1. Quantitative Chemistry

2. Activity:
Print out the keywords worksheet and add the following words, once you know the meanings fill them in.
Relative atomic mass
Molecule
Relative molecular mass
Mole
Relative formula mass
Isotope
Avogadro’s number
Diatomic
Element
Compound

3. Chemical Formulas and the Mole concept
Elements
Compounds
Mole concept and Avogadro's constant
Isotopes
Formulas of compounds
Empirical formula
Molecular formula
Structural formula

4. Elements All substances are made up of one or more __________.
An element ___________be broken down by any chemical process into ___________substances.
There are just over _____ known elements.
The smallest part of an element is called an __________.
Hand out IB periodic table/ data book that they must bring each lesson
cannot
100
elements
atom
simpler

5. Molecules and compounds
Some substances are made up of a single element although there may be more than one atom of the element in a particle of the substance.
For example, Oxygen is diatomic, that is, a molecule of oxygen containing two oxygen atoms.
A compound contains more than one element.
For example; a molecule of water contains two hydrogen atoms and one oxygen atom.
Water is a compound not an element because it can be broken down chemically in to it’s constituent elements: hydrogen and oxygen

6. Mole concept and Avogadro’s constant
A single atom of an element has an extremely small mass. For example, an atom of carbon–12 has a mass of x g.
This is far too small to weigh
Instead of weighing just one atom we can weigh a mole of atoms.
1 mole contains 6.02 x particles
6.02 x atoms of carbon–12 = ?
This number is known as Avogadro’s constant(NA or L)
1 mole of Carbon = 12g

7. Mole concept and Avogadro’s constant
Chemists measure amounts of substances in moles.
A mole is the amount that contains L particles of that substance.
The mass of one mole of any substance is known as the molar mass and has the symbol M. For example, Hydrogen atoms have one 12th the mass of carbon–12 atoms so one mole of hydrogen atoms contains x hydrogen atoms and has a mass of 1.01 g.
For diatomic molecules e.g. H2 there are 6.02 x molecules of hydrogen and therefore x 1023 atoms.
Try some molar calculations involving Avogadro’s const.

8. Relative Atomic mass
The actual atomic mass of an individual atom of an element is so small that we use a relative atomic mass as seen on the periodic table.
Since hydrogen is the lightest atom (it has an actual mass of x 10-25) we say that hydrogen has a relative atomic mass of 1.
Experiments have shown that an atom of carbon weighs 12 times as much as an atom of hydrogen. So the relative atomic mass of carbon is 12.
Relative atomic mass can be abbreviated to Ar or RAM and is found on the periodic table. It has no units.

9. Activity Complete the worksheet IB 1.1 – Using Avogadro’s Number
There are 20 questions to complete
Show all working

10. Isotopes and RMM/RFM
In reality elements are made up of a mixture of isotopes.
The relative atomic mass of an element Ar is the weighted mean of all the naturally occurring isotopes of the element relative to carbon–12.
This explains why the relative atomic masses given for elements on the periodic table are not whole numbers
The units of molar mass are g mol-1 (this means grams per mole) but relative molar masses Mr have no units.
For molecules relative molecular mass is used (RMM). For example the Relative Molecule Mass of glucose, C6H12O6 = (6 x 12.01) + (12 x 1.01) + (6 x 16.00) =
Recall what an isotope is/ look it up in text book
RMM is the Relative Molecular Mass for covalent molecules / RFM is the relative formula mass for any compounds it is safer to use RFM rather than RMM if you’re not sure what type of compound it is.
mol-1 means per

11. Calculating RFM
This is very simple but there are one or two rules to remember:
Numbers in subscript only refer to the number directly in front of it. E.g. SO4 there is one Sulphur atom and 4 Oxygen atoms RFM = 96 in total
Everything inside a bracket is multiplied by the small number directly outside of it on the right hand side. E.g. (NH4)2SO4 So the RFM of NH4 must be multiplied by 2 RFM = 132 in total
When a . is placed into a formula it means another compound is attached to the formula and must be included in the RFM e.g. CuSO4.5H2O means there are 5 water molecules added to this compound so you must include them in the total RFM = in total

