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35
1 How many protons, neutrons and electrons are in the following atoms and ions?(a) 115B; (b) 188O;(c) 3014Si; (d) 6430Zn;(e) 2311Na
+; (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) Cr2O72–;(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)
24.0V (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 met
hod in Worked example 1 to c
alculate the unknown concentr
ation below.
In a titration, 25.0 cm
3 of 0.125 mol dm–3 aqueous so
dium hydroxidereacted exactly
with 22.75 cm3 of hydrochloric
acid.
HCl(aq) NaOH(aq) NaCl(
aq) H2O(l)
Find the concentration of th
e hydrochloricacid.
2 Use the method in Worked
example 2 to calculate the mo
lar mass of theacid H2X.
A student dissolved 1.571 g of
an acid, H2X, inwater and mad
e the solution up to
250 cm3. She titrated 2
5.0 cm3 of this solution
against 0.125 mol dm–3 sodium
hydroxide, NaOH(aq). 21.30 cm
3 of NaOH(aq) were needed to
reach the end point.
The equationfor this reaction
is:
2NaOH(aq) H2X(aq) Na2X
(aq) 2H2O(l)
Module 1Atoms and reaction
s
Titrations
Notes
For (a), we use:
amount, n c V (in cm
3)________
1000
For (b), we use the balanced equat
ion to
work out the reacting quantities of
the
acid and alkali.
2 mol NaOH reacts with 1 mol H2S
O4
For (c), we rearrange: n c
V (in cm3)
________1000
Hence, c n 1000________
V
In a titration, 25.0 cm
3 of 0.150 mol dm–3 sodium hy
droxide NaOH(aq) reacted
exactly with 23.40 cm
3 of sulfuric acid,H2SO4(aq).
2NaOH(aq) H2SO4(aq)
Na2SO4(aq)2H2O(l)
(a) Calculate the amount, in m
ol, of NaOH that reacted.
n(NaOH) c
V_____1000
0.150 25.0_____1000
3.75 10–3 mol
(b) Calculate the amount, in m
ol, of H2SO4 that was used.
equation2NaOH(aq)
H2SO4(aq)
moles fromequation 2 m
ol 1mol
actual moles 3.75
10–3 mol 1.875 10
–3 mol
(c) Calculate the concentration
, in mol dm–3 of the sulfuric a
cid.
c(H2SO4) n 1000________
V
1.875 10–3 1000______________
_____23.40
8.0110–3 mol dm
–3
Worked example 1: Calcula
ting an unknown concentra
tion
A student dissolved 2.794 g of
an acid HX in water and made
the solution upto
250 cm3. The student ti
trated 25.0 cm3 of this solution
against 0.0614mol dm
–3
sodium carbonate Na2CO3(aq
). 23.45 cm3 of Na2CO3(aq)
were needed toreach
the end point.
The equation for this reaction i
s:
Na2CO3(aq)2HX(aq) 2
NaX(aq) CO2(g) H2O(l)
(a) Calculate the amount, in m
ol, of Na2CO3 that reacted.
n(Na2CO3) c
V_____1000
0.0614 23.45______1000
1.44 10–3 mol
(b) Calculate the amount, in m
ol, of HX that was used in the
titration.
equationNa2CO3 (aq)
2 HX(aq)
moles fromequation 1 m
ol 2mol
actual moles 1.4
4 10–3 mol 2.88 1
0–3 mol
(c) Calculate the amount, in m
ol, of HX that was used to mak
e up the 250 cm3
solution.
25.0 cm3 HX(aq) contain
s 2.88 10–3 mol
So, the 250cm3 solution co
ntains 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: Calcula
ting an unknown molar mas
s
Notes
For (a) and (b), the steps are the s
ame
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 exa
m.
For (c), we scale up by a factor of
10 to
find the amount, in mol, of HX in t
he
250 cm3 solution that wa
s 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 Titrations
By the end of this spread, you should be
able to . . .
✱ Perform acid–base titrations, and ca
rry out structured calculations.
Acid–base titrations
During volumetric analysis, you
measure the volume of one so
lution that reacts with a
measured volume of a second
, different solution.
An acid–base titration is a spe
cial type of volumetric analysis
, in which you react a
solution of an acid with a solut
ion of a base.
