Topic 1a – Atomic StructureRevision Notes

1) Fundamental particles

Atoms consist of protons, neutrons and electrons Protons and neutrons are found in the nucleus which contains most of the mass of

the atom and all of the positive charge The neutrons help to reduce repulsion between the positively charged protons The electrons are arranged in energy levels (shells) around the nucleus The electron arrangement of an element determines its chemical properties i.e. what

reactions it does

Relative mass Relative chargeProton 1 +1Neutron 1 0Electron 1/2000 -1

2) Mass number and isotopes

Atomic number = number of protons in the nucleus Mass number = number of protons and neutrons in the nucleus Number of neutrons = mass number – atomic number Number of electrons = number of protons (in a neutral atom)

9 Mass number = 9 Atomic number = 4

Be4 4 protons, 5 neutrons, 4 electrons

3) Isotopes

Isotopes have the same number of protons but different numbers of neutrons For example, chlorine has two isotopes 35Cl and 37Cl. Both have 17 protons but they

have 18 and 20 neutrons, respectively Isotopes of an element have the same chemical properties because they have the

same electron arrangement Isotopes of an element may have different physical properties, such as rate of

diffusion, because they have different masses

4) Ions

Ions are formed when atoms gain or lose electrons As an atom Cl has 17 electrons. A Cl- ion has gained one electron so it now has 18 As an atom Na has 11 electrons. An Na+ ion has lost one electron so it now has 10 Early models of atomic structure predicted that atoms and ions with noble gas

electron arrangements should be stable e.g. Cl- has the same electron arrangementas argon and Na+ has the same electron arrangement as neon

Topic 1b – Formulae & EquationsRevision Notes

1) Formulae

a) Elements

For most elements the formula is just the symbol e.g. Na for sodium, S forsulphur

The exceptions are the seven diatomic elements – H2, N2, O2, F2, Cl2, Br2 andI2

b) Ionic compounds

Compounds of a metal and a non-metal are made of ions. Metal ions have a positive charge and non-metal ions have a negative

charge. To work out the formula of an ionic compound

Write the formulae of the ions Adjust the number of each ion so that there is no overall

charge

Example 1 – magnesium bromide Example 2 – aluminium nitrate

Ions are Mg2+ and Br- Ions are Al3+ and NO3-

Need 2 x Br- to balance Mg2+ Need 3 x NO3- to balance Al3+

Formula is MgBr2 Formula is Al(NO3)3

The formulae for ions are given on the attached sheet. This sheet is notavailable in exams so the formulae will have to be learnt.

c) Covalent compounds

Some formulae for covalent compounds can be worked out from the name. The prefix mono- means one, di- means two and tri- means three. Therefore, carbon monoxide is CO, silicon dioxide is SiO2 and sulphur trioxide

is SO3 Other formulae have to be learnt e.g. ammonia is NH3 and methane is CH4

2) Equations

There are no word equations at A-level. An equation means a balancedsymbol equation.

To write a balanced symbol equation: Identify the reactants and products Write a word equation Write down the formula for each substance Balance the equation by putting numbers in front of formulae Add state symbols (s), (l), (g) or (aq)

Example – marble chips and hydrochloric acid

Reactants are calcium carbonate and hydrochloric acidProducts are calcium chloride, carbon dioxide and water

Calcium carbonate + hydrochloric acid calcium chloride + carbon dioxide + waterCaCO3 + HCl CaCl2 + CO2 + H2O

Ca 1 1C 1 1O 3 3H 1 2Cl 1 2

2 in front of HCl balances the equation

CaCO3 + 2HCl CaCl2 + CO2 + H2O

Adding state symbols

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

Topic 1c – CalculationsRevision Notes

1. Molar Mass

Molar mass is calculated by adding up the masses of the atoms in the formula The percentage of the total made up by a particular element can also be calculated

Example - sodium carbonate, Na2CO3

Na 2 x 23.0 = 46.0C 1 x 12.0 = 12.0O 3 x 16.0 = 48.0

Total = 106.0

% by mass of oxygen = 48.0 x 100/106.0= 45.3%

2. Empirical & Molecular Formulae

Write down mass or % of each element Divide each one by the atomic mass of that element Find the ratio of the numbers (divide them all by the smallest one)

In a substance containing only sodium, sulphur and oxygen, the compositionis found to be 32.4% sodium and 45.0% oxygen. Calculate the substance’sempirical formula.

