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AS A2OCRChemistry A

Course Guide

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AS Revision Guide

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This revision guide is tailored to the OCR specification and exclusively endorsed by OCR for GCE Chemistry A. It is written by experienced examiners and teachers, giving you:

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Series Editor

Rob Ritchie

35

1 How many protons, neutrons and electrons are in the following atoms and ions?(a) 11

5B; (b) 188O;

(c) 3014Si; (d) 64

30Zn;(e) 23

11Na+; (f) 7935Br–;

(g) 4521Sc3+.

2 Calculate the relative atomic mass, Ar, to 4 significant figures, for the following.(a) Lithium containing: 7.42%6Li and 92.58%7Li.(b) Lead containing: 204Pb, 1.1%; 206Pb, 24.9%;

207Pb, 21.7%; 208Pb, 52.3%.

3 Use Ar values from the Periodic Table to calculate the relative formula mass of the following:(a) Cr2O3; (b) Rb2S;(c) Zn(OH)2; (d) (NH4)2CO3;(e) Fe3(PO4)2.

4 Calculate the mass, in g, in:(a) 6 mol MnO2; (b) 0.25 mol Ga2O3;(c) 0.60 mol Ag2SO4.

5 Calculate the amount, in mol, in:(a) 8.823 g CaI2; (b) 48.55 g K2CrO4;(c) 7.50 g Cu(NO3)2.

6 Determine the empirical formula of the compound formed when 1.472 g of tungsten reacts with 0.384 g of oxygen.

7 Determine the molecular formula for the compound of carbon, hydrogen and oxygen atoms with the composition by mass: C, 40.0%; H, 6.7%; O, 53.3%; Mr 90.

8 (a) At RTP, what amount, in mol, of gas molecules are in: (i) 84 dm3; (ii) 300 cm3; (iii) 10 dm3?(b) At RTP, what is the volume, in dm3, of: (i) 12 mol CO2(g); (ii) 0.175 mol N2(g); (iii) 2.55 g NO(g)?

9 Find the amount, in mol, of solute dissolved in the following solutions:(a) 2 dm3 of a 0.225 mol dm–3 solution;(b) 25 cm3 of a 0.175 mol dm–3 solution.

10 Find the concentration, in mol dm–3, for the following solutions:(a) 0.75 moles dissolved in 5 dm3 of solution;(b) 0.450 moles dissolved in 100 cm3 of solution.

11 Balance each of the following equations:(a) P4(s) + O2(g) P2O5(s)(b) C5H12(l) + O2(aq) CO2(g) + H2O(l)(c) NaOH(aq) + H3PO4(aq) Na2HPO4(aq) + H2O(l).

12 Define the terms:(a) acid; (b) base; (c) alkali.

13 Write balanced equations for the following acid reactions:(a) nitric acid and iron(III) hydroxide.(b) sulfuric acid and copper(II) carbonate.(c) hydrochloric acid and aluminium.

14 What is the oxidation state of each species in the following?(a) Cu; (b) I2; (c) CH4;(d) MgSO4; (e) PbCO3; (f) Cr2O7

2–;(g) (NH4)3PO4.

15 From the experimental results below, work out the formula of the hydrated salt.Mass of ZnSO4·xH2O = 8.985 gMass of ZnSO4 = 5.047 g

16 Lithium carbonate decomposes with heat: Li2CO3(s) Li2O(s) + CO2(g) What volume of CO2, measured at RTP, is formed by the decomposition of 5.7195 g of Li2CO3?

17 Calcium carbonate reacts with nitric acid: Na2CO3(s) + 2HNO3 2NaNO3(aq) + H2O(l) + CO2(g) 0.371 g of Na2CO3 reacts with an excess of HNO3.The final volume of the solution is 25.0 cm3.(a) What volume of CO2, measured at RTP, is formed?(b) What is the concentration, in mol dm–3, of NaNO3formed?

18 In the redox reactions below, use oxidation numbers to find out what has been oxidised and what has been reduced.(a) N2 + 3H2 2NH3(b) 3Mg + 2Fe(NO3)3 3Mg(NO3)2 + 2Fe(c) MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

Practice questions

34

Salt CO2 H20

Salt H2

Salt H20

Metal

Carbonate Base

Acidproton donor

Reduction• Gain of electrons• Decrease in oxidation

number

Oxidation• Loss of electrons• Increase in

oxidation number

Redox

Base

Alkalisoluble base

OH–

Moles

Redox

Acids

n = mass, mmolar mass, M

Mass

Solutions

n = c V (in dm3) V (in cm3)1 000

= c

Gas volumes

n = V (in dm3)24.0

V (in cm3)24 000

=

Moles

1.1 Atoms and reactions summaryModule 1

Atoms and reactionsPractice questions

935 chemistry.U1 M1.indd 34-35

10/3/08 11:28:15 am

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28

29

Questions1 Use the method in Worked example 1 to calculate the unknown concentration below.

In a titration, 25.0 cm3 of 0.125 mol dm–3 aqueous sodium hydroxide reacted exactly

with 22.75 cm3 of hydrochloric acid.

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

Find the concentration of the hydrochloric acid.

