Q & A STPM 2022 Chemistry

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FEATURES TITLES IN THIS SERIES: Chemistry
• Based on the latest syllabus ■ Pengajian Am Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
• Arranged according to subtopics ■ Bahasa Melayu
• Tips to guide students in answering ■ Sejarah
questions effectively ■ Geografi
• Common Errors to show common ■ Ekonomi
mistakes frequently made by students ■ Pengajian Perniagaan STPM
while answering questions ■ Mathematics (T)
• Model Paper which follows the latest ■ Physics
STPM assessment format to prepare ■ Chemistry
students for the actual exam ■ Biology . .
Pengajian Am Ekonomi Physics Semester 1 . 2 . 3 Semester 1 2 3
Bahasa Melayu Pengajian Perniagaan Chemistry
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Chemistry
Geograf

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ISBN: 978-967-2898-87-0


PELANGI Lee Chwee Neo • Peter Yip




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CONTENTS
CONTENTS

STPM Scheme of Assessment ii

TERM 1
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CHAPTER Atoms, Molecules and Stoichiometry 1
1 1.1 Fundamental Particles of an Atom 1 2
1.2 Relative Atomic, Isotopic, Molecular and Formula Masses
1.3 The Mole and the Avogadro Constant 11


CHAPTER Electronic Structure of Atoms 21
2 2.1 Electronic Energy Level of Atomic Hydrogen 21
29
2.2 Atomic Orbitals: s, p and d
2.3 Electronic Configuration 31
2.4 Classification of Elements in the Periodic Table 39


CHAPTER Chemical Bonding 42
3 3.1 Ionic Bonding 42
45
3.2 Covalent Bonding
3.3 Metallic Bonding 58
3.4 Intermolecular Forces 62


CHAPTER States of Matter 70
4 4.1 Gases 70
4.2 Liquids
88
4.3 Solids 94
4.4 Phase Diagrams 99


CHAPTER Reaction Kinetics 105
5 5.1 Rate of Reaction 105
109
5.2 Rate Law
5.3 The Effect of Temperature on Reaction Kinetics 112
5.4 The Role of Catalysts in Reactions 118
5.5 Order of Reactions and Rate Constants 121


CHAPTER Equilibria 134
6 6.1 Chemical Equilibria 134
6.2 Ionic Equilibria
143
6.3 Solubility Equilibria 156
6.4 Phase Equilibria 165

STPM Model Paper 962/1 178


v




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TERM 2

CHAPTER Chemical Energetics 183
7 7.1 Enthalpy Changes of Reaction, ΔH 183
188
7.2 Hess’ s Law
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7.3 Born-Haber Cycle 195
7.4 The Solubility of Solids in Liquids 202

CHAPTER Electrochemistry 206
8 8.1 Half-cell and Redox Equations 206
208
8.2 Standard Electrode Potential
8.3 Non-standard Cell Potentials 218
8.4 Fuel Cells 221
8.5 Electrolysis 222
8.6 Applications of Electrochemistry 233


CHAPTER Periodic Table: Periodicity 236
9 9.1 Physical Properties of Elements of Period 2 and Period 3 236
9.2 Reactions of Period 3 Elements with Oxygen and Water
240
9.3 Acidic and Basic Properties of Oxides and Hydrolysis of Oxides 243


CHAPTER Group 2 252
10 10.1 Selected Group 2 Elements and Their Compounds 252
10.2 Anomalous Behaviour of Beryllium
262
10.3 Uses of Group 2 Compounds 269


CHAPTER Group 14 271
11 11.1 Physical Properties of Group 14 Elements 271
274
11.2 Tetrachlorides and Oxides of Group 14 Elements
11.3 Relative Stability of +2 and +4 Oxidation States of Group 14
Elements 282
11.4 Silicon, Silicone and Silicate 287
11.5 Tin Alloys 293


CHAPTER Group 17 295
12 12.1 Physical Properties of Selected Group 17 Elements 295
12.2 Reactions of Selected Group 17 Elements
298
12.3 Reactions of Selected Halide Ions 305
12.4 Industrial Applications of Halogens and Their Compounds 309




vi




0iv Con STPM Q&A Chem T1.indd 6 4/19/19 5:14 PM

CHAPTER Transition Elements 313
13 13.1 Physical Properties of First Row Transition Elements 313
315
13.2 Chemical Properties of First Row Transition Elements
13.3 Nomenclature and Bonding of Complexes 329
13.4 Uses of First Row Transition Elements and Their Compounds 332
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STPM Model Paper 962/2 335


TERM 3

CHAPTER Introduction to Organic Chemistry 342
14 14.1 Bonding of the carbon atoms: The Shapes of Ethane, Ethene, 342
Ethyne, and Benzene Molecules
14.2 General, Empirical, Molecular and Structural Formulae of Organic
Compounds 343
14.3 Functional Groups: Classification and Nomenclature 346
14.4 Isomerism: Structural and Stereoisomerism 347
14.5 Free Radicals, Nucleophiles and Electrophiles 350
14.6 Molecular Structure and its Effect on Physical Properties 353
14.7 Inductive and Resonance Effect 355


CHAPTER Hydrocarbons 360
15 15.1 Alkanes 360
15.2 Alkenes
365
15.3 Arenes 380

CHAPTER Haloalkanes 391
16 16.1 Physical Properties of Haloalkanes 391
16.2 Nucleophilic Subtitution of Haloalkanes
391
16.3 Elimination Reaction 406
16.4 Mechanism of Nucleophilic Substitution 410
16.5 Reactivity of Primary, Secondary and Tertiary Haloalkanes 413
16.6 Reactivity of Chlorobenzene and Chloroalkanes in Hydrolysis
Reactions 413
16.7 Organometallic Compounds 417
16.8 Uses of Haloalkanes 419
16.9 Effects of CFC on the Environment 419



CHAPTER Hydroxy Compounds 421
17 17.1 Introduction to Hydroxy Compounds 421
422
17.2 Alcohols
17.3 Phenols 442



vii




0iv Con STPM Q&A Chem T1.indd 7 4/19/19 5:14 PM

CHAPTER Carbonyl Compounds 451
18 18.1 Introduction to Carbonyl Compounds 451
18.2 Physical Properties of Carbonyl Compounds
451
18.3 Reactions of Carbonyl Compounds 452
18.4 Carbohydrates 471
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CHAPTER Carboxylic Acids and their Derivatives 472
19 19.1 Carboxylic Acid 472
481
19.2 Acyl Chlorides
19.3 Esters 488
19.4 Amides 493



CHAPTER Amines, Amino Acids and Proteins 499
20 20.1 Amines 499
20.2 Amino Acids
510
20.3 Protein 522


CHAPTER Polymers 523
21 21.1 Synthetic Polymers 523
524
21.2 Condensation Polymerisation
21.3 Addition Polymerisation 527
21.4 Classification of Polymers 532


STPM Model Paper 962/3 533


Answers 539






















viii




0iv Con STPM Q&A Chem T1.indd 8 4/19/19 5:14 PM

Chapter

1 Atoms, Molecules and

Stoichiometry



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1.1 Fundamental Particles of an Atom

Section A Multiple-choice Questions
Question 1 Term

A beam of protons, electrons and neutrons is passed between two electrically
charged plates. 1
+ + + + + +
X
Proton,
electron, Y
neutron Z
– – – – – –
Which of the following statements is not true?
A X has a lower mass than Z.
B Y is electrically neutral.
C X and Z have opposite charges.
D Y is larger in size than Z.

Answer: D
Based on the deflection, X is an electron (negatively charged), Y is a neutron
and Z is a proton (positively charged).
The size of a neutron and proton are similar.

Question 2

An atom X forms an ion with a charge of –2. Which of the following statements
is not correct if this ion has 18 electrons and 19 neutrons?
A X is an isotope of sulphur.
B X is found in Group 16 in the periodic table.
C X has a nucleon number of 35.
D X has high melting and boiling points.


Answer: D
Exam Tips
Exam Tips
Determine the number of subatomic particles in an atom of X.


1




01 STPM Q&A Chem T1 layout.indd 1 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Atom X has (18 – 2) = 16 electrons and hence, 16 protons. X is an element of
Group 16, which is sulphur. The nucleon number of X is (16 + 19) = 35. Since
X is a non-metal, it has low melting and boiling points.


Common Errors
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2–
For a charge of X , students wrongly minus two electrons from X instead of
adding two electrons. Hence, the proton number of X should be 16 and not 20.

