 35.45
Mr = = 36.46
C2H4
Ar(C) = 12.01
Ar (H) = 1.01
Mr = 2x x1.01
= 16.06
H2SO4
Ar(H) = 1.01
Ar (S) = 32.06
Ar(O) = 16.00
Mr = 2x x16.00
= 98.08
Mg(OH)2
Ar(Mg) = 24.31
Ar (H) = 1.01
Mr = x x1.01
= 58.33

22. Calculate Mr for:
Br2
C3H8
(NH4)2SO4
C6H12O6

23. Molar Mass, Mm This is the mass of one mole of something.
To calulate Mm, simply stick ‘g’ for grams on the end of Mr.
For example:
Mr(H2O) = 18.02
Mm(H2O) = g
Note: This is why the value of L was chosen to be what it was.

24. Relating ‘mass’ and ‘molar mass’
You need to be able to solve problems like:
How many moles of Y is mass X?
What is the mass of X moles of Y?
X moles of Y has a mass of Z, what is it’s molar mass?
Use this equation:

25. For example: How many moles of water are present in 27.03g?
Calc Mm(H2O):
Mm(H2O)= 2x
= 18.02
Find n(H2O):
n(H2O) = M / Mm
= / 18.02
= 1.50 mol
What is the mass of 4.40 mol of iron (III) oxide (Fe2O3)?
Calc Mm(Fe2O3):
Mm(H2O)= 2x x16.00
= g
Find M(Fe2O3):
M(Fe2O3) = n x Mm
= 4.40 /
= 703 g
1.30 moles of an unknown compound has a mass of 20.9g, what is it’s molar mass?
Mm(unknown)= M / n
= 20.9 / 1.30
= 16.1 g/mol

26. Questions
Calculate the mass of mol of benzene (C6H6)
Calculate the mass of mol of ammonium nitrate (NH4NO3)
What quantity of iron (III) oxide is present in a 1.0 kg sample?
What quantity of cobalt (II) chloride (CoCl2) is present in a g sample?
8.8 moles of a compound has a mass of kg. Calculate its molar mass.
0.010 mol of an oxide of hydrogen has a mass g. Deduce it’s formula.

27. Moles of Water of Crystallisation
Many compounds can incorporate water into their crystal structure, this is called the water of crystallisation.
CoCl2 is blue CoCl2.6H2O is pink
The ‘.’ means the water is ‘associated’ with the CoCl2
It is loosely bonded, but exactly how is unimportant
In this experiment you will calculate the moles of water of crystallisation of a compound

28. Key Points

29. Lesson 3
Solutions

30. 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 2.50 ± 0.02 g
Determine the amount, in mol, of salicylic acid, C6H4(OH)COOH, used.

31. We Are Here

32. Lesson 3: 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)

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

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

35. 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 % by weight, % by volume and molality.

36. 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

37. 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
= 4.00/40.0
= 0.100
Calculate volume of solution:
Volume = moles / concentration
= / 2.00
= dm3
= 50.0 cm3

38. Example 3:
It is found by titration that 25.0 cm3 of an unknown solution of sulfuric acid is just neutralised by adding cm3 of1.00 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)/1 = (1.00 x11.3)/2
C1 = ((1.00 x 11.3) / 2)/25.0) = mol dm-3
Where:
n = coefficient
C = concentration
V = volume
‘1’ refers to H2SO4
‘2’ refers to NaOH
𝐶 1 𝑉 1 𝑛 1 = 𝐶 2 𝑉 2 𝑛 2

39. 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 mass of copper (II) chloride (CuCl2) should be added to 240 cm3 water 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?
The titration of 50.0 cm3 of an unknown solution of barium hydroxide was fully neutralised by the addition of 12 cm3 of mol dm-3 hydrochloric acid solution. What concentration is the barium hydroxide solution?
Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O

40. Preparing a Standard Solution
Prepare standard solutions silver nitrate with concentrations of mol dm-3
You will use these in a future lesson so make them well!

41. 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.

42. Key Points

43. Alcohol is not the answer but it is a solution*!
Remember!
Alcohol is not the answer but it is a solution*!
*Of ethanol, water and various other bits and pieces

44. Lesson 4
Formulas

45. 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.

46. We Are Here

47. Lesson 4: Formulas Objectives:
Understand the difference between empirical and molecular formulas
Experimentally determine the empirical formula of a compound.
Solve problems involving empirical and molecular formulas

48. Formulas
You have one minute to write down as many different chemical formulas as you can.
Go!
What is the job of a formula?
No!!!!

