2020_Skema Bab 3 Konsep mol, Formula dan persamaan kimia

BAB Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia
3
KONSEP MOL, FORMULA DAN PERSAMAAN KIMIA

MOLE CONCEPT, FORMULAE AND CHEMICAL EQUATIONS

3.1 Jisim Atom Relatif (JAR) dan Jisim Molekul Relatif (JMR)

Relative Atomic Mass (RAM) and Relative Molecular Mass (RMM)

3.1.1 Jisim Atom Relatif, JAR

Relative Atomic Mass, RAM

1. Maksud Jisim purata satu atom unsur apabila dibandingkan dengan 1/12 daripada
jisim satu atom carbon –12
Meaning
The average mass of one atom of the element when compared wxith 1/12 of the
2. Formula mass of an atom of carbon-12.

Formula Jisim atom relatif, JAR satu element

3. Contoh Relative Atomic Mass, RAM of an element

Example = Jisim satu atom unsur @ Mass of one atom of the element

1/12 kali jisim satu atom karbon-12 1/12 x mass of one carbon-12 atom

JAR magnesium 24 = 24
1 12
RAM of magnesium = 12

= 24 kali lebih besar magnesium berbanding karbon-12

24 timers larger of magnesium than carbon-12

Latihan 1: Tulis Jisim Atom Relatif, JAR (rujuk Jadual Berkala Unsur)

Write the Relative Atomic Mass, RAM (refer The Periodic Table of Element)

Atom Jisim atom relatif Atom Jisim atom relatif

Atoms Relative atomic mass Atoms Relative atomic mass
7 65
a) Litium,Li e) Zink, Zn
39 20
Lithium, Li Zinc, Zn
24 23
b) Kalium,K f) Neon,Ne
16 40
Potassium,K Neon, Ne

c) Magnesium,Mg g) Natrium,Na

Magnesium,Mg Sodium,Na

d) Oksigen,O h) Kalsium,Ca

Oxygen,O Calcium,Ca

Latihan 2 : Jisim atom kuprum =
Jisim atom sulfur
a) Berapa kali atom kuprum lebih berat berbanding
satu atom sulfur? 64/32 = 2 kali/ times

How many times is copper atom heavier than one
sulphur atom?
[JAR/ RAM ; Cu, 64 ; S, 32]

b) Hitungkan bilangan atom karbon yang mempunyai 108/12 = 9 kali / times (2 kali)
jisim yang sama seperti 1 atom argentum? (9 kali)

How many carbon atoms that have the same mass as 1
silver atom?
[JAR/ RAM ; C, 12, Ag, 108]

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Bab 3_Formula dan Persamaan Kimia

Latihan 3 : selesaikan / Solve it

a) Berapa kali atom magnesium yang sama Mg = 2 x Ag
dengan dua atom argentum? = (2 x 108) / 24 = 9 kali

How many magnesium atoms have the same mass (9 kali)
as two silver atoms? Bilangan atom oksigen/ number of oxygen
= 64/ 16 = 4 kali /times
b) Kira bilangan atom oksigen yang sama
dengan jisim satu atom kuprum? (4 kali)
[JAR/ RAM ; Mg;24, Ag;108, Cu; 64, O;16]
Calculate the number of oxygen atoms that has
the same mass as one copper atom.

3.1.2 Jisim Molekul Relatif (JMR)

Relative Molecular Mass (RMM)

1. Maksud Jisim purata molekul apabila dibandingkan dengan 1/12 daripada jisim
satu atom karbon-12.
Meaning
The number of times the mass of a molecule of that substance is heavier than
2. Hubungan 1/12 of a carbon-12 atom.

Relatioship Jisim molekul relatif, JMR sebatian

3. Pengiraan Relative Molecular Mass, RMM of compound

Calculation = Jisim satu molekul sebatian @ Mass of one molecule of compound
1/12 x jisim satu atom karbon-12 1/12 x mass of one carbon-12 atom

Jumlah jisim atom relatif semua atom dalam molekul.

Sum of relative atomic masses of all the atoms in the molecule.

3.1.3 Jisim Formula Relatif (JFR)

Relative Formula Mass (RFM)

1. Hubungan Jisim molekul relatif, JMR sebatian

Relatioship Relative Formula Mass, RFM of compound

= Jisim satu formula sebatian @ Mass of one formulae of compound

1/12 x 1/12 x jisim satu atom karbon-12 mass of one carbon-12 atom

2. Pengiraan Jumlah jisim atom relatif semua atom dalam sebatian

Calculation Sum of relative atomic masses of all the atoms in the compound.

Latihan 4; Kira Jisim Molekul Relatif, JMR bagi setiap sebatian molekul

Calculate the relative molecular mass, RMM of the following molecular compound.
[JAR/ RAM ;N;14; C;12; Na;23, S;32; H ; 1, Cu; 64; O = 16]

Molekul Jisim molekul relatif, JMR

Molecules Relative Molecular Mass, RMM

a) Gas oksigen,O2 2 x 16 = 32 (32)
2 x 14 = 28 (28)
Oxygen gas, (1x14) + (3x1) = 17 (17)
72 + 12 + 96 = 180 (180)
b) Gas nitrogen,N2 12 + 32 = 44 (44)
14 + 32 = 46 (46)
Nitrogen gas

c) Ammonia, NH3

Ammonia, NH3

d) Glukosa, C6H12O6

Glucose,

e) Karbon dioksida, CO2

Carbon dioxide,

f) Nitrogen dioksida, NO2

Nitrogen dioxide, NO2

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Bab 3_Formula dan Persamaan Kimia

Latihan 5 : Kira Jisim Formula Relatif, (JFR) bagi sebatian ion berikut

Calculate the Relative Formulae Mass, (RFM) of the following ionic compound.

Sebatian Jisim formula relatif (JFR)

Compounds Relative Formulae Mass, (RFM)

a) Natrium hidroksida, NaOH (23x1) + (1x16) + (1x1) = 40

Sodium hydroxide, (40)

b) Aluminium klorida, AlCl3 (1x27) + (35.5x3) = 133.5

Aluminium chloride, (133.5)

c) Natrium karbonat, Na2CO3 (2x23) + (1x12) + (3x16) = 106

Sodium carbonate, (106)

d) Magnesium sulfat,MgSO4 (1x24) + (1x32) + (4x16) = 120

Magnesium sulphate, (120)

e) Ammonium karbonat (NH4)2CO3 (2x14) + (8x1) + (1x12) x (3x16) = 96

Ammonium carbonate, (96)

[JAR/ RAM ;Al,27,Cl,35.5;Na;23; S;32;N,14; H ; 1, C;12; O = 16]

Latihan 6 : Kira jisim formula relatif

Calculate the relative formulae mass

a) Natrium tiosulfat , Na2S2O3 . 5H2O = 64 + 48 + 10 + 80
= 248
Sodium thiosulphate, Na2S2O3 . 5H2O
(248)
b) Kuprum(II) sulfat, CuSO4 .5H2O
= (64x1) + (32x1) + (4x16) + (10x1) + (5x16)
Copper(II) sulphate, CuSO4. 5H2O
= 250

(250)
[JAR/ RAM ; Na;23, S;32; H ; 1, Cu; 64; O = 16]

Soalan 7; Hitung jisim molekul atau formula bagi setiap bahan berikut.

Calculate the relative molecular or formula masses of the following substances.
[JAR/ RAM ; Na,23, S, 32; H,1; C,12, O,16 ; N,14 ;Zn,65]

a) Etanol, C2H5OH. b) Zink nitrat, Zn(NO3)2

Ethanol Zinc nitrate

= (23x2) + (5x1) +(1x16) +(1x1) = (65x1) + (2x14) + (6x16)
= 46
= 189
(46) (189)

Soalan 8 ; Selesaikan = (39x3) + (56x1) + (12x6) + (14x6) (329)
= 329 (80)
Solve it

a) Formula kimia bagi kalium heksasianoferrat (III)
ialah K3Fe(CN)6 . Apakah jisim formula relatif ?

The chemical formula for potassium hexacyanoferrate
(III) is K3Fe(CN)6 . What is its relative formula mass?

[JAR/ RAM ; C, 12; N, 14; K, 39; Fe;56 ]

b) Formula kimia bagi ammonium nitrat adalah = (14x1) + (1x4) + (14x1) + (16x3)
NH4NO3. Apakah jisim formula relatifnya? = 80

The chemical formula for ammonium nitrate is
NH4NO3. What is its relative formula mass?

[JAR/ RAM ; H;1,N ;14 ,O ;16]

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Bab 3_Formula dan Persamaan Kimia

3.2 Konsep Mol

Mole Concept

1. Mol • Jumlah bahan yang mengandungi bilangan zarah yang sama dengan bilangan
atom yang terdapat di dalam 12 g karbon-12
Mole
An amount of substance that contains as many particles as the number of atoms in
exactly 12 g of carbon-12.

• Unit : mol

3.2.1 Bilangan mol dengan pemalar Avogadro, NA dan bilangan zarah

The numbers of mole with the Avogadro contant,NA and the number of particles

1. Pemalar • Jumlah zarah dalam satu mol bahan
Avogadro, NA
The number of particles in one mole of a substance
Avogadro
Constant, NA • Jumlah zarah dalam satu mol ialah 6.02 x 1023

The number of particles in one mole is 6.02 x 1023 particles.

2. Contoh Jenis zarah / Type of particles
Atom
Example
• 1 mol atom natrium mengandungi 6.02 x 1023 atom

1 mole of sodium atoms contains 6.02 x 1023 sodium atoms

Molekul

Molecule

• 1 mol molekul air mengandungi 6.02 x 1023 molekul H2O

1 mole of water molecules contains 6.02 x 1023 H2O molecules

Ion
✓ 1 mol ion magnesium mengandungi 6.02 x 1023 ion Mg2+

1 mole of magnesium ions contains 6.02 x 1023 Mg2+ ions

3. Formula  Bilangan zarah = bilangan mol,n x pemalar Avogadro,NA

Formulae Number of particles = number of moles,n x Avogadro constant, NA

Pemalar Avogadro, NA = 6.02 x 1023 zarah

Avogadro’s constant, NA = 6.02 x 10 23 particles

• Jenis zarah : Atom

Type of particle: Atom

Latihan 9 ; Hitungkan [Pemalar Avogadro/ Avogadro constant, NA = 6.02 x 1023 zarah]

Calculate

Atom Bilangan mol Bilangan atom

Atom Number of moles Number of atoms

a) Litium,Li = 3.0 x 6.02 x1023

Lithium, Li 3.0 = 1.806 x 1024

b) Kalsium,Ca = 1.204 x1023 = 0.2 mol
6.02 x1023
Calcium, Ca 1.204 x1023

c) Aluminium,Al = 0.5 x 6.02 x1023

Aluminium, Al 0.5 = 3.01 x 1023

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Bab 3_Formula dan Persamaan Kimia

Latihan 10 : Selesaikan

Solve it [Diberi Pemalar Avogadro/ Given Avogadro constant,NA = 6.02 x 10 23]

a) Berapakah mol atom magnesium yang b) Berapakah atom yang terdapat di dalam 1 mol
terdapat dalam 5.02 x 1021. atom hidrogen.