12. Activity
Complete the worksheet IB Calculation of the Molar Mass of Compounds
There are 60 questions to complete
Don’t forget to refer to the previously stated rules
Show all working

13. Using Molar masses
As we have seen 1g of Hydrogen contains many more atoms than 1g of carbon.
Calculate the number of atoms in 1g of hydrogen and 1g of carbon.
Suggest why we cannot use mass as a means of keeping a fair test during an experiment i.e. The reaction of substances with oxygen which would give out more energy 1g of hydrogen or 1g of carbon?
To keep a test fair we use molar quantities.
1g of hydrogen = 6 x 1023 atoms
1g of carbon = 0.5 x 1023
Not fair because there are far more hydrogen atoms to react than carbon atoms therefore will have more reactions and give out more energy.

14. Calculating molar quantities
As we have seen a mole is a fixed number of particles, 6 x 1023, we can find the molar mass of an element from the periodic table (mass number). The molar mass of a compound is just the sum of all the elements in the formula in their correct proportions.
E.g.
1 mole of Na = 23g
1 mole of NaOH = = 40g
1 mole of Na2SO4 = (2 x 23) (4 x 16) = 142g

15. Finding the number of moles
Mass ÷
Number of moles
RAM or RFM x
It is very rare to use exactly one mole of a substance in a reaction. So it is important to be able to find the number of moles of elements and compounds in different reactions.
We can do this using the following equation triangle.
You must learn this!!!!!!

16. Example 1; ÷ x Find the number of moles in 46g of sodium Mass
RAM or RFM x
Mass 46
Number of moles
________ = 2 23
RAM or RFM

17. Example 2; ÷ x Find the mass of 0.2 moles of sodium Mass
Number of moles
RAM or RFM x
Number of moles
RAM or RFM
4.6 g
0.2 x 23
Mass =

18. Activity
Complete the worksheet IB 1.3– Molar calculations and IB 1.4 Molar calculations (II)
There are 60 questions to complete
Don’t forget to refer to the previously stated rules
Show all working

19. Formula’s of compounds
Compounds can be described by different chemical formulas:
Empirical
Molecular
Structural

20. Empirical Formula
Literally this is the formula obtained by experiment.
This shows the simplest whole number ratio of atoms of each element in a particle of the substance
It can be obtained by either knowing the mass of each element in the compound or from the percentage composition by mass of the compound.
The percentage composition can be converted directly into mass by assuming 100g of the compound are taken.

21. Formula Triangle
You can use a formula triangle to help you rearrange the equation:
Mass ÷
Number of moles
RAM or RFM x
Mass
Number of moles
________ =
Practice questions on moles
RAM or RFM

22. Example
A compound contains 40.00% carbon, 6.73% hydrogen and 53.27% oxygen by mass, determine the empirical formula.
Amount / mol
ratio C H O
40.00/12.01 =
3.33 1
6.73/1.01 =
6.66 2
53.27/16.00 =
3.33 1
Empirical formula is therefore – CH2O

23. Worked example
An organic compound contains 60.00% carbon and 4.46% hydrogen by mass. Calculate the empirical formula of the compound.
As the percentages do not add up to 100% it can be assumed that the compound also contains 35.54% oxygen by mass.
Assume that there are 100g of compound, and calculate the amount of each element by dividing the mass of the element by its atomic mass Ar
Find the simplest ratio by dividing through by the smallest amount and then converting any decimals to give the simplest whole number ratio
Write the empirical formula, using subscripts for the numbers in the ratio.
The empirical formula of the compound is
Practice questions
C9H8O4

