6 For ionic compounds, the particles are ions. Conversion of the Number of Moles to the Number
For example, of Particles
1 mol of magnesium ions contains 6.02 1023
Mg2+ ions. 6 SPM
1 mol of potassium iodide, KI contains 6.02 ’09/P1
1023 K+ ions and 6.02 1023 I– ions.
1 mol of magnesium chloride, MgCl2 contains Calculate the number of particles in:
6.02 1023 Mg2+ ions and 2 6.02 1023 (a) 0.75 mol of aluminium atoms, A1,
Cl– ions. (b) 1.2 mol of chloride ions, Cl–,
(c) 0.07 mol of carbon dioxide molecules, CO2.
7 For covalent compounds, the particles are
molecules. For example, [Assume NA = 6 1023 mol–1]
1 mol of carbon dioxide contains 6.02 1023
CO2 molecules. Solution
Conversion of the Number of Moles to (a) 1 mol of aluminium contains 6 1023 Al atoms.
the Number of Particles and Vice Versa
3 0.75 mol of aluminium contains
1 Since one mole of any substance contains
6.02 1023 particles, n moles of the substance —0—.1—75—m—mo—lo—l 6 1023 Al atoms Make sure that the
will contain n 6.02 1023 particles. numerator and the
denominator have
2 Hence, = 4.5 1023 Al atoms the same unit.
number of particles = number of mole 3 NA (b) 1 mol of chloride ions contains 6 1023 Cl– ions.
(where NA = 6.02 3 1023)
1.2 mol of chloride ions contain
—1—1.2—m—m—oo—ll 6 1023 Cl– ions.
= 7.2 1023 Cl– ions
(c) 1 mol of carbon dioxide contains 6 1023 CO2
molecules.
0.07 mol of carbon dioxide contains
3 If 6.02 1023 particles are found in 1 mol, —0—.10—7m——mo—lo—l 6 1023 CO2 molecules
then one particle is found in
— 6—.—0—2——1—1—p0—a—2r3—tip—c—al—er—ti—c—l—es 1 mol
= —6—.—0—2—1——1——0—23 mol = 4.2 1022 CO2 molecules
Therefore, x particles are found in 7 SPM
—6—.0—2——x——1—0—23 mol ’11/P1
4 Thus, the
Calculate the number of atoms in:
number of moles = number of particles NA (a) 0.2 mol of sulphur dioxide gas, SO2,
(where NA = 6.02 1023) (b) 0.125 mol of methane gas, CH4.
5 Generally, NA mol [NA = 6 1023 mol–1]
NA
number of Solution
particles (a) 0.2 mol of SO2 contains 0.2 6 1023
A student need not memorise that the Avogadro constant molecules = 1.2 1023 molecules.
1 sulphur dioxide molecule (SO2) has 3 atoms
is 6.02 1023. It will be given in the examination. (one sulphur and two oxygen atoms).
Therefore the number of atoms
However in most cases, the value of NA given is = 3 1.2 1023
6 1023 for easy calculation. = 3.6 1023 atoms
(b) 0.125 mol of CH4 contains 0.125 6 1023
molecules = 7.5 1022 molecules.
1 methane molecule (CH4) has 5 atoms (one
carbon and four hydrogen atoms). Therefore the
number of atoms = 5 7.5 1022
= 3.75 1023 atoms
Chemical Formulae and Equations 44
Conversion of the Number of Particles to the 3 The mole-atom is the relative atomic mass of
Number of Moles SPM an atom expressed in gram.
’10/P1 [Relative atomic mass: C, 12; Al, 27; S, 32]
8
1 mole-atom of carbon Each sample
Calculate the number of moles of the following = 12 g contains
substances: 1 mole-atom of aluminium 6.02 1023
(a) 6 1021 iron atoms, = 27 g atoms
(b) 7.5 1023 carbon monoxide molecules. 1 mole-atom of sulphur
= 32 g
[NA = 6 1023 mol–1]
4 The mole-molecule is the relative molecular 3
Solution mass of a compound expressed in gram.
(a) 1 mol of iron contains 6 1023 atoms.
[Relative molecular mass: H2O, 18; CO2, 44;
Therefore 6 1021 iron atoms NH3, 17; C2H5OH, 46; CH4, 16]
= —66———11—00—22—13—aa—ttoo—mm— 1 mol
1 mole-molecule of water, Each sample
= 0.01 mol H2O = 18 g contains
(b) 1 mol contains 6 1023 molecules. 1 mole-molecule of carbon 6.02 1023
dioxide, CO2 = 44 g molecules
Therefore 7.5 1023 CO molecules contain 1 mole-molecule of ethanol,
—7—6.5———1—10—02—323—mm—o—ol—ele—c—cuu—le—les—s 1 mol C2H5OH = 46 g
= 1.25 mol
Conversion of the Number of Moles of a SPM
Substance to Its Mass ’07/P2
3.2 1 Since 1 mol of an element is the relative
atomic mass in gram, x mol of the element
1 Calculate the number of particles in has x relative atomic mass in gram.
(a) —61 mol of copper,
(b) 0.0625 mol of water molecule, H2O, Number of mole-atom
(c) 1.3 mol of sodium ions, Na+.
= ——r—e—la—mt—i—va—es—s—a—tion——mg——rica—m—m——a—s—s
[NA = 6 1023 mol–1]
2 Calculate the number of atoms in
(a) 0.012 mol o1suf0le2p3thhmuarnotelr–i1og]xaisd,eC, 2SHO63, . 2 Similarly, since 1 mol of a compound is the
(b) 1.1 mol of relative molecular mass in gram, x mol of
the compound has x relative molecular
[NA = 6 mass in gram.
3 Calculate the number of moles of the following
substances:
(a) 6 1022 sodium ions,
(b) 1.8 1024 1H022S3 molecules. Number of mole-molecule
[NA = 6 mol–1]
= ——r—e—l—a—ti—mv—e—a—ms—s—o—iln—e—cg—ur—la—am—r—m——a—s—s—
3.3 Relationship between the 3 Generally,
Number of Moles of a
Substance and Its Mass mole Ar or Mr mass in gram
4 Ar or Mr
1 The mass of a substance that contains one mole
of the substance is called the molar mass. number of moles
2 One mole of substance contains 6.02 1023
particles. Therefore the molar mass of any
substance contains 6.02 1023 particles.
45 Chemical Formulae and Equations
Conversion of the Number of Moles of a Conversion of the Number of Particles of a
Substance to Its Mass Substance to Its Mass and Vice Versa
9 1 Two steps are involved in the conversion of the
Determine the mass for each of the following mass of substance to the number of particles.
substances:
(a) —32 mol of aluminium atoms, Step 1: Mass in gram is converted to number
(b) 0.08 mol of ascorbic acid, C6H8O6,
(c) 0.125 mol of magnesium hydroxide, Mg(OH)2. of moles by dividing the mass by
[Relative atomic mass: H, 1; C, 12; O, 16; Mg, 24;
Al, 27; Cl, 35.5] the relative atomic mass or relative
Solution molecular mass.
(a) 1 mol of Al = 27 g
—23 mol of Al = —23 27 g = 18 g Step 2: Number of moles is converted to
(b) 1 mol of C6H8O6 = 6(12) + 8(1) + 6(16) g
number of particles by multiplying
= 176 g
0.08 mol C6H8O6 = 0.08 176 g the number of moles by the Avogadro
= 14.08 g constant.
(c) 1 mol of Mg(OH)2 = 24 + 2(16 + 1) g
3 2 Two steps are involved in the conversion of
= 58 g
0.125 mol of Mg(OH)2 = 0.125 58 g the number of particles to mass.
= 7.25 g Step 1: Number of particles is converted to
the number of moles by dividing the
number of particles by the Avogadro
constant.
Step 2: Number of moles is converted to mass
in gram by multiplying the number of
moles by the relative atomic mass or
relative molecular mass.
3 In general, Ar
or Mr
number of NA mole Ar mass in
particles NA or Mr gram
Conversion of the Mass of a Substance to the Conversion of the Mass of a Substance to the
Number of Moles Number of Particles
10 SPM
’08/P1
Calculate the number of moles of the following 11
substances:
(a) 23.5 g of copper(II) nitrate, Cu(NO3)2, Calculate the number of particles in:
(b) 0.97 g of caffeine, C8H10N4O2 (a stimulant).
[Relative atomic mass: H, 1; C, 12; N, 14; O, 16; Si, (a) 12.8 g of copper,
28; S, 32; Cu, 64]
(b) 8.5 g of ammonia, NH3.
[Relative atomic mass: H, 1; C, 12; N, 14; O, 16;
Solution Fe, 56; Cu, 64; NA = 6 1023 mol–1] Divide mass
Solution in gram by the
(a) 1 mol of Cu(NO3)2 = 64 + 2[14 + 3(16)] g ( a) 12.8 g of Cu = —16—24—.8 mol = 0.2 mol relative atomic
= 188 g mass to find the
number of moles
23.5 g of Cu(NO3)2 = —21——38—.85 1 mol
= 0.125 mol 1 mol contains 6 1023 atoms.
(b) 1 mol of C8H10N4O2 0.2 mol contains 0.2 6 1023 atoms
= 8(12) + 10 + 4(14) + 2(16) g = 194 g
= 1.2 1023 atoms Number of moles
0.97 g of C8H10N4 O2 (b) 1 mol of NH3 = 17 g is converted to
= —01—.99—47 1 mol 8.5 g of NH3 = —81—.75 mol = 0.5 mol number of particles
by multiplying the
= 0.005 mol number of moles
by the Avogadro
1 mol contains 6 1023 molecules. constant
0.5 mol contains 0.5 6 1023 molecules
= 3 1023 molecules
Chemical Formulae and Equations 46
Conversion of the Number of Particles of a 3.3
Substance to Its Mass
1 Calculate the mass of each of the following
12 substances:
(a) 1.25 mol of helium gas,
Calculate the mass of the following substances:
(a) 1.2 1022 zinc atoms, (b) —25 mol of cobalt, Co,
(b) 3 1023 ethanol (C2H5OH) molecules. (c) 0.15 mol of hydrated copper(II) sulphate,
CuSO4.5H2O,
[Relative atomic mass: H, 1; C, 12; O, 16; Zn, (d) 0.05 mol of potassium manganate(VII),
65; NA = 6 1023 mol–1] KMnO4.
[Relative atomic mass: H, 1; He, 4; O, 16; S,
Solution 32; K, 39; Mn, 55; Co, 59; Cu, 64]
(a) 1 mol contains 6 1023 atoms. 2 Calculate the number of moles of the following
substances:
1.2 1022 atoms Number of particles is (a) 2.8 g of iron, 3
= —1—6.2———1—1—0—02—322 mol converted to the number of (b) 4.05 g of nicotine, C10H14N2 (an addictive
moles by dividing the number substance in cigarette),
= 0.02 mol of particles by the Avogadro (c) 1.49 g of ammonium phosphate, (NH4)3PO4
constant (a fertiliser),
(d) 2.3 g of ethanol, C2H5OH.
1 mol of Zn = 65 g [Relative atomic mass: H, 1; C, 12; N, 14; O,
16; P, 31; Fe, 56]
0.02 mol of Zn = 0.02 65 g
3 Calculate the mass of the following substances:
= 1.3 g (a) 3 1023 titanium atoms,
(b) 1.2 1024 argon atoms,
(b) 1 mol contains Number of moles is (c) 7.5 1022 citric acid (C12H16O14) molecules.
6 1023 molecules. converted to mass in [Relative atomic mass: H, 1; C, 12; O, 16; Ar,
3 1023 C2H5OH molecules gram by multiplying 40; Ti, 48; NA = 6 1023 mol–1]
the number of moles
is contained in —36———11—00—2233 mol by the relative atomic 4 Calculate the number of particles in the following
mass substances:
(a) 4 g of sulphur,
= 0.5 mol (b) 2.24 g of cadmium,
(c) 36 g of glucose, C6H12O6.
1 mol of C2H5OH = 46 g [Relative atomic mass: H, 1; C, 12; O, 16;
0.5 mol of C2H5OH = 0.5 46 g = 23 g S, 32; Cd, 112; NA = 6 1023 mol–1]
2 ’04 5 Geranial is a compound found in lemon grass
(daun serai). Its molecular structure is shown
The relative atomic mass of X and Y is 64 and 16 below.
respectively. Which of the following is about the
atoms of X and Y? CH3 H H H CH3 H H
The mass of one atom of Y is 16 g. ⎮ ⎮ ⎮ ⎮ ⎮ ⎮
The number of protons in X is 64.