How many moles atoms of magnesium are there How many atoms are there in 1 mole of hydrogen
in 5.02x1021 atoms of magnesium? atoms?

mol = 5.02 x1021 Bilangan atom / number of atoms
6.02 x1023 = 1 x 6.02 x1023
= 6.02 × 10 23 atom
= 8.34 x 10-3 mol
( 6.02 × 10 23 atom)
(8.34x10-3 mol)

• Jenis zarah : Molekul

Type of particle: Molecule

Formula  i) Bilangan molekul = mol x NA

Number of molecules mol x NA

ii) Bilangan atom = mol x NA x bilangan atom

Number of atoms mol x NA x number of atom

(NA = 6.02 x 10 23 zarah/particles)

Latihan 11 : Selesaikan/ Solve it

Bahan Bilangan mol Bilangan molekul Bilangan atom

Substance Number of moles Number of molecules Number of atoms

Hidrogen, H2 ✓ Bil mol x NA x 1 ✓ Bil mol x NA x blgn atom
molekul
Hydrogen, H2 mol x NA x no of atom
mol x NA x 1 molecule ✓ 2 x 6.02 x 10 23 x 2
2.0 ✓ 2 x 6.02 x 1023 x 1 ✓ 2.408 x 10 24 atom
✓ 1.204 x 10 24 molekul
a) Gas klorin, 1.2 = 1.2 x 6.02 x1023 x 2
= 1.2 x 6.02 x1023
Chlorine gas, = 1.44 x 1024
Cl2 = 7.2x1023

b) Gas nitrogen, = 1.20 x1024 1.2x1024 = 2 x 6.02 x1023 x 2
6.02 x1023 = 2.408 x 1024
Nitrogen gas
N2 = 2.0 mol = 0.075 x 6.02 x1023 x 1 = 0.075 x 6.02 x1023 x 4

c) Ammonia 0.075

Ammonia, = 4.5x1022 = 1.8x1023
NH3

[Pemalar Avogadro/Avogadro constant,NA = 6.02 x 10 23]

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Bab 3_Formula dan Persamaan Kimia

Latihan 12 : Selesaikan

Solve it [Pemalar Avogadro/Avogadro constant,NA = 6.02 x 10 23]

a)Bilangan mol air yang hadir dalam molekul air, H2O = 9.02 x1023
dalam 9.02 x 1023 molekul air 6.02 x1023

How many moles of water, H2O are present in 9.02 x 1023

molecules of water?

b) Satu contoh gas oksigen,O2 mengandungi = 3.01 x 1022 (1.5 mol)
3.01 x 1022 molekul. 6.02 x1023 (0.05 mol)

Berapakah bilangan mol molekul oksigen,O2 dalam

contoh itu

A sample of oxygen gas contains 3.01 x 1022 molecules.
How many moles of oxygen, O2 molecules are there in the
sample?

c) Berapakah bilangan molekul dalam 1.0 mol = 1.0 x 6.02 x1023

ammonia,NH3 (6.02 x 1023)

What is the number of molecules in 1.0 mole of ammonia?
NA =6.02 x1023

Latihan 13 : Lengkapkan jadual

Complete the table [Pemalar Avogadro/Avogadro constant, NA = 6.02 x 10 23mol-1]

Sebatian Bilangan mol Bilangan ion dalam Bilangan setiap ion

Compound Number of sebatian Errata Number of particles ions
moles
Magnesium Number of ions in
klorida, MgCl2 2.0
compound
Magnesium 0.45
chloride = bilangan mol x NA i) Ion magnesium / Magnesium ion, Mg2+
MgCl2 = 2.0 x 6.02 x 1023 x 1
number of mol x NA = 1.2 08 x 1024
a) Natrium
oksida, = 2.0 x 6.02 x 10 23 x 3 ii) Ion klorida/ Chloride ion,Cl-
Na2O = 3.612 x 10 24

Sodium = 2.0 x 6.02 x 1023 x 2 =2.4 08 x 1024
oxide,
Na2O = 0.45 x 6.02 x1023 i) Ion natrium /Sodium ion,Na+

= 0.45 x 6.02 x1023 x 2

(5.4x1023)

ii) Ion oksida/Oxide, O2- ion

= 0.45 x 6.02 x1023 x 1

(8.13x1023) (2.71 x 1023 )

i) Ion aluminium/Aluminium ion, Al3+

b) Aluminium = 0.14 x 6.02 x1023 = 0.14 x 6.02 x1023 x 1
klorida,
AlCl3 0.14 (8.4x1022)

Aluminium ii) ion klorida /Chloride ion, Cl-
chloride,
AlCl3

= 0.14 x 6.02 x1023 x 3

(3.37x1023)

(2.52x1023)

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Bab 3_Formula dan Persamaan Kimia

Latihan 14 : Selesaikan

Solve it

a) Cari bilangan ion natrium yang hadir dalam 0.5 = 0.5 x 6.02 x1023 x 2
mol natrium oksida,Na2O = 6.02 x 1023 ion

What is the number of sodium ions present in 0.5 mol of
sodium oxide, Na2O.

b) 0.4 mol kuprum(II) klorida, CuCl2 dilarutkan dalam = 0.4 x 6.02 x1023 x 3 (6.02 x 1023 ions)
satu bikar yang mengandungi air. = 7.224 x 1023 ion ( 7.224 x 1023 ions)

Kira jumlah bilangan ion yang terdapat dalam bikar
0.4 mole of copper (II) chloride is dissolved in a beaker
of water. Calculate the total number of ions present.

c) Satu botol yang tertutup mengandungi 4 mol molekul oksigen,O2

A closed glass bottle contains 4 mol molecules of oxygen, O2.
[Pemalar Avogadro/Avogadro constant,NA = 6.02 x 10 23]

i) Apakah bilangan molekul oksigen dalam ii) Berapakah atom oksigen yang terdapat
botol? di dalam botol?

What is the number of oxygen molecules in How many oxygen atoms are there in the
the bottle? bottle?

= 4 x 6.02 x1023 = 4 x 6.02 x1023 x 2

(2.41 x 1024) (4.816 x 1024 )

3.2.2 Bilangan mol, jisim molar dan jisim bahan

Number of moles, molar mass and mass of substances

1. Jisim molar Jisim satu mol bagi sebarang bahan

Molar mass The mass of one mole of any substance

2. Unit Jisim 1 mol bahan

Unit the mass of 1 mole of the substances

unit; g mol-1

3. Konsep • 1 mol sebarang bahan : mengandungi 6.02 x 1023 zarah

Concept 1 mol of any substances : contains 6.02 x 1023 particles

• Maka jisim molar bahan adalah jisim 6.02 x 1023 zarah bahan.

So, the molar mass of the substances is the mass of 6.02 x 1023 particles of

substances.

Latihan 15 ; Lengkapkan jadual berikut .

Complete the table as below

Bahan Jisim relatif Jisim molar

Substances Relative mass Molar mass
17 gmol-1
b) Ammonia = 1(14) + 3(1)
Ammonia, NH3 = 17 (17 g mol-1)

c) Magnesium oksida, = 1(24) + 1(16) 40 gmol-1
= 40
Magnesium oxide, (40 g mol-1)
= 1(1) + 1(14) + 3(16)
MgO = 63 63 gmol-1

d) Asid nitrik (63)
[JAR/JMR: Mg;24, H ; 1, N; 14; O;16]
Nitric acid,
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HNO3

4. Formula  Jisim bahan = Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia
Formulae Mass of substances
Bilangan mol x jisim molar

Number of mole x molar mass

Latihan 16: Kira jisim molar, jisim bahan dan bilangan mol zarah

Calculate the molar mass, mass of element and number of mole particles

Bahan Mol Jisim molar (g mol-1) Jisim bahan (g)

Substances Mole Molar mass, (g mol-1) Mass of substances (g)

a) Aluminium, Al 2.0 mol

Aluminium,Al

= 27 moldm-3 = 2.0 x 27

b) Gas karbon = 11/ 27 = 44 moldm-3 11 g
dioksida,CO2 = 28.4/ 142

Carbon dioxide 28.4 g
gas,CO2
= 142 moldm-3
c) Natrium sulfat
Na2SO4

Sodium sulphate,
Na2SO4

[JAR/RAM : Al;27,S;32,C;12,O;16,Na;23]

5. Formula  Jisim bahan = Bilangan zarah x Jisim molar
Pemalar Avogadro
Formulae Mass of substances
Number of particles x Molar mass
Avogadro Constant,NA

NA = 6.02 x 1023 zarah/ particles

Latihan 17 : Selesaikan Jisim,g Bilangan zarah (atom/molekul/ion)

Solve it Mass (g) Number of particles (atom/molecule/ion)

Bahan 16 g Bilangan atom

Substance Number of atoms

a) Kuprum,Cu Mol =16/64 = 0.25
Bilangan atom = 0.25 x 6.02 x 1023
Copper, Cu Number of atoms

b) Berilium,Be = 9x 1023 = 1.5 mol (1.505x1023 atom)
6.02 x 1023
Beryllium, Be 9 x 1023 atom
= 1.5 x 9

= 13.5 g

c) Ammonia, (13.45 g)

Ammonia, Bilangan molekul

NH3 number of molecules

8.5g

Mol =8.5/17 = 0.5

Bilangan molekul = 0.5 x 6.02 x 1023 x 1

(3.01x1023 molekul)

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Errata

Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia

d) Bilangan ion
i) Kalium
Number of ions
oksida, K2O
28.2 g Mol =28.2/94 = 0.3
Potassium 28.2 g Bilangan atom = 0.3 x 6.02 x 1023 x 3
oxide
28.2 g (5.418 x1023 )
ii) Ion
kalium,K+ Bilangan ion kalium

potassium number of potassium ion
ion,
Mol =28.2/94 = 0.3
iii) Ion oksida,
O2- Bilangan atom = 0.3 x 6.02 x 1023 x 2

oxide ion, Number of atoms
(3.612x1023 )

Bilangan ion oksida

number of oxide ion

Mol =28.2/94 = 0.3

Bilangan atom = 0.3 x 6.02 x 1023 x 1

Number of atoms
(1.806x1023 )

[JAR/RAM ; Be;9; Cu,64; N,14; H,1; K,39; O,16; NA,6.02 x 1023]

Soalan 18 ; =0.32 /64 = 0.005 mol
= 0.005 x 6.02 x 1023
a) Berapakah atom kuprum,Cu yang terdapat = 3.0 x 1021
dalam 0.32 g kuprum?

