22 sundayherald 26 March 2006
EXAM GUIDE
CHEMISTRY
BY JOSEPH FARRELL
Joseph Farrell has been teaching
chemistry for 33 years. He currently
teaches in St Columba’s High, Gourock,
and has been a Higher chemistry marker
for six years, having previously marked
ordinary grades. He was invited to
become a member of the team making
up the Higher papers for the years
2002 and 2003 and has been a member
of the Higher chemistry examining team
since 2000.
C
HEMISTRY is not as difficult as
it at first appears. Once you
learn the basic concepts, further
work is more understandable. In
general, as with other subjects, chemistry
should be studied in short periods of time –
30 to 40 minutes is probably the length of
your concentration span – with breaks of
about 10 minutes in between.
No matter which chemistry course you
are studying, a useful technique is to use
lists of learning outcomes for each part of
the course and, using highlighters, colour
each learning outcome as green (fully
understood), amber or orange (not too
sure), or red (not understood at all).
In all chemistry exams, different parts of
the course can be asked about within one
question. This means “spotting” questions
and concentrating most of your efforts on a
few topics is unlikely to be of any value.
Familiarity with the whole course is
needed.
Some students commit ideas to memory
by making mind maps in which many facts
and concepts can be linked pictorially.
Some examples are given later for “Rusting”
(Standard Grade) and “The Faraday”
(Higher).
Since 10% of the Higher paper is devoted
to Standard Grade revision questions,
Higher candidates should read and think
about both parts of this study guide.
Another useful tip for studying is always
to have at hand a pen or pencil for jotting
down important facts as you read your
notes or book.
This way you can build up a set of
essential facts which you can revise in a
relatively short time.
STANDARD GRADE
You should learn the facts about the
structure of atoms (protons, electrons and
neutrons , and so on) and bonding (ionic
and covalent) right at the start since all
chemistry is based on these.
When you start studying the topics it is best
to keep a record of your studies such as the
study plan shown for Standard Grade (see
below).
CALCULATIONS
Many calculations are included under
Knowledge and Understanding because the
method for carrying out the calculation
should be learned and practised.
Chemists use the concept of “the mole”
in almost all calculations. You should
therefore become familiar with the following procedures:
a) Writing formulae (using valencies)
b) Using Relative Formula Masses from the
Data Book to calculate formula masses.
Formula Mass is the sum of the RAMs of all
of the atoms in the formula. For example.
Formula mass for H2SO4 is:
(1 x 2) + 32 + (16 x 4) = 98 amu
c) Finding one mole (the gram formula
STANDARD GRADE CHEMISTRY STUDY PLAN
Group
Reaction rates
Periodic Table
Bonding (covalent, ionic)
Hydrocarbons (including synthetic polymers)
Acids & alkalis
Metals & corrosion
Cells
Fertilisers
Cardohydrates
Times studied
1 2 3 4 5
6
7
8
Resources designed to suit the way you study
9
26 March 2006 sundayherald 23
CHEMISTRY
Calculate the mass of magnesium
chloride produced when 3g of magnesium
react with excess hydrochloric acid
according to the equation :
Mg + 2HCl
1 mole
24.5g
1g
3g
ACTIVITY SERIES
MgCl2 + H2
1 mole
24.5 + (2 x 35.5)g
95.5g
95.5g
24.5g
95.5g
x3
24.5g
= 11.7g
Note: The balanced equation tells us the
relationship between the number of moles
of reactant and the number of moles of
product. In the above example, 1 mole of
magnesium gives 1 mole of magnesium
chloride. Only these two substances need
be considered.
Since the masses of both compounds are
involved in the question both numbers of
moles are changed into grams and then a
proportion calculation is carried out.
h) Calculating the number of moles in a
solution = C x V(litres)
i) Using titration results to carry out a
calculation: use the equation above to find
the number of moles then use the chemical
equation to relate the numbers of moles of
each substance.
