Basic Revision Notes

7/21/2019 Basic Revision Notes

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Common Exam Terms and their meanings

Objective 1

Define: Give the precise meaning of a word, phrase or physical quantity.

Draw: Represent by means of pencil lines.

Label: Add labels to a diagram.

List: Give a sequence of names or other brief answers with no explanation.

Measure: Find a value for a quantity.

State: Give a specific name, value or other brief answer without explanation or calculation.

Objective 2

Annotate: Add brief notes to a diagram or graph.

Apply: Use an idea, equation, principle, theory or law in a new situation.

Calculate: Find a numerical answer showing the relevant stages in the working (unless instructed not to do

so).

Describe: Give a detailed account.

Distinguish: Give the differences between two or more different items.

Estimate: Find an approximate value for an unknown quantity.

Identify: Find an answer from a given number of possibilities.

Outline: Give a brief account or summary.

Objective 3

Analyse : Interpret data to reach conclusions.

Comment: Give a judgment based on a given statement or result of a calculation.

Compare : Give an account of similarities and differences between two (or more) items, referring

to both (all) of them throughout.

Construct: Represent or develop in graphical form.

Deduce: Reach a conclusion from the information given.

Derive: Manipulate a mathematical relationship(s) to give a new equation or relationship.

Design: Produce a plan, simulation or model.

Determine: Find the only possible answer.

Discuss: Give an account including, where possible, a range of arguments for and against the

relative importance of various factors, or comparisons of alternative hypotheses.

Evaluate: Assess the implications and limitations.

Explain: Give a detailed account of causes, reasons or mechanisms.

Predict: Give an expected result.

Show: Give the steps in a calculation or derivation.

Sketch: Represent by means of a graph showing a line and labelled but unscaled axes but withimportant features (for example, intercept) clearly indicated.

Solve: Obtain an answer using algebraic and/or numerical methods.

Suggest: Propose a hypothesis or other possible answer.


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Atoms and Electron Structure

Revision Notes

1) Atomic Structure

Relative mass Relative charge

Proton 1 +1

Neutron 1 0

Electron 1/2000 -1

The nucleus contains almost all of the mass of an atom because that is where the

protons and neutrons are found The nucleus of an atom contains all of the positive charge

The electrons are outside the nucleus and, therefore, so is the negative charge 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

2) Isotopes and ions

Isotopes are atoms with the same number of protons but different numbers ofneutrons (and different masses)

For example, chlorine has two isotopes35

Cl and37

Cl. Both have 17 protons but theyhave 18 and 20 neutrons, respectively

Isotopes of an element have the same chemical properties because they have the

same electron arrangement 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. A Na+ion has lost one electron so it now has 10.

3) Relative atomic mass

Relative atomic mass is the average mass of an atom of an element taking the

mixture of isotopes into account. However, learn the technical definition fromdefinitions sheet To calculate relative atomic mass, add together (mass number x percentage/100) for

each isotope

Example:75% of Cl atoms have a mass number of 35

25% of Cl atoms have a mass number of 37

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

For simple molecules, such as O2and H2O, the relative molecular mass is calculatedby adding the relative atomic masses of the elements involved, giving 32.0 for O2and

18.0 for H2O


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For giant structures, such as Na2S and SiO2, the relative formula mass is calculated

by adding the relative atomic masses of the elements involved, giving 68.1 for Na2Sand 60.1 for SiO2

4) Orbitals

An orbital is a region that can hold up to two electrons with opposite spins

Orbitals have different shapes called s, p, d, and f (but f orbitals are beyond oursyllabus)

S orbitals are spherical in shape and come in sets of one (which can hold up to 2electrons)

P orbitals are hour-glass or egg-timer shaped and come in sets of three (which can

hold up to 6 electrons)

5) Energy levels (or shells)

The first energy level (or shell) only contains an s orbital, labelled 1s The first shell can hold up to 2 electrons

The second energy level contains an s orbital (labelled 2s) and three p orbitals(labelled 2p)

The second shell can hold up to 8 electrons

The third energy level contains an s orbital, three p orbitals and five d orbitals The third shell can hold up to 18 electrons 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

Some examples of electronic structures are shown below.

Hydrogen 1 electron 1s1

Nitrogen 7 electrons 1s22s22p3Sodium 11 electrons 1s22s22p63s1Sulphur 16 electrons 1s22s22p63s23p4

Calcium 20 electrons 1s22s22p63s23p64s2

In a Cl-ion, the 18 electrons are arranged 1s22s22p63s23p6 In a Na+ion, the 10 electrons are arranged 1s22s22p6

The diagram over shows the relative energies of the orbitals from 1s to 4f


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Source: www.chemsheets.co.uk

6) Successive Ionisation Energies

Evidence that electrons are arranged in shells or energy levels can be obtained bymeasuring the successive ionisation energies of an element

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

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

Note - State symbols are essential in ionisation equations

In general, ionisation is easier if the nuclear charge is smaller, the electron is furtheraway from the nucleus and there is more shielding from inner electron shells.

For an element, successive ionisation energies get bigger because the remainingelectrons are held more tightly by the unchanged nuclear charge.

Jumps in ionisation energies occur when going from one energy level (shell) to

another. This tells you which group the element is in. The jump in energy occursbecause the new energy level is closer to nucleus and less shielded.

7) Blocks in the Periodic Table

An element can be assigned to the s, p or d block by working out which type of

orbital its outermost electron is in The s block is groups 1 and 2

The p block is groups 3 to 8 The d block is between the s and p blocks

4s of "lower"energy than 3d

Distance from nucleus

Energy

1s

2s

2

3s

3p

3d4s

4p

4d

4f

Ionisation energy


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Equations

Revision 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, Br2and

I2

b) Ionic compounds

Compounds of a metal and a non-metal are made of ions

Metal ions have a positive charge

Ions of Group 1 elements have a +1 charge, ions of Group 2 elements havea +2 charge

For transition elements, like copper and iron, the number after the name

gives the charge on the ion e.g. copper(II) oxide contains Cu2+ions, iron(III)oxide contains Fe3+ ions

Non-metal ions have a negative charge

Ions of Group 7 elements have a -1 charge, ions of Group 6 elements have a

-2 charge

You need to learn the formulas of five ions: nitrate, NO3-, carbonate, CO3

2-,sulphate, SO4

2-, hydroxide, OH-, and ammonium, NH4+

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

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 SiO2and sulphur trioxideis SO3