How many atoms of copper,Cu are there in
0.32 g of copper?

b) Hitungkan bilangan molekul dalam 16 g = 16 / 32 = 0.5 mol ( 3.0 x 10 21)
gas oksigen,O2 = 0.5 x 6.02 x 1023 ( 3.0 x 1023 )
= 3.0 x 1023 (3.01x 10 23)
Calculate the number of molecules in 16 g of
oxygen gas

c) Berapakah bilangan atom zink yang = 3.25 x / 65 = 0.05 mol
= 0.05 x 6.02 x 1023
terdapat dalam 32.5 g zink? = 3.01x 10 23

How many zinc atoms are there in 32.5 g of
zinc?
[NA = 6.02 x 1023 mol-1, JAR/RAM zink = 65]

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3.2.3 Bilangan Mol, isipadu molar dan isi padu gas

Number of Moles, molar volume and volume of gases

1. Isi padu molar Isi padu yang dipenuhi oleh satu mol gas.

Molar volume of gas The volume occupied by one mole of the gas

2. Konsep Satu mol bagi sebarang gas mengandungi isipadu yang sama di bawah
suhu dan tekanan yang sama.
Concept
One mol of any gas occupies the same volume under the same temperature
and pressure.

Isipadu molar sebarang gas adalah ;

The molar volumes of any gas;

• 22.4 dm3 pada suhu dan tekanan piawai /at standard temperature
and pressure (s.t.p)

(temperature = 00 C and 1 atmosphere pressure)

• 24 dm3 pada suhu bilik / at room conditions, (r.c)

(temperature = 250 C and 1 atmosphere pressure)

• Unit dm3 mol-1

3. Hubungan diantara mol dan isi padu molar

Relationship between moles and molar volumes

1 mol sebarang gas = 22.4 dm3 pada s.t.p

1 mole for any gas = 24 dm3 pada suhu bilik

• 22.4 dm3 = 22 400 cm3
• 24 dm3 = 24 000 cm3

4. Hubungan diantara mol dan bilangan zarah

Relationship between moles and number of particles

1 isi padu sebarang gas = 1 mol zarah

1 molar volume any gas = 1 mol particles

= 6.02 x 1023 particles

Contoh/ Examples ;

1 mol gas oksigen = 6.02x 1023 molekul/ molecule

1 mol oxygen gas

1 mol gas karbon dioksida = 6.02x 1023 molekul/ molecule

1 mol carbon dioxide gas

Latihan 19 ; Selesaikan

Solve it [isi padu molar gas pada STP/molar volume of gas at STP : 22.4 dm3 mol-1]

a) Hitungkan isipadu yang terisi oleh 3.5 mol gas = 3.5 x 22.4
ammonia pada suhu dan tekanan piawai. = 78.4 dm3

Calculate the volume occupied by 3.5 moles (78.4 dm3)
ammonia gas , NH3 at STP (0.224 dm3)
[isipadu molar/molar volume ; 22.4 dm 3 pada STP]

b) Apakah isi padu bagi 0.01 mol gas hidrogen = 0.01 x 22.4
pada suhu dan tekanan piawai. = 0.224 dm3

Calculate the volume of 0.01 mol hydrogen gas at
STP
[isipadu molar/molar volume ; 22.4 dm 3 pada STP]

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c) Apakah isi padu 1.2 mol gas ammonia pada = 1.2 x 22.4 Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
suhu dan tekanan piawai. = 26.88 dm3 Bab 3_Formula dan Persamaan Kimia

What is the volume of 1.2 mol of ammonia gas at = 0.6 x 24 (26.88 dm3)
STP? = 0.025 dm3
[isipadu molar/molar volume ; 22.4 dm 3 pada STP] (0.025 dm3)

d) Berapakah mol gas nitrogen yang terdapat
dalam 600 cm3 yang diukur pada suhu bilik

How many moles of nitrogen are present in 600
cm3 of the gas measured at room condition?
[isipadu molar/molar volume ; 24 dm 3 pada suhu bilik]

Soalan 20 ;

Jadual 20 menunjukkan formula empat jenis gas yang dilepaskan semasa letusan gunung berapi.

Diagram 20 shows the formulae of four types of gases released during the eruption of a volcano.

N2 CO2 H2S Wap air, H2O

Water vapour

Jadual 20/ Table 20

Berdasarkan rajah 20;

Based on diagram 20

a) Hitungkan jisim molar bagi setiap gas berikut. b) Adakah 0.9 g wap air,H2O mengandungi
bilangan molekul yang sama seperti dalam 2.2 g
Calculate the molar mass of each gas gas karbon dioksida, CO2. Buktikan
[JAR/RAM : H, 1; C, 12; O, 16; N, 14; S, 32]
Does 0.9 g of water vapour contain the same
i) N2 number of molecules as in 2.2 g of carbon dioxide.
Prove it
Jisim molar, N2 [JAR/RAM : H, 1; C, 12; O, 16; NA = 6.02 X 1023]
Molar mass

= 14 x2 = 28 Ya / Yes
Bilangan molekul, H2O
ii) CO2 Number of molecules
= 0.9/18 x 6.02 x 1023
Jisim molar, = 3.01 x 1023
Molar mass
= (12x1) + (16 x 2) = 44

iii) H2S Bilangan molekul, CO2
Number of molecules
Jisim molar,
Molar mass = 22/44 x 6.02 x 1023
= 3.01 x 1023
= (1x2) + (1x32) = 34

iv) H2O

Jisim molar,
Molar mass
= (2x1) + (1x16) = 18

[4 markah] [4 markah]
(28, 44, 34,18) (Ya, 3.01 x 1022 )

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Bab 3_Formula dan Persamaan Kimia

Latihan 21 : Selesaikan

Solve it

a) Hitungkan bilangan mol 720 cm3 oksigen,O2 pada = 720/1000 = 0.025 x 71
= 0.03 mol
keadaan bilik
( 0.03 mol)
Calculate the number of moles of 720 cm3 oxygen,O2 at (8.96 dm3)
room conditions.
[isipadu molar/molar volume ; 24 dm 3 pada suhu dan (1.775 g)
tekanan bilik]

b) Apakah isi padu 12.8 g gas oksigen,O2 pada STP = 12.8 / 32 = 0.4 x 22.4
= 8.96 dm3
What is the volume of 12.8g of oxygen gas, O2 at STP?
[isipadu molar/molar volume ; 22.4 dm 3 pada STP]

c) Apakah jisim bagi 0.6 dm3 gas klorin,Cl2 pada = 0.6/ 24 = 0.025
= 0.025 x 71
suhu dan tekanan bilik.? = 1.775 g

What is the mass of 0.6 dm 3 of chlorine gas, Cl2 at

room temperature and pressure?
[JAR: Cl, 35.5; Molar volume at room temperature and
pressure, 24 dm3]

3.3 Formula Kimia

Chemical Formula

1. Maksud Formula dimana perwakilan bahan kimia yang menggunakan huruf bagi unsur
dan bilangan pekali bagi jenis atom yang hadir dalam bahan itu.
Meaning
Formula that representation of chemical substance using letters for elements and
2. Terdiri subscript number for type of atoms present in the substance.

Consist a) unsur yang hadir dalam sebatian.

the elements present in the compound

b) nisbah atom bagi setiap unsur yang membentuk sebatian

the ratio of atom for each element to form the compound.

3.3.1 Formula Empirik dan Formula Molekul

Empirical Formula and Molecular Formula

1. Jenis a) Formula empirik
formula
Empirical formula
Type of
formulae • Formula kimia yang menunjukkan nisbah teringkas bagi atom setiap unsur di
dalam sebatian.

Chemical formula that show the simplest number ratio of atom of element in a
compound.

• Contoh : Glukosa : C H2O Propena : CH2

Example Glucose Propene

b) Formula molekul

Molecular formula

• Formula kimia yang menunjukkan bilangan sebenar atom bagi setiap unsur
di dalam satu molekul sebatian.

Chemical formula that shows the actual number of atoms of each element in one
molecule of the compound

• Contoh : Glukosa : C6H12O6 Propena : C3H6

Example Glucose Propene

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Latihan 22

Contoh ; Natrium klorida a) Magnesium bromida b) Kuprum(II) oksida
Magnesium bromide Copper (II) oxide
Sodium chloride

Simbol natrium Simbol klorin Magnesium bromine Kuprum oksigen

Sodium symbol Chlorine symbol MgBr2 CuO

NaCl Nisbah : Nisbah :
1 atom magnesium 1 atom kuprum
Nisbah : 1 atom Na ; 1 atom Cl : 2 atom bromin :1 atom oksigen

Ratio : 1 Na atom : 1 Cl atom

Latihan 23 ; Lengkapkan jadual berikut

Complete the table as following

Sebatian Unsur hadir Nisbah atom

Compound Element present Ratio of atom

a) Klorin, Cl2 Klorin 2 atom klorin
Chlorine 2 chlorine atom.
Chlorine,
2 atom hidrogen
b) Air, H2O Hidrogen/ hydrogen 1 atom oksigen/oxygen
Oksigen/ oxygen
Water H2O 2 atom hidrogen
1 atom sulfur/ sulphur
c) Asid sulfurik, Hidrogen/hydrogen 4 atom oksigen/oxygen
H2SO4 Sulfur/ sulphur
Oksigen/oxygen 1 atom magnesium
Sulphuric acid , 2 atom nitrogen
Magnesium
d) Magnesium nitrat Nitrogen 6 atom oksigen/oxygen
Mg(NO3)2
Oksigen/oxygen
Magnesium nitrate,

3.3.2 Penentuan Formula Empirik

Determination of an Empirical Formula

Pengiraaan  : Formula Empirik

Calculation Empirical Formula

Latihan 23: Unsur Kuprum,Cu Oksigen,O

a) Dalam satu experiment, 1.60 g Element Copper, Cu Oxygen, O
kuprum bertindak balas dengan 1.60 0.40
0.40 g oksigen membentuk Jisim (g)
kuprum (II) oksida. 1.6 / 64 0.4/16
Apakah formula empirik sebatian Mass (g) =0.025 =0.025
ini.
Bilangan 0.025/0.025 0.025/0.025
In an experiment, 1.60 g of copper mol =1 =1
reacts with 0.40 g of oxygen to form
copper oxide. Number of
What is the empirical formula of this moles
compound?
[JAR/RAM : O, 16 ; Cu, 64] Nisbah
ringkas mol
Formula empirik : CuO
Simplest
Empirical formula ratio of mole

(CuO)

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b) 1.69 g ferum berpadu dengan 0.72 Unsur Ferum Oksigen
g oksigen. 1.69 0.72
Apakah formula empirikal bagi Element
sebatian oksida ini. 1.69/56 0.72/16
Jisim (g) = 0.030 0.045
1.69 g of iron combines with 0.72 g
of oxygen. Mass (g) 0.030/0.030 0.045/0.030
What is the empirical formula of this =1 =1.5
oxide. Bilangan =1x2=2 = 1.5 x 2 =3
mol
[JAR/RAM : Fe,56 ; O,16]
Number of
Formula empirik : Fe2O3 moles

Empirical formula Nisbah
(Fe2O3) ringkas mol

Simplest
ratio of mole

c) Apakah formula empirikal bagi Unsur Karbon, C Hydrogen,H
pembentukan sebatian apabila 6 g 6 2
karbon berpadu dengan 2 g Element
hidrogen 6/12 = 0.5 2/1 = 2.0
Jisim (g)
What is the empirical formulae for the 0.5 / 0.5 2.0 / 0.5
formation compound when 6 g of Mass (g) =1 =4
carbon combined with 2 g of
hydrogen. Bilangan O
mol 0.79
[JAR/RAM : C,12 ; H, 1] = 0.79/ 16
Number of = 0.049
Formula empirik : CH4 moles
0.79/0.033
Empirical formula Nisbah =1.5
ringkas mol = 1.5 x 2 = 3