49. Types of Formula Empirical: Molecular:
The ratio of the atoms of each element in a compound in its lowest terms
Molecular:
The number of atoms of each element in a molecule
You will meet other types in the organic chemistry unit including structural, displayed and skeletal.

50. Examples *Why do these two not have an empirical formula? Oxygen Water
Molecular: O2
Empirical: O
Water
Molecular: H2O
Empirical: H2O
Ethane
Molecular: C2H6
Empirical: CH3
Glucose
Molecular: C6H12O6
Empirical: CH2O
Sodium Chloride
Molecular: n/a*
Empirical: NaCl
Copper Sulphate
Empirical: CuSO4
*Why do these two not have an empirical formula?

51. Your turn… Chlorine gas Propane Sulphuric acid Ethanol
Write empirical and molecular formulas for each of the following:
Chlorine gas
Propane
Sulphuric acid
Ethanol

52. Determining empirical formulas from % by mass data*
For example: a sample of a compound contains 20% hydrogen and 80% carbon by mass.
Empirical formula = CH3
*This may seem like an odd thing to do but this data is easily determined in the lab and so is very useful C: H
write out the elements as a ratio
80%
20%
write the % composition below
6.67 20
divide each % by the Ar of each element 1 3
divide each number by the smallest of the two numbers

53. Determining Molecular Formulas From Empirical Formulas
You need to know the molecular mass (can be found from mass spectrometry (see Atomic Structure unit).
Following on from the previous question. Your unknown molecule is found to have Mr = 30.06
Determine empirical formula mass:
Mr(CH3) = (3 x 1.01) = 15.03
Divide molecular mass by empirical formula mass to tell you the multiplier for the empirical formula:
Multiplier = / = 2
Multiply the empirical formula by the multiplier:
Molecular formula = CH3 x 2 = C2H6

54. Determining an empirical formula
You will be determining the empirical formula of an unknown compound of potassium, chlorine and oxygen.
Follow the instructions here

55. Practice Questions Begin the practice questions found here
Complete for homework.

56. Key points: Empirical formula gives the lowest ratio of atoms
The molecular formula is a whole multiple of the empirical formula
When using % by mass data, start by dividing each % by the Ar of that element.

57. Lesson 5
Equations

58. Refresh
A toxic gas, A, consists of 53.8 % nitrogen and 46.2 % carbon by mass and has a molar mass of g/mol.
Determine the empirical formula of A.
Determine the molecular formula of A

59. We Are Here

60. Lesson 5: Equations 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’

61. Word Equations Hydrogen + Oxygen  Water REACTANTS PRODUCTS
The arrow means: ‘makes’ or ‘becomes’ not ‘equals’

62. Symbol Equations 2 H2 + O2  2 H2O vs H2 + O2  H2O 4 H 4 H 2 H 2 H
2 O O O 1 O
The bold numbers are called coefficients and tell you the number of each molecule involved in the reaction
Required to balance the equation
Without them the equation does not balance – each side of the reaction would have different numbers of each atom – which would break physics
You can’t change the little numbers in the formulas as this changes the chemical
If there is no coefficient, it is ‘1’

63. Tips for balancing equations
Only change the coefficients
Only make one change at a time
Re-count the numbers of each atom after each change
Try keeping a tally-chart of the numbers of atoms on each side of the equation
If you get to a point where you just need a fractional amount of a compound, write it in and then multiply everything by the denominator of the fraction.
BE PATIENT!!!!

64. Construct equations and then balance each of the following:
Magnesium (Mg) reacts with hydrochloric acid (HCl) to make magnesium chloride (MgCl2) and hydrogen gas
Ethane (C2H6) reacts with oxygen gas to make carbon dioxide and water
Lead nitrate (Pb(NO3)2) reacts with aluminium chloride (AlCl3) to make aluminium nitrate (Al(NO3)3) and lead chloride (PbCl2)
Barium nitride (Ba3N2) reacts with water to make barium hydroxide (Ba(OH)2) and ammonia (NH3)
Extension: Write a flow chart that enables you to balance equations

65. 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

66. Mole Ratios cont…. 2 Al(OH)3 + 3 H2SO4  Al2(SO4)3 + 3 H2O
In the above equation, what quantity of Al(OH)3 in moles is required to produce 5.00 mol of H2O?
Divide the given moles by the given coefficient and multiply by the target coefficient:
n(Al(OH)3) = 5.00/3 x 2 = 3.33 mol*
*Dividing by ‘3’ and multiplying by ‘2’ uses the H2O:Al(OH)3 mole ratio of 3:2 to adjust the number of moles.