% sulphur = 100 – 32.4 – 45.0= 22.6%

Na S O

Composition 32.4 22.6 45.0R.a.m. 23.0 32.1 16.0Comp/r.a.m. 1.41 0.70 2.81 by smallest 2.01 1 4.01

Empirical formula is Na2SO4

Molecular formula is a multiple of empirical formula

Empirical formula = CH2O & Mr = 60. Find molecular formula.Empirical mass = 30 so molecular formula = 2 x empirical formula = C2H4O2

3) Atom economy

Atom economy = Molecular mass of desired product x 100%Molecular masses of all products

Example

Bromoethane, CH3CH2Br, reacts with sodium hydroxide to produce ethanol, CH3CH2OH.

CH3CH2Br + NaOH CH3CH2OH + NaBr

In the above example

Molecular mass of desired product = 46.0Molecular masses of all products = 46.0 + 102.9

= 148.9

Atom economy = 46.0/148.9 x 100%= 30.9%

Chemical processes with a high atom economy produce fewer waste materials Atom economy can be improved by finding a use for waste product

Topic 2a – Mass SpectrometryRevision Notes

1) Mass Spectrometry

The mass spectrum of an element gives the following information:

o Number of peaks = number of isotopeso M/z of peak = mass number of isotopeo Size of peak = relative abundance (i.e. percentage)

The mass spectrum for zirconium

The 5 peaks in the mass spectrum shows that there are 5 isotopes of zirconium -with relative isotopic masses of 90, 91, 92, 94 and 96 on the 12C scale.

In this case, the five species detected (with their relative abundances) are:

90Zr+ 51.5%, 91Zr+ 11.2%, 92Zr+ 17.1%, 94Zr+ 17.4%, 96Zr+ 2.8%

a) Operation

Four processes take place inside a mass spectrometer; ionisation, acceleration,deflection and detection

The sample is vaporised and then ionised using high energy electrons form anelectron gun. X(g) + e- X+(g) + 2e-

Ionisation is necessary to allow the remaining processes to take place. The energy ofthe electrons from the gun is controlled so that only one electron is knocked off eachatom or molecule in the sample i.e. only 1+ ions are formed

The positive ions are accelerated by a negatively charged plates which also focus theions into a beam

The ions are deflected by a magnetic field. The amount of deflection depends onmass:charge ratio (m/z). As the ions have the same charge, deflection depends onmass with heavier ions deflected less and lighter ions deflected more. The strength ofthe magnetic field can be altered so that ions of different m/z are detected

When an ion hits the detector a small current is produced which is recorded. The sizeof the current is proportional to relative abundance

b) Uses

A mass spectrum can be used to calculate the relative atomic mass of an element(see below)

Mass spectrometry can be used to identify elements from the mass numbers ofthe isotopes e.g. in space probes

On another planet, such as Mars, the number of isotopes and their massnumbers will be the same as on Earth. However, the relative abundances maywell be different so the relative atomic mass will be different as well.

Mass spectrometry can also be used to determine the relative molecular mass ofa covalent molecule (see Topic 16 – Analytical Techniques for more details)

2) Relative atomic mass

Relative atomic mass is the weighted average mass of an atom of an element takingthe mixture of isotopes into account. However, learn the technical definitionfrom definitions sheet

To calculate relative atomic mass, add together (mass number x abundance/totalabundance) for each isotope

Abundances can be given as percentages or may have to be worked out from the lineheights on the spectrum

Example 1:

Calculate the relative atomic mass of Cl from the following information.

75% of Cl atoms have a mass number of 3525% of Cl atoms have a mass number of 37

Total abundance = 100%

Average mass of a Cl atom = (mass no x percent/100) + (mass no x percent/100)= (35 x 75/100) + (37 x 25/100)= 35.5

Example 2:

Calculate the relative atomic mass of unknown metal Z from the following information.

M/z Abundance188 1.5189 2.5190 3.0192 4.5

Total abundance = 11.5

Average mass of a Z atom = (188 x 1.5/11.5) + (189 x 2.5/11.5) +(190 x 3.0/11.5) + (192 x 4.5/11.5)

= 24.52 + 41.09 + 49.57 + 75.13= 190.3

Example 3:

A mass spectrum of a sample of indium shows two peaks at m/z = 113 and m/z = 115. Therelative atomic mass of indium is 114.5. Calculate the relative abundances of these twoisotopes.