2 Use the method in Worked example 2 to calculate the molar mass of the acid H2X.

A student dissolved 1.571 g of an acid, H2X, in water and made the solution up to

250 cm3. She titrated 25.0 cm3 of this solution against 0.125 mol dm–3 sodium

hydroxide, NaOH(aq). 21.30 cm3 of NaOH(aq) were needed to reach the end point.

The equation for this reaction is:

2NaOH(aq) H2X(aq) Na2X(aq) 2H2O(l)

Module 1Atoms and reactions

Titrations

Notes

For (a), we use:

amount, n c V (in cm3)________

1000

For (b), we use the balanced equation to

work out the reacting quantities of the

acid and alkali.

2 mol NaOH reacts with 1 mol H2SO4

For (c), we rearrange: n c V (in cm3)________

1000

Hence, c n 1000________V

In a titration, 25.0 cm3 of 0.150 mol dm–3 sodium hydroxide NaOH(aq) reacted

exactly with 23.40 cm3 of sulfuric acid, H2SO4(aq).

2NaOH(aq) H2SO4(aq) Na2SO4(aq) 2H2O(l)

(a) Calculate the amount, in mol, of NaOH that reacted.

n(NaOH) c V_____

1000 0.150 2

5.0_____1000

3.75 10–3 mol

(b) Calculate the amount, in mol, of H2SO4 that was used.

equation 2NaOH(aq) H2SO4(aq)

moles from equation 2 mol 1 mol

actual moles 3.75 10–3 mol 1.875 10–3 mol

(c) Calculate the concentration, in mol dm–3 of the sulfuric acid.

c(H2SO4) n1000________V

1.875 10–3 1000___________________

23.40 8.01 10–3 mol dm–3

Worked example 1: Calculating an unknown concentration

A student dissolved 2.794 g of an acid HX in water and made the solution up to

250 cm3. The student titrated 25.0 cm3 of this solution against 0.0614 mol dm–3

sodium carbonate Na2CO3(aq). 23.45 cm3 of Na2CO3(aq) were needed to reach

the end point.


Na2CO3(aq) 2HX(aq) 2NaX(aq) CO2(g) H2O(l)

(a) Calculate the amount, in mol, of Na2CO3 that reacted.

n(Na2CO3) c V_____

1000 0.0614 2

3.45______1000

1.44 10–3 mol

(b) Calculate the amount, in mol, of HX that was used in the titration.

equation Na2CO3 (aq) 2 HX(aq)



(c) Calculate the amount, in mol, of HX that was used to make up the 250 cm3

solution.

25.0 cm3 HX(aq) contains 2.88 10–3 mol

So, the 250 cm3 solution contains 10 2.88 10–3 2.88 10–2 mol

(d) Calculate the concentration, in g mol–1, of the acid HX.

n m__M

so, M m__n

2.794__________

2.88 10–2 97.0 g mol–1

Worked example 2: Calculating an unknown molar mass

Notes

For (a) and (b), the steps are the same

as in the first worked example.

In this worked example, however, there

are two further steps. You would be

helped through these in an AS exam.

For (c), we scale up by a factor of 10 to

find the amount, in mol, of HX in the

250 cm3 solution that was made up.

For (d), we rearrange:

amount, n mass, m

____________ molar mass, M

Hence, M m__n

In this example,

the mass of HX is 2.794 g.

1.1 13 TitrationsBy the end of this spread, you should be able to . . .

✱ Perform acid–base titrations, and carry out structured calculations.

Acid–base titrationsDuring volumetric analysis, you measure the volume of one solution that reacts with a

measured volume of a second, different solution.

An acid–base titration is a special type of volumetric analysis, in which you react a

solution of an acid with a solution of a base.

• You must know the concentration of one of the two solutions. This is usually a

standard solution (see spread 1.1.7).

• In the analysis, you use this standard solution to find out unknown information about

the substance dissolved in the second solution.

The unknown information may be:

• the concentration of the solution

• a molar mass

• a formula

• the number of molecules of water of crystallisation.

You carry out a titration as follows.

• Using a pipette, you add a measured volume of one solution to a conical flask.

• The other solution is placed in a burette.

• The solution in the burette is added to the solution in the conical flask until the reaction

has just been completed. This is called the end point of the titration. The volume of the

solution added from the burette is measured.

You now know the volume of one solution that exactly reacts with the volume of the

other solution.

We identify the end point using an indicator.

• The indicator must be a different colour in the acidic solution than in the basic solution.

Table 1 lists the colours of some common acid–base indicators. It also shows the colour

at the end point. Notice that this end point colour is in between the colours in the acidic

and basic solutions.

Indicator Colour in acid Colour in base End point colour

methyl orange redyellow

orange

bromothymol blue yellowblue

green

phenolphthalein colourless pinkpale pink*

* This assumes that the aqueous base has been added from the burette to the aqueous acid. If acid is added to

base, the titration is complete when the solution goes colourless.

Table 1 Colours of some common acid–base indicators

Calculating unknowns from titration results

Analysis of titration results follows a set pattern, as shown in the worked examples:

• the first two steps are always the same;

• the third step may be different, depending on the unknown that you need to work out.

In AS chemistry, any calculations that you carry out will be structured similarly to the

examples below.

For A2, you may have to work out these steps yourself.