4 mol of Y have the same mass as 1 mol of X.
The density of one atom of X is 4 times that of
an atom of Y.
Solution C == C –– C –– C –– C == C –– C == O
The mass of one atom of Y is ———1—6———g.
6 1023 ⎮ ⎮
CH3 H H
(A is incorrect)
The nucleon number of X is 64. (B is incorrect) (a) Determine the mass of 1 mol of geranial.
(b) Determine the mass of 0.02 mol of geranial.
Mass of 4 mol of Y = 4 16 g (C is correct) (c) Determine the number of molecules present
= 64 g
in 7.6 g of geranial.
Mass of 1 mol of X is 64 g. (d) Determine the mass of 7.5 1022 geranial
Densities cannot be determined because the volume molecules.
[Relative atomic mass: H, 1; C, 12; O, 16;
of the atom is not given. (D is incorrect) NA = 6 1023 mol–1]
47 Chemical Formulae and Equations
3.4 Relationship between Conversion of the Number of Moles of SPM
the Number of Moles of a Gas to Its Volume and Vice Versa ’04/P2
a Gas and Its Volume
1 Since 1 mol of any gas occupies 22.4 dm3 at
Conversion of the Number of Particles of s.t.p. (or 24 dm3 at r.t.p.), n mol of the gas
a Substance to Its Volume will occupy n 22.4 dm3 at s.t.p. (or n
24 dm3 at r.t.p.)
3 1 The volume occupied by a gas depends on
2 Hence,
the temperature. As the temperature increases, volume of gas = number of moles of gas
molar volume
the gas expands and occupies a larger volume. where the molar volume is 22.4 dm3 at s.t.p.
or 24 dm3 at room temperature.
If the gas is cooled, the gas contracts and
3 If a volume of 22.4 dm3 (or 22 400 cm3) is
occupies a smaller volume. occupied by 1 mol of gas,
2 The volume occupied by a gas also depends
1 cm3 of gas is occupied by —2—2——14—0—0‑ mol.
on the pressure. If a gas is compressed (with 4 Conversely,
increased pressure), the volume of the gas n umber of moles of gas = —mv—o—ol——ul—a—m—r———ev——o—o—l—fu—g—ma—es‑
decreases. If the pressure is decreased, the A student need not memorise that molar volume is
22.4 dm3 at s.t.p. or 24 dm3 at r.t.p. It will be given in
volume of the gas increases. the examination.
3 At the same temperature and pressure,
Conversion of the Number of Moles to Volume
equal volumes of all gases contain the of Gas
same number of particles. Accordingly, one
13
mole of any gas (which contains 6.02 1023
Calculate the volume of 0.75 mol of nitrogen gas
particles) will occupy the same volume at a at s.t.p.
[1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.]
particular temperature and pressure.
4 It is found that one mole of any gas at Solution
room temperature (25°C) and pressure of 1
atmosphere occupies a volume of 24 dm3 (or
24 000 cm3).
5 At standard temperature and pressure
(s.t.p.), that is, at a temperature of 0 °C
and pressure of 1 atmosphere, one mole of
any gas occupies a volume of 22.4 dm3 (or
22 400 cm3).
6 The volume occupied by one mole of any gas
SPM is called the molar volume.
’11/P1 Example
• 1 mol of oxygen gas, occupies 24 1 mol of gas occupies a volume 22.4 dm3 at s.t.p.
O2(32 g) dm3 at room
temperature or 0.75 mol of N2 gas occupies Make sure that
• 1 mol of carbon occupies 22.4 —0——.17—5m—m—o—lo—l the numerator and
dioxide gas, CO2(44 g) dm3 at s.t.p. 22.4 dm3 = 16.8 dm3 denominator have the
same unit
[Relative molecular mass: O2, 32; CO2, 44] Conversion of Volume of Gas to Number of Moles
7 One mole of gas contains 6.02 1023 14
molecules and therefore at s.t.p.
Calculate the number of moles of the following
22.4 dm3 of oxygen gas, each contains gases at room temperature and pressure.
O2(32 g) 6.02 1023 (a) 4.8 dm3 of chlorine gas,
molecules (b) 1200 cm3 of methane gas.
22.4 dm3 of carbon
dioxide gas, CO2(44 g) [1 mol of gas occupies a volume of 24 dm3 at
room temperature]
Chemical Formulae and Equations 48
Solution Solution
1 mol of NH3 gas = 17 g
(a) 1 mol of gas occupies a volume of 24 dm3 at 3 .4 g of NH3 = —31—.7—4—gg 1 mol = 0.2 mol
room temperature. 1 mol of gas occupies a volume of 22.4 dm3 at s.t.p.
4.8 dm3 Cl2 gas contain —42—.48——dd—mm—33 1 mol 0.2 mol of ammonia gas occupies a volume of
0.2 22.4 dm3 = 4.48 dm3.
= 0.2 mol
(b) 1 mol of gas occupies a volume of 24 000 cm3 at
room temperature.
1200 cm3 of CH4 gas contain
—2—14—2—00—00—0c—cm—m—33 1 mol
Conversion of Volume of Gas to Mass
= 0.05 mol
16 3
Conversion of the Volume of Gases to SPM Calculate the mass of the following gases at room
’06/P2 temperature and pressure:
Mass and Vice Versa ’08/P2 (a) 7.2 dm3 of sulphur dioxide gas, SO2,
(b) 600 cm3 of methane gas, CH4.
1 Two steps are involved in the conversion of
volume of gas to mass. [Relative atomic mass: H, 1; C, 12; O, 16; S,
Step 1: Volume of gas is converted to 32; 1 mol of gas occupies a volume of 24 dm3
number of moles (by dividing the at room temperature]
volume of gas by the molar volume). Solution
Step 2: Number of moles is converted to
(a) 1 mol of gas occupies a volume of 24 dm3 at
mass in gram (by multiplying the
room temperature.
number of moles by the relative
7.2 dm3 of SO2 = —72——.4—2——dd——mm——33 1 mol = 0.3 mol
atomic mass or relative molecular
mass of the gas). 1 mol of SO2 = (32 + 32) g = 64 g
2 Two steps are involved in the conversion of 0.3 mol of SO2 = 0.3 64 g = 19.2 g
mass to volume of gas. (b) 1 mol of gas occupies a volume of 24 000 cm3 at
Step 1: Mass in gram is converted to
room temperature.
number of moles (by dividing the
600 cm3 of CH4 gas = —2—46—0—00—0—0c—mc—m3—3 1 mol
mass by the relative atomic mass or
relative molecular mass). = 0.025 mol
Step 2: Number of moles is converted 1 mol of CH4 = 16 g
0.025 mol of CH4 = 0.025 16 g = 0.4 g
to volume of gas (by multiplying
the number of moles by the molar
volume).
3 In general:
22.4 dm3 number Ar or Mr mass in Conversion of the Volume of Gases to SPM
of moles Ar or Mr gram
volume the Number of Particles and Vice Versa ’04,06
/P2
22.4 dm3 1 Two steps are involved in the conversion
of the volume of gas to the number of
(at s.t.p.) particles.
Step 1: Volume of gas is converted to
Conversion of Mass to Volume of Gas number of moles (by dividing the
15 volume of gas by the molar volume).
Step 2: Number of moles is converted to
Calculate the volume occupied by 3.4 g of ammonia
gas, NH3 at standard temperature and pressure. number of particles (by multiplying
[Relative atomic mass: H, 1; N, 14; 1 mol of gas
occupies a volume of 22.4 dm3 at s.t.p.] the number of moles by the Avogadro
constant).
2 Two steps are involved in the conversion of
the number of particles to volume of gas.
49 Chemical Formulae and Equations
Step 1: Number of particles is converted to Solution
number of moles (by dividing the 1 mol contains 6 1023 molecules.
number of particles by the Avogadro 1.5 1023 molecules
= —1—6.—5———11—00—232—3m—m—o—ol—el—ec—cu—ule—les—s 1 mol = 0.25 mol
constant).
Step 2: Number of moles is converted 1 mol of gas occupies a volume of 24 dm3 at room
temperature.
to volume of gas (by multiplying 0.25 mol of gas occupies a volume of 0.25 24 dm3
the number of moles by the molar = 6 dm3
volume).
3 In general:
volume 22.4 dm3 number NA number of
of gas of moles NA particles
3 22.4 dm3
(at s.t.p.)
Conversion of Volume of Gas to Number of volume 22.4 dm3 number of Mr mass in
Particles (or 24 dm3) moles Mr gram
22.4 dm3
(or 24 dm3)
17 (61023) (61023)
Calculate the number of molecules present at s.t.p. in number of particles
(a) 0.28 dm3 of N2 gas,
(b) 448 cm3 of carbon monoxide, CO gas. 3 ’03
[1 mol of gas occupies a volume of 22.4 dm3 at Which of the following gases contain 6 1022
s.t.p., NA = 6 1023 mol–1] molecules?
[Relative atomic mass: H, 1; C, 12; N, 14; O, 16;
Solution Avogadro constant = 6 1023 mol–1]
I 1.0 g of hydrogen gas
(a) 1 mol of gas occupies a volume of 22.4 dm3at s.t.p. II 2.8 g of nitrogen gas
III 4.4 g of carbon dioxide
0.28 dm3 of nitrogen gas = —02—.22—.84——dd—mm—33 1 mol IV 1.8 g of water vapour
I, II and III only II, III and IV only
I, III and IV only I, II, III and IV
= 0.0125 mol
1 mol of N2 contains 6 1023 molecules.
0.0125 mol of N2 contains
0.0125 6 1023 molecules = 7.5 1021 molecules
(b) 1 mol of gas occupies a volume of 22 400 cm3 at
s.t.p. Solution
448 cm3 of CO gas = —2—24—44—80—0—cm—c—m3—3 1 mol 6 1022 molecules = —66———11——00—2223‑ 1 mol = 0.1 mol
= 0.02 mol I 1 mol of H2 = 2 g
1 g of H2 = —21 mol = 0.5 mol
1 mol of CO contains 6 1023 molecules. (I is incorrect)
0.02 mol of CO contains
0.02 6 1023 molecules = 1.2 1022 molecules II 1 mol of N2 = 28 g
2.8 g of N2 = —22—.88 mol = 0.1 mol (II is correct)
Conversion of Number of Particles to Volume of
Gas III 1 mol of CO2 = 44 g
4.4 g of CO2 = —44—.44 mol = 0.1 mol (III is correct)
18
IV 1 mol of H2O = 18 g
Calculate the volume of 1.5 1023 molecules of 1.8 g of H2O = —11—.88 mol = 0.1 mol (IV is correct)
ethane, C2H6 gas at room temperature and pressure.
[1 mol of gas occupies a volume of 24 dm3 at r.t.p., Answer
NA = 6 1023 mol–1]
Chemical Formulae and Equations 50
4 ’05 3.5 Chemical Formulae SPM
The activity of microorganisms on waste products ’05/P2
at dump sites produces methane gas. If 180 dm3 ’08/P1
of methane gas is collected, calculate the mass of
methane obtained. 1 A chemical formula is used to represent a 3
[Relative atomic mass: H, 1; C, 12; 1 mol of gas chemical compound.
occupies a volume of 24 dm3 at room temperature The chemical formula shows
and pressure] (a) the elements, denoted by their symbols,
present in the compound.
Solution (b) the relative numbers, indicated by
1 mol of gas occupies a volume of 24 dm3 at room subscripts written after the symbols, of
temperature. each element present in the compound.
1 80 dm3 of CH4 gas = 1—28—40—d—dm—m33 1 mol 2 For example, the chemical formula of
= 7.5 mol sulphuric acid is H2SO4. The chemical formula
indicates that
1 mol of CH4 = 16 g (a) the elements present in sulphuric acid are
7.5 mol of CH4 = 7.5 16 g hydrogen, sulphur and oxygen.
(b) two hydrogen atoms, one sulphur atom
= 120 g and four oxygen atoms combine to form
the compound.
3 Table 3.1 shows the chemical formulae of
some covalent compounds.