• You must know the concen
tration of one of the two soluti
ons. This is usually a
standard solution (see spread
1.1.7).
• In the analysis, you use this
standard solution to find out u
nknown information about
the substancedissolved in the
second solution.
The unknown information may
be:
• the concentration of the sol
ution
• a molar mass
• a formula
• the numberof molecules of
water of crystallisation.
You carry out atitration as follo
ws.
• Using a pipette, you add a m
easured volume of one solutio
n to a conical flask.
• The other solution is placed
in a burette.
• The solutionin the burette is
added to the solution in the c
onical flask until the reaction
has just been completed. This
is called the end point of the t
itration. The volume of the
solution addedfrom the burett
e is measured.
You now knowthe volume of o
ne solution that exactly reacts
with the volumeof the
other solution.
We identify theend point using
an indicator.
• The indicator must be a diff
erent colour inthe acidic solut
ion than in thebasic solution.
Table 1 lists thecolours of som
e common acid–base indicator
s. It also showsthe colour
at the end point. Notice that th
is end point colour is in betwe
en the coloursin the acidic
and basic solutions.
IndicatorColour in acid
Colour in baseEnd point colour
methyl orangered
yelloworange
bromothymol blueyellow
bluegreen
phenolphthaleincolourless
pinkpale pink*
* This assumes that the aqueous bas
e has been added from the burette t
o the aqueous acid. If acid is added
to
base, the titration is complete when
the solution goes colourless.
Table 1 Colours of some com
mon acid–baseindicators
Calculating unknowns from titration re
sults
Analysis of titration results follo
ws a set pattern, as shown in
the worked examples:
• the first twosteps are alway
s the same;
• the third stepmay be differen
t, depending onthe unknown th
at you need towork out.
In AS chemistry, any calculatio
ns that you carry out will be st
ructured similarly to the
examples below.
For A2, you may have to work
out these stepsyourself.
Figure 1 Thisacid–base titra
tion is using
methyl orangeas an indicator
. Methyl
orange is coloured red in acidi
c solutions
and yellow in basic solutions.
The end
point is the colour in between –
orange.
The solution inthe conical flas
k above
has reached the end point
935 chemistry.U1 M1.indd
28-29
10/3/08 11:27:26 am
4
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
Exclusively endorsed by OCR for GCE Chemistry A
AS
In Exclusive Partnership
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) 115B; (b) 188O;(c) 3014Si; (d) 6430Zn;(e) 2311Na
+; (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) Cr2O72–;(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)
24.0V (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
28
29
Questions1 Use the met
hod in Worked example 1 to c
alculate the unknown concentr
ation below.
In a titration, 25.0 cm
3 of 0.125 mol dm–3 aqueous so
dium hydroxidereacted exactly
with 22.75 cm3 of hydrochloric
acid.
HCl(aq) NaOH(aq) NaCl(
aq) H2O(l)
Find the concentration of th
e hydrochloricacid.
2 Use the method in Worked
example 2 to calculate the mo
lar mass of theacid H2X.
A student dissolved 1.571 g of
an acid, H2X, inwater and mad
e the solution up to
250 cm3. She titrated 2
5.0 cm3 of this solution
against 0.125 mol dm–3 sodium
hydroxide, NaOH(aq). 21.30 cm
3 of NaOH(aq) were needed to
reach the end point.
The equationfor this reaction
is:
2NaOH(aq) H2X(aq) Na2X
(aq) 2H2O(l)
Module 1Atoms and reaction
s
Titrations
Notes
For (a), we use:
amount, n c V (in cm
3)________
1000
For (b), we use the balanced equat
ion to
work out the reacting quantities of
the
acid and alkali.
2 mol NaOH reacts with 1 mol H2S
O4
For (c), we rearrange: n c
V (in cm3)
________1000
Hence, c n 1000________
V
In a titration, 25.0 cm
3 of 0.150 mol dm–3 sodium hy
droxide NaOH(aq) reacted
exactly with 23.40 cm
3 of sulfuric acid,H2SO4(aq).