1.2 Relative Atomic, Isotopic, Molecular and Formula Masses
Term
Section A Multiple-choice Questions
1
Question 1

When 2.40 g of metal X is added to an aqueous solution containing Y ions,
2+
3+
15.53 g metal Y is precipitated. In this reaction, X ions are produced. What is
the relative atomic mass of X if the relative atomic mass of Y is 207?
A 27.0
B 48.0
C 55.6
D 63.5


Answer: B
2X + 3Y : 3Y + 2X 3+
2+
Mass
15.53
The number of moles of Y = ————————– = ——– = 0.075
Relative atomic mass 207
2
The number of moles of X = — × 0.075 = 0.05 Exam Tips
Exam Tips
3
2.40
Relative atomic mass of X = —— = 48.0 Write a balanced ionic
0.05 equation for yhe reaction
Question 2
The mass of nitrogen in 5.22 g of a compound X(NO 3 ) 2 is 0.56 g. What is the
relative atomic mass of X? [Relative atomic mass: C = 12, N = 14, O = 16]
A 65 C 137
B 84 D 207


Answer: C
0.56
Exam Tips
The number of moles of nitrogen = —— Exam Tips
14
= 0.04 Determine the number of
moles of nitrogen in the
compound.

2




01 STPM Q&A Chem T1 layout.indd 2 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
The formula of the compound indicates that the number of moles of the
compound is half that of nitrogen.
1
The number of moles of compound = —(0.04)
2
= 0.02
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5.22
Relative molecular mass of compound = ——
0.02
= 261
X + 2(14) + 6(16) = 261
Relative atomic mass of X = 137 Term

Question 3 1

121.7 g of an acid, HXO 4 is dissolved in distilled water and made up to 1 dm 3
solution. 25.0 cm of an aqueous solution of acid HXO 4 is exactly neutralised by
3
19.80 cm of 0.80 mol dm sodium hydroxide. What is the relative atomic mass
–3
3
of X? [Relative atomic mass: H = 1, O = 16]
A 77 C 108
B 91 D 127
Answer: D
Exam Tips
Exam Tips
The chemical formula of the acid indicates that it is a monoprotic acid. Hence,
the number of moles of acid is equal to the number of moles of the alkali.

HXO 4 + NaOH : NaXO 4 + H 2 O
Using the equation,
a
——– = — where M a, M b = molarity of acid and base respectively
M aV a
b
M b V b a, b = number of moles of acid and base in the equation
19.80(0.80)
Molarity of HXO 4 = ————— = 0.6336 mol dm
–3
25.0
121.7
Relative molecular mass of HXO 4 = ——— = 192
0.6336
Relative atomic mass of X = 192 – [1 + 4(16)] = 127
Question 4
The chemical formula of an oxide is M 2 O 3 . When 2.28 g of M 2 O 3 was reduced by
aluminium, the mass of the metal M left is 1.56 g. What is the relative atomic
mass of M? [Relative atomic mass: O = 16]
A 52 C 152
B 118 D 194



3




01 STPM Q&A Chem T1 layout.indd 3 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Answer: A
M 2 O 3 + 2Al : 2M + Al 2 O 3
Mass of oxygen in the oxide = 2.28 – 1.56
= 0.72
0.72
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The number of moles of oxygen = ——
16
= 0.045
From the chemical formula,
2
the number of moles of M = — × the number of moles of oxygen
3
Term
2
= — × 0.045
1 3
= 0.03
Mass
Relative atomic mass of M = ———————–
Number of moles
1.56
= ——
0.03
= 52

Common Errors
Students wrongly calculate the mole ratio of M in the oxide.




Question 5
The density of a gaseous compound WCl 4 is r g cm at room conditions. What is
–3
the relative atomic mass of W if the volume of one mol of gas at room conditions
is 24.0 dm ? [Relative atomic mass: Cl = 35.5]
3
A 24 000r – 142
B 12 000r – 142
C 24 000r – 71
D 6000r – 142

Answer: A
The mass of 1 mol of gas (24 dm ) = 24 000r
3
Relative molecular mass of WCl 4 = 24 000r
Relative atomic mass of W = 24 000r – 142







4




01 STPM Q&A Chem T1 layout.indd 4 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 6

The following mass spectrum is obtained for an element X which exists as a
molecule X 2. X has two isotopes P and Q with relative atomic mass a and b.

Relative
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abundance
4 28 49

2a a + b 2b m/e Term
What is the relative abundance of isotope P and Q?
A 1 : 2 C 2 : 3 1
B 1 : 3 D 2 : 7


Answer: D
Let the abundance ratio of P and Q is x : y.
y
x
Probability of obtaining P and Q is ——— : ———
m + a a (x + y) (x + y)
— of X 2 : P P = (a + a) = 2a
e
x
x

 

Probability = ——– × ——–
x + y
x + y
m + a b b a Common Errors
— of X 2 : P P or P P = (a + b)
e
m
x
x
e
 


Probability = ——– × ——– × 2 For the — value of (a + b), students
did not multiply the relative
x + y
x + y
m + b b abundance by two (There are two
— of X 2 : P P = (b + b) = 2b
possible combinations of P P and
a b
e
x
x
a
 


Probability = ——– × ——– b P P.).
x + y
x + y
From the diagram,
x
x
m


 
at — 2a, probability = ——– × ——– = 4
x + y
x + y
e
x
x
m

 

at — (a + b), probability = ——– × ——– × 2 = 28
x + y
e
x + y
x
x
m

 

at — 2b, probability = ——– × ——– = 49
x + y
x + y
e
Since the denominator is the same in all probabilities,
x = 4 and y = 49
2
2
Hence,
1
—
relative abundance of isotope P = (4) = 2
2
1
—
relative abundance of isotope Q = (49) = 7
2

5
01 STPM Q&A Chem T1 layout.indd 5 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 7

An aqueous solution of a substituted ethanoic acid, X 3 CCOOH contains
16.0 g dm of the acid. 25.0 cm of the acid solution exactly neutralises
–3
3
24.50 cm of 0.10 mol dm sodium hydroxide.
–3
3
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(a) (i) Write a chemical equation for the reaction between the acid and sodium
hydroxide.
(ii) Write an ionic equation for the reaction.
(b) (i) Determine the relative molecular mass of the acid, X 3 CCOOH.
(ii) Calculate the relative atomic mass of X.
(c) Suggest an identity for X.
Term
1
Answer:
Exam Tips
Exam Tips
The chemical formula of the acid indicates that it is a monoprotic acid. Hence,
one mole of acid reacts with one mole of the alkali.
(a) (i) X 3 CCOOH + NaOH : X 3 CCOONa + H 2 O
(ii) H + OH : H 2 O
–
+
a
(b) (i) ——– = — where M a , M b = molarity of acid and base respectively
M a V a
M b V b b a, b = number of moles of acid and base in the equation
24.50(0.10)
–3
Molarity of acid = ————–– = 0.098 mol dm
25.0
Mass
Relative molecular mass of acid = ——–————–
Number of moles
16.0
= ——–
0.098
= 163.3
(ii) X 3 CCOOH = 163.3
3X + 2(12) + 2(16) + 1 = 163.3
X = 35.4
Relative atomic mass of X = 35.4
(c) Chlorine














6




01 STPM Q&A Chem T1 layout.indd 6 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 8

(a) The following mass spectrum was obtained when a gaseous sample of an
organic acid was analysed.

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Relative
abundance


12 15 16 17 28 43 60 61 m/e Term
m
(i) Suggest possible ions responsible for peaks with — values of 17, 28 1
e
and 43.
(ii) Draw a possible structural formula for this acid.
(iii) What is the relative molecular mass of the acid?