Figure 1 This acid–base titration is using

methyl orange as an indicator. Methyl

orange is coloured red in acidic solutions

and yellow in basic solutions. The end

point is the colour in between – orange.

The solution in the conical flask above

has reached the end point


10/3/08 11:27:26 am

4



AS


We listen to teachers’ needs...

Text is structured in line with the new OCR specification by Unit and Module.

Sample pages from OCR AS Chemistry A Student Book.

Learning objectives are taken from the specification to highlight what students need to know and understand.

Questions students should be able to answer after studying each spread.

Worked examples show students how calculations should be set out.

35

1 How many protons, neutrons and electrons are in the following atoms and ions?(a) 11

5B; (b) 188O;

(c) 3014Si; (d) 64

30Zn;(e) 23

11Na+; (f) 7935Br–;

(g) 4521Sc3+.

2 Calculate the relative atomic mass, Ar, to 4 significant figures, for the following.(a) Lithium containing: 7.42%6Li and 92.58%7Li.(b) Lead containing: 204Pb, 1.1%; 206Pb, 24.9%;

207Pb, 21.7%; 208Pb, 52.3%.

3 Use Ar values from the Periodic Table to calculate the relative formula mass of the following:(a) Cr2O3; (b) Rb2S;(c) Zn(OH)2; (d) (NH4)2CO3;(e) Fe3(PO4)2.

4 Calculate the mass, in g, in:(a) 6 mol MnO2; (b) 0.25 mol Ga2O3;(c) 0.60 mol Ag2SO4.

5 Calculate the amount, in mol, in:(a) 8.823 g CaI2; (b) 48.55 g K2CrO4;(c) 7.50 g Cu(NO3)2.

6 Determine the empirical formula of the compound formed when 1.472 g of tungsten reacts with 0.384 g of oxygen.

7 Determine the molecular formula for the compound of carbon, hydrogen and oxygen atoms with the composition by mass: C, 40.0%; H, 6.7%; O, 53.3%; Mr 90.

8 (a) At RTP, what amount, in mol, of gas molecules are in: (i) 84 dm3; (ii) 300 cm3; (iii) 10 dm3?(b) At RTP, what is the volume, in dm3, of: (i) 12 mol CO2(g); (ii) 0.175 mol N2(g); (iii) 2.55 g NO(g)?

9 Find the amount, in mol, of solute dissolved in the following solutions:(a) 2 dm3 of a 0.225 mol dm–3 solution;(b) 25 cm3 of a 0.175 mol dm–3 solution.

10 Find the concentration, in mol dm–3, for the following solutions:(a) 0.75 moles dissolved in 5 dm3 of solution;(b) 0.450 moles dissolved in 100 cm3 of solution.

11 Balance each of the following equations:(a) P4(s) + O2(g) P2O5(s)(b) C5H12(l) + O2(aq) CO2(g) + H2O(l)(c) NaOH(aq) + H3PO4(aq) Na2HPO4(aq) + H2O(l).

12 Define the terms:(a) acid; (b) base; (c) alkali.

13 Write balanced equations for the following acid reactions:(a) nitric acid and iron(III) hydroxide.(b) sulfuric acid and copper(II) carbonate.(c) hydrochloric acid and aluminium.

14 What is the oxidation state of each species in the following?(a) Cu; (b) I2; (c) CH4;(d) MgSO4; (e) PbCO3; (f) Cr2O7

2–;(g) (NH4)3PO4.

15 From the experimental results below, work out the formula of the hydrated salt.Mass of ZnSO4·xH2O = 8.985 gMass of ZnSO4 = 5.047 g

16 Lithium carbonate decomposes with heat: Li2CO3(s) Li2O(s) + CO2(g) What volume of CO2, measured at RTP, is formed by the decomposition of 5.7195 g of Li2CO3?

17 Calcium carbonate reacts with nitric acid: Na2CO3(s) + 2HNO3 2NaNO3(aq) + H2O(l) + CO2(g) 0.371 g of Na2CO3 reacts with an excess of HNO3.The final volume of the solution is 25.0 cm3.(a) What volume of CO2, measured at RTP, is formed?(b) What is the concentration, in mol dm–3, of NaNO3formed?

18 In the redox reactions below, use oxidation numbers to find out what has been oxidised and what has been reduced.(a) N2 + 3H2 2NH3(b) 3Mg + 2Fe(NO3)3 3Mg(NO3)2 + 2Fe(c) MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

Practice questions

34

Salt CO2 H20

Salt H2

Salt H20

Metal

Carbonate Base

Acidproton donor

Reduction• Gain of electrons• Decrease in oxidation

number

Oxidation• Loss of electrons• Increase in

oxidation number

Redox

Base

Alkalisoluble base

OH–

Moles

Redox

Acids

n = mass, mmolar mass, M

Mass

Solutions

n = c V (in dm3) V (in cm3)1 000

= c

Gas volumes

n = V (in dm3)24.0

V (in cm3)24 000

=

Moles

1.1 Atoms and reactions summaryModule 1

Atoms and reactionsPractice questions


10/3/08 11:28:15 am

28

29

Questions1 Use the method in Worked example 1 to calculate the unknown concentration below.

In a titration, 25.0 cm3 of 0.125 mol dm–3 aqueous sodium hydroxide reacted exactly

with 22.75 cm3 of hydrochloric acid.

