qdoc.tips_nilam-publication-module-chemistry-form-4-answer

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CONTENTS

KANDUNGAN

1 THE STRUCTURE OF ATOMS 1
STRUKTUR ATOM 22
49
2 CHEMICAL FORMULA AND EQUATIONS 72
FORMULA DAN PERSAMAAN KIMIA 88
114
3 PERIODIC TABLE 139
JADUAL BERKALA 168

4 CHEMICAL BOND
IKATAN KIMIA

5 ELECTROCHEMISTRY
ELEKTROKIMIA

6 ACID AND BASES
ASID DAN BES

7 SALT
GARAM

8 MANUFACTURED SUBSTANCES IN INDUSTRY
BAHAN KIMIA DALAM INDUSTRI

m Publicat Chemistry Form 4 • MODULE hd.

1 THE STRUCTURE OF ATOMS
STRUKTUR ATOM

MATTER / JIRIM
• PARTICLE THEORY OF MATTER / TEORI ZARAH JIRIM

– To state the particle theory of matter
Menyatakan teori zarah jirim

– To differentiate and draw the three types of particles i.e. atom, ion and molecule
Membezakan dan melukis tiga jenis zarah jirim iaitu atom, ion dan molekul

– To describe the laboratory activity to investigate the diffusion of particles in gas, a liquid and a solid. (To prove that matter is
made up of tiny and discrete particles)
Menghuraikan aktiviti makmal untuk mengkaji resapan zarah dalam gas, cecair dan pepejal (Untuk membuktikan bahawa jirim terdiri daripada
zarah-zarah yang halus dan diskrit)

• KINETIC THEORY OF MATTER / TEORI KINETIK JIRIM
– To state the kinetic theory of matter

Menyatakan teori kinetik jirim
– To relate the change of physical states of matters with energy change

Menghubungkaitkan perubahan keadaan jirim dengan perubahan tenaga
– To relate the change of energy in the particles with kinetic particle theory of matter

Menghubungkaitkan perubahan tenaga dalam zarah dengan perubahan tenaga kinetik zarah

THE STRUCTURE OF ATOMS / STRUKTUR ATOM
• HISTORY OF ATOMIC MODELS DEVELOPMENT / SEJARAH PERKEMBANGAN MODEL ATOM

– To state the contribution of scientists in the atomic structure model such as the scientists who discovered electron, proton,
nucleus, neutron and shell
Menyatakan sumbangan ahli sains kepada perkembangan model struktur atom dan ahli sains yang menemui elektron, proton, nukleus, neutron dan
petala

• SUBATOMIC PARTICLES / ZARAH-ZARAH SUBATOM
– To compare and differentiate subatomic particles i.e. proton, neutron and electron from the aspect of charge, relative mass
and location

Membanding dan membezakan zarah-zarah subatom iaitu proton, neutron dan elektron dari segi cas, jisim relatif dan kedudukan
– To state the meaning of proton number and nucleon number based on the subatomic particle

Menyatakan maksud nombor proton dan nombor nukleon berdasarkan zarah subatom
– To write the symbol of elements with proton number and nucleon number

Menulis simbol unsur yang mengandungi nombor proton dan nombor nukleon

• ISOTOPE / ISOTOP
– To state the meaning, examples and the use of isotopes

Menyatakan maksud isotop, contoh-contoh isotop dan kegunaan isotop

• ELECTRON ARRANGEMENT / SUSUNAN ELEKTRON
– To know the number of electron shells and number of electrons in the 1st, 2nd and 3rd shell

Mengetahui bilangan petala elektron serta bilangan elektron yang diisi dalam petala 1, 2 dan 3
– To write the electron arrangement of atoms based on proton number or number of electrons and state the number of valence

electron
Menulis susunan elektron bagi suatu atom berdasarkan nombor proton atau bilangan elektron dan seterusnya menyatakan bilangan elektron valens

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MODULE • Chemistry Form 4

MATTER / JIRIM

Matter is any substance that has mass and occupies space.
Jirim adalah sebarang bahan yang mempunyai jisim dan memenuhi ruang.

The Particle Theory of Matter / Teori Zarah Jirim
1 Matter is made up of tiny and discrete particles. Three types of tiny particles are atoms , ions and molecules .

Jisim terdiri daripada zarah yang halus dan diskrit. Tiga jenis zarah tersebut ialah atom , ion dan molekul .
2 Matter can be classified as element or compound. / Jirim boleh dikelaskan sebagai unsur atau sebatian.
3 Complete the following: / Lengkapkan yang berikut:

MATTER / JIRIM

ELEMENT / UNSUR COMPOUND / SEBATIAN
A substance made from only satu type of atom. A substance made from two or more different
elements which are bonded together.
Bahan yang terdiri daripada satu jenis atom sahaja. Bahan yang terdiri daripada dua atau lebih

unsur berbeza yang terikat secara kimia.

Types of particles / Jenis zarah Types of particles / Jenis zarah

Atom / Atom Molecule / Molekul Molecule / Molekul Ion / Ion
The smallest neutral particle A neutral particle consists A neutral particle consists Positively or negatively
of an element (Normally pure of similar non-metal atoms of different non-metal atoms charged particles, which
metals, noble gases and a which are covalently-bonded. which are covalently-bonded. are formed from metal
few non-metal elements such Zarah neutral terdiri daripada Zarah neutral terdiri daripada atom and non-metal atom
as carbon and silicon). atom-atom bukan logam serupa atom-atom bukan logam berlainan respectively. The force of
Zarah neutral yang paling kecil terikat secara ikatan kovalen. terikat secara ikatan kovalen. attraction between the two
bagi suatu unsur (Biasanya logam oppositely charged ions
tulen, gas adi dan beberapa unsur Example: Example: forms an ionic bond.
bukan logam seperti karbon dan Contoh: Contoh: Zarah bercas positif atau negatif
silikon). terbentuk dari logam dan bukan
Oxygen gas, O2 Carbon dioxide gas, CO2 logam terikat secara ikatan ion.
Example: Gas oksigen, O2 Gas karbon dioksida, CO2 Daya tarikan antara dua ion yang
Contoh: berlawanan cas membentuk ikatan
Sodium metal, Na OO OO OCO ion.
Logam natrium, Na OCO
OO Example:
Na Na Na Na Na OCO Contoh:
Na Na Na Na Na Na Hydrogen gas, H2
Gas hidrogen, H2 Water, H2O Sodium chloride, NaCl
Na Na Na Na Na Air, H2O Natrium klorida, NaCl
HH HH
Neon gas, Ne HOH Na+ Cl – Na+ Cl – Na+
Gas Neon, Ne HH HOH
Cl – Na+ Cl – Na + Cl –
Ne Ne HOH
Na+ Cl – Na+ Cl – Na+
Ne
Calcium oxide, CaO
Kalsium oksida, CaO

Ca2+ O 2– Ca2+ O 2– Ca2+

O 2– Ca2+ O 2– Ca2+ O 2–

Ca2+ O 2– Ca2+ O 2– Ca2+

Nila tion Sdn.– Elements can be identified as metal or non-metal by referring to the Periodic Table.
Unsur boleh dikenal pasti sebagai logam atau bukan logam dengan merujuk kepada Jadual Berkala Unsur.

– Formation of molecule and ion will be studied in Chapter 4 (Chemical Bond).
Pembentukan molekul atau ion akan dipelajari dalam Tajuk 4 (Ikatan Kimia).

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Chemistry Form 4 • MODULE

4 Determine the type of particles in the following substances:
Tentukan jenis zarah bagi setiap bahan berikut:

Substances Type of particle Substances Type of particle Substances Type of particle
Bahan Jenis zarah Bahan Jenis zarah Bahan Jenis zarah

Hydrogen gas (H2) Molecule Sulphur dioxide Molecule TeTtertarachklloorroommeettahnaan(eC (CCCl4l)4) Molecule
Gas hidrogen (H2) Ion Sulfu(rSdOio2)ksida
Atom
Copper(II) sulphate (SO2)
(CuSO4)
Iron (Fe) Atom ZinkZ cihnlkokrildoeri d(ZanCl2) Ion
Kuprum(II) sulfat Ferum (Fe) (ZnCl2 )
(CuSO4 )
Carbon (C) Atom HHiyddrrooggeennp peerorokxsiiddae ((HH22OO22)) Molecule
Argon (Ar) Karbon (C)
Argon (Ar)

5 Diffusion
Resapan
(a) The tiny and discrete particles that made up matter are constantly moving. In gases, these particles are very far apart
from each other, in liquids, the particles are closer together and in solids, they are arranged closely packed.
Jirim terdiri daripada zarah-zarah halus dan diskrit yang sentiasa bergerak. Dalam gas, susunan zarah-zarahnya adalah berjauhan

antara satu sama lain, dalam cecair, zarah-zarahnya disusun lebih rapat dan dalam pepejal, zarah-zarahnya disusun dengan sangat

padat dan teratur.
(b) Diffusion occurs when particles of a substance move between the particles of another substance.

Resapan berlaku apabila zarah-zarah suatu bahan bergerak di antara zarah-zarah bahan lain.
(c) Diffusion occurs in a solid, liquid and gas. Complete the following table:

Resapan berlaku dalam pepejal, cecair dan gas. Lengkapkan jadual berikut:

Diffusion in a gas Diffusion in a liquid Diffusion in a solid
Resapan dalam gas Resapan dalam cecair Resapan dalam pepejal

Experiment A few drops After few Water After a Gel After a
Eksperimen of bromine minutes Air few hours Agar-agar day
liquid Selepas Copper(II)
Beberapa titis beberapa Selepas sulphate Selepas
cecair bromin beberapa jam Kuprum(II) sulfat sehari
minit

Potassium manganate(VII)
Kalium manganat(VII)

The brown colour of bromine vapour, The purple colour of solid potassium The blue colour of copper(II) sulphate,
CuSO4 spreads very slowly
Br2 spreads far throughout masnlgoawnlayte(V tIIh)r, oKuMghnoOu4 ts tphree awdas ter. throughout the gel.
the two jars.
Observation Warna ungu pepejal kalium manganat(VII), Warna biru kuprum(II) sulfat,
Pemerhatian Warna perang wap bromin, Br2 merebak perlahan CuSO4 merebak sangat perlahan
dengan cepat memenuhi kedua-dua KMnO4 merebak dengan di dalam agar-agar.
Explanation di dalam air.
Penerangan balang gas.

uBpro omf ine vmaoploecuur,l eBsr2 an. d air are made Potassium manganate(VII) is uCopp opfe cr(oIpI)p seur(lIpI)h ate,i oCnusSO4 iasn md ade

Wap bromin, Br2 dan udara terdiri made up of potassium ions and sulphate ions . The ions
molekul . manganate(VII) ions. The ions
daripada diffuse slowly between close diffuse very slow between

Bromine molecules diffuse space of water particles which is in closely packed space of gel particles

quickly between large liquid form. which is in solid form.

space of air particles which is in gas Kalium manganat(VII) terdiri daripada Kuprum(II) sulfat, CuSO4 terdiri daripada
ion kuprum(II) dan ion
form. ion kalium dan ion manganat(VII).
Ion-ion ini meresap perlahan
Molekul bromin meresap pantas rapat sulfat. Ion-ion ini meresap dengan
antara ruang zarah air sangat perlahan antara ruang
melalui ruang besar antara zarah-
zarah udara yang berbentuk gas. yang berbentuk cecair. padat zarah agar-agar yang

berbentuk pepejal.

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MODULE • Chemistry Form 4

(d) Conclusions:
Kesimpulan:

(i) Diffusion occurs faster in gas than in liquid. There is larger space in between the particles of

a gas than a liquid. Particles in a gas are further apart. The particles in the liquid

are closer together.

Resapan berlaku lebih cepat di dalam gas berbanding di dalam cecair. Terdapat ruang yang lebih besar

antara zarah-zarah gas berbanding dengan cecair. Zarah-zarah gas adalah berjauhan

antara satu sama lain. Zarah-zarah cecair adalah lebih rapat antara satu sama lain.

(ii) Diffusion occurs faster in a liquid than in solid. There is larger space in between the particles

of a liquid than a solid. The particles in the solid are very close together.

Resapan berlaku lebih cepat di dalam cecair berbanding di dalam pepejal. Terdapat ruang yang lebih besar

antara zarah-zarah cecair berbanding dengan pepejal. Zarah-zarah pepejal tersusun sangat rapat

dan padat antara satu sama lain.

(iii) Bromine gas, potassium manganate(VII) and copper(II) sulphate are made up of tiny and discrete
dan diskrit
particles that are constantly moving/constant motion .

Gas bromin, kalium manganat(VII) dan kuprum(II) sulfat terdiri daripada zarah-zarah halus

yang sentiasa bergerak .

The Kinetic Theory of Matter / Teori Kinetik Jirim

1 Matter exists in three different states which are solid , liquid and gas .

Jirim wujud dalam tiga keadaan iaitu pepejal , cecair dan gas .

2 Matter that made up of tiny and discrete particles which are always in constantly moving .
Jirim terdiri daripada zarah-zarah
halus dan diskrit yang sentiasa bergerak .

3 As the temperature increases, the kinetic energy of particles increases and the particles move faster .
Apabila suhu meningkat, tenaga kinetik zarah-zarah akan bertambah dan zarah-zarah akan bergerak dengan lebih cepat .

4 Particles in different states of matter have different arrangement, strength of forces between them, movement and
energy content.

Zarah-zarah dalam keadaan jirim yang berbeza mempunyai susunan, daya tarikan antara zarah, pergerakan dan kandungan

tenaga yang berbeza.

5 Complete the following table: / Lengkapkan jadual di bawah:

State of matter Solid Liquid Gas
Keadaan jirim Pepejal Cecair Gas

Draw the particles arrangement.
Each particle (atom/ ion/
molecule) is represented by
Lukis susunan zarah. Setiap zarah

(atom / ion / molekul) diwakili dengan

‘’

Particles arrangement The particles are arranged The particles are arranged The particles are very
Susunan zarah closely packed in closely packed but not in widely separated from
orderly manner. orderly manner .
Particles movement each other.
Pergerakan zarah Zarah-zarah tersusun padat Zarah-zarah tersusun padat Zarah-zarah terpisah jauh
dan teratur . tetapi tidak teratur . antara satu sama lain.
m Publica
Particles can only vibrate Particles can vibrate , Particles can vibrate ,
4 and rotate about their rotate and move rotate and move
fixed position.
Zarah bergetar dan berputar throughout the liquid. freely.
Zarah bergetar , berputar dan Zarah bergetar , berputar dan
pada kedudukan tetap.
bergerak dalam cecair. bergerak bebas.

Nila tion Sdn.

Bhd.

Chemistry Form 4 • MODULE

Attractive forces between the Very strong forces Strong forces between the Weak forces between
particles between the particles. particles but weaker than the perticles
Daya tarikan antara zarah Daya tarikan yang sangat kuat the forces in the solid. Daya tarikan yang lemah
antara zarah-zarah. Daya tarikan yang kuat
Energy content of the particles antara zarah-zarah.
Kandungan tenaga zarah Energy content is very low . antara zarah-zarah tetapi
Kandungan tenaga sangat lebih lemah berbanding di

rendah . dalam pepejal.

Energy content is higher Energy content is very
than solid but less than in a high.
gas. Kandungan tenaga sangat
Kandungan tenaga lebih tinggi
tinggi.
daripada pepejal tetapi
lebih rendah daripada gas.

6 Changes in the state of matter heat energy is absorbed or released/lose :
Perubahan keadaan jirim

(a) Matter undergoes change of state when

Jirim mengalami perubahan keadaan apabila tenaga haba di serap atau di bebaskan :

(i) When heat energy is absorbed by the matter (it is heated), the kinetic energy of the particles
kinetik zarah bertambah
increases and they vibrate faster.

Apabila tenaga haba diserap oleh jirim (semasa dipanaskan), tenaga

dan zarah tersebut bergerak dengan lebih cepat.

(ii) When matter releases heat energy (it is cooled), the kinetic energy of the particles decreases and

they vibrate less vigorously.

Apabila tenaga haba dibebaskan oleh jirim (semasa disejukkan), tenaga kinetik zarah berkurang dan

zarah tersebut bergerak kurang cergas.

(b) Inter - conversion of the states of matter:
Perubahan keadaan jirim:

Solid Melting / Peleburan Liquid Boiling/Evoporation / Pendidihan/Penyejatan Gas
Pepejal Freezing / Pembekuan Cecair Condensation / Kondensasi Gas

7 Determination of melting and freezing points of naphthalene
Penentuan takat lebur dan takat beku naftalena

Materials / Bahan: Naphthalene powder, water

Apparatus / Radas: Boiling tube, conical flask, beaker, retort stand, thermometer 0 – 100°C, stopwatch,
Bunsen burner and wire gauze

Procedure / Prosedur:

I. Heating of naphthalene / Pemanasan naftalena
Set-up of apparatus: / Susunan radas:

Thermometer / Termometer

Boiling tube / Tabung didih
Water / Air
Naphthalene / Naftalena

Heat
Haba

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MODULE • Chemistry Form 4

(a) A boiling tube is filled 3 - 5 cm height with naphthalene powder and a thermometer is
placed into it.
Tabung didih diisi dengan serbuk naftalena setinggi 3 – 5 cm dan termometer diletakkan
di dalamnya.

(b) The boiling tube is immersed in a water bath as shown in the diagram so that the water level in the water bath
is higher than naphtalene powder in the boiling tube.
Tabung didih dimasukkan ke dalam kukus air seperti di dalam gambar rajah dan pastikan aras air dalam kukus air lebih tinggi

daripada aras naftalena dalam tabung didih.

