OCR Chemistry AS.
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2.1.1 Atomic Structure
Details of the three Sub-atomic (fundamental) Particles
Particle Position Relative Mass Relative Charge
Proton Nucleus 1 +1
Neutron Nucleus 1 0 There are various
Electron Orbitals 1/1800 -1 models for atomic
structure
An atom of Lithium (Li) can be represented as follows:
Mass Number 7 Li Atomic Symbol
Atomic Number 3
The atomic number, Z, is the number of protons in the nucleus.
The mass number ,A, is the total number of protons and neutrons in the atom.
Number of neutrons = A - Z
Isotopes Isotopes are atoms of same element with the same number of protons, but different numbers of neutrons.
Isotopes have similar chemical properties because they have the same electronic
structure. They may have slightly varying physical properties because they have
different masses.
DEFINITION: Relative Isotopic mass is the mass of one isotope compared to
one twelfth of the mass of one atom of carbon-12
DEFINITION: Relative atomic mass is the average mass of one atom
compared to one twelfth of the mass of one atom of carbon-12
DEFINITION: Relative molecular mass is the average mass of a molecule
compared to one twelfth of the mass of one atom of carbon-12
Calculating the Relative Atomic Mass of an Element
The relative atomic mass quoted on the periodic table is a weighted average of all the isotopes
Percentage Abundance 78.7 10.13 11.17
R.A.M = (isotopic mass x % abundance)
Relative Isotopic mass 24.00 25.00 26.00
100
Isotope Mg24 Mg25 Mg26
For above example of Mg
R.A.M = [(78.7 x 24) + (10.13 x 25) + (11.17 x 26)] /100 = 24.3
R.A.M = (isotopic mass x relative abundance)
If relative abundance is used instead of
total relative abundance percentage abundance use this equation
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,Example: Calculate the relative atomic mass of Tellurium from the following abundance data:
124-Te relative abundance 2; 126-Te relative abundance 4; 128-Te relative abundance 7;
130-Te relative abundance 6
Example: Copper has two isotopes 63-Cu and 65-Cu. The relative atomic mass of copper is 63.5.
Calculate the percentage abundances of these two isotopes.
63.55 = yx63 + (1-y)x65
63.55 = 63y +65 -65y
63.55 = 65 -2y
2y = 1.45
y = 0.725
%abundance 63-Cu =72.5% %abundance 65-Cu = 27.5%
Mass spectra for Cl2 and Br2
Cl has two isotopes Cl35 (75%) and Cl37(25%) Br has two isotopes Br79 (50%) and Br81(50%)
These lead to the following spectra caused by the diatomic molecules
Br79Br81 +
Br81Br79 +
Cl35Cl35 + relative
relative abundance
abundance Cl35Cl37 + Br79Br79 + Br81Br81 +
Cl37Cl37 +
70 72 74 m/z m/z
158 160 162
The 160 peak has double the abundance of the other
two peaks because there is double the probability of
160 Br79-Br81 + as can be Br79-Br81 and Br81-79
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,2.1.2 Compounds, formulae and equations
Summary of most important ions to know
+1 +2 +3 -3 -2 -1
Group 1 Group 2 Group 3 Group 5 Group 6 Group 7
Hydrogen Zinc Iron (III) Phosphate Carbonate Nitrate (NO3-)
Silver Copper (II) (PO43-) (CO32-) Hydroxide
Gold Iron (II) Sulphate (OH-)
Ammonium Tin (SO42-)
(NH4+) Lead
How to work out the formula from the ionic charge
What is the formula of Lithium Sulphate?
1. Identify the ionic charges of the two ions
Lithium is in group 1 so has a +1 charge, Li+
sulphates have a -2 charge, SO42-
2. Combine the ions together to get a neutral compound. i.e. combine so
that the total +ve charge cancels out the total –ve charge
We need two lithium ions to cancel out
Li + Li+ SO42-
the -2 charge on the sulphate
The formula is therefore Li2SO4
What is the formula of Calcium phosphate?
