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Transcript of Struktur Atom
Bab 2: Struktur Atom
A Jirim
1. Jirim adalah sebarang bahan yang mempunyai _______________ dan memenuhi _________. Jirim terdiri dalam tiga keadaan – pepejal, ____________, dan _________
2. Jirim terdiri daripada zarah yang halus dan ______________. 3. Jirim terdiri daripada tiga jenis zarah: atom, _____________, dan _________4. Jirim juga boleh dikelaskan kepada unsur dan sebatian.5. Unsur: Bahan yang terdiri daripada ___________ jenis atom sahaja6. Sebatian: Bahan yang terdiri daripada _________ atau _____________ unsur berbeza yang terikat
secara kimia.7. Resapan berlaku apabila zarah-zarah dalam sesuatu bahan bergerak memenuhi ruang antara zarah
pada bahan lain. 8. Resapan berlaku paling cepat dalam __________, diikuti __________ dan _____________.
Teori Kinetik Jirim.i) Perubahan haba mengubah keadaan jirim.ii) Apabila bahan dipanaskan, suhu akan meningkat, tenaga kinetic zarah-zarah akan bertambah dan
zarah-zarah akan bergerak dengan lebih ____________.
Keadaan Jirim Pepejal Cecair Gas
Susunan zarah Zarah-zarah tersusun sangat padat dan teratur.
Zarah-zarah tersusun _________ tetapi tidak __________
Zarah-zarah terpisah __________ antara satu sama lain
Daya tarikan Sangat ____________
___________ antara zarah tetapi lebih lemah berbanding ___________
_________
Pergerakan zarah Zarah ___________ pada kedudukan tetap
Zarah ___________, __________, dan bergerak bebas dalam cecair
Zarah ____________, _________, dan bergerak secara rawak.
Kandungan tenaga Sangat ___________ Lebih __________ berbanding pepejal tetapi lebih __________ berbanding gas
Sangat ____________
1
Change of States of Matter
2
SOLID LIQUID GAS
3
B C : Pepejal dan cecairSuhu adalah tetap kerana tenaga haba yang diserap oleh zarah-zarah dalam pepejal digunakan untuk mengatasi daya tarikan antara zarah-zarah supaya pepejal berubah menjadi cecair.
D E: Cecair dan gasSuhu adalah tetap kerana tenaga haba yang diserap oleh zarah-zarah dalam cecair digunakan untuk mengatasi daya tarikan antara zarah-zarah supaya cecair berubah menjadi gas
4
GAS LIQUID SOLID
ED: Gas dan cecairSuhu adalah tetap kerana tenaga haba yang diserap oleh zarah-zarah dalam gas digunakan untuk mengatasi daya tarikan antara zarah-zarah supaya gas berubah menjadi cecair.
C B: Cecair dan pepejalSuhu adalah tetap kerana haba dibebaskan ke persekitaran oleh zarah-zarah dalam cecair diimbangi oleh tenaga haba yang terbebas apabila zarah-zarah tertarik antara satu sama lain untuk membentuk pepejal.
5
B Struktur Atom
Sejarah pembentukan model atom
Saintis Model Atom
1. John Dalton
- Membayangkan atom sebagai sebiji bola halus -Atom tidak boleh dicipta, dimusnah atau dibahagi
2.J.J. Thomson
-Atom adalah sfera yang bercas positif yang mengandungi zarah bercas negative dipanggil electron. .
3. Ernest Rutherford
-Menjumpai nucleus yang merupakan pusat bagi atom dan bercas positif. -Proton adalah sebahagian daripada nucleus-Elektron bergerak sekeliling nukleus
4.Neils Bohr
-Menjumpai kewujudan petala electron-Elektron bergerak di dalam petala mengelilingi nukleus
5.James Chadwick
-Menjumpai kewujudan neutron.-Nukleus mengandungi zarah-zarah neutral dipanggil neutron dan zarah-zarah bercas positif dipangil proton.
6
Proton, neutron, dan eektron adalah zarah subunit bagi atom.
i) Atom adalah neutral ii) Bilangan proton adalah sama dengan bilangan elektroniii) Nombor proton sesuatu unsur adalah bilangan proton yang terdapat dalam sesuatu atomiv) Nombor nukleon sesuatu unsur adalah jumlah bilangan proton dan neutron di dalam nucleus
sesuatu atom.
v) Setiap unsur mempunyai nama dan symbol tersendiri.
Proton number
Element symbol Proton number
Element symbol
1 Hydrogen H 11 Sodium Na2 Helium He 12 Magnesium Mg3 Lithium Li 13 Aluminium Al4 Beryllium Be 14 Silicon Si5 Boron B 15 Phosphorus P6 Carbon C 16 Sulphur S7 Nitrogen N 17 Chlorine Cl8 Oxygen O 18 Argon Ar9 Flourine F 19 Potassium K10 Neon Ne 20 Calcium Ca
Atom boleh ditulis dalam bentuk: A
X Z
Di mana A ialah nombor nukleon, X ialah symbol unsur, Z ialah nombor proton.
