GCSE Revision Unit C5
The hardest one.
Reacting amounts - basics
C5
• You can count atoms using the mole - “gram-molecular mass”
• The “molar mass” of substance is its relative formula mass in grams.
• Need to calculate the molar mass of a substance from its formula
(without brackets) by adding the appropriate relative atomic masses.
• Mass is conserved in a chemical reaction. REMEMBER
• Carry out simple subtractions, based on the idea that mass is never
lost:
– mass of gas or water lost during thermal decomposition or heating;
– mass of gas gained during reaction;
– determine a reacting amount for a simple reaction given all the other
reacting amounts.
• Determine the mass of an element in a known mass of compound
given the masses of the other elements present.
(i.e. just subtract what you know from the total)
• E.g. in 100g of CaCO3, there is 40g of calcium & 12g of carbon. What
mass of oxygen is there?
More:
C5
• Add the appropriate relative atomic
masses to get molar mass of a substance
from its formula (with brackets).
• Empirical formula:
“The simplest whole number ratio of each
type of atom in a compound”.
• Can deduce the empirical formula of a
compound from its chemical formula.
Higher tier
C5
• Number of moles = mass ÷ molar mass;
– determine the number of moles of an element from the mass of
that element;
– determine the number of moles of a compound from the mass of
that compound;
– determine the masses of the different elements present in a
given number of moles of a compound.
• The relative atomic mass of an element is the average
mass of an atom of the element compared to the mass
of 1/12th of an atom of carbon-12.
• Calculate mass of products and / or reactants using the
mole concept from a given balanced equation and the
appropriate relative atomic masses,
• Calculate empirical formula of a compound given the:
– percentage composition by mass;
– mass of each element in a sample of the compound.
Example
C5
• NaOH has a mass of 40 (23+16+1)
• Out of this, 16/40 is oxygen.
• If I had 100g of NaOH, then 16/40 of this is
oxygen, which is 100x16/40 = 40g
• In this 100g, 100x1/40 is hydrogen, 2.5g
• The other 57.5g must be sodium.
Another example
• In CaCO3, the total mass is 100
(40+12+48)
• So oxygen is 48/100 of the compound
• Carbon is 12/100 of the compound
• Potassium is 40/100 of the total
• So if there is 25g of calcium carbonate,
potassium is 25x40/100 = 10g
C5
Or:
C5
• If the calcium carbonate is being heated to make
carbon dioxide and calcium oxide:
• CaCO3  CaO + CO2
• 100
56
44 (relative masses)
• the CO2 is 44/100 of the starting materials so it
must also be 44/100 of the total at the end.
• If you start with 200g of calcium carbonate, you
must make 200x44/100 grams of CO2 and
200x56/100g of CaO
• Therefore you make 88g of CO2 and 112g of
CaO
Electrolysis - basics
C5
• Need to be able to label the apparatus used to electrolyse solutions in
a school laboratory:
– anode, cathode, d.c. power supply, probably in a beaker.
• Electrolysis is the decomposition of a compound by passing an
electric current through it or a solution of it.
• Positive ions go to the negative electrode and negative ions go to the
positive electrode obviously
• In the electrolysis of copper(II) sulphate solution using copper
electrodes:
– the negative electrode is plated with copper;
– the positive electrode gets smaller and dissolves.
– this is used to purify copper
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More or less time and current (amps) gives more or less product.
Melted Al2O3(l) gives oxygen and aluminium,
Melted PbBr2(l) gives lead and bromine,
Melted PbI2(l) gives lead and iodine
Melted KCl(l) gives potassium and chlorine
More:
C5
• Know electrolysis gives:
– K2SO4(aq) – hydrogen and oxygen;
– KNO3(aq) – hydrogen and oxygen
– H2SO4 – hydrogen and oxygen
• Describe electrolysis in terms of flow of charge and discharge of
ions.
• Describe the changes in mass of the copper electrodes used in the
electrolysis of copper(II) sulphate solution:
– the negative electrode gains mass;
– the positive electrode loses mass;
– the gain in mass of the negative electrode is equal to the loss in mass of
the positive electrode.
• Describe the factors that affect the amount of substance produced
during electrolysis:
– amount made increases as time increases and as current increases.
• Ionic substances contain ions which are in fixed positions when solid
but can move when in solution or when melted.
Higher tier
C5
• In the electrolysis of aqueous solutions it may be easier to
discharge ions from the water rather than from the solute.