3.4 Table 3.1 The chemical formulae of some covalent
compounds
1 Calculate the volume of 0.55 mol of oxygen gas at Name of Chemical Number of each element
room temperature and pressure.
[1 mol of gas occupies a volume of 24 dm3 at compound formula in the compound
room temperature]
Oxygen O2 2 oxygen atoms
2 Calculate the number of moles of 672 cm3 of Water H2O 2 hydrogen atoms and
carbon dioxide gas at s.t.p.
[1 mol of gas occupies a volume of 22.4 dm3 at 1 oxygen atom
s.t.p.]
Ammonia NH3 1 nitrogen atom and
3 hydrogen atoms
3 Calculate the volume occupied by 1.4 g of ethene Sulphuric H2SO4 2 hydrogen atoms,
acid 1 sulphur atom and
g[Raes,laCti2vHe4 at room temperature and pressure. of 4 oxygen atoms
atomic mass: H, 1; C, 12; 1 mol
gas occupies a volume of 24 dm3 at room
temperature] 4 The chemical formula of an ionic compound
can be written if the charge of the cation
4 Calculate the mass of each of the following gases (positively-charged ion) and the anion
at standard temperature and pressure: (negatively-charged ion) forming the ionic
(a) 16672.80dcmm33ooffmcaertbhoannem, oCnHo4x, ide gas, CO. compound are known.
(b) 5 Table 3.2 shows the charges of some ions.
[Relative atomic mass: H, 1; C, 12; O, 16; 1
mol of gas occupies a volume of 22.4 dm3 at
s.t.p.] Table 3.2 Charges of some cations and anions
5 Calculate the number of molecules present in: Charge Cation Symbol
(a) 3.6 dm3 of N2 gas, +1
(b) 1200 cm3 of ammonia gas at room Sodium ion Na+
temperature and pressure. Potassium ion K+
[1 mol of gas occupies a volume of 24 dm3 Lithium ion Li+
at room temperature, NA = 6 1023 mol–1] Silver ion Ag+
Copper(I) ion Cu+
6 Calculate the volume of 9 1021 molecules of Hydrogen ion H+
CO at standard temperature and pressure. Ammonium ion NNHi+ 4+
[1 mol of gas occupies a volume of 22.4 dm3 at Nickel(I) ion
s.t.p., NA = 6 1023 mol–1]
51 Chemical Formulae and Equations
Charge Cation Symbol must be equal to the total negative charge of
+2
+3 Magnesium ion Mg2+ the anion.
Calcium ion Ca2+
Zinc ion Zn2+ 7 To write the chemical formula of an ionic
Iron(II) ion Fe2+
Copper(II) ion Cu2+ compound, the following steps can be used:
Manganese(II) ion Mn2+
Lead(II) ion Pb2+ (a) Write the formulae of the ions involved in
Nickel(II) ion Ni2+
forming the compound and their charges.
Iron(III) ion Fe3+
Aluminium ion Al3+ (b) Then, balance the positive and negative
Chromium(III) ion Cr3+
charges. This can be done by writing the
numerical charge of the cation next to the
anion as a subscript and the numerical
charge of the anion next to the cation as a
subscript.
3 (c) Finally write the chemical formula of the
ionic compound without the charges.
Charge Anion Symbol 8 Generally, if the ionic compound is formed by
–1
Fluoride ion F– the ions Xm+ and Yn–, then the chemical formula
–2 Chloride ion Cl–
–3 Bromide ion Br– of the compound is XnYm.
Iodide ion I– If m = n, then the chemical formula is XY.
Hydroxide ion OH–
Nitrate ion NO3– Examples,
Nitrite ion NO2–
Bicarbonate ion HCO3– (a) Chemical formula of sodium sulphate
Permanganate ion MnO4–
Hydride ion H– Charge of ion +1 –2
Oxide ion O2– Formula of ion Na+ SO42–
Sulphide ion S2– Ratio 2 1
Sulphate ion SO42–
Sulphite ion SO32– Chemical formula is Na2SO4.
Carbonate ion CO32– (b) Chemical formula of iron(III) chloride
Thiosulphate ion S2O32–
Chromate(VI) ion CrO42– Charge of ion +3 –1
Dichromate(VI) ion Cr2O72–
Formula of ion Fe3+ Cl–
Phosphide ion P3–
Phosphate ion PO43– Ratio 1 3
Nitride ion N3–
Chemical formula is FeCl3.
(c) Chemical formula of magnesium oxide
Charge of ion +2 –2
Formula of ion Mg2+ O2–
Ratio 2 2
Chemical formula is MgO.
Note that if the charges of the ions are the
same, the ratio of the ions that combine
is 1 : 1.
9 In general, the chemical formulae of
6 Since a chemical compound is always electrically compounds formed between the ions are
neutral, the total positive charge of the cation
summarised in Table 3.3.
Table 3.3 Formulae of ionic compounds
Cation Anion Chemical formula Examples
X+ Y– XY Sodium chloride, NaCl
X+ Y 2– X2Y Potassium dichromate(VI), K2Cr2O7
X+ Y 3– X3Y Ammonium phosphate, (NH4)3PO4
X 2+ Y– XY2 Calcium chloride, CaCl2
X 2+ Y 2– XY Copper(II) sulphate, CuSO4
X 2+ Y 3– X3Y2 Magnesium nitride, Mg3N2
X 3+ Y– XY3 Iron(III) chloride, FeCl3
X 3+ Y 2– X2Y3 Chromium(III) sulphate, Cr2(SO4)3
X 3+ Y 3– XY Aluminium nitride, AlN
Chemical Formulae and Equations 52
10 Transition elements can form ions with Compound Molecular Simplest Empirical
different charges. In the IUPAC (International formula ratio of the formula
Union of Pure and Applied Chemistry) system, elements
the standardised chemical nomenclature is
denoted by Roman numerals in brackets to Glucose C6H12O6 C : H : O CH2O
indicate the charge of the ion. For example, =1:2:1
iron(II) represents Fe2+ ion and iron(III)
represents Fe3+ ion. Table 3.4 shows examples Quinine C20H24N2O2 C : H : N : O C10H12NO
of some compounds of transition elements. = 10: 12 : 1 : 1
Table 3.4 Formulae of some compounds of transition 4 The following steps can be used to determine
elements
the empirical formula of a compound:
Cation Anion Name of compound Chemical Step 1: Write the mass or percentage of each 3
formula
element in the compound.
Cu+ O2– Copper(I) oxide Cu2O Step 2: Calculate the number of moles of
Cu2+ O2– Copper(II) oxide CuO
each element in the compound by
Fe2+ Cl– Iron(II) chloride FeCl2
Fe3+ Cl– Iron(III) chloride FeCl3 dividing the mass or percentage of the
Mn2+ O2– Manganese(II) oxide MnO element by the relative atomic mass of
Mn4+
O2– Manganese(IV) oxide MnO2 the element.
Step 3: Next, divide each number by the
All transition elements are metals. Many of them
like iron, manganese, copper, silver, gold, nickel and smallest number to obtain the
titanium are of major technological importance. A large simplest ratio.
number of the transition elements combine with each Step 4: Finally write the empirical formula
other to form useful alloys.
of the compound based on the ratio
of the elements.
5 ’09
The table shows the mass of elements M and O in
an oxide, and relative atomic mass of M and O.
Empirical and Molecular Formulae Element MO
1 The empirical formula of a compound shows Mass (g) 2.4 1.6
SPM the simplest ratio of the atoms of the elements
’07/P2 that combine to form the compound. Relative atomic mass 48 16
2 The molecular formula of the compound
What is the empirical formula of this compound?
shows the actual numbers of the atoms of the
Solution
elements that combine to form the compound.
3 Table 3.5 shows the molecular and empirical Element M O
formulae of some compounds.
Table 3.5 The molecular and empirical formulae SPM Step 1 Mass 2.4 g 1.6 g
Step 2
of some compounds ’09/P1
Step 3
Compound Molecular Simplest Empirical Number of —24—.84 mol —11—.66 mol
formula ratio of the formula moles = 0.1
elements
= 0.05
Water H2O H:O=2:1 H2O Simplest —00—..00—55 —00—..01—05
Ethene C2H4 C:H=1:2 CH2 ratio =1 =2
Butane C4H10 C:H=2:5 C2H5
Ethane C2H6 C:H=1:3 CH3 Empirical formula of compound is MO2.
53 Chemical Formulae and Equations
6 ’05 Simplest —14—..68 —11—..—66 —41—.8.—64
ratio =1 =3
2.5 g of X combined with 4 g of Y to form a =3
compound with the formula XY2. If the relative
atomic mass of Y is 80, determine the relative Empirical formula of boric acid is H3BO3.
atomic mass of X.
Solution
Assume the relative atomic mass of X is a.
Element XY 20 SPM
Mass
Number of moles 2.5 4 ’04/P2
3 Simplest ratio —2—a.5 mol —8—40 mol A gaseous hydrocarbon X contains 85.7% of carbon
= 0.05 by weight. 4.2 g of the gas X occupies a volume of
3.36 dm3 at standard temperature and pressure.
12 [Relative atomic mass: H, 1; C, 12; 1 mol of gas
occupies a volume of 22.4 dm3 at s.t.p.]
Since the empirical formula is XY2, the ratio of X : (a) Determine the empirical formula of X.
Y=1:2 (b) Determine the relative molecular mass of X.
— 0——2.—a0.—5 5 = —21 (c) What is the molecular formula of X?
—2a—.5 = —0—.20—5 Solution
a = —2—.05—.—0—5—2 Element C H
= 100 Percentage 85.7%
100 – 85.7%
Number of —8—15—2.—7 mol = 14.3%
moles
—1—41—.3 mol
= 7.14 = 14.3
19 Simplest ratio —77—..11—44 —71—4.1—.—43
=1 =2
Boric acid is used to preserve prawns and fish.
Chemical analysis of the compound shows that it (a) Empirical formula of X is CH2.
contains 4.8% hydrogen, 17.7% boron and the rest (b) 4.2 g of the gas has a volume of 3.36 dm3 at s.t.p.
is oxygen. Determine the empirical formula of boric
acid. [Relative atomic mass: H, 1; B, 11; O, 16] 1 mol of the gas (22.4 dm3) has a mass of
— 32—.23—.64——dd—mm—33 4.2 g
Solution
= 28 g
Element H B O
Therefore the relative molecular mass of X is 28.
Percentage 4.8% 17.7% 100 – 4.8 – 17.7% (c) Assume the molecular formula of X is (CH2)n.
= 77.5%
The relative molecular mass of X is
Mass in —14—0—.80 100 g —11—70—.07100 g —71—70—.05 100 g (12 + 2)n = 28
100 g of n = —12—48
compound = 4.8 g = 17.7 g = 77.5 g =2
Number —41—.8 mol —1—17—1.7 mol —7—176—.5 mol Thus the molecular formula of X is C2H4.
of moles
= 4.8 = 1.6 = 4.84
Chemical Formulae and Equations 54
To determine the empirical formula of magnesium oxide by
experiment
Apparatus Crucible with lid, tripod stand, combustion of the magnesium. Care is taken
sandpaper, Bunsen burner, clay pipe to prevent white smoke (magnesium oxide
Materials triangle, electronic balance and tongs. powder formed) from escaping by closing the lid
Procedure immediately after it is lifted.
A magnesium ribbon of about 20 6 When the magnesium is completely burnt, the
cm in length and oxygen. crucible is cooled and is weighed again with its
lid. The weight is recorded.
7 The heating, cooling and weighing process is
repeated until a constant mass is obtained.
Results Activity 3.2 & 3.3 3
Mass of crucible + lid = a gram
Mass of crucible + lid + magnesium = b gram
Mass of crucible + lid + magnesium oxide = c gram
1 The 20 cm length of magnesium ribbon is polished Calculations = (b – a) gram
with sandpaper to remove the oxide layer on its = (c – b) gram
surface. Mass of magnesium used
Mass of oxygen that combined with
2 The magnesium ribbon is rolled into a loose coil. magnesium
3 An empty crucible with its lid is first weighed
Element Mg O
and its weight is recorded. (c – b)
4 The rolled magnesium ribbon is placed in the Mass (g) (b – a) ( —c—1–—6—b) mol
crucible. The crucible is weighed again and its Number of (—b—2—–4—a)mol y
weight recorded. moles
5 The crucible is placed on a clay pipe triangle and
then heated strongly. The lid is lifted at intervals Simplest ratio x
to allow the oxygen from the air to enter for the
Conclusion
The empirical formula is MgxOy.