Simplest
ratio of mole

[CH4]

d) 0.90 g aluminium terbakar dalam Unsur Al
udara menghasilkan 1.7 g 0.91
aluminium oksida. Element = 0.91 / 27
Apakah formula bagi aluminium = 0.033
oksida Jisim (g)

0.90 g of aluminium burns in air to Mass (g)
form 1.7g of aluminium oxide.
What is the formula of aluminium Bilangan
oxide? mol

[JAR/RAM : Al; 27 ; O, 16] Number of
moles

Formula empirik : Al2O3 Nisbah 0.033/0.033
ringkas mol =1
Empirical formula =1x2 =2
Simplest
ratio of mole

(Al2O3)

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Latihan 24 ; Selesaikan Unsur X O
1.08 0.96
Solve it; Element 1.08/x 0.96 / 16
=0.06
a) 1.08 g unsur X bertindak balas dengan Jisim (g)
oksigen membentuk 2.04 g sebatian 1.08 /x = 0.06
dengan formula empirik X2O3. Mass (g) 23
Berapakah jisim atom relatif X.
Bilangan x = 1.08 / 0.04
g of element X reacts with oxygen to form mol = 27
2.04 g of a compound with an empirical
formula of X2O3. Number of Jisim atom relatif X = 27
What is the relative atomic mass of X? moles Relative atomic mass
[JAR/RAM : O, 16]
Nisbah
b) 20.8 g unsur M bergabung dengan 9.6 ringkas mol
g oksigen membentuk oksida dengan
formula empirik M2O3 Simplest ratio
Hitungkan jisim atom relatif unsur M of mole

20.8 g of an element M combines with 9.6 (27)
g of oxygen to form an oxide with
empirical formula of M2O3. Unsur M O
Determine the relative atomic mass of 20.8
element M. Element 20.8/ x 9.6
[JAR/RAM : O, 16]
Jisim (g) 9.6/ 16
=0.6
Mass (g)
20.8/x = 0.6
Bilangan 23
mol
x = 20.8 / 0.4
Number of = 52
moles
Jisim atom relatif M = 52
Nisbah ringkas Relative atomic mass
mol

Simplest ratio of
mole

( 52)

Pengiraaan  : Formula Empirik - peratus mengikut jisim %

Calculation Empirical Formula - percentage by mass %

Konsep : peratus mengikut jisim (100% = 100 g)

Concept percentage by mass

Latihan 25 :

a) Asid laktik mengandungi 40.00 % karbon, 6.67 % hidrogen dan 53.33 % oksigen mengikut jisim.
Apakah formula empirik sebatian itu

Lactic acid contains 40.00 % carbon, 6.67 % hydrogen and 53.33 % oxygen by the mass.
What is the empirical formula of the compound?
[ JAR/RAM : H = 1, C =12, O = 16]

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Element
40 HO
Peratus jisim % 40/12 = 3.334
6.67 53.33
Mass percentage %
6.67 53.33
Jisim (g)/ Mass (g) 6.67/1 =6.67 53.33 / 16 = 3.33

Bilangan mol

Number of moles

Nisbah ringkas mol 3.334/3.33 =1 6.67/ 3.33 =2 3.33/3.33 =1

Simplest ratio of mole

Formula empirik : CH2O ( CH2O)

Empirical formula

b) Satu metal T klorida tertentu mengandungi 67.19% klorida dengan jisim.
Tentukan formula empirik bagi metal T klorida ini.

A certain chloride of metal T contains 67.19% of chloride by mass.
Determine the empirical formula of this chloride of metal T.
[JAR/RAM: T, 52; Cl, 35.5]

Unsur T Cl

Element 32.81 67.19

Peratus jisim% 32.81 67.19
32.81/52 67.19/35.5
Mass percentage % = 0.63 = 1.89
0.63/0.63 =1
Jisim (g) / Mass (g) 1.89/0.63 = 3

Bilangan mol

Number of moles

Nisbah ringkas mol

Simplest ratio of mole

Formula empirik : TCl3 (TCl3)

Empirical formula

c) Berikut adalah peratus komposisi kalsium karbonat

The following is the percentage composition of calcium carbonate.

Ca = 40%, C = 12% , O = 48%

Tentukan formula empirik bagi kalsium karbonat

Determine the empirical formula of calcium carbonate.
[JAR/RAM ; Ca;40, C;12, O; 16]

Unsur Ca C O
40 12 48
Element 40 12 48
40/40 =1 12/12 =1 48/ 16 =3
Peratus jisim
1 1 3
Mass percentage %

Jisim (g)/ Mass (g)

Bilangan mol

Number of moles

Nisbah ringkas mol

Simplest ratio of mole

Formula empirik : CaCO3 (CaCO3)

Empirical formula

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d) Maklumat menunjukkan komposisi Unsur C H O
suatu sebatian organik Q Elements 13.51 21.62
Peratus jisim 64.87 13.51 21.62
The information shows the composition of Percentage by mass 64.87 13.51 /1 21.62/ 16
an organic compound Q Jisim (g) = 13.51 = 1.351
Mass (g) 64.87/12
• Karbon - 64.87% Bilangan mol = 5.406 10 1
Number of mol
Carbon 4 C4H10O (C4H10O)
Nisbah ringkas mol
• Hidrogen - 13.51% Simplest ratio of
mole
Hydrogen Formula empirik
Empirical formula
• Oksigen - 21.62%

Oxygen

Tentukan formula empirik bagi
sebatian Q

Determine the empirical formula of
compound Q
[JAR/RAM: H, 1; C, 12 ; O, 16]

[3 markah]

Pengiraaan  : Formula Empirik

Calculation Empirical Formula

Latihan 26

Jadual 26 menunjukkan keputusan eksperimen untuk menentukan formula empirik oksida logam M.
M kurang reaktif daripada hidrogen.

Table 26 shows the results of an experiment to determine the empirical formula of oxide of metal M.
M is less reactive than hydrogen.

Jisim tabung pembakaran + piring porselin 52.34 g
105.86 g
Mass of combustion tube + porcelain dish 102.02 g

Jisim tabung pembakaran + piring porselin + oksida M

Mass of combustion tube + porcelain dish + oxide of M

Jisim tabung pembakaran + piring porselin + M

Mass of combustion tube + porcelain dish + M

Jadual 26 / Table 26

Tentukan formula empirik oksida M

Determine the empirical formula of the oxide of M
[JAR/RAM O = 16, M = 207]

Unsur M O

Element 102.02 - 52.34 = 49.68 105.86 - 102.02 = 3.89
Jisim (g)
Mass (g) 49.68/207 3.89 / 16
= 0.24 = 0.24
Bilangan mol
0.24/0.24 0.24/0.24
Number of moles =1 =1

Nisbah ringkas mol

Simplest ratio of mole

Formula empirik : MO

Empirical formula

[jaw;MO]

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Latihan 27

Jadual 27 menunjukkan keputusan bagi satu eksperimen untuk menentukan formula empirik bagi
magnesium oksida.

Table 27 shows the results for the experiment to determine the empirical formula of magnesium oxide.

Jisim mangkuk pijar + penutup 28.24 g

Mass of crucible + lid

Jisim mangkuk pijar + penutup + pita magnesium 30.64 g

Mass of crucible + lid + magnesium ribbon

Jisim mangkuk pijar + penutup + magnesium oksida 32.24 g

Mass of crucible + lid + magnesium oxide

Jadual 27 / Table 27

Berdasarkan keputusan dalam jadual 27, tentukan formula empirik magnesium oksida

Based on the results in Table 27, determine the empirical formula of magnesium oxide.
[JAR/RAM : Mg = 24 , O = 16 ]

Unsur Mg O

Element 30.64 – 28.24 = 2.4 32.24 – 30.64 = 1.6

Jisim (g) 2.4/24 = 0.1 1.6/16 = 0.1
Mass (g)
0.1/0.1 = 1 0.1/0.1 = 1
Bilangan mol
MgO
Number of moles

Nisbah ringkas mol

Simplest ratio of mole
Formula empirik
Empirical formula

(MgO) [3 markah]

Latihan 28

Jadual 28 menunjukkan keputusan satu eksperimen untuk menentukan formula empirik kuprum(II) oksida
oleh seorang pelajar. Maklumat di bawah menunjukkan keputusan eksperimen

Table 28 shows the results of an experiment to determine the empirical formula of copper (II) oxide by a student
Information below shows the results of the experiment.

Huraian Jisim (g)

Description Mass (g)

Jisim tabung pembakaran + piring porselin 30.24

Mass of combustion tube + porcelain dish

Jisim tabung pembakaran + piring porselin + kuprum(II) oksida 32.26

Mass of combustion tube + porcelain dish + copper(II) oxide

Jisim tabung pembakaran + piring porselin + kuprum 31.86

Mass of combustion tube + porcelain dish + copper

Jadual 28 / Table 28

a) hitungkan bilangan mol kuprum 31.84 – 30.24 = 1.62/64 = 0.025 mol

calculate the number of moles of copper (0.025mol)
[JAR/RAM : Cu = 64]
32.26 – 31.86 = 0.4 / 16 = 0.025 mol
b) hitungkan bilangan mol oksigen
(0.025mol)
Calculate the number of moles of oxygen
[JAR/RAM : O = 16] CuO

c) Tentukan formula empirik kuprum(II) 0.025 mol;0.025 mol ;CuO
oksida

Determine the empirical formula of
copper(II) oxide

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Bab 3_Formula dan Persamaan Kimia

Latihan 29;

Satu eksperimen dijalankan dalam makmal sekolah untuk menentukan formula empirik oksida kuprum
oleh hidrogen. Jadual 29 menunjukkan keputusan yang dicapai.

An experiment was carried out in the school laboratory to determine the empirical formula of an oxide of copper by
hydrogen gas. Table 29 shows the results obtained.

Huraian Jisim (g)

Description Mass (g)

Jisim tabung pembakaran + piring porselin 36.20
39.40
Mass of combustion tube + porcelain dish 38.76

Jisim tabung pembakaran + piring porselin + kuprum(II) oksida 39.40 – 38.76 = 0.64
( 2.56g,0.64g)
Mass of combustion tube + porcelain dish + copper (II) oxide
( 1:1)
Jisim tabung pembakaran + piring porselin + kuprum ( CuO)

Mass of combustion tube + porcelain dish + copper

Jadual 29/Table 29

Berdasarkan keputusan yang diperolehi ;

Based on the results obtained,

a) Hitungkan jisim kuprum dan oksigen yang 38.76 - 36.20 = 2.56
bertindak balas
Cu : O
calculate the mass of copper and oxygen that 3 2.56/64 : 0.64 /16
have reacted. 0.04/0.04 : 0.04/ 0.04

b) tentukan nisbah mol bagi atom kuprum 11
kepada atom oksigen

determine the ratio of moles of copper atom to
oxygen atom
[JAR/RAM : Cu, 64; O, 16]

c) tentukan formula empirik kuprum(II) oksida CuO
determine the empirical formula of copper (II)
oxide

Latihan 30 :

Rajah 30 menunjukkan penyusunan radas bagi suatu eksperimen untuk menentukan formula empirik bagi
magnesium oksida.