(c) The water is heated and the naphthalene is stirred slowly with thermometer .

Air dipanaskan dan naftalena dikacau perlahan-lahan dengan termometer .

(d) When the temperature of naphthalene reaches 60°C , the stopwatch is started. The temperature of
naphthalene is recorded at 30 seconds intervals until the temperature of naphthalene reaches 90°C .

Apabila suhu naftalena mencapai 60°C , mulakan jam randik. Suhu naftalena dicatat setiap 30 saat

sehingga suhunya mencapai 90°C .

II. Cooling of naphthalene / Penyejukan naftalena

Naphthalene
Naftalena

(a) The boiling tube and its content is removed from the water bath and put into a conical flask as shown

in the diagram. kelalang kon seperti
Tabung didih dan kandungannya dikeluarkan daripada kukus air dan dipindahkan ke dalam

dalam gambar rajah.

(b) The content in the boiling tube is stirred constantly with thermometer throughout cooling

process to avoid supercooling (the temperature of cooling liquid drops below freezing point, without

the appearance of a solid).

Kandungan dalam tabung didih dikacau perlahan-lahan dengan termometer sepanjang proses penyejukan untuk

mengelakkan penyejukan lampau (Suhu cecair yang disejukkan turun melepasi takat beku tanpa pembentukan
pepejal).

(c) The temperature of naphthalene is recorded every 30 seconds interval until the temperature drops

to 60°C .

Suhu naftalena dicatat setiap 30 saat sehingga suhunya mencapai 60°C .

(d) A graph of temperature against time is plotted for the heating and cooling process respectively.
Graf suhu melawan masa dilukis untuk proses pemanasan dan penyejukan.

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Chemistry Form 4 • MODULE

The Explanation of the Heating Process of Matter / Penerangan Proses Pemanasan

1 The heating curve of naphthalene:
Lengkung pemanasan naftalena:

Temperature/°C
Suhu/°C

F

DE

BC

A

Time/s
Masa/s

2 When a solid is heated, the particles absorb heat and move faster as its energy content increases. As the heat

energy is absorbed , the state of matter will change.

Apabila pepejal dipanaskan, zarah-zarah menyerap haba dan bergerak lebih cepat disebabkan kandungan tenaga bertambah.

Tenaga haba diserap menyebabkan perubahan keadaan jirim.

Point State of Matter Explanation
Titik Keadaan jirim Penerangan
A to B Solid
A ke B Heat energy is absorbed by the particles in the solid naphthalene causing their
Solid and kinetic energy to
B to C Liquid increase and vibrate faster . The temperature
B ke C
Liquid increases.
C to D
C ke D Liquid and Tenaga haba diserap oleh zarah-zarah pepejal naftalena menyebabkan tenaga kinetik akan
Gas bertambah dan zarah bergetar dengan lebih cepat . Suhu semakin meningkat .
D to E
D ke E Gas Heat energy absorbed by the particles in the liquid naphthalene is used to
overcome solid . The
E to F forces between particles so that the turn to liquid
E ke F temperature
remains constant .

Tenaga haba yang diserap oleh zarah-zarah dalam pepejal naftalena digunakan untuk mengatasi

daya tarikan antara zarah-zarah supaya pepejal berubah menjadi cecair . Suhu adalah tetap .

Heat energy absorbed by the particles in the liquid naphthalene causing their

kinetic energy to increase and move faster . The temperature

increases .

Tenaga haba diserap oleh zarah-zarah cecair naftalena menyebabkan tenaga kinetik akan

bertambah dan zarah-zarah bergerak dengan lebih cepat . Suhu semakin meningkat .

Heat energy absorbed by the particles in the liquid naphthalene is used to
overcome
the forces of attraction between particles. The particles begin to move freely
to form a
gas . The temperature remains constant .

Tenaga haba diserap oleh zarah-zarah dalam cecair naftalena digunakan untuk mengatasi

daya tarikan antara zarah-zarah. Zarah-zarah mula bergerak bebas untuk membentuk gas . Suhu

adalah tetap .

Heat energy is absorbed by the particles in the gas causing their kinetic
increases .
energy to incerease and move faster . The temperature

Tenaga haba diserap oleh zarah-zarah gas naftalena menyebabkan tenaga kinetik akan bertambah

dan zarah-zarah bergerak dengan lebih cepat . Suhu semakin meningkat . ion Sdn. B

7m Publicat hd.

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MODULE • Chemistry Form 4

3 The constant temperature at which a solid completely changes to become a liquid is called the melting point .

During the melting process, the temperature remains unchanged because heat energy absorbed by the particles

is used to overcome the forces between particles so that the solid change to turn into a liquid .

Suhu tetap di mana suatu pepejal berubah kepada keadaan cecair dipanggil takat lebur .

Semasa proses peleburan, suhu tidak berubah kerana haba yang diserap oleh zarah-zarah digunakan untuk

mengatasi daya tarikan antara zarah supaya pepejal berubah menjadi cecair .

4 The constant temperature at which a liquid completely changes to become a gas is called the boiling point .

During the boiling process, the temperature remains unchanged because heat energy absorbed by the particles

is used to overcome the forces between particles so that the liquid change to turn into a gas.

Suhu tetap di mana suatu bahan dalam keadaan cecair berubah kepada keadaan gas dipanggil takat didih .

Semasa proses pendidihan, suhu tidak berubah kerana haba yang diserap oleh zarah-zarah digunakan untuk

mengatasi daya tarikan antara zarah supaya cecair berubah menjadi gas.

The Explanation for the Cooling Process of Matter: / Penerangan Proses Penyejukan Bahan:

1 The cooling curve of naphthalene:
Lengkung penyejukan naftalena:
Temperature/°C

Suhu/°C

P

QR

S

Time/s
Masa/s

2 When the liquid is cooled, the particles in the liquid release energy and move slower as its energy content

decreases. As the energy is released to the surrounding, the state of matter will change.

Apabila cecair disejukkan, zarah cecair membebaskan tenaga dan bergerak semakin perlahan. Keadaan jirim

berubah semasa tenaga dibebaskan ke persekitaran.

Point State of matter Explanation
Titik Keadaan jirim Penerangan

P to Q Liquid Heat is released/given out to the surrounding by the particles in the liquid naphthalene.
P ke Q The particles in the liquid lose their kinetic energy and move slower. The
temperature decreases .

Haba dibebaskan ke persekitaran oleh zarah-zarah dalam cecair naftalena. Zarah-zarah dalam
cecair kehilangan tenaga kinetik dan bergerak semakin perlahan. Suhu semakin menurun .

The heat released to the surrounding by the particles in liquid naphthalene is balanced
by the heat energy released as the particles attract one another to form a solid .

Q to R Liquid and The temperature remains constant .
Q ke R Solid
Haba dibebaskan ke persekitaran oleh zarah-zarah dalam cecair naftalena diimbangi oleh

tenaga haba yang terbebas apabila zarah-zarah tertarik antara satu sama lain untuk membentuk

pepejal . Suhu adalah tetap .

The particles in the solid naphthalene releases heat and vibrate slower . The temperature
lebih perlahan .
R to S Solid decreases .
R ke S
Zarah-zarah dalam pepejal naftalena membebaskan tenaga dan bergetar dengan

Suhu semakin menurun .

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Chemistry Form 4 • MODULE

3 The constant temperature at which a liquid changes to a solid is called freezing point . During the freezing

process, the temperature remains unchanged because the heat released to the surrounding is balanced by the

heat released when the liquid particles rearrange themselves to become a solid .

Suhu tetap di mana suatu cecair berubah kepada keadaan pepejal dipanggil takat beku . Semasa proses

pembekuan, suhu tidak berubah kerana haba yang dibebaskan ke persekitaran diimbangi oleh haba yang terbebas

apabila zarah-zarah cecair menyusun semula untuk membentuk pepejal .

Keadaan Fizik Bahan pada Sebarang Suhu: / Physical State Of A Substance At Any Given Temperature:

1 A substance is in solid state if the temperature of the substance is below melting point

Suatu bahan berada dalam keadaan pepejal jika suhu bahan tersebut lebih rendah daripada takat leburnya.

2 A substance is in liquid state if the temperature of the substance is between melting and boiling points.

Suatu bahan berada dalam keadaan cecair jika suhu bahan tersebut berada antara takat lebur dan takat didihnya.

3 A substance is in gas state if the temperature of the substance is above boiling point.
Suatu bahan berada dalam keadaan gas jika suhu bahan tersebut lebih tinggi daripada takat didihnya.

EXERCISE / LATIHAN

1 The table below shows substances and their chemical formula.
Jadual di bawah menunjukkan bahan dan formula kimia masing-masing.

Substance / Bahan Chemical formula / Formula kimia Type of particle / Jenis zarah
Silver / Argentum Ag Atom
K2O Ion
Potassium oxide / Kalium oksida NH3
Ammonia / Ammonia Cl2 Molecule
Chlorine / Klorin Molecule

(a) State the type of particles that made up each substance in the table.
Nyatakan jenis zarah yang membentuk bahan dalam jadual di atas.

(b) Which of the substances are element? Explain your answer.
Yang manakah antara bahan tersebut merupakan suatu unsur? Jelaskan jawapan anda.

Silver and chlorine. Silver and chlorine are made up of one type of atom

(c) Which of the substance are compound? Explain your answer.
Yang manakah antara bahan tersebut merupakan suatu sebatian? Jelaskan jawapan anda.

Potassium oxide and ammonia. Potassium oxide and ammonia are made up of two different elements

2 The table below shows the melting and boiling points of substance P, Q and R.
Jadual di bawah menunjukkan takat lebur dan takat didih bagi bahan P, Q dan R.

Substance / Bahan Melting point / Takat lebur / °C Boiling point / Takat didih / °C
P –36 6
Q –18 70
R 98
230

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MODULE • Chemistry Form 4

(a) (i) What is meant by ‘melting point’?
Apakah yang dimaksudkan dengan ‘takat lebur’?
The constant temperature at which a solid charges to a liquid at particular pressure

(ii) What is meant by ‘boiling point’?
Apakah yang dimaksudkan dengan ‘takat didih’?
The constant temperature at which a liquid changes to a gas at particular pressure

(b) Draw the particles arrangement of substances P, Q and R at room condition.
Lukis susunan zarah P, Q dan R pada keadaan bilik.

Substance P / Bahan P Substance Q / Bahan Q Substance R / Bahan R

(c) (i) What is the substance that exist in the form of liquid at 0°C.
Nyatakan bahan yang wujud dalam keadaan cecair pada suhu 0°C.

P, Q

(ii) Give reason to your answer.
Jelaskan jawapan anda.

The temperature 0°C is above the melting point of Q and below the boiling point of Q

(d) (i) Substance Q is heated from room temperature to 100°C. Sketch a graph of temperature against time for the
heating of substance Q.
Bahan Q dipanaskan dari suhu bilik hingga 100°C. Lakarkan graf suhu melawan masa bagi pemanasan bahan Q terhadap masa

untuk pemanasan bahan Q.

Temperature/°C

70

Time/s

(ii) What is the state of matter of substance Q at 70°C?
Apakah keadaan fizik bahan Q pada 70°C?
Liquid and gas

(e) Compare the melting point of substances Q and R. Explain your answer.
Bandingkan takat lebur bahan Q dan R. Terangkan jawapan anda.
The melting point of substance R is higher than subtance Q. The attraction force between particles in substance R
is stronger than Q. More heat is needed to overcome the force between particles in substance R.

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Chemistry Form 4 • MODULE

3 The melting point of acetamide can be determined by heating solid acetamide until it melts as shown in the diagram
below. The temperature of acetemide is recorded every three minutes when it is left to cool down at room temperature.
Takat lebur asetamida boleh ditentukan dengan memanaskan pepejal asetamida sehingga lebur seperti dalam rajah di bawah. Suhu asetamida
dicatatkan setiap tiga minit semasa disejukkan pada suhu bilik.

Thermometer / Termometer
Boiling tube / Tabung didih

Water / Air
Acetamide / Asetamida

(a) What is the purpose of using water bath in the experiment?
Apakah tujuan menggunakan kukus air dalam eksperimen ini?
To ensure even heating of acetemide. Acetamide is easily combustible.

(b) State the name of another substance which its melting point can also be determined by using water bath as shown
in the above diagram.
Namakan satu bahan lain yang mana takat leburnya boleh ditentukan dengan menggunakan kukus air seperti rajah di atas.
Naphthalene

(c) Sodium nitrate has a melting point of 310°C. Can the melting point of sodium nitrate be determined by using the
water bath as shown in the diagram? Explain your answer.
Natrium nitrat mempunyai takat lebur 310°C. Bolehkah takat lebur natrium nitrat ditentukan dengan menggunakan kukus air seperti
yang ditunjukkan dalam rajah di atas? Jelaskan jawapan anda.
No, because the melting point of water is 100°C which is less than the melting point of sodium nitrate.

(d) Why do we need to stir the acetemide in the boiling tube in above experiment?
Mengapakah asetamida dalam tabung didih itu perlu dikacau sepanjang eksperimen?
To make sure the heat is distributed evenly

(e) The graph of temperature against time for the cooling of liquid acetamide is shown below.
Rajah di bawah menunjukkan graf suhu melawan masa untuk penyejukan cecair asetamida.

Temperature / Suhu/ °C

T3 R
T2 Q

T1 Time / Masa/s

(i) What is the freezing point of acetamide?
Apakah takat beku asetamida?
T2°C

(ii) The temperature between Q and R is constant. Explain.
Suhu antara titik Q dan R adalah tetap. Jelaskan.
The heat lost to the surrounding is balanced by the heat released when the liquid particles

rearrange themselves to become solid.

(f) Acetemide exists as molecules. State the name of another compound that is made up of molecules.
Asetamida wujud sebagai molekul. Namakan sebatian lain yang terdiri daripada molekul.
Water/naphthalene

(g) What is the melting point of acetamide?
Apakah takat lebur asetamida?
T2°C

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MODULE • Chemistry Form 4

The Atomic Structure / Struktur Atom

1 History of the development of atomic models:
Sejarah perkembangan model atom:

Scientist Atomic Model Discovery
Saintis Model atom Penemuan

Dalton (i) Matter is made up of particles called atoms . .
Jirim terdiri daripada zarah-zarah dipanggil atom .

(ii) Atoms cannot be created , destroyed or divided
Atom tidak boleh dicipta , dimusnah atau dibahagi .

(iii) Atoms from the same element are identical .
Atom daripada unsur sama adalah sama .

Thomson Positively charged sphere (i) Discovered the electrons , the first subatomic particle.
Rutherford Sfera bercas positif Menjumpai elektron , zarah subatom yang pertama.
Neils Bohr
Electron charges negative (ii) Atom is sphere of positive charge which embedded with
Elektron bercas negatif
negatively charged particles called electrons .
Atom adalah sfera yang bercas positif yang mengandungi zarah
bercas negatif dipanggil elektron .

Electron moves (i) Discovered the nucleus as the centre of an atom and
outside the nucleus
positively charged .
Elektron bergerak di luar Menjumpai nukleus yang merupakan pusat bagi atom dan
nukleus
bercas positif .
Nmuecnlgeaunsd tuhnagti
contain proton (ii) Proton is a part of the nucleus.
Nukleus mengandungi
Proton adalah sebahagian daripada nukleus.
proton
(iii) Electron move outside the nucleus.

Elektron bergerak di sekeliling nukleus.

(iv) Most of the mass of the atom found in the nucleus .

Nukleus mempunyai hampir semua jisim atom.

Shell (i) Discovered the existence of electron shells .

Nucleus that Menjumpai kewujudan petala elektron.
contain proton
Nukleus mengandungi (ii) Electrons move in the shells around the nucleus.

proton Elektron bergerak di dalam petala mengelilingi nukleus .

Electron

Shell (i) Discovered the existence of neutron .
Menjumpai kewujudan neutron .
Nucleus that contain
proton and neutron (ii) Nucleus of an atom contains neutral particles called

James Nukleus mengandungi neutron and positively charged particles called
Chadwick proton dan neutron
proton .
m Publica Electron
Nukleus mengandungi zarah-zarah neutral dipanggil neutron dan
12
zarah-zarah bercas positif dipanggil proton .

(iii) The mass of a neutron and proton is almost the same.

Jisim neutron dan proton adalah hampir sama.

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Chemistry Form 4 • MODULE

2 The structure of an atom: / Struktur Atom:

Shell / Petala

Nucleus that contain proton and neutron
Nukleus yang mengandungi proton dan neutron

Electron / Elektron

(a) An atom has a central nucleus and electrons that move in the shells around the nucleus.
Atom mempunyai nukleus di tengahnya dan elektron bergerak di dalam
petala mengelilingi nukleus tersebut.

(b) The nucleus contains protons and neutrons.

Nukleus mengandungi proton dan neutron.

(c) Each proton has charge of +1 . Each electron has an electrical charge of –1 . The neutron has no

charge (it is neutral ). An atom has the same number of protons and electrons, so the overall charge

of atom is zero . Atom is neutral . (If an atom loses or gains electrons it is called an ion – formation
of ion will be studied in Chapter 4)

Setiap proton bercas +1 . Setiap elektron bercas –1 . Neutron tidak mempunyai cas (ianya adalah neutral ).