To determine the empirical formula of copper oxide SPM
’09/P2
The empirical formula of copper oxide is determined 1 The combustion tube with a porcelain dish is
by reducing the copper oxide using hydrogen gas. weighed. The weight is recorded.
Apparatus Porcelain dish, combustion tube and 2 A spatula of copper oxide powder is placed in
Materials electronic balance. the porcelain dish. The combustion tube with its
contents is weighed again. The weight is recorded.
Copper oxide powder and dry
hydrogen gas. 3 Dry hydrogen gas is passed through the tube for
a few minutes to expel all the air.
Procedure
4 The copper oxide is then heated strongly and
the hydrogen gas passing through the end of the
combustion tube is lit.
5 When all the copper oxide is reduced to copper
metal (the black copper oxide has all become
brown), heating is stopped.
6 A continuous stream of hydrogen gas is allowed
to pass through the tube until it is cooled.
55 Chemical Formulae and Equations
7 The combustion tube with its contents is weighed Precautions taken
and the weight recorded.
1 Hydrogen gas must be allowed to pass through
8 The heating, cooling and weighing process is SPM the combustion tube for a few minutes before the
repeated until a constant weight is obtained. ’11/P1 copper oxide is heated. This is to remove the air
Results in the tube. A mixture of air and hydrogen can
explode when lit.
Mass of combustion tube + empty porcelain dish 2 The flow of hydrogen gas must be continued
throughout heating. This is to ensure that air
= 54.31 g does not enter the combustion tube. Hence
the hydrogen gas must be seen to be burning
Mass of combustion tube + porcelain dish + copper continuosly at the end of the combustion tube.
3 The hot copper metal is allowed to cool in a
oxide = 62.32 g stream of hydrogen gas. This is to ensure that
oxygen from the air does not oxidise the hot
Mass of combustion tube + porcelain dish + copper copper to copper oxide again.
4 The heating, cooling and weighing process is
= 60.71 g repeated until a constant weight is obtained. This
is to ensure that all the copper oxide has been
3 Calculations reduced.
Mass of copper obtained = (60.71 – 54.31)g = 6.40 g Conclusion
Mass of oxygen that combined with the copper
= (62.32 – 60.71)g = 1.61 g The empirical formula of copper oxide is CuO.
Element Cu O
Mass (g) 6.40 1.61
Number of —66——.44 mol = 0.1 —1—1.6—61 mol = 0.1
moles
Simplest ratio —00—..—11 = 1 —00——..11 = 1
3.5 4 10.2 g of vanadium metal combined with 8 g of
oxygen to form a compound with empirical formula
1 Write down the chemical formulae of the following V2O5. Determine the relative atomic mass of
ionic compounds: vanadium. [Relative atomic mass: O, 16]
(a) Silver nitrate
(b) Sodium thiosulphate 5 Hydrazine is used as a rocket fuel. It contains 87.5%
(c) Ammonium phosphate nitrogen and 12.5% hydrogen. [Relative atomic
(d) Calcium hydroxide mass: H, 1; N, 14]
(e) Magnesium carbonate (a) Determine the empirical formula of the
(f) Zinc phosphide compound.
(g) Iron(III) hydroxide (b) If the relative molecular mass of hydrazine is 32,
(h) Aluminium oxide determine its molecular formula.
(i) Chromium(III) chloride
(j) Copper(II) sulphate 6 x g of iron combined with 5.04 g of oxygen to form
(k) Nickel(I) chloride
(l) Magnesium nitride an oxide with empirical formula Fe2O3. Determine
the value of x. [Relative atomic mass: O, 16; Fe, 56]
2 Silicon hydride contains 87.5% silicon by mass.
Determine the empirical formula of silicon hydride. 7 A gaseous hydrocarbon Q contains 20% hydrogen.
[Relative atomic mass: H, 1; Si, 28] 6 g of this hydrocarbon occupies a volume of 4.48
dm3 at s.t.p. [Relative atomic mass: H, 1; C, 12; 1
3 Reduction of 7.55 g of tin oxide using hydrogen gas mol of gas occupies 22.4 dm3 at s.t.p.]. Determine:
yields 5.95 g of tin metal. Determine the empirical (a) The empirical formula of Q
formula of tin oxide. [Relative atomic mass: Sn, 119; (b) Relative molecular mass of Q
O, 16] (c) Molecular formula of Q
Chemical Formulae and Equations 56
3.6 Chemical Equations
Reactants and Products of Chemical Remember that a chemical equation cannot be
Equations balanced by altering the formulae of the reactants or
products. You can only balance the equation by putting
1 A chemical reaction can be represented by a a number in front of each of the formulae.
chemical equation. For example, when sodium
hydroxide reacts with sulphuric acid, sodium Writing Chemical Equations
sulphate and water are produced. The reaction
can be represented symbolically by the equation: 1 Lets’s look at a chemical reaction to practice
these rules. During combustion, magnesium
2NaOH + H2SO4 → Na2SO4 + 2H2O reacts with the oxygen in air to produce a
white powder, magnesium oxide. You might
start by writing an equation for this reaction 3
by writing the symbols for the reactants and
(a) The reactants are the chemicals that products.
are reacting and they are written on the reactants yield product
Mg + O2 ⎯→ MgO
left-hand side of the equation. Hence,
2 On the left-hand side of the equation there
sodium hydroxide and sulphuric acid are is 1 atom of magnesium and there is also
1 atom of Mg on the product side of the
the reactants. equation. However, there are 2 oxygen atoms
on the left and only 1 on the right. To balance
(b) The products are the chemicals formed in the number of atoms of each type you will
need to add coefficients. If you multiply Mg on
a reaction and they are written on the right- the left-hand side of the equation by 2 then
you must do the same to the Mg atom on the
hand side of the equation. Hence, sodium right which will mean doubling the number
of oxygen atoms on the right-hand side of the
sulphate and water are the products. equation too since you may not change the
ratio of atoms within a molecule because it
(c) 2 mol of NaOH react with 1 mol of H2SO4 will change the identity of the substance. The
to produce 1 mol of Na2SO4 and 2 mol of new equation will look like this:
H2O.
2 Some symbols which appear in a chemical 2Mg + O2 ⎯→ 2MgO
equation are: 3 Now the equation is balanced. There are the same
number of Mg and O atoms on both sides of the
Symbol Meaning equation. A balanced equation is the source of a
∆ Heating of substance great deal of information about both products
and reactants during a chemical change.
↑ or (g) Gas evolved
↓ or (s) Precipitate formed
Reversible reaction
3 When writing a chemical equation, the Examples of Chemical Equations
following steps are followed:
(a) Write the correct formulae of all reactants 1 (a) When green copper(II) carbonate is heated,
on the left-hand side of the equation. it decomposes to form black copper(II)
(b) Write the correct formulae of all products oxide and carbon dioxide gas which turns
on the right-hand side of the equation. limewater cloudy.
(c) The equation is then balanced. This
involves making sure that the number of CuCO3(s) ⎯→ CuO(s) + CO2(g)
atoms of each element before and after reactant products
the reaction are the same.
(d) Finally the physical state of each of the
reactants or products is written as:
(s) – represents solid state
(l) – represents liquid state
(g) – represents gaseous state
(aq) – represents aqueous state, that is,
the substance is dissolved in water.
57 Chemical Formulae and Equations
(b) When calcium carbonate is reacted with If 0.12 g of magnesium is added to excess
hydrochloric acid, the products formed are hydrochloric acid, calculate
calcium chloride, carbon dioxide and water. (a) the mass of magnesium chloride salt formed.
(b) the volume of hydrogen gas evolved at room
CaCO3(s) + 2HCl(aq) →
CaCl2(aq) + CO2(g) + H2O(l) temperature and pressure.
[Relative atomic mass: Mg, 24; Cl, 35.5; 1 mol
(c)W hen an aqueous solution of potassium iodide is of gas occupies a volume of 24 dm3 at r.t.p.]
added to an aqueous solution of lead(II) nitrate,
a yellow precipitate of lead(II) iodide is formed. Solution
(a) Referring to the equation, 1 mol of Mg (24 g)
2KI(aq) + Pb(NO3)2(aq) →
PbI2(s) + 2KNO3(aq) produces 1 mol of MgCl2(95 g).
Hence 0.12 g of Mg will produce
3 — 02—.1—4—2——g—g——M—M——g—g 95 g MgCl2
Interpreting Chemical Equations SPM = 0.475 g of MgCl2
Qualitatively and Quantitatively (b) 1 mol of Mg (24 g) produces 1 mol of H2 gas
’11/P1
(24 dm3) at r.t.p.
1 From a chemical equation, we can obtain Hence 0.12 g of Mg will produce
information on —0—2.1—42 24 dm3 of H2
(a) the reactants taking part in the reaction = 0.12 dm3 of H2
(on the left-hand side of the equation). = 120 cm3 of H2
(b) the products formed in the reaction (on
the right-hand side of the equation). 8 ’04
(c) the number of moles of each substance
taking part in the reaction and the 3.2 g of copper(II) oxide powder is reacted with
excess dilute nitric acid.
number of moles of the products formed (a) Write a chemical equation for the reaction.
(from the coefficients/numbers before (b) Calculate the mass of copper(II) nitrate salt
the substances).
(d) the physical states of all the reactants and formed in the reaction.
products (from the symbols in bracket [Relative atomic mass: N, 14; O, 16; Cu, 64]
after the substance).
Solution
For example: (a) CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l)
(b) 1 mol of CuO produces 1 mol of Cu(NO3)2.
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
The chemical equation shows that: 80 g of CuO produces 188 g of Cu(NO3)2.
Hence 3.2 g of CuO will produce
2 mol of solid sodium react with 2 mol — 38—.02—gg—CCu—uOO— 188 g of Cu(NO3)2
= 7.52 g of Cu(NO3)2
of liquid water to form 2 mol of aqueous
21
sodium hydroxide and 1 mol of hydrogen
Ethanol burns in air as represented by the equation
gas. C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l)
Calculate the mass of ethanol burnt if 2.4 dm3 of
Solving Numerical Problems Using SPM carbon dioxide is produced at room temperature.
Chemical Equations [Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of
’08/P1 gas occupies a volume of 24 dm3 at room temperature]
The quantity of reactants and products can be
calculated using stoichiometric equations when
numerical information is given (stoichiometry).
7 ’09
Magnesium reacts with hydrochloric acid as
represented by the equation
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Chemical Formulae and Equations 58
Solution (d) Displacement reaction between zinc metal 3
Referring to the equation, 1 mol of C2H5OH and copper(II) sulphate solution to form
(46 g) produces 2 mol (2 24 dm3) of CO2. copper metal and zinc sulphate salt.
Therefore 2.4 dm3 of CO2 is produced from
— 2——2—.42—d4—m—d—m3—3 46 g of C2H5OH (e) Decomposition of zinc carbonate to zinc oxide
= 2.3 g of ethanol, C2H5OH and carbon dioxide gas when heated.
22 (f) Reduction of solid lead(IV) oxide, PbO2 by
hydrogen gas to form lead metal and water.
The chemical name of baking powder is sodium
bicarbonate (NaHCO3). When sodium bicarbonate 3 Iron metal reacts with excess hydrochloric acid to
is heated, it decomposes to sodium carbonate, produce iron(II) chloride and hydrogen gas.
carbon dioxide and water. (a) Write a balanced chemical equation for the
(a) Write a balanced chemical equation for the reaction.
(b) If 2.8 g of iron metal is used in the reaction,
decomposition of sodium bicarbonate when it is calculate
heated.
(b) If 2.1 g of sodium bicarbonate is heated, (i) the maximum mass of iron(II) chloride
calculate the volume in cm3 of CO2 produced at formed.
room temperature.
[Relative atomic mass: H, 1; C, 12; O, 16; Na, 23; (ii) the volume of hydrogen gas produced at
1 mol of gas occupies a volume of 24 dm3 at room room conditions.
temperature] [Relative atomic mass: Cl, 35.5; Fe, 56;
1 mol of gas occupies a volume of 24 dm3
Solution at r.t.p.]