Diagram 30 shows the set-up of apparatus for an experiment to determine the empirical formula of magnesium
oxide.

Rajah 30 / Diagram 30

Berdasarkan Rajah 30

Based on Diagram 30

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Bab 3_Formula dan Persamaan Kimia

a) Apakah maksud formula empirik?

What is the meaning of empirical formula?

Formula yang menunjukkan nisbah teringkas bagi setiap atom setiap unsur dalam sebatian
Formula that shows the simplest ration of each atom of the elements in a compound

[1markah]

b) Mengapakah penutup mangkuk pijar dibuka sekali sekala semasa eksperimen?

Why was the crucible lid opened once in a while during the experiment?

Membenarkan udara masuk ke dalam mangkuk pijar dan melengkapkan pembakaran [1markah]
To allow the movement of air into crucible and complete the reaction

c) Jadual 30 menunjukkan data yang diperoleh daripada eksperimen itu

Table 30 shows the data obtained from the experiment.

Penerangan Jisim (g)

Description Mass (g)

Jisim mangkuk pijar + penutup 24.00

Mass of crucible + lid

Jisim mangkuk pijar + penutup + jalur magnesium 26.40

Mass of crucible + lid + magnesium ribbon

Jisim mangkuk pijar + penutup + magnesium oksida 28.00

Mass of crucible + lid + magnesium oxide

Jadual 30

Berdasarkan keputusan di Jadual 30 ,

Based on results in Table 30 ;

i) Hitung jisim bagi ;

Calculate the of ;

Magnesium : 26.40 - 24.00 = 2.4 g Oksigen : 28.00 – 26.40 = 1.6 g........................

[2markah]

ii) Kira nisbah mol bagi atom magnesium kepada atom oksigen.

Calculate the mole ratio of magnesium atoms to oxygen atoms
[JAR/RAM ; O; 16, Mg; 24]

Magnesium : oksigen

2.4/24 : 1.6/16

0.1 0.1

11

[1markah]

iii) Tentukan formula empirik bagi magnesium oksida.

Determine the empirical formula of magnesium oxide.

MgO

[1markah] [2.4g/1.6g/0.1:0.1/MgO]

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Latihan 31 ;

Rajah 31 menunjukkan susunan radas untuk menentukan formula empirik bagi X oksida.

Diagram 31shows the apparatus set-up to determine the empirical formula of X oxide.

Gas hidrogen Pembakaran lebihan gas

Hydrogen gas Burning of excess hydrogen
gas

hidrogXenoksida

X oxide

Piring porselin

Porcelain dish

Kalsium klorida kontang

Anhydrous calcium chloride

Rajah 31 / Diagram 31

a) Apakah maksud formula molekul?

What is the meaning of chemical formula?

Formula yang menunjukkan nisbah sebenar bilangan atom setiap unsur dalam sebatian [1 markah]
Formula that shows the actual ratio number of atoms in the element of the compound

b) Nyatakan fungsi kalsium klorida kontang

State the function of anhydrous calcium chloride.

Mengeringkan gas hidrogen/ to dry the hydrogen gas

[1 markah]

Penerangan Jisim (g)

Description Mass (g)

Tiub pembakaran + piring porselin 112.30

Combustion tube + porcelain dish 172.10

Tiub pembakaran + piring porselin + oksida X 160.10

Combustion tube + porcelain dish + X oxide

Tiub pembakaran + piring porselin + X

Combustion tube + porcelain dish + X

Jadual 31/ Table 31

c) i) Berdasarkan Jadual 31, hitung jisim bagi:

Based on Table 31, calculate the mass of:

X: 160.10 - 112.30 = 47.8 g Oksigen : 172.10 – 160.10 = 12 g

Oxygen

[2 markah]

ii) Hitung bilangan mol atom X dan oksigen dalam sampel.

Calculate the number of mole of X atom and oxygen atom
[JAR/RAM; X = 64]

X oksigen
47.8 / 64 = 0.75 mol 12/16 = 0.75 mol

0.7547/0.75 = 1 0.75/0.75 = 1

[2 markah]
(0.75 mol/0.75mol)

iii) Tentukan formula empirik bagi X oksida.

Determine the empirical formula of X oxide

Formula empirik/ empirical formula : XO

[1 markah]

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c) Nyatakan bagaimana untuk menentukan bahawa tindak balas antara X oksida dengan hidrogen telah
lengkap.

State how to determine that the reaction between X oxide with hydrogen gas has completed.

Proses pemanasan, penyejukkan dan penimbangan diulangi sehingga mendapat jisim yang tetap.
Repeat heating, cooling and weighing until get the constant of mass

[1 markah]

d) Namakan satu logam oksida yang formula empiriknya boleh ditentukan menggunakan kaedah yang
sama.
Jelaskan jawapan anda

Name a metal oxide that its empirical formula can be determined using the same methode. Explain your answer.

Argentum kerana argentum kurang reaktif daripada hidrogen
Silver because silver less reactive than hydrogen

[2 markah]

Latihan 32

Rajah 32 menunjukkan susunan radas suatu eksperimen untuk menentukan formula empirik kuprum(II)
oksida

Diagram 32 shows the set -up apparatus of an experiment to determine the empirical formula of copper (II)
oxide.

Gas hidrogen Kuprum(II) oksida

Hydrogen gas Copper(II) oxide

Piring porselain

Porcelain dish

Panaskan

Heat

Kalsium klorida kontang

Anhydrous calcium chloride

Rajah 32/ Diagram 32

a) i) Namakan bahan tindak balas bagi penyediaan gas hidrogen.

Name two reactants for the preparation of hydrogen gas.

✓ Asid hidrklorik logam zink/magnesium

[1 markah]

ii) Tuliskan persamaan kimia bagi tindak balas di (a) (i).

Write a chemical equation for the reaction in (a) (i)
Mg + 2HCl → MgCl2 + H2

[1 markah]

b) Nyatakan satu langkah berjaga-jaga yang perlu diambil sebelum kupum(II) oksida dipanaskan

State one precaution that must be taken before the copper (II) oxide is heated.

Alirkan gas hidrogen beberapa minit sebelum tindak balas dijalankan [1 markah]
Flow the gas a few minute before the reaction

c) Jadual 32 menunjukkan keputusan eksperimen yang dijalankan oleh seorang pelajar

Table 32 shows the results of an experiment carried out by a student.

Jisim tiub pembakaran + mangkuk porselin 30.24 g
32.26 g
Mass of combustion tube + porcelain dish 31.86 g

Jisim tiub pembakaran + mangkuk porselin + kuprum(II) oksida

Mass of combustion tube + porcelain dish + copper (II) oxide

Jisim tiub pembakaran + mangkuk porselin + kuprum

Mass of combustion tube + porcelain dish + copper

Jadual 32/ Table 32

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Bab 3_Formula dan Persamaan Kimia

i) Kirakan bilangan mol kuprum dan oksigen dalam tindak balas ini.

Calculate the number of moles of copper and copper in this reaction
[JAR/RAM ; Cu = 64, O = 16]

Mol kuprum = 31.86 - 30.24 = 1.62 g Mol oksigen = 32.26 – 31.86 = 0.4 g

= 1.62/64 = 0.0253 mol = 0.4/16 = 0.025 mol

[1 markah]

ii) Tentukan formula empirik kuprum (II) oksida.

Determine the empirical formula of copper (II) oxide.

Formula empirik = CuO

[2 markah]
(0.025 mol/0.025 mol/CuO)

d) Namakan suatu oksida logam yang lain di mana formula empiriknya ditentukan dengan kaedah
yang sama

Name another metal oxide which uses the same method to determine the empirical formula.

Argentum oksida

[1 markah]

3.3.3 Penentuan Formula Molekul

Determine the molecular formula.

1. Pengiraan

Calculation

Langkah pengiraan Contoh 1

Calculation steps Example 1

 Kira nilai n Formula empirik etana adalah CH3 dan jisim
formula relatif adalah 30.
Find value of n Apakah formula empirik sebatian ini

JMR formula molekul The empirical formula of ethane is CH3 and relative
= (JMR formula empirik) x n molecular mass of 30. What is the molecular formula
of this compound?
RMM of molecular formula
= ( RMM of empirical formula ) x n n = 30 n= 2

n = JMR formula molekul (1 x 12) + (3 x 1)
JMR formula empirik

RMM of molecular formula Formula molekul = (formula empirik) x n
RMM of empirical formula
Molecular formula = ( empirical formula ) x n
 Cari formula molekul
= (CH3 ) x 2
Find molecular formula = C2H6

Formula molekul = (formula empirik) x n

Molecular formula ( empirical formula ) x n

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Bab 3_Formula dan Persamaan Kimia

2. Banding dan beza formula empirik dengan formula molekul

Compare and contrast empirical formula with molecular formula

Sebatian Formula molekul Formula empirik Nisbah teringkas bagi atom

Compounds Molecular formula Empirical formula Simplest ratio for atoms

a) Karbon dioksida CO2 CO2 1

Carbon dioxide H2O H2O 1

b) Air NH3 NH3 1

Water C2H6 CH3 2

c) Ammonia C2H4O2 CH2O 2

Ammonia

d) Etana

Ethane

e) Asid etanoik

Ethanoic acid

Soalan 33;

1. Formula empirik benzena adalah CH. Jika jisim

molekul relatif adalah 78, apakah formula molekul n = 78 / 13 = 6

sebatian ini. Formula molekul = (CH) x 6 = C6H6
Molecular formula
The empirical formula of benzene is CH. If its relative
molecular mass is 78, what is it molecular formula of the

compound?

[ JAR/RAM C = 12, H = 1 ]

(2 markah)

2. Maklumat di bawah tentang hidrokarbon J ( C6H6)

The information below is about hydrocarbon J n = 28 / 14 = 2
Formula molekul = (CH2) x 2 = C2H4
a) Formula empirik J adalah CH2 Molecular formula

Empirical formula of J is CH2

b) Jisim 1 mol J = 28 g

Mass of 1 mole of J = 28 g

Tentukan formula molekul bagi hidrokarbon (C2H4)

Determine the molecular formula for hydrocarbon J
[JAR/RAM; C = 12, H = 1 ]

(2 markah)

Soalan 34 ;

Satu sebatian organik Q mengandungi 64.87% karbon, 13.51% hidrogen dan 21.62% oksigen.