24. Activity;
Read through the examples 16, 17 and 18 on pages 31 and 32 in ‘Calculations in AS/A Level Chemistry’ – Jim Clark
Complete problems 15 and 16 part a only.
Answers are at the back of the book page 279
Complete worksheet 1.5 Formulae and Percentage composition

25. Molecular formula
For molecules this is much more useful as it shows the actual number of atoms of each element in a molecule of the substance.
It can be obtained from the empirical formula if the molar mass of the compound is known.
Methanal CH2O (Mr = 30), ethanoic acid C2H4O2 (Mr = 60) and glucose C6H12O6 (Mr = 180) are different substances with different molecular formulas but all with the same empirical formula CH2O.
Note that the subscripts are used to show the number of atoms of each element in the compound.

26. Activity;
Using your answers from part a (from last lesson)complete problems 15 and 16 part b only.
Answers are at the back of the book page 279
Complete worksheet; Worksheet IB1.6 Section B-Molecular Formulae

27. Structural formula
This shows the arrangement of atoms and bonds within a molecule and is particularly useful in organic chemistry.
The three different formulas can be illustrated using ethene:
CH2
C2H4
CH2CH2
Empirical
Molecular
Structural

28. Structural formula
Two compounds with the same molecular formula but different structural formulas are known as isomers.
For example, both 1,1-dichloroethane and 1,2-dichloroethane have the same empirical formula, CH2Cl, and the same molecular formula, C2H4Cl2, but their structural formulas are different.
In 1,1-dichloroethane both the chlorine atoms are bonded to the same carbon atom, whereas in 1,2-dichloroethane each of the two carbon atoms has one chlorine atom bonded to it Cl C H H C Cl
1,1-dichloroethane
1,2-dichloroethane

29. Structural formulas
The use of structural formulas is particularly important in organic chemistry.
Compounds with the same molecular formula but with different structural formulas are called structural isomers.
Structural isomers often have very different physical and chemical properties. H C O
For example, methoxymethane, CH3OCH3, and ethanol, C2H5OH, both have the same molecular formula, C2H6O but, unlike methoxymethane, ethanol is completely miscible with water and is generally much more reactive chemically.
methoxymethane H C O
ethanol

30. Quick Questions
What is the empirical formula of the following compounds? You could try to name them too.
A liquid containing 2.0g of hydrogen, 32.1 g sulphur, and 64g oxygen.
A white solid containing 0.9g beryllium, 3.2g oxygen, and 0.2g hydrogen.
A white solid containing 0.234g magnesium and 0.710g chlorine.
3.888g magnesium ribbon was burnt completely in air and 6.488g of magnesium oxide were produced.
How many moles of magnesium and of oxygen are present in 6.488g of magnesium oxide?
What is the empirical formula of magnesium oxide?
More to follow…

31. Just a few more!
What are the empirical formulae of the following molecules?
Cyclohexane C6H12
Dichloroethene C2H2Cl2
Benzene C6H6
Mr for ethane-1,2-diol is It is composed of carbon, hydrogen and oxygen in the ratio by moles of 1:3:1. What is its molecular formula?

32. Answers 1 C2H6O2 H2SO4 sulphuric acid
BeO2H2 which is Be(OH)2, beryllium hydroxide
MgCl2 magnesium chloride
0.16 moles of each
MgO
CH2
CHCl CH
C2H6O2

33. Working out formulae for ionic compounds
You can’t write equations until you can write formulae.
People tend to remember the formulae for common covalent substances like water or carbon dioxide or methane, and will rarely need to work them out.
That is not true of ionic compounds. You need to know the symbols and charges of the common ions and how to combine them into a formula.

34. The need for equal numbers of “pluses” and “minuses”
Ions are atoms or groups of atoms which carry electrical charges, either positive or negative.
Compounds are electrically neutral.
In an ionic compound there must therefore be the right number of each sort of ion so that the total positive charge is exactly the same as the total negative charge.
Obviously, then, if you are going to work out the formula, you need to know the charges on the ions.