Nombor nukleon = bilangan proton + bilangan neutron
7
Subatomic particle
Symbol Relative mass
Charge Location
Proton p 1 +1
Electron e 1/1840 -1
Neutron n 1 0
Lengkapkan jadual di bawah
Lengkapkan jadual di bawah
Unsur bilangan proton Bilangan neutron Simbol Lithium
(Li)Neon (Ne)Zinc (Zn)
8
symbol atom 27 Al 13
19 F 7
23 Na 11
No proton
No nukleon
Bilangan proton
Bilangan elektron
Bilangan neutron
2.3 Isotop dan Kepentingan
1. Isotop ialah atom-atom unsur yang mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza.
2. Conto isotop
Unsur IsotopHidrogen
11 H
1 proton0 neutron
12 H
1 proton 1 neutron
13 H
1 proton 2 neutron
Karbon 6
12 C 6 proton 6 neutron
613 C
6 proton7 neutron
614 C
6 proton8 neutron
Oksigen 8
16 O 8 proton 8 neutron
817 O
8 proton9 neutron
818 O
8 proton10 neutron
Sulfur 16
32 S16 proton16 neutron
1634 S
16 proton18 neutron
-
Bromin 35
79 Br35 proton44neutron
3581 Br
35 proton46 neutrons
-
3. Kegunaan Isotop
9
Lapangan Kegunaan
perubatan Kobalt-60: memusnahkan sel barah Kobalt-60: membunuh mikroorganisma
semasa proses pensterilan
Arkeologi Karbon-14 digunakan untuk mengesan umur artifak
pertanian Fosforus-32: mengesan kadar penyerapan
baja fosfat oleh tumbuhan.
Industri Natrium-24:mengesan kebocoran paip bawah tanah
Petala pertama: 2 elektron
Petala kedua: 8 elektron
Petala ketiga: 8 elektron
Nukleus mengandungi proton dan neutron
Elektron pada petala terluar:elektron valens.
2.4 Susunan Elektron
Valence electron: ________ Valence electron: ________
Valence electron: ________ Valence electron: ________
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4. The electron arrangement of elements with proton number 1 to 20.( must know how to memorize)
Element Number of neutrons
Number of protons
Number of electrons
Number of nucleon
Electron arrangement
Number of valence
electronsHydrogenHeliumLithiumBerylliumBoronCarbonNitrogenOxygenFluorineNeonSodiumMagnesiumAluminumSiliconPhosphorusSulphurChlorineArgonPotassiumCalcium
11
Chapter 3: Chemical Formulae and Equations
Subtitle 3.1: Relative Atomic Mass and Relative Molecular Mass
Concept:We can determine the mass of an atom relative to a standard atom
A Helium atom is 4 times heavier compare to a hydrogen atom.Helium is said to have relative atomic mass of 4
Important !!!Define:
Define
Important!!!!Relative mass does not have any unit.
Numerical problems
How to measure mass of an atom?
Standard atom1. hydrogen 2. oxygen
3. carbon-12✔ Solid & easy to
handle✔ Also used as
standard for mass spectrometer
Check to:page 176 of text book Look at Ar of all elements listed in
periodic table
from periodic table: Ar of Nitrogen atom is 14.The average mass of a nitrogen atom is 14 times larger than 1/12 of a carbon-12 atom.
Mr of Water Molecule is 18The average mass of one water molecule is 18 times larger than 1/12of a carbon-12 atom
Relative molecular mass of an element= The average mass of one molecule 1/12 x the mass of an atom of carbon -12
✄ not use any more because gasseous form are difficult to handle
Relative molecular mass, Mr-of a molecule is the average mass of the molecule when compared with 1/12 of the mass of an atom of carbon-12
Relative atomic mass, A r
- of an element is the average mass of one atom of the element when compared with 1/12 of the mass of an atom of carbon-12
Relative atomic mass of an element= The average mass of one atom of an element 1/12 x the mass of an atom of carbon -12
Hydrogen as standard atom
How many helium atoms are here?????
helium atom
12
A. About Relative Atomic Mass
1. How many times is copper atom heavier than two helium atom? Solution: Mass of a copper atom = Ar of copperMass of 2 helium atom 2 x Ar of helium = 64 2 x 4 = 8 times
2. How many magnesium atom have the same mass as two silver atoms ? Solution:
Lets the number of magnesium atoms = n Mass of n magnesium atoms = mass of 2 silver atoms So, n x Ar of magnesium = 2 x __________ n x 24 =
=
Do It Yourself
1. How many times is one atom of silicon heavier than one atom of lithium
2. Calculate the number of atoms of lithium that have the same mass as two atoms of nitrogen
3. The mass of one atom Y is A times larger than the mass of one nitrogen .Calculate the relative atomic mass of Y.
B. About Relative Molecular Mass
To determine Relative Molecular Mass, Mr
Get Ar value from periodic table
Form 4 text bookQuick reviewpage 30
13
Molecular substance
Relative Molecular Mass
Carbon dioxide, CO2
Ar of C + ( 2 x Ar of O) = 12 + (2 x 16 ) = 44
Nitrogen gas, N2 2 x Ar of N = 2 x 14 = 28
Relative formula mass is used to replace Mr for ionic substances
Ionic substance Relative formula mass
Sodium Hydroxide, NaOH Ar of Na + Ar of O + Ar of H= 23 + 16 + 1 = 40
Aluminium sulphate, Al2 (SO4)3
2 x Ar of Al +3 ( Ar of S + 4 x Ar of O )=
Hydrated copper(II) sulphate, CuSO4. 5H2O
Ar of Cu + Ar of S + 4 x Ar of O + 5 ( Mr of H2O)=
Do it yourself
1. Calculate the relative molecular mass of
a) Bromine, Br2
c) Ammonia, NH3
b) Methane, CH4 d) Glucose, C6H12O6
2. Calculate relative formula mass of
a) Zinc oxide, ZnO c) Copper(II) hydroxide, Cu(OH)2
Get Ar value from periodic table
14
Practical Book Activity 3.2, page 17
b) Magnesium nitrate, Mg(NO)3 d) Hydrated sodium carbonate, Na2CO3.10H2O
B. The Mole and the Number of Particles
Definition of mole
The word ‘pair’ and ‘dozen’ represent a fixed number of objects.