An organic compound Q consists of 64.87% of carbon, 13.51% of hydrogen and 21.62% of oxygen.
[JAR/ RAM : H,1 ; C,12 ;O, 16]

a) Tentukan formula empirik sebatian Q b) Jika jisim molekul relatif sebatian Q adalah
74, cari formula molekul sebatian Q
Determine the empirical formula of compound Q.
[3 markah] If the relative molecular mass of compound Q is
74, find the molecular formula of compound Q.
Unsur C H O
[2 markah]

Jisim (g) 64.87 13.51 21.62 n = 74 / (12x4) + (1x10) + 16
= 74 / 74 = 1
Bilangan 64.87/12 13.51/1 21.62/16
mol = 5.408 = 13.51 = 1.351 Formula molekul = (C4H10O)1 = C4H10O
Molecular formula
Nisbah 5.408/1.351 13.51/1.351 1.351/1.351
ringkas = 4 = 10 = 1 (C4H10O ; C4H10O)
mol C4H10O
Formula
empirik

|61 @azemi/kimiaT4SPM 4541

Soalan 35; Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia
a)Suatu sebatian mempunyai formula empirik CH2.
Jisim molekul sebatian tersebut ialah 70. b)Suatu sebatian mempunyai formula empirik
Apakah formula molekul sebatian tersebut? CH2O. Jisim molekul sebatia tersebut ialah 180.
Apakah formula molekul sebatian tersebut?
A compound has the empirical formula of CH2 .
The relative molecular mass of the compound is 70. A compound has the empirical formula of CH2O.
What is the molecular formula of the compound? The relative molecular mass of the compound is 180.
[JAR/RAM : H,1 ; C,12 ] What is the molecular formula of the compound?
[JAR/RAM: C,12 ; H,1 ; O; 16]

n = 70 /12 = 5 n = 180/30 = 6
Formula molekul = (C H2O) x 6 = C6H2O6
Formula molekul = (C H2) x 5 = C5H10 Molecular formula
Molecular formula
(C6H12O6)
(C5H10)

3.3.4 Formula kimia sebatian ion

Chemical formulae of ionic compounds

1. Formula ion / Ionic formulae

• Ion positif (Kation) / Positive ions: (cation)

Ion dengan 1+ Ion dengan 2+ Ion dengan 3+

Ions with 1+ Ions with 2+ Ions with 3+

Ion litium Li+ Ion magnesium Mg2+ Ion aluminium Al3+
Lithium ion Na+ Magnesium ion Ca2+ Aluminium ion Fe3+
Ion natrium K+ Ion kalsium Zn2+ Ion ferum(III) Cr3+
Sodium ion H+ Calcium ion Cu2+ Iron(III) ion
Ion kalium NH4+ Ion zink Ion kromium
Potassium ion Ag+ Zinc ion Chromium ion
Ion hidrogen Ion kuprum(II)
Hydrogen ion Copper (II) ion Fe2+
Ion ammonium Ion ferum(II)
Ammonium ion Iron (II) ion Pb2+
Ion argentum Ion Plumbum(II)
Silver ion Lead(II) ion

• Ion negatif (anion) / Negative ions; (anion)

Ion dengan 1- Ion dengan 2- Ion dengan 3-

Ions with 1- Ions with 2- Ions with 3-

Ion fluorida F- Ion oksida O2- Ion fosfat PO43-

Fluoride ion Oxide ion Phosphate ion

Ion klorida Cl- Ion sulfida S2-

Chloride ion Sulphide ion

Ion bromida Br - Ion karbonat CO32-

Bromide ion Carbonate ion

Ion nitrat NO3- Ion sulfat SO42-

Nitrate ion Sulphate ion

Ion hidroksida OH- Ion tiosulfat S2O32-

Hydroxide ion Thiosulphate ion

Ion iodida I-

Iodide ion

Ion etanoat CH3COO-

Ethanoat ion

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Bab 3_Formula dan Persamaan Kimia

Latihan 36 ; Tulis formula bagi sebatian ion

Write the formulae for ionic compounds

Ion Klorida,Cl- Sulfat,SO2-4 Nitrat, NO3- Karbonat,CO2-3

Ions Chloride, Cl- Sulphate,SO2-4 Nitrate, NO3-, Carbonate,CO2-3

1. Ion kalium, K+ Nama Kalium klorida Kalium sulfat Kalium nitrat Potassium
Potassium Potassium Potassium carbonate
Potassium, K+ Name sulphate sulphate nitrate
Formula KCl K2SO4 KNO3 K2CO3
2. Ion argentum, Argentum
Ag+ Nama klorida Argentum sulfat Argentum nitrat Argentum
Silver sulphate Silver nitrate karbonat
Silver ion, Ag+ Name Silver chloride AgNO3 Silver carbonate
AgCl Ag2SO4 Ag2CO3
formula

3. Ion magnesium, Nama Magnesium Magnesium Magnesium Magnesium
klorida sulfat nitrat karbonat
Mg2+ Name
Magnesium Magnesium Magnesium Magnesium
Magnesium,Mg2+ formula chloride sulphate nitrate carbonate
MgCl2 MgSO4 MgCO3
4. Ion kuprum(II) Nama Kuprum(II) Mg(NO3)2 Kuprum(II)
Cu2+ Kuprum(II) sulfat Kuprum(II) nitrat karbonat
Name klorida Copper(II) Copper(II)
Copper(II) sulphate Copper(II) carbonate
ion,Cu2+ Formula Copper(II) CuSO4 nitrate
chloride CuCO3
CuCl2 Cu(NO3)2

Latihan 37 ; Tuliskan formula kimia bagi sebatian berikut

Write the chemical formulae for the compound

Sebatian Formula kimia Sebatian Formula kimia

Compound Chemical formulae Compound Chemical formulae
AgBr
a) Argentum bromida j) Natrium nitrat NaNO3
MgSO4
Silver bromide Sodium nitrate
PbO
b) Magnesium sulfat k) Barium sulfat BaSO4
Al(NO3)3
Magnesium sulphate Barium sulphate
ZnCO3
c) Plumbum(II) oksida HCl l) Ferum(II) klorida Fe(NO3)2
HNO3
Lead(II) oxide Iron (II) chloride

Errata d) Aluminium nitrat m) Magnesium klorida MgCl2

Aluminium nitrate Magnesium chloride

e) Zink karbonat n) Kuprum(II) sulfat CuSO4

Zinc carbonate Copper (II) sulphate

f) Asid hidroklorik o) Asid sulfurik H2SO4
CH3COOH
Hydrochloric acid Sulphuric acid

g) Asid nitrik p) Asid etanoik

Nitric acid Ethanoic acid

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Bab 3_Formula dan Persamaan Kimia

Latihan 38 ; Tuliskan formula bagi sebatian kovalen berikut

Write the formula of covalent compound

Sebatian Formula Sebatian Formula Sebatian Formula

Compound Formulae Compound Formulae Compound Formulae
Cl2 H2 SO2
a) Klorin e) Hidrogen i) Sulphur dioksida
Br2 N2 H2S
Chlorine Hydrogen Sulphur dioxide
I2 H2O HCl
b) Bromin f) Nitrogen j) Hidrogen sulfida
O2 NH3 CO2
Bromine Nitrogen Hydrogen sulphide

c) Iodin g) Air k) Hidrogen klorida

Iodine Water Hydrogen chloride

d) Oksigen h) Ammonia l) Karbon dioksida

Oxygen Ammonia Carbon dioxide

3.4 Persamaan Kimia

Chemical Equations

3.4.1 Menulis persamaan kimia dan seimbangkan

Writing the chemical equation and balanced

a) Maksud Suatu simbol yang mewakili tindak balas di antara bahan tindak balas dan
hasilnya.
Meaning
A symbolic representation of a reaction between reactant and their product.
b) Persamaan
kimia Bahan tindak balas Hasil tindak balas

Chemical Reactant Products
equation
Magnesium dan oksigen Magnesium oksida

Magnesium and oxygen Magnesium oxide

Latihan 39 ; Tuliskan persamaan perkataan dan persamaan kimia

Write the word equation and chemical equation. (* Nota tak perlu seimbangkan)

Magnesium bertindak balas dengan klorin menghasilkan magnesium klorida

Magnesium reacts with chlorine to form magnesium chloride

Persamaan perkataan magnesium + klorin ⎯→ magnesium klorida

Word equations magnesium + chlorine magnesium chloride
Mg + Cl2 ⎯→ MgCl2
Persamaan kimia

Chemical equation

a) Kalium bertindak balas dengan oksigen menghasilkan kalium oksida

Potassium reacts with oxygen to form potassium oxide

Persamaan perkataan Kalium + oksigen ⎯→ kalium oksida
Potassium + oxygen potassiumoxide
Word equations
K + O2 ⎯→ K2O
Persamaan kimia

Chemical equation

b) Zink bertindak balas dengan asid hidroklorik menghasilkan zink klorida dan gas hidrogen

Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas.

Persamaan perkataan Zink + asid hidroklorik ⎯→ zink klorida + hidrogen
Zinc + hydrochloric acid zinc chloride + hydrogen
Word equations
Zn + 2HCl ⎯→ ZnCl2 + H2
Persamaan kimia

Chemical equation

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c) Natrium bertindak balas dengan air menghasilkan natrium hidroksida dan gas hidrogen

Sodium reacts with water to produce sodium hydroxide and hydrogen gas

Persamaan perkataan Natrium + air ⎯→ natrium hidroksida + hydrogen
Sodium water sodium hydroxide hydrogen
Word equations
Na + H2O ⎯→ NaOH + ½ H2 // 2Na + 2H2O ⎯→ 2NaOH + H2
Persamaan kimia

Chemical equation

d) Apabila larutan argentum nitrat ditambahkan kepada larutan zink klorida, mendakan putih,
argentum klorida dan larutan zink nitrat terbentuk

When silver nitrate solution is added to zinc chloride solution, white precipitate, silver chloride and zinc
nitrate solution is formed.

Persamaan perkataan Argentum nitrat + zink klorida ⎯→ argentum klorida + zink nitrat

Word equations Silver nitare + zinc chloride silver chloride + zinc nitrate

Persamaan kimia AgNO3 + ZnCl2 ⎯→ AgCl + Zn(NO3)2

Chemical equation

e) Apabila pepejal kalsium ditambahkan ditambahkan kepada asid nitrik, larutan kalsium nitrat
dan gas hidrogen terbebas.