• Half equations for the electrode processes that happen
during the electrolysis of the following given the formula of
the ions in the electrolyte: K2SO4(aq) and KNO3(aq)
• Half equations for electrode processes that happen during
the electrolysis of CuSO4(aq) using copper electrodes:
– (Anode) positive electrode
– (Cathode) negative electrode
Cu - 2e-  Cu2+
Cu2+ + 2e-  Cu
• Relationship between charge transfer, current and time:
Q = It
• Perform simple calculations based on current, charge and
the amount of substance produced in electrolysis.
• They might expect half-equations for what happens at
each electrode during the electrolysis of:
• Al2O3(l), PbBr2(l), PbI2(l), KCl(l)
Concentration - basics
• Concentration of solutions may be measured in g/dm3 (g
per dm3).
• Concentration of solutions may be measured in mol/dm3
(mol per dm3).
• Volume is measured in dm3 (litre) or cm3
1000 cm3 equals 1 dm3 (litre)
• Describe how to dilute a concentrated solution
• You must get concentration right by diluting food
preparations, medicine and baby milk.
• Dangers of getting the dilution wrong (obvious ones):
food consistency wrong; overdoses, very poorly babies
• Food packaging, recommended daily allowances.
C5
More:
• The more concentrated a solution the more
crowded the solute particles.
• Volume in cm3 is 1000 times the volume in dm3
• Volume in dm3 is 1000th of the volume in cm3
• Perform calculations involving concentration for
• Simple dilutions of solutions e.g. how to dilute a
1.0 mol/dm3 solution into a 0.1 mol/dm3 solution
C5
Higher tier
C5
• To convert concentration in g/dm3 into mol/dm3 divide by
the molecular mass
• To do the opposite, multiply by molecular mass
• Calculate the concentration of a solution given
appropriate information about the mass or number of
moles of solute in a particular volume of solution.
• Perform simple calculations involving concentration,
number of moles and volume of solution.
• Interpret more complex food packaging information and
its limitations e.g:
• • convert amounts of sodium to amounts of salt;
• • understand that an ion may come from several
sources, so the above conversion will be inaccurate.
Titrations - basics
•
C5
Remember changes in pH when an acid reacts with an alkali:
– pH decreases when acid is added to alkali;
– pH increases when alkali is added to a acid;
– They expect you to interpret a simple pH curve, find the pH at a particular volume
added or read volume at a pH – major grid lines (whole numbers).
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You can estimate the pH of a solution with universal indicator.
Apparatus used in an acid-base titration:
– burette and conical flask; pipette and pipette filler.
•
Describe the procedure for carrying out a simple acid-base titration:
– acid in burette, alkali in pipette (or vice versa);
– acid slowly added to alkali (or vice versa) until end point is reached;
– end point detected by the change in colour of an indicator.
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Calculate the titre given appropriate information from tables or diagrams.
Indicators in acids and alkalis:
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universal indicator, - range from red/yellow/green/turquoise/dark blue
Litmus – red in acid, blue in alkali
Phenolphthalein – colourless in acid, pink in alkali, changes around 9
screened methyl orange – red in acid, yellow in alkali, changes/orange around pH 5
More:
C5
• There is a sudden change in pH at the “end point” of a
titration, as the “balance” tips the indicator
– neutralisation;
– acid + alkali  salt + water
• Interpret a simple pH curve:
– determine the volume of acid or alkali at neutralisation;
– determine the pH when a particular volume added or vice versa.
• Need several consistent titre readings in titrations,
otherwise you can’t be sure....
• A single indicator such as litmus - sudden colour change
• Mixed indicator (like universal) has a continuous colour
change – less useful in titration
C5
Higher tier
• Sketch a pH titration curve for the titration of an acid or an alkali.
• Exam questions will be a one to one molar ratio (one mole acid : one
mole alkali).
• Use equation concentration1 x volume1 = concentration2 x volume2 –
you know three of these, just change the subject.
• State and use the relationship between
moles, concentration and volume:
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–
–
–
moles = concentration × volume in dm3
concentration = moles ÷ volume in dm3
volume in dm3 = moles ÷ concentration
Think of the unit for concentration – moles
per litre, means conc. is moles divided by
volume then make a triangle
• Why is an acid-base titration better with
a single indicator rather than a
mixed indicator?
moles
concentration
solvent
volume
3
(dm )
Gas volumes - basics
C5
• Possible apparatus to collect and measure the volume of a gas from a
reaction:
– gas syringe;
– upturned measuring cylinder;
– upturned burette.
• If you can measure the mass of the reaction mixture, it will get lighter
as gas is lost to the air.
• A reaction stops when one of the reactants is all used up.
• The more amount of reactant you have, the greater volume of gas is
made.