1. Identify the ionic charges of the two ions
Calcium is in group 2 so has a +2 charge, Ca2+
Phosphates have a -3 charge, PO43-
2. Combine the ions together to get a neutral compound. i.e. combine so
that the total +ve charge cancels out the total –ve charge
We need to multiply up to get the same charge. Three calcium ions would produce +6 charge to
cancel out the -6 charge on two phosphate ions
The formula is therefore Ca3(PO4)2
Only use brackets when there is more
than one of the compound ion in the e.g. Copper nitrate is Cu(NO3)2
formula. Calcium hydroxide is Ca(OH)2
Ammonium sulphate is (NH4)2SO4
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,Writing Ionic equations
We usually write ionic equations to show the key Spectator ions are ions that are not
changes in a reaction. Ionic equations only show the • Not changing state
ions that are reacting and leave out spectator ions. • Not changing oxidation number
Take full equation Pb(NO3)2 (aq) + 2NaCl (aq) PbCl2 (s) + 2 NaNO3 (aq)
Separate (aq) solutions Pb2+(aq) + 2NO3-(aq) + 2Na+ (aq)+ 2Cl-(aq) PbCl2 (s) + 2 Na+(aq)+ 2NO3- (aq)
into ions
Cancel out spectator ions
Pb2+(aq) + 2Cl-(aq) PbCl2 (s)
leaving ionic equation
N Goalby chemrevise.org 4
,2.1.3 Amount of substance
The mole is the key concept for chemical calculations
DEFINITION: The mole is the amount of substance in grams that has the same number of particles as there are
atoms in 12 grams of carbon-12.
DEFINITION: Relative atomic mass is the average mass of one atom compared to one twelfth of the mass of
one atom of carbon-12
DEFINITION: Molar Mass is the mass in grams of 1 mole of a substance and is given the unit of g mol-1
Molar Mass for a compound can be calculated by adding
up the mass numbers(from the periodic table) of each
element in the compound
eg CaCO3 = 40.1 + 12.0 +16.0 x3 = 100.1
molar gas volume (gas volume per mole, units dm3 mol–1) . This is the volume of 1 mole of a gas at a
given temperature and pressure. All gases have this same volume. At room pressure (1atm) and room
temperature 25oC the molar gas volume is 24 dm3 mol–1
Avogadro's Constant
There are 6.02 x 1023 atoms in 12 grams of carbon-12. Therefore explained in simpler terms 'One mole of any
specified entity contains 6.02 x 1023 of that entity':
1 mole of copper atoms will contain 6.02 x 1023 atoms
Avogadro's Constant can be
1 mole of carbon dioxide molecules will contain 6.02 x 1023 molecules
used for atoms, molecules and
1 mole of sodium ions will contain 6.02 x 1023 ions
ions
For pure solids and gases Example 1: What is the amount, in mol, in 35.0g of
CuSO4?
amount = mass
amount = mass/Mr
Mr
= 35/ (63.5 + 32 +16 x4)
Unit of Mass: grams = 0.219 mol
Unit of amount : mol
Significant Figures
Give your answers to the same
number of significant figures as the
number of significant figures for the
data you given in a question. If you
are given a mixture of different
significant figures, use the smallest
N Goalby chemrevise.org 1
,Empirical formulae
Definition: An empirical formula is the simplest ratio of atoms of each element in the compound.
General method
The same method can be
Step 1 : Divide each mass (or % mass) by the atomic mass of the element used for the following types
Step 2 : For each of the answers from step 1 divide by the smallest one of of data:
those numbers. 1. masses of each element
in the compound
Step 3: sometimes the numbers calculated in step 2 will need to be multiplied
up to give whole numbers. 2. percentage mass of each
element in the compound
These whole numbers will be the empirical formula.
Example 2 : Calculate the empirical formula for a compound that contains 1.82g of
K, 5.93g of I and 2.24g of O
Step1: Calculate amount, in mol, by dividing each mass by the atomic mass of the element
K = 1..1 I = 5.93/126.9 O = 2.24/16
= 0.0465 mol = 0.0467mol = 0.14 mol
Step 2 For each of the answers from step 1 divide by the smallest one of those numbers.
K = 0.0465/0.0465 I = 0.0467/0.0465 O = 0..0465
=1 =1 =3
Empirical formula =KIO3
Molecular formula from empirical formula
Definition: A molecular formula is the actual number of atoms of each element in the compound.
From the relative molecular mass (Mr) work out how many times
the mass of the empirical formula fits into the Mr.
Example 3. work out the molecular formula for the
compound with an empirical formula of C3H6O and The Mr does not need to be exact to turn an
a Mr of 116 empirical formula into the molecular formula
because the molecular formula will be a
C3H6O has a mass of 58 whole number multiple of the empirical
The empirical formula fits twice into Mr of 116 formula
So molecular formula is C6H12O2
N Goalby chemrevise.org 2
,Hydrated salt
A Hydrated salt contains water of crystallisation Example 4
Na2SO4 . xH2O has a molar mass of 322.1, Calculate
.