Difference between mass numbers = 115 - 113= 2

Relative atomic mass lies 1.5/2 along this difference i.e. 75% of the way

Abundances are m/z =113, 25%, m/z 115, 75%

Topic 2b – Electronic StructureRevision Notes

1) Orbitals

An orbital is the region in which an electron can be found Each orbital can hold up to two electrons of opposite spin Orbitals have different shapes called s, p, d, and f S orbitals are spherical in shape and come in sets of one

P orbitals are hour-glass or egg-timer shaped and come in sets of three (which canhold up to 6 electrons)

D orbitals come in sets of five and f orbitals come in sets of seven (which can hold upto 10 and 14 electrons, respectively)

2) Energy levels (or shells)

The first energy level (or shell) only contains an s orbital, labelled 1s The second energy level contains an s orbital and three p orbitals, labelled 2s and 2p The third energy level contains an s orbital, three p orbitals and five d orbitals,

labelled 3s, 3p and 3d The order in which the orbitals are filled is as follows: 1s 2s 2p 3s 3p 4s 3d 4p Note that the 4s fills before the 3d

4s of "lower"energy than 3d

Distance from nucleus

Energy

1s

2s

2p

3s

3p

3d4s

4p

4d

4f

Ionisation energy

Some examples of electronic structures are shown below.

Hydrogen 1 electron 1s1

Nitrogen 7 electrons 1s2 2s2 2p3

Sodium 11 electrons 1s2 2s2 2p6 3s1

Sulphur 16 electrons 1s2 2s2 2p6 3s2 3p4

Calcium 20 electrons 1s2 2s2 2p6 3s2 3p6 4s2 or [Ar] 4s2

Iron 26 electrons 1s2 2s2 2p6 3s2 3p6 4s2 3d6 or [Ar] 4s2 3d6

There are 2 exceptions to the pattern: chromium and copper Cr is 1s2 2s2 2p6 3s2 3p6 4s1 3d5 and Cu is 1s2 2s2 2p6 3s2 3p6 4s1 3d10

These electronic structures are more stable than the alternative structures that followthe pattern

When transition metals, like iron and copper, form ions they lose their 4s electronsbefore their 3d electrons

Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6 or 1s2 2s2 2p6 3s2 3p6 4s0 3d6

There are various models to illustrate atomic structure. The model chosen dependson what we are trying to explain e.g. bonding can be explained by representing Naas 2.8.1 and Cl as 2.8.7 with both seeking to gain a full outer shell. Explaining whycopper(II) sulphate is blue requires us to use the sub-shell model

Topic 3 – MolesRevision Notes

1. Moles

In Chemistry, amounts are measured in moles A mole contains 6.02 x 1023 particles. Particles can be atoms, molecules, ions or

electrons For a solution, moles = concentration x volume/1000 (volume in cm3)

2. Reacting Mass Calculations

Step 1 - Find the number of moles of the thing you are told aboutStep 2 – Use the equation to find out the moles of the thing you are asked about.Step 3 – Find the mass of the thing you are asked about.

Example

Work out the mass of HCl formed from 6.0g of hydrogen

H2 + Cl2 2HCl

Step 1: Moles H2 = 6.0 2.0 = 3.0 (mass molar mass)Step 2: Moles HCl = 3.0 x 2/1 (from equation) = 6.0Step 3: Mass HCl = 6.0 x molar mass = 6 x 36.5 = 219g (moles x molar mass)

3. Titration Calculations

Concentration is usually measured in moles of solute per cubic decimetre of solution,mol dm-3

A cubic decimetre, 1dm3, has the same volume as a litre i.e. 1000cm3

The volume of a solution is often measured in cm3. This needs to be converted todm3 by dividing by 1000 before calculating a concentration in mol dm-3

Step 1 - Find the number of moles of the thing you know the concentration and volume of.Step 2 – Use the equation to find out the moles of the thing you are asked about.Step 3 – Find the unknown concentration or molar mass

Example

25 cm3 of NaOH needed 21.5 cm3 of 0.1 mol dm-3 H2SO4 for neutralisation. Calculate theconcentration of the NaOH solution.