Other formulae have to be learnt e.g. ammonia is NH3and methane is CH4


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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 acid

Products are calcium chloride, carbon dioxide and water

Calcium carbonate + hydrochloric acid !calcium chloride + carbon dioxide + water

CaCO3+ HCl !CaCl2+ CO2 + H2O

Ca 1 1C 1 1O 3 3

H 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)

4) Ionic equations

Ionic equations leave out ions that are unchanged in a reaction. They give a clearer

picture 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

Example

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)

Na+(aq) + Cl-(aq) + H2O(l)

H+(aq) + OH-(aq) H2O(l)

Ag+(aq) + Cl-(aq) AgCl(s)


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5) Simple Equations

A salt is formed when the H+of an acid is replaced by a metal ion or NH4+

Salts are formed when acids react with bases and carbonates

Acid + carbonate salt + water + carbon dioxide

e.g. 2HCl(aq) + MgCO3(s) MgCl2(aq) + H2O(l) + CO2(g)Observations effervescence (fizzing), carbonate dissolves/disappears

Acid + base salt + water

e.g. 2HCl(aq) + MgO(s) MgCl2(aq) + H2O(l)

Observations base dissolves/disappears

Acid + alkali salt + water

e.g. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Observations no visible change unless indicator added

e.g. Sulphuric acid + ammonia ammonium sulphate

H2SO4(aq) + 2NH3(aq) (NH4)2SO4(aq)

e.g. Methane + Oxygen Carbon Dioxide + Water

CH4(g) + 2O2(aq) CO2(g) + H2O (g)

e.g. 2HCl(aq) + Mg (s) MgCl2(aq) + H2(g)

Observations base dissolves/disappearsMASH


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MolesRevision Notes

1. Relative molecular mass and relative formula mass

In Topic 1 we met relative isotopic mass and relative atomic mass. Remember that

relativemeans compared with 12C The relative mass of a simple covalent substance, like H2O or O2, is called its relative

molecular mass

The relative mass of a giant ionic or giant covalent substance, like NaCl or SiO2, iscalled its relative formula mass

Relative masses do not have units

Relative molecular masses and relative formula masses are calculated by adding uprelative atomic masses

Example relative formula mass of sodium carbonate, Na2CO3

Na 2 x 23.0 = 46.0C 1 x 12.0 = 12.0

O 3 x 16.0 = 48.0Total = 106.0

2. Empirical & Molecular Formulae

The empirical formula is the simplest whole number ratio of the atoms of each

element in a compound Write down mass or % of each element Divide each one by the relative atomic mass of that element

Find the ratio of the numbers (divide them all by the smallest one)

Example Find the empirical formula of a compound which is found to contain 1.40g ofnitrogen and 0.30g of hydrogen

N HComposition 1.40 0.30Divide by r.a.m. 14.0 = 0.1 1.0 = 0.3

Divide by smallest 1 3Empirical formula NH3

The molecular formula is the actual number of atoms of each element in a compound

Molecular formula is a multiple of empirical formula

Example Find the molecular formula of the compound whose empirical formula is CH2Oand whose relative molecular mass is 60.0

Mass of empirical formula = (1 x 12.0) + (2 x 1.0) + (1 x 16.0) = 30.0

60/30 = 2 so molecular formula = 2 x empirical formula = C2H4O2


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3. The mole

In Chemistry amounts of substance are measured in moles A mole contains 6.02 x 1023particles (atoms, molecules, ions or electrons)

There are 4 ways of calculating a number of moles

For a number of particles, moles = number of particles/6.02 x 10

23

Given a mass (in grams), moles = mass/molar mass

Given a gas volume, moles = volume in dm3/24 or moles = volume in cm3/24000

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

4. Molar mass

Molar mass is the mass of one mole of a substance

Its units are g mol-1

For an atom, the molar mass is the relative atomic mass expressed in g mol-1e.g.

23.0 g mol-1for Na

For a simple molecule, the molar mass is the relative molecular mass expressed in gmol-1e.g. (2 x 16.0) = 32.0 g mol-1for O2

For a giant ionic or giant covalent substance, the molar mass is the relative formula

mass expressed in g mol-1e.g. (23.0 x 35.5) = 58.5 g mol-1for NaCl

5. 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.

Work out the mass of HCl formed from 6 g of hydrogen

H2+ Cl2 2HCl

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

6. Gas Volume Calculations

First 2 steps same as reacting mass calculations but in step 3 use 24 dm3per mole of gas.

Work out the volume of CO2formed from 3.99 kg of iron (III) oxide

Fe2O3+ 3CO 2Fe + 3CO2

Step 1: Moles Fe2O3= 3990 159.6 = 25 (mass molar mass)Step 2: Moles CO2= 25 x 3 (from equation) = 75Step 3: Volume CO2= 75 x 24 = 1800 dm

3 (moles x 24)


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7. Titration Calculations

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

25 cm3of NaOH needed 21.5 cm3of 0.1 mol dm-3H2SO4for neutralisation. Calculate the

concentration 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 10-3x 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)

8) Water of crystallisation

Hydrated salts, like copper (II) sulphate crystals, contain water of crystallisation aspart of their structure (they are not damp!)

The water of crystallisation is shown in the formula by . which effectively means +e.g. CuSO4.5H2O

The water of crystallisation can be driven off by strong heating to leave an anhydrous

salt e.g.