In chemistry, we use the unit ‘mole’ to measure the amount of substance. The symbol of mole is mol. 1 mol of substance = the number of particles in 12 g of carbon-12.
= 6.02 x 1023 particles.
The value of 6.02 x 1023 is called as the Avogadro constant (NA). To determine the number of moles or the number of particles:
Example 1:i. 1 mol of iron atom = 6.02 x 1023 iron atomsii. 1 mol of hydrogen molecule = 6.02 x 1023 hydrogen moleculesiii. 1 mol of sodium chloride = 6.02 x 1023 formula units of sodium chloride
Form 4 practical bookTry this 3.1page 17
Number of particles = Number of moles¿ 6.02 x 1023
Number of moles =Number of particles
6 . 02 x 1023
15
Example 2:A closed glass bottle contains 0.5 mol of oxygen gas, O2.i. How many oxygen molecules, O2 are there in the bottle?ii. How many oxygen atoms are there in the bottle?
Solution:i. Number of oxygen molecules = 0.5 x 6.02 x 1023
= 3.01 x 1023
ii. 1 oxygen molecule, O2 has 2 oxygen atoms.Therefore, number of oxygen atoms = number of oxygen molecules ¿ 2= 3.01¿ 1023¿ 2= 6.02¿ 1023
Example 3:Find the number of moles of molecules in a sample containing 9.03 × 1023 molecules of carbon dioxide, CO2.
Solution:
Number of moles =
9 .02×1023
6 .02×1023
= 1.5 mol.
Do it yourself
[Avogadro constant = mol-1]
1 Define a mole?
A mole is the amount of substance which has the same number of particles as there in 12 g carbon -12.
2 Calculate the number of atoms in 2 mol carbon.
Number of atoms = 2 × 6.02 x 1023 = 1.2 ×1024 atoms.
3 How many ions are there in 1.5 mol sodium chloride, NaCl?
1 formula unit sodium chloride, NaCl has 2 ions which are 1 sodium ion and 1 chloride ion.Thus, number of ions = number of formula units x 2
= 1.5 × 6.02 x 1023 × 2 = 1.806 × 1024 ions.
4 Calculate the number of moles of bromine molecules which consists 1.5 × 1022 of bromine molecules.
Number of moles = 1.5 ×10 22 6.02 × 1023
16
= 0.025 mol.
5 How many atoms are there in 2 mol of ammonia, NH3?
1 ammonia molecule, NH3 has 4 atoms which are 1 nitrogen atom and 3 hydrogen atoms.Thus, number of atoms = number of molecules x 4
= 2 × 6.02 x 1023 × 4 = 4.8 × 1024 atoms.
C. The mole and the mass of substances
Molar mass is
Unit of molar mass is g mol-1 or grams per mole.
The molar mass of a substance = the mass of 1 mol of the substance= the mass of NA number of particles= the mass of 6.02 x 1023 particles
Example:
Element/ Compound Relative mass Mass of 1 mol
Molar mass
Lithium, Li 7 7g 7g mol-1
Iron, Fe 56 56gMagnesium oxide, MgO 24+16=40 40g mol-1
Carbon dioxide, CO2 12+16x2=44
*1 : The value of molar mass of an element is equal to its relative atomic mass*2 : The value of molar mass of a compound is equal to its relative molecular or formula mass
Formula: Number of moles = mass
Relative atomic mass
(or relative molecular mass or relative formula mass)
Example:
1. Calculate the number of moles found in 20g of magnesium oxide, MgO. (Relative atomic mass: Mg, 24; O, 16)
*1
*2
17
Solution:
Number of moles = mass Relative formula mass = 20 24 + 16 = 0.5 mol
2. Calculate the mass in gram found in 0.2 mol of magnesium oxide, MgO. (Relative atomic mass: Mg, 24; O, 16) Solution:
Number of moles = mass
Relative formula mass
Mass = number of moles x relative formula mass = 0.2 x (24 + 16)g
= 8g
3. How many magnesium ions are there in 30g of magnesium oxide, MgO. (Relative atomic mass: Mg, 24; O, 16. Avogradro constant: 6.02 x 1023) Solution:
The relative formula mass of magnesium oxide, MgO = 24 + 16 = 40
Therefore, the molar mass of magnesium oxide, MgO = 40g mol-1
Number of moles of 30g magnesium oxide, MgO = mass of MgO
Relative formula mass of MgO = 30g
40 g mol-1 = 0.75 mol
The number of formula units of MgO = 0.75x 6.02 x 1023
= 4.515 x 1023
Each formula units of MgO has 1 magnesium ions.