Setiap atom mempunyai bilangan proton dan elektron yang sama, oleh itu cas keseluruhan bagi atom adalah sifar . Atom

adalah neutral . (Suatu atom akan membentuk ion apabila ia kehilangan atau menerima elektron – pembentukan ion akan

dipelajari dalam Tajuk 4.)

(d) The relative mass of a neutron and a proton which are in the nucleus is 1. The mass of an atom is obtained mainly

from the number of proton and neutron .

Jisim relatif proton dan neutron di dalam nukleus ialah 1. Jisim suatu atom diperoleh daripada jumlah bilangan proton

dan bilangan neutron .

(e) The mass of an electron can be ignored as the mass of an electron is about 1 times the size of a proton or
neutron. 1 840

Jisim elektron boleh diabaikan kerana ia terlalu kecil iaitu 1 daripada jisim proton dan neutron.
1 840

3 Complete the following table:
Lengkapkan jadual di bawah:

Subatomic particles Symbol Charge Relative atomic mass Position
Zarah subatom Simbol Cas Jisim atom relatif Kedudukan
1 8140 = 0 In the shells
Electron/Elektron e – (negative) 1 In the nucleus
Proton/Proton p + (positive)

Neutron/Neutron n neutral 1 In the nucleus

4 Atom is the smallest neutral particle of an element.
Atom adalah zarah neutral paling kecil dalam suatu unsur.

Complete the following diagram: / Lengkapkan yang berikut:

Na Na Na Na Na
Na Na Na Na Na

Na Na Na Na

Sodium element Sodium element Sodium element Sodium atom
Unsur natrium Unsur natrium Unsur natrium Atom natrium

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MODULE • Chemistry Form 4

5 Proton number of an element (Refer to Periodic table of an element)
Nombor proton sesuatu unsur (Rujuk Jadual Berkala Unsur)

(a) Proton number of an element is the number of proton in its atom .

Nombor proton sesuatu unsur adalah bilangan proton yang terdapat dalam atom .

(b) The number of proton of an atom is also equal to the number of electrons in the atom because atom is neutral .
neutral .
Bilangan proton sesuatu atom adalah sama dengan bilangan elektron dalam atom kerana atom adalah

(c) Every element has its own proton number:

Setiap unsur mempunyai nombor protonnya tersendiri:

– Proton number of potassium, K is 19. Potasium atom has 19 protons in the nucleus and 19 electrons

in the shells. Atom 19 proton

Nombor proton untuk kalium, K ialah 19. kalium mempunyai di dalam nukleus dan

19 elektron di dalam petala.

– Proton number of oxygen, O is 8. Oxygen atom has 8 protons in the nucleus and 8 electrons

in the shells. Atom 8 proton

Nombor proton untuk oksigen, O ialah 8. oksigen mempunyai di dalam nukleus dan

8 elektron di dalam petala.

6 Nucleon number of an element (Refer to Periodic table of an element)
Nombor nukleon sesuatu unsur (Rujuk Jadual Berkala Unsur)

(a) Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom .
.
Nombor nukleon sesuatu unsur adalah jumlah bilangan proton dan neutron di dalam nukleus sesuatu atom

(b) Nucleon number is also known as a mass number.

Nombor nukleon juga dikenali sebagai nombor jisim.

(c) Nucleon number = number of proton + number of neutron.

Nombor nukleon = bilangan proton + bilangan neutron.

Symbol of Element And Standard Representation For An Atom of Element
Simbol Unsur dan Perwakilan Piawai bagi Atom Sesuatu Unsur

1 The symbol of an element is a short way of representing an element. If the symbol has only one letter, it must be a capital
letter. If it has two letters, the first is always a capital letter, while the second is always a small letter.
Simbol unsur adalah cara mudah untuk mewakilkan unsur. Jika simbol hanya terdiri daripada satu huruf, maka ia mesti ditulis dengan huruf
besar. Tetapi jika simbol terdiri daripada dua huruf, maka huruf pertama merupakan huruf besar dan huruf kedua merupakan huruf kecil.

Example: / Contoh:

Element Symbol Element Symbol Element Symbol
Unsur Simbol Unsur Simbol Unsur Simbol

Oxygen/Oksigen O Nitrogen/Nitrogen N Calcium/Kalsium Ca

Magnesium/Magnesium Mg Sodium/Natrium Na Copper/Kuprum Cu

Hydrogen/Hidrogen H Potassium/Kalium K Chlorine/Klorin Cl



The first letter of each element is capitalised to show that it is a new element. This is helpful when writing a chemical formula.

For example KCl. There are two elements chemically bonded in KCl because there are two capital letters represent potassium and

chlorine.

Huruf yang pertama bagi setiap unsur ditulis dengan huruf besar untuk menunjukkan ia adalah unsur yang baru. Ini sangat berguna semasa menulis formula
kimia. Contohnya KCl. Terdapat dua unsur yang terikat secara kimia dalam KCl kerana adanya dua huruf besar yang mewakili kalium dan klorin.

2 Standard representation symbol represents one atom of an element. It can be written as:
Simbol perwakilan piawai mewakili satu atom sesuatu unsur. Ianya boleh ditulis sebagai:

Nucleon number/Nombor nukleon AX Symbol of an element/Simbol unsur
Proton number/Nombor proton
Z

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Example: / Contoh:

27 A1
13


– The element is Aluminium.

Unsur itu adalah Aluminium.

– The nucleon number of Aluminium is 27 .

Nombor nukleon Aluminium adalah 27 .

– The proton number of Aluminium is 13 .

Nombor proton Aluminium adalah 13 .

– Aluminium has 13 protons , 14 neutrons and 13 electrons.

Atom Aluminium mempunyai 13 proton , 14 neutron dan 13 elektron.

3 Isotope / Isotop
(a) Isotopes are atoms of the same element with same number of protons but different number of neutrons.

Isotop ialah atom-atom unsur yang mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza.

Or / Atau proton number but different nucleon number.
proton
Isotopes are atoms of the same element with same yang sama tetapi nombor nukleon yang

Isotop ialah atom-atom unsur yang mempunyai nombor
berbeza.
Example: / Contoh:

1 H 2 H
1 1

Nucleon number/Nombor nukleon = 1 Nucleon number/Nombor nukleon = 2

Proton number/Nombor proton = 1 Proton number/Nombor proton = 1

Number of neutron/Bilangan neutron = 0 Number of neutron/Bilangan neutron = 1

– Hydrogen-1 and Hydrogen-2 are isotopes. Hydrogen-1 and Hydrogen-2 atoms have the same proton number or the same

number of protons but different in nucleon number because of the difference in the number of neutron .

AtomHidrogen-1 dan Hidrogen-2 mempunyai nombor proton atau bilangan proton yang sama tetapi nombor nukleon yang berbeza

kerana perbezaan bilangan neutron .

– Isotopes have the same chemical properties but different physical properties because they have the same electron

arrangements.

Isotop mempunyai sifat kimia yang sama kerana mempunyai susunan elektron yang sama tetapi sifat fizik yang berbeza.

(b) Examples of the usage of isotopes:

Contoh kegunaan isotop:

i. Medical field

Bidang perubatan
– To detect brain cancer.

Untuk mengesan barah otak.
– To detect thrombosis (blockage in blood vessel).

Untuk mengesan trombosis (saluran darah tersumbat).
– Sodium-24 is used to measure the rate of iodine absorption by thyroid gland.

Untuk mengukur kadar penyerapan iodin oleh kelenjar tiroid. Contoh: Natrium-24
– Cobalt-60 is used to destroy cancer cells.

Untuk memusnahkan sel barah. Contoh: Kobalt-60
– To kill microorganism in the sterilising process.

Untuk membunuh mikroorganisma semasa proses pensterilan.

ii. In the industrial field

Bidang industri
– To detect wearing out in machines.

Untuk mengesan kehausan enjin.
– To detect any blockage in water, gas or oil pipes.

Untuk mengesan saluran paip air, gas atau minyak yang tersumbat.

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MODULE • Chemistry Form 4

– To detect leakage of pipes underground.
Untuk mengesan kebocoran paip bawah tanah.

– To detect defects/cracks in the body of an aeroplane.
Untuk mengesan keretakan atau kecacatan pada badan kapal terbang.

iii. In the agriculture field
Bidang pertanian
– To detect the rate of absorption of phosphate fertilizer in plants.

Untuk mengesan kadar penyerapan baja fosfat oleh tumbuhan.
– To sterile insect pests for plants.

Untuk memandulkan serangga perosak tumbuhan.

iv. In the archeology field
Bidang arkeologi
– Carbon-14 can be used to estimate the age of artifacts.

Karbon-14 untuk menentukan usia sesuatu artifak.

4 Electron Arrangement
Susunan elektron
(a) The electrons are filled in specific shells. Every shell can be filled only with a certain number of electrons. For the
elements with atomic numbers 1-20:
Elektron diisi dalam petala tertentu. Setiap petala hanya boleh diisi dengan bilangan elektron tertentu. Bagi unsur-unsur yang

mempunyai nombor proton 1–20:

– First shell can be filled with a maximum of 2 electrons.
Petala pertama boleh diisi dengan bilangan maksimum 2 elektron.

– Second shell can be filled with a maximum of 8 electrons.

Petala kedua boleh diisi dengan bilangan maksimum 8 elektron.

– Third shell can be filled with a maximum of 8 electrons.
Petala ketiga boleh diisi dengan bilangan maksimum 8 elektron.

First shell is filled with 2 electrons (duplet)
Petala pertama diisi 2 elektron (duplet)

Second shell is filled with 8 electrons (octet)
Petala kedua diisi 8 elektron (oktet)

Third shell is filled with 8 electrons (octet)
Petala ketiga disi 8 elektron (oktet)

(b) Valence electrons are the electrons in the outermost shell of an atom.
Elektron valens: Elektron yang diisi dalam petala paling luar suatu atom.

5 Complete the following table:
Lengkapkan jadual berikut:
(a) Draw the electron arrangement and complete the description for each element:
Lukis susunan elektron bagi atom dan penerangan bagi setiap unsur berikut:

Standard Electron arrangement Description
representation of an atom Penerangan
of an element
Perwakilan piawai Lukiskan susunan elektron

unsur bagi atom

Hydrogen Atom Number of protons/Bilangan proton 1
Atom Hidrogen Number of eletrons/Bilangan elektron 1
0
1 H Number of neutrons/Bilangan neutron 1
1 H 1
1
Proton number/Nombor proton

Nucleon number/Nombor nukleon

Electron Arrangement/Susunan elektron

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23 Na Sodium Atom Number of protons/Bilangan proton Chemistry Form 4 • MODULE
11 Atom Natrium Number of electrons/Bilangan elektron
Number of neutrons/Bilangan neutron 11
Na Proton number/Nombor proton 11
Nucleon number/Nombor nukleon 12
Electron Arrangement/Susunan elektron 11
23
2.8.1

(b) Choose the correct statement for the symbol of element X.
Pilih pernyataan yang betul bagi simbol unsur X.

23 Na
11

Statement Tick ( 3 / 7 )
Pernyataan Tanda ( 3 / 7 )

Element X has 11 proton number. 7
Unsur X mempunyai 11 nombor proton. 3
3
The proton number of element X is 11. 7
Nombor proton unsur X ialah 11. 3
3
The proton number of atom X is 11. 3
Nombor proton atom X ialah 11. 7
7
The number of proton of element X is 11. 7
Bilangan proton unsur X ialah 11. 3
7
The number of proton of atom X is 11.
Bilangan proton atom X ialah 11.

Nucleon number of element X is 23.
Nombor nukleon unsur X ialah 23.

Nucleon number of atom X is 23.
Nombor nukleon atom X ialah 23.

Number of nucleon of element X is 23.
Bilangan nukleon unsur X ialah 23.

Atom X has 23 nucleon number.
Atom X mempunyai 23 nombor nukleon.

Neutron number of atom X is 12.
Nombor neutron atom X ialah 12.

Number of neutron of atom X is 12.
Bilangan neutron atom X ialah 12.

Number of neutron of element X is 12.
Bilangan neutron unsur X ialah 12.

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MODULE • Chemistry Form 4

EXERCISE / LATIHAN

1 Complete the following table:
Lengkapkan jadual berikut:

Element Symbol of Number of Number of Number of Proton Nucleon Electron Number
Unsur element proton electron neutron number number arrangement of valence
Simbol unsur Bilangan Bilangan Bilangan Nombor Nombor
proton elektron neutron proton nukleon Susunan electron
Bilangan
1 1 0 1 1 elektron atom
elektron valens
2 2 2 2
Hydrogen 1 H 11
Hidrogen 1 5 5 6 5

Helium 4 He 6 6 6 6 422
Helium 2
7 7 7 7
Boron 11 B 11 2.3 3
Boron 5 10 10 10 10

Carbon 12 C 11 11 12 11 12 2.4 4
Karbon 6
12 12 12 12
Nitrogen 14 N 14 2.5 5
Nitrogen 7 20 20 20 20

Neon 20 Ne 20 2.8 8
Neon 10

Sodium 23 Na 23 2.8.1 1
Natrium 11

Magnesium 24 Mg 24 2.8.2 2
Magnesium 12

Calcium 40 Ca 40 2.8.8.2 2
Kalsium 20

2 The diagram below shows the symbol of atoms P, R and S.
Rajah di bawah menunjukkan simbol atom P, R dan S.

35 P 12 R 37 S
17 6 17

(a) What is meant by nucleon number / Apakah maksud nombor nukleon?
Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom

(b) What is the nucleon number of P / Apakah nombor nukleon atom P?
35

(c) State the number of neutron in atom P / Nyatakan bilangan neutron atom P.
18

(d) State number of proton in atom P / Nyatakan bilangan proton atom P.
17

(e) (i) What is meant by isotope / Apakah maksud isotop?
Isotopes are atoms of the same element with same number of proton but different number of neutrons

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Chemistry Form 4 • MODULE

(ii) State a pair of isotope in the diagram shown / Nyatakan sepasang isotop dalam rajah yang ditunjukkan.
P and S

(iii) Give reason for your answer in (e)(ii) / Berikan sebab bagi jawapan di (e)(ii).
Atom P and S have same proton number but different nucleon number//number of neutron

(f) An isotope of R has 8 neutron. Write the symbol for the isotope R.
Isotop bagi atom R mempunyai 8 neutron. Tuliskan simbol bagi isotop R.

14 R
6

3 The table below shows the number of proton and neutron of atoms of elements P, Q and R.
Jadual di bawah menunjukkan bilangan proton dan neutron bagi atom unsur P, Q dan R.

Element Number of proton Number of neutron
Unsur Bilangan proton Bilangan neutron

P 1 0

Q 1 1

R 6 6

(a) Which of the atoms in the above table are isotope? Explain your answer.
Berdasarkan jadual di atas, atom yang manakah merupakan isotop? Terangkan jawapan anda.

P and Q. Atom P and Q have same number of proton but different number of neutron // nucleon number.

(b) (i) Write the standard representation of element Q.
Tuliskan perwakilan piawai untuk unsur Q.

2 Q
1

(ii) State three information that can be deduced from your answer in (b)(i).
Nyatakan tiga maklumat yang boleh didapati daripada jawapan anda di (b)(i).

The proton number of element Q is 1 // Number of proton of atom Q is 1

Nucleon number of element Q is 2 // Atomic mass of atom Q is 2

Number of neutron of atom Q is 1

Nucleus of atom Q contains 1p and 1n

(c) (i) Draw atomic structure for atom of element R.
Lukiskan struktur atom bagi atom unsur R.

6 protons + 6 neutrons

(ii) Describe the atomic structure in (c)(i). m Publicat hd.
Huraikan struktur atom di (c)(i).
– The atom consists of 2 parts: the centre part called nucleus and the outer part called electron shell.

– The nucleus consists of 6 protons which are positively charged and 6 neutrons which are neutral.

– The electrons are in two shells, the first shell consists of two electrons and the second shell consists of

four electrons.

– Electrons move around nucleus in the shells.

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MODULE • Chemistry Form 4

(d) Element R react with oxygen and to produce liquid Z at room temperature. The graph below shows the sketch of the
graph when liquid Z at room temperature, 27°C is cooled to –5°C.
Unsur R bertindak balas dengan oksigen dan menghasilkan cecair Z pada suhu bilik. Rajah di bawah menunjukkan lakaran graf
apabila cecair Z pada suhu bilik, 27°C disejukkan sehingga –5°C.
Temperature /°C

Suhu /°C

0 Time /s
t1 t2 Masa /s

−5

(i) tW1 htoa tt 2i.s the state of matter of liquid Z from t1 to t2? Explain why is the temperature remain unchanged from
Apakah keadaan jirim Z daripada t1 hingga t2? Terangkan mengapa suhu tidak berubah daripada t1 hingga t2.

Liquid and solid. Heat lost to the surrounding is balanced by the heat released when the particles at 0 °C

(ii) Draw the arrangement of particles of Z at 20°C.
Lukiskan susunan zarah-zarah Z pada suhu 20°C.

(iii) Describe the change in the particles movement when Z is cooled from room temperature to –5°C.
Nyatakan perubahan dalam pergerakan zarah-zarah apabila cecair Z disejukkan daripada suhu bilik ke –5°C.

The particles move slower

Objective Questions / Soalan Objektif

1 The diagram shows the arrangement of particles for a type 3 The diagram below shows the heating curve for substance X.

of matter that undergoes a change in physical state through Rajah di bawah menunjukkan lengkung pemanasan bahan X.

process X. Temperature / Suhu °C
Rajah di bawah menunjukkan susunan zarah sejenis bahan yang
U
mengalami perubahan keadaan fizik melalui proses X.