CuSO4.5H2O CuSO4+ 5H2OBlue (hydrated) white (anhydrous)


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Bonding and StructureRevision Notes

1) Introduction

Atoms form bonds to get a full outer shell of electrons There are three types of bonding: ionic, covalent and metallic

The structures produced by forming bonds are either giant or simple

The possible combinations of structure and bonding are giant ionic, simple covalent,

giant covalent and giant metallic Simple covalent is sometimes called simple molecular 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). Technically this is called a mobile charged species

To dissolve in a particular solvent the substance must interact with the solvent

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 provides both of the shared pair of electrons e.g.formation of an ammonium ion, NH4

+, from ammonia, NH3, and H+


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! 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

Most are soluble in polar solvents like water. The ions interact with thedipoles in the solvent molecules

o Tend not to dissolve in non-polar organic solvents like cyclohexane. The ionsdo not interact with non-polar solvents

b) Simple covalent 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 betweenmoleculesare easily broken)

o

Do not conduct (no mobile charge carriers)o Most are insoluble in polar solvents, like water, because they do not interact

with the dipoles in the solvent. Alcohols, however, can hydrogen bond towater molecules

o Tend to dissolve in non-polar organic solvents, like cyclohexane, because the

solvent can interact with the simple covalent substance

c) Giant metallic lattices e.g. magnesium, copper

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

metal ions and free electrons need to be broken).o Conduct when solid (free electrons).

o Insoluble in all solvents (some react with water)

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 doesnt conduct (no mobile charge carriers). Graphite is the onlynon-metal that conducts as a solid (structure contains delocalised electrons)

o

Insoluble in polar solvents, like water, because they do not interact with thedipoles in the solvent in water


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4) Electronegativity and bond polarity

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

If 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 electronegative atomcreating a polar covalent bond (the bond has a permanent dipole)

For example, chlorine is more electronegative than hydrogen so the H-Cl bond ispolar

!+ !-

HCl

Polar molecules have permanent dipoles that dont cancel out (e.g. H2O) because the

dipoles 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

Three types of weak force hold simple covalent molecules together.

a) Van der Waals

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

induces dipole in another molecule.o The more electrons, the stronger the van der Waals forceso Occur in all simple covalent substances

b) Dipole-dipole

o Attraction between molecules with permanent dipoleso !+ ends attracted to !- ends

c) Hydrogen bonds

o Need H attached to N/O/F (highly electronegative elements).

o Exposed H nucleus is strongly attracted to lone pair on N/O/Fo Diagram must show lone pairs, dipoles and H-bond shown by dotted line e.g. water

d) Anomalous properties of water

Water has some unusual properties due to the presence of hydrogen bonding Ice is less dense than water because ice has an open structure caused by hydrogen

bonding

Water has a higher melting and boiling point than expected due to the strength ofhydrogen bonds that have to be broken


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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.

107in ammonia compared with the tetrahedral bond angle of 109.5in methane

Number of pairs Examples Name of shape Bond angle

3 bonding pairs(repel equally)

BF3 Trigonal planar 120

4 bonding pairs(repel equally)

CH4, NH4+ Tetrahedral 109.5

6 bonding pairs

(repel equally)

SF6 Octahedral 90

3 bonding, 1 lone(lone pair repels

more than bonds)

NH3 Pyramidal 107

2 bonding , 2 lone(lone pair repels

more than bonds)

H2O Non-linear 104.5

2 double bonds(repel equally)

CO2 Linear 180

OCR seems quite keen on SO2where S has 2 double bonds and 1 lone pair. Repulsion is

roughly equal for double bonds and lone pairs so bond angle is 120, shape non-linear


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Basic ConceptsRevision 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 molecule

e.g. C2H6O for ethanol

Empirical formula is the simplest whole number ratio of the atoms of each element

in a molecule e.g. CH2for ethene (from molecular formula C2H42)

General formula is the simplest algebraic formula for a member of a homologous

series e.g. CnH2n+2for alkanes

Structural formula is the minimum detail that shows the arrangement of the atoms

in a molecule e.g. CH3CH2OH for ethanol

Displayed formula shows the relative positioning of atoms and the bonds between

them e.g. for ethanol:

All bonds should be shown. Do notput OH for the alcohol group

Skeletal formula shows just the carbon skeleton and functional groups e.g. for

ethanol (see also Appendix 1)

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

The rules for naming organic compounds 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-

dimethylhexane5)

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


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3) Structural isomers

Structural isomershave the same molecular formula but different structural formulae

e.g. the molecular formula C4H19Br can produce four different structures

Differences between structural isomers arise from the position of the functional group

and/or the arrangement of the carbon chain e.g. C4H9Br has four isomers

1-bromobutane 2-bromobutane

2-bromo-2-methylpropane 1-bromo-2-methylpropane

4) Cyclic Alkanes

Carbon can form rings as well as chains Cyclic alkanes have general formula CnH2n(same as alkenes)

Cyclopentane is 5 CH2s in a ring with molecular formula C5H10. Skeletal formula is

a pentagon

Cyclohexane is 6 CH2s in a ring. Skeletal formula is a hexagon

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, theremay 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


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Example

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

percentage 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 mol

Percentage yield = 0.0163/0.020 x 100%= 82%

6) Atom economy

Atom economy = Molecular mass of desired products x 100%

Sum of molecular masses of all products

Example

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 A reaction may have a high percentage yield but a low atom economy (as in the

above example)


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6) More on naming organic compounds

This section covers naming compounds with more than one alkene or alcoholfunctional group

a) Dienes

Compounds containing two alkene groups are called dienes e.g.

The name of this compound is buta-1,3-diene

b) Diols and triols

Compounds containing two alcohol groups are called diols while three OH groupsmakes a triol e.g.

Ethane-1,2-diol propane-1,2,3-triol (glycerol)


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Appendix 1 - Skeletal Formulae

Skeletal formulae show carbon-carbon bonds and functional groups

Alkane e.g. hexane

Alkene e.g. hex-3-ene

Alcohol e.g. ethanol

Halogenoalkane e.g. 2-chloro-2-fluoropentane

Aldehyde e.g. butanal

Ketone e.g. propanone

Carboxylic acid e.g. 3-methylbutanoic acid

Ester e.g. methyl propanoate

Amine e.g. ethylamine

Benzene

or


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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

C-H bonds are non-polar (C and H have similar electronegativities) so:

o The only intermolecular forces in alkanes are Van der Waals forces

o Alkanes do not attract charged species (nucleophiles and electrophiles)

In alkanes, the bonds round each carbon are tetrahedral in shape. Each C has 4 bondpairs in its outer shell which repel each other and get as far apart as possible.

2) Boiling points

Boiling point increases with chain length more electrons, more Van der Waals

forces Boiling point decreases as branching increases branched alkanes have less surface

area in contact so intermolecular forces are weaker (or straighter chains can packcloser, more Van der Waals forces)

3) Reactions

a) Combustion

! Reaction with oxygen producing CO2and H2O! Exothermic (produces heat)

! Gas volume increases (used to drive pistons in engines)

! Alkanes are used as fuels in industry, in the home and in transport

b) Substitution

H replaced by Cl or Br e.g.