When solid calcium is added to nitric acid, calcium nitrate solution and hydrogen gas are formed

Persamaan perkataan Kalsium + asid nitrik ⎯→ kalsium nitrat + hidrogen
Calcium + nitric acid calcium nitrate + hydrogen
Word equations
Ca + 2HNO3 ⎯→ Ca(NO3)2 + H2
Persamaan kimia

Chemical equation

Latihan 40 : Seimbangkan persamaan kimia f) Li2O + H2O ⎯→ 2LiOH

Balance the following chemical equations. g) Cu + 2AgNO3 ⎯→ 2Ag + Cu(NO3)2
a) 2Mg + O2 ⎯→ 2MgO
h) Zn + 2HCl ⎯→ ZnCl2 + H2
b) 2Li + 2H2O ⎯→ 2LiOH + H2
i) Pb + Cu(NO3)2 ⎯→ Pb(NO3 )2 + Cu
c) C + O2 ⎯→ CO2
j) 4Fe + 3O2 ⎯→ 2Fe2O3
d) 2K + Cl2 ⎯→ 2KCl

e) 2Na + 2H2O ⎯→ 2NaOH + H2

Latihan 41 ; Tuliskan persamaan kimia dan seimbangkan

Write the chemical equation and balanced the equation

a) Natrium + klorin ⎯→ natrium klorida ✓ Li + O2 ⎯→ Li2O
C + O2 ⎯→ CO2
Sodium + Chlorine ⎯→ sodium chloride Fe + Br2 ⎯→ FeBr3
Na2O + H2O ⎯→ 2NaOH
b) Karbon + oksigen ⎯→ karbon dioksida ✓ K + H2O ⎯→ KOH + H2
Mg + Cl2 ⎯→ MgCl2
Carbon + oxygen ⎯→ carbon dioxide gas

c) Ferum + bromin ⎯→ ferum(III) bromida ✓
Iron + bromine ⎯→ iron (III) bromide

d) Natrium oksida + air ⎯→ natrium hidroksida ✓
Sodium oxide + water ⎯→ sodium hydroxide

e) Kalium + air ⎯→ kalium hidroksida + hidrogen ✓
Potassium + water ⎯→ potassium hydroxide + hydrogen

f) Magnesium + klorin ⎯→ magnesium klorida ✓
Magnesium + chlorine gas ⎯→ magnesium chloride

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g) Zink + asid sulfurik ⎯→ zink sulfat + hidrogen Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Zinc + sulphuric acid ⎯→ zinc sulphate + hydrogen Bab 3_Formula dan Persamaan Kimia

✓ Zn + 2HCl ⎯→ ZnCl2 + H2

h) Karbon bertindak balas dengan plumbum(II) oksida menghasilkan karbon dioksida dan plumbum

Carbon reacts with lead (II) oxide to produces carbon dioxide and lead

✓ C + PbO ⎯→ CO2 + Pb

i) Barium klorida bertindak balas dengan zink sulfat menghasilkan barium sulfat dan zink klorida

Barium chloride reacts with zinc sulphate to produces barium sulphate and zinc chloride

✓ BaCl2 + ZnSO4 ⎯→ BaSO4 + ZnCl2

3.4.2 Mentafsir persamaan kimia secara kualitatif dan kuantitatif.

Interpreting chemical equations qualitatively and quantitatively

a) Aspek Jenis bahan dan hasil tindak balas yang terlibat dalam tindak balas dan
kualitatif keadaan fizikal bagai setiap bahan dan hasil tindak balas.

Qualitative Type of reactants and products involved in the chemical reaction and the state of
aspect each reactant and product.

b) Aspek Bilangan mol bahan yang bertindak balas dan hasil tindak balas yang
kuantitatif terbetuk iaitu pekali bagi setiap formula bahan dan hasil tindak balas
dalam persamaan kimia
Quantitative
aspect Number of moles of reactants and products involved in the chemical reaction that
is the coefficients involved in a balanced equation of the formulae of reactants and
products.

Contoh: Persamaan untuk tindak balas adalah seperti berikut : 2 Li (p) + Cl2 (g) ⎯→ 2 LiCl (p)

The equation for the reaction is as follow: 2 Li (s) + Cl2 (s) ⎯→ 2 LiCl (s)

Nyatakan maklumat yang boleh disimpulkan daripada persamaan itu.

State the information that can be deduced from the equation.

Aspek kualitatif a) Bahan tindak balas : Litium dan klorin

Qualitative aspect Reactants Lithium and chlorine

Hasil tindak balas : Litium klorida

Product Lithium chloride

b) Li adalah pepejal , Cl2 adalah gas dan LiCl adalah pepejal

Li is a solid, Cl2 is gas and LiCl is a solid.

Aspek kuantitatif c) 2 mol Li bertindak balas dengan 1 mol Cl2 menghasilkan 2 mol LiCl

Quantitative aspect 2 mol Li reacts with 1 mol Cl2 to produces 2 mol LiCl

Latihan 42 :

a) Persamaan untuk tindak balas adalah seperti berikut : C (p) + O2 (g) ⎯→ CO2 (g)

The equation for the reaction is as follow: C (s) + O2 (g) ⎯→ CO2 (g)

Nyatakan maklumat yang boleh disimpulkan daripada persamaan itu.

State the information that can be deduced from the equation.

Aspek kualitatif ✓ Bahan tindak balas : karbon dan oksigen
Hasil tindak balas : karbon dioksida
Qualitative aspect

✓ karbon adalah pepejal,oksigen adalah gas dan karbon dioksida adalah gas

Aspek kuantitatif ✓ 1 mol karbon bertindak balas dengan 1 mol oksigen menghasilkan 1 mol

Quantitative aspect karbon dioksida

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b) Persamaan untuk tindak balas adalah seperti berikut : Mg (p) + H2SO4 (ak) ⎯→ MgSO4 (p) + H2 (g)

The equation for the reaction is as follow : Mg (s) + H2SO4 (aq) ⎯→ MgSO4 (s) + H2 (g)

Nyatakan maklumat yang boleh disimpulkan daripada persamaan itu.

State the information that can be deduced from the equation.

Aspek kualitatif ✓ Bahan tindak balas : magnesium dan asid sulfurik

Qualitative aspect Hasil tindak balas : magnesium sulfat dan gas hidrogen

✓ Magnesium adalah pepejal, asid sulfurik adalah larutan, magnesium sulfat

adalah pepejal dan hydrogen adalah gas

Aspek kuantitatif ✓ 1 mol magnesium sulfat bertindak balas dengan 1 mol asid sulfurik

Quantitative aspect menghasilkan 1 mol magnesium sulfat dan 1 mol hydrogen

c) Persamaan untuk tindak balas adalah seperti berikut : NH3 (g) + HCl (g) ⎯→ NH4Cl (p)

The equation for the reaction is as follow : NH3 (g) + HCl (g) ⎯→ NH4Cl (s)

Nyatakan maklumat yang boleh disimpulkan daripada persamaan itu.

State the information that can be deduced from the equation.

Aspek kualitatif ✓ Bahan tindak balas : ammonia dan hydrogen klorida
✓ Hasil tindak balas : ammonium klorida
Qualitative aspect

✓ Ammonia adalah gas, hidrogen klorida adalah gas, ammonium klorida

adalah pepejal

Aspek kuantitatif ✓ 1 mol ammonia bertindak balas dengan 1 mol hydrogen klorida

Quantitative aspect menghasilkan 1 mol ammonium klorida

Soalan 43 :

Seorang pelajar memanaskan magnesium karbonat dengan kuat. Ia terurai mengikut persamaan di bawah.
A student heats magnesium carbonate strongly. It decomposes according to the equation below.

MgCO3 ⎯→ MgO + CO2

a) Nyatakan nama bagi yang berikut:

State the name of the following:

Bahan tindak balas : Magnesium karbonat/ magnesium carbonate

Reactant :

Hasil-hasil tindak balas : Magnesium oksida dan karbon dioksida

Products : magnesium oxide and carbon dioxide

[2 markah]

b) Nyatakan jenis-jenis zarah yang terkandung dalam bahan-bahan di bawah.

State the type of particles present in the substances below.

Bahan Jenis zarah

Substance Type of particle
Ion
MgO

CO2 Molekul

[1 markah]

c) Terangkan persamaan kimia tersebut dari aspek kuantitatif.

Describe the chemical equation in terms of quantitative aspect.

✓ 1 mol magnesium karbonat menghasilkan 1 mol magnesium oksida dan 1 mol karbon dioksida
[1 markah]

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Soalan 44 :

Rajah 44 menunjukkan satu persamaan kimia yang seimbang.

Diagram 44 shows a balance chemical equation.

NaCl(ak) + AgNO3 (ak) ⎯→ AgCl(p) + NaNO3(ak)
NaCl(aq) + AgNO3 (aq) ⎯→ AgCl (s) + NaNO3 (aq)

Berdasarkan persamaan itu, [1 markah]

Based on the equation,

a) Nyatakan jenis zarah dalam NaCl : ion

State the type of particles in NaCl

b) Nyatakan nama bagi bahan tindak balas dan hasil tindak balas.

State the name of reactants and products.

Bahan tindak balas Natrium klorida dan argentum nitrat
Sodium chloride and silver nitrate
Reactant
Argentum klorida dan natrium nitrat
Hasil-hasil tindak balas Silver chloride and sodium nitrate

Products

c) Apakah keadaan fizik bagi AgCl? : pepejal / solid [2 markah]
[1 markah]
What is the physical state of AgCl?

d) Hitung jisim formula relatif bagi AgNO3

Calculate the relative formula mass of AgNO3
[JAR/RAM : Ag; 108 , N;14, O;16]

= 1(108) + 1(14) + 3(16) = 179 [179]

e) Nyatakan dua maklumat lain yang boleh disimpulkan daripada persamaan itu.

State two information that can be deduced from the question.

✓ Keadaan fizik NaCl adalah cecair, AgNO3 adalah cecair, AgCl adalah pepejal dan NaNO3 adalah

cecair

The physical state of NaCl is liquid, AgNO3 is liquid, AgCl is solid and NaNO3 is liquid

✓ 1 mol NaCl bertindak balas dengan AgNO3 menghasilkan 1 mol AgCl dan 1mol NaNO3

1 mol NaCl reacts with AgNO3 to produces 1 mol AgCl and 1 mol NaNO3

[2 markah]

3.4.3 Penyelesaian Berangka Menggunakan Persamaan Kimia

Numerical Problems Using Chemical Equations

Latihan 45 :
Logam litium, Li bertindak balas dengan air, H2O menghasilkan litium hidroksida, LiOH dan gas
hidrogen.

Lithium metal, Li reacts with water, H2O to form lithium hydroxide and hydrogen gas

2Li + 2H2O ⎯→ 2LiOH + H2
Logam litium mempunyai 0.5 mol,

Lithium atom has 0.5 mol,;

apakah bilangan mol bagi ;

what is the number of mol for

|68 @azemi/kimiaT4SPM 4541

a) air,H2O / water,H2O Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia

b) gas hidrogen,H2 / hydrogen,H2 gas

(0.5 mol) (0.25 mol)

a) Maklumat(Info) Soalan (Q) b) Maklumat(Info) Soalan (Q)
L1 Litium/Lithium(Li) Air/water(H2O) L1 Litium/Lithium(Li hidrogen/hydrogen(H
Data Data
Mol = 0.5 mol Mol ? ) 2)
L2 Mol = 0.5 mol - L2 Mol = 0.5 mol Mol ?
Mol Mol Mol = 0.5 mol
L3 L3 -
Nisba Nisba
h Daripada persamaan; h Daripada persamaan;
2 mol Li : 2 mol H2O 2 mol Li : 1 mol H2
L4 0.5 mol Li : 0.5 x 2 L4 0.5 mol : 0.5 x 1
Soala 2 Soala
= 0.5 mol H2O Li 2
n Bilangan mol n = 0.25 mol H2
- air,H2O
= 0.5 mol Bilangan mol H2
- = 0.25 mol

Latihan 46 ;
Zink,Zn bertindak balas dengan oksigen,O2 menghasilkan zink oksida,ZnO.