35. Some charges you may remember!
Can you put the correct charges into the table?
Charge
Group 1 metals
Group 2 metals
Group 3 metals
Group 5 non-metals
Group 6 non-metals
Group 7 non-metals +3 +1 -1 +2 -2 -3

36. Cases where the name tells you the charge on an ion
A name like lead(II)oxide tells you that the charge on the lead (a metal) is +2.
Iron(III)chloride contains 3+ iron ion.
Copper(II) sulphate contains a Cu2+ ion
Notice that all metals form positive ions.
Ions that need to be learnt:
Positive ions
Negative ions
Zinc
Nitrate
Silver
Hydroxide
Hydrogen
Hydrogencarbonate
ammonium
Carbonate
Lead
Sulphate
Phosphate
Zn2+
NO3-
Ag+
OH- H+
HCO3-
NH4+
CO32-
Pb2+
SO42-
PO43-

37. Confusing endings Do not confuse ions like sulphate with sulphide.
Try naming the following compounds:
A name like sodium sulphide means that it contains sodium and sulphur only.
Mg3N2
Mg(NO3)2
CaC2
CaCO3
Once you have an “ate” ending it means that there is something else there as well – often, but not always, oxygen

38. Activity
Fill in the worksheet; Worksheet IB1.6 Important ions for IB calculations how many can you complete without looking them up?
Click here for answers.

39. Working out the formula of an ionic compound
Example 1.
To find the formula for sodium oxide, first find the charges on the ions
Sodium is in Group 1, so the ion is Na+
Oxygen is in Group 6, so the ion is O2-
To have equal numbers of positive and negative charges, you would need 2 sodium ions for each oxide ion
The formula is therefore Na2O

40. Working out the formula of an ionic compound
Example 2.
To find the formula for barium nitrate, first find the charges on the ions
Barium is in Group 2, so the ion is Ba2+
Nitrate ions are NO3-
To have equal numbers of positive and negative charges, you would need 2 nitrate ions for each barium ion
The formula is therefore Ba(NO3)2
Notice the brackets around the nitrate group. Brackets must be written if you have more than one of these complex ions (ions containing more than one type of atom).

41. Working out the formula of an ionic compound
Example 3.
To find the formula for iron(III)sulphate, first find the charges on the ions
Iron(III) tells us that the ion is Fe3+
Sulphate ions are SO42-
To have equal numbers of positive and negative charges, you would need 2 iron (III) ions for every 3 sulphate ions
The formula is therefore Fe2(SO4)3

42. Activity
Complete the worksheet; Worksheet IB1.7 Writing formulae from names
For answers click here

43. Chemical Reactions and equations
Properties of chemical reactions
Chemical equations
State symbols
One way or reversible reactions
Ionic equations

44. Properties of chemical reactions
Once the correct formulas of all the reactants and products are known, it is possible to write a chemical equation to describe a reaction taking place.
New substances are formed.
Bonds in the reactants are broken and bonds in the products are formed, resulting in an energy change between the reacting system and the surroundings.
There is a fixed relationship between the number of particles of reactants and products, resulting in no overall change in mass.

45. Chemical Equations
In a chemical equation the reactants are written on the left hand side, and the products are written on the right hand side.
As there is no overall change in mass, the total amount of each element must be the same on the two side of the equation.
For example, consider the reaction between zinc metal and hydrochloric acid to produce zinc chloride and hydrogen gas.
The correct formulas for all the reacting species and products are first written down. Zn +
HCl →
ZnCl2 + H2
REACTANTS
PRODUCTS

46. Chemical Equations
The equation is then balanced by adding the correct coefficients
When the correct coefficients are in place, the reaction is said to be stoichiometrically balanced.
The stoichiometry tells us that in this reaction two moles of hydrochloric acid react with one mole of zinc to form one mole of zinc chloride and one mole of hydrogen. Zn + 2
HCl →
ZnCl2 + H2
REACTANTS
PRODUCTS