CH4+ Br2 CH3Br + HBr

Requires u/v light to break Br-Br bond

The bond breaking is homolytic fission because two radicals are produced, each

having an unpaired electron

A mechanism shows the detailed steps by which the reactants turn into theproducts

The mechanism here is called radical substitution

Initiation step e.g. Br2!2Br

Propagation e.g. CH4+ Br!CH3+ HBr } as a

CH3+ Br2!CH3Br + Br } pair

Termination step e.g. 2CH3!CH3CH3


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Further substitution can occur with more Hs being replaced by Brs

CH3Br + Br2 CH2Br2+ HBr etc

The reaction produces a mixture of products (CH3Br, CH2Br2, CHBr3and CBr4).

This means that radical substitution is not a good way of making a particular

product (the reaction has limited use in synthesis which means making adesired product in a number of steps)

FuelsRevision Notes

1) Fractional Distillation

Crude oil 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

Vaporising the crude oil

o Passing the vapour into a column that is hot at the bottom and cool at the

topo

The vapour rising and condensing at the appropriate levelo Separation is based on the different boiling points of alkanes

2) Processing of fractions

a) Cracking

Crude oil contains no alkenes and more long chain alkanes than are needed.

Cracking converts long chain alkanes into shorter chain alkanes and an alkene

e.g.C11H24 C8H18+ C3H6

Cracking requires heat and a catalyst. It is a thermal decomposition reaction.

The alkenes from cracking are used to make polymers and alcohols

b) Isomerisation

! Isomerisation turns straight chain alkanes into branched chain alkanes e.g.

Octane!2,2,4-trimethylpentane


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!

Branched chain hydrocarbons make better fuels as their lower boiling points

mean they burn more efficiently! Isomerisation increases the octane number of the hydrocarbon

! It needs a catalyst and heat

c) Reforming

Reforming turns straight chain alkanes into cyclic alkanes and arenes (and

hydrogen) e.g.

C6H14 C6H12+ H2Hexane cyclohexane

C6H14 C6H6+ 4H2Hexane benzene

Arenes contain a benzene ring (see A2 - Topic 1). Benzene can be represented inseveral ways. The third one will make more sense next year:

Arenes make better fuels than straight chain alkanes as they have higher octane

numbers and burn more efficiently

Reforming needs heat and a catalyst

3) Fossil fuels and biofuels

Fossil fuels (coal, gas and oil) are very useful to us as sources of energy and as afeedstock for making petrochemicals

However, they are non-renewable as they take millions of years to form andincreased CO2levels from burning fossil fuels are leading to global warming andclimate change

We will eventually need to replace fossil fuels with renewable energy sources

such as biofuels e.g. alcohol made from sugar cane and biodiesel made from

grain

Biofuels are renewable as plant material takes only a short time to grow.

4) Octane Number Octane number or Octane rating is a value used to indicate the

resistance of a motor fuel to knock. Octane numbers are based on ascale on which octane is 100 (minimal knock) and heptane is 0 (bad

knock). Examples:A fuel with an octane number of 92 has the same knock as

a mixtureof 92% octaneand 8% heptane.


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AlkenesRevision Notes

1) General

Alkenes are unsaturated hydrocarbons with general formula CnH2nCycloalkenes are unsaturated hydrocarbons with general formula CnH2n-2

Unsaturated = contain a double C=C bond

Hydrocarbon = contains C and H only

2) Bonding in Alkenes

o The carbons at either end of the double bond and the 4 atoms they are bonded to

are all in a plane (flat). These 6 atoms are joined by single bonds

o The bonds around each C in the C=C bond are trigonal planar in shape. There are 3areas of electrons round each C (1 double bond and 2 single bonds) which repel

equally to give a bond angle of 120

o

The double bond is formed by sideways overlap of p orbitals producing a !bond(sausage-shaped clouds of electrons above and below the plane of the single bondframework)

3) E/Z isomers

Isomers

_________"__________# #

Structural Stereoisomers

Isomers ______"_______# #

E/Z isomers Optical isomers (A2 only)

Stereoisomershave the same structural formula but with a different arrangement in

space

E/Z isomerismis one type of stereoisomerism arising from restricted rotation about adouble bond when two different groups are attached to each carbon of the C=Cgroup

E is short the German word entgegenwhich means opposite i.e. on opposite sides of

the double bond Z is short for the German word zusammenwhich means together i.e. on the same

side of the double bond

1-bromo-2-chloropropene displays E/Z isomerism. This is because one C of the C=Cbond is attached to an H and a Br, which are different, and the other C of the C=C is

attached to a Cl and a CH3which are also different


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Z-1-bromo-2-chloropropene E-1-bromo-2-chloropropene

For an example like this you only need to say that it will have E and Z isomers, not

which one is which But-2-ene is a simpler example where it is more straightforward to say which is E and

which is Z

Z-but-2-ene E-but-2-ene(CH3s are together) (CH3s on opposite sides of the double bond)

Cis/trans isomerism is a special case of E/Z isomerism in which two of the substituent

groups are the same. Cis corresponds to Z and trans corresponds to E

Sticking with but-2-ene as the example

Cis-but-2-ene trans-but-2-ene

4) Addition Reactions of Alkenes

In addition reactions, 2 molecules join to form 1 molecule.

Alkenes undergo addition reactions because they have a double bond.

a) Addition of hydrogen produces an alkane

e.g. CH2=CH2+ H2 CH3CH3ethane

Needs a Ni catalyst

One mole of hydrogen needed per double bond

This reaction is used to produce margarine from unsaturated vegetable oils bycatalytic hydrogenation

b) Addition of halogens produces a dihalogenoalkane

e.g. CH2=CH2+ Br2 CH2BrCH2Br1,2-dibromoethane

The colour change in this reaction is from orange to colourless

This is used as a test for unsaturation (to show the presence of a double bond)


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c) Addition of hydrogen halides produces a halogenoalkane

e.g. CH2=CH2+ HBr CH3CH2Brbromoethane

If this reaction is done with an unsymmetrical alkene, two isomeric products maybe formed e.g.