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

Find the concentration of the hydrochloric acid.

2 Use the method in Worked example 2 to calculate the molar mass of the acid H2X.

A student dissolved 1.571 g of an acid, H2X, in water and made the solution up to

250 cm3. She titrated 25.0 cm3 of this solution against 0.125 mol dm–3 sodium

hydroxide, NaOH(aq). 21.30 cm3 of NaOH(aq) were needed to reach the end point.


2NaOH(aq) H2X(aq) Na2X(aq) 2H2O(l)

Module 1Atoms and reactions

Titrations

Notes

For (a), we use:

amount, n c V (in cm3)________

1000

For (b), we use the balanced equation to

work out the reacting quantities of the

acid and alkali.

2 mol NaOH reacts with 1 mol H2SO4

For (c), we rearrange: n c V (in cm3)________

1000

Hence, c n 1000________V

In a titration, 25.0 cm3 of 0.150 mol dm–3 sodium hydroxide NaOH(aq) reacted

exactly with 23.40 cm3 of sulfuric acid, H2SO4(aq).

2NaOH(aq) H2SO4(aq) Na2SO4(aq) 2H2O(l)

(a) Calculate the amount, in mol, of NaOH that reacted.

n(NaOH) c V_____

1000 0.150 2

5.0_____1000

3.75 10–3 mol

(b) Calculate the amount, in mol, of H2SO4 that was used.

equation 2NaOH(aq) H2SO4(aq)



(c) Calculate the concentration, in mol dm–3 of the sulfuric acid.

c(H2SO4) n1000________V

1.875 10–3 1000___________________

23.40 8.01 10–3 mol dm–3

Worked example 1: Calculating an unknown concentration

A student dissolved 2.794 g of an acid HX in water and made the solution up to

250 cm3. The student titrated 25.0 cm3 of this solution against 0.0614 mol dm–3

sodium carbonate Na2CO3(aq). 23.45 cm3 of Na2CO3(aq) were needed to reach

the end point.


Na2CO3(aq) 2HX(aq) 2NaX(aq) CO2(g) H2O(l)

(a) Calculate the amount, in mol, of Na2CO3 that reacted.

n(Na2CO3) c V_____

1000 0.0614 2

3.45______1000

1.44 10–3 mol

(b) Calculate the amount, in mol, of HX that was used in the titration.

equation Na2CO3 (aq) 2 HX(aq)



(c) Calculate the amount, in mol, of HX that was used to make up the 250 cm3

solution.

25.0 cm3 HX(aq) contains 2.88 10–3 mol

So, the 250 cm3 solution contains 10 2.88 10–3 2.88 10–2 mol

(d) Calculate the concentration, in g mol–1, of the acid HX.

n m__M

so, M m__n

2.794__________

2.88 10–2 97.0 g mol–1

Worked example 2: Calculating an unknown molar mass

Notes

For (a) and (b), the steps are the same

as in the first worked example.

In this worked example, however, there

are two further steps. You would be

helped through these in an AS exam.

For (c), we scale up by a factor of 10 to

find the amount, in mol, of HX in the

250 cm3 solution that was made up.

For (d), we rearrange:

amount, n mass, m

____________ molar mass, M

Hence, M m__n

In this example,

the mass of HX is 2.794 g.

1.1 13 TitrationsBy the end of this spread, you should be able to . . .

✱ Perform acid–base titrations, and carry out structured calculations.

Acid–base titrationsDuring volumetric analysis, you measure the volume of one solution that reacts with a

measured volume of a second, different solution.

An acid–base titration is a special type of volumetric analysis, in which you react a

solution of an acid with a solution of a base.

• You must know the concentration of one of the two solutions. This is usually a

standard solution (see spread 1.1.7).

• In the analysis, you use this standard solution to find out unknown information about

the substance dissolved in the second solution.

The unknown information may be:

• the concentration of the solution

• a molar mass

• a formula

• the number of molecules of water of crystallisation.

You carry out a titration as follows.

• Using a pipette, you add a measured volume of one solution to a conical flask.

• The other solution is placed in a burette.

• The solution in the burette is added to the solution in the conical flask until the reaction

has just been completed. This is called the end point of the titration. The volume of the

solution added from the burette is measured.

You now know the volume of one solution that exactly reacts with the volume of the

other solution.

We identify the end point using an indicator.

• The indicator must be a different colour in the acidic solution than in the basic solution.

Table 1 lists the colours of some common acid–base indicators. It also shows the colour

at the end point. Notice that this end point colour is in between the colours in the acidic

and basic solutions.

Indicator Colour in acid Colour in base End point colour

methyl orange redyellow

orange

bromothymol blue yellowblue

green

phenolphthalein colourless pinkpale pink*

* This assumes that the aqueous base has been added from the burette to the aqueous acid. If acid is added to

base, the titration is complete when the solution goes colourless.

Table 1 Colours of some common acid–base indicators

Calculating unknowns from titration results

Analysis of titration results follows a set pattern, as shown in the worked examples:

• the first two steps are always the same;

• the third step may be different, depending on the unknown that you need to work out.

In AS chemistry, any calculations that you carry out will be structured similarly to the

examples below.