Therefore, the number of magnesium ions = the number of formula units of MgO x 1 = 4.515 x 1023 x 1 = 4.515 x 1023
4. Calculate the mass in gram of 3 x 1022 units of magnesium oxide, MgO. (Relative atomic mass: Mg, 24; O, 16. Avogradro constant: 6.02 x 1023) Solution:
Number of moles = number of particles NA
18
Mass = number of particlesRelative formula mass NA
Mass = number of particles x relative formula mass NA
Mass of 3x1022 units of magnesium oxide, MgO= 3 x 10 22 x (24+16) 6 X 1023
= 0.05 X 40 = 2 g
Do It Yourself
1. Calculate the number of moles found in 9.5g of magnesium chloride, MgCl2. (Relative atomic mass: Mg, 24; Cl, 35.5)
2. Calculate the mass in gram found in 0.3 mol of magnesium chloride, MgCl2.
(Relative atomic mass: Mg, 24; Cl, 35.5)
3. How many chloride ions are there in 19g of magnesium chloride, MgCl2. (Relative atomic mass: Mg, 24; Cl, 35.5. Avogradro constant: 6.02 x 1023)
4. Calculate the mass in gram of 3 x 1022 units of magnesium chloride, MgCl2 . (Relative atomic mass: Mg, 24; Cl, 35.5. Avogradro constant: 6.02 x 1023)
E. Chemical Formulae
A chemical formula is a representation of a chemical substance using letters for atoms and subscript numbers to show the numbers of each type of atoms that are present in the substance. Examples : (a) Glucose
Form 4 TextBookWork This Out 3.2Page 35
Quick Review CPage 35
19
C6 H12O6
(b) Sodium hydroxide
Mg (OH)2
(1) Empirical Formulae
(i) The empirical formula of a compound gives the simplest whole number ratio of atoms of each element present in the compound.
(ii) Steps in determining the empirical formula of a compound.i. find the mass of each element in the compoundii. convert the masses to the numbers of moles of atomsiii. find the simplest ratio of moles of the elements
Example : 2.24 g of iron combines chemically with 0.96g of oxygen to form an oxide. What is the empirical formula of the oxide ?
[ Relative atomic mass: O, 16; Fe, 56 ]
Element Iron, Fe Oxygen, O
Mass (g) 2.24 0.96
Number of moles of atoms 2.24 = 0.0456
0.96 = 0.0616
Ratio of moles 0.04 =10.04
0.06 =1.50.04
Simplest ratio of moles 1 × 2 = 2 1.5 × 2 = 3
The empirical formula of the oxide is Fe2O3.
Do it Yourself
1. The table below shows the relative atomic mass and the mass of elements V and O in an oxide.Element V ORelative Atomic Mass 56 16Mass(g) 5.6 2.4
What is the empirical formula of this compound ?
Show the symbols for carbon, hydrogen and oxygen
Show the numbers of carbon, hydrogen and oxygen
Show the symbols for magnesium, oxygen and hydrogen.
Show the numbers of magnesium, oxygen
and hydrogen.
20
element V Oxygen, OMass (g) 5.6 2.4Number of moles of atomsRatio of moles
Simplest ratio of moles
The empirical formula of the oxide is ……………………
2. Copper (II) iodide constains 20.13% of copper by mass. Find its empirical formula. [ Relative atomic mass : Cu,64 ; I, 127 ]
Based on its percentage composition, 100g of copper(II) iodine contains 20.13g of copper. So, taking 100g of the compound.
element K ClMass (g)Number of moles of atomsRatio of moles
Simplest ratio of moles
The empirical formula of the oxide is ………………….
3. A potassium compound has a percentage composition as the following K, 31.84% ; Cl, 28.8% ; O, 39.18% What is the empirical formula of the potassium compound ? [ Relative atomic mass : O, 16; Cl,35.5; K, 39 ]
Based on its percentage composition, 100g of compound contains 31.84g of potassium, 28.98g of chlorine and 39.18g of oxygen. So, by taking 100g of the compound:
element K Cl OMass (g)Number of moles of atomsRatio of moles
Simplest ratio of moles
1 mole of potassium atoms combines with 1 mole of chlorine atoms and 3 moles of oxygen atoms.Therefore, the empirical formula of the potassium compound is KClO3.
(2) Molecular Formulae
(i) The molecular formula of a compound gives the actual number of atoms of each element present in a molecule of the compound.
(ii) The molecular formula of a compound is a multiple of its empirical formula.
Molecular formula = ( Empirical formula )n
Form 4 TextBook Work this out 3.7Page 42
21
(iii) Relating empirical formula to molecular formula Compound Empirical formula Molecular formula n
Water H2O H2O = (H2O)1 1Ethene CH2 C2H4 = (CH2)2 2Ethane CH3 C2H6 = (CH3)2 2propane CH2 C3H9 = (CH3)3 3glucose CH2O C6H12O6 = (CH2O)2 6
(iv) Calculation involving molecular formulae
Example :The empirical formula of a compound is CH. Its relative molecular mass is 78. Find its molecular formula. [ Relative atomic mass : H, 1; C, 12 ]
Let the molecular formula be (CH)n.
The relative molecular mass = n[ 12 + 1 ] = 13n
However, its molar mass is 78. Therefore, 13n = 78 n = 78/13 = 6
Hence, the molecular formula of the compound is (CH)6 or C6H6.
Do it yourself
1. A carbon compound has an empirical formula of CH2 and a relative molecular mass of 70. Find the molecular formula of the compound. [ Relative atomic mass : H, 1; C, 12 ]
Hence, the molecular formula of the compound is (CH2)5 or C5H10.2. 2.07 g of element Z reacts with bromine to form 3.67g of a compound with an empirical formula of
ZBr2. Find the relative atomic mass of element Z. [ Relative atomic mass: Br, 80 ]
element Z BrMass (g)
22
Number of moles of atoms
Simplest ratio of moles (from the emp for given)
Based on the empirical formula ZBr2 , the ratio of atoms of Z : Br is 1 : 2 herefore,
2.07 : 0.02 = 1 : 2 z 2.07/0.02z = ½ z = 207
The atomic mass of the element Z is 207.