S
QT
XR

P Time (m)
Masa (m)

What is process X? Which region of the graph does boiling process occur?
Apakah proses X ?
A Melting Bahagian manakah pada graf berlaku proses pendidihan?

Peleburan C Freezing A PQ C ST
Pembekuan B QR D TU

B Boiling D Sublimation 4 Which of the following information is true?
Pendidihan Pemejalwapan Antara pernyataan berikut, yang manakah adalah betul?

2 Which of the following substances can undergo sublimation Change of state Process Heat energy
Perubahan keadaan Proses Tenaga haba
when heated?
Released
Antara bahan berikut, yang manakah mengalami pemejalwapan apabila A Solid → Liquid Melting Dibebaskan
Pepejal → Cecair Peleburan
dipanaskan?

A Sulphur C Glucose
Sulfur Glukosa
B Liquid → Gas Evaporation Released
B Ammonium chloride D Sodium chloride Cecair → Gas Penyejatan Dibebaskan
Ammonium klorida Natrium klorida
C Gas → Solid Sublimation Released

Gas → Pepejal Pemejalwapan Dibebaskan

Nila m Publication Sdn. D Gas → Liquid Condensation Absorbed
Gas → Cecair Kondensasi Diserap
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Chemistry Form 4 • MODULE

5 The diagram below shows the graph of temperature against Substance Melting point/°C Boiling point/°C
time when a liquid Y is cooled. Bahan Takat lebur/°C Takat didih/°C
Rajah di bawah menunjukkan graf suhu melawan masa apabila cecair Y
disejukkan. S –182 –162

Temperature / Suhu °C T –23 77

t3 P U –97 65
V 41 182

t2 Q R W 132 290

Which substance exists as liquid at room temperature?

t1 S Time (m) Bahan yang manakah wujud sebagai cecair pada suhu bilik?
Masa (m)
A S only C T and U only
Which of the following statements are true about the curve? S sahaja T dan U sahaja
Antara pernyataan berikut, yang manakah adalah betul tentang lengkung
B S and T only D V and W only
S dan T sahaja V dan W sahaja

itu? 8 The diagram below shows standard representation of an atom
I At Q, liquid Y begins to freeze. copper.

Pada Q, cecair Y mula membeku. Rajah di bawah menunjukkan perwakilan piawai atom kuprum.
II At PQ, particles in Y absorb heat from the surroundings.
64
Pada PQ, zarah dalam Y menyerap haba dari persekitaran. 29 Cu
III Liquid Y freezes completely at S.

Cecair Y membeku dengan lengkap pada S. Which of the following is correct based on the symbol the
diagram?
IV The freezing point of Y is t2°C.
Takat beku bagi Y adalah t2C°C. Antara berikut, yang manakah betul berdasarkan rajah di atas?
I and III only
A II and III only

I dan III sahaja II dan III sahaja Proton number Nucleon number Number of electron
B I and IV only D II and IV only Nombor proton Nombor nukleon Bilangan elektron

I dan IV sahaja II dan IV sahaja

6 The diagram below shows the graph of temperature against A 29 64 29
time when solid Z is heated. B 35 29 64

Rajah di bawah menunjukkan graf suhu melawan masa apabila pepejal Z C 64 35 29
dipanaskan. D 29 64 35

Temperature / Suhu °C

9 The diagram below shows the standard representation of

beryllium atom.

80 Rajah di bawah menunjukkan perwakilan piawai atom berillium.

9 Be
4

Time (m) What is the number of valence electrons of beryllium atom?
Masa (m)
01 2 3 4 5 6 7 8 9 Apakah bilangan elektron valens bagi atom berillium?

A 2 C 4
B 3 D7
Which of the following is true during the fourth minute?

Antara berikut, yang manakah adalah benar pada minit keempat? 10 The table below shows the proton number and the number of
neutrons for atoms of elements W, X, Y and Z.
A All the molecules are in random motion. Jadual di bawah menunjukkan nombor proton dan bilangan neutron bagi
Semua molekul bergerak secara rawak.
atom unsur W, X, Y dan Z.
B All the molecules are closely packed and in random
motion. Element Proton number Number of neutrons
Atom Nombor proton Bilangan neutron
Semua molekul sangat rapat dan bergerak secara rawak.
C All the molecules are vibrating at fixed positions. W 7 7

Semua molekul bergetar pada kedudukan tetap. X 8 8
D Some of the molecules are vibrating at fixed positions but
some are in random motion. Y8 9

Sebahagian molekul bergetar pada kedudukan tetap dan Z9 10

sebahagian bergerak secara rawak. Which of the following pair of elements is isotope?

7 The table shows the melting points and boiling points of Antara pasangan berikut, yang manakah adalah isotop?
substances S, T, U, V and W.
Jadual di bawah menunjukkan takat lebur dan takat didih bahan S, T, U, A W and X C X and Y
W dan X X dan Y
V dan W.
B W and Y D Y and Z
W dan Y Y dan Z

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Nila tion Sdn.MODULE • Chemistry Form 4

2 CHEMICAL FORMULA AND EQUATIONS
FORMULA DAN PERSAMAAN KIMIA

RELATIF MASS / JISIM RELATIF
• RELATIVE ATOMIC MASS / JISIM ATOM RELATIF (JAR)

– To state the meaning of relative mass and solve numerical problems
Menyatakan maksud jisim atom relatif dan menyelesaikan masalah pengiraan

• RELATIVE FORMULA MASS / JISIM FORMULA RELATIF (JFR)
– To state the meaning of RAM, RMM and RFM based on carbon-12 scale

Menyatakan maksud JAR, JMR dan JFR berdasarkan skala karbon-12

• RELATIVE MOLECULAR MASS / JISIM MOLEKUL RELATIF (JMR)
– To calculate RAM, RMM and RFM using the chemical formulae of various substances

Menghitung JAR, JMR dan JFR menggunakan formula kimia beberapa bahan

MOLE CONCEPT / KONSEP MOL
• MOLE AND THE NUMBER OF PARTICLES / MOL DAN BILANGAN ZARAH

– To solve numerical problems involving mole and the number of atoms/ ions/ molecules
Menyelesaikan masalah pengiraan melibatkan mol dan bilangan atom, ion dan molekul

• MOLE AND THE MASS OF SUBSTANCES / MOL DAN JISIM BAHAN
– To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole
concept

Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol

• MOLE AND THE VOLUME OF GAS / MOL DAN ISIPADU GAS
– To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole
concept

Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol

CHEMICAL FORMULA AND EQUATIONS / FORMULA DAN PERSAMAAN KIMIA
• EMPIRICAL FORMULA / FORMULA EMPIRIK

– Stating the purpose and describe the empirical formula laboratory activities to determine the formula empirical
Menyatakan maksud formula empirik dan menghuraikan aktiviti makmal untuk menentukan formula empirik

• MOLECULAR FORMULA / FORMULA MOLEKUL
– Solve calculation problems involving empirical formula

Menyelesaikan masalah pengiraan melibatkan formula empirik

• CHEMICAL FORMULAE / FORMULA KIMIA
– To write formula of anion and cation and to write chemical formula for ionic compounds

Menulis formula kation dan anion dan menulis formula kimia untuk sebatian ion

• CHEMICAL EQUATIONS / PERSAMAAN KIMIA
– Write a balanced chemical equation and solve problems arrangements involving the mole concept

Menulis persamaan kimia seimbang dan menyelesaikan masalah pengiraan yang melibatkan konsep mol

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Chemistry Form 4 • MODULE

RELATIVE ATOMIC MASS (RAM) / JISIM ATOM RELATIF (JAR)

1 A single atom is too small and light and cannot be weighed directly.
Satu atom adalah terlalu ringan, kecil dan tidak dapat ditimbang secara langsung.

2 The best way to determine the mass of a single atom is to compare its mass to the mass of another atom of an element
that is used as a standard.
Cara yang paling sesuai untuk menentukan jisim satu atom ialah dengan membandingkan jisimnya dengan jisim suatu atom unsur lain
yang dianggap sebagai piawai.

3 Hydrogen was the first element to be chosen as the standard for comparing mass because the hydrogen atom is the
lightest atom with a mass of 1.0 a.m.u (atomic mass unit).
Hidrogen adalah unsur pertama dipilih sebagai piawai untuk membandingkan jisim kerana atom hidrogen adalah unsur yang paling
ringan dengan jisim 1.0 u.j.a (unit jisim atom).
Example:
Contoh:

• The mass of one helium atom is four times larger than one hydrogen atom.
Jisim satu atom Helium adalah 4 kali lebih besar daripada satu atom hidrogen.

• RAM for He is 4.
JAR untuk He ialah 4.

4 On the hydrogen scale, the relative atomic mass of an element means the mass of one atom of the element compared to
the mass of a single hydrogen atom:

Pada skala hidrogen, jisim atom relatif suatu unsur ditakrifkan sebagai jisim satu atom unsur berbanding jisim satu atom hidrogen:

Relative atomic mass of an element (RAM) / Jism atom relatif suatu unsur (JAR)

= The average mass of one atom of the element / Jisim purata satu atom unsur
Mass of one hydrogen atom / Jisim satu atom hidrogen

• RAM has no unit.
JAR tiada unit.

• The new standard used today is the carbon-12 atom.

Piawai yang digunakan sekarang adalah berdasarkan atom karbon-12.
• JRaAtAoRMmb eboradf ascesaadrr bkooannn t-sh1ka2el :acaartbomonk-a1r2b osnc-a1l2e aidsa tlhahe jmisiamsss aotuf oatnoem autnosmur obfe rtbhaen deilnegmdeenntg acon m112pajirseidm wsaittuha t1o12m
of the mass of an
karbon-12:

• Relative atomic mass of an element (RAM) / Jisim atom relatif suatu unsur (JAR)

= The average mass on one atom of the element / Jisim purata satu atom unsur

1 × The mass of an atom of carbon-12 / Jisim satu atom karbon-12
12

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MODULE • Chemistry Form 4

RELATIVE MOLECULAR MASS (RMM) / RELATIVE FORMULA MASS (RFM)
JISIM MOLEKUL RELATIF (JMR) / JISIM FORMULA RELATIF (JFR)

1 RMM / JMR = The average mass on one atom of the element / Jisim purata satu molekul

1 × The mass of an atom of carbon-12 / Jisim satu atom karbon-12
12

2 RMM is obtained by adding up the RAM of all the atoms that are present in the molecule.
JMR diperoleh dengan menambahkan JAR semua atom yang terdapat dalam satu molekul.

Molecular substance Molecular formula Relative molecular mass
Bahan molekul Formula molekul Jisim molekul relatif

Oxygen / Oksigen O2 2 × 16 = 32

Water / Air H2O 2 × 1 + 16 = 18

Carbon dioxide / Karbon dioksida CO2 12 + 2 × 16 = 44

Ammonia / Ammonia NH3 14 + 3 × 1 = 17

[Relative atomic mass / Jisim atom relatif : O = 16, H = 1, C = 12, N = 14] Relative molecular mass
Jisim formula relatif
3 For ionic substances, RMM is replaced with Relative Formula Mass (RFM). 23 + 35.5 = 58.5
Untuk sebatian ion, JMR digantikan dengan Jisim Formula Relatif (JFR).

Substance Chemical formula
Bahan Formula kimia

Sodium chloride / Natrium klorida NaCl

Potassium oxide / Kalium oksida K2O 2 × 39 + 16 = 94

Copper(II) sulphate / Kuprum(II) sulfat CuSO4 64 + 32 + 4 × 16 = 160

Ammonium carbonate / Ammonium karbonat (NH4)2CO3 2 [14 + 4 × 1] + 12 + 3 × 16 = 96

Aluminium nitrate / Aluminium nitrat Al(NO3)3 27 + 3 [14 + 3 × 16] = 213

Calcium hydroxide / Kalsium hidroksida Ca(OH)2 40 + 2 [16 + 1] = 74

Lead(II) hydroxide / Plumbum(II) hidroksida Pb(OH)2 207 + 2 [16 + 1] = 241

Hydrated copper(II) sulphate / Kuprum(II) sulfat terhidrat CuSO45H2O 64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250

[Relative atomic mass / Jisim atom relatif : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12, Al = 27,
Ca = 40, Pb = 207]

(i) mTheet aflo Mrm?ula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of

Oksida logam M mempunyai formula M2O3. Jisim formula relatif ialah 152. Apakah jisim atom relatif logam M?

M = RAM for M

2 M + 3 × 16 = 152

M = 52

(ii) Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x.
Fosforus membentuk sebatian klorida dengan formula PClx. Jisim molekul relatifnya adalah 208.5. Hitungkan nilai x.

[Relative atomic mass / Jisim atom relatif : P = 31, Cl = 35.5]

31 + x × 35.5 = 208.5
35.5x = 208.5 – 31
35.5x = 177.5
x = 5


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Chemistry Form 4 • MODULE

MOLE CONCEPT / KONSEP MOL

Mole and the Number of Particles / Bilangan Mol dan Bilangan Zarah

1 To describe the amount of atoms, ions or molecules, mole is used.
Untuk menyatakan jumlah atom, ion atau molekul, unit mol digunakan.

2 A mole is an amount of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12.
Satu mol ialah jumlah bahan yang mengandungi bilangan zarah seperti mana yang terdapat dalam 12 g atom karbon-12.

3 A mole of a substance is the amount of substance which contains a constant number of particles (atoms, ions,
molecules), which is 6.02 × 1023.
Satu mol bahan adalah jumlah bahan yang mengandungi bilangan zarah yang tetap (atom, molekul, ion) iaitu 6.02 × 1023.

4 The number 6.02 × 1023 is called the Avogadro Constant or Avogadro Number (NA).
Nombor 6.02 × 1023 dikenali sebagai Pemalar Avogadro atau Nombor Avogadro (NA ).

5 For compounds that exist as molecules/ions, the number of atoms/ions in that compound must be known.
Bagi sebatian yang wujud dalam bentuk molekul/ion, bilangan atom/ion dalam sebatian itu mestilah diketahui.

6 The symbol of mole is mol.
Simbol untuk mol ialah mol.

7 Complete the following table:
Lengkapkan jadual berikut:

Substance Formula Type of Model / Figure Number of atom per
Bahan Formula particles Model / Rajah molecule/ Number of
Jenis zarah positive and negative ion
Bilangan atom per molekul/
Bilangan ion positif dan negatif

Chlorine / Klorin Cl2 Molecule Cl Cl Cl : 2

Water / Air H2O Molecule HOH H : 2
O : 1

Ammonia / Ammonia NH3 Molecule H N : 1
HNH H : 3

Sulphur dioxide / Sulfur dioksida SO2 Molecule OSO S : 1
O : 2

Magnesium chloride / Magnesium klorida MgCl2 Ion [Cl]– [Mg]2+ [Cl]– Mg2+ : 1
Cl– : 2

Aluminium oxide / Aluminium oksida Al2O3 Ion [O]2– [A1]3+ [O]2– [A1]3+ [O]2– Al3+ : 2

O2– : 3

8 Relationship between number of moles and number of particles (atoms/ions/molecules):
Hubungan bilangan mol dan bilangan zarah (atom/ion/molekul):

Number of moles × Avogadro Constant / Pemalar Avogadro Number of particles
Bilangan mol ÷ Avogadro Constant / Pemalar Avogadro Bilangan zarah

9 Complete the following: [Differentiate between “mole” dan “molecule”]
Lengkapkan yang berikut: [Bezakan antara “mol” dan “molekul”]

(a) 1[C mhloolr ionfe C gl2as] 6.02 × 1023 molecules of chlorine, Cl2 / molekul klorin, Cl2
2 × 6.02 × 1023 atoms of chlorine, Cl / atom klorin, Cl

1 mol Cl2
[Gas klorin]

(b) 1[A mmoml oonf iNa Hg3as] 6.02 × 1023 molecules of ammonia, NH3 / molekul ammonia, NH3
1 mol of nitrogen atom, N / mol atom nitrogen, N
4 mol atoms / mol atom
1 mol NH3
[Gas ammonia] 3 mol of hydrogen atoms, H / mol atom hidrogen, H

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MODULE • Chemistry Form 4

(c) 1 mol of NH3 0.25 × 6.02 × 1023 molecules of ammonia, NH3 / molekul ammonia, NH3
4 0.25 mol of N atoms / mol atom N,
[Ammonia gas] 1 mol of atoms number of N atoms / bilangan atom N = 0.25 × 6.02 × 1023
1 mol atom 0.75 mol of H atoms / mol atom H,
1 mol NH3 number of H atoms / bilangan atom H = 0.75 × 6.02 × 1023
4

[Gas ammonia]

(d) 2[M maognl oesf iMumgC cl2hloride] 2 mol of Mg2+ ions / mol ion Mg2+,
number of Mg2+ ions / bilangan ion Mg2+ = 2 × 6.02 × 1023

2 mol MgCl2 4 mol of Cl– ions / mol ion Cl–,
[Magnesium klorida] number of Cl- ions / bilangan ion Cl– = 4 × 6.02 × 1023

(e) 2[S muloplh oufr SdOio2xide] 2 × 6.02 × 1023 molecules of SO2 / molekul SO2
2 mol of S atoms / mol atom S,

2 mol SO2 3 × 2 = 6 mol of atoms number of S atoms / bilangan atom S = 2 × 6.02 × 1023
[Sulfur dioksida]

3 × 2 = 6 mol atom 4 mol of O atoms / mol atom O,
number of O atoms / bilangan atom O = 4 × 6.02 × 1023

10 Complete the table below:
Lengkapkan jadual berikut:

Number of moles Number of particles
Bilangan mol Bilangan zarah

0.5 mole of carbon, C 3.01 × 1023 atoms of carbon
0.5 mol atom karbon, C
0.2 moles of hydrogen gas, H2 3.01 × 1023 atom karbon
0.2 mol gas hidrogen, H2
(i) 0.2 × 6.02 × 1023 molecules of hydrogen / molekul hidrogen
(ii) 2 × 0.2 × 6.02 × 1023 atoms of hydrogen / atom hidrogen

1 mole of carbon dioxide molecules, CO2 6.02 × 1023 molecules of carbon dioxide contains:
1 mol molekul karbon dioksida, CO2
6.02 × 1023 molekul karbon dioksida mengandungi:

6.02 × 1023 atoms of C and 2 × 6.02 × 1023 atoms of O.