CH3CH=CH2+ HBr CH3CHBrCH32-bromopropane

and CH3CH=CH2+ HBr CH3CH2CH2Br1-bromopropane

d) Addition of steam produces an alcohol

e.g. CH2=CH2+ H2O CH3CH2OHethanol

Needs a strong acid catalyst e.g. phosphoric acid, H3PO4

Temperature must be above 100C so that H2O is in the form of steam

e) Mechanism for Electrophilic Addition

! The high electron density in the !bond makes alkenes attractive to electrophiles(electron pair acceptors, such as Br2and HBr)

! As the bromine molecule approaches the !bond, the electrons in the Br-Br bondare repelled. A dipole is induced in the bond with the Br nearer the alkene being!+

! The Br-Br bond undergoes heterolytic fission to produce a cation (positive ion)and an anion (negative ion), Both electrons from the bond go to the negative ion

Br2 Br++ :Br-

! The Br+is the electrophile that accepts an electron pair from the !bond toproduce an intermediate, CH2BrCH2

+! The intermediate is a carbocation (it has a carbon with a positive charge)

! In organic mechanisms, a curly arrow represents the movement of a pair ofelectrons

! Curly arrows should start from a bond or lone pair and finish at the atom where abond or lone pair is being formed


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5) Polymerisation of Alkenes

Alkenes can join together to form addition polymers

Monomer = small molecule that can be polymerised

Polymer = long chain molecule formed by joining many monomers together (many

means several thousand) The !bond breaks and forms single bonds that join the monomers together

a) ExamplesEquation 2 repeat units of polymer

Ethene poly(ethene)

Chloroethene poly(chloroethene)

b) Processing of waste polymers (e.g. plastic bottles)

Mechanical RecyclingWaste polymers can be melted down and re-used

following separation into types (PTFE etc). However it is expensive to collect andsort waste polymers

Combustion for energy production Waste polymers can be burnt as a fuel

but this may produce toxic waste products (see below)

Feedstock recycling Waste polymers can be cracked and the products used tomake fuels and other polymers (after separation)

d) Minimising environmental damage in polymer disposal

Chemists and chemical processes can minimise the environmental damage caused bydisposing of polymers

Combustion for energy production Burning halogenated plastics, such asPVC, produces toxic products like HCl. However, the HCl can be removed by gasscrubbers which by dissolve it in a spray of alkali

Developing new polymersAddition polymers are not biodegradable (theydont rot down). Chemists have developed biodegradable and compostable

polymers that will rot down e.g. from isoprene (methyl-1,3-butadiene), maizeand starch


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Enthalpy ChangesRevision Notes

1) Introduction

An enthalpy change is a change in heat energy measured at constant pressure. Enthalpy changes refer to the chemicals notthe surroundings.

The symbol for an enthalpy change is !H (!= change, H = heat energy)

The units for enthalpy change are kJ mol-1

2) Exothermic Reactions

In exothermic reactions, the chemicals lose energy so !H is negative. Thesurroundings gain energy so feel hotter.

In terms of a reaction with oxygen, oxidationis an exothermic process.

Examples include:

Combustion burning fuels for heating and in engines e.g.

CH4+ 2O2 CO2+ 2H2O

Respiration oxidation of carbohydrates in living things

C6H12O6+ 6O2 6CO2+ 6H2O

3) Endothermic Reactions

In endothermic reactions, the chemicals gain energy so !H is positive. Thesurroundings lose energy so feel cooler.

Endothermic reactions require an input of heat energy or they will stop.

Examples of endothermic processes include:

Thermal decomposition of calcium carbonate

CaCO3 CaO + CO2

Photosynthesis

6CO2+ 6H2O C6H12O6+ 6O2

4) Measuring Enthalpy Changes

o Measuring enthalpy changes by experiment is called calorimetryo Calorimetry works by using the energy released or absorbed in a reaction to

change the temperature of a known mass of watero Calculating an enthalpy change from experimental results involves a two step

process

o

Firstly: q = -mc

T/1000

Where m = mass of water in g

c= specific heat capacity of water (4.18 Jg-1K-1)

!T = change in temperature

o Secondly: H = q/n

Where n= number of moles of reactant (mass/molar mass or conc xvol/1000)

o Enthalpy changes measured from calorimetry are smaller than the expected

values because of heat loss to the apparatus and the environmento Other reasons for differences from standard values are non-standard conditions

and evaporation of water


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o In the case of measuring enthalpies of combustion, differences from standard

values can occur through incomplete combustion and evaporation of the fuel (if itis a liquid)

Example

The combustion of 0.15g of ethanol, C2H5OH, in a spirit burner increased the temperature of

75 cm3of water by 12.5C. Calculate the enthalpy of combustion of ethanol in kJ mol-1.

q = -mc!T/1000= -75 x 4.18 x 12.5/1000

= -3.919 kJ

n = 0.15/46

= 3.26 x 10-3mol

!H = -3.919/3.26 x 10-3= -1202 kJ mol-1

5) Enthalpy Changes

! The standard conditions for measuring enthalpy changes are a pressure of 100kPa and a temperature of 298K. Standard enthalpy changes are indicated by the

symbol

! Enthalpy change of reaction, HR, is the enthalpy change when the reactionoccurs in the molar quantities shown in the chemical equation

!

Enthalpy change of combustion, Hc, is the enthalpy change when one mole of asubstance is completely burnt under standard conditions e.g.

C5H12(l)+ 8O2(g) 5CO2(g) + 6H2O(l)

! Some substances cannot be burnt and have zero enthalpy of combustion e.g. O2,CO2, H2O

! Enthalpy change of formation, Hf, is the enthalpy change when one mole of asubstance is formed under standard conditions e.g.

5C(s) + 6H2(g) C5H12(l)

! !Hffor elements is zero! It is difficult to measure !Hffor hydrocarbons because carbon and hydrogen do

not react readily and other hydrocarbons will be formed in addition to the one of

interest


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7) Enthalpy Profile Diagrams

These diagrams show the difference in enthalpy between the reactants and products.In the space below, draw in an example of an exo and endothermic reaction.

