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
Chemical Formulas and the
Mole concept
 Elements
 Compounds
 Mole concept and Avogadro's constant
 Isotopes
 Formulas of compounds
 Empirical formula
 Molecular formula
 Structural formula
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
__________.
cannot
100
elements
atom
simpler
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
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 1.993 x 10-23 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 10 23 particles
6.02 x 10 23 atoms of carbon–12 = ?
This number is known as Avogadro’s constant(NA or L)
1 mole of Carbon = 12g
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
6.02 x 10 23 hydrogen atoms and has a mass of 1.01 g.
 For diatomic molecules e.g. H2 there are 6.02 x 1023
molecules of hydrogen and therefore 12.04 x 1023 atoms.
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 8.275 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.
Activity
 Complete the worksheet IB 1.1 – Using Avogadro’s
Number
 There are 20 questions to complete
 Show all working
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) = 180.18
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 = 249.5 in total
Activity
 Complete the worksheet IB 1.2 - 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
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.
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 = 23 + 16 + 1 = 40g
 1 mole of Na2SO4 = (2 x 23) + 32 + (4 x 16) = 142g
Finding the number of
moles
 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!!!!!!
Mass
÷
Number
of moles
x
RAM or
RFM
Example 1;
 Find the number of
moles in 46g of
sodium
Number
of moles
Mass
46
2
= ________
23
RAM or
RFM
Mass
÷
Number
of moles
x
RAM or
RFM
Example 2;
 Find the mass of 0.2
moles of sodium
Mass
Number
0.2 g
Mass = of4.6
moles
x
RAM
23 or
RFM
÷
Number
of moles
x
RAM or
RFM
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
Formula’s of compounds
 Compounds can be described by different
chemical formulas:
 Empirical
 Molecular
 Structural
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.
Formula Triangle
 You can use a
formula triangle to
help you rearrange
the equation:
Number
of
moles
Mass
÷
Mass
= ________
RAM or
RFM
Number
of moles
x
RAM or
RFM
Example
A compound contains 40.00% carbon, 6.73% hydrogen and
53.27% oxygen by mass, determine the empirical formula.
Amount / mol
C
40.00/12.01 =
H
6.73/1.01 =
O
53.27/16.00 =
ratio
3.33
6.66
3.33
Empirical formula is therefore – CH2O
1
2
1
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 C9H8O4
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
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.
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 BMolecular Formulae
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
Empirical
C2H4
CH2CH2
Molecular
Structural
Structural formula
 Two compounds with the same molecular formula but
different structural formulas are known as isomers.
 For example, both 1,1-dichloroethane and 1,2dichloroethane 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 1,2dichloroethane each of the two carbon atoms has one
chlorine atom bonded to it
Cl
Cl
H
C
C
H
H
H
1,1-dichloroethane
H
Cl
Cl
C
C
H
H
H
1,2-dichloroethane
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
H
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.
H
C
O
C
H
H
H
methoxymethane
H
H
H
C
C
H
H
ethanol
O
H
Quick Questions
1.
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.
b) A white solid containing 0.9g beryllium, 3.2g oxygen, and
0.2g hydrogen.
c) A white solid containing 0.234g magnesium and 0.710g
chlorine.
a)
2. 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?
b) What is the empirical formula of magnesium oxide?
More to follow…
a)
Just a few more!
3) What are the empirical formulae of the
following molecules?
a) Cyclohexane C6H12
b) Dichloroethene C2H2Cl2
c) Benzene C6H6
4) Mr for ethane-1,2-diol is 62.0. It is composed of
carbon, hydrogen and oxygen in the ratio by
moles of 1:3:1. What is its molecular formula?
Answers
1) 1
H2SO4 sulphuric acid
b) BeO2H2 which is Be(OH)2, beryllium hydroxide
c) MgCl2 magnesium chloride
a)
2) 1
0.16 moles of each
b) MgO
a)
3) 1
CH2
b) CHCl
c) CH
a)
4) C2H6O2
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.
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.
Some charges you may remember!
Can you put the correct charges into the table?