Answer:
+ m
m
+ m
16
1
16
1
12
16
12
(a) (i) — 17: O H , — 28: C O , — 43: C 2 H 3 O +
e
e
e
(ii) H O (iii) 60
& '
H9C9C9O9H
&
H
Question 9
(a) The relative atomic mass of X is 35.50.
(i) Why is the relative atomic mass of X not a whole number?
(ii) In nature, X contains two isotopes with nucleon numbers of 35 and 37.
Calculate the percent abundance of the isotopes X and X.
37
35
(iii) If X exists as a diatomic element, sketch the mass spectrum of molecule
X.
Answer:
(a) (i) X has more than one isotope.
(ii) Assume the relative abundance of X : X as x% and (100 – x)%.
37
35
35(x) + 37(100 – x)
Relative atomic mass of X = ————————–
100
35x + 3700 – 37x
35.50 = ———————–
100
7




01 STPM Q&A Chem T1 layout.indd 7 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Solving, x = 75.0%
Percent abundance of X = 75.0%
35
Percent abundance of X = 25.0%
37
3
m
m
3
9
(iii) — 70: X X + — 70: — × — = —–
35
35
e
e
4
4
16
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m
1
m
6
3
— 72: X X + — 72: 2 × — × — = —–
37
35
e
e
4
16
4
1
m
1
m
1
— 74: X X + — 74: — × — = —–
37
37
e
e
16
4
4
Therefore, the mass spectrum of molecule X is as below:
10
Term
8
1 Relative 6
abundance 4 9
2 6
1
70 72 74 m/e
Section C Essay Questions
Question 10
(a) The four naturally occurring stable isotopes of iron are given in the table
below.
Relative atomic mass 53.94 55.93 56.94 57.93
Relative abundance/% 5.82 91.80 2.10 0.28
Determine the relative atomic mass of iron.
(b) Due to the differing proportions of isotopes, the relative atomic mass of
iron depends on its source. Two samples of iron were analysed. The mass
spectrum of the first sample was taken and found to be the same as in the
above table. When 1.12 g of the second sample was converted to iron(III)
chloride, the mass of the product is 3.25 g. [Relative atomic mass: Cl = 35.5]
(i) Write an equation for the conversion of iron to iron(III) chloride.
(ii) How many moles of chlorine was used in the reaction?
(iii) Determine the relative atomic mass of iron from this experiment.
(iv) Compare the relative atomic mass of iron with the value obtained in (a).
Decide whether the two samples have identical isotopic compositions.
Answer:
(a) Relative atomic mass of Fe
53.94(5.82) + 55.93(91.80) + 56.94(2.10) + 57.93(0.28)
= ————————–—————————————–
100
= 55.84
8
01 STPM Q&A Chem T1 layout.indd 8 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
(b) (i) 2Fe + 3Cl 2 : 2FeCl 3
(3.25 – 1.12)
(ii) The number of moles of Cl 2 = —————
71.0
= 0.03

(iii) The number of moles of Fe = 0.02
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1.12
Relative atomic mass of Fe = ——
0.02
= 56.0
(iv) The two samples have different isotopic compositions because the Term
relative atomic masses of the two samples are vary.

1
Question 11

(a) Combustion of 1.18 g sample of an organic compound P yields 1.76 g CO 2
and 0.54 g H 2 O.
(i) Determine the empirical formula of P.
(ii) If the relative molecular mass of P is 118, what is the molecular formula
of P?


Answer:
12
(a) (i) Proportionate mass of carbon in CO 2 = —– × 1.76
44
= 0.48 g
2
Proportionate mass of hydrogen in H 2 O = —– × 0.54
18
= 0.06 g
Mass of oxygen = 1.18 – 0.48 – 0.06
= 0.64 g
0.48
0.06
0.64
Mole ratio of C : H : O = —— : —— : ——
12 1 16
= 1 : 1.5 : 1
= 2 : 3 : 2
Empirical formula of P is C 2 H 3 O 2 .
(ii) (C 2 H 3 O 2 ) n = 118
59n = 118
n = 2
Molecular formula of P is C 4H 6O 4.






9




01 STPM Q&A Chem T1 layout.indd 9 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 12

m
The mass spectrum of dichloromethane consists of peaks at — values of 84, 86
e
and 88. Only H, C, Cl and Cl isotopes are present in the molecule.
1
35
12
37
(a) Write the formula of ions responsible for the above peaks.
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(b) Calculate the percent abundance of the above peaks if the natural abundance
ratio of Cl and Cl is 3 : 1.
35
37
(c) Sketch and label a mass spectrum for 1,1-dichloroethane showing the above
peaks.
Answer:
Term
1 Exam Tips
Exam Tips
Write all the possible combination of chlorine isotopes in the dichloromethane
mass
molecule, then add up their respective atomic mass to get the —–—– ratio,
m charge
—–.
e
m 12 1 35 + m 12 1 35 37 + m 12 1 37 +
e
e
e
(a) — 84: C H 2 Cl 2 , — 86: C H 2 Cl Cl , — 88: C H 2 Cl 2
m 3 3
(b) — 84: — × — = 0.5625
e
4
4
m
1
3
— 86: 2 × — × — = 0.375
e
4
4
m
1
1
— 88: — × — = 0.0625
e
4
4
m
m
m
— 84: — 86: — 88 = 56.25% : 37.50% : 6.25%
e
e
e
(c) Percent abundance
56.25
37.50
6.25
84 86 88 m/e

10




01 STPM Q&A Chem T1 layout.indd 10 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
1.3 The Mole and the Avogadro Constant

Section A Multiple-choice Questions

Question 1
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Which of the following gases occupies the largest volume at standard
temperature and pressure? [Relative atomic mass: H = 1, C = 12, N = 14,
O = 16, Cl = 35.5]
A 8.0 g methane C 20.0 g nitrogen
B 16.0 g oxygen D 30.0 g chlorine Term


Answer: C 1
Exam Tips
Exam Tips
Determine the number of moles of each compound. Number of moles of gas
is directly proportional to its volume (n ∝ V ).
8.0
In A, the number of moles of CH 4 = —– = 0.5
16
16.0
In B, the number of moles of O 2 = —— = 0.50
32
20.0
In C, the number of moles of N 2 = —— = 0.71
28
30.0
In D, the number of moles of Cl 2 = —— = 0.42
71
Question 2
0.78 g potassium is completely burnt in excess chlorine vapour. The solid product
is then dissolved in 200 cm of 0.20 mol dm sodium chloride solution. What is
3
–3
the concentration of chloride ions in the solution?
[Relative atomic mass: K = 39, Cl = 35.5]
A 0.02 mol dm C 0.20 mol dm
–3
–3
B 0.05 mol dm D 0.30 mol dm –3
–3
Answer: D
Exam Tips
Exam Tips
Write a balanced equation for the reaction between potassium and chlorine.
2K + Cl 2 : 2KCl
Number of moles of Cl ions formed from KCl
–
0.78
= number of moles of K = —— = 0.02
39
11




01 STPM Q&A Chem T1 layout.indd 11 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
0.20(200)
MV
Number of moles of Cl ions from NaCl = ——– = ————– = 0.04
–
1000 1000
Total number of moles of Cl ions = (0.02 + 0.04) = 0.06 mol
–
0.06 mol
–
Concentration of Cl ions = ————– = 0.30 mol dm –3
0.20 dm 3
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Question 3
3
10 cm of a gaseous hydrocarbon X is mixed with 90 cm of oxygen and burned
3
completely. When the mixture is cooled, the total volume of gas is 70 cm .
3
After the gaseous mixture has been passed through concentrated potassium
hydroxide solution, 40 cm of gas remains. What is the molecular formula of W?
3
Term
A C 2H 6 C C 3H 8
1 B C 3H 6 D C 4H 8
Answer: C
y
y
C x H y + (x + — ) O 2 : xCO 2 + — H 2 O
10 cm 3 4 2
Volume of excess O 2 + volume of CO 2 formed = 70 cm 3
Since KOH absorbs an acidic gas, the residual gas is O 2 .
Volume of excess O 2 = 40 cm
3
Volume of CO 2 formed = 70 – 40 = 30 cm
3
Since 1 mole of C x H y produces x moles of CO 2 ,
30
x = —– = 3
10
y
Since 1 mole of C x H y reacts with (x + — ) moles of O 2 ,
4
Volume of O 2 used = 90 – 40 = 50 cm Exam Tips
3
Exam Tips
y 50
(x + — ) O 2 = —– = 5
4 10 Compare the volume ratio of
Substituting x = 3 into the equation, y = 8 gases to obtain the mole ratio.
Molecular formula of X is C 3 H 8 .

Question 4
What is the Na ion concentration in the solution formed by mixing 130 cm
3
+
of 0.20 mol dm NaCl solution with 70 cm of 0.80 mol dm Na 2 SO 4 solution?
3
–3
–3
A 0.41 mol dm –3
B 0.50 mol dm
–3
C 0.69 mol dm –3
–3
D 1.00 mol dm
12




01 STPM Q&A Chem T1 layout.indd 12 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Answer: C
Exam Tips
Exam Tips
Calculate the number of moles of Na ions in both solutions, then find the
+
total number of moles of Na ions.
+
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MV
Number of moles = ——–
1000
(0.13 × 200)
2(0.07 × 0.8)
The number of moles of Na ion = ————–– + ————–—
+
1000 1000
= 0.138 Term
Total volume of solution = 130 + 70 = 200 cm = 0.20 dm 3
3
Number of moles
Concentration of Na ion = ————–——– 1
+
Volume (dm )
–3
0.138
= ——–
0.20
= 0.69 mol dm –3
Common Errors
Students forget that each mole of sodium sulphate has two moles of sodium
ions.