Note: In titration calculations the “rough
titre” is not used to calculate the average
volume used in the titration. Only the “concordant” results (within 0.2ml of each
other) are used.
j) Calculating the concentration (C) of a
solution
C = number of moles
Volume (in litres)
mass). This is the formula mass in grams.
e.g. One mole of H2SO4 is 98g.
d) Finding the number of moles in a mass
of a substance = mass (g)
gfm
e.g. 49g of H2SO4 = 49(g) = 0.5moles
98(g)
e) Calculating the mass of a number of
moles = no. of moles x gfm
e.g 0.1 moles of H2SO4 = 0.1 x 98g = 9.8g
Q14. Standard Grade Chemistry Credit 1997
Q14, Pure gold is known as 24-carat gold. In
jewellery, it is often mixed with other metals
to make it harder.
One common mixture is 9-carat gold. Its
composition by mass is shown in the pie-chart.
31.5%
31%
gold
silver
copper
37.5%
f) calculating percentage composition
% composition = mass of element in compound(g) x100
gram formula mass (g)
e.g. % by mass of S in H2S04 = 32(g) x 100 = 32.7%
98(g)
Sometimes the question will not be
straightforward. For example, in the question from 1997’s Standard Grade Credit
(right) the candidate has to realise that gold
forms 37.5% of the total mass of the alloy.
g) Calculating mass of a product or reactant
knowing the balanced equation. This is best
shown using an example:
Revise
(a) What name is given to a mixture of metals
such as 9-carat gold?
Answer: Alloy
(b) A 9-carat gold ring weighs 7.88g
Calculate the number of moles of gold in
the ring.
Show your working clearly.
Answer: Mass of gold in ring = 37.5% of 7.88g
= 37.5 x 7.88g = 2.96g
100
Mass of 1 mole of gold = 197g
Therefore, number of moles
of gold in ring
= 2.96g
197g =0.015moles
/ Practise
= mol/1
k) calculating the empirical formula of a
compound.
Elements Mass or %
Divide by relative atomic mass
Divide all by smallest
Ratio
Empirical Formula
PROBLEM SOLVING
Often the information needed to solve the
problem can be found in the question.
Remember to read back through the whole
question since very often the information
needed to solve a problem in the last part of
the question has been given in the first part
of the question.
GRAPHS
You should become familiar with and
practise drawing line graphs, bar graphs,
pie charts and tables of information as well
as obtaining information from these types
of presentation.
When drawing graphs or bar charts you
must label the axes identifying the variable
each represents and giving the units in
which it is measured. A regular scale must
be used – the same distance on the scale
should always represent the same quantity.
For a bar graph, each bar should be labelled
at the bottom or along its length showing
what it represents. The graph can be used to
estimate some value and to do this you
should use a ruler to draw lines from the
Test
Police
Potassium
Sergeant
Sodium
Charlie
Calcium
M
Magnesium
A
Aluminium
Z
Zinc
I
Iron
N
Nickel
T
Tin
L
Lead
Caught
Copper
Me
Mercury
Stealing
Silver
Gold
Gold
Plate
Platinum
Some questions about the Activity Series
or reactions in cells will require you to
write an ion-electron equation for
reduction or oxidation. A reliable rule to
follow when dealing with ion-electron
equations is that for reduction the
electrons appear on the left-hand side of
the equation but for oxidation they
appear on the right-hand side.
axes to the graph. You may have to extend a
graph to predict values.
INTERMEDIATE 1
You should split the course up into several
small sections of work and study each until
you feel confident enough to answer
questions on it.
The exam lasts for one-and-a-half hours
and consists of 20 multiple choice
questions (20 marks) and extended answer
questions worth 40 marks.
INTERMEDIATE 2
This course is designed as an intermediate
between Standard Grade and Higher and
therefore consists of mostly credit level
work with some parts of the Higher course.
The exam lasts for two hours and consists
of section A: 30 multiple choice questions
worth 30 marks and section B: extended
answer questions worth 50 marks.