47. State symbols
The physical state that the reactants and products are in can affect both the rate of the reaction and the overall energy change so it is good practice to include the state symbols in the equation.
(s)
solid
(aq)
In aqueous solution
(l)
liquid
(g)
gas Zn
(s) + 2
HCl
(aq) →
ZnCl2
(aq) + H2
(g)
REACTANTS
PRODUCTS

48. Activity Complete the worksheet; Worksheet IB1.8 Balancing equations
For answers click here

49. One way or reversible
A single arrow → is used if the reaction goes to completion.
Sometimes the reaction conditions are written on the arrow:
Reversible arrows are used for reactions where both the reactants and products are present in the equilibrium mixture:
Ni Catalyst, 180oC
C2H4(g) +
H2(g)
C2H6(g)

50. Ionic equations
Ionic compounds are completely dissociated in solution so it is sometimes better to use ionic equations to describe their reactions.
For example, when silver nitrate solution is added to sodium chloride solution a precipitate of silver chloride is formed.
Na +(aq) and NO3 – (aq) are spectator ions and do not take part in the reaction. So the ionic equation becomes:
AgNO3(aq) +
NaCl(aq)
NaNO3(aq) +
AgCl(s)
Ag+(aq) +
Cl-(aq)
AgCl(s)

51. Activity
Carry out Experiment 1.1 Precipitation reactions

52. Precipitation Reactions
Metal compound
Metal ion
Product (
Observation
MgCl2(aq)
Mg2+
Mg(OH)2(s)
White solid
FeSO4(aq)
Fe2+
Fe(OH)2(s)
Dirty green solid
CaCl2(aq)
Ca2+
Ca(OH)2(s)
FeCl3(aq)
Fe3+
Fe(OH)3(s)
Rusty brown solid
Cu(NO3)2(aq)
Cu2+
Cu(OH)2(s)
Blue solid
To each of the following solutions add NaOH solution drop-wise to see a precipitate form

53. Ionic equations Mg2+(aq) MgCl2(aq) 2NaOH(aq) 2OH-(aq) + Mg(OH)2(s) +
2NaCl(aq)
FeSO4(aq)
Fe2+(aq) +
2OH-(aq)
2NaOH(aq)
Fe(OH)2(s) +
Na2SO4(aq)
Ca2+(aq)
CaCl2(aq) +
2NaOH(aq)
2OH-(aq)
Ca(OH)2(s)
2NaCl(aq) +
Fe(OH)3(s) +
3NaCl(aq)
FeCl3(aq)
Fe3+(aq)
3OH-(aq)
3NaOH(aq) +
Cu2+(aq)
Cu(NO3)2(aq)
2OH-(aq)
2NaOH(aq)
Cu(OH)2(s) + +
2NaNO3(aq)

54. Measurements and calculations
Error and uncertainty
Measurements of molar quantities
Solids
Solutions
Liquids
Gases
Worked examples

55. Error and uncertainty
In the laboratory moles can conveniently be measured using either mass or volume depending on the substances involved.
But with any measurement there is always uncertainty and error.

56. Systematic errors Quantitative chemistry involves measurement.
A measurement is a method which some quantity or property of a substance is compared with a known standard.
If the instrument used to take the measurements has been calibrated wrongly or if the person using it consistently misreads it then the measurements will always differ by the same amount.
Such an error is known as a systematic error.
An example might always be reading a pipette from the sides of the meniscus rather than from the middle of the meniscus.

57. Random uncertainties
Random uncertainties occur if there is an equal probability of the reading being too high or too low from one measurement to the next.
These might include variations in the volume of glassware due to temperature fluctuations or the decision on exactly when an indicator changes colour during an acid base titration.

58. Precision and accuracy.
Precision refers to how close several experimental measurements of the same quantity are to eachother.
Accuracy refers to how close the readings are to the true value. This may the standard value,

59. Solids Solids are normally measured by weighing to obtain the mass
Masses are measured in grams and kilograms remember kg = 1000g
When weighing a substance the mass should be recorded to show the accuracy of the balance.
For example, exactly 16 g of a substance would be recorded as g on a balance weighing to + or – 0.01 g but as g on a balance weighing to + or – g.