Question 5
Hydrogen sulphide gas has the odour of rotten eggs. It burns in air to form
sulphur dioxide.
2H 2 S(g) + 3O 2 (g) : 2H 2 O(l) + 2SO 2 (g)
If 6 dm hydrogen sulphide is burnt in 12 dm oxygen gas, what is the final
3
3
volume of the gaseous mixture?
A 6 dm
3
B 9 dm
3
C 12 dm 3
D 21 dm 3
Answer: B
From the equation, 2 mol of H 2 S reacts with 3 mol of O 2 to produce 2 mol of
SO 2 .
Since the volume of gas is directly proportional to the amount of gas (V ∝ n),
6 dm of H 2 S reacts with 9 dm of O 2 to produce 6 dm of SO 2 .
3
3
3
Volume of excess O 2 gas = 12 – 9 = 3 dm
3
Total volume of gas = 3 + 6 = 9 dm 3
13




01 STPM Q&A Chem T1 layout.indd 13 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 6

3
The volume of 50 drops of carbon tetrachloride, CCl 4 is V cm . Each drop of CCl 4
contains n molecules. If the density of CCl 4 is 1.60 g cm , what is the expression
–3
for the Avogadro’s constant? [Relative molecular mass of CCl 4 is 154.]
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1.60 × 50 × n 154 × 50 × n
A ————–—– C ————–—–
154 × V 1.60 × V
1.60 × 50 × V 154 × 50 × V
B ————–—– D ————–—–
154 × n 1.60 × n
Answer: C V
Term
1 Volume of 1 drop = —–
50
V
Volume of n molecules = —–
50
V
50
Volume of 1 mole of molecules = —– ÷ n × N A
Mass of 1 mole of molecules = 154 g
Mass
Since density = ———–
Volume
Mass of 1 mole of molecules = Density × Volume
V
154 = 1.60 × —– × —–
N A
50 n
154 × n × 50
N A = —–———––
1.60 × V
Question 7

Using chlorine as an example, which of the following formulae can be used to
determine Avogadro’s constant?
Mass of 1 mol gas Volume of 1 mol gas
A ————–——–—– C ————–——–—––
Mass of 1 atom Mass of 1 atom
Mass of 1 mol gas Density of 1 mol gas
B ————–——–—– D ————–——–——–
Mass of 1 molecule Density of 1 molecule


Answer: B
Mass of 1 mol Cl 2 gas = m g Volume of 1 mol Cl 2 gas =V
m
V
Mass of 1 Cl 2 molecule = —– Volume of 1 Cl 2 molecule = —–
N A N A
m
V
Mass of 1 Cl atom = —– – Volume of 1 Cl atom = —––
2N A 2N A


14




01 STPM Q&A Chem T1 layout.indd 14 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
mass of 1 mol gas m
m
For A, ————–——–— = —– — = 2N A
mass of 1 atom —––
2N A
mass of 1 mol gas m
m
For B, ————–——–—– = —–— = N A
mass of 1 molecule —–
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N A
volume of 1 mol gas V
V
For C, ————–——–—–– = —– — = 2N A
mass of 1 atom —––
2N A
For D, density of a substance does not depend on the number of particles. Term
1
Section B Structured Questions
Question 8
A 100 cm gaseous mixture of ethane, C 2 H 6 and butene C 4 H 8 is burnt in 500 cm 3
3
oxygen (excess) for complete combustion. After the mixture has been cooled,
the total volume of gas is 331 cm . Only 55 cm of gas remains when the gaseous
3
3
mixture is passed through concentrated NaOH solution.
(a) Write two separate equations for the combustion of both gases.
(b) Determine the volume of each gas in the mixture.


Answer:
7
(a) C 2 H 6 + —O 2 : 2CO 2 + 3H 2 O
2
C 4 H 8 + 6O 2 : 4CO 2 + 4H 2 O
(b) Assume that the volume of ethane gas = x cm ,
3
volume of butene gas = (100 – x) cm 3
7
C 2 H 6 + —O 2 : 2CO 2 + 3H 2 O
2
x cm 3 2x cm 3
C 4 H 8 + 6O 2 : 4CO 2 + 4H 2 O
(100 – x) cm 3 4(100 – x) cm 3
Since NaOH absorbs CO 2 gas (acidic), volume of excess, unreacted O 2
= 55 cm
3
Total volume of CO 2 formed = 331 – 55 = 276 cm
3
Total volume of CO 2 formed = 2x + 4(100 – x) = 276
Volume of C 2 H 6 , x = 62 cm 3
Volume of C 4 H 8 = 38 cm
3





15




01 STPM Q&A Chem T1 layout.indd 15 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 9

When an aqueous solution of hydrogen peroxide is added to an aqueous
solution of potassium manganate(VII), KMnO 4 , oxygen gas is evolved. At the
same time, KMnO 4 is decolourised. The incomplete half-equations for the
reactions are shown below.
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MnO 4 + H + e : Mn + H 2 O
+
–
2+
–
H 2 O 2 : H + O 2 + e
–
+
In an experiment, it was found that 22.80 cm of acidified KMnO 4 was required
3
to react completely with 25.0 cm of 0.20 mol dm hydrogen peroxide.
–3
3
(a) Complete the two half-equations above.
Term
(b) Write a balanced chemical equation to represent the reaction above.
1 (c) Calculate the concentration of the acidified KMnO 4 solution.
(d) What is the volume of gas evolved at s.t.p.?
Answer:
(a) MnO 4 + 8H + 5e : Mn + 4H 2 O
–
+
2+
–
H 2 O 2 : 2H + O 2 + 2e –
+
– + 2+
(b) 2MnO 4 + 6H + 5H 2 O 2 : 2Mn + 8H 2 O + 5O 2
a
(c) ——– = —
M a V a
M b V b b
M a = Concentration of MnO 4
–
V a = Volume of MnO 4
–
M b = Concentration of H 2 O 2
V b = Volume of H 2 O 2
M a (22.80) 2
——–—— = —
0.2(25.0) 5
Concentration of KMnO 4 , M a = 0.088 mol dm –3
(d) The number of moles of O 2 = The number of moles of H 2 O 2
MV
= ——–
1000
0.20 × 25.0
= ——–——–
1000
= 0.005
Volume of O 2 = 0.005 × 22 400
= 112 cm 3





16




01 STPM Q&A Chem T1 layout.indd 16 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 10

In an experiment, 19.6 g of concentrated sulphuric acid is found to react
completely with 17.0 g of sodium nitrate, NaNO 3 .
[Relative atomic mass: H = 1, N = 14, O = 16, Na = 23, S = 32]
(a) Calculate the number of moles of sulphuric acid and sodium nitrate used.
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(b) Deduce a chemical equation for the reaction.
Answer:
(a) Relative molecular mass of H 2 SO 4 = [2(1) + 32 + 4(16)] = 98 Term
Mass 19.6
The number of moles of H 2 SO 4 = = = 0.20
Relative molecular mass 98
Relative formula mass of NaNO 3 = [23 + 14 + 3(16)] = 85 1
Mass 17.0
The number of moles of NaNO 3 = = = 0.20
Relative molecular mass 85
(b) Since 0.20 mol H 2 SO 4 reacts with 0.20 mol NaNO 3 ,
1 mol H 2 SO 4 reacts with 1 mol NaNO 3 .
H 2 SO 4 + NaNO 3 : NaHSO 4 + HNO 3
Question 11
A student was told that the white solid in a beaker is a mixture of barium oxide
and calcium oxide. The mass of the white solid is 10.0 g. The student found that
the white solid dissolves in exactly 100.0 cm of 2.0 mol dm hydrochloric acid.
3
–3
[Relative atomic mass: O = 16, Ca = 40, Ba = 137]
(a) Write two separate chemical equations for the reaction of both oxides with
hydrochloric acid.
(b) Determine the percentage of barium oxide in the mixture.