(a) 2NaHCO3(s) ⎯∆→ Na2CO3(s) + CO2(g) + H2O(l)
(b) 2 mol of NaHCO3 (2 84 g) produces 1 mol of 4 Acetylene gas (C2H2) is used in metal welding. This
gas can be prepared by reacting calcium carbide
CO2 (24 dm3) at r.t.p. with excess water as represented by the equation
2.1 g of NaHCO3 will produce
— 2—2——.—18—4 24 dm3 of CO2 = 0.3 dm3 of CO2 CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)
= 300 cm3 of CO2 If 4.8 g of calcium carbide is reacted with excess
water, calculate
(a) the volume of acetylene gas evolved at room
temperature.
(b) the mass of calcium hydroxide formed.
[Relative atomic mass: H, 1; C, 12; O, 16;
Ca, 40; 1 mol of gas occupies a volume of 24
dm3 at r.t.p.]
5 Hydrogen gas is prepared by reacting methane gas
with steam using platinum as catalyst. The reaction
is represented by the equation
CH4(g) + H2O(g) → CO(g) + 3H2(g)
3.6 If 60 dm3 of hydrogen gas is produced at room
temperature and pressure, calculate
1 Balance the following chemical equations: (a) the mass of methane that is used in the
(a) Na(s) + HHC2Ol((alq))→→NZanOCHl2((aaqq)) + HH22((gg)) reaction.
(b) Zn(s) + +
(b) the number of carbon monoxide molecules
(c) CuCO3(s) + HNO3(aq) → released.
Cu(NO3)2(aq) + CO2(g) + H2O(l) [Relative atomic mass: H, 1; C, 12; 1 mol
(d) HCl(aq) + Na2S2O3(aq) → of gas occupies a volume of 24 dm3 at r.t.p;
NaCl(aq) + SO2(g) + S(s) + H2O(l) NA = 6 1023 mol–1]
( e) P b(NO3)2(s) ⎯⎯∆ → PbO(s) + NO2(g) + O2(g)
6 Zinc spuhbosstapnhcideeca(nZnb3eP2p)reispaureseddbyasreaacrtiantgpeoxicseosns.
This
2 Write balanced equations for the following reactions:
(a) Reaction between sulphur dioxide and oxygen zinc powder with phosphorus.
to form sulphur trioxide.
(b) Neutralisation reaction between potassium (a) Write a balance chemical equation for the
hydroxide and sulphuric acid to form
potassium sulphate and water. reaction.
(c) Burning of ethane (C2H6) in air to form carbon
dioxide and water. (b) Calculate the mass of phosphorus needed to
react with the excess zinc to produce 51.4 kg
of zinc phosphide.
[Relative atomic mass: P, 31; Zn, 65]
59 Chemical Formulae and Equations
3.7 Scientific Attitudes and Values in Investigating Matter
3 1 In the 19th century, scientists introduced a to produce 2 mol of magnesium oxide, we
simple way to represent an element. need to burn 2 mol of magnesium in air.
Each element is represented by a letter or If we need to produce 1 mol of MgO (40 g),
two letters of the alphabet; the first letter is a then we have to burn 1 mol of Mg (24 g) in
capital letter and the second letter is a small air.
letter. For example, the reaction between Similarly, to produce 40 kg of MgO, the
carbon and oxygen to form carbon dioxide required amount of Mg needed is 24 kg.
can be represented by the following equation: Any excess amount of magnesium used is
considered a wastage.
C(s) + O2(g) → CO2(g) 5 The use of relative mass by scientists enabled
the mass of atoms and molecules to be
2 The symbols of elements using letters of determined although they are too small to
the alphabet are universal in that they are be measured by any weighing machines. The
agreed upon by all scientists, regardless of the development of the standard to be used in the
countries of origin. They become a tool of determination of relative atomic mass and
communication between chemists although relative molecular mass started with the use
the spoken languages may be different. of the hydrogen atom, followed by the oxygen
atom, and finally carbon-12 which is now
3 Many chemical reactions involve the formation used as the standard.
of compounds. However, the names of some 6 The Italian scientist, Amedeo Avogadro
compounds are long. This makes the writing proposed the hypothesis that equal volumes of
of equations in words cumbersome. all gases at the same temperature and pressure
For example, contain the same number of molecules. With
perseverance and ingenuity, he calculated that
Sulphuric acid + sodium hydroxide → 22.4 dm3 of any gas contains 6.02 1023
sodium sulphate + water particles at standard temperature and pressure.
The number of 6.02 1023 is now known as
To simplify the writing of the chemical Avogadro’s number or the Avogadro constant
equations, scientists used chemical formulae in honour of him.
to represent compounds. The above reaction 7 Some products can be produced by different
can then be represented by processes. For example, calcium oxide can be
produced by
H2SO4(aq) + 2NaOH(aq) → (a) burning calcium in air:
Na2SO4(aq) + 2H2O(l)
2Ca(s) + O2(g) → 2CaO(s) or
Thus the writing of chemical equations is (b) heating calcium carbonate:
made easy.
4 In the formation of a desired product through CaCO3(s) → CaO(s) + CO2(g)
a chemical reaction, it is important to calculate
the correct amount of reactants required so as 8 The ideal process is one in which
to prevent wastage. This is made possible (a) a high percentage yield is obtained.
by the mole concept used by chemists in the For example, a process which produces
quantitative calculations of chemical reactions. 80% yield is superior to one which
For example, in the reaction between magnesium produces only 45% yield.
and oxygen, The mole concept makes it possible to
calculate the percentage yield.
2Mg + O2 → 2MgO (b) the cost of reactants used is cheap.
In general, the cheaper the reactants, the
better the process is because the cost of
producing the product is lower. Thus, the
product can be priced competitively.
Chemical Formulae and Equations 60
3.7 2 Explain the meaning of the following chemical
1 Write the chemical formulae for the following equations:
compounds:
(a) Hydrobromic acid (a) CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
(b) Zinc hydroxide (b) 2Mg(NO3)2(s) ⎯∆→ 2MgO(s) + 4NO2(g) +
(c) Potassium nitrate O2(g)
(d) Sodium sulphate
(e) Silver nitrate ((dc)) CHCaa2(CCgOl2)(3+a(qsP))b++O2C(sHO)C2→(l(ga)Pq+b)(→Hs)2O+(Hl)2O(l)
(f) Magnesium chloride
3
1 The relative atomic mass of an element is the 6 The molar volume of a gas is the volume occupied
by one mole of the gas.
number of times one atom of the element is heavier At standard temperature and pressure, one mole of
gas occupies a volume of 22.4 dm3.
than one-twelfth the mass of one carbon-12 atom.
2 The relative molecular mass of a compound is the 7 The interconversion between mole, mass, volume
and number of particles can be summarised
number of times one molecule of the compound is below:
heavier than one-twelfth the mass of one carbon-12 Volume 22.4 dm3 Number 3 RAM Mass in
3 22.4 dm3 of or RMM gram
atom.
3 One mole is the quantity of substance which moles RAM
or RMM
contains the same number of particles as there are
in 12.00 grams of carbon-12.
4 One mole of element is the relative atomic mass
of the element expressed in gram.
For example,
1 mol of C = 12 g All contain 3 (6 3 1023) (6 3 1023)
1 mol of Na = 23 g 6.02 3 1023 atoms
1 mol of S = 32 g
Number
5 One mole of compound is the relative molecular of
mass of the compound expressed in grams. For
example, particles
1 mol of H2O = 18 g All contain
1 mol of CO2 = 44 g 6.02 3 1023 molecules
1 mol of C2H6 = 30 g
61 Chemical Formulae and Equations
3
Multiple-choice Questions
3.1 Relative Atomic Mass [Relative atomic mass: H, 1; 3.2 Relationship between the
and Relative Molecular Number of Moles and
Mass C, 12; O, 16] the Number of Particles
A 136
3 1 The relative formula mass of B 140 8 Which of the following samples
C 151
Mg(XO3)2 is 148. Determine the D 152 contain 3.0 3 1022 molecules?
relative atomic mass of element
[Relative atomic mass: H,1;
X. [Relative atomic mass: O, 16; 5 Three elements are represented C, 12; N, 14; O,16; S, 32; 1 mol
Mg, 24] by the letters X, Y and Z. One contains 6 3 1023 molecules]
A 10 C 14 ’08 atom of Z is two times heavier I 0.9 g gooffofCHN22HOH43
II 0.85
B 12 D 18 than one atom of Y. One atom III 1.4 g
2 A compound with formula of Y is three times heavier than IV 1.6 g of SO2
A I, II and III only
fMor2mS2uOla3.5mHa2sOs has a relative one atom of X. If the relative
of 248.
atomic mass of X is 39, what is B I, III and IV only
What is the relative atomic mass the relative atomic mass of Z? C II, III and IV only
of M? [Relative atomic mass: A 78 C 195 D I, II, III and IV
H, 1; O, 16; S, 32] B 117 D 234
A 23 C 27 6 Which compound in the table 9 Calculate the number of molecules
below is correctly matched with
B 24 D 39 its relative formula mass? in 0.88 g of vitamin C, C6H8O6.
[Relative atomic mass: H, 1; [Relative atomic mass: H, 1; C, 12;
3 Veronal is a barbiturate used C, 12; N, 14; O, 16; Na, 23;
P, 31; S, 32; Cl, 35.5; Ca, 40; O, 16; NA =6 3 1023 mol–1]
to induce sleep in psychiatric Co, 59] A 3.0 3 1021
patients. The molecular formula B 3.0 3 1022
of veronal is C4H2N2O3(C2H5)2. C 7.5 3 1021
Determine the relative molecular
D 7.5 3 1022
mass of veronal. [Relative
atomic mass: H, 1; C, 12; N, 14; Relative 10 Calculate the number of atoms
formula mass
O, 16] Compound in 0.96 g of titanium.
A 160 C 186 [Relative atomic mass: Ti, 48;
B 184 D 196 I Ca3(PO4)2 310 AN A 1=.26 3 1023 mol–1]
II C14H18N2O5 294 3 1021
4 The diagram shows the molecular III C17H35COONa 306
structure of vanillin molecule IV CoCl2.6H2O 170 B 1.5 3 1021
which gives vanilla its taste.
C 1.2 3 1022
H O D 1.5 3 1022
C 11 Which of the following has the
|
A I and III only most number of molecules?
H C H B II and IV only
C I, II and III only [Relative atomic mass: H, 1;
D I, III and IV only
C, 12; O, 16; Br, 80]
C C A 3.6 g of water, H2O
B 4.0 g of methane, CH4
H C 19.6 g of sulphuric acid, H2SO4
| D 13.2 g of ethanoic acid,
7 Calculate how many beryllium
C C–O–C–H
| atoms that will have the same CH3COOH
H C H mass as one quinine molecule,
| Cdr2u0Hg)2.4N[R2eOla2 t(ivaen anti-malarial 12 The relative atomic mass of
O–H atomic mass:
oxygen and sulphur are 8 and
H,1; Be,9; C,12; N,14; O,16]
32 respectively. Which of the
Determine the relative molecular A 35 C 37 following statements is/are true
mass of vanillin.
B 36 D 38 about oxygen and sulphur?
Chemical Formulae and Equations 62
I 4 g of oxygen and 8 g of A I and III only Calculate the number of moles
sulphur contain the same B I and IV only in 4.86 g allicin.
number of atoms. C II and III only [Relative atomic mass: H, 1;
D II and IV only C, 12; O, 16; S, 32]
II The sulphur atom has four A 0.02 mol C 0.03 mol
times more neutrons than an 17 Calculate the mass of 7.5 3 1021 B 0.025 mol D 0.05 mol
oxygen atom. [aRsepliaritniv,eCa9Hto8mO4icmmoalescsu: lHes,.1;
C, 12; O, 16; NA = 6 1023 23 Cocaine, Cth1e7Hm21aOs4sNofis1a.2d3rug1. 021
III The sulphur atom is four times mol–1] Calculate
denser than the oxygen atom. 3
IV 8 g of oxygen contain two A 1.25 g C 2.25 g cocaine molecules.
times more atoms than 8 g
of sulphur. B 1.44 g D 3.00 g [Relative atomic mass: H, 1;
A I and III only C, 12; N, 14; O, 16;
B II and IV only AN A
C II and III only 18 Caffeine is found in coffee beans. = 6 3 1023 mol–1] g
D IV only IAtspmillocleocnutlaairnfsor0m.0u5lamisoCl 4oHf5N2O. 0.202 g C 0.404
B 0.303 g D 0.606 g
3
caffeine. Determine the mass
13 What is the total number of the compound in the pill. 3.4 Relationship between
[Relative atomic mass: H, 1; the Number of Moles of
of atoms present in 6.05 g C, 12; N, 14; O, 16] a Gas and Its Volume
C1C2l;2F2.
dichlorodifluoromethane, A 1.94 g C 4.85 g
[Relative atomic mass: C, B 2.42 g D 9.70 g 24 1.0 g of calcium carbonate is
NA
F, 19; Cl, 35.5; = 6 3 1023 added into excess hydrochloric
mol–1]
A 1.2 3 1022 C 1.2 3 1023 19 A cockroach repellent has the acid.
fDoermteurmlainCeHt3h(CeHm2)a5sCsHoCfHCHO.