(3) Ionic Formulae
(i) Ionic compounds are compounds consisting of anions and cations.
(ii) The formulae of some common cations
Cation ( positive ion ) Formula of cation Charge of cationSodium ion Na+ +1Potassium ion K+ +1Silver ion Ag+ +1Hydrogen ion H+ +1Ammonium ion NH4
+ +1Copper (II) ion Cu2+ +2Calcium ion Ca2+ +2Magnesium ion Mg2+ +2Zinc ion Zn2+ +2Barium ion Ba2+ +2Iron (II) ion Fe2+ +2Copper (I) ion Cu+ +1Tin (II) ion Sn2+ +2Lead (II) ion Pb2+ +2Aluminium ion Al3+ +3Iron (III) ion Fe3+ +3Chromium (III) ion Cr3+ +3
(iii) The formulae of some common anions
Anion ( negative ion ) Formula of anion Charge of anionFluoride ion F- -1Chloride ion Cl- -1Bromide ion Br- -1
FORM 4 Textbook Work this out 3.8Page 44
23
Iodide ion I- -1Hydroxide ion OH- -1Nitrate ion NO3
- -1Nitrite ion NO2
- -1Hydride ion H- -1Oxide ion O2- -2Phosphate ion PO4
3- -3Carbonate ion CO3
2- -2Sulphate ion SO4
2- -2Chromate (VI) ion Cr2O7
2- -2
(iv) The chemical formulae of ionic compounds are electrically neutral because the total of positive charges are equal to the total of negative charges
(v) The chemical formula of an ionic compound can be constructed as the following :i. identify and write down the formula of its cation and anionii. determine the number of cations and anions by balancing the positive and
negative charges.iii. Write the formula of the compoundiv. The number of cations and anions are written as subscript numbers.
MgCl2
Do it yourself
1. magnesium chloride
2. aluminium oxide
Magnesium chloride
Magnesium ion, Mg2+ Chloride ion, Cl-
1 magnesium ion, Mg2+
Total of positive charges=1 (+2)=+2
2 chloride ions, Cl-
Total of negative charges= 2 (-1)= -2
24
3. aluminiuim hydroxide
4. sodium sulphate
(4) Naming of chemical compounds
1. Chemical compounds are named systematically according to the guidelines given by the International Union of Pure and Applied Chemistry (IUPAC).
2. For ionic compounds, the name of the cation comes first, followed by the name of anion.
cation anion Name of ionic compoundSodium ion Chloride ion Sodium chloride
Magnesium ion Oxide ion Megnesium oxideAluminium ion Oxide ion Aluminium oxide
Zinc ion Sulphate ion Zinc sulphate
3. Transition metals can form more than one ions, Roman numerals ( such as I, II, III ) are used to differentiate the ions.
Fe2+ - iron (II) ion Fe3+ - iron (III) ion
4. For simple molecular compounds, the name of the first element is maintained. However, the name of the second element is added with an “ –ide “.
Examples : HCl – hydrogen chloride HF - hydrogen flouride 5. Greek prefixes are used to show the number of atoms of each element in a compound. Examples : CO – carbon monoxide CO2 – carbon dioxide CCl4 – carbon tetrachloride / tetrachloromethane SO3 – sulphur trioxide
6. Table below shows the meaning of the prefixes.
prefix meaning prefix meaningMono- 1 Hexa- 6
di- 2 Hepta- 7Tri- 3 Octa- 8
Tetra- 4 Nona- 9Penta- 5 Deca- 10
F. CHEMICAL EQUATION
A) Qualitative aspect of chemical equation
♣ A chemical equation is a shorthand description of a chemical reaction.♣ The starting substances are called reactants.
Form 4 TextbookWork This Out 3.9Page 46
Form 4 TextbookWork This Out 3.10Page 47
25
♣ The new substances formed are called products.
♣ The reactants are written at the left-hand side of the equation.♣ The products are written at the right-hand side of the equation.
♣ A chemical equation also shows the states of each substance.
Symbol Physical states of substancess Solidℓ Liquidg Gasaq Aqueous solution
Example :
Do It Yourself 3f
Identify the reactants, products and the state of each substance. Present your answer in the form of a table.
Solution :
Reactants Products123
B) Writing chemical equation
A chemical equation must be balanced. There must always be the same number of atom of each element on each side of the equation.
Example :
Reactants Products
C (s) + O2 (g) CO2 (g)
Zn (s) + Cl2 (g) ZnCl2 (s)
1. HCl (aq) + NaOH (aq) NaCl (aq) + H2O (i)
2. CuCO3 (s) CuO (s) + CO2 (g)
3. HCl (g) + NH3 (g) NH4Cl (s)
Form 4 TextbookWork This Out 3.11Page 49
26
Magnesium reacts with dilute hydrochloric acid, HCl to produce magnesium chloride, MgCl2 and hydrogen gas, H2. Write an equation to represent the reaction.
Solution :
Do It Yourself 3.f B
Write a chemical equation for each of the following reactions.
1. A solution of silver nitrate is added to a solution of sodium chloride. A precipitate of silver chloride and a solution of sodium nitrate are produced.
2. Nitrogen gas reacts with hydrogen gas to produce ammonia gas.
3. When solid lead (II) carbonate is heated strongly, it decomposes into solid lead (II) oxide and carbon dioxide gas is released.
C) Quantitative aspect of chemical equation
The coefficients in a balanced equation tell us the exact proportions of reactants and products in a chemical reaction.