H2SO4 + 2NaOH 2NaCl + 2H2O

Step 1: Moles H2SO4 = 0.1 x 21.5 1000 = 2.15 x 10-3 (conc x vol 1000)Step 2: Moles NaOH = 2.15 x 10-3 x 2 (from equation) = 4.30 x 10-3

Step 3: Conc NaOH = 4.30 x 10-3 (25 1000) = 0.172 mol dm-3 (moles volume in dm3)

4. Ideal Gas Equation

The ideal gas equation is:

PV = nRTWhere: P = pressure in Pa

V = volume in m3 (1 m3 = 103 dm3 = 106 cm3)n = number of molesR = gas constant (8.31 J K-1 mol-1)T = Kelvin temperature (C + 273)

Example

0.166 mol of oxygen is in a sealed container whose volume is 1725 cm3. The temperature is300 K. Calculate the pressure of the oxygen inside the container.(The gas constant R = 8.31 J K-1 mol-1)

PV = nRTP = nRT/V

T has correct units but V is in cm3 rather than m3

1725 cm3 = 1725 x 10-6 m3

= 1.725 x 10-3 m3

P = 0.166 x 8.31 x 300/(1.725 x10-3)= 239906 Pa= 240 kPa

For a fixed number of moles of gas, the ideal gas equation reduces to:

P1V1/T1 = P2V2/T2

This version can be used to calculate the effect of changes in P, V or T on such asample.

The ideal gas equation can be combined with n = m/Mr or with (density) = m/V

PV = mRT/Mr

P = RT/Mr

If these versions are used, mass must be in grams and density in g m-3

5) Percentage yield

Most organic reactions do not give 100% conversion of reactant to product Reasons for this include the fact that most organic reactions are reversible, there

may be side products and there will be loss of the desired product during purification

% yield = Actual moles of product x 100%Possible moles of product

Example

In the following reaction, 2.18g of bromoethane produce 0.75g of ethanol. Calculate thepercentage yield.

CH3CH2Br + NaOH CH3CH2OH + NaBr

Moles of reactant (bromoethane) = mass/molar mass= 2.18/109= 0.020 mol

Possible moles of ethanol = 0.020 mol (from equation)Actual moles of ethanol = 0.75/46.0

= 0.0163 molPercentage yield = 0.0163/0.020 x 100%

= 82%

6) Ionic equations

Ionic equations leave out ions that are unchanged in a reaction. They give a clearerpicture of what is happening in a reaction

To go from a symbol equation to an ionic equation:o Split up anything that is (aq) and ionic (acids, alkalis and salts)o Cancel ions that are on both sides

ExampleSymbol equation: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Split up into ions: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l)

Cancel Na+ & Cl-: H+(aq) + OH-(aq) H2O(l)

Topic 4a – BondingRevision Notes

1) Introduction

Atoms form bonds to get a full outer shell of electrons Elements in Group 0 (He, Ne, Ar, Kr, Xe, Rn) do not normally undergo chemical

reactions as they have full outer shells and, so, do not need to form bonds to otherelements

There are three types of bonding: ionic, covalent and metallic Crystals (solids) have one of four types of structure: giant ionic, simple molecular,

giant covalent and giant metallic. Giant structures have millions of atoms or ions heldtogether in 3 dimensions. Simple molecular substances consist of a countable numberof atoms e.g. H2O, C12H22O11

Giant covalent is sometimes called giant molecular or macromolecular To melt a substance the forces holding the particles together need to be broken To conduct electricity there must something charged that can move (ions or

electrons).

2) Types of Bonding

Ionic bonding – metals transfer electrons to non-metals producing positive metal ionsand negative non-metal ions. An ionic bond is an electrostatic attraction betweenoppositely charged ions. Dot-cross diagrams show outer electrons only e.g. NaCl

Covalent bonding – A covalent bond is a shared pair of electrons. Only non-metalscan get a full shell by sharing electrons. The bond is the attraction of the sharedelectrons for the two nuclei. Dot-cross diagrams show outer electrons only e.g. Cl2

In dative covalent bonds, one atom donates a lone pair of electrons to form thebond. The atom accepting the electrons has an empty orbital

Examples of species that donate lone pairs include NH3, H2O, Cl- Examples of species that accept lone pairs include H+ and BF3

Metallic bonding – metals lose their outer shell electrons to produce a lattice ofpositive metal ions surrounded by delocalised (free) electrons.