For A2, you may have to work out these steps yourself.

Figure 1 This acid–base titration is using

methyl orange as an indicator. Methyl

orange is coloured red in acidic solutions

and yellow in basic solutions. The end

point is the colour in between – orange.

The solution in the conical flask above

has reached the end point


10/3/08 11:27:26 am

5

Sample pages from OCR AS Chemistry A Student Book.


A2



Worked examples show students how calculations should be set out.

Don’t forget

our A2 resources

coming in the

Autumn term!

End-of-module summary pages help students link together all the topics within each module.

Practice exam questions provided at the end of each module. Answers are in the back of the book.

In our unique Exam Café, students will find plenty of support to help them prepare for their exams. They can Relax and prepare with handy revision advice, Refresh their memories by testing their understanding and Get That Result through practicing exam-style questions, accompanied by lots of hints and tips.

An Exam Café CD-ROM is included FREE in the back of every Student Book.

6

Sample screen from OCR AS Chemistry A Exam Café CD-ROM.

Sample questions and answers to each module with tips on how to improve, examiner tips and advice to students on practical skills.

Study and revision skills to support students making the transition from GCSE to A Level.

Questions to test understanding. Also vocabulary tests, a glossary, and revision flashcards.

Links to how students can use New Scientist to reinforce their learning.

What do students think about Exam Café?

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Three stages allow students to see questions against student answers and examiner feedback.

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The AS and A2 Teacher Support CD-ROMs help you plan and deliver the new specification with confidence. Each provide you with:

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Sample lesson plan from OCR AS Teacher Support

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01865 888118 In Exclusive Partnership

ISBN 978 0 435961 83 7

Teacher Resource CD

Baldwin • Wood

Series Editor: Rob Ritchie

Sarah Baldwin

Christopher Wood


9

Sample screen from OCR AS Chemistry A Teacher Support CD-ROM.

A2


01865 888118


ISBN 978 0 435961 93 6

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Baldwin • WoodSeries Editor: Rob Ritchie

Sarah BaldwinChristopher Wood Series Editor: Rob Ritchie

Split into 30 weekly plans.

Easy-to-use content menu and search box to help you locate your required content.

24

25

H2O

The oxygen atom has four pairs of electrons, so these will take up a basically tetrahedral

shape. But two of the electron pairs are lone pairs, so the shared pairs are repelled more

than in NH3, and the bond angle in H2O is less than in NH3. The final bond angle is

104.5°. The final shape of the molecule is described as non-linear.Shapes of molecules –

electron-pair repulsion theory

1UNIT

Key words

electron-pair

repulsioncovalent bond

lone pair

Electron pairs repel each other so they are as far apart as possible. In a covalent

compound or ion, the number of electron pairs around the central atom determines the

shape of the molecule.

Mod

ule

2

Module 2

Shapes of molecules – electron-pair repulsion theory

Quick check 1✔

WOrked exaMple

To work out the shape of an ion, for example H3O+:

STep 1 There are two ways to approach this

problem.

either draw a dot-and-cross diagram of the

molecule (see diagram opposite);

or count up the number of electrons at the

central atom.

STep 2 See what shape the electron pairs will adopt. They will be arranged

tetrahedrally because there are four pairs of electrons.

STep 3 Now see if there are any lone pairs. Yes, one lone pair.

STep 4 Electron-pair repulsion theory tells you that the lone pair will repel the

bonding pairs a little more strongly than they repel each other, and the

molecule should have the same shape and bond angle as NH3.

H3O+ has a pyramidal shape with a bond angle of 107°.

atomNumber of

electrons

Oxygen 6

Three hydrogens 3

Positive charge –1

Total number 8

× ×××

×

×OH H

H

1 Sketch the shapes and predict the bond angles in the following molecules:

a PCl3

b OCl2

c SeF6 (note that Se is in Group 6) d CCl4

2 Sketch the shapes and predict the bond angles in the following ions:

a PCl4+

b PCl6–

c NH2–

d NH4+

3 The dot-and-cross diagram for SO2 is shown in the diagram opposite.

a What is the shape of the molecule?

b Give the approximate bond angle.

QUICk CHeCk QUeSTIONS ?We know the different shapes of molecules by working out how the electron pairs

arrange themselves in space. The rules of electron-pair repulsion theory tell us that:

electron pairs repel each other so they are as far apart as possible

an electron pair shared between two atoms is called a bonding pair (BP)

a non-bonding pair on one atom only (not shared) is called a lone pair (LP)

LP–LP repulsion > LP–BP repulsion > BP–BP repulsion.

These rules explain the shapes of the ammonia, NH3, and water, H2O, molecules.

NH3

The nitrogen atom has four pairs of electrons, so these will take up a basically

tetrahedral shape. But one electron pair is a lone pair, so it repels the shared pairs more

than a bonding pair would, and the bond angle is decreased from 109.5° to 107°. The

final shape of the molecule is described as pyramidal.

Examiner tip

Don’t confuse the

shape of the molecule

with the shape the

electron pairs adopt.