Example :
STEP 1 Write the equation in words. The reactants are written on the left whereas the products are written on the right.
STEP 2 Write the correct chemical formula for each reactants and products.
STEP 3 Balance the equation. You just need to adjust the coefficients in front of the chemical formulae and not the subscripts in the formulae.
STEP 4 Put the state symbols in the equation.
Form 4 TextbookWork This Out 3.12Page 50
STEP 1 Magnesium + hydrochloric acid Magnesium chloride + hydrogen gas
Reactants Products
STEP 2 Mg + HCl MgCl2 + H2
STEP 3 Mg + 2HCl MgCl2 + H2
STEP 4 Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
27
The equation tell us that 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water.
Or
The equation tell us that 2 molecules of hydrogen reacts with 1 molecule of oxygen to produce 2 molecules of water.
D) Numerical problems involving chemical equation
Stoichiometry is a study of quantitative composition of substances involved in chemical reactions.We can always make use of the stoichiometric coefficients in a chemical equation to solve various numerical problems.
Generally the steps involved in stoichiometric calculations are as follows.
Example :
Copper (II) oxide, CuO reacts with aluminium according to the following equation.
Calculate the mass of aluminium required to react completely with 12 g of copper (II) oxide, CuO.[Relative atomic mass : O, 16 ; Al, 27 ; Cu, 64]
Solution :
The number of moles of 12g of Copper (II) oxide, CuO = 12 g (64 + 16) g mol-1
= 12 g = 0.15 mol 80 g mol-1
Based on the chemical equation, 3 mole of Copper (II) oxide, CuO requires 2 mole of aluminium. Therefore, the number of aluminium required by 0.15 mole of Copper (II) oxide, CuO
= 0.15 mole x 2 mole = 2 mole 3 mole
2H2 (g) + O2 (g) 2H2O (l)
STEP 1 Write the balanced equation of the reaction.
STEP 2 Compare the mole ratio.
STEP 3 Identify the information given and you want to find.
STEP 4 Calculate the number of moles.
3CuO (s) + 2Al (s) Al2O3 (s) + 3Cu (s)
3CuO (s) + 2Al (s) Al2O3 (s) + 3Cu (s)
3 mole 2 mole
28
Thus the mass of aluminium required
= 0.1 mol x 29 g mol-1 = 2.7 g
Do It Yourself 3f (D)
How many moles of potassium are needed to reacts with 0.5 mole of bromine gas ?
Solution :
Information : ? mole 0.5 mole
Based on the equation, 1 mole of bromine gas reacts with 2 moles of potassium.
Therefore, 0.5 mole of bromine gas will react with
2 x 0.5 = 1 mole of potassium.
2. 1.35 g of aluminium reacts with excessive copper (II) oxide powder to produce aluminium oxide powder and copper. Find the number of copper atoms produced.[Relative atomic mass : Al, 27 ; Avogadro constant : 6.02 x 1023 mol-1]
3.
What is the mass of zinc needed to produce 2.4 dm3 of hydrogen gas at room conditions ?[Relative atomic mass : Zn, 65 ; Molar volume 24 dm3 mol-1 at room conditions]
1. 2K (s) + Br2 (g) 2KBr (s)
Zn (s) + 2HNO3 (aq) Zn(NO3)2 (aq) + H2 (g)
2K (s) + Br2 (g) 2KBr (s)
2 mole 1 mole
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More Exercises:
1. CuCO3 CuO + CO2
In this reaction, 3.1 g of copper(II) carbonate are heated in a laboratory. Find :
(a) the mass of copper (II) oxide that being produced.(b) the volume of carbon dioxide gas produced at s.t.p
2. CaCO3 CaO + CO2
In this reaction, 300 cm3 gas carbon dioxide are produced at room temperature, when calcium carbonate are heated. Find:
(a) the mass of calcium carbonate used.(b) mass of calcium oxide produced.
3. 2Na + 2H2O 2NaOH + H2
When 0.23 g of sodium is added to water, the metal will react vigorously at the surface of the water, find(a) the mass sodium hydroxide produced.(b) volume of hydrogen gasses being produced at temperature room.
4. 2Mg + O2 2MgO
A strip of magnesium has a weight of 1.2 g are being burn with sufficient oxygen to produced magnesium oxide. Find:
(a) the mass magnesium oxide being produced.(b) the mass of oxygen that needed for this reaction.
5. C3H8 + 5O2 3CO2 + 4H2O
Propane gas was burned in oxygen follow as equation above. If 3.36 dm3 of carbon dioxide gas are produced in this reaction at s.t.p, find
(a) the mass of propane burned
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(b) volume of oxygen gas that reacted
6. 2Al + 3CuO Al2O3 + 3Cu
1.35g of aluminium powder and copper (II) oxide was heated strongly in laboratory to produced aluminium oxide and copper. Find
(a) the mass of copper (II) oxide reacted(b) the mass of aluminium oxide produced.(c) the mass of copper produced.
Chapter 4.
PERIODIC TABLE OF ELEMENTSA. Historical Development of the Periodic Table
31
Group I Group II Group III Group IV
Oxygen NitrogenHidrogenLightHeat
SulphurPhosporusCarbon ChlorinFluorin
ArsenicBismutCobaltLeadZincNikelStanumArgentum
Potassium oxideBarium oxideSilicon(IV) oxsideMagnesium oxideAluminium oxide
Scientist like to find patterns. In the 18th and 19th centuries, scientist discovered many elements. The elements found were classified through many stages of hard work by scientist. This led to the development of the Periodic Table of Elements that we use today.