Answer:
(a) BaO + 2HCl : BaCl 2 + H 2 O
CaO + 2HCl : CaCl 2 + H 2 O
(b) Let the mass of BaO = x g; mass of CaO = 10.0 – x
BaO + 2HCl : BaCl 2 + H 2 O
x —–
2x
—–
153 153
CaO + 2HCl : CaCl 2 + H 2 O
2(10 – x)
10 – x —–——–
—–—
56 56
2.0 × 100.0
Total number of moles of HCl = ——–——
1000
2x
2(10 – x)
—– + ——–—– = 0.20
153 56
Solving, x = 6.94
6.94
Percentage of BaO = —— × 100% = 69.4%
10
17
01 STPM Q&A Chem T1 layout.indd 17 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Question 12

1 dm sample of air containing carbon dioxide is passed through aqueous
3
calcium hydroxide, Ca(OH) 2 . A white precipitate is formed.
(a) (i) Name the precipitate.
(ii) Write a balanced equation for the reaction.
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(b) If 0.45 g of precipitate is formed, determine
(i) the number of moles of CO 2 present in the sample
(ii) the percentage by volume of CO 2 in the sample at s.t.p.
[Relative molecular mass of CaCO 3 = 100; Molar volume of gas = 22 400 cm
3
at s.t.p.]
Term
1 Answer:
(a) (i) Calcium carbonate
(ii) Ca(OH) 2 + CO 2 : CaCO 3 + H 2 O
(b) (i) Number of moles of CO 2 = Number of moles of CaCO 3
0.45
= —— = 4.5 × 10
–3
100
(ii) Volume of CO 2 = 4.5 × 10 × 22 400 = 100.8 cm 3
–3
100.8
Percentage of CO 2 = ——– × 100% = 10.08%
1000
Question 13
In the presence of H ions, hydroxyammonium ions NH 3 OH reduce Fe ions
+
+
3+
2+
to Fe ions. In the reaction, NH 3 OH is oxidised to dinitrogen oxide, N 2 O.
+
In an experiment, an aqueous solution, labelled as A is prepared by boiling
5.00 g of hydroxyammonium sulphate, (NH 3 OH) 2 SO 4 per dm with excess iron(III)
3
ammonium sulphate and dilute sulphuric acid. An aqueous solution labelled as
B, contains 3.0 g KMnO 4 per dm solution. 25.0 cm of H 2 SO 4 is added to 25.0 cm
3
3
3
3
of solution A. The mixture completely reacts with 31.60 cm solution B.
[Relative atomic mass: H = 1, N = 14, O = 16, S = 32, K = 39, Mn = 55, Fe = 56]
(a) Calculate the molarity of manganate(VII) ions in solution B.
(b) Calculate the concentration of iron(II) ions in solution A.
(c) Determine the number of moles of iron(III) ions required to oxidise 1 mol
of hydroxyammonium ions.
(d) Write a balanced redox equation between NH 3 OH ions and Fe ions.
3+
+
Answer:
(a) Molar mass of KMnO 4 = 39 + 55 + 4(16) = 158
Mass 3.0
The number of moles of KMnO 4 = ——–——– = —— = 0.019
Molar mass 158
Concentration of KMnO 4 = 0.019 mol dm –3
18




01 STPM Q&A Chem T1 layout.indd 18 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Exam Tips
Exam Tips
Change the concentration of KMnO 4 from g dm to mol dm .
–3
–3

(b) 5Fe + MnO 4 + 8H : 5Fe + Mn + 4H 2 O
3+
2+
–
2+
+
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MV Fe 2+
——–— = 5
–
MV MnO 4
5 × 0.019(31.60)
Concentration of Fe = ——–——–——
2+
25.0
= 0.120 mol dm Term
–3
(c) In the reaction, NH 3 OH reduce Fe ions to Fe ions. The amount of Fe
+
2+
2+
3+
ions formed is determined by titrating with manganate(VII) ions. 1
Hence, in 1 dm of solution, Fe (that has reacted with NH 3 OH )
+
3+
3
= the number of moles of Fe (formed) = 0.120
2+
Molar mass of (NH 3 OH) 2 SO 4 = 164
In 1 dm of solution, the number of moles of (NH 3 OH) 2 SO 4
3
Mass
= ——–——–
Molar mass
5.00
= ——
164
= 0.030
+ 2–
(NH 3 OH) 2 SO 4 : 2NH 3 OH + SO 4
In 1 dm solution, the number of moles of NH 3 OH = 2(0.030) = 0.060
+
3
In 1 dm solution, the number of moles of Fe that has reacted
3
3+
= the number of moles of Fe formed = 0.120
2+
0.060 mol NH 3 OH reacts with 0.120 mol Fe .
3+
+
Hence, 1 mol NH 3 OH reacts with 2 mol Fe .
+
3+
(d) Exam Tips
Exam Tips
Write two separate half-equations for the redox reaction, then add up the
two equations in order to obtain the overall equation.
Fe + e : Fe 2+
3+
–
2NH 3 OH : N 2 O + H 2 O + 6H + 4e –
+
+
Overall: 2NH 3 OH + 4Fe : N 2 O + H 2 O + 4Fe + 6H
2+
+
3+
+
Common Errors
Students must multiply the half-equations with the appropriate coefficients
so that the number of electrons in each half-equation is the same.
19
01 STPM Q&A Chem T1 layout.indd 19 4/19/19 9:28 AM

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Section C Essay Questions

Question 14
Titanium carbide, TiC, is the hardest of the known metal carbides. It can be made
by heating titanium(IV) oxide, TiO 2 , with carbon black to 2200°C. When 4.00 g of
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TiO 2 is heated with excess carbon, 2.8 g carbon monoxide gas is formed.
[Relative atomic mass: C = 12, O = 16, Ti = 47.8]
(a) Calculate the number of moles of titanium(IV) oxide and carbon monoxide
gas formed in the reaction.
(b) Construct an equation for the reaction of titanium(IV) oxide and carbon.
(c) If 5 kg of titanium(IV) oxide is reacted with 10 kg of carbon, what is the
Term
mass of titanium carbide formed?
1
Answer:
(a) Molar mass of TiO 2 = 47.8 + 2(16) = 79.8
Mass
4.00
The number of moles of TiO 2 = ——–——– = ——– = 0.050
Molar mass 79.8
Molar mass of CO = 12 + 16 = 28
Mass 2.8
The number of moles of CO = ——–——– = —— = 0.100
Molar mass 28
(b) 0.050 mol of TiO 2 produces 0.100 mol of CO.
1 mol of TiO 2 produces 2 mol of CO.
TiO 2 + 3C : TiC + 2CO
Mass 5000
(c) The number of moles of TiO 2 = ——–——– = ——– = 62.7
Molar mass
79.8
Mass 10 000
The number of moles of C = ——–——– = ——–– = 833.33 (in excess)
Molar mass
12
When determining the amount of product, carbon is disregarded because
the amount of carbon used is in excess.
Since one mole titanium(IV) oxide produces one mole titanium carbide,
the number of moles of titanium carbide formed = 62.7
Molar mass of titanium carbide, TiC = 47.8 + 12 = 59.8
Mass of TiC formed = 62.7 mol × 59.8 = 3749 g = 3.75 kg


Common Errors
Students did not change the mass of the substances from kilogram to gram
when they determine the number of moles of the substances.







20




01 STPM Q&A Chem T1 layout.indd 20 4/19/19 9:28 AM

Chapter
Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
14 Introduction to Organic

Chemistry



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14.1 Bonding of the Carbon Atoms: The Shapes of Ethane, Ethene, Ethyne
and Benzene Molecules


Section A Multiple-choice Questions

Question 1
Which of the following combinations is correct?
Molecule Type of hybridisation of the carbon atom(s)
A CH 3CH 2CH 3 s p
2
B CH 3 !CH"CH 2 sp
3
C C 6H 6 sp and sp 2
D CH 3 !C#CH sp and sp
3



Answer: D
There are only three types of hybridisation involving the s and p orbitals: sp,
sp and sp .
3
2
Type of carbon-carbon bond Hybridisation
Single, C!C sp 3
Double, C"C sp 2
Triple, C#C sp


Molecule Type of hybridisation of carbon atom(s)
sp 3
CH 3 CH 2 CH 3
sp and sp 2
3
CH 3 !CH"CH 2
Term
3 C 6 H 6 sp 2
CH 3 !C#CH sp and sp
3


342




14 Q&A STPM Chem T3 Layout.indd 342 4/19/19 5:29 PM

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

Which atomic orbitals overlap to form the carbon-carbon π-bond in ethene,
CH 2 "CH 2 ?
A 2p and 2p
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B 2s and 2p
C sp and sp
2
2
D 2p and sp 2
Answer: A
Only the lateral overlap of two p orbitals will produce a π-bond.
π-bond:
Exam Tips
π bond Exam Tips
σ-Bonds are formed by the overlap of
H H hybridised orbitals. π-Bonds are formed
C C
H H by the overlap of non-hybridised orbitals.
σ bond

14.2 General, Empirical, Molecular and Structural Formulae of Organic
Compounds
Section A Multiple-choice Questions

Question 1
Compound X (molar mass = 54.0 g mol has the following composition by
–1
mass:
C, 89%; H, 11%
What is the molecular formula of X?
A C 9 H 11
B C 2 H 3
C C 4 H 6
D C 4 H 8

Answer: C
Mole ratio of C : H = 89 : 11 = 7.4 : 11 = 1 : 1.5 = 2 : 3
12 1
Let the molecular formula be (C 2 H 3 )n. Term
Relative molecular mass = [2(12) + 3(1)]n
= 54 3
n = 2
\ The molecular formula is C 4 H 6 .