8) Average Bond Enthalpy! This is the energy needed to break one mole of gaseous bonds

! Example equation:

H2(g) 2H(g)

! For a bond such as C-H that is found in many compounds, the value given in thedata book is an average over many compounds containing the bond

! Average bond enthalpies have a positive sign because energy is needed to breaka bond

9) Calculating Enthalpy Changes

Hesss Law = enthalpy change is independent of route

Enthalpy changes can be calculated in three ways, based on Hesss Law. The three ways can be used for to calculate any enthalpy change formation,

combustion, other types of reaction

The data provided determines which method to use

If the data is enthalpy changes of formation, use:

H = # Hf(products) - # Hf(reactants)

If the data is enthalpy changes of combustion, use:

H = # Hc(reactants) - # Hc(products)

If the data is bond enthalpies, use:

H =#

(bonds broken) -#

(bonds formed)

Using average bond enthalpies gives less accurate results than the other two

methods because bond enthalpies are not specific to the compounds involved inthe reaction (they are average values from many different compounds)

The reaction is exothermic if more energy is released when new bonds are

formed than is needed to break the old bonds

The reaction is endothermic if more energy is needed to break bonds in the

reactants than is released when bonds are formed in the products

Endo Exo


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Example data is enthalpy of formation

Calculate the enthalpy change for the following reaction.

Li2CO3(s) Li2O(s) + CO2(g)

Enthalpies of formation (kJ mol-1

) Li2CO3(s) -1216, Li2O(s) -596, CO2(g) -394

!H = #!Hf(products) - #!Hf(reactants)

!H = (-596 + (-394)) (-1216)= -990 + 1216

= 226 kJ mol-1

Example data is enthalpy of combustion


3C(s) + 4H2(g) C3H8(g)

Enthalpies of combustion (kJ mol-1) C(s) -394, H2(g) -286, C3H8(g) -2220

!H = #!Hc(reactants) - #!Hc(products)

!H = (3 x -394) + (4 x -286) (-2220)= -2326 + 2220= -106 kJ mol-1

Example data is bond energies


2HI(g) H2(g) + I2(g)

Bond enthalpies (kJ mol-1) H-I 299, H-H 436, I-I 151

!H = #(bonds broken) - #(bonds formed)

!H = (299 x 2) (436 + 151)= 598 - 587

= 11 kJ mol

-1


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FOR EXAMINERS USE

Qu. Max. Mark

1 21

2 12

3 22

4 20

TOTAL 75

ADVANCED SUBSIDIARY GCE

CHEMISTRY (SALTERS) 2850/01Chemistry for Life

*

O

C

E

T

7

6

0

7

6

*

INSTRUCTIONS TO CANDIDATES Write your name clearly in capital letters, your Centre Number and Candidate Number in the boxes above.

Use black ink. Pencil may be used for graphs and diagrams only.

Read each question carefully and make sure that you know what you have to do before starting your answer.

Answer allthe questions.

Do notwrite in the bar codes.

Write your answer to each question in the space provided, however additional paper may be used if

necessary.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ]at the end of each question

or part question.

The total number of marks for this paper is 75.

You may use a scientific calculator.

A copy of theData Sheet for Chemistry (Salters)is provided as an insert

with this question paper.

You are advised to show all the steps in any calculations.

This document consists of 16pages. Any blank pages are indicated.

* 2 8 5 0 0 1 *

OCR is an exempt Charity

Turn over OCR 2009 [R/100/3429]

SPA (NH/CG) T76076/6

Candidates answer on the question paperA calculator may be used for this paper

OCR Supplied Materials: Data Sheet for Chemistry (Salters)

(inserted)

Other Materials Required: Scientific calculator

Wednesday 3 June 2009

Morning

Duration:1 hour 15 minutes

THIS IS A LEGACY SPECIFICATION


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2

OCR 2009

Answer allthe questions.

1 To produce maximum power in a car engine, the fuel must ignite at the correct stage of the enginecycle. Auto-ignition causes a loss in power.

(a) What name is given to the number which is used as a measure of the tendency of a fuel toauto-ignite?

.............................................................................................................................................. [1]

(b) The branched hydrocarbon 2,2,4-trimethylpentane is an isomer of octane and has a lowtendency to auto-ignite.

(i) Draw the skeletalformulae for octane and 2,2,4-trimethylpentane.

octane

2,2,4-trimethylpentane

[2]

(ii) Explain why octane and 2,2,4-trimethylpentane are structural isomers.

...........................................................................................................................................

...................................................................................................................................... [1]

(iii) Draw the fullstructural formula of one otherisomer of octane.

Name the isomer you have drawn.

name .............................................................................[2]


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3

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(c) The process of reforming producescycloalkanes such as cyclooctane as the main products.Cyclooctane is notan isomer of octane.

(i) Give the molecular formula for octane and cyclooctane.

octane.. cyclooctane.. [2]

(ii) Reforming can also produce benzene. What type of hydrocarbon is benzene?

...................................................................................................................................... [1]

(iii) Reforming reactions take place in the presence of a heterogeneous catalyst.

Describe the fourmain stages by which heterogeneous catalysis works.

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...................................................................................................................................... [4]


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4

OCR 2009

(d) Molecules that contain an oxygen atom, such as alcohols, also have a low tendency toauto-ignite and can be added to petrol blends.

(i) What is the general name given to oxygen-containing molecules added to petrol blends?

...................................................................................................................................... [1]

(ii) In the box below, write a balanced equation to show the complete combustion ofmethanol, CH3OH.

[2]

(iii) The equation for the complete combustion of methane is shown below.

CH4

+ 2O2

CO2

+ 2H2

O

In the combustion chamber of a car engine, alcohols burn more completely than alkanesthat contain the same number of carbon atoms.

Use the equation above and your answer to (ii) to suggest why alcohols burn morecompletely in the combustion chamber.

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...................................................................................................................................... [2]

(iv) Name a product of incomplete combustion of alkanes and say why it is a polluting gas.

...........................................................................................................................................

...................................................................................................................................... [2]

(v) Another oxygen-containing compound, (CH3)3COCH

3, was used in petrol blends until

recently. Name the functional group in this compound.

...................................................................................................................................... [1]

[Total: 21]


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6

OCR 2009

2 Radioactive isotopes are used in a wide variety of industrial applications.

Radioactive isotopes decay to form new elements. There are three types of radioactive emission:alpha (), beta () and gamma ().

(a) Complete the following table to show the type of emission described.

descriptiontype of emission

(, or )

The atomic number of the decayingatom increases by one

Radiation of a very short wavelength

The particle emitted has a negativecharge

The particle emitted has a mass offour units

The particle emitted is stronglydeflected in an electric field

[5]

(b) The radioactive isotope of thallium (Tl), thallium-204, is used to monitor the thickness of thinfilms.