Here is the history ofPeriodic Table of Elements.
Antoine Lavoisier First chemist who classify the element into 4 group. The 4 group consisted of gases, metal, non-metal and metal
oxide. Element in the group is classify into metal and non-metal.
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Element in triadLithium
LiSodium
NaPotassium
K
Average atomic mass
Li and K
Relative atomic mass
7 23 39
7 + 39 = 23 2
Element in triadChlorine
ClBromine
BrIodine I
Average atomic mass Li and K
Relative atomic mass
35 80 12735 + 127 = 2 81
Ca Sr Ba (40 + 137) ÷ 2 = 88 Li Na K Cl Br I 7 23 39 35 80 127
Johann W. Dobereiner
Classify the element with same chemical properties into a few group Each of group consisted from 3 element called triad. He found that the relative atomic mass of the element in the middle of
each triad is approximately equal to the average atomic mass of other two elements.
Triad law show the relationship between the relative atomic mass of elements with it chemicals properties.
This law cannot be use to most of the other element.
John Newlands Arranged 62 known elements in order of increasing nucleon
number (atomic weights ) in horizontal rows. He noted that after interval of eight elements similar
physical/chemical properties reappeared. He was the first to formulate the concept of periodicity in the
properties of the chemical elements. He proposed the Law of Octaves:
Elements exhibit similar behavior to the eighth element following it in the table.
He was not successful because;i. Was only accurate for the first 16 elements (from
hydrogen to potassium)ii. There were no gaps allocated from the elements yet
to be discovered.
33
Lothar Meyer
Determine the volume of an atom of an element. Formula;
Volume of an atom = mass of one mole-atom of the element Density of the element
He plotted a graph of volume of atoms of elements against their relative atomic masses to produce meyer’s atomic volume curve.
From the graph he found elements occupying the corresponding positions of the curve exhibit similar chemical properties. example
(a) Li, Na, K, Rb : located at the peak of the curve
(b) Be, Mg, Ca, Sr : located after the maximum point
Like Newlands, Meyer showed the properties of the elements recured periodically.
Dimitri Mendeleev
Arranged the elements in order of increasing atomic weights and properties.
He left gaps for elements yet to be discovered.He arranged the element that have the same properties in group.
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Henry Mosely
He was able to derive the relationship between x-ray frequency and number of protons. and obtained a straight line graph.
When Moseley arranged the elements according to increasing atomic numbers and not atomic masses, some of the inconsistencies associated with Mendeleev's table were eliminated.
The modern periodic table is based on Moseley's Periodic Law (atomic numbers/proton number). He suggest proton number determine the position of elements in periodic table. He arranged elements in periodic table in order of increasing proton number. He also left gaps for the elements yet to be discovered.
35
Henry Mosely Periodic Table
Arrangements of elements in the Periodic Table
Elements in the Periodic Table are arranged in an increasing order of proton number.
What is the basic principle applied in arranging the elements in the Periodic Table today?
36
X40
20
Number of proton = 20Number of electron = 20Number neutron = 20
Electron arrangement = 2.8.8.2The number of valence electrons = 2The number of shells = 4
Group in the Periodic Table = 2Period in the Periodic Table = 4
Elements with similar chemical properties are placed in the same vertical column. There are 18 vertical column of elements in the Periodic Table. Each column is called group. The
vertical columns are known as Group 1 to Group 18. There are 7 horizontal rows of elements in the Periodic Table. Each of these horizontal rows of
elements is called a period. The horizontal rows are known as Period 1 to Period 7.
1. The number of valence electrons in an atom decides the position of the group of an element.
The number of valence electron
1 2 3+10 4+10 5+10 6+10 7+10 8+10
Group in The Periodic Table
1 2 13 14 15 16 17 18
2. The number of shells occupied with electrons in its atom decides the period number of an element.
Example 1;
Example 2;
Do you know how the electron arrangement of the atom of an element related to its group and period?????
37
Y16
8
Number of proton = 8Number of electron = 8Number neutron = 8
Electron arrangement = 2.6The number of valence electrons = 6The number of shells = 2
Group in the Periodic Table = 16Period in the Periodic Table = 2
Z40
18
Number of proton = 18Number of electron = 18Number neutron = 22
Electron arrangement = The number of valence electrons = The number of shells =
Group in the Periodic Table = Period in the Periodic Table =
Example 3;
Hw: WTO 4.3 pg. 62 no. 1,2,3
PERIODIC TABLE OF ELEMENTS
B. GROUP 18 ELEMENTS
http://
periodictable.com/1.The elements in Group 18 are
Helium 2
GROUP 18
38
Neon 2.8Argon 2.8.8Krypton 2.8.18.8Xenon 2.8.18.18.8Radon 2.8.18.32.18.8
1. They are also known as noble gases, which are chemically unreactive. Noble gases are monoatomic.
2. Helium has two valence electrons. This is called duplet electron arrangement.
3. Other noble gases have eight valence electrons. This is called octet electron arrangement. 4. Duplet and octet electron arrangements are very stable because the outermost occupied shells are full.
5. All nobles gases are inert which means chemically unreactive.
BECAUSE THE OUTERMOST OCCUPIED SHELLS ARE FULL
Physical Properties of Group 18 Elements1. Group 18 elements have very small atom.