Cu(NO3)2 6H2O the value of x
hydrated copper (II) nitrate(V). Molar mass xH2O = 322.1 – (23x2 + 32.1 + 16x4)
= 180
Cu(NO3)2 X = 180/18
Anhydrous copper (II) nitrate(V). =10
This method could be used for measuring mass loss in various
Heating in a crucible thermal decomposition reactions and also for mass gain when
reacting magnesium in oxygen.
The lid improves the accuracy of the
The water of crystallisation in calcium sulphate crystals can be
experiment as it prevents loss of solid
removed as water vapour by heating as shown in the following
from the crucible but should be loose
equation.
fitting to allow gas to escape.
CaSO4.xH2O(s) → CaSO4(s) + xH2O(g)
Method.
•Weigh an empty clean dry crucible and lid .
•Add 2g of hydrated calcium sulphate to the crucible and
weigh again
•Heat strongly with a Bunsen for a couple of minutes
•Allow to cool
•Weigh the crucible and contents again
•Heat crucible again and reweigh until you reach a constant
mass ( do this to ensure reaction is complete).
Large amounts of hydrated calcium sulphate, such as
50g, should not be used in this experiment as the
decomposition is like to be incomplete.
Small amounts the solid , such as
0.100 g, should not be used in
The crucible needs to be dry otherwise a wet crucible this experiment as errors in
would give an inaccurate result. It would cause mass loss weighing are too high.
to be too large as water would be lost when heating.
Example 5. 3.51 g of hydrated zinc sulphate were heated and 1.97 g
of anhydrous zinc sulphate were obtained.
Use these data to calculate the value of the integer x in ZnSO4.xH2O
Calculate the mass of H2O = 3.51 – 1.97 = 1.54g
Calculate moles Calculate moles = 1.54
= 1.97 of H2O
of ZnSO4 161.5 18
= 0.0122 mol =0.085 mol
Calculate ratio of mole
of ZnSO4 to H2O = 0.0122 = 0.085
0.0122 0.0122
=7
=1
X=7
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,Calculation of reacting masses, gas volumes and mole concentrations
Commonly in questions converting between quantities of substances reacting we will use more than just mass data. We
might have the volume and concentration of a solution, or the volume of a gas. We need to adapt our existing method or
reacting masses to include other quantities. Any of the equations below can be used to convert quantities into moles.
Learn these equations carefully and what units to use in them.
1. For pure solids, liquids and gases 2. For Gases 3. For solutions
amount = mass Gas Volume (dm3)= amount x 24 Concentration = amount
MolarMass volume
This equation give the volume of a
gas at room pressure (1atm) and Unit of concentration: mol dm-3 or M
Unit of Mass: grams
room temperature 25oC. Unit of Volume: dm3
Unit of amount : mol
Or use the ideal gas equation to
work out gas volumes at other Converting volumes
It is usually best to give temperatures and pressures
your answers to 3sf cm3 dm3 ÷ 1000
PV = nRT
cm3 m3 ÷ 1000 000
dm3 m3 ÷ 1000
Simple mole calculations
Some Simple calculations using above equations
Example 6: What is the concentration of solution made by Example 7 : What is the volume in dm3 at room temperature
dissolving 5.00g of Na2CO3 in 250 cm3 water? and pressure of 50.0g of Carbon dioxide gas ?
amount = mass/Mr amount = mass/Mr
= 5 / (23 x2 + 12 +16 x3) = 50/ (12 + 16 x2)
= 0.0472 mol = 1.136 mol
conc= amount/Volume Gas Volume (dm3)= amount x 24
= 0..25 = 1.136 x 24
= 0.189 mol dm-3 = or 27.3 dm3 to 3 sig fig
Ideal Gas Equation
The ideal gas equation is a slightly more difficult way of working PV = nRT
out a gas volume. It can calculate the volume for any
temperature and pressure though so is a more useful method. Unit of Pressure (P):Pa
Unit of Volume (V): m3
The biggest problems students have with this equation is choosing Unit of Temp (T): K
and converting to the correct units, so pay close attention to the units. n= moles
R = 8.31 JK–1mol–1
Example 8: What is the mass of Cl2 gas that has a
Converting temperature
pressure of 100kPa, temperature 293K, volume 500cm3.