(i) Thallium-204 undergoes beta decay.

Write a nuclear equation for this process.

[3]

(ii) In the above application the beta particles pass through the film and cause a smallcurrent to flow in a detector.

Suggest why an alpha emitter would notbe suitable for monitoring the thickness of thinfilms.

...................................................................................................................................... [1]


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(c) Nuclear fusion reactions occur in the Sun.

Suggest why it is very difficult to reproduce fusion reactions here on Earth.

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

.............................................................................................................................................. [3]

[Total: 12]


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8

OCR 2009

3 Until recently, solder containing lead was used to join metals together. Chemists can use massspectrometry to distinguish between various types of solder by the ratio of the isotopes of leadthey contain.

(a) (i) Use the following terms to label the diagram of the mass spectrometer below.

electric field ionisation chamber magnetic field sample inlet

to a pump which maintains

low pressure

[4]


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(ii) The analysis of one sample of solder showed the existence of three isotopes of lead. Themass spectrum is shown below.

Calculate the relative atomic mass, Ar, for this sample of lead from the data on thespectrum.

Give your answer to foursignificant figures.

206

relative intensity

207

mass

208

2522

53

Ar= ................................................... [4]

(iii) Complete the following table, with the help of your Data Sheet, to show the number ofprotons, neutrons and electrons in the three isotopes.

isotope number of protons number of neutrons number of electrons

lead-206

lead-207

lead-208

[3]

(b) In the past, solder was an alloy of tin and lead. 100 g of one type of solder contained 0.500 molof tin. Calculate the mass of lead in 100 g of the solder.

Ar: Sn, 119

mass of lead = .............................................................. g [2]


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10

OCR 2009

(c) Solder can be used for electrical connections because both lead and tin are good conductorsof electricity. This is because of the nature of metallic bonding.

(i) Draw a labelled diagram below to illustrate metallic bonding.

[3]

(ii) Use your diagram to suggest how metals conduct electricity.

...........................................................................................................................................

...................................................................................................................................... [1]

(d) Tin and lead are in the same group of the Periodic Table.

For the metals in Group 2of the Periodic Table, explain the similarities and trends in the waythe elements react, in terms of their electronic configurations.

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

...................................................................................................................................................

.............................................................................................................................................. [5]

[Total: 22]


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OCR 2009

4 Known crude oil reserves are likely to be exhausted long before coal reserves. Methods of obtaininga variety of hydrocarbons from coal are increasing in importance.

(a) The first stage of one process involves the reaction of the carbon in coal with steam andoxygen. The products are carbon monoxide and hydrogen in the ratio 3:1 by moles.

Write a balanced equation for this reaction. Include state symbols.

[3]

(b) The bonding in carbon monoxide can be represented as shown.

C O

Name the bond represented by the arrow in this formula and explain how it arises.

...................................................................................................................................................

...................................................................................................................................................

.............................................................................................................................................. [2]


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OCR 2009

(c) The equation for the production of heptane from carbon monoxide and hydrogen is givenbelow.

(i) Complete the enthalpy cycle by filling in the box.

Use the following enthalpy changes of formation to calculate the enthalpy change for this

reaction.

compound Hf/kJ mol1

CO(g) 111

C7H16(g) 224

H2O(g) 242

7CO(g) + 15 H2(g) C

7H

16(g) + 7H

2O(g)

elements:

Enthalpy change = .............................................................. kJ mol-1 [5]

(ii) Explain why this reaction is accompanied by a decrease in entropy.

...........................................................................................................................................

...................................................................................................................................... [2]

Turn over


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14

OCR 2009

(d) Coal is used as a fuel in some power stations. Sulphur impurities present in coal react andproduce the toxic gas sulphur dioxide, SO2.

The sulphur dioxide can be removed by reacting it with calcium hydroxide in the presence ofoxygen, as shown below.

2SO2(g) + 2Ca(OH)2(s) + O2(g) 2CaSO4(s) + 2H2O(l)

Calculate the volume of sulphur dioxide (in dm3) at room temperature and pressure that couldbe removed by reaction with 1.0kg of calcium hydroxide in the presence of excess oxygen.

1 mol of any gas at room temperature and pressure occupies 24 dm3.

Ar: Ca,40; O, 16; H,1.0

volume = ...................................... dm3 [3]

(e) At the high temperatures generated when coal is burnt in a power station, nitrogen in the airreacts with oxygen to form nitrogen oxides. The high temperature is needed because of thelarge bond enthalpy of N2.

(i) Explain the term bond enthalpy.

...........................................................................................................................................

...................................................................................................................................... [2]


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OCR 2009

(ii) Describe the bond-breaking and bond-making processes that occur when nitrogen reactswith oxygen to form nitrogen oxides.

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...................................................................................................................................... [3]

[Total: 20]

END OF QUESTION PAPER


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2850

MarkScheme

June2009

21

2850

ChemistryforLife

Questio

n

ExpectedAnswers

M

arks

AdditionalGuidance

1

a

octane(number)(1);

1

allowoctanerating

ignorereferencestop

re-ignition/auto-ignition/knocking

b

i

(1);

(1);

2

allowanyskeletalfor1mark(nodots)

allowvariationsinang

leoflinesrepresentingbranch

esin2,2,4-

trimethylpentane

notanyfullstructural

ii

same(molecular)formula

ebutdifferent

structural(1);

1

allowsamenumberandtypeofatomsdifferentstruc

tureAW

notshapes/arrangementonown

iii

fullstructural(1);

correctnameforexample

chosen(1);

2

allbondsmustbesho

wnconnectingcorrect.

allowcorrectnameifformulaisskeletalornotfullstructural,not

othercompounds

c

i

octaneC8H18(1);cyclooctaneC8H16(1);

2

allowHandCatomsreversed

ii

arene/aromatic(hydrocarbon)(1);

1

allowunsaturated

iii

reactantsadsorbedonsu

rface(1);

bondsbreak(1);

newbondsform(1);

productsdiffuseaway/desorb/comeofffrom

surface(1);

4

allows

tickto/bondto/chemisorbonsurface

bondsbreakbetween

reactantsisacon,butallown

ewbonds

formingbetweenreactantsincontext

d

i

oxygenates(1);

1

Notoxidiser


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2850

MarkScheme

June2009

Questio

n

ExpectedAnswers

M

arks

AdditionalGuidance

ii

CH3OH(g)+1O2(g)CO2(g)+2H2O(g)

formulae(1);

balancing(1);