343




14 Q&A STPM Chem T3 Layout.indd 343 4/19/19 5:29 PM

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

Compound X has the following composition by mass.
C, 60.0%; O, 26.8%; H, 13.2%
Which of the following compound is definitely not X?
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A CH 3 CH(OH)CH 3
B HOCH 2CH 2CH 2OH
C CH 3OCH 2CH 3
D CH 3 CH 2 CH 2 OH
Answer: B
Exam Tips
Exam Tips
Compounds A, C and D are isomeric with one oxygen atom in their structures,
whereas compound B has two oxygen atoms. So, Compound B is the odd one
out.
An alternative is to calculate the empirical formula of X.
C : O : H = 60.0 : 26.8 : 13.2
12 16
= 5 : 1.7 : 13.2
= 3 : 1 : 8
Empirical formula is C 3 H 8 O.
So, the simplest molecular formula is (C 3 H 8 O) 2 = C 6 H 16 O 2 , which does not
match compound B.



Section B Structured Questions
Question 3
An organic compound, Y, contains carbon, hydrogen and oxygen only. When
6.0 g of Y is burned completely in excess oxygen, 13.2 g of carbon dioxide and
7.2 g of water is produced. Determine the empirical formula of Y.

Answer:
44 g of CO 2 contains 12 g of C.
\ Mass of C in 13.2 g of CO 2 = 12 × 13.2 g = 3.6 g
44
18 g of H 2 O contains 2 g of H.
Term
2
3 \ Mass of H in 7.2 g H 2 O = 18 × 7.2 = 0.8 g
\ Mass of oxygen in 6.0 g of Y = 6.0 – 3.6 – 0.8 = 1.6 g



344




14 Q&A STPM Chem T3 Layout.indd 344 4/19/19 5:29 PM

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Secondly, determine the mole ratio of C : H : O
C : H : O = 3.6 : 0.8 : 1.6 Exam Tips
12 16 Exam Tips
= 0.3 : 0.8 : 0.1 First, determine the mass of
= 3 : 8 : 1 C, H and O in 6.0 g of Y.
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Empirical formula of Y is C 3 H 8 O.



Section C Essay Questions

Question 4
An organic compound, X, has the following composition by mass:

C, 35.0%; H, 6.6%; Br, 58.4%
When 1.36 g of X is vaporised, it occupies a volume of 389.2 cm at 200°C and
3
101 kPa.
(a) Determine the empirical and molecular formula of X.
(b) Draw and name all possible isomers for X and determine which of them
(if any) is/are chiral.


Answer:
(a) C : H : Br = 35 : 6.6 : 58.4
12 80
= 2.9 : 6.6 : 0.73
= 4 : 9 : 1
Empirical formula of X is C 4 H 9 Br.

Using pV = nRT,
(101 × 10 )(389.2 × 10 ) = n × 8.31 × (200 + 273)
3
–6
n = 0.010
Common Errors
Relative molecular mass of X = 1.36 When using the equation
0.010 pV = nRT, students
= 136 forget to convert the
temperature from °C to
3
Let the molecular formula of X be (C 4 H 9 Br) n , K, volume from cm to
3
[(12 x 4) + 9 + 80]n = 136 m and pressure from Term
kPa to Pa.
136n = 136
n = 1 3
The molecular formula of X is C 4 H 9 Br.



345




14 Q&A STPM Chem T3 Layout.indd 345 4/19/19 5:29 PM

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
(b) CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane
*
CH 3 CH 2 !CH!CH 3 2-Bromobutane is chiral
&
Br
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CH 3
&
CH 3 —CH—CH 2 Br 2-Methyl-1-bromopropane

CH 3
&
2-Bromo-2-methylpropane
CH 3 —C—CH 3
&
Br


14.3 Functional Groups: Classification and Nomenclature

Section A Multiple-choice Questions

Question 1

The structural formula of cholesterol is shown below:

H C CH
3
C 3
CH 3 CH 3
CH
3


HO

Which of the following statements is not correct regarding cholesterol?
A The hybridisation of the carbon atoms includes sp and sp only.
3
2
B Its molecular formula is C 27 H 46 O 2 .
C It contains the hydroxy and alkene functional groups.
D The molecule is not planar.


Answer: B
The structural formula shows only one oxygen atom. Thus, B cannot be correct.
Term
For A, there are only two types of carbon-carbon bonds in the molecules. They
3 are C—C (sp ) and C=C (sp ).
3
2
For D, molecules with sp hybridised carbon atoms are tetrahedral (three
3
dimensional).
346




14 Q&A STPM Chem T3 Layout.indd 346 4/19/19 5:29 PM

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
14.4 Isomerism: Structural and Stereoisomerism

Section A Multiple-choice Questions


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Question 1
Which of the following is not true for the following molecule?


HO! !CH!CH"CH! !COOH
&
CHO

A It has cis-trans isomers.
B It is optically active.
C It is soluble in water.
D It has more than one functional group.


Answer: C
The presence of three hydrophobic benzene rings renders the compound
insoluble in water although the !OH and !COOH groups can form
hydrogen bonds with water molecules.

The chiral centre and the various functional groups are shown below:
*
HO! !CH!CH"CH! !COOH
&
CHO

The presence of the C"C double bond (where no two identical atoms/groups
are attached to the unsaturated carbon atoms) enables it to exhibit cis-trans
isomerism.
COOH
H H H
C"C C"C CHO
HO! !CH HO! !CH H
& COOH &
CHO
Term
Cis–isomer Trans–isomer
3





347




14 Q&A STPM Chem T3 Layout.indd 347 4/19/19 5:29 PM

Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

The following compound is used to treat skin problems. How many chiral
centres does it have?
O
H H C HO OH
3
HO
H C H
3
H H
O
A 3 C 7 OHRights Reserved.
B 5 D 8


Answer: C
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HO O
18
HO 11 12 17 20 21
1 19 9 13 16
2 14
10 H 8 15
O 3 4 5 6 7

The chiral centes are at C-8, C-9, C-10, C-11, C-13, C-14 and C-17.


Section B Structured Questions
Question 3
(a) What do you understand by structural isomerism? Draw all the possible
structural isomers for chlorobutane, C 4 H 9 Cl and give their IUPAC names.
(b) Which of the above isomers is/are optically active? Mark the chiral centres
with an asterisk (*).


Answer:
(a) Structural isomerism refers to the occurrence of two or more compounds
having the same molecular formula but with different connectivity of
their atoms.
Term
CH 3CH 2 CH 2 CH 2Cl 1-Chlorobutane
3 CH 3 CH 2 !CH!CH 3 2-Chlorobutane
&
Cl


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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry

CH 3
&
2-Chloro-2-methylpropane
CH 3 !C!CH 3
&
Cl
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(b) 2-Chlorobutane
H
&
CH 3 CH 2 !C*!CH 3
&
Cl




Section C Essay Questions

Question 4
Compound X has the following structure:

CH 3
&
CH 3 !CH 2 !C!CH"CH!CH 3
&
NH 2
X exhibits geometrical isomerism and optical isomerism.
(a) Name X according to the IUPAC convention.

(b) State what do you understand by the terms:
(i) geometrical isomerism
(ii) optical isomerism
(c) Draw the:
(i) geometrical isomers of X
(ii) optical isomers of X
(d) How would you differentiate the two optical isomers of X? Name the
instrument used and state what quantity does it measure.


Answer:
(a) 3-Amino-3-methyl-4-hexene Term
(b) (i) Geometrical isomerism is the occurrence of different compounds
having the same molecular formula, the same structural formula, but 3
are different due to the different spatial arrangement of atoms/group
of atoms on either side of a C"C double bond or a cyclic structure.


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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
(ii) Optical isomerism is the occurrence of two compounds having the
same molecular formula, the same structural formula, but are mirror
images of one another and cannot be superimposed.
(c) (i) H H H CH 3
C"C C"C
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CH 3 CH 2 !C!CH 3 CH 3 CH 3 CH 2 !C!CH 3 H
& &
NH 2 NH 2
Cis-isomer Trans-isomer

(ii) CH"CHCH 3 CH 3 CH"CH
& &
C C
CH 3 H 3 C CH 2 CH 3
CH 3 CH 2
H 2 N
NH 2
(d) The optical isomers can be differentiated using a polarimeter which
measures the specific rotation (α) of the plane-polarised light. One isomer
would rotate the plane-polarised light clockwise (where α is positive)
and the other isomer would rotate the light anticlockwise (where α is
negative).