• Interpret graphs or tables of data about the volume of gas produced
during the course of a reaction (limited to major grid lines on graphs):
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–
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–
• deduce total volume of gas produced;
• deduce when the reaction has stopped;
• deduce volume of gas at a particular time and vice versa;
• deduce when the reaction is at its fastest..
More:
C5
• “Describe an experimental method to measure the volume of gas
produced in a reaction given appropriate details about the reaction”
• “Describe an experimental method to measure the mass of gas
produced in a reaction given appropriate details about the reaction”
• State that the limiting reactant is the one that is used up first of all.
• Explain why a reaction stops in terms of the limiting reactant present
given appropriate qualitative (no numbers) information.
• Describe that the amount of gas produced is directly proportional to
the amount of the limiting reactant present.
• Interpret graphs as above, but also:
– read from minor or estimate from major lines;
– deduce the volume of gas produced with different amounts of limiting
reactant.
Higher tier
C5
• One mole of gas molecules occupy a volume of 24 dm3
at room temperature and pressure So we can calculate
the volumes of samples of gases.
• E.g. 5 moles has a volume of 120dm3.
• Don’t get cm3 and dm3 confused, for example,
250cm3 is 0.25 dm3 and
1.2 dm3 is 1200 cm3
• You may be asked to sketch a graph to show how the
volume of gas produced during the course of a reaction
changes, given appropriate details:
– change in the rate of the reaction;
– total volume of gas collected;
– when the reaction stops.
Equilibria - basics
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C5
Many chemical reactions are easily reversed, giving a forward and a
backward reaction.
It is possible for both forward and backward reactions to be proceeding at the
same time.
The Ý symbol is used to show that a reaction is reversible, e,g:
N2 + 3H2 Ý 2NH3
Interpret data in the form of tables or graphs (using major grid-lines) about the
equilibrium composition:
– • composition at particular temperatures;
– • composition at particular pressures;
– • effect of temperature and pressure on composition.
•
The raw materials to make sulphuric acid by the Contact Process:
– • sulphur;
– • air;
– • water.
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State that the production of sulphuric acid by the Contact Process involves the
reversible reaction between sulphur dioxide and oxygen:
sulphur dioxide + oxygen Ý sulphur trioxide
More:
•
C5
Some reversible reactions may reach an equilibrium where:
– the rate of the forward reaction equals the rate of the backward reaction;
– the concentrations of the reactants and the products do not change and have no
reason to be equal.
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If the concentration of product is greater than the concentration of reactant,
we say the position of equilibrium is “to the right”
And vice-versa
A change in temperature, pressure or concentrations of reactant (LHS) or
product (RHS) may change the position of equilibrium.
“Interpret data in the form of tables or graphs about the equilibrium
composition”:
– composition at particular temperatures;
– composition at particular pressures;
– effect of temperature and pressure on composition.
•
Conditions used in the Contact Process:
– V2O5 catalyst;
– around 450°C;
– atmospheric pressure.
•
Recall that sulphur dioxide needed for the Contact Process often comes from
burning sulphur:
– sulphur + oxygen Ý sulphur dioxide
Higher tier
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C5
Explain why a reversible reaction may reach an equilibrium:
– a closed system;
– rate of forward reaction decreases;
– rate of backward reaction increases;
– eventually rate of forward equals rate of backward reaction.
Explain in simple qualitative terms factors that affect the position of equilibrium:
– removing a product moves the position of equilibrium to the right or vice versa;
– adding extra reactant moves the position of equilibrium to the right or vice versa;
– increasing the pressure moves the position of equilibrium to the side with the least
number of moles of gas molecules
Equations for the manufacture of sulphuric acid by the Contact Process:
S + O2  SO2
2SO2 + O2 Ý 2SO3
SO3 + H2O  H2SO4
Conditions used in the Contact Process:
– high temperature decreases yield and increases rate of reaction so an optimum is
used;
– Catalysts increase rate but do not change position of equilibrium; “same amounts of
products but faster”
– Position of equilibrium is already on right so high pressure is expensive and is not
needed.
Acid strength - basics
• Ethanoic acid is an example of a weak acid.
• Hydrochloric, nitric and sulphuric acids are strong acids.
• Strong acids have a lower pH (about 1-3) than weak acids
(about 3-6) of the same concentration.
• Any acid will react with magnesium to give hydrogen.
• Any acid will react with calcium carbonate to give carbon
dioxide.
• These two reactions are slower with weak acids like ethanoic
acid than with hydrochloric acid of the same concentration.
• The same amount of same concentration strong and weak
acids produce the same volume of gas in these reactions.
• Ethanoic acid has a lower electrical conductivity than
hydrochloric acid of the same concentration.