2

donotallowbalancin

gmarkifformulaeincorrect.

ignorestatesymbols

allowanymultiple

iii

samneedlessoxygen/1

to2(1);

forforsameamount/no

ofmoles/molecules(of

fuel)(1);forequivalenta

lkane

2

allowreverseargument

iv

carbonmonoxide(1);

toxic/poisonous/photochemicalsmog(1);(ignore

referencestogreenhousegas)

secondmarkdependson

first

2

allowC

O

allowdescriptionofna

tureofprobleme.g.irreversiblybondsto

haemoglobinAW

donotallowcarbono

runburnthydrocarbon

v

ether(1);

1

allowa

lkoxy(alkane)

Total

21

22


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2850

MarkScheme

June2009

Questio

n

ExpectedAnswers

M

arks

AdditionalGuidance

2

a

Typeof

radiation

(,or)

Theparticleemittedresultsinthe

atomicnumberincreasingbyone.

Radiationofaveryshor

t

wavelength

Theparticleemittedhas

anegative

charge

Theparticleemittedhas

amassof

fourunits.

Theparticleemittediss

trongly

deflectedinanelectricfield.

one

mark

each

5

allowwords(alphaetc)

b

i

Tl

204

81

+

/

Pb

204

82

e01

onemarkforeachcorrec

tspecies

3

donotallowe

-

numbersonrightcon

mark(once)symbolmustbeco

rrecte.g.notPB

(missingelectronscores2max)

wrongtypeofdecay/anyadditionalspeciesscoresze

ro

ii

cannotpenetratefilmorlayer/doesnotpass

throughtodetector/stop

pedbypaper-AW(

1);

1

notionising/notdeflec

ted

c

anythreef

rom:

havetoovercome(nuclear)repulsion(1);

very/extremelyhightemp

sneeded(1);

very/extremelyhighpressuresneeded(1);

difficult/specialistequipmentneededtocontain

fusionprocess(1);

highgravitationalforcesneeded(1);

3

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MarkScheme

June2009

Questio

n

Expected

Answers

Marks

AdditionalGuidance

3

a

i

oneeachcorrectlylabelled

;i.e.goinganticlockwise

fromtopright

electricfield(1);

magneticfield(1);

ionisationchamber(1);

sampleinlet(1);

4

ii

isotopemassxabundance

(1);

added(=20728)(1);

dividedby100(1);(207.28);

foursigfigs(1);

(207.3-scoresallfour)(1);ecfsonfirsttwomarking

points(207.2onown3)

4

ignoreanswerswithn

oworkingunlesscorrect

sfmarkisastandaloneprovidingitfollowsfrompre

vious

calculationandiscorrect.Lookfor207.2wrong

iii

Isotope

Number

of

protons

Number

of

neutrons

Number

of

electrons

lead-206

82

124

82

lead-207

82

125

82

lead-208

82

126

82

one

mark

for

each

row

3

b

massoftin=0.5x119(59.5)(1);

massoflead=100-59.5(40

.5scoresbothmarks)(1);

2

ecfonsecondmark

25


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June2009

Questio

n

Expected

Answers

Marks

AdditionalGuidance

c

i

delocalised/freeelectron

ure)

(1);

s

(mustbewithinstruct

esidue,NOTatomsor

y/AW(1);

cations

cations/positivemetallicr

nuclei(1);

regulararra

Lookingfortworowsof

minimumforregulararray

3

allow(mobile)seaofelectrons

allow

+/e

(-)orjust(-)

aslabels

allowregulararraymarkfromdiagram

allowm

ixtureofPb2+andSn

2+

ii

free/delocalisedelectrons

move/flow/carrythe

;

1

chargethroughstructure/

betweencationsAW(

1)

d

Reactivitydependsonouter(shell)electrons/

5

electronslostfromoutershell/outershellimplie

2/Samenumberofoutere

lectrons/formationof2+

d(1);

sfurtherfrom

ghtly/attractiontonucleusless

eac

tivity/getmorereactive

ion(resultinsimilarreactiv

ity)(1);

ongoingdowngroup,oute

relectron

nucleus/moreelectronsh

ells/more

screening/shielding(1);

electronsheldlessti

/lostmoreeasily(AW)(1);

causesanincreaseinr

downgroup(ORA)(1);

Total

22

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MarkScheme

June2009

Question

Expecte

dAnswers

Marks

AdditionalGuidance

4

a

3C(s)+H2O(g)+O2(g)

3CO(g)+H2(g)ss(1);

balancing(1);

formulae(1);

3

ssstandaloneforrecognisablespecies

neitheroflasttwoma

rkingpointsscoreifformulaeincorrect

b

dative(covalentbond)(1);

pairofelectronsbothcom

efromtheoxygen

atom(1);

2

allowco-ordinatebon

d

c

i

sumofproductsminussu

mofreactants(1);

correctnumbers(1);

correctanswerwithsign(1);ecfsasappropriate

5

allowcorrectbalancin

gifOandHinsteadofO2and

H2

donotallowmultiples/halvesforbalancingmark

firstmarkoflatterthre

emarksisprocessmark

{(7x-242)+(-224)}

(7x-111)}(777-1918)

-1141(-355scores1

nomultiples)(+1141(2)1141(1

))

ii

decreaseinmoles/molecules(ofgas)(1);

thereforelessdisordered(ORA)/fewerwaysof

arrangement/fewerways

ofarrangingenergy/less

random(1);

2

notparticlesoratomonfirstmark

d

formulamassofCa(OH)2

=74(1);

molesofCa(OH)2=1000/74(13.51)(1);

volume=13.51x24(320

/324/324.3/324.32)(1);

3

ecfspossibleafter74

correctanswerscores

3

e

i

bondenthalpyisenthalpy

changethatoccurswhen

1moleofbondbroken(1)

;ingaseousstate(1);

2

elements:

7C+3O2+15H2

formulaeofelements(1

);balancing(1);

27


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MarkScheme

June2009

Expecte

dAnswers

Marks

AdditionalGuidance

NandOObondsbreak(1);

ecifytypesofbondNN

tripleOOdouble(AW)(1);

Nbondsformed(1);

3

conintramolecular

tal

20

28

Basic Revision Notes

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