B 3.0 3 1022 D 1.5 3 1023 CaCO3 + 2HCl →
CaCl2
14 Pyrethrin is an insecticide with 0.02 mol of this substance. + H2O + CO2
Cmaolcleuclautlearthfoermnuumla boefr Co1f9H26O3. [Relative atomic mass: H, 1; Determine the volume of
C, 12; O, 16] carbon dioxide gas evolved at
pyrethrin molecules contained A 1.40 g C 3.50 g room temperature.
in a spray with 15.1 g of the B 2.80 g D 5.60 g [Relative atomic mass: C, 12;
compound. 20 Acetaminophen is a medicine O, 16; Ca, 40; 1 mol of gas
[Relative atomic mass: H, 1; used to relieve pain. 0.0002 occupies a volume of 24 dm3
C, 12; O, 16; NA = 6 3 1023 mol of acetaminophen has a at r.t.p.] C 180 cm3
mol–1] A 120 cm3
mass of 0.0302 g. Which of B 150 cm3 D 240 cm3
the following is the molecular
A 1.5 3 1022 C 1.5 3 1023 formula of acetaminophen?
B 3.0 3 1022 D 3.0 3 1023 25 Which of the following gases will
[Relative atomic mass: H, 1;
15 If m is the number of atoms in C, 12; N, 14; O, 16] occupy the same volume as 2.42
3 g of carbon, the number of AB CC88HH98NNOO22 DC CC88HH99NNO2O g dichlorodifluoromethane, C1C;l2F2?
[Relative atomic mass: H,
atoms in 3 g of magnesium in
terms of m is [Relative atomic C, 12; N, 14; O, 16; F, 19;
mass: C, 12; Mg, 24] 21 Ethyl ethanoate is a liquid used S, 32; Cl, 35.5]
A —21m as nail varnish remover. If 0.025 I 0.34 g of caamrbmonondiaio, xNidHe,3 CSOO22
mol of ethyl ethanoate has a II 0.88 g of
C m III
mass of 2.20 g, determine the 1.28 g of sulphur dioxide,
IV
B —31m D 2m relative molecular mass of ethyl A 0.40 g of methane, CH4
I, II and III only
ethanoate. B I, II and IV only
A 44 C 77
B 55 D 88 C I, III and IV only
3.3 Relationship between the D II, III and IV only
Number of Moles of a 22 The diagram shows the molecular
Substance and Its Mass structure of allicin which is a 26 Which of the following gases
compound obtained from garlic.
16 Which of the following Allicin have powerful antibiotic occupy a volume of 600 cm3 at
substances contain the same and antifungal properties.
room temperature and pressure?
’09 number of atoms as in 36 gram
of carbon? [Relative atomic [Relative atomic mass: H, 1;
mass: H, 1; C, 12; O, 16; S, 32]
I 3 g of hydrogen O C, 12; O, 16; S, 32; 1 mol of
II 64 g of sulphur i
III 18 g of water gas occupies a volume of 24
IV 22 g of carbon dioxide
C S C C dm3 r.t.p.]
H H2 H2 I 0.64 g eootffhcoaaxnryebg,oeCnn2,mHO6o2noxide,
C C S C II 0.75 g
III 0.56 g
H2 H2 H
allicin CO
63 Chemical Formulae and Equations
IV 1.15 g nitrogen dioxide, NO2 3.5 Chemical Formulae [Molar volume: 24 dm3 mol–1 at
A I and III only room conditions; Relative atomic
B II and IV only mass: Zn, 65]
C I, II and III only 31 1.04 g of element X reacted A 0.325 g
B 0.650 g
D II, III and IV only with 0.48 g of oxygen to form an C 0.975 g
D 4.333 g
oxide with the empirical formula
37 0.31 gram of copper(II)
27 Which of the following gases X2O3. Determine the relative carbonate is heated. Determine
atomic mass of element X. the volume of carbon dioxide
has the heaviest mass at room gas released at room conditions.
[Relative atomic mass: O, 16] [Relative atomic mass: C, 12; O,
temperature and pressure? 16; Cu, 64; Molar volume : 24
A 24 C 52 dm3 at room conditions]
[Relative atomic mass: H, 1; A 60 cm3
B 48 D 56 B 120 cm3
C, 12; N, 14; O, 16; S, 32; C 240 cm3
D 360 cm3
1 mol of gas occupies a volume 32 When 3.64 g of a metal oxide
of 24 dm3 r.t.p.] 38 Calculate the mass of aluminium
A 391656ddmmddmm33 o33offoonsffiutmhrlopyegdhetruhnoragdndeieiono,x,xiCidHdHee2,4,NSOO22 of M is reduced, 2.04 g of the oxide, Al2O3 formed if 5.4 g of
B aluminium is heated in air.
C metal is obtained. [Relative atomic mass: O, 16;
D K, 39]
3 Determine the empirical formula A 8.4 g
B 8.6 g
of the metal oxide. C 10.2 g
D 11.8 g
[Relative atomic mass: O, 16;
39 Magnesium oxide reacts with
28 Which of the following gases M, 51] nitric acid to form magnesium
nitrate and water.
occupies the greatest volume at A MO2 C M3O2 If 8.0 g of magnesium oxide
B M2O3 D M2O5 is reacted with excess nitric
room temperature and pressure? acid, calculate the mass of salt
formed.
[Relative atomic mass: H, 1; 33 2.75 g of metal M combines [Relative atomic mass: N, 14;
O, 16; Cu, 64]
C, 12; N, 14; O, 16; 1 mol of with 1.6 g of oxygen to form an A 14.8 g
’10 oxide with the empirical formula B 17.2 g
gas occupies a volume of 24 dm3 C 20.4 g
aotfoMmOic2.mDaestseromf iMne. the relative D 29.6 g
r.t.p.]
A 40 Iron reacts with chlorine
B 0.64 g cnoaixtryrbogogenenn,d,OioN2x 2ide, CO2 [Relative atomic mass: O, 16] according to the equation
C 0.70 g ammonia, NH3
D 1.10 g A 48 C 55 ’09 below:
0.51 g
B 52 D 56 2Fe + 3Cl2 → 2FeCl3
29 What is the number of atoms in 34 x gram of antimony (Sb) If 1.68 gram of iron burns
completely in chlorine, calculate
0.05 mol of ammonia g3as1, 0N2H3 3? combines with 0.48 g of the mass of product formed.
[Avogadro number : 6 [Relative atomic mass: Cl, 35.5,
oxygen to form an oxide with Fe, 56]
mol–1] A 2.438 g
the empirical formula of Sb2O3. B 4.875 g
A 1.2 3 1022 Determine the value of x. C 6.555 g
D 9.750 g
B 3.0 3 1022 [Relative atomic mass: O, 16;
C 1.2 3 1023 Sb, 122]
A 1.22 g
D 3.0 3 1023 B 2.44 g
C 3.66 g
30 Which of the statements below D 4.88 g
are true?
I Sv1oO.1lu20mggaesoeafstCsoO.ct.c2pu.apnydth1e.25samg oef 35 An element E forms a fluoride
II v0Co.O4lu22mggaesoeafstNso.2tc.pcaun. pdy0t.h6e6 g of compound with the formula
same
EEFb3,ywmhaicshs. contains 16.2 % of
What is the relative
III 2.24 odthfmeS3Osoa2mfgCaes2Hems6aaastns.ds.t1.p.1. 2 atomic mass of E?
dm3
have [Relative atomic mass; F, 19]
A 11
IV 4.48 odthfmeC3Hsoa4mfgOaes2emasnaasdts.8s..t9.p6. B 24
dm3
have C 27
D 31
[Relative atomic mass: H, 1; C, 12;
N, 14; O, 16; S, 32; 1 mol of 3.6 Chemical Equations
gas occupies 22.4 dm3 at s.t.p.] 36 Calculate the mass of zinc
required to react with excess
A I and III only
’09 nitric acid to produce 360 cm3
B II and IV only of hydrogen at room conditions.
C I, II and IV only
D II, III and IV only
Chemical Formulae and Equations 64
Structured Questions
1 A hydrocarbon X contains 82.76% carbon by mass. (i) Determine the empirical formula of M oxide.
[2 marks]
2.9 g of hydrocarbon X occupies a volume of 1.2
(ii) Write a chemical equation for the reduction
’10 dm3 at room temperature and pressure. of M oxide to metal M using hydrogen gas.
[2 marks]
[Relative atomic mass: H, 1; C, 12; O, 16; 1 mol of
(e) Can the empirical formula of magnesium oxide
gas occupies a volume of 24 dm3 at r.t.p.; be determined using the same arrangement of
apparatus as above? Explain your answer. [2 marks]
N(aA) =W6hat is1a0h23ymdroolc–a1]rbon? [1 mark]
(b) Determine the empirical formula of hydrocarbon 3 Table 1 shows the positive and negative ions in three
X. [2 marks]
(c) Calculate the relative molecular mass of salt solutions.
hydrocarbon X. [2 marks] ’05
(d) Determine the molecular formula of hydrocarbon X. Name of salt Positive ion Negative ion 3
[2 marks]
Copper(II) sulphate Cu2+ SO42–
(e) Combustion of X in air produces carbon dioxide Potassium iodide K+ I–
and water. Write a chemical equation for the Lead(II) nitrate Pb2+ NO3–
reaction. [2 marks]
(f) If 11.6 g of X is burnt, calculate Table 1
(i) the mass of water formed. [1 mark]
(ii) the number of carbon dioxide molecules (a) What is another name for a positively-charged
produced at room temperature. [1 mark] ion? [1 mark]
2 The apparatus shown is used to determine the (b) Write the formula of lead(II) nitrate. [1 mark]
empirical formula of the oxide of metal M by reducing
(c) When aqueous lead(II) nitrate solution is added
’07 the metal oxide with dry hydrogen gas.
[Relative atomic mass: O, 16; M, 55] to aqueous potassium iodide solution, a yellow
precipitate is formed.
(i) Write a chemical equation for the reaction.
[2 marks]
(ii) Describe the chemical equation in (i).
[1 mark]
(iii) Name the yellow precipitate. [1 mark]
(iv) If 0.04 mol of aqueous potassium iodide
solution is added to excess lead(II) nitrate
Diagram 1 solution, calculate the maximum mass of
the yellow precipitate formed. [Relative
(a) State one precaution that must be taken when atomic mass: I, 127; Pb, 207] [2 marks]
carrying out the experiment. [1 mark] 4 Table 2 shows the descriptions and observations of
(b) How can you ensure that all the oxide of metal two experiments, I and II.
M has been reduced? [1 mark] ’05
(c) (i) Name two chemicals used to prepare Experiment Description Observation
hydrogen gas in the laboratory. [1 mark]
I Combustion Magnesium burns
(ii) Write an equation for the reaction in (i).
of 1.2 g of brightly and a
[1 mark]
(iii) Name a chemical used to dry hydrogen gas. magnesium white powder is
[1 mark] powder in formed
excess oxygen
(d) The information below shows the results of the II Heating Black solid X is
experiment:
copper(II) formed and a
carbonate gas P which turns
Mass of combustion tube + asbestos paper strongly in a limewater cloudy
= 39.25 g test tube is evolved
Mass of combustion tube + asbestos paper + Table 2
oxide of metal M before heating = 47.95 g [Relative atomic mass: C, 12; O, 16; Mg, 24; Cu, 64;
1 mol of gas occupies a volume of 24 dm3 at room
Mass of combustion tube + asbestos paper + temperature and pressure]
metal M after heating = 44.75 g
65 Chemical Formulae and Equations
(a) Based on Experiment I: (a) Write a balanced chemical equation for the
decomposition of sodium bicarbonate on heating.