oC K add 273
(R = 8.31 JK–1mol–1)
100 kPa = 100 000 Pa
moles = PV/RT 500 cm3 = 0.0005 m3
= 100 000 x 0.0005 / (8.31 x 293)
= 0.0205 mol
Mass = moles x Mr
= 0.0205 x (35.5 x2)
= 1.46 g
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,Avogadro's Constant
The mole is the amount of substance in Avogadro's Constant
grams that has the same number of There are 6.02 x 1023 atoms in 12 grams of carbon-12. Therefore
particles as there are atoms in 12 grams explained in simpler terms 'One mole of any specified entity
of carbon-12. contains 6.02 x 1023 of that entity':
1 mole of copper atoms will contain 6.02 x 1023 atoms
Avogadro's Constant can be
1 mole of carbon dioxide molecules will contain 6.02 x 1023 molecules
used for atoms, molecules and
1 mole of sodium ions will contain 6.02 x 1023 ions
ions
No of particles = amount of substance (in mol) X Avogadro's constant
Example 9 : How many atoms of Tin are Example 10 : How many chloride ions are there in a 25.0
there in a 6.00 g sample of Tin metal? cm3 of a solution of magnesium chloride of concentration
amount = mass/Ar 0.400 moldm-3 ?
= 6/ 118.7 amount= concentration x Volume
= 0.05055 mol MgCl2 = 0.400 x 0.025
Number atoms = amount x 6.02 x 1023 = 0.0100 mol
There are two moles of
= 0.05055 x 6.02 x 1023 Amount of chloride ions = 0.0100 x2 chloride ions for every
= 3.04 x1022 = 0.0200 one mole of MgCl2
Number ions of Cl- = amount x 6.02 x 1023
= 0.0200 x 6.02 x 1023
= 1.204 x1022
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, Converting quantities between different substances using a balanced equation Typically we are given a quantity of
one substance and are asked to
N2 + 3H2 2NH3 work out a quantity for another
substance in the reaction. Any of
The balancing (stoichiometric) numbers are mole ratios
the above three equations can be
e.g. 1 mol of N2 reacts with 3 mol of H2 to produce 2mol of NH3 used.
Step 3
Step 1:
Convert amount, in mol, of second
Use one of the above 3 equations to Step 2:
substance into quantity question
convert any given quantity into amount Use balanced equation to convert
asked for using relevant equation
in mol amount in mol of initial substance
e.g. amount ,Mr mass
Mass amount into amount in mol of second
Amount gas vol gas
Volume of gas amount substance
amount, vol solution conc
Conc and vol of solution amount
Example 12: 23.6cm3 of H2SO4 neutralised 25.0cm3 of 0.150M
Example 11: What mass of Carbon dioxide would be
NaOH. What is the concentration of the H2SO4?
produced from heating 5.50 g of sodium
H2SO4 + 2NaOH Na2SO4 +2H2O
hydrogencarbonate?
2NaHCO3 Na2CO3 + CO2 + H2O
Step 1: work out amount, in mol, of sodium hydrogencarbonate Step 1: work out amount, in mol, of sodium hydroxide
amount = mass / Mr amount = conc x vol
= 5.5 /84 = 0.150 x 0.025
= 0.0655 mol = 0. 00375 mol
Step 2: use balanced equation to give amount in mol of CO2 Step 2: use balanced equation to give moles of H2SO4
2 moles NaHCO3 : 1 moles CO2 2 moles NaOH : 1 moles H2SO4
So 0.0655 HNO3 : 0.0328mol CO2 So 0.00375 NaOH : 0.001875 mol H2SO4
Step 3: work out mass of CO2
Step 3 work out concentration of H2SO4
Mass = amount x Mr
= 0.0328 x 44.0 conc= amount/Volume
=1.44g = 0..0236
= 0.0794 mol dm-3
Example 13: What volume in cm3 of oxygen gas would be Example 14: What mass of Copper would react completely with
produced from the decomposition of 0.532 g of potassium 150 cm3 of 1.60M nitric acid?
chlorate(V)? 3Cu + 8HNO3 3Cu(NO3 )2 + 2NO + 4H2O
2KClO3 2 KCl + 3O2
Step 1: work out amount, in mol, of potassium chlorate(V)? Step 1: work out amount, in mol, of nitric acid
amount = mass / Mr amount = conc x vol
= 0.532 /122.6 = 1.60 x 0.15
= 0.00434 mol = 0.24 mol
Step 2: use balanced equation to give amount in mol of O2 Step 2: use balanced equation to give moles of Cu
2 moles KClO3 : 3 moles O2 8 moles HNO3 : 3 moles Cu
So 0.00434 HNO3 : 0.00651mol O2 So 0.24 HNO3 : 0.09 (0.24 x 3/8) mol Cu
Step 3: work out volume of O2 Step 3: work out mass of Cu
Gas Volume (dm3)= amount x 24 Mass = amount x Mr
= 0.09 x 63.5
= 0.00651 x 24 =5.71g
= 0.156 dm3
= 156 cm3
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