• Electrolysis of both ethanoic acid and hydrochloric acid
makes hydrogen at the negative electrode.
• Weak acids are sometimes used as descalers, some occur in
foods.
C5
More:
C5
• All acids ionise in water to produce H+ ions.
• A strong acid completely ionises in water. HCl  H+ + Cl• The ionisation of a weak acid is an
example
of a reversible reaction.
+
For a weak acid HA:
HA Ý H + A
• This gives an equilibrium mixture.
• Ethanoic acid reacts slower than hydrochloric acid because there
are fewer hydrogen ions and so fewer collisions with hydrogen ions.
• But the volume of hydrogen gas formed in a metal reaction is
determined by the amounts of reactants present not the acid
strength.
• Ethanoic acid is less conductive than hydrochloric acid of the same
concentration because there are fewer hydrogen ions available to
move.
• Hydrogen is produced during the electrolysis of ethanoic acid and of
hydrochloric acid. Hydrogen ion + electron  hydrogen gas
• Strong acids are a bad idea for descaling agents, as they would
corrode metals.
Higher
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C5
The pH of a weak acid is much higher than the pH of a strong acid of the
same concentration because there are fewer hydrogen ions per unit volume
The difference between acid strength and acid concentration:
– acid strength (strong or weak) is a measure of the degree of ionisation of the acid;
– concentration of an acid is a measure of how many moles of acid in one dm3.
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That is, strong acids separate to give more H+(aq) ions than weak ones do
Equations for examples of the ionisation of weak and strong acids:
CH3COOH Ý CH3COO- + H+
HCl  H+ + Cl(these are the only ones they will ask for)
Ethanoic acid reacts more slowly than hydrochloric acid of the same
concentration because:
– ethanoic acid is weak and hydrochloric acid is strong;
– greater concentration of hydrogen ions;
– greater collision frequency with hydrogen ions.
•
ethanoic acid is less conductive than hydrochloric acid of the same
concentration because:
– ethanoic acid is weak and hydrochloric acid is strong;
– greater concentration of hydrogen ions to carry the charge.
•
“Explain why a weak acid may be more useful than the more dilute strong
acid.”
Ionic reactions - basic
C5
• A precipitation reaction involves two solutions reacting together to
make an insoluble substance.
• Most precipitation reactions involve ions from one solution reacting
with ions from another solution.
• A nitrate solution (usually silver nitrate) can be used to test for halide
ions, giving:
– white precipitate with Cl– cream precipitate with Br– yellow precipitate with I
• Describe that barium nitrate or barium chloride solution can be used
to test for sulphate ions (both form a white precipitate).
• You need to be able to identify the reactants and the products in an
ionic equation.
• You need to be able to recognise and use the state symbols (aq), (s),
(g) and (l).
• You need to be able to label apparatus used during the preparation of
an insoluble compound by precipitation (basically, a beaker for the
reaction, followed by a filter and funnel to get the solid.)
More:
C5
• Ionic substances contain ions which are in fixed
positions in the solid but can move when in solution or
when melted.
• Ions must collide with other ions if they are to react. This
happens when solutions are mixed.
• Use word equations for simple precipitation reactions
e.g. for the reaction between solutions of barium chloride
and sodium sulphate.
• Preparation of a dry sample of an insoluble compound
by precipitation,
(you will be given the names of the reactants):
– mix solutions of reactants;
– filter;
– wash and dry the residue.
Highest level
C5
• Most precipitation reactions are extremely fast reactions
between ions.
• They occur because the forces attracting the ions to
make a precipitate are stronger than those holding the
ions to water molecules.
• Heat energy is released as the ionic lattice forms, but
you can’t usually detect it.
• Ionic equations involve just the ions that react.
• ‘Spectator ions’ are the ions that stay in solution and
haven’t formed a precipitate.
• E.g. in NaBr + AgNO3  AgBr + NaNO3
• Solubility rules tell us what should happen, because Na+
and NO3- ions always stay dissolved, so the Ag+ and Brions can be the only precipitate:
• Ag+(aq) + Br-(aq)  AgBr(s)
Simplified solubility rules
C5
• Group 1 metal and ammonium (NH4+)
compounds always dissolve
• Nitrates (NO3-) always dissolve
• Chlorides, bromides and iodides dissolve,
except those with silver, mercury and lead
• Metallic sulphates usually dissolve, but not those
of calcium, strontium, barium, silver, mercury
and lead
• Metallic oxides, carbonates and hydroxides are
insoluble except where the first rule applies.
• All metallic hydroxides (OH-) are insoluble
except where the first rule applies and calcium,
strontium and barium in Group II.