14.5 Free Radicals, Nucleophiles and Electrophiles

Section A Multiple-choice Questions

Question 1
Which of the following is not a nucleophile?
A H 2 O
B NH 4
+
C Cl –
D Br 2

Answer: B
Exam Tips
Exam Tips
A positive species cannot be a nucleophile.
Term
3 Nucleophiles are either negatively charged or are species with a lone pair of
electrons. They attack centres of low electron density (with partial positive
charge).


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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

Which of the following statements regarding electrophiles is not correct?
A It is formed by the homolytic fission of a covalent bond.
B It has an electron-deficient site.
C It is a Lewis acid.
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D It acts as an electron-pair acceptor.
Answer: A
+
Electrophiles are usually represented by E . It is electron deficient. It accepts a
lone pair of electrons from another species and thus is also a Lewis acid.
Homolytic fission of a covalent bond results in the production of free radicals.


Question 3
Which order shows the correct relative stability of a free radical?
A 1°> 2° > 3° C 2° > 3° > 1°

B 1° > 3° > 2° D 3° > 2° > 1°

Answer: D
Free radicals are stabilised by electron-releasing groups. Alkyl groups are
electron-releasing groups. As a result, the stability of the free radicals increases
with the number of alkyl groups in the species.

Section B Structured Questions

Question 4
(a) What do you understand by stereoisomerism?
(b) Name the two main types of stereoisomerism.
(c) Draw all possible stereoisomers of the following compound.
CH 3 CH 2 !CH!CH"CH!COOH
&
OH


Answer:
(a) Different compounds having the same molecular formula, the same
structural formula, but have different spatial arrangement of the atoms/
groups. Term

Common Errors
3
Stereoisomers cannot have different structural formula. Otherwise they will
be called structural isomers.


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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
(b) Cis-trans (geometrical) isomerism and optical isomerism.
(c) Optical isomers:
CH"CH!COOH COOH!CH"CH
& &
C C
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OH HO CH 2 CH 3
CH 3 CH 2
H H
Cis-trans-isomers:
OH OH
CH 3 CH 2 CH COOH CH 3 CH 2 CH H
C"C C"C
H H H COOH
Cis-isomer Trans-isomer


Section C Essay Questions

Question 5
(a) Using suitable examples explain what is meant by the following species.
(i) Free radical (ii) Nucleophile (iii) Electrophile
(b) Name the three classes of free radicals and explain the factors that affect
their stability.


Answer:
(a) (i) A free radical is a species that has one unpaired electron in its
structure. It is usually formed via the homolytic fission of a covalent
bond. Examples are:
.
Cl 2 !: 2Cl
. .
C 2 H 5 Br !: C 2 H 5 + Br
(ii) A nucleophile is a species that has a lone pair of electrons that can be
donated to other species to form a coordinate bond. A nucleophile is
–
an electron-rich species. Examples are :NH 3 and :OH .
(iii) An electrophile is an electron-deficient species that accepts a lone
pair of electrons from another species to form a coordinate bond.
Examples are NO 2 , Cl and SO 3 .
+
+
(b) The three classes of free radicals are: Primary, secondary and tertiary free
Term
3 radicals.
Free radicals can be considered as electron-deficient species. They are
stabilised by electron-donating groups such as alkyl groups.


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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Thus, the more alkyl groups attached to the carbon atom that carries the
lone electron the more stable is the free radical. Take the example of the
.
butyl free radicals, C 4 H 9 below:
. . .
CH 3 CH 2 CH 2 —C—H CH 3 CH 2 —C—CH 3 CH 3 —C—CH 3
& & &
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H H CH 3
1° free radical 2° free radical 3° free radical
In the primary free radical there is one electron-donating group,
CH 3 CH 2 CH 2 !. In the secondary free radical there are two electron-
donating groups, CH 3 CH 2 ! and CH 3 !. In the tertiary free radical there
are three electron-donating groups, three CH 3 ! groups. As a result, the
stability of the free radical increases in the order: 1° < 2° < 3°.


14.6 Molecular Structure and Its Effect on Physical Properties

Section A Multiple-choice Questions

Question 1
The boiling point of pentane is higher than its isomer, 2,2-dimethylpropane.
Why?
A The bonds in pentane is more polar.
B A pentane molecule is linear while 2,2-dimethylpropane is branched.
C The molar mass of pentane is higher than 2,2-dimethylpropane.
D 2,2-dimethylpropane is optically active but pentane is not.


Answer: B
The strength of van der Waals force increases with the total number of electrons
in the molecule as well as the total surface area of the molecule.
Both pentane and 2,2-dimethylpropane are hydrocarbons. Their molecules are
non-polar.
Both the isomers have the same molar mass.
There is no chiral centre in 2,2-dimethylpropane. Hence, it is not optically
active.
Being linear, pentane has a larger surface area than 2,2-dimethylpropane,
which is more spherical. This increases the strength of the van der Waals force
between pentane molecules. Term


3


Pentane 2,2-Dimethlpropane

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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

Arrange the following alcohols in the order of decreasing boiling point.
CH 3 CHCH 3 ; CH 3 CH 2 CH 2 OH; HOCH 2 CH 2 OH
&
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OH
I II III
A I, II, III
B I, III, II
C II, I, III
D III, II, I


Answer: D
Generally, diols have higher boiling points than monohydric alcohols because
they can form more hydrogen bonding.
II is linear but I is branching. Branching decreases the strength of the van der
Waals force between the molecules.

Question 3
The table below lists the boiling point of some haloalkanes.

Compound C 2 H 5 Cl C 2 H 5 Br C 2 H 5 I
Boiling point/°C 12 37 73

Which statement explains the difference in their boiling points?
A The strength of the hydrogen bond increases in the order C 2 H 5 Cl, C 2 H 5 Br,
C 2H 5I.
B The strength of the carbon-halogen bond decreases in the order C 2 H 5 Cl,
C 2 H 5 Br, C 2 H 5 I.
C The size of the molecule increases in the order C 2 H 5 Cl, C 2 H 5 Br, C 2 H 5 I.
D The polarity of the carbon-halogen bond increases in the order C 2 H 5 Cl,
C 2 H 5 Br, C 2 H 5 I.



Answer: C
Exam Tips
Exam Tips
Boiling point depends on the strength of the intermolecular forces only.
Term
3 The strength of the van der Waals force increases with the size of the molecule.
Haloalkanes cannot form hydrogen bonds.


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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 4

Ibuprofen, a nonsteroid anti-inflammatory drug used for pain-relief has the
following structure.
CH 3
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COOH
CH 3
H 3 C

Which of the following is not correct regarding ibuprofen?
C It is insoluble in water.
A Its molecular formula is C 13 H 16 O 2 .
B It is a carboxylic acid. D It has one chiral centre.

Answer: A
The molecular formula is C 13 H 18 O 2 .
The !COOH is a carboxylic acid group.
The chiral centre is marked with an asterisk.

CH 3
* COOH
CH 3
H 3 C

The presence of large hydrophobic hydrocarbon groups makes it insoluble in
water.


14.7 Inductive and Resonance Effect

Section A Multiple-choice Questions

Question 1
Arrange the following compounds in the order of increasing K a values.
CH 3 COOH; ClCH 2 COOH; Cl 2 CHCOOH
I II III
A I, II, III B I, III, II C II, III, I D III, II, I


Answer: A
K a is a measure of the strength of an acid. The stronger the acid, the higher the Term
acid strength.
Chlorine is an electron-withdrawing group (negative inductive effect). It 3
weakens the O!H bond in the !COOH group thus increasing the acid
strength. The more chlorine there is the weaker the O!H bond.



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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
Question 2

The acid dissociation constants, K a for benzoic acid and 4-nitrobenzoic acid are
given below:

Acid Benzoic acid 4-Nitrobenzoic acid
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K a /mol dm 6.4 × 10 –5 4.0 × 10 – 4
-3
With reference to the above, which of the following is correct?
A 4-Nitrobenzoic acid is a stronger acid than benzoic acid.
B 4-Nitrobenzoic acid is more stable than benzoic acid.
C The benzoate ion is more stable than the 4-nitrobenzoate ion.
D Benzoate ion is a weaker base compared to 4-nitrobenzoate ion.