6.02 × 1023 atom C dan 2 × 6.02 × 1023 atom O.

NUMBER OF MOLES AND MASS OF SUBSTANCE / BILANGAN MOL DAN JISIM BAHAN

1 Molar mass / Jisim molar
(a) Molar mass is the mass of one mole of any substance / Jisim molar adalah jisim satu mol sebarang bahan.
(b) Molar Mass is the relative atomic mass, relative molecular mass and relative formula mass of a substance in
g mol–1.

Jisim molar adalah jisim atom relatif, jisim molekul relatif dan jisim formula relatif suatu bahan dalam g mol–1.

(c) Molar mass of any substance is numerically equal to its relative mass (Relative atomic mass/ relative formula
mass/relative molecular mass).

Jisim molar sebarang bahan mempunyai nilai yang sama dengan jisim relatif (Jisim atom relatif/ jisim formula relatif/ jisim molekul

relatif).

2 Example / Contoh:

Molar mass of H2O = 18 g mol–1
Jisim molar H2O = 18 g mol–1
Mass of 1 mol of H2O = 18 g × RAM/ /RFM/RMM
Jisim 1 mol H2O = 18 g Number of × JAR/JFR/JMR Mass in gram
Mass of 2 mol of H2O = 2 mol × 18 g mol–1 = 36 g moles Jisim dalam gram
÷ RAM/ /RFM/RMM
Jisim 2 mol H2O = 2 mol × 18 g mol–1 = 36 g Bilangan mol ÷ JAR/JFR/JMR

Mass of 2.5 mol of H2O = 45 g
Jisim 2.5 mol H2O = 45 g
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Chemistry Form 4 • MODULE

3 Complete the following table:
Lengkapkan jadual berikut:

Element/ Chemical RAM/RMM/RFM Calculate
Compound formula JAR/JMR/JFR Penghitungan
Unsur/Sebatian Formula kimia
Copper RAM/JAR = 64 (a) Mass of 1 mol of Cu / Jisim 1 mol Cu : 1 mol × 64 g mol–1 = 64 g
Kuprum Cu
RFM/JFR = 40 (b) Jisim 2 mol / Jisim 1 mol : 2 mol × 64 g mol–1 = 128 g
Sodium hydroxide NaOH
Natrium hidroksida (c) Jisim 1 mol / Jisim 1 mol: 1 mol × 64 g mol–1 = 32 g
2 2 2

(d) Mass of 3.01 × 1023 Cu atoms / Jisim 3.01 × 1023 atom Cu: 32 g

(a) Mass of 3 mol of sodium hydroxide: 120 g

Jisim 3 mol natrium hidroksida: 120 g
(b) Number of moles in 20 g sodium hydroxide: 0.5 mol

Bilangan mol natrium hidroksida dalam 20 g: 0.5 mol

Oxygen gas O2 RMM/JMR = 32 (a) Mass of 2.5 mol of oxygen gas: 2.5 mol × 32 g mol–1 = 80 g
Gas oksigen
Jisim 2.5 mol gas oksigen: 2.5 mol × 32 g mol–1 = 80 g

(b) Number of moles is 1.5 mol oxygen gas:

Bilangan molekul dalam 1.5 mol gas oksigen:

1.5 mol × 6.02 × 1023
BNiulamngbaenr moof lmekoulledcaulalems i21n m21o lmgaosl ookfs iogxeny:gen gas:
(c)


0.5 mol × 6.02 × 1023

(d) Number of atoms in 2 mol of oxygen gas:

Bilangan atom dalam 2 mol gas oksigen:
2 × 2 × 6.02 × 1023

Sodium chloride NaCl RFM/JFR = 58.5 Mass of 0.5 mol of NaCl / Jisim bagi 0.5 mol NaCl:
Natrium klorida 0.5 mol × 58.5 g mol–1 = 29.25 g

Zinc nitrate Zn(NO3)2 RFM/JFR = 189 Number of moles in 37.8 g of zinc nitrate:
Zink nitrat Bilangan mol dalam 37.8 g zink nitrat:

37.8 g/189 g mol–1 = 0.2 mol

[Relative atomic mass / Jisim atom relatif: Cu = 64, Na = 23, O = 16, H = 1, Cl = 35.5, Zn = 65, N = 14]

NUMBER OF MOLES AND VOLUME OF GAS / BILANGAN MOL DAN ISI PADU GAS m Publicat hd.

1 Molar volume of a gas: Volume occupied by one mole of any gas is 24 dm3 at room conditions and 22.4 dm3 at
standard temperature and pressure (STP).
Isi padu molar gas: Isipadu yang dipenuhi oleh satu mol sebarang gas iaitu 24 dm3 pada keadaan bilik dan 22.4 dm3 pada suhu dan
tekanan piawai (STP).

2 The molar volume of any gas is 24 dm3 at room conditions and 22.4 dm3 at STP.
Isi padu molar sebarang gas adalah 24 dm3 pada keadaan bilik dan 22.4 dm3 pada STP.

3 Generalisation: One mole of any gas always occupies the same volume under the same temperature and pressure:
Umumnya: satu mol sebarang jenis gas menempati isi padu yang sama pada suhu dan tekanan yang sama.
Example / Contoh:
(i) 1 mol of oxygen gas, 1 mol ammonia gas, 1 mol helium gas dan 1 mol sulphur dioxide gas occupy the same
volume of 24 dm3 at room conditions.
1 mol gas oksigen, 1 mol gas ammonia, 1 mol gas helium dan 1 mol gas sulfur dioksida menempati isi padu yang sama iaitu 24 dm3
pada keadaan bilik.
(ii) 2 mol of carbon dioxide gas occupies 44.8 dm3 pada STP.
2 mol gas karbon dioksida menempati 44.8 dm3 pada STP.

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MODULE • Chemistry Form 4

(iii) 16 g of oxygen gas = 0.5 mol of oxygen gas. Therefore, 16 g of oxygen gas occupies a volume of 12 dm3
at room conditions [Relative atomic mass: O =16]

16 g gas oksigen = 0.5 mol gas oksigen. Oleh itu, 16 g gas oksigen menempati isi padu 12 dm3 pada keadaan bilik.
[Jisim atom relatif; O = 16]

Number of moles of gas × 24 dm3 mol–1/ 22.4 dm3 mol–1 Volume of gas in dm2
Bilangan mol gas ÷ 24 dm3 mol–1/ 22.4 dm3 mol–1 Isi padu gas dalam dm3

Formula for conversion of unit:
Formula untuk penukaran unit:

Volume of gas in dm3
Isi padu gas dalam dm3

÷ 24 dm3 mol–1/ 22.4 dm3 mol–1 × 24 dm3 mol–1/ 22.4 dm3 mol–1

Mass in gram (g) ÷ (RAM/ /RFM/RMM) g mol–1 Number of ÷ (6.02 × 1023) Number of particles
Jisim dalam gram (g) ÷ (JAR/JFR/JMR) g mol–1 moles × (6.02 × 1023) Bilangan zarah
Bilangan
× (RAM/ /RFM/RMM) g mol–1
× (JAR/JFR/JMR) g mol–1 mol

EXERCISE / LATIHAN

1 Relative atomic mass of calcium is 40 based on the carbon-12 scale.

Jisim atom relatif kalsium berdasarkan skala karbon-12 ialah 40.

(a) State the meaning of the statement above.

Nyatakan maksud penyataan di atas. 1 mass of carbon-12 atom.
Mass of calcium atom is 4 times greater than 12

(b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16]
Berapa kalikah satu atom kalsium lebih berat daripada satu atom oksigen? [JAR: O = 16]

Relative atomic mass of calcium = 40 = 2.5 times
Relative atomic mass of oxygen 16

(c) How many calcium atoms have the same mass as two atoms of bromine? [RAM Br = 80]
Berapakah bilangan atom kalsium yang mempunyai jisim yang sama dengan dua atom bromin? [Jisim atom relatif: Br = 80]

Number of calcium atom × 40 = 2 × 80
Number of calcium atom = 2 × 80 = 4

40

2 A sampel of chlorine gas weighs 14.2 g. Calculate / Suatu sampel gas klorin berjisim 14.2 g. Hitungkan:
[Relative atomic mass / Jisim atom relatif : Cl = 35.5]
(a) Number of moles of chlorine atoms / Bilangan mol atom klorin.

Number of mol of chlorine atoms, Cl = 14.2 = 0.4 mol
35.5

(b) Number of moles of chlorine molecules (Cl2) / Bilangan mol molekul klorin (Cl2).
14.2
Number of mol of chlorine molecule, Cl2 = 71 = 0.2 mol

(c) Volume of chlorine gas at room conditions / Isi padu gas klorin pada keadaan bilik.
[Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure]

[Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan piawai]

Volume of chlorine gas = 0.2 mol × 24 dm3 mol–1

= 4.8 dm3

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Chemistry Form 4 • MODULE

3 (a) Calculate the number of atoms in the following substances / Hitungkan bilangan atom yang terdapat dalam bahan berikut:
[Relative atomic mass: N = 14; Zn = 65; Avogadro Constant = 6.02 × 1023]

[Jisim atom relatif: N = 14; Zn = 65; Pemalar Avogadro = 6.02 × 1023]

(i) 13 g of zinc / 13 g zink
Number of mol of zinc atom = 13 = 0.2 mol

65

Number of zinc atom = 0.2 × 6.02 × 1023
= 1.204 × 1023

(ii) 5.6 g of nitrogen gas / 5.6 g gas nitrogen
Number of mol of N atom = 5.6 = 0.4 mol

14

Number of N atom = 0.4 × 6.02 × 1023
= 2.408 × 1023

(b) Calculate the number of molecules in the following substances / Hitungkan bilangan molekul dalam bahan berikut:

[Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant = 6.02 × 1023]

[Jisim atom relatif: N = 14, H = 1, Cl = 35.5, Pemalar Avogadro = 6.02 × 1023]

(i) 8.5 g of ammonia gas, NH3 / 8.5 g gas ammonia, NH3
8.5 × 6.02 × 1023
17

= 2.408 × 1023

(ii) 14.2 g of chlorine gas, Cl2 / 14.2 g gas klorin, Cl2
14.2 × 6.02 × 1023
71

= 1.2 × 1023

4 A gas jar contains 240 cm3 of carbon dioxide gas. Calculate:

Suatu balang gas berisi 240 cm3 gas karbon dioksida. Hitungkan:

[Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions]

[Jisim atom relatif: C = 12, O = 16; Isi padu molar gas: 24 dm3 mol–1 pada keadaan bilik]

(a) Number of moles of carbon dioxide gas / Bilangan mol gas karbon dioksida:

Number of moles of CO2 = 240 = 0.01 mol
24 000

(b) Number of molecules of carbon dioxide gas / Bilangan molekul gas karbon dioksida:

Numb er of molecules of CO2 == 60..0021 ×× 160.0221 × 1023


(c) Mass of carbon dioxide gas / Jisim gas karbon dioksida:

Ma ss of CO2 == 00..4041 gmol × [12 + 2 × 16] g mol–1


5 What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as that found in 3.6 g of water? m Publicat hd.
Berapakah jisim molekul klorin (Cl2 ) yang mengandungi dua kali ganda bilangan molekul yang terdapat dalam 3.6 g air?
[Relative atomic mass / Jisim atom relatif : H = 1, O = 16, Cl = 35.5]
Number of moles of chlorine molecule = 2 × no of mol in H2O
= 2 × 3.6 = 0.4 mol
18
Mass of Cl2 = 0.4 × 71= 28.4 g

6 Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium.
Hitungkan jisim karbon yang mempunyai bilangan atom yang sama seperti yang terdapat dalam 4 g magnesium.
[Relative atomic mass / Jisim atom relatif : C = 12, Mg = 24]
2 g

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MODULE • Chemistry Form 4

7 Compare the number of molecule in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your answer.
B[Raenldaintigvkea natboilmanicg amn amsosl e/k Juilsdimalaatmom32reglastuilff:u rSd =iok 3si2d,a O(S =O2 )1d6e,n Nga =n 7 1g4g] as nitrogen (N2 ). Terangkan jawapan anda.

Number of moles of molecules in 32 g SO2 = 32 = 0.5 mol
64

Number of moles of molecules in 7 g N2 = 7 = 0.25 mol
28

Number of molecule in 32 g SO2 is two times more than 7 g N2.
Number of mole in sulphur dioxide molecule is two times more than number of mole of nitrogen molecule.

8 Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer.

Bandingkan bilangan atom dalam 1.28 g oksigen dengan bilangan atom dalam 1.3 g zink. Terangkan jawapan anda.

[Relative atomic mass / Jisim atom relatif : O = 16, Zn = 65]

Number of mol of O atoms in 1.28 g SO2 = 1.28 = 0.08 mol
16

Number of mol of Zn atoms in 1.3 g Zn = 1.30 = 0.04 mol
65

Number of oxygen atoms in 1.28 g oxygen is 2 times more than number of zinc atoms in 1.3 g zinc.

Number of mol of oxygen atom is 2 times more than zinc atom.

CHEMICAL FORMULAE AND CHEMICAL EQUATIONS / FORMULA KIMIA DAN PERSAMAAN KIMIA

1 Symbol of elements – use capital letters for the first alphabet and use small letters if there is a second alphabet.
Simbol unsur – gunakan huruf besar untuk huruf pertama dan huruf kecil jika ada huruf kedua.

Example / Contoh: Potassium / Kalium – K, Sodium / Natrium – Na
Nitrogen / Nitrogen – N
Calcium / Kalsium – Ca, Fluorine / Fluorin – F

Iron / Ferum – Fe,

Chemical Formula – A set of chemical symbols for atoms of elements in whole numbers representing chemical
substances.

Formula kimia – Satu set simbol kimia bagi atom-atom unsur dengan gandaan nombor bulat yang mewakili bahan kimia.

Chemical substance Chemical formula Notes
Bahan kimia Formula kimia Catatan
H2O
Water NH3 2 atoms of H combines with 1 atom of O.
Air C3H8 2 atom H bergabung dengan 1 atom O.

Ammonia 3 atoms of H combines with 1 atom of N.
Ammonia 3 atom H bergabung dengan 1 atom N.

Propane 3 atoms of C combines with 8 atoms of H.
Propana 3 atom C bergabung dengan 8 atom H.

2 Information that can be obtained from the chemical formula / Maklumat yang diperoleh daripada formula kimia:
(i) All the elements present in the compound / Jenis unsur yang terdapat dalam sebatian,
(ii) Number of atoms of each element in the compound / Bilangan atom setiap unsur yang terdapat dalam sebatian,
(iii) Calculation of RMM/RFM of the compound / Pengiraan JMR/JFR bagi sebatian.

3 Two types of chemical formula / Dua jenis formula kimia:
(i) Empirical formula / Formula empirik,
(ii) Molecular formula / Formula molekul.

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Chemistry Form 4 • MODULE

EMPIRICAL FORMULA / FORMULA EMPIRIK

1 A formula that shows the simplest whole number ratio of atoms of each element in a compound.
Formula yang menunjukkan nisbah nombor bulat teringkas bagi bilangan atom setiap unsur yang terdapat dalam sebatian.

2 The formula can be determined by calculating the simplest ratio of moles of atoms of each element in the compound.
Formula itu boleh ditentukan dengan menghitung nisbah bilangan mol atom bagi setiap unsur yang terdapat dalam sebatian.

3 Experiments to determine empirical formula of metal oxide / Formula empirik bagi oksida logam diperoleh dengan cara:

Empirical formula of magnesium oxide Empirical formula of copper(II) oxide
Formula empirik magnesium oksida Formula empirik kuprum(II) oksida

Set-up of apparatus / Susunan radas: Set-up of apparatus / Susunan radas:

Magnesium Copper(II) oxide
Magnesium Kuprum(II) oksida

Hydrogen gas
Gas hidrogen

Heat Heat
Panaskan Panaskan

Reaction occurs / Tindak balas yang berlaku: Reaction occurs / Tindak balas yang berlaku:
Magnesium is burnt in a crucble to react with oxygen to form Hydrogen gas is passed through heated copper(II) oxide.
magnesium oxide. Hydrogen reduces copper(II) oxide to form copper and water.
Magnesium dipanaskan dengan kuat di dalam mangkuk pijar untuk Gas hidrogen dilalukan melalui kuprum(II) oksida yang dipanaskan.

bertindak balas dengan oksigen membentuk magnesium oksida. Hidrogen menurunkan kuprum(II) oksida kepada kuprum dan air.