2NaOH(aq) H2SO4(aq)
Na2SO4(aq)2H2O(l)
(a) Calculate the amount, in m
ol, of NaOH that reacted.
n(NaOH) c
V_____1000
0.150 25.0_____1000
3.75 10–3 mol
(b) Calculate the amount, in m
ol, of H2SO4 that was used.
equation2NaOH(aq)
H2SO4(aq)
moles fromequation 2 m
ol 1mol
actual moles 3.75
10–3 mol 1.875 10
–3 mol
(c) Calculate the concentration
, in mol dm–3 of the sulfuric a
cid.
c(H2SO4) n 1000________
V
1.875 10–3 1000______________
_____23.40
8.0110–3 mol dm
–3
Worked example 1: Calcula
ting an unknown concentra
tion
A student dissolved 2.794 g of
an acid HX in water and made
the solution upto
250 cm3. The student ti
trated 25.0 cm3 of this solution
against 0.0614mol dm
–3
sodium carbonate Na2CO3(aq
). 23.45 cm3 of Na2CO3(aq)
were needed toreach
the end point.
The equation for this reaction i
s:
Na2CO3(aq)2HX(aq) 2
NaX(aq) CO2(g) H2O(l)
(a) Calculate the amount, in m
ol, of Na2CO3 that reacted.
n(Na2CO3) c
V_____1000
0.0614 23.45______1000
1.44 10–3 mol
(b) Calculate the amount, in m
ol, of HX that was used in the
titration.
equationNa2CO3 (aq)
2 HX(aq)
moles fromequation 1 m
ol 2mol
actual moles 1.4
4 10–3 mol 2.88 1
0–3 mol
(c) Calculate the amount, in m
ol, of HX that was used to mak
e up the 250 cm3
solution.
25.0 cm3 HX(aq) contain
s 2.88 10–3 mol
So, the 250cm3 solution co
ntains 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: Calcula
ting an unknown molar mas
s
Notes
For (a) and (b), the steps are the s
ame
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 exa
m.
For (c), we scale up by a factor of
10 to
find the amount, in mol, of HX in t
he
250 cm3 solution that wa
s 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 Titrations
By the end of this spread, you should be
able to . . .
✱ Perform acid–base titrations, and ca
rry out structured calculations.
Acid–base titrations
During volumetric analysis, you
measure the volume of one so
lution that reacts with a
measured volume of a second
, different solution.
An acid–base titration is a spe
cial type of volumetric analysis
, in which you react a
solution of an acid with a solut
ion of a base.
• You must know the concen
tration of one of the two soluti
ons. This is usually a
standard solution (see spread
1.1.7).
• In the analysis, you use this
standard solution to find out u
nknown information about
the substancedissolved in the
second solution.
The unknown information may
be:
• the concentration of the sol
ution
• a molar mass
• a formula
• the numberof molecules of
water of crystallisation.
You carry out atitration as follo
ws.
• Using a pipette, you add a m
easured volume of one solutio
n to a conical flask.
• The other solution is placed
in a burette.
• The solutionin the burette is
added to the solution in the c
onical flask until the reaction
has just been completed. This
is called the end point of the t
itration. The volume of the
solution addedfrom the burett
e is measured.
You now knowthe volume of o
ne solution that exactly reacts
with the volumeof the
other solution.
We identify theend point using
an indicator.
• The indicator must be a diff
erent colour inthe acidic solut
ion than in thebasic solution.
Table 1 lists thecolours of som
e common acid–base indicator
s. It also showsthe colour
at the end point. Notice that th
is end point colour is in betwe
en the coloursin the acidic
and basic solutions.
IndicatorColour in acid
Colour in baseEnd point colour
methyl orangered
yelloworange
bromothymol blueyellow
bluegreen
phenolphthaleincolourless
pinkpale pink*
* This assumes that the aqueous bas
e has been added from the burette t
o the aqueous acid. If acid is added
to
base, the titration is complete when
the solution goes colourless.
Table 1 Colours of some com
mon acid–baseindicators
Calculating unknowns from titration re
sults
Analysis of titration results follo
ws a set pattern, as shown in
the worked examples:
• the first twosteps are alway
s the same;
• the third stepmay be differen
t, depending onthe unknown th
at you need towork out.
In AS chemistry, any calculatio
ns that you carry out will be st
ructured similarly to the
examples below.