Answer: A
The higher the K a value, the stronger is the acid, and the less stable is the acid
molecule.
The stronger the acid, the weaker is the conjugate base, and the more stable is
the conjugate base.
For example,
O 2 NC 6 H 4 COOH L O 2 NC 6 H 4 COO + H K a = 6.4 × 10 mol dm –3
–5
+
–
(Acid) (Conjugate base)
C 6 H 5 COOH L C 6 H 5 COO + H K a = 4.0 × 10 mol dm –3
+
–4
–
(Acid) (Conjugate base)
4-Nitrobenzoic acid is stronger than benzoic acid. That is, the degree of
dissociation of 4-nitrobenzoic acid is higher. This means that 4-nitrobenzoic
acid is less stable compared to benzoic acid.
On the other hand, the 4-nitrobenzoate ion is a weaker base than the benzoate
ion. This means that the tendency for 4-nitrobenzoate ion to combine with H
+
is less than that of benzoate ion. That is, the 4-nitrobenzoate ion is more stable
than the benzoate ion.
Section B Structured Questions

Question 3
Consider the four carboxylic acids below:

CH 3 COOH !COOH HO! !COOH O 2 N! !COOH
Term
3 I II III IV





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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry

(a) Which is a stronger acid, I or II. Explain your answer.
(b) Arrange acids II, III and IV in order of increasing acid strength. Explain your
answer.
(c) How would the strength of acid IV change if the !NO 2 group is replaced
by !NH 2 group?
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Answer:
(a) Acid II > acid I
The benzene ring in acid II is an electron-withdrawing group that weakens
the O!H bond in !COOH.

Exam Tips
Exam Tips
Unsaturated acids or aromatic acids are stronger acids than aliphatic acids.
(b) III < II < IV
!OH is an electron-donating group that strengthens the O!H bond
in the !COOH group. On the other hand, !NO 2 is an electron-
withdrawing group that weakens the O!H bond.
(c) The acid strength will decrease. This is because —NH 2 is an electron-
donating group.

Section C Essay Questions

Question 4
Using the concept of resonance explain the differences in
(a) the acidity between C 2 H 5 OH and C 6 H 5 OH
(b) the basicity between CH 3 NH 2 and C 6 H 5 NH 2
Answer:
(a) Ethanol and phenol act as acids via the following dissociation process:
C 2 H 5 O!H L C 2 H 5 O + H +
–
Ethoxide ion
OH O –
& &
L + H +

Phenol Phenoxide ion Term
The strength of the acids depends on the degree of dissociation. This in
turn depends on the stability of the anions formed. The more stable the
anions, the higher the degree of dissociation. Due to the presence of the 3
benzene ring, the negative charge of the oxygen atom in the phenoxide
ion can delocalise into the benzene ring via resonance:


357




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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry


O – O –


This delocalisation decreases the negative charge of the oxygen atom thus
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stabilising it.
On the other hand, the negative charge of the oxygen atom in the ethoxide
ion cannot undergo such delocalisation making it less stable than the
phenoxide ion.
As a result, phenol is a stronger acid than ethanol.
The delocalisation of the negative charge in the phenoxide ion occurs via
the following resonance structures:
:O – O O O

– –

–
(b) Methylamine and phenylamine are proton acceptors. They are Bronsted-
Lowry bases.
..
+
CH 3 NH 2 + H L CH 3 NH 3 +
..
!NH 2 + H !: !NH 3 +
+
The molecules make use of their lone pair electrons on the nitrogen atom
to form a coordinate bond with H . The strength of the base depends on
+
its readiness to combine with H . The higher the tendency, the stronger
+
the base.
The methyl group in methylamine is an electron-donating group. It
increases the electron density of the lone pair electrons on the nitrogen
atom making them more ready to combine with H . On the other hand,
+
the lone pair electrons in phenylamine can delocalise into the benzene
ring, via resonance, and make them less available to combine with H .
+
H H
N
H H
As a result, methylamine is a stronger base than phenylamine.

Term
3





358




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Chemistry Term 3 STPM Chapter 14 Introduction to Organic Chemistry
The resonance structures of phenylamine is shown below:
+ + +
:NH 2 NH 2 NH 2 NH 2
– –
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–
Question 5

Given the following three acids:
Acid Structural formula
Propanoic acid CH 3 CH 2 COOH
2-Chloropropanoic acid CH 3 CH(Cl)COOH

3-Chloropropanoic acid CH 2 (Cl)CH 2 COOH
(a) Arrange the three acids in order of increasing acid strength. Explain your
answer using the concept of induction.
(b) Which of the above acids has the strongest conjugate base? Explain your
answer.


Answer:
(a) CH 3 CH(Cl)COOH > CH 2 (Cl)CH 2 COOH > CH 3 CH 2 COOH
The chloropropanoic acids are stronger than propanoic acid. This is
because chlorine is an electron-withdrawing group (negative inductive
effect). This helps to weaken the O!H bond in the molecule and increases
its degree of dissociation. 3-Chloropropanoic acid is a weaker acid than
2-chloropropanoic acid because the inductive effect of the chlorine atom
decreases when distance between the chlorine atom and the carboxyl
group increases.
CH 3 —CH—COOH Cl;CH 2 !CH 2 !COOH
p
Cl
(b) CH 3 CH 2 COOH would produce the strongest conjugate base.
A conjugate base is formed when an acid loses a hydrogen ion. For example:
HA L H + A –
+
Acid Conjugate base
If the acid is relatively strong, the above equilibrium will lie more to the Term
right-hand side. Then the tendency of the conjugate base to accept H is
+
low. The conjugate base is weak and vice versa. 3
Hence, the weakest acid would produce the strongest conjugate base.



359




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Jawapan




Kertas Model STPM 962/1
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Section A
1. B 2. D 3. A 4. D 5. B
6. B 7. A 8. B 9. D 10. B
11. C 12. A 13. D 14. C 15. C
Section B
16. (a) Solid Giant Giant Simple
type metallic Giant ionic covalent covalent
van der
Waals and/
Bonding Metallic Ionic Covalent
or Hydrogen
bonds
Sodium Silicon
Example Copper Water
chloride dioxide
(b) (i) Cl N Cl Cl B Cl

Cl Cl
(ii) NCl : Trigon pyramidal
3
BCl : Trigonal planar
3
17. (a) NH ion donates a proton to NH .
–
+
4
2
NH is a Brønsted-Lowry acid and NH is a Brønsted-Lowry base.
–
+
4
2
(b) 39 g NaNH reacts with 50.5 g NH Cl.
4
2
4.12 g will react with 4.12 × 50.5 g NH Cl = 5.33 g NH Cl
39 4 4
The limiting reagent is NaNH .
2
4.12
Volume of NH produced = × 2 × 22.4 dm 3
3 39
= 4.73 dm 3
Section C
18. (a) The assumptions are:
The size of the gas molecules is negligible compared to the volume of
the container where the gas is placed. There is no intermolecular forces
(attractive or repulsive) between the gas molecules.
These assumptions are valid for a real gas only if the gas is under very low
pressure and/or very high temperature.
539


08 Ans STPM Q&A Chem T1 Layout.indd 539 4/19/19 10:48 AM

Chemistry STPM Answers
When the pressure on the gas is high, the molecules are pushed very close
to one another and the volume occupied by the gas is small. In such case
the total volume of the gas molecules cannot be ignored when compared
to the volume of the container where the gas is placed. However, at very
low pressure, the volume occupied by the gas is so large that the total
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volume of the gas molecules is negligible.
At low temperatures, the kinetic energy of the gas molecules is low
and the intermolecular forces between the molecules become significant.
However, at high temperatures, the kinetic energy of the gas molecules is
very high and the intermolecular forces between them can be ignored.
(b) (i) Composition by mass of nitrogen = 100 – 12.5
= 87.5%
87.5 12.5
Mol ratio of N : H = :
14 1
= 6.25 : 12.5
= 1 : 2
Empirical formula of X is NH .
2
Let the molecular formula be (NH )n.
2
Equation of decomposition:
n
(NH )n ➝ N + nH 2
2
2
2
From the equation:
(
12.5 cm of X would produce 12.5 n + n ) cm product
3
3
2
(
12.5 n + n ) = 37.5
2
n = 2
Molecular formula of X is N H .
2
4
(ii) The oxidation number of H in X = +1.
Let the oxidation number of N = y.
2y + (1 × 4) = 0
y = –2
(iii) Lewis diagram of X is:
H H
H N N H
(c) Toluene is a hydrocarbon. The intermolecular forces are the van der
Waal’s forces.
Ethanol on the other hand contains the highly polar O—H bond.
CH — CH — O — H
3 2
The intermolecular forces are the van der Waal’s forces and hydrogen
bonds.


540




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