3) Types of Structure

a) Giant ionic lattices e.g. sodium chloride

o Lattice of oppositely charged ions.o High melting and boiling points (strong forces of attraction between ions

need to be broken).o Do not conduct when solid (ions not free to move).o Conduct when molten or dissolved in water (ions then free to move).o Are brittle (layers of ions cannot slide over each other without repelling)

Note – should be Na+ and Cl-

b) Molecular lattices e.g. iodine and ice

o Consists of molecules held together by weak intermolecular forces (seesection 5 below)

o Low melting and boiling points (weak forces of attraction betweenmolecules are easily broken)

o Do not conduct (no free electrons)

Structure of ice ( is hydrogen, is oxygen)

c) Giant metallic lattices e.g. magnesium

o Lattice of metal ions surrounded by delocalised electronso High melting and boiling points usually (strong forces of attraction between

metal ions and free electrons need to be broken)o Conduct when solid as delocalised electrons can moveo Are malleable (bendy) and ductile (can be drawn into wires) because the

delocalised electrons allow the layers of ions to slide over each other withoutrepelling

d) Giant covalent lattices e.g. diamond, graphite

o Lattice of non-metal atoms joined by strong covalent bondso Very high melting and boiling points usually (many strong covalent bonds to

be broken)o Diamond doesn’t conduct (no free electrons)o In graphite each C forms covalent bonds to 3 other C’s within the layers.

There are weak forces between the layers allowing them to slide over eachother (hence use in pencils and as a lubricant). Fourth outer shell electron isdelocalised and can move between the layers allowing graphite to conductelectricity

Diamond graphite

4) Electronegativity and bond polarity

Electronegativity is the ability of an atom to attract the electrons in a covalent bond. When there is a big difference in electronegativity between the atoms at either end

of a covalent bond the electrons will be pulled towards the more electronegativeatom (shared unequally). This is a polar covalent bond (the molecule has apermanent dipole)

For example, fluorine is more electronegative than hydrogen so the H-F bond is polar

δ+ δ-HF

Polar molecules have permanent dipoles that don’t cancel out (e.g. H2O) because thedipoles are at an angle

Non-polar molecules either have no dipoles (e.g. Cl2) or dipoles that cancel out (e.g.CO2) because the dipoles are at 180

5) Intermolecular Forces

Intermolecular forces (IMF) only occur in simple molecular substances.Mentioning molecules or IMF in questions about ionic, metallic or giantcovalent substances is a chemical error (CE) and will lose all marks for thatquestion

There are 2 types of intermolecular force

a) Van der Waal’s

o Arise from temporary dipole (uneven distribution of electrons) in one molecule thatinduces dipole in another molecule

o The more electrons, the stronger the van der Waal’s forces between moleculeso Van der Waal’s forces occur in all simple molecular substances

b) Hydrogen bonds

o Only occur when hydrogen is bonded to one of the 3 most electronegative elements:N, O and F. Other elements are not electronegative enough

o There is a large difference in electronegativity between H and N/O/F. This makes Hδ+ and N/O/F δ-

o δ+ H is strongly attracted to lone pair on N/O/F in another moleculeo Diagram must show lone pairs, dipoles and H-bond shown by dotted line (see

example for NH3 below. (Note that N in NH3 has 1 lone pair, O in H2O has 2 lonepairs and F in HF has 3 lone pairs)

H

NH

H

H

NH

H

o Water, ammonia and HF have higher melting and boiling points than expected due tothe strength of the hydrogen bonds that have to be broken

o Ice is less dense than water because, in ice, the molecules are held further apart sothere is more space in the structure

6) Shapes of Molecules

The following procedure allows the shape of a molecule to be worked out: Draw a dot-cross diagram Count number of electron pairs round the central atom Pairs of electrons repel each other and get as far apart as possible Lone pairs repel more than bonding pairs so bonds are pushed closer together

e.g. 107 in ammonia compared with the tetrahedral bond angle of 109.5 inmethane

Number of pairs(explanation)

Examples Name of shape Bond angle

2 bonding pairs(repel equally)

BeCl2 Linear 180

3 bonding pairs(repel equally)

BF3 Trigonal planar 120

4 bonding pairs(repel equally)

CH4, NH4+, PCl4+ Tetrahedral 109.5

5 bonding pairs(repel equally)

PCl5 Bipyramidal 90 and 120

6 bonding pairs(repel equally)

SF6, PCl6- Octahedral 90

3 bonding, 1 lone(lp repels more than bps)

NH3 Pyramidal 107

2 bonding , 2 lone(lps repels more than bps)

H2O, NH2- V-shaped 104.5

4 bonding, 2 lone(lps repels more than bps)

XeF4 Square Planar 90

This can be done mathematically e.g. PCl4+

Outer shell electrons (P) 5Bonds formed 4Adjust for charge (lost e-) -1Total for P 8Pairs for P 4Lone pairs (pairs-bonds) 0

tetrahedral, bond angle 109.5

Topic 4b – PeriodicityRevision Notes

1) Blocks in the Periodic Table

An element can be assigned to the s, p or d block by working out which sub-level itsoutermost electron is in:

Na is 1s2 2s2 2p6 3s1 so is in the s-block O is 1s2 2s2 2p4 so is in the p-block Sc is 1s2 2s2 2p6 3s2 3p6 4s2 3d1 so is in the d-block

2) Ionisation Energies

a) First ionisation energy

Evidence that electrons are arranged in shells or energy levels can be obtainedby measuring ionisation energies

The first ionisation energy of an element is the energy needed to remove onemole of electrons from one mole of gaseous atoms i.e.

M(g) M+(g) + e-

b) Successive Ionisation Energies of an element

Second and subsequent ionisation energies of an element can also be measuredi.e.

M+(g) M2+(g) + e-

Second ionisation is greater than first ionisation energy as it is harder to removean electron from a positive ion than from a neutral atom

Jumps in ionisation energies occur when going from one energy level (shell) toanother. The jump occurs because the new energy level is closer to nucleus andless shielded

c) Trends Down Group 2 (Be-Ba)

1st ionisation energy decreases down Group 2 (and all other groups) With each successive element, there is an extra electron shell, so the outer

electron is further from the nucleus and more shielded. Less attractionbetween the nucleus and the outer shell

This is evidence for electrons being arranged in energy levels (shells)

3) Trends Across Period 3 (Na-Ar)

Be able to describe and explain trends in the following:

1st ionisation energy Describe – general increaseExplain – bigger nuclear charge, same shieldingDip from Mg to Al because Mg is losing 3s electron, Al is losing 3p.3p is higher in energy, easier to remove.Dip from P to S because P is 3p3, S is 3p4. Mutual repulsion of pairedelectrons in S make electron easier to remove than in P.

Atomic radius Describe – it decreasesExplain – bigger nuclear charge, same shielding

Electronegativity Describe – it increasesExplain – bigger nuclear charge, same shielding, stronger attractionbetween nucleus and shared pair of electrons

Melting & boiling Na, Mg and Al have metallic bonding.points Attraction between positive ions and delocalised free electrons is

strong so melting points are high.Melting point increases from Na to Mg to Al because metal ion hasgreater charge and there are more free electrons per ion so metallicbonding is stronger.Si has very high melting point. Giant covalent structure has manystrong covalent bonds to be broken.P4, S8 and Cl2 have low melting points. These are simple covalentmolecules held together by weak Van der Waal’s forces. Van derWaal’s forces increase with molecular mass so S8 has highest meltingpoint, then P4 then Cl2.Ar has simple atomic structure. Fewest electrons, weakest Van derWaal’s forces between atoms

Periodic trends are repeated across different rows of the Periodic Table. Any trend inproperties across Period 3 will also be shown across Period 2 (and period 4, 5 etc)

Topic 5a – Introduction to Organic ChemistryRevision Notes

1) Formulae

Be able to recognise and use the different ways of showing organic compounds:

Molecular formula is the actual number of atoms of each element in a moleculee.g. C2H6O for ethanol

Empirical formula is the simplest whole number ratio of the atoms of each elementin a molecule e.g. CH2 for ethene (from molecular formula C2H4 2)

General formula is the simplest algebraic formula for a member of a homologousseries e.g. CnH2n+2 for alkanes

Structural formula is the minimum detail that shows the arrangement of the atomsin a molecule e.g. CH3CH2OH for ethanol

Displayed formula shows the relative positioning of atoms and the bonds betweenthem e.g. for ethanol:

All bonds should be shown. Do not put –OH for the alcohol group

2) Functional groups and naming organic compounds

Be able to recognise and use the following terms:

A homologous series is a series of organic compounds having the same functionalgroup with successive members differing by CH2

Alkanes, alkenes, alcohols and halogenoalkanes are all homologous series A functional group is a group of atoms responsible for the characteristic reactions

of a compound e.g. C=C for alkenes and –OH for alcohols

The rules for naming organic compounds were devised by IUPAC (International Union of Pureand Applied Chemistry). They are as follows.