Balanced equation / Persamaan kimia seimbang: Balanced equation / Persamaan kimia seimbang:
2Mg + O2 → 2MgO CuO + H2 → Cu + H2O

This method can also be used to determine the empirical This method can also be used to determine the empirical
formulae of reactive metals such as aluminium oxide and zinc formulae of less reactive metals such as lead(II) oxide and
oxide. tin(II) oxide.
Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida
logam reaktif seperti aluminium oksida dan zink oksida. logam kurang reaktif seperti plumbum(II) oksida and stanum(II) oksida.

4 Experiment to Determine Empirical Formula of Magnesium Oxide
Eksperimen untuk Menentukan Formula Empirik Magnesium Oksida
In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide:
Semasa eksperimen ini, magnesium bertindak balas dengan oksigen dalam udara untuk membentuk asap putih, magnesium oksida:

Magnesium + Oxygen → Magnesium oxide
Magnesium + Oksigen → Magnesium oksida

Material / Bahan: Magnesium ribbon, sand paper
Apparatus / Radas: Crucible with lid, tongs, Bunsen burner, tripod stand and balance

Set-up of apparatus / Susunan radas:

Magnesium ribbon

Heat

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MODULE • Chemistry Form 4

Procedure / Langkah:

(a) A crucible and its lid are weighed.

Mangkuk pijar dengan penutup ditimbang.

(b) 10 cm of magnesium ribbon is cleaned with sand paper .

10 cm pita magnesium dibersihkan dengan menggunakan kertas pasir ..

(c) The magnesium ribbon is coiled loosely and placed in the crucible.

Pita magnesium di gulung dan diletakkan dalam mangkuk pijar.

(d) The crucible together with the lid and magnesium ribbon are weighed again.
bersama dengan penutup dan pita magnesium ditimbang.
Mangkuk pijar

(e) The apparatus is set up as shown in the diagram.
Radas disusun seperti dalam gambar rajah.

(f) The crucible is heated strongly without its lid . When the magnesium starts to

burn , the crucible is covered with its lid .

Mangkuk pijar dipanaskan dengan kuat tanpa penutup . Apabila pita magnesium mula

terbakar , mangkuk pijar ditutup dengan penutup .

(g) The lid of the crucible is lifted from time to time using a pair of tongs.

Penutup dibuka sekali sekala dengan menggunakan penyepit.

(h) When the magnesium ribbon stops burning , the lid is removed and the crucible is
heated strongly for another 2 minutes.

Apabila pita magnesium berhenti terbakar , penutup dibuka dan mangkuk pijar dipanaskan dengan
kuat selama 2 minit lagi.

(i) The crucible lid and its content are allowed to cool down to room temperature .

Mangkuk pijar , penutup dan kandungannya dibiarkan sejuk ke suhu bilik .

(j) The crucible , lid and its content are weighed again .

Mangkuk pijar , penutup dan kandungannya ditimbang sekali lagi .

(k) The process of heating , cooling and weighing are repeated until a

constant mass is obtained.

Proses pemanasan , penyejukan dan penimbangan diulang beberapa kali sehingga jisim

tetap diperoleh.

Observation / Pemerhatian:

Magnesium burns brightly to release white fumes and white solid is formed.
dan kemudiannya membentuk
Magnesium terbakar dengan terang membebaskan wasap putih pepejal putih .

Inference / Inferens:

Magnesium is a reactive metal.

Magnesium adalah logam yang reaktif .

Magnesium reacts with oxygen in the air to form magnesium oxide .

Magnesium bertindak balas dengan oksigen dalam udara membentuk magnesium oksida .

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Chemistry Form 4 • MODULE

Precaution steps / Langkah berjaga-jaga:

Step taken / Langkah yang diambil Purpose / Tujuan

Magnesium ribbon is cleaned with sand paper . To remove the oxide layer on the surface of the magnesium
Pita magnesium perlu digosok dengan kertas pasir . ribbon.

Untuk membuang lapisan oksida pada permukaan magnesium oksida.

The crucible lid is lifted from time to time. To allow oxygen from the air to react with magnesium .
Penutup mangkuk pijar dibuka sekali sekala. Untuk membenarkan oksigen masuk dan bertindak balas dengan magnesium .

The crucible lid then replaced quickly. To prevent fumes of magnesium oxide from escaping.
Penutup mangkuk pijar kemudian ditutup semula dengan cepat. Untuk mengelakkan wasap magnesium oksida dari terbebas.

The process of heating , cooling and weighing are To ensure magnesium react completely with oxygen to
repeated until a constant mass is obtained. for magnesium oxide .
Untuk memastikan semua magnesium telah bertindak balas lengkap
Proses pemanasan , penyejukan dan penimbang
diulang beberapa kali sehingga jisim tetap diperoleh. dengan oksigen untuk membentuk magnesium oksida .

Result / Keputusan:

Description / Penerangan Mass (g) / Jisim (g)
x
Mass of crucible + lid y
Jisim mangkuk pijar + penutup z

Mass of crucible + lid + magnesium
Jisim mangkuk pijar + penutup + magnesium

Mass of crucible + lid + magnesium oxide
Jisim mangkuk pijar + penutup + magnesium oksida

Calculation / Pengiraan: Mg O
Element / Unsur y–x z–y
y–x z–y
Mass (g) / Jisim (g)
24 16
Number of mole of atoms / Bilangan mol atom p q

Simplest ratio of moles / Nisbah mol teringkas

Empirical formula of magnesium oxide is MgpOq .

Formula empirik magnesium oksida ialah MgpOq .

5 Experiment to Determine Empirical Formula of Copper(II) Oxide
Eksperimen untuk Menentukan Formula Empirik Kuprum(II) Oksida

Copper(II) Oxide + Hidrogen → Copper + Water
Kuprum(II) oksida + Hidrogen → Kuprum + Air

Set-up of apparatus / Susunan radas: Copper(II) oxide
Hydrogen gas Burning of hydrogen gas

Combustion tube

Heat

Anhydrous calcium chloride, CaCl2 ion Sdn. B

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MODULE • Chemistry Form 4

Observation / Pemerhatian:
The black colour of copper(II) oxide turns brown .

Warna hitam kuprum(II) oksida menjadi perang .

Inference / Inferens:
Copper(II) oxide reacts with hydrogen to produce the brown copper metal .

Kuprum(II) oksida bertindak balas dengan hidrogen untuk menghasilkan logam kuprum yang berwarna perang.

Precaution steps / Langkah berjaga-jaga: Purpose / Tujuan
Step taken / Langkah yang ambil

Hydrogen gas is passed through anhydrous calcium Anhydrous calcium chloride absorb water vapour to dry
chloride.
hydrogen gas.
Gas hidrogen dialirkan melalui kalsium klorida kontang. Kalsium klorida kontang menyerap wap air untuk mengering gas hidrogen.

Dry hydrogen is passed through the combustion To remove all the air in the combustion tube.
tube for 5 to 10 minutes. (The mixture of hydrogen gas and air explodes when lighted).

Gas hidrogen kering dialirkan melalui tabung pembakaran Untuk mengeluarkan semua udara dalam tabung pembakaran.
selama 5 hingga 10 minit. (Campuran hidrogen dan udara menghasilkan letupan apabila dinyalakan)

The gas that comes out from the small hole is collected If the gas burns quietly without ‘pop’ sound , all the air
in the test tube. Then, a lighted wooden splinter is has been removed from the combustion tube.
Jika gas terbakar tanpa bunyi ‘pop’ , semua gas telah dikeluarkan
placed at mouth of the test tube.
Gas yang keluar daripada lubang kecil dikumpul dalam sebuah daripada tabung pembakaran.

tabung uji. Kayu uji menyala di letakkan di mulut tabung
uji.

The flow of hydrogen gas must be continuous To prevent hot copper from reacting with oxygen to form
throughout the experiment. copper(II) oxide again.

Gas hidrogen dialirkan secara berterusan sepanjang eksperimen. Untuk mengelakkan kuprum panas daripada bertindak balas dengan

oksigen dan membentuk kuprum(II) oksida .

The process of heating , cooling and weighing are To ensure all copper(II) oxide has changed to copper .

repeated until a constant mass is obtained. Untuk memastikan semua kuprum(II) oksida telah bertukar kepada kuprum .

Proses pemanasan , penyejukan dan penimbang

diulang beberapa kali sehingga jisim tetap diperoleh.

Result / Keputusan: Mass (g) / Jisim (g)
Description / Penerangan x
y
Mass of combustion tube + porcelain dish z
Jisim tabung pembakaran + piring tanah liat

Mass of combustion tube + porcelain dish + copper(II) oxide
Jisim tabung pembakaran + piring tanah liat + kuprum(II) oksida

Mass of combustion tube + porcelain dish + copper
Jisim tabung pembakaran + piring tanah liat + kuprum

Calculation / Pengiraan: Cu O
Element / Unsur z–x y–z
z–x y–z
Mass (g) / Jisim (g)
64 16
Number of mole of atoms / Bilangan mol atom p q

Simplest ratio of moles / Nisbah mol teringkas

Empirical formula of copper(II) oxide is CupOq .

Formula empirik kuprum(II) oksida ialah CupOq .

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Chemistry Form 4 • MODULE

6 Explain why the set-up of apparatus to determine the empirical formula in both the experiments is different.
Terangkan mengapa susunan radas untuk menentukan formula empirik dalam kedua-dua eksperimen itu berbeza.

(a) Magnesium is reactive metal (above hidrogen in reactivity series). Magnesium reacts easily to form

magnesium oxide .

Magnesium adalah logam reaktif (terletak di atas hidrogen dalam siri kereaktifan. Magnesium mudah teroksida

membentuk magnesium oksida .

(b) Copper is below hydrogen in the metal reactivity series. Oxygen in copper(II) oxide can be reduced/removed

by hydrogen gas to form copper and water.

Kuprum di bawah hidrogen dalam siri kereaktifan. Kuprum(II) okida boleh diturunkan/disingkirkan oleh

gas hidrogen untuk membentuk kuprum dan air.

7 To calculate the empirical formula of a compound, use the following table:
Untuk menghitung formula empirik suatu sebatian, jadual di bawah boleh digunakan sebagai panduan:

Element / Unsur Calculation steps / Langkah pengiraan:
Mass of element (g) / Jisim unsur (g) (a) Calculate the mass of each element in the compound.
Number of mole of atom / Bilangan mol atom
Simplest ratio of moles / Nisbah mol teringkas Hitungkan jisim setiap unsur dalam sebatian.
(b) Convert the mass of each element to number of mole of atom.

Tukar jisim setiap unsur kepada bilangan mol atom.
(c) Calculate the simplest ratio of moles of atom of the elements.

Hitungkan nisbah bilangan mol atom teringkas unsur-unsur tersebut.

EXERCISE / LATIHAN

1 When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical formula
of metal X oxide.

Apabila 11.95 g oksida logam X diturunkan oleh hidrogen, 10.35 g logam terhasil. Hitungkan formula empirik bagi oksida logam X.

[RAM / JAR: X = 207, O = 16]

Element / Unsur X O
Mass of element (g) / Jisim unsur (g) 10.35 1.6

Number of mole of atoms / Bilangan mol atom 0.05 0.1

Ratio of moles / Nisbah mol 12

Simplest ratio of moles / Nisbah mol teringkas 1 2

Empirical formula / Formula empirik: XO2 .

2 A certain compound contains the following composition / Satu sebatian mengandungi komposisi unsur seperti berikut:
Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass / Jisim atom relatif: O = 16, Na = 23, Br = 80]

(Assume that 100 g of substance is used / Anggap 100 g bahan digunakan)

Element / Unsur Na Br O
Mass of element (g) / Jisim unsur (g) 15.23 52.98 31.79

Number of mole of atoms / Bilangan mol atom 0.66 0.66 1.99

Ratio of moles / Nisbah mol 1 1 3.01

Simplest ratio of moles / Nisbah mol teringkas 113

Empirical formula / Formula empirik: NaBrO3 .

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MODULE • Chemistry Form 4

3 a2t.0om8 igc omf aeslesm oef netl eXm ceonmt Xbi.n [eRsA wMit: hY 4=.2 63 5g. 5o]f element Y to form a compound with formula XY3. Calculate the relative
2.08 g unsur X bergabung dengan 4.26 g unsur Y untuk membentuk sebatian dengan formula XY3 . Hitung jisim atom relatif unsur X.
[JAR: Y = 35.5]

Element / Unsur X Y x = relative atomic mass of X

Mass of element (g) 2.08 4.26 Mol X = 1
Jisim unsur (g) Mol Y 3

Number of mole of atoms 2.08 4.26 = 0.12 2.08
Bilangan mol atom x 35.5
x = 1
Simplest ratio of moles 0.12 3
Nisbah mol teringkas 1 3 x = 52

4 r2e.l0a7t igv eo af teolmemice nmta Zs sr eoaf cetlse mwietnht bZr.o m[RiAnMe t: oB fro =rm 8 30.]67 g of a compound with empirical formula ZBr2. Calculate the
2.07 g unsur Z bertindak balas dengan bromin membentuk 3.67 g sebatian dengan formula empirik ZBr2. Hitung jisim atom relatif bagi
unsur Z. [JAR: Br = 80]

Element / Unsur Z Br z = relative atomic mass of Z

Mass of element (g) 2.07 1.6 Mol Z = 1
Jisim unsur (g) Mol Br 2

Number of mole of atoms 2.07 1.6 = 0.02 2.08
Bilangan mol atom z 80
z = 1
Simplest ratio of moles 0.02 2
Nisbah mol teringkas 1 2 z = 207

5 The statement below is about compound J / Pernyataan berikut adalah mengenai sebatian J.

• It is black solid / Merupakan pepejal hitam.
• Contains 12.8 g copper and 0.2 mol of oxygen / Mengandungi 12.8 g kuprum dan 0.2 mol oksigen.

[Relative atomic mass / Jisim atom relatif : Cu = 64]

(a) What is meant by empirical formula / Apakah maksud formula empirik?
A formula that shows the simplest whole number ratio of atoms of each element in a compound.

(b) (i) Calculate the number of mol of copper atom / Hitung bilangan mol atom kuprum.

12.8 = 0.2 mol
64

(ii) What is the empirical formula of compound J / Apakah formula empirik sebatian J ?
0.2 mol Cu : 0.2 mol O.
1 mol Cu : 1 mol O.
Empirical formula of Compound J is CuO.

(c) Compound J reacts completely with hydrogen to produce copper and compound Q.
Sebatian J bertindak balas lengkap dengan hidrogen menghasilkan kuprum dan sebatian Q.

(i) State one observation for the reaction / Nyatakan satu pemerhatian daripada tindak balas tersebut.

Black solid change to brown

(ii) Name two the substances that can be used to prepare hydrogen gas.
Namakan dua bahan yang digunakan untuk menyediakan gas hidrogen.

Zinc/magnesium and hydrochloric acid/nitric acid/sulphuric acid.

(iii) Name compound Q / Nama sebatian Q.
Water

m Publica (iv) Write a balanced equation for the reaction.
Tuliskan persamaan kimia yang seimbang bagi tindak balas tersebut.
36 CuO + H2 → Cu + H2O

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(d) Draw a labelled diagram of the set-up of apparatus for the experiment. Chemistry Form 4 • MODULE
Lukiskan gambar rajah berlabel susunan radas bagi tindak balas tersebut.
Compound J
Gas hidrogen

(e) (i) Heat

Why is hydrogen gas passed through the combustion tube after heating has stpopped?
Mengapakah gas hidrogen dilalukan melalui tiub pembakaran selepas pemanasan dihentikan?

To avoid copper produced react with oxygen to form copper(II) oxide.

(ii) State how to determine that the reaction between compound J and hydrogen has completed.
Nyatakan bagaimana menentukan tindak balas antara sebatian J dengan hidrogen telah lengkap.

By repeating the process of heating, cooling and weighing until constant mass is obtained.

(f) (i) Can the empirical formula of magnesium oxide be determined by the same method? Explain your answer.
Bolehkah formula empirik bagi magnesium oksida ditentukan dengan cara yang sama? Jelaskan jawapan anda.

Cannot. Magnesium is more reactive than hydrogen. Hydrogen cannot reduce magnesium oxide to form

magnesium.

(ii) Magnesium can reduce copper oxide to copper. Explain why the empirical formula of the copper oxide cannot
be determined by heating the mixture of copper oxide and magnesium powder.
Magnesium boleh menurunkan kuprum oksida kepada kuprum. Terangkan mengapa formula empirik kuprum oksida tidak boleh
ditentukan dengan pemanasan campuran kuprum oksida dengan serbuk magnesium.

Magnesium oxide and copper produced are in solid form, copper cannot be separated from magnesium oxide.

The mass of copper cannot be weighed.

MOLECULAR FORMULA / FORMULA MOLEKUL

1 Molecular formula of a compound shows the actual number of atoms of each element that are present in a molecule of
the compound.

Formula molekul suatu sebatian menunjukkan bilangan sebenar atom bagi setiap unsur yang terdapat dalam satu molekul sebatian.

Molecular Formula = (empirical formula)n, where n is a integer.
Formula molekul = (Formula empirik)n, di mana n adalah integer.

2 Example / Contoh: Molecular formula Empirical formula Value of n
Compound Formula molekul Formula empirik Nilai n

Sebatian

Water / Air H2O H2O 1

Carbon dioxide / Karbon dioksida CO2 CO2 1

Sulphuric acid / Asid sulfurik H2SO4 H2SO4 1

Ethene / Etena C2H4 CH2 2

Benzene / Benzena C6H6 CH 6

Glucose / Glukosa C6H12O6 CH2O 6

The molecular formula and the empirical formula of a compound may be the same if the value of n = 1 but different if the
value is n > 1.
Formula molekul dan formula empirik suatu sebatian akan sama sekiranya nilai n = 1 tetapi akan berbeza sekiranya nilai n > 1.