For A2, you may have to work
out these stepsyourself.
Figure 1 Thisacid–base titra
tion is using
methyl orangeas an indicator
. Methyl
orange is coloured red in acidi
c solutions
and yellow in basic solutions.
The end
point is the colour in between –
orange.
The solution inthe conical flas
k above
has reached the end point
935 chemistry.U1 M1.indd
28-29
10/3/08 11:27:26 am
5
Sample pages from OCR AS Chemistry A Student Book.
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
A2
Exclusively endorsed by OCR for GCE Chemistry A
In Exclusive Partnership
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.
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7
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Links directly to the module and specification.
Student answer activities focus on improving subject knowledge.
Three stages allow students to see questions against student answers and examiner feedback.
Teacher Support CD-ROM
The AS and A2 Teacher Support CD-ROMs help you plan and deliver the new specification with confidence. Each provide you with:
l weekly teaching plans and guidance sheets to help save time
l customisable student practical sheets with accompanying teacher and technician notes
l a media bank of all the diagrams and learning objectives in the Student Book, all ready to use in PowerPoint format.
8
Sample lesson plan from OCR AS Teacher Support
CD-ROM.
AS
Exclusively en
dorsed by OCR
for GCE Biolo
gy A
01865 888118
In Exclusive
Partnership
ISBN 978 0 435
961 83 7
Teacher Resource C
D
Baldwin • Wood
Series Editor: Rob Ritchie
Sarah Baldwin
Christopher Wood
Series Editor: Rob R
itchie
9
Sample screen from OCR AS Chemistry A Teacher Support CD-ROM.
A2
Exclusively endorsed by OCR for GCE Biology A
01865 888118
In Exclusive Partnership
ISBN 978 0 435961 93 6
Teacher Resource CD
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 shar
ed pairs are repelled more
than in NH3, and the bond angl
e in H2O is less than in NH3. Th
e final bond angle is
104.5°. The final shape of the m
olecule is described as non-line
ar.Shapes of molecules –
electron-pair repulsion theory1
UNIT
Key words
electron-pair
repulsioncovalent bond
lone pair
Electron pairs repel each other s
o they are as far apart as possib
le. In a covalent
compound or ion, the number o
f electron pairs around the cent
ral atom determines the
shape of the molecule.
Mod
ule
2
Module 2
Shapes of molecules – electr
on-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-c
ross diagram of the
molecule (see diagram opposite
);
or count up the numbe
r of electrons at the
central atom.
STep 2 See what shape the ele
ctron pairs will adopt. They wil
l be arranged
tetrahedrally because there are
four pairs of electrons.
STep 3 Now see if there are an
y lone pairs. Yes, one lone pair.
STep 4 Electron-pair repulsion
theory tells you that the lone p
air will repel the
bonding pairs a little more stro
ngly than they repel each other
, and the
molecule should have the sam
e shape and bond angle as NH3.
H3O+ has a pyramid
al shape with a bond angle of
107°.
atomNumber of
electrons
Oxygen6
Three hydrogens 3
Positive charge –1
Total number8
× ×××
×
×OH H
H
1 Sketch the shapes and pred
ict the bond angles in the follow
ing molecules:
a PCl3 b
OCl2
c SeF6 (note that Se is in Grou
p 6) dCCl4
2 Sketch the shapes and pred
ict the bond angles in the follow
ing ions:
a PCl4+
bPCl6
–
c NH2–
dNH4
+
3 The dot-and-cross diagram
for SO2 is shown in the diagram
opposite.
a What is the shape of the mo
lecule?
b Give the approximate bond
angle.
QUICk CHeCk QUeSTIONS ?We know the different shapes of molecules by worki
ng out how the electron pairs
arrange themselves in space. Th
e rules of electron-pair repulsio
n theory tell us that:
electron pairs repel each other s
o they are as far apart as possib
le
an electron pair shared betwee
n two atoms is called a bondin
g pair (BP)
a non-bonding pair on one atom
only (not shared) is called a lon
e pair (LP)
LP–LP repulsion > LP–BP repuls
ion > BP–BP repulsion.
These rules explain the shapes
of the ammonia, NH3, and wate
r, H2O, molecules.