×× ×

××

H

HH

H

HN

N

H

107

Bond angle 107

×× ×

××

×H

H

HH

O

O

104.5

Bond angle 104.5

Number of pairs Shapeexample

2 electron pairs linearCO2

(in fact two double bonds)

3 electron pairs trigonalBF3

4 electron pairs tetrahedralCH4

NH+4

6 electron pairs octahedralSF6

××

××

××××

××××

O OC

O OC

180

Bond angle 180

×× ××

×× ×

××× ×

×××

F

F ×× ××

× ××F

F

F FB

B

120

Bond angle 120

×××

×H

H

H

HH

H

C

H HC

Bond angle 109.5

109.5

FF

F

F

F F

Bond angle 90

What about multiple bonds?

If a molecule has double (or triple) bonds, the bonding electron pairs are all located

between the bonding atoms and therefore they cound as one bonding pair of electrons

around the central atom for working out the shape.

CO2

A dot-and-cross diagram of CO2 shows that the bonding is O=C=O (see table). The

electron pairs in the double bonds will repel each other as much as possible, so the final

shape of the molecule is linear.

What about a non-linear molecule with double bonds? A good example is sulfur dioxide

(see quick check question 3).

Quick check 2✔

Quick check 3✔

Revision Guides

l Clearly written and well designed to aid revision.

l Written by experienced examiners and tailored to the new specification.

l Packed with examiner tips.

l Targeted at ensuring understanding with quick-check questions on each topic and end of unit exam-style questions.

10

Sample pages from OCR AS Chemistry Revision Guide.

Mike Wooster and Helen EcclesSeries editor: Rob Ritchie


This revision guide is tailored to the OCR specification and exclusively

endorsed by OCR for GCE Chemistry A. It is written by experienced

examiners and teachers, giving you: complete coverage of the specification for the exams

content organised by module and unit to follow the structure of the

specification and exams bite-sized chunks of information to make it easier to organise your

revision time quick-check revision questions so that you can test your own

knowledge easily hints and tips from examiners to help you avoid common errors

lots of practice exam-style questions for each unit all the answers to questions so that you can check that you’re on the

right track.Titles in this series:OCR AS Chemistry student book and exam café CD-ROM 978 0 435691 81 3

OCR A2 Chemistry student book and exam café CD-ROM 978 0 435691 98 1

OCR AS Chemistry Teacher Support CD-ROM978 0 435691 83 7

OCR A2 Chemistry Teacher Support CD-ROM978 0 435691 93 6

OCR AS Chemistry revision guide

978 0 435583 71 2

OCR A2 Chemistry revision guide

978 0 435583 74 3


AS AS

Second Edition

01865 888118

www.heinemann.co.uk

In Exclusive PartnershipI S B N 978-0-435583-71-2

9 7 8 0 4 3 5 5 8 3 7 1 2


Clearly linked to the specification.

Provides students with lots of opportunities to see problems worked through and the answers are provided step by step.

Hints and tips help students avoid common errors.

Enables students to check their knowledge and understanding. Answers are provided at the back of the book.

iv

v

IntroductionviUNIT 1 Atoms, bonds and groups (F321)

Module 1 Atoms and reactions 21 The changing atom42 Atomic structure63 Atomic masses84 Amount of substance and the mole 105 Types of formula

126 Moles and gas volumes 147 Moles and solutions168 Chemical equations189 Moles and reactions2010 Acids and bases2211 Salts2412 Water of crystallisation2613 Titrations2814 Oxidation number3015 Redox reactions32Summary and practice questions 34End-of-module examination questions 36Module 2 Electrons, bonding and structure

381 Evidence for shells402 Shells and orbitals423 Sub-shells and energy levels 444 Electrons and the Periodic Table 465 An introduction to chemical bonding 486 Ionic bonding

507 Ions and the Periodic Table 528 Covalent bonding549 Further covalent bonding 5610 Shapes of molecules and ions 5811 Electronegativity and polarity 6012 Intermolecular forces

6213 Hydrogen bonding6414 Metallic bonding and structure 6615 Structure of ionic compounds 6816 Structures of covalent compounds 70Summary and practice questions 72End-of-module examination questions 74

Module 3 The Periodic Table 761 The Periodic Table: searching for order 782 The Periodic Table: Mendeleev and beyond 803 The modern Periodic Table 824 Periodicity: ionisation energies and atomicradii845 Periodicity: boiling points 856 Group 2 elements: redox reactions 887 Group 2 compounds: reactions 908 Group 7 elements: redox reactions 929 Group 7 elements: uses and halide tests 94Summary and practice questions 96End-of-module examination questions 98

UNIT 2 Chains, energy and resources (F322)Module 1 Basic concepts and hydrocarbons

1001 Introduction to organic chemistry 1022 Naming hydrocarbons 1043 Naming compounds with functionalgroups1064 Formulae of organic compounds 1085 Structural and skeletal formulae 1106 Skeletal formulae and functional groups 1127 Isomerism

1148 Organic reagents and their reactions 1159 Hydrocarbons from crude oil 11810 Hydrocarbons as fuels 12011 Fossil fuels and fuels of the future 12212 Substitution reactions of alkanes 12413 Alkenes12614 Reactions of alkenes12815 Further addition reactions of alkenes 13016 Alkenes and bromine13217 Industrial importance of alkenes 13418 Polymer chemistry13619 Polymers – dealing with our waste 13820 Other uses of polymer waste 140Summary and practice questions 142End-of-module examination questions 144