1) The functional group gives the ending of the name e.g. –ol for an alcohol2) The number of carbons gives the first part of the name e.g. prop- or propan- for

3 carbons3) Number the carbon chain to give the functional group carbon the lowest number4) Any side chains (branches) or halogens go at the front of the name with commas

between numbers and dashes between numbers and words e.g. 2,2-dimethylhexane

5) With more than 1 side chain or halogen, use alphabetical order e.g. 1-bromo-2-methylbutane

3) Structural isomers

Structural isomers have the same molecular formula but different structural formulae There are 3 types of structural isomers: chain, position and functional group Position isomers differ in the location of the functional group e.g.

1-bromobutane 2-bromobutane

Chain isomers have different arrangements of the carbon chain e.g.

Pentane 2-methylbutane

A molecular formula can be common to compounds from different families. These arefunctional group isomers e.g.o C3H6O can be either propanal, CH3CH2CHO, or propanone, CH3COCH3o C3H6 can be either propene or cyclopropaneo C3H6O2 can be propanoic acid, methyl ethanoate or ethyl methanoateo C2H6O can be either ethanol or methoxymethane, CH3OCH3 (an ether)

For CHEM1, the only pair of functional group isomers needed is alkenesand cyclic alkanes (both of which have general formula CnH2n)

Topic 6b – AlkanesRevision Notes

1) General

Alkanes are saturated hydrocarbons with general formula CnH2n+2 Saturated = only single C-C bonds Hydrocarbon = contains C and H only Alkane molecules are non-polar so the only intermolecular forces are Van der Waal’s

forces

2) Boiling points

Boiling point increases with chain length – more electrons, more Van der Waal’sforces between molecules

Boiling point decreases as branching increases – branched alkanes have less surfacearea in contact so intermolecular forces are weaker (or straighter chains can packcloser, more Van der Waal’s forces between molecules)

The first four alkanes are gases and are used as fuels (methane for domestic heatingand cooking, propane as LPG and in canisters for camping/caravanning, butane forcigarette lighters and in canisters)

Petrol consists of liquid alkanes with between 5 and 8 carbons

3) Fractional Distillation

Crude oil (petroleum) is a mixture of many compounds, most of which are alkanes Crude oil is separated into fractions, many of which can be used directly as fuels The separation process is called fractional distillation. This involves:

o Separation is based on the different boiling points of alkaneso The boiling points of alkanes depend on the size of the molecule (and

strength of Van der Waals forces)o Vaporised crude oil is fed into a column that is hot at the bottom and cool at

the topo Smaller molecules with lower boiling points come out at the top of the

column. Larger molecules with higher boiling points come out at the bottom

4) Cracking

Crude oil contains more long chain alkanes than are needed. Cracking breaksthese alkanes down into products for which there is higher demand

Cracking involves the breaking of C-C bonds in alkanes and this requires a hightemperature

Example: C10H22 C8H18 + C2H4Decane octane ethene

Thermal cracking gives a high proportion of alkenes. It needs a high temperature(400-900C) and a high pressure (7000 kPa)

The alkenes from thermal cracking are used to make polymers and alcohols Catalytic cracking produces motor fuels and aromatic hydrocarbons. It needs a

zeolite catalyst, a high temperature (450C) and a slight pressure

5) Combustion

a) Introduction

Complete combustion requires a plentiful supply of oxygen e.g.

C5H12 + 8O2 5CO2 + 6H2O

Combustion of fossil fuels, including alkanes, results in the release of carbondioxide into the atmosphere

In a limited supply of air, incomplete combustion occurs forming CO or C (soot)

C5H12 + 5.5O2 5CO + 6H2OC5H12 + 3O2 5C + 6H2O

Incomplete combustion wastes petrol meaning more fuel is needed

b) Catalytic converters

The pollutants produced by car engines include carbon, C, carbon monoxide, CO,sulphur dioxide, SO2, oxides of nitrogen, NOx and unburnt hydrocarbons

Carbon monoxide and carbon come from the incomplete combustion of thehydrocarbons in petrol and are toxic. Carbon is also a respiratory irritant

Sulphur dioxide is produced when traces of sulphur in the fuel react with oxygen.SO2 causes acid rain

Oxides of nitrogen are produced when oxygen and nitrogen from the air reacttogether due to the very high temperatures reached inside car engines. NO2 istoxic, triggers asthma attacks and forms HNO3 (i.e. acid rain) when it reacts withwater and oxygen

Small amounts of hydrocarbons pass straight through a car engine without beingburnt

Catalytic converters consist of a honeycomb of ceramic material coated withplatinum, palladium or rhodium (Pt/Pd/Rh)