67. The Limiting Reagent
In a reaction, there is 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
You should use the limiting reactant when doing all further calculations including:
Determining amounts of products formed
Determining amounts of other reactants used

68. Use the idea of limiting reactants to identify the limiting reactant and solve the problem.
Mg + 2 HCl  MgCl2 + H2
0.15 mol Mg reacting with 0.25 mol HCl. How many moles of H2 is formed?
2 C2H O2  4 CO2 + 6 H2O
0.55 mol C2H6 reacting with 3.50 mol O2. How many moles of CO2 is formed?
3 Pb(NO3)2 + 2 AlCl3  2 Al(NO3)3 + 3 PbCl2
2.60 mol Pb(NO3)2 reacting with 3.00 mol AlCl3. Which element is in excess and how much remains after the reaction?
Ba3N2 + 6 H2O  3 Ba(OH)2 + 2 NH3
1.45 mol Ba3N2 reacting with 10.0 mol H2O. Which reactant is present in excess and how much remains after the reaction?

69. 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

70. Lesson 6
Theoretical Yields

71. 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

72. We Are Here

73. Lesson 6: Theoretical Yields
Objectives:
Use the idea of limiting reactant to determine the proportions of an unknown mixture
Use stoichiometry to calculate theoretical yields

74. 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

75. Determining Theoretical Yield
The theoretical yield is the amount of product you expect to make if all of your limiting reactant fully reacts.
It can be calculated by following these steps:
All this does is string together all the calculations you have already met.
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.

76. Some problems
Have a go at the problems found here.

77. 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.

78. Lesson 7
Molar Volumes of Gases

79. 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.

80. We Are Here

81. Lesson 7: Equations 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.

82. The Molar Volume of an Ideal Gas
At standard temperature and pressure (STP):
Molar Volume of Gas = 2.24x10-2 m3 mol-1
= 22.4 dm3 mol-1
T = 273K, P = 1.01x105 Pa
At room temperature and pressure (RTP):
Molar Volume of Gas = 2.45x10-2 m3 mol-1
= 24.5 dm3 mol-1
This is true (ish) no matter what the gas!...we will look at why next lesson

83. 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) /2 x 1 = 0.050/1 x 1 = 0.025
V(H2) = x 22.4 = 0.56 dm3

84. More calculations
At RTP, 30 cm3 ethane reacts with 60 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: 30/2 = 15
O2: 60/10 = 6
therefore O2 limiting, ‘6’ 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 x 2 = 12 cm3
V(C2H6 remaining) = 30 – 12 = 18 cm3
Note: there is no need to convert to moles as they are all in a ratio of moles as you would divide by 24.0 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 x 4 = 24 cm3

85. CuO(s) + H2(g)  Cu(s) + H2O(l)
Time to practice
What is the minimum volume of H2 gas required to fully reduce 10.0 g copper (II) oxide to copper (assume RTP…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 RTP?
2 NaN3(s)  2 Na(s) + 3 N2(g)
500 cm3 carbon monoxide reacts with 300 cm3 oxygen to produce carbon dioxide. What are the final volumes of each of the three gases on completion of the reaction?
CO(g) + O2(g)  CO2(g)

86. 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
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!)

87. 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
Molar Volume at RTP = 24.5 dm3

88. Lesson 8
Ideal Gases

89. 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

90. We Are Here

91. Lesson 8: 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

92. 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

93. 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

94. Ideal Gas Assumptions Particles occupy no volume
Particles have zero intermolecular forces
These are not always valid, particularly at:
Low temperature
High pressure

95. Study the equation to make some predictions:
How would does temperature change if you decrease pressure at fixed volume?
How would volume change if you heat something at fixed pressure?
How would pressure change if you decrease the volume at fixed temperature?

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

97. Some calculations:
1.048 g of unknown gas A, occupies 846 cm3. at 500K What is it’s molecular mass?
Mm(A) = mass / moles = / = 60.0 g/mol

98. 54.0 converted to Kelvin by adding 273
More calculations
The volume of an ideal gas at 54.0 °C is increased from 3.00 dm3 to 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
54.0 converted to Kelvin by adding 273

99. Time to practice
Complete the questions found here

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

101. Lesson 9
Test

102. Good Luck
You have 80 minutes!

103. Lesson 10
Test Debrief

104. 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.

105. 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

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

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