2. They are colourless gases a room temperature and pressure.
3. They have low melting and boiling point.
4. They have low densities.Elements/ symbol
Electron arrangement Atomic radius (nm)
Melting points (°C)
Boiling points (°C)
Density(g cm-3)
Helium 2 0.050 -270 -269 0.17Neon 2.8 0.070 -248 -246 0.84Radon 2.8.8 0.094 -189 -186 1.66Krypton 2.8.18.8 0.109 -156 -152 3.45Xenon 2.8.18.18.8 0.130 -112 -107 5.45Radon 2.8.18.32.18.8 - -71 -62 -
Table 1: Physical Properties of Group 1 Elements
4. From Table 1, when going down the group, atomic size and density increase.5. When going down the group, melting points and boiling points decrease
Uses of Group 18 elements
Helium Used to fill airships and weather balloons, because the gas is very light. The diver’s oxygen tank contains a mixture of helium (80%) and oxygen (20%).
Neon Advertising lights. Television tubes.
Why noble gases exist as monoatomic gases and are chemically unreactive?
39
Airport landing bulb to help aero plane landing safely.
Argon To fill light bulbs, it can last longer To provide inert atmosphere for welding at high temperature.
Krypton Used in lasers to repair the retina of the eye. To fill photographic flash lamps.
Radon Used in treatment of cancer.
Xenon Used in bubble chambers in atomic energy reactors.
Hw: QR B pg. 65 no. 1,2
C. GROUP 1 ELEMENTS
http://periodictable.com/
6. The elements in Group 1 are
Lithium 2.1 Sodium 2.8.1Potassium 2.8.8.1Rubidium 2.8.18.8.1Caesium 2.8.18.18.8.1Francium 2.8.18.32.18.8.1
7. They are also known as alkali metals which react with water to form alkaline solutions.
8. All Group 1 elements have one valence electron in their outermost occupied shells.
Physical Properties of Group 1 Elements1. Group 1 elements are soft metals with low densities and low melting points as compared to other metals such as iron and copper.
2. They have silvery and shiny surfaces .
3. They are good conductor of heat and electricity.
GROUP 1
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Elements/ symbol Electron arrangement Atomic radius (nm)
Melting points (°C)
Boiling points (°C)
Density(g cm-3)
Lithium, Li 2.1 0.15 180 1336 0.57Sodium, Na 2.8.1 0.19 98 883 0.97Potassium, K 2.8.8.1 0.23 64 756 0.86Rubidium, Rb 2.8.18.8.1 0.25 39 701 1.53
Table 1: Physical Properties of Group 1 Elements
6. From Table 1, when going down the group, atomic size and density increase.
7. When going down the group, melting points and boiling points decrease
Chemical Properties of Group 1 Elements
Lithium, sodium and potassium have similar chemical properties but differ in reactivity.
Let us carry out this Experiment! Practical Book Experiment 4.1, page 35 Activity 4.3, page 38
1. Alkali metals react vigorously with water to produce alkaline metal hydroxide solutions and hydrogen gas.
[Video]
Chemical equation;
2Li + 2H2O → 2LiOH + H2
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Lithium Water Lithium Hydrogen hydroxide gas
2Na + 2H2O → 2NaOH + H2 Sodium Water Sodium Hydrogen
hydroxide gas
2K + 2H2O → 2KOH + H2 Potassium Water Potassium Hydrogen
hydroxide gas
2. Alkali metals react rapidly with oxygen gas, to produce white solid metal oxides. Chemical equations;
4Li + O2 → 2Li2O Lithium Oxygen Lithium
gas oxide 4Na + O2 → 2Na2O
Sodium Oxygen Sodium gas oxide
4K + O2 → 2K2O Potassium Oxygen Potassium
gas oxide
3. Alkali metals burn in chlorine gas to form white solid metal chlorides.
Chemical reaction;
2Li + Cl2 → 2LiCl Lithium Chlorine Lithium
gas chloride
2Na + Cl2 → 2NaCl Sodium Chlorine Sodium
gas chloride
2K + Cl2 → 2KCl Potassium Chlorine Lithium
gas chloride
4. Alkali metals burn in bromine gas to form metal bromides.
For example,
2Li + Br2 → 2LiBrLithium Bromine Lithium
gas bromide
2Na + Br2 → 2NaBrSodium Bromine Sodium
gas bromide
2K + Br2 → 2LiBrPotassium Bromine Potassium
42
gas bromide
5. Therefore, alkali metals have similar chemical properties.
Alkali metals have one valence electron in their outermost occupied shells.
Each of them reacts by donating one electron from its outermost occupied shell to form an ion with a charge of +1, thus achieving the stable electron arrangement of the atom of noble gas.
Li Li+ + 1e-
2.1 2
Na Na+ + 1e- 2.8.1 2.8
K K+ + 1e- 2.8.8.1 2.8.8
6. The reactivity of Group 1 elements increases down the group.
Going down Group 1, the atomic size (atomic radius) increases.
The single valence electron in the outermost occupied shell becomes further away from the nucleus
Hence, the attraction between the nucleus and the valence electron becomes weaker
Therefore, it is easier for the atom to donate the single valence electron to achieve the stable electron arrangement.
Safety precautions in handling Group 1 elementsAlkali metals are very reactive. Safety precautions must be taken when handling alkali metals.
Why alkali metals have similar chemical properties?
Why The reactivity of Group 1 elements increases down the group?
43
The elements must be stored in paraffin oil in bottles Do not hold alkali metals with your bare hands Use forceps to handle them Wear safety goggles Wear safety gloves Use a small piece of alkali metal when conducting experiments
Hw: QR C pg. 69 no. 1,2,3
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