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MODULE • Chemistry Form 4

EXERCISE / LATIHAN

1 Tcohme peomupnirdi cXa.l [fRoremlautilvae o aft coommicp omuansds :X H i s= C H1;2 Ca n=d 1re2l]ative molecular mass is 56. Determine the molecular formula of
Formula empirik sebatian X adalah CH2 dan JMR adalah 56. Tentukan formula molekul sebatian X. [Jisim atom relatif: H = 1; C = 12]

(12 + 2)n = 56
n = 56 = 4

14
Molecular formula = (CH2)4 = C4H8

2 2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86.
2.58 g suatu hidrokarbon mengandungi 2.16 g karbon. Jisim molekul relatif bagi hidrokarbon ini ialah 86.

[Relative atomic mass / Jisim atom relatif : H = 1; C = 12]
(i) Calculate the empirical formula of the hydrocarbon / Hitungkan formula empirik bagi hidrokarbon ini.

Element CH

Mass of element (g) 2.16 0.42

Number of mole of atoms 0.18 0.42

Ratio of moles 1 2 1 = 7
3 3

Simplest ratio of moles 37

Empirical formula = C3H7

(ii) Determine the molecular formula of the hydrocarbon / Tentukan formula molekul hidrokarbon tersebut.

(12 × 3 + 7 × 1)n = 86

n = 86 = 2
43
Molecular formula = (C3H7)2 = C6H14

3 The diagram below shows the structural formula for benzene molecule.
Rajah di bawah menujukkan formula struktur bagi benzena.
H

HCCCH

HCCCH

H
(a) Name the element that make up benzene / Namakan unsur yang membentuk benzena.

Carbon and hydrogen

(b) What are the molecular formula and empirical formula for benzene?
Apakah formula molekul dan formula empirik bagi benzena?

Molecular formula / Formula molekul: C6H6

Empirical formula / Formula empirik: CH

(c) Compare and contrast the molecular formula and empirical formula for benzene.
Banding dan bezakan formula molekul dan formula empirik bagi benzena.

• Both empirical formula and molecular formula shows benzene is made up of carbon and hydrogen
elements.
Kedua-dua fomula molekul dan formula empirik menunjukkan benzena terdiri dari unsur karbon dan hidrogen .

• Molecular formula shows the actual number of carbon atoms and hydrogen atoms in benzene

molecule . Each benzene molecule consists of 6 carbon atoms and 6 hydrogen atoms.

Formula molekul menunjukkan bilangan sebenar bagi atom karbon dan atom hidrogen dalam molekul

benzena. Setiap molekul benzena terdiri daripada 6 atom karbon dan 6 atom

Nila m Publica hidrogen .

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Chemistry Form 4 • MODULE

• Empirical formula shows the simplest ratio of number carbon atoms to hydrogen atoms, the simplest

ratio of number of carbon atoms to hydrogen atoms in benzene is 1 : 1 .

Formula empirik benzena menunjukkan nisbah paling ringkas bilangan atoms karbon kepada atom hidrogen .

Nisbah paling ringkas bilangan atom karbon kepada hidrogen adalah 1:1 .

PERCENTAGE COMPOSITION BY MASS OF AN ELEMENT IN A COMPOUND
PERATUS KOMPOSISI UNSUR MENGIKUT JISIM DALAM SEBATIAN

Total RAM of the element in the compound × 100%
Jumlah JAR unsur dalam suatu sebatian × 100%
1 % composition by mass of an element = RMM/RFM of compound/JMR/JFR sebatian
% komposisi unsur mengikut jisim

2 Example / Contoh:

Calculate the percentage composition by mass of nitrogen in the following compounds:
Hitungkan peratusan nitrogen mengikut jisim dalam sebatian berikut:
[Relative atomic mass / Jisim atom relatif : N = 14, H = 1, O = 16, S = 32, K = 39]
(i) (NH4)2SO4
%N = 2 × 14 × 100%

132

= 21.2%

(ii) KNO3
%N = 14 × 100%

101

= 13.9%

CHEMICAL FORMULA FOR IONIC COMPOUNDS / FORMULA KIMIA BAGI SEBATIAN ION

1 Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is by exchanging the charges on each ion.
The formula obtained will be XmYn.
Formula kimia sebatian ion yang mengandungi ion X m+ dan Y n– boleh diperoleh melalui pertukaran bilangan cas setiap ion. Formula yang

diperoleh ialah XnYm.

2 Example / Contoh:
(i) Sodium oxide / Natrium oksida

Ion / Ion Na+ O2–

Charges / Bilangan cas +1 –2

Exchange of charges / Pertukaran bilangan cas 21

Smallest ratio / Nisbah teringkas 21

Number of combining ions / Bilangan ion yang bergabung 2 Na+ O2–

Formula / Formula Na2O
(ii) Copper(II) nitrate / Kuprum(II) nitrat (iii) Zinc oxide / Zink oksida

Cu2+ N–O13– Zn2+ O2–
+2 +2 –2

1 2 (Ratio / Nisbah) 2 2
1 (Ratio / Nisbah)
⇒ Cu(NO3)2 1
⇒ ZnO

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ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS
Aktiviti 1: TULIS FORMULA KIMIA DAN NAMA BAGI BAHAN KIMIA BERIKUT
Bhd.tion Sdn. MODULE • Chemistry Form 4

m Publica O2–, CCaOr3b2–o, nat ion SSuOl4p2–h, ate ion Cl–, Br–, I–, OH–, NNOitr3–a,te ion
Oxide ion Ion karbonat Ion sulfat Chloride ion Bromide ion Iodide ion Hydroxide ion Ion nitrat
40 Ion oksida Ion klorida Ion bromida Ion iodida Ion hidroksida

K+ KPo2Otassium oxide KPo2CtaOs3sium carbonate PKo2StaOs4sium sulphate KCl KBr KI KOH KPoNtOas3sium nitrate
Potassium ion Potassium chloride Potassium bromide Potassium iodide Potassium hydroxide

Ion kalium

Na+ SNoad2Oium oxide NSoad2CiuOm3 carbonate SNoad2SiuOm4 sulphate NaCl NaBr NaI NaOH NSoadNiOum3 nitrate
Sodium ion Sodium chloride Sodium bromide Sodium iodide Sodium hydroxide

Ion natrium

H+ CHa2rCbOo3nic acid SHu2SlpOh4uric acid HCl HBr HI NHiNtrOic3 acid
Hydrogen ion
Hydrocloric acid Hydrobromic acid Hydroiodic acid
Ion hidrogen

Ag+ ASiglv2Oer oxide SAiglv2CeOr 3carbonate SAiglv2SeOr 4sulphate AgCl AgBr AgI AgOH ASiglvNeOr 3nitrate
Silver ion Silver chloride Silver bromide Silver iodide Silver hydroxide

Ion argentum

ANmH4m +onium ion A(NmHm4)o2CnOiu3m A(NmHm4)o2SnOiu4m ANmH4mClonium NAmH4mBor nium ANmH4mI onium ANmH4mNoOn3ium
Ion ammonium carbonate sulphate chloride bromide iodide nitrate

Ca2+ CaO CCaaClcOiu3m carbonate CCaalScOiu4m sulphate CCaalCcli2um chloride CCaaBlcri2um bromide CCaalIc2 ium iodide CCaa(lcOiuHm)2 hydroxide CCaa(lcNiuOm3 ) 2nitrate
Calcium ion Calcium oxide

Ion kalsium

Cu2+ CuO CCuopCpOe3r(II) carbonate CCoupSpOe4r(II) sulphate CCoupCpl2er(II) chloride CCuopBpr2er(II) bromide CCoupI2per(II) iodide CCuop(OpHer)(2II) hydroxide CCuop(NpOer3( )II2) nitrate
Copper(II) ion Copper(II) oxide

Ion kuprum(II)

Mg2+ MgO MMgaCgOne3sium MMagSgnOe4sium MMagCgln2esium MMgaBgnr2esium MMagIg2nesium iodide MMga(gOnHes)i2um MMga(gNnOes3 i)u2m nitrate
carbonate sulphate chloride bromide hydroxide
Magnesium ion Magnesium
Ion magnesium oxide

Zn2+ ZnO ZZninCcO c3arbonate ZZinnScO s4ulphate ZZinnCcl 2chloride ZZninBcr 2bromide ZZinnIc2 iodide ZZnin(cO hHy)2droxide ZZnin(cN nOi3t )r2ate
Zinc ion Zinc oxide

Ion zink

Pb2+ PbO PLbeaCdO(3II) carbonate LPbeaSdO(4II) sulphate LPbeaCdl2(II) chloride PLbeaBdr2(II) bromide LPbeaI2d(II) iodide PLbea(OdH(II))2 hydroxide PLbea(NdO(II3) ) 2nitrate
Lead(II) ion Lead(II) oxide

Ion plumbum(II)

Al 3+ AAllu2Om3 inium oxide AAllu2(mCOin3 i)u3 m carbonate AAllu2(mSOin4 i)u3 m AAlluClm3 inium AAllBumr3 inium bromide AAlluI3minium iodide AAll(uOmHin)3ium AAll(uNmOi3n)i3um nirate
Aluminium ion sulphate chloride hydroxide

Ion aluminium

ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULAE OF THE FOLLOWING COMPOUNDS
AKTIVITI 2: TANPA MERUJUK KEPADA JADUAL AKTIVITI 1, TULISKAN FORMULA KIMIA BAGI SEBATIAN BERIKUT

Oxide ion Carbonat ion Sulphate ion Chloride ion Bromide ion Iodide ion Hydroxide ion Nitrate ion
Ion oksida Ion karbonat Ion sulfat Ion klorida Ion bromida Ion iodida Ion hidroksida Ion nitrat

Potassium ion K2O K2CO3 K2SO4 KCl KBr KI KOH KNO3
Ion kalium

Sodium ion Na2O Na2CO3 Na2SO4 NaCl NaBr NaI NaOH NaNO3
Ion natrium

Hydrogen ion H2CO3 H2SO4 HCl HBr HI HNO3
Ion hidrogen

Silver ion Ag2O Ag2CO3 Ag2SO4 AgCl AgBr AgI AgOH AgNO3
Ion argentum

Ammonium ion (NH4 )2CO3 (NH4 )2SO4 NH4Cl NH4 Br NH4 I NH4 NO3
Ion ammonium

Calcium ion CaO CaCO3 CaSO4 CaCl2 CaBr2 CaI2 Ca(OH)2 Ca(NO3 )2
Ion kalsium

Copper(II) ion CuO CuCO3 CuSO4 CuCl2 CuBr2 CuI2 Cu(OH)2 Cu(NO3 )2
Ion kuprum(II)

Magnesium ion MgO MgCO3 MgSO4 MgCl2 MgBr2 MgI2 Mg(OH)2 Mg(NO3 )2
Ion magnesium

Zinc ion ZnO ZnCO3 ZnSO4 ZnCl2 ZnBr2 ZnI2 Zn(OH)2 Zn(NO3 )2
Ion zink
Chemistry Form 4 • MODULE
Lead(II) ion PbO PbCO3 PbSO4 PbCl2 PbBr2 PbI2 Pb(OH)2 Pb(NO3 )2
ion Sdn. BIon plumbum(II)

41
m Publicat Aluminium ion Al2O3 Al2(CO3)3 Al2(SO4 )3 AlCl3 AlBr3 AlI3 Al(OH)3 Al(NO3 )3
Ion aluminium

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MODULE • Chemistry Form 4

ACTIVITY 3: WRITE THE CHEMICAL FORMULAE AND TYPE OF PARTICLES FOR THE FOLLOWING ELEMENT/COMPOUND
AKTIVITI 3: TULIS FORMULA KIMIA DAN JENIS ZARAH UNTUK UNSUR/SEBATIAN BERIKUT

Compound / Element Formula Type of particles Compound / Element Formula Type of particles
Sebatian/Unsur Formula Jenis zarah Sebatian/Unsur Formula Jenis zarah

Sodium sulphate Na2SO4 Ion Zinc carbonate ZnCO3 Ion
Natrium sulfat Zink karbonat

Ammonium carbonate (NH4 )2CO3 Ion Ammonium carbonate (NH4 )2CO3 Ion
Ammonium karbonat Ammonium karbonat

Magnesium nitrate Mg(NO3 )2 Ion Silver chloride AgCl Ion
Magnesium nitrat Argentum klorida

Hyrochloric acid HCl Ion Sulphuric acid H2SO4 Ion
Asid hidroklorik Asid sulfurik

Potassium oxide K2O Ion Copper(II) nitrate Cu(NO3 )2 Ion
Kalium oksida Kuprum(II) nitrat

Magnesium oxide MgO Ion Hydrogen gas H2 Molecule
Magnesium oksida Gas hidrogen

Lead(II) carbonate PbCO3 Ion Carbon dioxide gas CO2 Molecule
Plumbum(II) karbonat Gas karbon dioksida

Iron(III) sulphate Fe2(SO4)3 Ion Oxygen gas O2 Molecule
Ferum(III) sulfat Gas oksigen

Magnesium chloride MgCl2 Ion Aluminium sulphate Al2(SO4 )3 Ion
Magnesium klorida Aluminium sulfat

Zinc sulphate ZnSO4 Ion Lead(II) chloride PbCl2 Ion
Zink sulfat Plumbun(II) klorida

Silver nitrate AgNO3 Ion Potassium iodide KI Ion
Argentum nitrat Kalium iodida

Ammonium sulphate (NH4 )2SO4 Ion Copper(II) carbonate CuCO3 Ion
Ammonium sulfat Kuprum(II) karbonat

Zinc oxide ZnO Ion Potasium carbonate K2CO3 Ion
Zink oksida Kalium karbonat

Nitric acid HNO3 Ion Sodium hydroxide NaOH Ion
Asid nitrik Natrium hidroksida

Ammonia gas NH3 Molecule Aqueous ammonia NH3(aq) Ion and molecule
Gas ammonia Ammonia akueus

Magnesium Mg Atom Ammonium chloride NH4Cl Ion
Magnesium Ammonium klorida

Zinc Zn Atom Nitrogen dioxide gas NO2 Molecule
Zink Gas nitrogen dioksida

Copper(II) sulphate CuSO4 Ion Sodium chloride NaCl Ion
Kuprum(II) sulfat Natrium klorida

Iodine I2 Molecule Silver Ag Atom
Iodin Argentum

Chlorine Cl2 Molecule Bromine Br2 Molecule
Klorin Bromin

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Chemistry Form 4 • MODULE

CHEMICAL EQUATIONS / PERSAMAAN KIMIA

1 Two types of equation / Dua jenis persamaan:

• Equation in words / Persamaan perkataan
– using names of reactants and products / menggunakan nama bahan tindak balas dan hasil tindak balas;

• Equation using symbols / Persamaan menggunakan simbol

– reactants and products are represented by chemical formulae and have certain meanings

menggunakan formula kimia untuk mewakili bahan tindak balas dan hasil tindak balas serta menggunakan pelbagai jenis simbol yang

membawa makna tertentu.

Symbol / Simbol Meaning / Maksud Symbol / Simbol Meaning / Maksud
(g)
+ Separating 2 reactants / products (g) Gaseous state
Mengasingkan 2 bahan / hasil (aq) Keadaan gas
(ak)
Produces Aqueous state
Menghasilkan ∆ Keadaan akueus

Reversible reaction Gas released
Tindak balas berbalik Gas terbebas

(s) Solid state Precipitation
(p) Keadaan pepejal Bahan termendap

(l) Liquid state Heating / Heat energy is given
(ce) Keadaan cecair Pemanasan / Haba dibekalkan

2 Information obtained from chemical equation using symbols / Maklumat yang diperoleh daripada persamaan kimia bersimbol:
(a) Qualitative aspect / Aspek kualitatif : type of reactants and products involved in the chemical reaction and the state
of each reactant and product.

jenis bahan / hasil tindak balas yang terlibat dalam tindak balas dan keadaan fizikal bagi

setiap bahan / hasil tindak balas.

(b) Quantitative aspect / Aspek kuantitatif : number of moles of reactants and products involved in the chemical reaction
that is the coeffficients involved in a balanced equation of the formulae of
reactants and products.

bilangan mol yang bertindak balas dan hasil tindak balas yang terbentuk iaitu pekali bagi

setiap formula bahan dan hasil tindak balas.

Example / Contoh: Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g)

Zn (p) + 2HCl (ak) ZnCl2 (ak) + H2 (g)

1 mol 2 mol 1 mol 1 mol

Interpretation / Tafsiran: 1 mol of zinc reacts with 2 mol of hydrochloric acid to produce 1 mol of zinc chloride and
1 mol of hydrogen.

1 mol zink bertindak balas dengan 2 mol asid hidroklorik menghasilkan 1 mol zink klorida dan 1 mol hidrogen.

3 Writing balanced chemical equations / Menulis persamaan kimia seimbang:

Step 1 / Langkah 1 : Write the correct chemical formulae for each reactant and product.
Tulis formula kimia bagi setiap bahan dan hasil tindak balas.

Step 2 / Langkah 2 : Detemine the number of atoms for each element / Tentukan bilangan atom setiap unsur.

Step 3 / Langkah 3 : Balance the number of atoms for each element by adjusting the coefficients in front of the chemical
formulae.

Imbangkan bilangan atom setiap jenis unsur dengan menambahkan pekali di hadapan setiap formula kimia.