60. Liquids Pure liquids may be weighed or the volume recorded.
The density of the liquid = mass / volume and is usually expressed in g cm-3 or kg dm-3
In the laboratory, volume is usually measured using different apparatus depending on how precisely the volume is required
For very approximate volumes a beaker or conical flask can be used.
Measuring cylinders are more precise, but still have a large amount of uncertainty.
For fixed volumes, volumetric flasks or pipettes are used for precise measurements, and for variable volumes a burette or graduated pipette is used.

61. Measuring Liquids
The uncertainty associated with the burette, pipette or volumetric flask can vary.
In schools, grade B equipment is usually used, but more expensive grade A (analytical) equipment can be accurate to ± 0.01cm3.

62. Solutions Volume is usually used for solutions.
1.000 litre = dm3 = 1000cm3
Concentration is the amount of solute (dissolved substance) in a known volume of solution (solute plus solvent).
It is expressed either in g dm-3 or more usually in mol dm-3
We will work out how to calculate concentrations and carry out volumetric calculations later in the topic.

63. We can use this formula…..
You can use a formula triangle to help you rearrange the equation:
moles ÷
moles
concentration =
Students complete exercises
Vol in dm3 x
concentration
Vol in dm3

64. Do you still remember it????
You can use a formula triangle to help you rearrange the equation:
Mass ÷
Mass
Number of moles
________ =
Practice questions on moles
Number of moles
RAM or RFM
RAM or RFM x

65. Concentrations of solutions
Concentrations can be measured in:
g dm-3
mol dm-3
And you have to be able to convert between them
You have already practiced converting moles into grams and vice versa.
When you are doing the conversions in concentration sums, the only thing that is different is the amount of substance you are talking about happens to be dissolved in 1dm3 of solution. This does not affect the sum in any way.

66. Calculations involving solutions
A solution of sodium hydroxide, NaOH, had a concentration of 4g dm-3. What is its concentration in mol dm-3?
(H = 1; O = 16; Na = 23)
1 mole of NaOH weighs 40g
4g is 4/40 moles = 0.1 mol
4 g dm-3 is therefore 0.1 mol dm-3

67. Calculations involving solutions
What is the concentration of a mol dm-3 solution of sodium carbonate, Na2CO3, in g dm-3?
(C = 12; O = 16; Na = 23)
1 mol Na2CO3 weighs 106g
mol weighs x 106g = 5.30g
So mol dm-3 is therefore 5.30g dm-3

68. Calculations involving concentrations
Recall ionic charges and names of ionic formulae.
Know how to solve problems involving concentration and amount of solute.

69. Basic calculations from equations involving solutions.
Example:
What mass of barium sulphate would be produced by adding excess barium chloride to 20.0cm3 of copper(II)sulphate solution of concentration 0.100mol dm-3 (O = 16, S = 32, Ba = 137)
BaCl2(aq) + CuSO4(aq) → BaSO4(aq) + CuCl2(aq)
From this equation we can see that 1 mole of CuSO4 gives 1 mole BaSO4
With the information given you can only work out the number of moles of copper(II)sulphate that you are starting with.

70. Volume of CuSO4 in Qn 20
Number of moles of CuSO4
0.100 = x
Concentration of CuSO4 given in qn.
1000
Converts cm3 to dm3
= mol

71. Going back to the question!!
If 1 mole of CuSO4 gives 1 mole of BaSO4
therefore mol of CuSO4 will give mol of BaSO4
RFM of 1 mol of BaSO4 = 233
Therefore mol of BaSO4 = x
233
= g of BaSO4

72. Next example:
What is the maximum mass of calcium carbonate which will react with 25cm3 of 2.00mol dm-3 hydrochloric acid? (C = 12, O = 16, Ca = 40)
Write the balanced symbol equation.
Calculate the number of moles of hydrochloric acid.
Work out the ratio of moles that HCl reacts with CaCO3
Calculate the number of moles of Calcium chloride
Work out the RFM of calcium carbonate
What is the maximum mass which could react with HCl?