NH3The nitrogen at
om has four pairs of electrons,
so these will take up a basically
tetrahedral shape. But one elec
tron pair is a lone pair, so it repe
ls the shared pairs more
than a bonding pair would, and
the bond angle is decreased fro
m 109.5° to 107°. The
final shape of the molecule is de
scribed 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 Sha
peexample
2 electron pairslinear
CO2(in fact two dou
ble bonds)
3 electron pairstrigonal
BF3
4 electron pairstetrahedral
CH4NH+4
6 electron pairsoctahedral
SF6
××
××
××××
××××
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 trip
le) bonds, the bonding electron
pairs are all located
between the bonding atoms an
d therefore they cound as one b
onding pair of electrons
around the central atom for wo
rking out the shape.
CO2A dot-and-cross
diagram of CO2 shows that the
bonding is O=C=O (see table).
The
electron pairs in the double bon
ds will repel each other as much
as possible, so the final
shape of the molecule is linear.
What about a non-linear molec
ule 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
In Exclusive Partnership
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
Exclusively endorsed by OCR for GCE Chemistry A
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
Exclusively endorsed by OCR for GCE Chemistry A
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 rea
ctions 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 volumes14
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 3
4
End-of-module examination questions 3
6
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 6
0
12 Intermolecular forces62
13 Hydrogen bonding64
14 Metallic bonding and structure 6
6
15 Structure of ionic compounds 68
16 Structures of covalent compounds 70
Summary and practice questions 7
2
End-of-module examination questions 7
4
Module 3 The Periodic Table 76
1 The Periodic Table: searching for order 7
8
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 8
5
6 Group 2 elements: redoxreactions 88
7 Group 2 compounds: reactions 90
8 Group 7 elements: redoxreactions 92
9 Group 7 elements: uses and halide tests 94
Summary and practice questions 9
6
End-of-module examination questions 9
8
UNIT 2 Chains, energy and
resources (F322)
Module 1 Basic concepts and
hydrocarbons100
1 Introduction to organic chemistry 102
2 Naming hydrocarbons104
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 ofthe future 12
2
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 13
4
18 Polymer chemistry136
19 Polymers – dealing withour waste 138
20 Other uses of polymer waste 140
Summary and practice questions 14
2
End-of-module examination questions 14
4
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 16
8
12 Mass spectrometry170
13 Mass spectrometry in organic chemistry 17
2
14 Mass spectrometry: fragmentation patterns 174
15 Reaction mechanisms176
Summary and practice questions 17
8
End-of-module examination questions 18
0
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 fromHc
198
9 Enthalpy changes fromHf
200
10 Rates of reaction – collision theory 20
2
11 Catalysts204
12 Economic importance of catalysts 206
13 The Boltzmann distribution 20
8
14 Chemical equilibrium210
15 Equilibrium and industry 212
Summary and practice questions 21
4
End-of-module examination questions 21
6
Module 4 Resources218
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 pollution230
7 Green chemistry232
8 CO2 – villain and saviour
234
Summary and practice questions 23
6
End-of-module examination questions 23
8
Answers240
Glossary258
Periodic Table/Data Sheet262
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 shar
ed pairs are repelled more
than in NH3, and the bond angl
e in H2O is less than in NH3. Th
e final bond angle is
104.5°. The final shape of the m
olecule is described as non-line
ar.Shapes of molecules –
electron-pair repulsion theory1
UNIT
Key words
electron-pair
repulsioncovalent bond
lone pair
Electron pairs repel each other s
o they are as far apart as possib
le. In a covalent
compound or ion, the number o
f electron pairs around the cent
ral atom determines the
shape of the molecule.
Mod
ule
2
Module 2
Shapes of molecules – electr
on-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-c
ross diagram of the
molecule (see diagram opposite
);
or count up the numbe
r of electrons at the
central atom.
STep 2 See what shape the ele
ctron pairs will adopt. They wil
l be arranged
tetrahedrally because there are
four pairs of electrons.
STep 3 Now see if there are an
y lone pairs. Yes, one lone pair.
STep 4 Electron-pair repulsion
theory tells you that the lone p
air will repel the
bonding pairs a little more stro
ngly than they repel each other
, and the
molecule should have the sam
e shape and bond angle as NH3.