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MODULE • Chemistry Form 4

EXERCISE / LATIHAN

Write a balanced chemical equation for each of the following reactions:
Tulis persamaan kimia seimbang bagi setiap tindak balas yang berikut:

1 Zinc carbonate Zinc oxide + Carbon dioxide / Zink karbonat Zink oksida + Karbon dioksida
ZnCO3 ZnO + CO2

2 Sulphuric acid + Sodium hydroxide Sodium sulphate + Water / Asid sulfurik + Natrium hidroksida Natrium sulfat + Air
H2SO4 + 2NaOH Na2SO4 + 2H2O

3 Silver nitrate + Sodium chloride Silver chloride + Sodium nitrate
Argentum nitrat + Natrium klorida Argentum klorida + Natrium nitrat

AgNO3 + NaCl AgCl + NaNO3

4 Copper(II) oxide + Hydrochloric acid Copper(II) chloride + Water
Kuprum(II) oksida + Asid hidroklorik Kuprum(II) klorida + Air

CuO + 2HCl CuCl2 + H2O

5 Magnesium + Oxygen Magnesium oxide / Magnesium + Oksigen Magnesium oksida
2Mg + O2 2MgO

6 Sodium + Water Sodium hydroxide + Hydrogen / Natrium + Air Natrium hidroksida + Hidrogen
2Na + 2H2O 2NaOH + H2

7 Potassium oxide + Water Potassium hydroxide / Kalium oksida + Air Kalium hidroksida
K2O + H2O 2KOH

8 Zinc oxide + Nitric acid Zinc nitrate + Water / Zink oksida + Asid nitrik Zink nitrat + Air
ZnO + 2HNO3 Zn(NO3 )2 + H2O

9 Lead(II) nitrate Lead(II) oxide + Nitrogen dioxide + Oxygen
Plumbum(II) nitrat Plumbum (II) oksida + Nitrogen dioksida + Oksigen

2Pb(NO3 )2 2PbO + 4NO2 + O2

10 Aluminium nitrate Aluminium oxide + Nitrogen dioxide + Oxygen
Aluminium nitrat Aluminium oksida + Nitrogen dioksida + Oksigen

4Al(NO3 )3 2Al2O3 + 12NO2 + 3O2

NUMERICAL PROBLEMS INVOLVING CHEMICAL EQUATIONS / PENGHITUNGAN BERKAITAN PERSAMAAN KIMIA

Calculation steps / Langkah perhitungan:
S1 / L1 : Write a balanced equation / Tulis persamaan kimia seimbang.
S2 / L2 : Write the information from the question above the equation / Tulis maklumat daripada soalan di atas persamaan.
S3 / L3 : Write the information from the chemical equation below the equation (information about the number of moles of
reactants/products).
Tulis maklumat daripada persamaan kimia di bawah persamaan (Maklumat perhubungan bilangan mol bahan/hasil tindak balas

terlibat).

S4 / L4 : Change the information in S2 into moles by using the method shown in the chart below.
Tukarkan maklumat L2 kepada mol menggunakan carta di bawah.
S5 / L5 : Use the relationship between number of moles of substance involved in S3 to find the answer.
Gunakan perhubungan bilangan mol bahan terlibat dalam L3 untuk mencari jawapan.
S6 / L6 : Change the information to the unit required using the chart below.

Tukar maklumat kepada unit yang dikehendaki dengan menggunakan carta di bawah.

Mass (g) ÷ (RAM/FRM/RMM) g mol–1 No. of moles (n) × 24 dm3 mol–1 / 22.4 dm3 mol–1 Volume of gas (dm3)
Jisim (g) × (RAM/FRM/RMM) g mol–1 Bilangan mol (n) ÷ 24 dm3 mol–1 / 22.4 dm3 mol–1 Isipadu gas (dm3)

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Chemistry Form 4 • MODULE

EXERCISE / LATIHAN

1 The equation shows the reaction between zinc and hydrochloric acid.
Persamaan menunjukkan tindak balas antara zink dengan asid hidroklorik.

Zn + 2HCl ZnCl2 + H2

Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at room
conditions. [Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions]
Hitungkan jisim zink yang perlu ditindakbalaskan dengan asid hidroklorik berlebihan untuk menghasilkan 6 dm3 gas hidrogen pada
keadaan bilik. [Jisim atom relatif: Zn = 65, Cl = 35.5, 1 mol gas menempati 24 dm3 pada suhu bilik]

Mol of H2 = 6 dm3 = 0.25 mol
24 dm3 mol–1

From the equation,

0 M.a2s15s mm oofo llZ oonff =HH2 20 ::. 2105 .m2 ×5o lm6 o5of l = Zon f1 Z6n.2 g



2 The equation shows the reaction between potassium and oxygen.
Persamaan berikut menunjukkan tindak balas antara kalium dengan oksigen.

4K + O2 2K2O

Calculate the mass of potassium required to produce 23.5 g of potassium oxide. [Relative atomic mass: K = 39, O = 16]
Hitungkan jisim kalium yang diperlukan untuk menghasilkan 23.5 g kalium oksida. [Jisim atom relatif: K = 39, O = 16]

Mol of K2O = 23.5 = 23.5 = 0.25 mol
(2 × 39 + 16) 94

From the equation,

M 0.a2s52s mmofoo llK oo =ff KK 022OO.5 ::m 04o. 5ml ×oml o3ol9f o Kgf Km ol–1 = 19.5 g



3 The equation shows the decomposition of hydrogen peroxide.
Persamaan menunjukkan penguraian hidrogen peroksida.

H2O2 H2O + O2

aBta SlaTnPc. e[ Rthelea teiqvuea Atitoonm aicb omvaes. sC:a Hlc u=la 1te, Oth =e n 1u6m, mbeorl aorf mvoolulems eo fo Hf g2Oa2s t=ha 2t 2d.e4c odmmp3 omsoesl– 1i fa 1t 1S.T2P d] m3 oxygen gas is collected
Seimbangkan persamaan di atas. Hitung bilangan mol H2O2 yang telah terurai sekiranya 11.2 dm3 gas oksigen dikumpulkan pada STP.
[Jisim atom relatif: H = 1, O = 16, isi padu molar gas = 22.4 dm3 mol–1 pada STP]

Mol of O2 = 11.2 dm3 = 0.5 mol
22.4 dm3 mol–1

From the equation,

0.15 mmooll ooff OO22 :: 21 .m0 oml oolf o Hf 2HO22O 2


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MODULE • Chemistry Form 4

4 8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate
produced. [Relative atomic mass: N = 14, O = 16, Cu = 64]
8.0 g serbuk kuprum(II) oksida dicampurkan kepada asid nitrik cair yang berlebihan dan dihangatkan. Hitungkan jisim kuprum(II) nitrat

yang terhasil. [Jisim atom relatif: N = 14, O = 16, Cu = 64]

CuO + 2HNO3 Cu(NO3 )2 + H2O

Mol of CuO = 8 g = 0.1 mol
(64 + 16)g mol–1

From the equation,
M as s1 o mf0 C.o1ul (ooNffO CC3uu)2OO = :: 010. .1m1 omml ooollf o ×Cfu C1(uN8(8ON 3gO) 23m)2ol–1 = 18.8 g

5 1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the
volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1at STP]
1.3 g zink bertindak balas dengan asid sulfurik cair yang berlebihan. Hasil tindak balas adalah zink sulfat dan hidrogen. Hitungkan isi padu

hidrogen yang terbebas pada STP. [Jisim atom relatif: Zn = 65, isipadu molar gas = 22.4 dm3 mol–1 pada STP]

Answer/Jawapan: 448 cm3

6 0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate the
volume of chlorine gas that has reacted. [Relative atomic mass: Na = 23, Molar volume of gas = 24 dm3 mol–1 at room
conditions]

0.46 g natrium terbakar lengkap dalam gas klorin pada keadaan bilik menghasilkan natrium klorida. Hitungkan isi padu klorin yang

diperlukan untuk bertindak balas lengkap. [Jisim atom relatif: Na = 23, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

Answer/Jawapan: 0.24 dm3

7 The equation shows the combustion of propane gas.
Persamaan menunjukkan pembakaran gas propana.

C3H8 + 5O2 3CO2 + 4H2O

[7R2e0l actmiv3e o aft pormoipca mnea sgsa:s C ( C=3H 182) ,a Ot r=oo 1m6 ,c oMnodliatrio vnosl ubmuren osf i nga esx =ce s2s2 o.4x ydgmen3 .m Coall–c1u alat treo othme cmoansdsi toiof ncsa]rbon dioxide formed.
720 cm3 gas propana (C3H8) pada keadaan bilik terbakar dalam oksigen berlebihan. Hitungkan jisim karbon dioksida yang terbentuk.
[Jisim atom relatif: C = 12, O = 16, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]

Answer/Jawapan: 3.96 g

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Chemistry Form 4 • MODULE

Objective Questions / Soalan Objektif

1 The mass of one atom of element Y is two times more 5 The table below shows the relative atomic mass of neon,
than an atom of oxygen. What is the relative atomic carbon, oxygen and calcium.

mass of element Y? [Relative atomic mass: O = 16] Jadual berikut menunujukkan jisim atom relatif bagi neon,

Jisim satu atom unsur Y adalah dua kali lebih daripada karbon, oksigen dan kalsium.

satu atom oksigen. Apakah jisim atom relatif bagi unsur Y? Element/Unsur Relative atomic mass/Jisim atom relatif
[Jisim atom relatif: O = 16]

A 12 Neon / Neon 20

B 24 Carbon / Karbon 12
C 32 Oxygen / Oksigen 16
D 36

2 fTohlelo cwhienmg isctaalt efomrmenutsla a froer t bruuet aanbeo uist Cb4uHta10n. eW? hich of the Calcium / Kalsium 40
[Relative atomic mass: H = 1, C =12 and O =16,
Avogadro Constant = 6 × 1023 mol–1] Which of the following statements is true?
Formula kimia bagi butana ialah C4H10. Antara pernyataan [Avogadro constant = 6.0 × 1023 mol–1]
berikut, yang manakah adalah benar tentang butana?
Antara pernyataan berikut, yang manakah adalah benar?
[Jisim atom relatif: H = 1, C = 12 dan O = 16, Pemalar Avogadro =
[Pemalar Avogadro = 6.02 × 1023 mol–1]
6 × 1023 mol–1] A Mass of one calcium atom is 40 g
I The empirical formula for butane is CH2.
Jisim satu atom kalsium ialah 40 g
Formula empirik butana ialah CH2.
II Each butane molecule is made up of 4 carbon atoms B Mass of 1 mol of neon is 20 g
Jisim 1 mol neon ialah 20 g

C 16 g of oxygen contains 6.02 × 1023 oxygen
molecule

and 10 hydrogen atoms. 16 g oksigen mengandungi 6.02 × 1023 molekul oksigen
Setiap molekul butana terdiri dari 4 atom karbon dan 10 D Mass of one oxygen atom is 16 times bigger than

atom hidrogen. one carbon atom
III 1 mol of butane contains a total of 8.4 × 1024
Jisim satu atom oksigen adalah 16 kali lebih besar daripada

atoms. satu atom karbon

Jumlah bilangan atom dalam 1 mol butana adalah 8.4 ×
1024. 6 A bulb is filled with 1 800 cm3 of argon gas at room
IV One butane molecule has a mass of 84 times higher conditions. What is the number of argon atom in the

than the mass of 1 hydrogen atom. bulb?

Satu molekul butana mempunyai jisim 84 kali lebih daripada [Molar volume of gas = 24 dm3 mol–1 at room conditions,

jisim satu atom hidrogen. Avogadro constant = 6.02 × 1023 mol–1]
A I and II only Sebuah belon diisi dengan 1 800 cm3 gas argon pada keadaan

I dan II sahaja bilik. Berapakah bilangan atom argon dalam belon itu?
B II and III only [Isipadu molar gas = 24 dm3 mol–1 pada keadaan bilik,

II dan III sahaja Pemalar Avogadro = 6.02 × 1023 mol–1]
C II, III and IV only A 4.515 × 1022 C 8.03 × 1022

II, III dan IV sahaja B 4.515 × 1023 D 8.03 × 1021

D I, II, III and IV

I, II, III dan IV 7 What is the number of hydrogen atom in 0.1 mol of

3 A bottle contains 3.01 × 1023 of gas particles. What is water? [Avogadro constant: 6.02 × 1023 mol–1]
the number of moles of the gas in the bottle? Berapakah bilangan atom oksigen dalam 0.1 mol air?

Sebuah botol mengandungi 3.01 × 1023 zarah gas. Berapakah [Pemalar Avogadro = 6.02 × 1023 mol–1]

bilangan mol zarah gas dalam botol itu? A 6.02 × 1022 C 6.02 × 1023
B 60.2 × 1023 D 3.01 × 1023
A 0.5 mol C 3.0 mol
B 1.0 mol D 6.0 mol
8 5 g of element X reacted with 8 g of element Y to form
a compound with the formula X[RYe2.l aWtihvea t aisto tmheic r emlaatisvse:
4 Which of the following gases contains 0.4 mol of atoms atomic mass of element X?

at room temperature and pressure? [Molar volume of Y = 80]

gas = 24 dm3 mol–1 at room temperature and pressure] 5 g unsur X bertindak balas dengan 8 g unsur Y membentuk

Antara gas berikut, yang manakah mengandungi 0.4 mol atom sebatian dengan formula XY2. Apakah jisim atom relatif unsur X?
[Jisim atom relatif: Y = 80]
pada suhu dan tekanan bilik? [Isi padu molar gas = 24 dm3 mol–1

pada suhu dan tekanan bilik] A 25 C 50

A 4.8 dm3 Ne C 44..88 ddmm33 CNOH23 B 40 D 100
B 4.8 dm3 O2 D

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MODULE • Chemistry Form 4

9 The diagram below shows the set-up of apparatus to 11 The equation shows a decomposition of magnesium
determine the empirical formula of an oxide metal X. nitrate when heated.
Rajah di bawah menunjukkan susunan radas bagi menentukan Persamaan di bawah menunjukkan penguraian nitrat apabila
dipanaskan.
formula empirik oksida logam X.
2Mg(NO3)2 2MgO + 4NO2 + O2
Metal X
Logam X What is the number of oxygen molecules is produced
when 7.4 g magnesium nitrate decomposed when
Heat heated.
A[Rveolgaatidvreo fcoornmsutalan tm =as 6s .o0f2 M ×g (1N0O233 )m2 =ol– 11]48;
Panaskan Berapakah bilangan molekul oksigen apabila 7.4 g magnesium
nitrat terurai apabila dipanaskan?
Which of the following is metal X? [Jisim formula relatif Mg(NO3)2 = 148; Pemalar Avogadro =
6.02 × 1023 mol–1]
Antara berikut, yang manakah mungkin bagi logam X? A 1.505 × 1022
B 3.010 × 1022
A Zinc C Tin C 1.505 × 1023
D 3.010 × 1023
Zink Stanum
12 The equation below shows the chemical equation of the
B Lead D Copper combustion of ethanol in excess oxygen.
Persamaan di bawah menunjukkan persamaan kimia pembakaran
Plumbum Kuprum etanol dalam oksigen berlebihan.

10 The following equation shows the decomposition reaction 2C2H5OH + 6O2 4CO2 + 6H2O
of lead(II) nitrate when heated at room temperature and
pressure. What is the volume of carbon dioxide gas released when
Persamaan tindak balas di bawah menunjukkan penguraian 9.20 g ethanol burnt completely?
plumbum(II) nitrat apabila dipanaskan pada suhu dan tekanan [Relative atomic mass of H = 1, C = 12, O = 16, 1 mol
bilik. of gas occupies 24 dm3 at room condition]
Apakah isi padu gas karbon dioksida dibebaskan apabila 9.20 g
2Pb(NO3)2 2PbO + 4NO2 + O2 etanol terbakar lengkap?
[Jisim atom relatif: H = 1, C = 12, O = 16, 1 mol gas menempati
Which of the following are true when 0.1 mol of lead(II) 24 dm3 pada keadaan bilik]
nitrate is decomposed? A 4.8 cm3
[Relative atomic mass: N = 14, O = 16, Pb = 207 B 9.6 cm3
and 1 mol gas occupies the volume of 24 dm3 at room C 96.0 cm3
temperature and pressure] D 9 600 cm3
Antara berikut, yang manakah adalah benar apabila 0.1 mol
plumbum(II) nitrat terurai? 13 What is the percentage by mass of nitrogen content in
[Jisim atom relatif: N = 14, O = 16, Pb = 207 dan 1 mol gas Hur =ea ,1 CaOn(dN OH =2)2 1? 6[]Relative atomic mass: C = 12, N = 14,
menempati isipadu 24 dm3 pada suhu dan tekanan bilik] Apakah peratus kandungan nitrogen mengikut jisim dalam urea,
I 66.2 g of lead(II) oxide is formed CO(NH2)2? [Jisim atom relatif: C = 12, N = 14, H = 1 dan
66.2 g plumbum(II) oksida terbentuk O = 16]
II 22.3 g of lead(II) oxide is formed A 23.3%
22.3 g plumbum(II) oksida terbentuk B 31.8%
III 2.4 dm3 of oxygen gases is given off C 46.7%
2.4 dm3 gas oksigen dibebaskan D 63.6%
IV 4 800 cm3 of nitrogen dioxide given off
4 800 cm3 nitrogen dioksida dibebaskan
A I and III only

I dan III sahaja
B I and IV only

I dan IV sahaja
C II and III only

II dan III sahaja
D II and IV only

II dan IV sahaja

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