Charge
Group 1 metals
+3
Group 2 metals
+1
Group 3 metals
-1
Group 5 non-metals
+2
Group 6 non-metals
-2
Group 7 non-metals
-3
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
Zn2+
Nitrate
Silver
Ag+
NO3-
Hydroxide
OH-
HCO3-
Hydrogen
H+
Hydrogencarbonate
ammonium
NH4+
Carbonate
Lead
Pb2+
Sulphate
SO42-
Phosphate
PO43-
CO32-
Confusing endings
 Do not confuse ions like  Try naming the
sulphate with sulphide.
 A name like sodium
sulphide means that it
contains sodium and
sulphur only.
 Once you have an “ate”
ending it means that
there is something else
there as well – often,
but not always, oxygen
following compounds:
 Mg3N2
 Mg(NO3)2
 CaC2
 CaCO3
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.
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
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).
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
Activity
 Complete the worksheet; Worksheet IB1.7
Writing formulae from names
 For answers click here
Chemical Reactions and equations
 Properties of chemical reactions
 Chemical equations
 State symbols
 One way or reversible reactions
 Ionic equations
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.
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
REACTANTS
→
ZnCl2
+
PRODUCTS
H2
Chemical Equations
 The equation is then balanced by adding the
correct coefficients
Zn
+
2 HCl
REACTANTS
→
ZnCl2
+
H2
PRODUCTS
 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.
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) + 2HCl (aq) → ZnCl2 (aq) +
REACTANTS
H2(g)
PRODUCTS
Activity
 Complete the worksheet; Worksheet IB1.8
Balancing equations
 For answers click here
One way or reversible
 A single arrow → is used if the reaction goes to
completion.
 Sometimes the reaction conditions are written on
the arrow:
Ni Catalyst, 180oC
C2H4(g) + H2(g)
C2H6(g)
 Reversible arrows are used for reactions where
both the reactants and products are present in
the equilibrium mixture:
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.
AgNO3(aq) + NaCl(aq)
 Na
NaNO3(aq) + AgCl(s)
and NO3 – (aq) are spectator ions and do not
take part in the reaction. So the ionic equation
becomes:
+
(aq)
Ag+(aq) +
Cl-(aq)
AgCl(s)
Activity
 Carry out Experiment 1.1 Precipitation reactions
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)
White solid
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
Ionic equations
Mg2+(aq)
MgCl
2(aq)
+
2OH-(aq)(aq)
2NaOH
Mg(OH)2(s) + 2NaCl(aq)
FeSO
Fe2+(aq)
4(aq) +
2OH
2NaOH
(aq)(aq)
Fe(OH)2(s) +
Ca2+(aq)
CaCl
2(aq)
+
2OH
2NaOH
(aq)(aq)
Ca(OH)2(s)
+
3OH
3NaOH
(aq)(aq)
Fe(OH)3(s) + 3NaCl(aq)
Fe3+(aq)
FeCl
3(aq)
2+
Cu
Cu(NO
(aq) 3)2(aq) + 2OH
2NaOH
(aq)(aq)
Cu(OH)2(s)
Na2SO4(aq)
+ 2NaCl(aq)
+ 2NaNO3(aq)
Measurements and
calculations
 Error and uncertainty
 Measurements of molar quantities
 Solids
 Solutions
 Liquids
 Gases
 Worked examples
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.
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.
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.
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,
Solids
 Solids are normally measured by weighing to
obtain the mass
 Masses are measured in grams and kilograms
remember 1.000 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 16.00 g on a balance weighing to + or –
0.01 g but as 16.000 g on a balance weighing to +
or – 0.001 g.
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.
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.
Solutions
 Volume is usually used for solutions.
1.000 litre =
1.000dm3 =
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.
We can use this formula…..
 You can use a
formula triangle to
help you rearrange
the equation:
concentration =
moles
÷
moles
Vol in dm3
concentration
x
Vol in dm3
Do you still remember
it????
 You can use a
formula triangle to
help you rearrange
the equation:
Number
of
moles
Mass
÷
Mass
= ________
RAM or
RFM
Number
of moles
x
RAM or
RFM
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.
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
Calculations involving solutions
What is the concentration of a 0.0500 mol dm-3 solution of
sodium carbonate, Na2CO3, in g dm-3?