73. 1 mol CaCO3 reacts with 2 mole HCl
Here it is…..
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Number of moles of HCl = 25.0 / 1000 x 2.00
= mol
1 mol CaCO3 reacts with 2 mole HCl
= mol
1 mol CaCO3 weighs
Therefore mol weighs x 100 =
So the max mass of CaCO3 which you could react with this amount of hydrochloric acid is 2.50g
100g
2.50 g

74. Here it is….. CaCO3 + 2HCl → CaCl2 + H2O + CO2
Number of moles of HCl = 25.0 / 1000 x 2.00
= mol
1 mol CaCO3 reacts with 2 mole HCl
0.5 mol CaCO3 reacts with 1 mole HCl
= mol
1 mol CaCO3 weighs
100g
Therefore mol weighs x 100 =
2.50 g
So the max mass of CaCO3 which you could react with this amount of hydrochloric acid is 2.50g

75. Last one….
What is the minimum volume of mol dm-3 sulphuric acid needed to react with 0.240g of magnesium? (Mg = 24)
Write a balanced symbol equation.
Work out the ratio which magnesium reacts with sulphuric acid.
Calculate the number of moles of magnesium
Calculate the number of moles of sulphuric acid
Calculate the volume of sulphuric acid needed.

76. Here it is… Mg + H2SO4 → MgSO4 + H2
1 mole of Mg reacts with 1 mol of H2SO4
Number of moles of Mg
0.240 / 24 =
mol =
So number of moles of H2SO4 =
mol

77. Next we can use this formula…..
So the volume in dm3
moles =
Moles / Concentration
/ 0.500 ÷ =
dm3 =
Students complete exercises x
concentration
Vol in dm3
Which is what in cm3

78. Calculations involving concentrations
Know how to solve problems involving concentration and amount of solute.
Know how to solve simple titration equations

79. Simple volumetric calculations
25.0cm3 of mol dm-3 NaOH solution required 23.5cm3 of dilute hydrochloric acid for neutralisation. Calculate the concentration of the hydrochloric acid.
Write the balanced symbol equation
Work out the ratio of reactants and products
What volumes of substances are known?
What concentrations are known?
What are you trying to calculate?

80. NaOH + HCl → NaCl + H2O 1 : 1 : 1 : 1 ratio 25cm3 23.5cm3 volumes
?????
concentrations
0.100M
The only thing we can do with the numbers given is to calculate the number of moles of NaOH…..
Moles of NaOH =
Vol (in dm3) x concentration =
25/1000 x 0.100 =
mol

81. NaOH + HCl → NaCl + H2O 1 : 1 : 1 : 1 ratio 25cm3 23.5cm3 volumes
What else do we now know?
?????
concentrations
0.100M
No. of moles
mol
mol
So now we can calculate the concentration of HCl using the formula:
concentration =
Mole / volume in dm3 =
/ (23.5/1000) =
0.106 mol dm3

82. Mass and Gaseous volume relationships
Calculate theoretical yields from chemical equations.
Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given
Solve problems involving theoretical, experimental and percentage yield.

83. Molar volume of a gas Avogadro’s law states that:
‘equal volumes of different gases at the same temperature and pressure will contain the same number of moles’
From this it follows that one mole of any gas will occupy the same volume at the same temperature and pressure.
This is known as the molar volume of gas.
At 273K (0oC) and x 105 Pa (1atm) pressure this volume is 2.24 x 10-2m3 (22.4 dm3 or cm3)

84. Calculations from equations
Write down the correct formulae for all the reactants and products
Balance the equation to obtain the correct ratio of the reactants to products
If the amounts of all reactants are known work out which are in excess and which one is the limiting reagent. By knowing the limiting reagent the maximum yield of any of the products can be determined.
Work out the number of moles of the substance required.
Convert the number of moles into the mass or volume.
Express the answer to the correct number of significant figures and include the appropriate units.