Although the course is a bridge between
Standard Grade and Higher and, therefore,
contains some Higher material, the exam
does not have the same emphasis on
problem solving and so knowledge and
understanding of the course work is tested
more thoroughly.
Review
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24 sundayherald 26 March 2006
CHEMISTRY
MIND MAPS AND MNEMONICS
Fe3+
Coating with
another metal
Electroplating
Turns ferroxyl
indicator blue
Fe2+
Painting
Physical
protection
Galvanising
Zn
If the gas is soluble in water it should be
collected in a syringe with volume markings.
OH–
Chemical
protection
Polymers
Remember that to draw the structure of
part of a polymer formed by addition
polymerisation between three monomers
you have to draw the monomers in the
form of an H with the double bond represented by the horizontal line and nothing
else on this line.
In the example, three molecules of
phenylethene (old name: styrene) are
drawn so that it can be easily seen how they
can form bonds between the carbons by
addition across the double bonds to form a
part of the polymer polyphenylethene (old
name: polystyrene)
Turns ferroxyl
indicator
red
Sacrificial
protection
Magnesium
attached to
pipeline
Hydrocarbons: Monkeys Eat Peanut Butter
E
T
R
U
T
H
O
T
H
A
P
A
A
N
A
N
N
E
N
E
E
E
Redox: O
x
i
d
a
t
i
o
n
I
s
DRAWING APPARATUS
In all chemistry courses pupils should learn
how to draw apparatus for dealing with
gases.
PASSING A GAS THROUGH A LIQUID
The tube bringing the gas must dip into the
liquid but the tube removing the gas must
not dip into the liquid.
Gas collection
If the gas is insoluble it can be collected
over water. If the volume is to be measured
L
o
s
s
R
e
d
u
c
t
i
o
n
I
s
G
a
i
n
HIGHER
As well as knowing the information given
above on polymerisation, Higher candidates have to know some factual material
about the properties of some important
polymers and their uses. A table, such as
the one below could prove a helpful means
of listing several connected facts in a
compact fashion.
H H
C=C
0 H
H H
C=C
0 H
addition
From previous page
The questions in the Intermediate 2 exam
relating to Higher learning outcomes are presented in a straightforward fashion, as can
be seen in the example given on the right.
Gas out
Liquid
Electrons
gained by 02
+ H20
RUSTING
Attaching
to negative
terminal
Gas in
Iron atoms
lose electrons
Greasing
Plastic
coating
Sometimes mnemonics
can be very helpful in
remembering lists and
the order in which
certain chemicals are
arranged and mind
maps can help to show
how certain items are
connected. Here are
some examples
it must be collected in a vessel with volume
markings on the side. The end of the
delivery tube must be under the collecting
vessel.
See diagram for gas collection over water.
H
C
H
H
CH3 H
C =C
H
C
H
0
H
H
C
0
H
C
H
H H
C=C
0 H
polymerisation
H
C
0
H
C
H
H
C
0
H
C
H
The above molecule is an example of
A
a saturated alcohol
B
an unsaturated alcohol
C
a saturated carboxylic acid
D
an unsaturated carboxylic acid
EXAMINATION TECHNIQUE
Some questions extend over one page. Each
question should be read from start to finish
so that an overall picture can be formed in
your mind. Then answer the question
carefully, taking into account the information given in all parts of the question.
Supporting students from Standard Grade to Advanced Higher
26 March 2006 sundayherald 25
CHEMISTRY
POLYMER NAME SPECIAL PROPERTY
Kevlar
strength
Poly(ethenol)
soluble in water
CHEMICAL FEATURE
hydrogen bonds
has many OH groups
Poly(ethyne)
conducts electricity
Polyvinyl carbazole light sensitivity
to move
Biopol
biodegradable
delocalised electrons
light causes electrons
OH group
USES
bulletproof vests
laundry bags for
hospitals and
stitching of
internal wounds
speakers
photocopiers
internal stitching
but expensive
Grade must be known except “empirical
formula” type. The unit called the mole is
considered in more detail. One mole of any
substance contains 6.02 x 1023 “elementary
entities” of that substance. We encounter a
lot of different types of “entities’’, i.e.
o
6.02 X 1023 entities
(molecules, atoms, formula units, ions)
gram formula mass(gfm)
ie: the mass of one mole in grams
o
CALCULATIONS AND MOLES
Units 1 and 3 of the H course contain all of
the calculations except for percentage yield
calculations.