H3O+ has a pyramid
al shape with a bond angle of
107°.
atomNumber of
electrons
Oxygen6
Three hydrogens 3
Positive charge –1
Total number8
× ×××
×
×OH H
H
1 Sketch the shapes and pred
ict the bond angles in the follow
ing molecules:
a PCl3 b
OCl2
c SeF6 (note that Se is in Grou
p 6) dCCl4
2 Sketch the shapes and pred
ict the bond angles in the follow
ing ions:
a PCl4+
bPCl6
–
c NH2–
dNH4
+
3 The dot-and-cross diagram
for SO2 is shown in the diagram
opposite.
a What is the shape of the mo
lecule?
b Give the approximate bond
angle.
QUICk CHeCk QUeSTIONS ?We know the different shapes of molecules by worki
ng out how the electron pairs
arrange themselves in space. Th
e rules of electron-pair repulsio
n theory tell us that:
electron pairs repel each other s
o they are as far apart as possib
le
an electron pair shared betwee
n two atoms is called a bondin
g pair (BP)
a non-bonding pair on one atom
only (not shared) is called a lon
e pair (LP)
LP–LP repulsion > LP–BP repuls
ion > BP–BP repulsion.
These rules explain the shapes
of the ammonia, NH3, and wate
r, H2O, molecules.
NH3The nitrogen at
om has four pairs of electrons,
so these will take up a basically
tetrahedral shape. But one elec
tron pair is a lone pair, so it repe
ls the shared pairs more
than a bonding pair would, and
the bond angle is decreased fro
m 109.5° to 107°. The
final shape of the molecule is de
scribed 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 Sha
peexample
2 electron pairslinear
CO2(in fact two dou
ble bonds)
3 electron pairstrigonal
BF3
4 electron pairstetrahedral
CH4NH+4
6 electron pairsoctahedral
SF6
××
××
××××
××××
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 trip
le) bonds, the bonding electron
pairs are all located
between the bonding atoms an
d therefore they cound as one b
onding pair of electrons
around the central atom for wo
rking out the shape.
CO2A dot-and-cross
diagram of CO2 shows that the
bonding is O=C=O (see table).
The
electron pairs in the double bon
ds will repel each other as much
as possible, so the final
shape of the molecule is linear.
What about a non-linear molec
ule with double bonds? A good
example is sulfur dioxide
(see quick check question 3).
Quick check 2✔
Quick check 3✔
11
NewScientistIn partnership with Heinemann
Free, up-to-date news articles......and guidance on how NewScientist.com can suppport the delivery of the new science specifications.
Summary of contents
Samples pages from OCR AS Chemistry A Student Book.
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
A2
Exclusively endorsed by OCR for GCE Chemistry A
In Exclusive Partnership
Dave Gent and Rob RitchieSeries Editor: Rob Ritchie
Exclusively endorsed by OCR for GCE Chemistry A
AS
In Exclusive Partnership
In Exclusive Partnership
Dave Gent and Rob R
itchie
Series Editor: Rob Ri
tchie
Exclusively endo
rsed by OCR for
GCE Chemistry
A
AS
In Exclusive
PartnershipDave Gent and Rob RitchieSeries Editor: Rob Ritchie
A2
Exclusively endorsed by OCR for GCE Chemistry A
In Exclusive Partnership
Evaluation PacksEach OCR Chemistry A Evaluation Pack is available on 60 days free evaluation and contains:
l Student Book and Exam Café CD-ROMl Teacher Support CD-ROMl Revision Guide (New Edition)
Course componentsOCR AS Chemistry A Student Book and CD-ROM 978 0 435691813 | £17.99* | January 2008
OCR A2 Chemistry A Student Book and CD-ROM 978 0 435691981 | £17.99* | December 2008
OCR AS Chemistry A Teacher Support CD-ROM 978 0 435691837 | £149.00* (+VAT) | April 2008
OCR A2 Chemistry A Teacher Support CD-ROM 978 0 435691936 | £149.00* (+VAT) | December 2008
OCR AS Revise Chemistry A (New Edition) 978 0 435583712 | £6.99* (+VAT) | April 2008
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08ST
T002
8