Module 2 Alcohols, halogenoalkanes and analysis1461 Making and using alcohol 1482 Properties of alcohols1503 Combustion and oxidation of alcohols 1524 Esterification and dehydration of alcohols 1545 Introduction to halogenoalkanes 1566 Reactions of halogenoalkanes 1587 Halogenoalkanes and the environment 1608 Percentage yield

1629 Atom economy16410 Infrared spectroscopy16611 Infrared spectroscopy: functional groups 16812 Mass spectrometry17013 Mass spectrometry in organic chemistry 17214 Mass spectrometry: fragmentation patterns 17415 Reaction mechanisms 176Summary and practice questions 178End-of-module examination questions 180

Module 3 Energy1821 Enthalpy1842 Exothermic and endothermic reactions 1863 Enthalpy profile diagrams 1884 Standard enthalpy changes 1905 Determination of enthalpy changes 1926 Enthalpy change of combustion 1947 Bond enthalpies

196

ContentsContents

8 Enthalpy changes from Hc 1989 Enthalpy changes from Hf 20010 Rates of reaction – collision theory 20211 Catalysts20412 Economic importance of catalysts 20613 The Boltzmann distribution 20814 Chemical equilibrium

21015 Equilibrium and industry 212Summary and practice questions 214End-of-module examination questions 216Module 4 Resources

2181 The greenhouse effect – global warming 2202 Climate change2223 Solutions to the greenhouse effect 2244 The ozone layer2265 Ozone depletion2286 Controlling air pollution 2307 Green chemistry2328 CO2 – villain and saviour 234Summary and practice questions 236End-of-module examination questions 238

Answers240Glossary258Periodic Table/Data Sheet 262Index264

935 chemistry.prelims.indd 4-5

10/3/08 11:23:32 am

iv

v

Introductionvi

UNIT 1 Atoms, bonds and

groups (F321)Module 1 Atoms and reactions 2

1 The changing atom4

2 Atomic structure6

3 Atomic masses8

4 Amount of substance and the mole 10

5 Types of formula12

6 Moles and gas volumes 14

7 Moles and solutions16

8 Chemical equations18

9 Moles and reactions20

10 Acids and bases22

11 Salts24

12 Water of crystallisation26

13 Titrations28

14 Oxidation number30

15 Redox reactions32

Summary and practice questions 34

End-of-module examination questions 36

Module 2 Electrons, bonding and

structure38

1 Evidence for shells40

2 Shells and orbitals42

3 Sub-shells and energy levels 44

4 Electrons and the Periodic Table 46

5 An introduction to chemical bonding 48

6 Ionic bonding50

7 Ions and the Periodic Table 52

8 Covalent bonding54

9 Further covalent bonding 56

10 Shapes of molecules and ions 58

11 Electronegativity and polarity 60

12 Intermolecular forces62

13 Hydrogen bonding64

14 Metallic bonding and structure 66

15 Structure of ionic compounds 68

16 Structures of covalent compounds 70



Module 3 The Periodic Table 76

1 The Periodic Table: searching for order 78

2 The Periodic Table: Mendeleev and beyond 80

3 The modern Periodic Table 82

4 Periodicity: ionisation energies and atomic

radii84

5 Periodicity: boiling points 85

6 Group 2 elements: redox reactions 88

7 Group 2 compounds: reactions 90

8 Group 7 elements: redox reactions 92

9 Group 7 elements: uses and halide tests 94



UNIT 2 Chains, energy and

resources (F322)