For Higher, all calculations for Standard
ONE MOLE
o
o
1 litre of solution of
concentration 1mol/l
o
Molar volume (volume
of 1 mole of any
gas if temperature
and pressure are same)
the electrons are on left then the equation
shows reduction.
If you are asked to write an ion-electron
equation remember that the total number
of charges on each side must be the same.
Moles are once again used in redox
calculations, as shown in this example:
In this question, an average of the
concordant readings from the burette is
used for the volume of iodine.
particles, in the H course so there are
several ways to apply the term “one mole”.
TABLE OF POLYMERS
96,500 coulombs = charge
in 1 mole of electrons
See the layout for “one mole” below right. In
general, only if you are asked for an actual
number of particles do you use Avogadro’s
Number, 6.02 x 1023.
In many calculations, molar quantities
are used. This is seen in questions involving
/\H value or molar volumes of gases such
as the one below.
A standard solution of iodine can be used to
determine the mass of vitamin C in orange
juice.
Iodine reacts with vitamin C as shown by
the following equation.
Respiration provides energy for the body
through “combustion”of glucose.
The equation for the enthalpy of combustion of glucose is:
C6H1206(s) + 602(g)
C6H806(aq) + I2(aq)
(vitamin C)
6CO2(g) + 6H20( )
H = -2807kJmol-1
(a) Calculate the volume of oxygen, in litres,
required to provide 418 kJ of energy.
(Take the molar volume of oxygen to be 24
litres mol-1.)
Show your working clearly.
In an investigation using a carton containing 500cm3 of orange juice, separate 50.0cm3
samples were measured out. Each sample
was then titrated with a 0.0050 mol -1
solution of iodine.
In this calculation we have to relate the
volume of oxygen to the number of kilojoules of heat produced. However, the
equation tells us that six moles of oxygen
produce 2807kJ and so we have to use the
molar volume to change the number of
moles of oxygen to a certain volume.
(iii) An average of 21.4cm3 of the iodine
solution was required for the complete
reaction with the vitamin C in 50.0cm3 of
orange juice.
Use this result to calculate the mass of vitamin C, in grams, in the 500cm3 carton of
orange juice.
Show your working clearly.
Earlier in the question we are told the
concentration of iodine solution. Therefore,
we can use the equation n = C x V(litres) to
find the number of moles of iodine used in
the titration. The chemical equation tells us
Redox calculations involving moles
Chemical reactions can often be broken
down to an oxidation step and a reduction
step. These are often shown as ion-electron
equations.
In an ion-electron equation, if the
electrons appear on the right-hand side of
the equation then it represents oxidation. If
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C6H606(aq) + 2H+(aq) + 2I-(aq)
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26 sundayherald 26 March 2006
CHEMISTRY
From previous page
that this is equal to the number of moles of
vitamin C in 50cm3 of orange juice and by
multiplying by 10 we can find the number
of moles in 500cm3. Using the equation:
mass of vit.C = (no. of moles) x (mass of 1
mole).
Reactants in excess
Moles can again be used to find out which
reactant is in excess.
Questions which have been asked since
the year 2000 are those asking the candidate to calculate which reactant is in excess.
Usually, one reactant is in the form of a
solid while the other is in solution. The
number of moles of each reactant should
be found using the equations:
number of moles = mass (g)
gfm
and
n = concentration x volume (litres)
and then, using the equation for the reaction (usually given), the numbers of moles
can be compared to see which one is in
excess. If the amount of product is then
required this can be calculated from the
reactant not in excess.