The honeycomb produces a large surface area on which reactions can occur Catalytic converters reduce the emission of CO and NO by allowing them to react

together to make harmless products

2CO(g) + 2NO(g) 2CO2(g) + N2(g)

c) Global Warming

In the troposphere (lowest level of the atmosphere), various gases absorbinfrared radiation and keep the atmosphere warm

Infrared radiation is absorbed by C=O bonds in CO2, O-H bonds in H2O and C-Hbonds in methane. The absorbed energy makes the bonds vibrate

Increased concentrations of greenhouse gases, like CO2, may contribute to globalwarming because of the increased absorption of IR radiation

d) Acid Rain

Combustion of fuels containing sulphur produces sulphur dioxide Acid rain is formed when SO2 dissolves in water SO2, which is acidic, can be removed from flue gases using CaO, which is a base.

This is a neutralisation reaction.

SO2 + CaO CaSO3

CHEM1 - Foundation ChemistryDefinitions to Learn

1. Formulae & Equations

Atomic number number of protons in the nucleus of an atom

Mass number sum of the protons and neutrons in the nucleus of an atom

Isotopes atoms with the same number of protons but differentnumbers of neutrons

Empirical formula simplest whole number ratio of the atoms of each element ina compound

Molecular formula actual number of atoms of each element in a molecule

Atom economy mass of desired product x 100%total mass of reactants

2. Mass Spectrometry & Electronic Structure

Relative atomic mass average mass of an atom relative to 1/12 of the mass of acarbon-12 atom

Relative molecular mass mass of a molecule relative to 1/12 of the mass of a carbon-12 atom

3. Moles

Mole unit for amount of substance

Avogadro constant, NA number of particles present in a mole (6.02 x 1023 mol-1)

4. Bonding & Periodicity

Lattice a regular 3-dimensional array

Ionic bond electrostatic attraction between oppositely charged ions in alattice

Covalent bond a shared pair of electrons

Dative covalent bond covalent bond formed by donation of a lone pair

Metallic bond a lattice of positive ions surrounded by delocalised electrons

Electronegativity ability of an atom to attract the electrons in a covalent bond

Polar bond electrons are shared unequally (due to difference inelectronegativity of atoms at either end)

1st ionisation energy energy change when one mole of electrons is removed fromone mole of gaseous atoms

5. Alkanes

Homologous series a series of organic compounds with: the same general formula difference of CH2 between each member a trend in physical properties similar chemical properties

Functional group a group of atoms responsible for the characteristic reactionsof a compound

Structural isomers same molecular formula, different structures

Hydrocarbon a compound that contains hydrogen and carbon only

Saturated contains only single C-C bonds

Fractional distillation separates due to differences in boiling point

Fraction mixture of compounds of similar boiling point

Cracking breaking a long chain alkane into a shorter chain alkane andan alkene

CHEM1 - Foundation ChemistryCalculations

1. Formulae & Equations

Molar mass* Sum of masses of elements in formulae.g. 106.0 for Na2CO3 (always give to 1 dp)

Percentage by mass Mass of specified element x 100%Molar masse.g. 43.4% for Na in Na2CO3

Empirical formula For each element, divide percentage by relative atomic massDivide through by smallestIf result is, say, 1 1 1.5, then double to get whole numbers.Do not round 1.5 to 2

Molecular formula Divide relative molecular mass by mass of empirical formulato get multiplier. Multiply empirical formula by multiplier.

Atom economy Mass of desired product x 100%Total mass of reactants

* (also applies to relative atomic mass, relative molecular mass & relative formula mass)

2. Mass Spectrometry & Electronic Structure

Relative atomic mass For each isotope, calculate mass number x abundance.Add results together.Mass number may come from m/z scale of graph.Abundance can be %/100 or line height/total line height.

3. Moles

Moles If given mass in g, moles = mass/molar massIf given concentration in mol dm-3 and volume in dm3, moles= concentration x volumeIf given concentration in mol dm-3 and volume in cm3, moles= concentration x volume/1000

Percentage yield Actual moles x 100%Possible moles

Percentage purity Mass of pure or desired x 100%Mass of impure or mixture

Concentration in g dm-3 Mol dm-3 x molar mass

Dissolved mass Concentration in g dm-3 x volume in dm3

Ideal gas equation PV = nRT

P=pressure in Pa, V=volume in m3, n=number of moles,R=gas constant (8.31 J K-1 mol-1), T=temperature in K1kPa = 1000 Pa, 1 cm3 = 10-6 m3, 1 dm3 = 10-3 m3,K = C+273