85. Example
Calculate the volume of hydrogen gas evolved at 273K and 1 atm pressure when 0.623g of magnesium reacts with 27.3 cm3 of 1.25 moldm-3 hydrochloric acid.

86. Equation and amounts Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
RAM for Mg = 24.31
Amount of Mg present = mass / RAM
= / 24.31
= 2.56 x 10-2mol
Amount of HCl present = vol / 1000 x concn
= 27.3 / 1000 x 1.25
= 3.41 x 10-2 mol

87. Excess and limiting
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq) Ratio: 1 : 2 : 1 : 1 Amounts: 2.56 x 10-2mol 3.41 x 10-2 mol So 2 x 2.56 x 10-2 mol of HCl will react completely with the Mg present = 5.12 x 10-2 mol So what is in excess? Magnesium So the Hydrochloric acid is the limiting reagent

88. Theoretical yield of Hydrogen
The maximum no. of moles of hydrogen that can be produced = 3.41 x 10-2 / 2
= x 10-2 mol
So the volume of the hydrogen at 273K and 1atm
= mol x 22.4
= (1.705 x 10-2) x 22.4
= dm3
To convert this to cm3 we need to multiply by 1000
= 382cm3

89. Gases Mass or volume may be used for gases.
Normally it is easier to measure the volume of a gas. However, as well as the amount of gas present, the volume of a gas also depends on the pressure and the temperature.
The physical behaviour of all gases is governed by the ideal gas equation:
where: p represents the pressure.
P V = n R T
V represents the volume measured in m3
N represents the amount of gas in moles = mass of gas/ Mr
T represents the absolute temperature measured in kelvin K.
R represents the gas constant.

90. Gases
The units of the gas constant R can be derived from the equation. p x V =
N m-2 x m3 = J
mol K n x T =
So the SI units of R must be J K-1 mol -1
R has the value J K -1 mol -1, and is one of the best known constants in science.

91. Ideal Gases The gas equation pV = nRT is only true for an ideal gas.
Unlike ideal gases, real gases such as oxygen or hydrogen do not obey the law equation exactly.
This is because there are still some weak attractive forces between the molecules in the gas, and the molecules themselves occupy some space even though most of the volume of a gas is empty space.
However, for practical purposes we can use this ideal gas equation to describe the behaviour of real gases.

92. Avogadro's constant and molar volume of a gas.
Consider two gases, A and B:
and
If the gases occupy equal volumes, and the temperature and pressure are the same, then
and therefore

93. Calculations involving gases
Find the volume occupied by 2.20g of carbon dioxide, CO2, at 298K and a pressure of 100kPa.
( C = 12, O = 16, R = 8.31J K-1 mol-1
It helps to rearrange the ideal gas equation before you start:
pV = nRT
pV = mass (g) / mass of 1 mole (g) x RT
So; V = mass (g) / mass of 1 mole (g) x RT /p

94. Calculations involving gases.
Substitute in the numbers.
Remember to check the units.
Everything is ok except for the pressure.
This has to be entered as Pa.
The mass of one mole of CO2 is 44g.
V = mass (g) / mass of 1 mole (g) x RT / p
Students need to practice questions on gas law
So; V = 2.20 (g) / 44 (g) x (8.31 x 298) /
1.24 x m3

95. Mole Calculations
Mole calculations form the basis of many of the calculations that you will meet in IB, they include calculating:
number of moles of material in a given mass of that material.
mass of material in a given number of moles of that material.
concentrations of solutions.
volume of a given number of moles of a gas
number of moles of gas in a given volume of that gas
volume of a given mass of a gas
mass of a given volume of gas
molar mass of a gas from the mass and the volume data for the gas.