(C = 12; O = 16; Na = 23)
1 mol Na2CO3 weighs 106g
0.0500 mol weighs 0.0500 x 106g = 5.30g
So 0.0500 mol dm-3 is therefore 5.30g dm-3
Calculations involving
concentrations
 Recall ionic charges and names of ionic formulae.
 Know how to solve problems involving
concentration and amount of solute.
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.
Volume of CuSO4 in Qn
Number of moles
of CuSO4
20
= --------1000
x
0.100
Converts cm3
to dm3
= 0.00200 mol
Concentration of
CuSO4 given in
qn.
Going back to the question!!
 If 1 mole of CuSO4 gives 1 mole of BaSO4
 therefore 0.00200mol of CuSO4 will give
0.00200mol of BaSO4
 RFM of 1 mol of BaSO4 = 233
 Therefore 0.00200mol of BaSO4
=
0.00200
x
=
0.466g of BaSO4
233
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?
Here it is…..
CaCO3 + 2HCl → CaCl2 + H2O + CO2
 Number of moles of HCl = 25.0 / 1000 x 2.00
= 0.0500 mol
1 mol CaCO3 reacts with 2 mole HCl
= 0.0250mol
 1 mol CaCO3 weighs 100g
 Therefore 0.0250 mol weighs 0.0250 x 100 = 2.50 g
 So the max mass of CaCO3 which you could react
with this amount of hydrochloric acid is 2.50g
Here it is…..
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Number of moles of HCl = 25.0 / 1000 x 2.00
= 0.0500 mol
1 mol CaCO3 reacts with 2 mole HCl
0.5 mol CaCO3 reacts with 1 mole HCl
= 0.0250mol
1 mol CaCO3 weighs 100g
Therefore 0.0250 mol weighs 0.0250 x 100 =
2.50 g
So the max mass of CaCO3 which you could react with this
amount of hydrochloric acid is 2.50g
Last one….
 What is the minimum volume of 0.500 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.
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
=
0.0100 mol
So number of moles of H2SO4 =
0.0100 mol
Next we can use this formula…..
So the volume in dm3
=
Moles / Concentration
=
0.0100 / 0.500
=
0.0200 dm3
Which is what in
cm3
moles
÷
concentration
x
Vol in dm3
Calculations involving
concentrations
 Know how to solve problems involving
concentration and amount of solute.
 Know how to solve simple titration equations
Simple volumetric
calculations
 25.0cm3 of 0.100 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?
NaOH + HCl → NaCl + H2O
1
ratio
:
1
:
volumes
25cm3
23.5cm3
concentrations
0.100M
?????
1
:
1
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
= 0.00250 mol
NaOH + HCl → NaCl + H2O
ratio
1
:
1
:
volumes
25cm3
23.5cm3
concentrations
0.100M
?????
1
:
1
What else do
we now know?
0.00250 mol 0.00250 mol
No. of moles
So now we can calculate the concentration of HCl using the
formula:
concentration = Mole / volume in dm3 = 0.0025 / = 0.106 mol dm3
(23.5/1000)
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.
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 1.013 x 105 Pa (1atm) pressure
this volume is 2.24 x 10-2m3 (22.4 dm3 or 22
400cm3)
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.
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.
Equation and amounts
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
 RAM for Mg = 24.31
 Amount of Mg present
= mass / RAM
= 0.623 / 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
Excess and limiting
Mg(s)
1
+
2HCl(aq) → H2(g) + MgCl2(aq)
2
: 1
: 1
Ratio:
:
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
Theoretical yield of
Hydrogen
 The maximum no. of moles of hydrogen that can
be produced
= 3.41 x 10-2 / 2
= 1.705 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
= 0.382 dm3
To convert this to cm3 we need to multiply by 1000
= 382cm3
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.
PV=nRT
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.
Gases
The units of the gas constant R can be derived from the
equation.
p
x
V
=
N m-2 x m3 =
n
x
T
=
mol K
J
So the SI units of R must be J K-1 mol -1
R has the value 8.314 J K -1 mol -1, and is one of the best
known constants in science.
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.
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
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
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 100 000Pa.
The mass of one mole of CO2 is 44g.
V = mass (g) / mass of 1 mole (g) x RT / p
So; V = 2.20 (g) / 44 (g) x (8.31 x 298) / 100 000
1.24 x 10 -3 m3
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.