Zinc,Zn reacts with oxygen, O2 to form zinc oxide, ZnO

2Zn + O2 ⎯→ 2ZnO

Zink oksida mempunyai 3.0 mol, berapakah bilangan mol bagi ;

Zinc oxide has 3.0 mol, what is the number of mol for ;

a) zink,Zn / zinc,Zn b) oksigen / oxygen

(3.0 mol) (1.5 mol)

a) zink,Zn b) oksigen

L1 Maklumat(Info) Soalan (Q) L1 Maklumat(Info) Soalan (Q)
Data Zink oksida,ZnO Zink,Zn Data Zink oksida,ZnO Oksigen,O2
Zinc oxide, ZnO Zinc(Zn) Zinc oxide, ZnO Oxygen(O2)
L2 Mol ? L2
Mol Mol = 3.0 mol - Mol Mol = 3.0 mol Mol ?
L3 Mol = 3.0 mol L3 Mol = 3.0 mol -
Nisba Nisba
h Daripada persamaan; h Daripada persamaan;

L4 2 mol ZnO : 2 mol Zn L4 2 mol ZnO : 1 mol O2
Soala Soala
2mol ZnO : 3.0 x 2 . 2 mol ZnO : 3.0 x 1
n n
2 2

= .3.0 mol Zn = 1.5 mol O2

Bilangan mol Zn Bilangan mol O2

- = 3.0 mol - = 1.5 mol

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Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia

Latihan 47 ; Selesaikan masalah berangka

Solve the numerical problems

Zink,Zn bertindak balas dengan asid hidroklorik,HCl menghasilkan zink klorida dan gas hidrogen,H2

Zinc,Zn reacts with hydrochloric acid,HCl to form zinc chloride,ZnCl2 and hydrogen gas,H2

Zn + 2HCl ⎯→ ZnCl2 + H2

a) berapakah mol asid hidroklorik, HCl yang b) berapakah mol gas hydrogen yang dihasilkan
diperlukan bertindak balas dengan 0.2 mol daripada 0.6 mol asid hidroklorik
zink
How many moles of hydrogen gas are is produced from
How many moles of hydrochloric acid, HCl are 0.6 mol hydrochloric acid.
needed to react with 0.2 mol zinc.
(0.3 mol)
(0.4 mol)

a) b)

L1 Maklumat(Info) Soalan (Q) L1 Maklumat(Info) Soalan (Q)

Data Zink HCl Data HCl H2

mol = 0.2 mol Mol ? mol = 0.6 mol Mol ?

L2 mol = 0.2 mol - L2 mol = 0.6 mol -

Mol Mol

L3 Daripada persamaan;; L3 Daripada persamaan;;

Nisbah 1 mol Zn : 2 mol HCl Nisbah 2 mol HCl : 1 mol HCl

0.2mol ZnO : 0.2 x 2 0.6 mol HCl : 0.6 x 1
1 2

= .0.4 mol HCl = .0.3 mol HCl

L4 Bilangan mol HCl L4 Bilangan mol

Soalan - = 0.4 mol Soalan - HCl

= 0.3 mol

Latihan 48 ;

Zink,Zn bertindak balas dengan oksigen,O2 menghasilkan zink oksida,ZnO.

Zinc,Zn reacts with oxygen, O2 to form zinc oxide,ZnO

4 K + O2 ⎯→ 2 K2O

11.7 g kalium terbakar dalam oksigen menghasilkan kalium oksida . Kira

11.7 g of potassium was burnt in oxygen to produce potassium oxide. Calculate
[JAR/RAM K, 39; O,16, isi padu molar gas pada suhu bilik / molar volume at room temperature = 24 dm3 mol-1]

a) Jisim kalium oksida,K2O yang dihasilkan b) isi padu oksigen,O2 yang bertindak balas pada
suhu bilik
mass of potassium oxide being produced
Volume of oxygen reacts at room temperature.
(14.1 g)
(1.8 dm3)

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Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia

a) Maklumat(Info) Soalan (Q) b) Maklumat(Info) Soalan (Q)
L1 L1
Data Kalium/Kalium,K Kalium Data Kalium/Kalium,K Oksigen/oxygen,O2

L2 Jisim = 11.7 g oksida,K2O L2 Jisim = 11.7 g Isi padu/volume,
Mol Mol
Jisim/mass ? V?
L3 L3
Nisba Mol = Jisim /JAR Nisbah Mol = Jisim /JAR
h
Mass/RAM - L4 Mass/RAM -
L4 Soalan
Soala = 11.7 / 39 = 11.7 / 39

n = 0.3 mol = 0.3 mol

Daripada persamaan; Daripada persamaan;

From the equation ; From the equation ;

4 mol K : 2 mol K2O 4 mol K : 1 mol O2

0.3 mol K : 0.3 x 2 0.3 mol K : 0.3 x 1

2 4

= 0.15. mol K2O = 0.075 mol O2

Jisim/mass of K2O = mol x JFR Isi padu gas,O2 = mol x isi padu molar

= 0.15 x [(2 x 39)+( 1x16)] Volume O2 = mol x molar

volume

= 0.15 x 94 = 14.1 g = 0.075. x 24 = 1.8 dm3

(1.8 dm3)

Latihan 49 ;

1. Persamaan berikut menunjukkan tindak balas penguraian hidrogen peroksida,H2O2

The following equation shows the decomposition of hydrogen peroxide

2H2O2 ⎯→ 2H2O + O2

Apabila 0.02 mol hidrogen peroksida , H2O2 terurai, kira

When 0.02 mole of hydrogen peroxide, H2O2 decomposes, calculate
[JAR/RAM ; H,1 ; O,16]

a) jisim air, H2O yang terbentuk b) isipadu gas oksigen, O2 yang terbebas pada

the mass of water formed s.t.p

(0.72 g) the volume of oxygen gas released at s.t.p
[1 mol gas menempati 22.4 dm3 pada s.t.p]
a)  2 mol H2O
1 mol H2O2 0.448 dm3)
b)

1 mol H2O2  1 mol O2

0.02 mol  0.02 x 2 0.02 mol  0.02 x 1
1 1

= 0.04 mol H2) = 0.02 mol O2

Jisim H2O = 0.04 x 18 Isi padu O2 = 0.02 x 22.4
Mass = 0.72 mol Volume = 0.0.448 dm3

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Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
Bab 3_Formula dan Persamaan Kimia

3. Persamaan berikut menunjukkan tindak balas di antara natrium dan air.

The following equation shows the reaction between sodium and water.

2Na + 2H2O ⎯→ 2NaOH + H2

a) bilangan mol natrium,Na bertindak balas b) hitungkan jisim natrium,Na yang diperlukan
dengan 3.0 mol air,H2O .
untuk menghasilkan 3.01 x 1023 molekul
how many moles of sodium will reacts with 3.0
moles of water? hidrogen,H2

[JAR/RAM ; Na,23 ;NA ,6.02 x 1023] calculate the mass of sodium needed to produce

(3 mol) 3.01 x 1023 hydrogen molecules.

[JAR/RAM ; Na,23 ;NA ,6.02 x 1023]

(23 g)

a)  2 mol H2O b)
2 mol Na Mol H2
 3.0 x 2 = 3.01 x 1023
3.0 mol Na 2 6.02 x 1023
=0.5 mol
= 3.0 mol H2O 1 mol H2
0.5 mol H2
 2 mol Na
Jisim Na
Mass  0.5 x 2
1

= 1.0 mol Na
= 1.0 x 23

23 g

Soalan 50 ;

Persamaan berikut menunjukkan tindakan haba terhadap garam plumbum(II) nitrat

The equation below shows the action of heat on the nitrate salt of plumbum

2Pb(NO3)2 ⎯→ 2PbO + 4NO2 + O2

7.282 g garam plumbum(II) nitrat telah dipanaskan dengan kuat.

7.282 g of lead (II) nitrate salt are heated strongly.
[JAR/RAM ; Pb, 207; N,14 ; O,16; Isi padu molar gas pada stp/ molar volume of gas at stp = 22.4 dm3 mol-1 ]

Hitung

Calculate :

a) Bilangan mol garam plumbum(II) nitrat b) jisim plumbum(II) oksida,PbO

number of moles of lead(II) nitrat mass of lead (II) oxide,PbO

Mol Pb(NO3)2 Mol Pb(NO3)2
= 7.282 /331
= 0.022 mol = 7.282 /331

=0.022 mol

2 mol Pb(NO3)2  2 mol PbO

0.022 mol Pb(NO3)2  0.022 x 2
2

(0.022 mol) = 0.022 mol PbO

Jisim PbO = 0.022 x 223

= 4.9 g

(4.9g)

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c) isi padu nitrogen dioksida,NO2 yang Modul Pengajaran & Pembelajaran Kimia_Tingkatan 4
dibebaskan Bab 3_Formula dan Persamaan Kimia

volume of nitrogen dioxide, NO2 released. d) isi padu gas oksigen, O2 yang terhasil dalam
eksperimen

volume of oxygen gas produced in the experiment

Mol Pb(NO3)2  4 mol NO2 Mol Pb(NO3)2  1 mol O2
= 7.282 /331 = 7.282 /331  0.022 x 1
=0.022 mol  0.022 x 4 =0.022 mol
2 mol Pb(NO3)2 2 2 mol Pb(NO3)2 2
0.022 mol Pb(NO3)2 0.022 mol Pb(NO3)2 = 0.011 mol O2
= 0.044 mol = 0.011 x 22.4
Isi padu NO2 NO2 Isi padu O2 0.25 dm3
Volume = 0.044 x 22.4 volume
0.99 dm3 (0.25 dm3)

(0.99 dm3)

Soalan 51 ;

Gas hidrogen disediakan oleh tindak balas di antara gas metana dengan stim, H2O.
Tindak balas diwakili dengan persamaan ditunjukkan di bawah.

Hydrogen gas is prepared by reacting methane gas with steam. The reaction is represented by the equation
shown below:

CH4 + H2O ⎯→ CO + 3H2

Jika 60.0 dm3 gas hydrogen, H2 dihasilkan pada suhu dan tekanan bilik, hitungkan

If 60 dm3 of hydrogen gas are produced at room temperature and pressure, calculate
[JAR/RAM ; H; 1, C;12,O;16; Isi padu molar gas pada stp/ molar volume of gas at stp = 24 dm3 mol-1]

a) jisim metana,CH4 yang digunakan dalam tindak b) bilangan molekul gas karbon monoksida,CO
balas yang dibebaskan

The mass of methane that is used in the reaction The number of carbon monoxide,CO molecules
released

Mol H2 Mol H2
= 60 / 24
= 2.5 mol = 60 / 24
3 mol H2
2.5 mol H2  1 mol CH4 = 2.5 mol

Jisim CH4  2.5 x 1 3 mol H2  1 mol CO
3
2.5 mol H2  2.5 x 1
= 0.833 mol CH4 3

= 0.833 x 16 = 0.833 mol CO

= 13.33 g Bilangan molekul = 0.833 x 6.02 x 1023

CO = 5.01 x 1023

(13.33g) (5.01 x 1023)
[1 markah] [1 markah]

********* Tamat Bab 3 **********

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