(i) The white powder formed is magnesium [2 marks]
oxide. Write the chemical equation for the
reaction that takes place. [2 marks] (b) State a chemical test for carbon dioxide gas.
[2 marks]
(ii) Calculate the mass of magnesium oxide
formed if 3 g of magnesium is completely (c) If 8.4 g of sodium bicarbonate decomposes,
calculate
burnt in excess oxygen. [2 marks]
(i) the volume of carbon dioxide gas envolved
(iii) State one use of magnesium oxide. [1 mark] at room temperature and pressure. [3 marks]
(iv) The magnesium oxide is basic and reacts (ii) the mass of sodium carbonate formed.
with nitric acid (HNO3) to form magnesium [3 marks]
nitrate and water. Write a chemical equation [Relative atomic mass: H, 1; C, 12; O, 16;
Na, 23; 1 mol of gas occupies a volume of
for this reaction. [2 marks] 24 dm3 at room temperature and pressure]
(b) Based on Experiment II:
3 (i) Name the black solid X and gas P formed
when copper(II) carbonate is heated strongly. (d) Sodium bicarbonate reacts with nitric acid (HNO3) to
form sodium nitrate, carbon dioxide and water.
[2 marks]
(ii) Write the chemical equation for the reaction (i) Write a balanced chemical equation for this
that takes place. [1 mark] reaction. [2 marks]
(iii) If 6.2 g of copper(II) carbonate had reacted, (ii) Calculate the mass of sodium bicarbonate
calculate that reacts with the excess acid to produce
(a) the mass of solid X formed. [1 mark] 17 g of sodium nitrate. [3 marks]
(b) the volume of gas P formed at room (iii) State one use of sodium nitrate. [2 marks]
temperature and pressure. [1 mark] [Relative atomic mass: H, 1; C, 12; N, 14;
O, 16; Na, 23]
5 When sodium bicarbonate (NaHCO3) is heated, it (e) (i) Name another chemical that reacts with
decomposes to sodium carbonate, carbon dioxide nitric acid to form sodium nitrate. [1 mark]
and water.
(ii) Write an equation for this reaction. [2 marks]
Essay Questions
1 (a) Using a suitable example, explain the meaning of (d) Describe a laboratory method of determining
the following terms: the empirical formula of lead oxide. Your answer
(i) Empirical formula [2 marks] should include
(ii) Molecular formula [2 marks] (i) the procedure of the experiment [4 marks]
(b) State a suitable method that can be used to (ii) tabulation of result [3 marks]
determine the empirical formula of lead(II) oxide. (iii) calculation of the results obtained. [4 marks]
[2 marks] [Relative atomic mass: O, 16; Pb, 207]
(c) Can the same method in (b) be used to
determine the empirical formula of magnesium
oxide? Explain your answer. [3 marks]
Experiment
1 You are required to plan an experiment to determine the empirical formula of magnesium oxide. [17 marks]
Your explanation should include the following:
(a) Statement of the problem
(b) All the variables
(c) List of materials and apparatus
(d) Procedure
(e) Tabulation of data
Chemical Formulae and Equations 66
4CHAPTER FORM 4
THEME: Matter Around Us
Periodic Table of Elements
SPM Topical Analysis
Year 2008 2009 2010 2011
Paper 1 2 31 2 31 2 31 2 3
Section ABC ABC ABC ABC
Number of questions 4 —12 – – 1 3 —41 – – – 3 1 – – 1 6 1 – – –
ONCEPT MAP
PERIODIC TABLE
Historical development of the Transition elements
Periodic Table
Metallic Special
J. W. Dobereiner properties characteristics
John Newlands of transition
Lothar Meyer elements of transition
Dmitri Mendeleev elements
Henry G. J. Moseley
Group Period
Physical and chemical properties Identifying the group and period of • Changes in the properties of
of elements in: an element based on the electron the elements and their oxides
Group 1 arrangement of the element across Period 3
Group 17
Inert property and uses of • Uses of semi-metals such as
Group 18 elements silicon and germanium in the
microelectronics industry
4 4.1 Periodic Table of 3 This systematic method of classification of the
Elements
elements will enable us to
Historical Development of the Periodic (a) study and generalise the chemical and
Table
physical properties of elements in the
1 Many of the elements known today were disco
vered from the years 1800 to 1900. Chemists same group.
noted that certain elements have similar chemi (b) predict the position of an element in the
cal properties. For example: chlorine, bromine
and iodine; potassium and sodium; magnesium Periodic Table from its properties.
and calcium have similar chemical properties. (c) identify and compare elements from
2 Chemists then tried to group the elements with different groups.
the same chemical properties together. This led (d) predict the chemical and physical proper
to the development of the Periodic Table.
ties of new elements in the same group.
4 Chemists such as Lavoisier, Dobereiner,
Newl ands, Meyer, Mendeleev and Moseley
contributed to the development of the
Periodic Table in use today.
Contribution of Scientists to the Development of the Periodic Table
Antoine Lavoisier
1 Antoine Lavoisier, a Frenchman, was the first chemist who attempted to
classify the elements into four groups as in Table 4.1.
2 The four groups consisted of gases, metals, non-metals and metal oxides.
3 His classification was not accurate as light and heat, which are not elements,
were includ ed. Furthermore, some elements in each group did not have the
same chemical properties.
Table 4.1 Lavoisier’s Periodic Table
Group I Group II Group III Group IV Antoine Lavoisier
(1743–1794)
Oxygen Sulphur Arsenic Calcium oxide
Nitrogen Phosphorus Bismuth Barium oxide
Hydrogen Carbon Cobalt Silicon(IV) oxide
Light Chlorine Lead Magnesium oxide
Heat Fluorine Zinc, Nickel, Tin, Silver Aluminium oxide
Johann W. Dobereiner
1 Dobereiner classified the elements with the same chemical properties into
groups of three called triads. Example:
2 Dobereiner discovered the
relationship between the relative Element in the Relative atomic
atomic mass (r.a.m.) of elements in triad mass (r.a.m.)
each triad. He found that the r.a.m.
Lithium (Li) 7
of the element in the middle of each
Sodium (Na) 23
triad is approximately equal to the
average of the total r.a.m. of the Potassium (K) 39 Johann W. Dobereiner
(1780–1849)
other two elements. Average of the total r.a.m. of Li and K is
3 However, this relationship did ⎯7 +⎯23⎯9 = 23
not apply to most of the other
elements.
Periodic Table of Elements 68
John Newlands
1 John Newlands arranged the elements in order of increasing nucleon number
(mass number) in horizontal rows. Each row consisted of seven elements.
2 He found that the chemical properties of every eighth element are similar.
This was known as the law of octaves.
3 Table 4.2 shows the arrangement of elements by John Newlands.
Table 4.2 The classification of elements by Newlands (law of octaves)
Li Be B C N O F John Newlands
(1837–1898)
Na Mg Al Si P S Cl
K Ca 4
4 The classification of elements by Newlands was not successful because:
(a) The law of octaves was only accurate for the first 16 elements (from Li to Ca).
(b) There were no positions allocated for elements yet to be discovered.
5 Nevertheless, Newlands was the first scientist who discovered the existence of periodicity in the
elements.
Lothar Meyer SPM
1 Meyer, a German chemist, Volume of an atom ’09/P1
calculated the volume of
= ⎯m⎯as⎯s o⎯f od⎯nene⎯smi⎯tyo⎯loef-⎯athto⎯eme⎯leom⎯f teh⎯net⎯ele⎯m⎯en⎯t Lothar Meyer
an atom of an element (1830–1895)
using the formula:
2 He plotted a graph of
volume of atoms of elements against their relative atomic masses. The
graph obtained is shown in Figure 4.1.
Figure 4.1 Lothar Meyer’s atomic volume curve
3 From the shape of the atomic volume curve, Meyer showed that elements occupying the
corresponding positions of the curve exhibit similar chemical properties. For example:
(a) Li, Na, K and Rb which are located at the peak of the curve show similar chemical properties.
(b) Be, Mg, Ca and Sr which are located at positions after the maximum points also show similar
chemical properties.
4 Just like Newlands, Meyer showed that the properties of the elements recur periodically.
69 Periodic Table of Elements
Dmitri Mendeleev
1 Dmitri Mendeleev, a Russian chemist, arranged the elements in order of
increasing atomic mass.
2 Table 4.3 below shows the Periodic Table suggested by Mendeleev.
Table 4.3 The partial Periodic Table by Mendeleev
I II III IV V VI VII VIII
4 1H B CNO F Fe, Co, Ni Dmitri Mendeleev
Al Si P S Cl Ru, Rh, Pd (1834–1907)
2 Li Be () Ti V Cr Mn
3 Na Mg () ( ) As Se Br
4 K Ca Y Zr Nb Mo ( )
5 Cu Zn
6 Rb Sr
( ) represents unknown elements yet to be discovered
3 Dmitri Mendeleev was more successful for several reasons.
(a) First, he left gaps for elements yet to be discovered. He even used the table to predict the
existence and properties of undiscovered elements. Mendeleev correctly predicted the properties
of the elements gallium, scandium and germanium which were only discovered much later.
(b) Secondly, although the elements were arranged in order of increasing atomic mass, he changed
the order of the elements if the chemical properties are not similar.
Henry G. J. Moseley
1 Moseley was a British physicist. He bomb arded different elements with high
energy electrons and measured the frequency (f) of the X-ray emitted by the
2 element. the square-root of the frequency of the X-ray (⎯f ) against
He then plotted
the proton number (atomic number) of the element and obtained a straight
line graph (Figure 4.2).
3 Thus from the square-root of the frequency of the X-ray emitted by an
unknown element, he could determine its proton number.
4 After obtaining the proton number of the elements, Moseley went on to Henry G. J. Moseley
(1887–1915)
arrange the elements in the Periodic Table in order of increasing proton
Figure 4.2
number (atomic number).
5 Just like Mendeleev did, Moseley left gaps ( ) for elements yet to be
discovered. He reasoned that there should be an element corresponding to
each proton number.
6 Moseley successfully predicted the existence of four undiscovered elements
from the atomic numbers. These elements were later determined to be
technetium, promethium, hafnium and rhenium. With this prediction,
Moseley was able to prove the separate existence of each element in the
lanthanide series.
Modern Periodic Table 2 In the modern Periodic Table, the elements
are arranged in order of increasing proton
1 There are 117 discovered elements now. Most of number. This order is also related to the
these elements are naturally occurring elements. electron arrangement of the elements.
However, a few of these elements are made
artificially in nuclear reactors.
Periodic Table of Elements 70
3 The Periodic Table is a classification of (b) Periods 2 and 3 have eight elements each. 4
elements whereby elements with the same The first three periods are called the short
chemical properties are placed in the same periods.
group. This makes the study of the chemistry
(c) Periods 4 and 5 have 18 elements each.
of these elements easier and more systematic. They are called the long periods.
4 The modern Periodic Table is shown on page
(d) Period 6 has 32 elements. Not all the
78. elements can be listed on the same
5 The vertical columns of the Periodic Table horizontal row. The elements with proton
numbers 58 to 71 are separated and are
are called groups. There are 18 groups in the grouped below the Periodic Table. These
elements are called the Lanthanide Series.
Periodic Table.
6 Each member of a group shows similar (e) Period 7 has 31 elements and not all can
be listed on the same horizontal row. The
chemical properties although their physical elements with proton numbers 90 to 103
properties such as density, melting point and are grouped below the Periodic Table.
colour may show a gradual change when They are called the Actinide Series.
descending the group. The Electron Arrangement of Elements in the
• Group 1 elements (Li, Na, K, Rb, Cs, Fr) are
Periodic Table
called alkali metals.
• Group 2 elements (Be, Mg, Ca, Sr, Ba, Ra) 1 Table 4.4 shows the electron arrangement of
the elements with proton numbers 1–20.
are called alkaline earth metals.
• Group 17 elements (F, Cl, Br, I, At) are 2 All members of the same group have the
same number of valence electrons. Valence
called halogens. electrons are electrons in the outermost shell.