A student added 50cm3 of 4.0 mol l -1
hydrochloric acid to 4.0g of magnesium
ribbon.
(a) The balanced equation for the reaction
is:
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
large an area as possible. The tube removing the gas must be out of the water so that
the gas can enter it.
PPAs
National Qualifications test your practical
work by including a number of marks
related to the PPAs carried out during the
course. In Higher chemistry there are nine
PPAs (three per unit) and six marks of part
B of the exam paper are more likely to be
obtained by those who have done the PPAs
and learned them in detail. Often the types
of questions asked refer to the experimental procedures – such as indicating which
readings have to be taken or the meaning of
an instruction or even which solution has
to be used.
PPA reports have to be kept by the school
in case of moderation but each candidate
should keep a note of the details of the experiments including an evaluation of results.
Extended writing answer
Since 2002, the paper for the Higher has
contained a question requiring extended
writing in the answer for four marks. These
require the candidate to link certain ideas
and to explain the connection.
Oxidation:
electrons removed
1 mole of electrons
This answer illustrates important practical points made earlier about passing gases
through liquids – the tube delivering the gas
to the water must dip into the water so that
the gas and water come into contact over as
Discharge
l mole ions charge n +_
n
1 Faraday
96,500
coulombs
Show by calculation which reactant was in
excess. Show your working clearly.
(b) The hydrogen produced in the reaction
can be contaminated with small quantities
of hydrogen chloride vapour.
This vapour is very soluble in water.
Complete the diagram (below) to show how
the hydrogen chloride can be removed before
the hydrogen is collected.
Reduction:
electrons added
Electrolysis
d.c. to keep
products
constant
Q = it
Current in
amps
Time in
seconds
A convenient way of studying for questions of this type is making mind maps
which show how ideas can be connected. See the example of a mind map
involving the Faraday, above.
Carbon chemistry – homologous series
Candidates must be familiar with what is
meant by a homologous series – compounds that have similar chemical properties and can be represented by a general
formula are usually the accepted criteria –
and must know the information shown in
the table:
TABLE OF HOMOLOGOUS SERIES
HOMOLOGOUS
SERIES
NAME
ENDING
FUNCTIONAL
GROUP
Alkenes
–ene
C=C
Alkynes
–yne
C C
Alcohols
–ol
–C–OH
Aldehydes
–al
–C=O
hydrochloric
acid
H
Ketones
–one
0
C–C–C
gas syringe to
collect hydrogen
Carboxylic acids
–C=O
OH
Esters
hydrochloric
acid +
magnesium
–oic acid
–yl ; –oate
0
–C–O–C–
Amines
Answer: The tube delivering the gas to the water must dip into the water.
The tube removing the gas must be out of the water
Prefix: amino– –NH2
Much practice is needed in drawing
structural formulae of the organic
molecules. It is worthwhile to practise
drawing the structures of molecules from
all homologous series. Remember that
every carbon must have 4 bonds, hydrogen
1 bond and oxygen 2 bonds, the same as
their valencies.
Using information from questions
In many problem-solving questions
information needed to solve the problem is
contained within the question. In the following question, part (b) can be answered
by using information in the question.
2005 Question 8
In the table, the H3O+ acts as an acid – it
releases H+ and turns into a base because it
can now accept an H+. The reverse is true
for HS-.
In part (b) (ii) the water becomes OH- .So
it must release H+ i.e. act as an acid.
These answers are applications of the
information in the question.
CHEMISTRY EXAM
TIMETABLE
Level/Paper
Monday May 8
Standard Grade
General
Credit
Tuesday May 30
Intermediate 1
Intermediate 2
Higher
Advanced Higher
Time
9am-10.30am
10.50am-12.20pm
9am-10.30am
9am-11am
9am-11.30am
9am-11.30am
Helping more Scottish students to achieve their goals