Module 1 Basic concepts and

hydrocarbons100

1 Introduction to organic chemistry 102

2 Naming hydrocarbons 104

3 Naming compounds with functional

groups106

4 Formulae of organic compounds 108

5 Structural and skeletal formulae 110

6 Skeletal formulae and functional groups 112

7 Isomerism114

8 Organic reagents and their reactions 115

9 Hydrocarbons from crude oil 118

10 Hydrocarbons as fuels120

11 Fossil fuels and fuels of the future 122

12 Substitution reactions of alkanes 124

13 Alkenes126

14 Reactions of alkenes128

15 Further addition reactions of alkenes 130

16 Alkenes and bromine132

17 Industrial importance of alkenes 134

18 Polymer chemistry136

19 Polymers – dealing with our waste 138

20 Other uses of polymer waste 140



Module 2 Alcohols, halogenoalkanes

and analysis146

1 Making and using alcohol 148

2 Properties of alcohols150

3 Combustion and oxidation of alcohols 152

4 Esterification and dehydration of alcohols 154

5 Introduction to halogenoalkanes 156

6 Reactions of halogenoalkanes 158

7 Halogenoalkanes and the environment 160

8 Percentage yield162

9 Atom economy164

10 Infrared spectroscopy166

11 Infrared spectroscopy: functional groups 168

12 Mass spectrometry170

13 Mass spectrometry in organic chemistry 172

14 Mass spectrometry: fragmentation patterns 174

15 Reaction mechanisms 176



Module 3 Energy182

1 Enthalpy184

2 Exothermic and endothermic reactions 186

3 Enthalpy profile diagrams 188

4 Standard enthalpy changes 190

5 Determination of enthalpy changes 192

6 Enthalpy change of combustion 194

7 Bond enthalpies196

Contents

Contents

8 Enthalpy changes from Hc198

9 Enthalpy changes from Hf200

10 Rates of reaction – collision theory 202

11 Catalysts204

12 Economic importance of catalysts 206

13 The Boltzmann distribution 208

14 Chemical equilibrium210

15 Equilibrium and industry 212



Module 4 Resources 218

1 The greenhouse effect – global warming 220

2 Climate change222

3 Solutions to the greenhouse effect 224

4 The ozone layer226

5 Ozone depletion228

6 Controlling air pollution 230

7 Green chemistry232

8 CO2 – villain and saviour 234



Answers240

Glossary258

Periodic Table/Data Sheet 262

Index264

935 chemistry.prelims.indd 4-5

10/3/08 11:23:32 am

24

25

H2O

The oxygen atom has four pairs of electrons, so these will take up a basically tetrahedral

shape. But two of the electron pairs are lone pairs, so the shared pairs are repelled more

than in NH3, and the bond angle in H2O is less than in NH3. The final bond angle is

104.5°. The final shape of the molecule is described as non-linear.Shapes of molecules –

electron-pair repulsion theory

1UNIT

Key words

electron-pair

repulsioncovalent bond

lone pair

Electron pairs repel each other so they are as far apart as possible. In a covalent

compound or ion, the number of electron pairs around the central atom determines the

shape of the molecule.

Mod

ule

2

Module 2

Shapes of molecules – electron-pair repulsion theory

Quick check 1✔

WOrked exaMple

To work out the shape of an ion, for example H3O+:

STep 1 There are two ways to approach this

problem.

either draw a dot-and-cross diagram of the

molecule (see diagram opposite);

or count up the number of electrons at the

central atom.

STep 2 See what shape the electron pairs will adopt. They will be arranged

tetrahedrally because there are four pairs of electrons.

STep 3 Now see if there are any lone pairs. Yes, one lone pair.

STep 4 Electron-pair repulsion theory tells you that the lone pair will repel the

bonding pairs a little more strongly than they repel each other, and the

molecule should have the same shape and bond angle as NH3.

H3O+ has a pyramidal shape with a bond angle of 107°.

atomNumber of

electrons

Oxygen 6

Three hydrogens 3

Positive charge –1

Total number 8

× ×××

×

×OH H

H

1 Sketch the shapes and predict the bond angles in the following molecules:

a PCl3

b OCl2

c SeF6 (note that Se is in Group 6) d CCl4

2 Sketch the shapes and predict the bond angles in the following ions:

a PCl4+

b PCl6–

c NH2–

d NH4+

3 The dot-and-cross diagram for SO2 is shown in the diagram opposite.

a What is the shape of the molecule?

b Give the approximate bond angle.

QUICk CHeCk QUeSTIONS ?We know the different shapes of molecules by working out how the electron pairs

arrange themselves in space. The rules of electron-pair repulsion theory tell us that:

electron pairs repel each other so they are as far apart as possible

an electron pair shared between two atoms is called a bonding pair (BP)

a non-bonding pair on one atom only (not shared) is called a lone pair (LP)

LP–LP repulsion > LP–BP repulsion > BP–BP repulsion.

These rules explain the shapes of the ammonia, NH3, and water, H2O, molecules.

NH3

The nitrogen atom has four pairs of electrons, so these will take up a basically

tetrahedral shape. But one electron pair is a lone pair, so it repels the shared pairs more

than a bonding pair would, and the bond angle is decreased from 109.5° to 107°. The

final shape of the molecule is described as pyramidal.

Examiner tip

Don’t confuse the

shape of the molecule

with the shape the

electron pairs adopt.

×× ×

××

H

HH

H

HN

N

H

107

Bond angle 107

×× ×

××

×H

H

HH

O

O

104.5

Bond angle 104.5

Number of pairs Shapeexample

2 electron pairs linearCO2

(in fact two double bonds)

3 electron pairs trigonalBF3

4 electron pairs tetrahedralCH4

NH+4

6 electron pairs octahedralSF6

××

××

××××

××××

O OC

O OC

180

Bond angle 180

×× ××

×× ×

××× ×

×××

F

F ×× ××

× ××F

F

F FB

B

120

Bond angle 120

×××

×H

H

H

HH

H

C

H HC

Bond angle 109.5

109.5

FF

F

F

F F

Bond angle 90

What about multiple bonds?

If a molecule has double (or triple) bonds, the bonding electron pairs are all located

between the bonding atoms and therefore they cound as one bonding pair of electrons

around the central atom for working out the shape.

CO2

A dot-and-cross diagram of CO2 shows that the bonding is O=C=O (see table). The

electron pairs in the double bonds will repel each other as much as possible, so the final

shape of the molecule is linear.

What about a non-linear molecule with double bonds? A good example is sulfur dioxide

(see quick check question 3).

Quick check 2✔

Quick check 3✔

11

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Summary of contents

Samples pages from OCR AS Chemistry A Student Book.


A2





AS



Dave Gent and Rob Ritchie



AS

In Exclusive PartnershipDave Gent and Rob RitchieSeries Editor: Rob Ritchie

A2



Evaluation PacksEach OCR Chemistry A Evaluation Pack is available on 60 days free evaluation and contains:

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