• Group 18 elements (He, Ne, Ar, Kr, Xe, Rn)
For example:
are called noble gases. (a) Group 1 elements (Li, Na, K) each has
7 A block of elements called the transition one valence electron.
(b) Group 2 elements (Be, Mg, Ca) each has
elements separates Group 2 and Group 13. two valence electrons.
(c) Group 17 elements (F, Cl, Br, I, At) each
(a) The elements in Groups 1, 2, 13 and the has seven valence electrons.
transition elements are metals. 3 (a) The number of valence electrons of Group
1 and Group 2 elements is the same as its
(b) The elements in Groups 15, 16 and 17 are group number.
non-metals.
(c) Group 14 consists of non-metals at the
top of the group, followed by semi-metals
(or metalloids) and metals lower down in
the group.
8 The horizontal rows are called periods. There
are seven periods.
(a) Period 1 has two elements only: hydrogen
and helium.
Table 4.4 The electron arrangement of the first 20 elements in the Periodic Table SPM
Group 1 2 13 14 15 16 17 ’11/P1,
P2
Period
18
1H He
1 2
2 Li Be B C N O F Ne
2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
3 Na Mg Al Si P S Cl Ar
2.8.1 2.8.2 2.8.3 2.8.4 2.8.5 2.8.6 2.8.7 2.8.8
4 K Ca
2.8.8.1 2.8.8.2
71 Periodic Table of Elements
4 (b) Except for helium, the elements with more 3
than 2 valence electrons (Groups 13 to The proton number of element X is 20. Which of
the following statements are true concerning the
18), the group number = 10 + (number of element X?
I X can conduct electricity.
valence electrons). II X belongs to Group 2 of the Periodic Table.
4 The period number is indicated by the number III X belongs to Period 4 of the Periodic Table.
SPM of filled electron shells. For example: IV X belongs to Period 3 of the Periodic Table.
’05/P2 (a) Elements in Period 1 (H and He) each A II and III only
B II and IV only
has only one electron shell filled with C I, II and III only
electrons. D I, II and IV only
(b) Elements in Period 2 (Li, Be, B, C, N, O,
F, Ne) each has two electron shells filled Comment
with electrons. X has 20 electrons. The electronic configuration of
(c) Elements in Period 3 (Na, Mg, Al, Si, P, X is 2.8.8.2.
S, Cl, Ar) each has three electron shells X belongs to Group 2 of the Periodic Table because
filled with electrons. it has two valence electrons. (II is correct)
(d) Elements in Period 4 (K and Ca) each has
four electron shells filled with electrons. Group 2 elements are metals and can conduct
5 All elements in the same period have the same electricity. (I is correct)
number of filled electron shells. X belongs to Period 4 of the Periodic Table because
it has four electron shells filled with electrons. (III
1 ’05 is correct, IV is incorrect)
Which of the following represents the electron Answer C
arrangement of an element in Group 17?
AB C D
4
Answer D (The element has seven valence electrons) Element Q R S
2.8.6 2.8.8 2
Electron
2 ’03 arrangement
In the Periodic Table, Y is below Z in the same Determine the group in which Q, R and S belong to
in the Periodic Table.
group. If the proton number of atom Z is 11, what is
Solution
the electron arrangement of atom Y ? For elements with one or two valence electrons,
A 2.2 C 2.8.3 Group number = Number of valence electrons
B 2.7 D 2.8.8.1
Comment For elements with more than two valence electrons,
Group number = 10 + Number of valence electrons
In an atom, the number of protons is equal to the
number of electrons. Therefore element Q belongs to Group 16 and
Thus Z has 11 electrons. The electron arrangement element R belongs to Group 18 of the Periodic
of Z is 2.8.1. Table. S has two electrons in the first electron shell.
The elements in the same group have the same It is helium and belongs to Group 18 of the Periodic
number of valence electrons. Table.
Therefore the electron arrangement of Y is 2.8.8.1
because each atom of Y and Z has one valence
electron.
Answer D
Periodic Table of Elements 72
1 ’04
Element QR T X Z Element QRT X Z
3 15 18 19 35
Proton Electron 2.1 2.8.5 2.8.8 2.8.8.1 2.8.18.7
number arrangement
The proton numbers of elements Q, R, T, X and Z Comment 4
Elements Q and X belong to the same group in the
are given in the table above. Which of the following Periodic Table because they have the same number
statements are true? of valence electrons. (Statement I is correct)
Element T belongs to Period 3 (it has 3 electron
I Elements Q and X belong to the same group in shells), whereas element X belongs to Period 4 (it
has four electron shells). (Statement II is wrong)
the Periodic Table. Elements in Groups 1, 2 and 13 have one, two
and three valence electrons and they are metals.
II Elements T and X belong to the same period. (Statement III is correct)
Elements with five, six, seven or eight valence
III Elements Q and X are metals. electrons are non-metals. (Statement IV is correct)
IV Elements R and Z are non-metals. Answer C
A I and III only
B II and IV only
C I, III and IV only
D II, III and IV only
4.1
1 Explain the term valence electrons. 3 Arsenic is represented by the symbol 75 As. In which
State the number of valence electrons of the 33
group and period does arsenic belong to in the
elements in the following groups in the Periodic
Table. Periodic Table? Name two elements that have the
(a) Group 1
(b) Group 2 same chemical properties as arsenic. Explain your
(c) Group 15
(d) Group 17 answer.
4 The following table shows the proton numbers of 10
elements represented by the letters A to K:
2 WX Y Z Element ABCDE F GH J K
Element
Proton number 2 9 13 19 18 16 7 20 17 6
Electron
arrangement 2.8.5 2.1 2.8.8.3 2.8.18.32.18.7 Electron
arrangement
The table shows the electron arrangement of four
elements. Group number
(a) State the group of each element:
(i) W Period number
(ii) X
(iii) Y (a) Write the electron arrangement of each element.
(iv) Z Then state the group number and period
number of each element.
(b) State the period of each element:
(i) W (ii) X (iii) Y (iv) Z (b) Pick a pair of elements which have the same
chemical properties.
73 Periodic Table of Elements
4.2 Group 18 Elements ’05
General
He • The noble gases are also known as the inert gases.
Ne They are elements of Group 18, at the far right of the
Ar
Kr Periodic Table (Figure 4.3).
Xe • The group consists of six elements: helium (He), neon
Rn
(Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon
Figure 4.3 The elements of Group 18 in the (Rn).
Periodic Table • The atomic radius of the noble gases increases down
the group. This is because as the number of filled
4
electron shells increases down the group, the valence
electron is further from the nucleus.
Physical Properties
Physical properties of noble gases: This is because as the size of the atoms increases
• All noble gases are insoluble in water and do down the group, the van der Waals forces of
not conduct electricity or heat. attraction become stronger.
• All noble gases have low melting points and • The noble gases have very low densities. This is
low boiling points. This is because the noble because the atoms are very far apart. The density
gases exist as monatoms which are attracted by of the noble gases increases when descending
very weak van der Waals forces of attraction. the group. This is due to the increase in the
The melting points and boiling points of the relative atomic mass of the element.
noble gases increase when descending the group.
Chemical properties
Chemical properties of noble gases: stable duplet electron arrangement.
• Noble gases are chemically inert which means – The other noble gases have eight electrons in
that they are unreactive in nature. They do not their outermost electron shells. They are said
react with any other elements. to have attained the stable octet electron
• Noble gases are the only elements which exists arrangement.
as monatoms (as single atoms). • Therefore the noble gases do not need to
• Noble gases are unreactive because they all accept, donate or share electrons with other
have filled outer shells of electrons which are elements.
a stable electron arrangement. (All chemical reactions involve either gaining,
– Helium has only one filled electron shell and losing or sharing electrons.)
it has exactly two electrons in this outermost
electron shell. It is said to have attained the
Table 4.5 The proton numbers, relative atomic mass, electron arrangement and physical properties of inert gases
Element Proton Relative atomic Electron Atomic Melting Boiling point Density
number mass arrangement radius (nm) point (°C) (°C) (g dm–3 )
He 2 4 2 0.06 –270 –269 0.17
Ne 10 20 2.8 0.07 –248 –246 0.84
Ar 18 40 2.8.8 0.094 –189 –186 1.66
Kr 36 84 2.8.18.8 0.109 –156 –152 3.54
Xe 54 131 2.8.18.18.8 0.130 –112 –107 5.45
Rn 86 222 2.8.18.32.18.8 – –71 –62 –
Periodic Table of Elements 74
2 ’04 4.2 4
A car distributor wants to use colourful electric 1 2P, 11Q, 13R, 18S, 20T.
lamps to attract customers. Which of the following The above is a set of elements with their proton
substances A, B, C or D in the Periodic Table is
suitable for use in the lamps? numbers. Choose two elements that exist as
monatomic gases. Explain your answer.
Comment
B is neon gas. Neon-filled electric bulbs produce 2 Inert gases are the elements of Group 18 of the
an attractive bright red light which is used in Periodic Table. Name two inert gases and state
advertising. one use of each.
Answer B
3 All the inert gases have low melting and boiling
points. Explain why as we go down Group 18, the
melting points and boiling points increase.
4 One of the characteristics of Group 18 gases is that
they exist as monatoms. Explain why neon does
not form compounds with other elements.
4.3 Group 1 Elements General
Figure 4.4 The elements of Group 1 in the • Group 1 elements are also known as the alkali metals.
Periodic Table • The elements in Group 1 are lithium (Li), sodium
(Na), potassium (K), rubidium (Rb), caesium (Cs)
and francium (Fr).
• All Group 1 elements are soft metals. The metals are
grey in colour and are silvery when the surface is freshly
cut, before being exposed to air.
Physical Properties
Physical properties of Group 1 elements: between the atoms becomes weaker down the
• All Group 1 elements are metals; hence they group as the atomic radius increases.
• The electropositivity of the metals increases
are good conductors of heat and electricity. down the group. Electropositivity is a measurement
• The atomic radius increases down the group. of the ability of an atom to lose an electron
and form a positive ion.
The reason is that as the number of filled
electron shells increases when descending M → M+ + e– (M = Li, Na, K, Rb, Cs, Fr)
the group, the distance between the outermost
As the atomic radius becomes larger down
electron shell and the nucleus increases. the group, the force of attraction between
• The density increases down the group. The the nucleus and the single valence electron
becomes weaker. Hence the elements lose
densities of Li, Na and K is lower than that the single valence electron more easily when
descending the group.
of water, and hence these metals can float on
water.
• The melting point decreases when descending
the group. This is because the metallic bond
75 Periodic Table of Elements
Reactivity SPM
’07/P2, ’11/P2
Reactivity of Group 1 elements M(s) → M+ + e– (M = Li, Na, K, Rb, Cs, Fr)
• All Group 1 elements are very reactive. The Li(s) → Li+ + e–
reactivity increases down the group. 2.1 2
• The elements in Group 1 have one valence
Na(s) → Na+ + e–
electron each. During a chemical reaction, 2.8.1 2.8 reactivity increases
an atom of a Group 1 element will donate a
valence electron to form a univalent positive K(s) → K+ + e–
ion to attain the stable duplet or octet in its 2.8.8.1 2.8.8
electron arrangement.
• The reactivity of Group 1 elements depends on how easily it can donate its valence electron. The
Experiment 4.1 atomic radius of Group 1 elements increases down the group. This causes the force of attraction
4
between the nucleus and the valence electron to become weaker, making it easier for a metal
lower in the group to donate its valence electron. Therefore, the reactivity of Group 1 elements
increases when descending the group.
Chemical Properties
Chemical properties of Group 1 elements – cold water to produce hydrogen gas and
• The elements in Group 1 have the same chemical alkalis,
properties because each has one valence – oxygen to produce metal oxides,
electron. – halogen to produce metal halides.
• Group 1 elements react with
Table 4.6 Some physical properties of Group 1 elements SPM
’06/P2, ’08/P1
Element Colour Electron Atomic Density Electrical Melting Electropositivity
arrangement radius (nm) (g dm–3) conductivity point (°C)
Li Silvery
Na Silvery 2.1 0.15 0.53 Conductor 181 ⏐
K Silvery 2.8.1 0.16 0.97 Conductor 98 ⏐
Ru Silvery 2.8.8.1 0.23 0.86 Conductor 63 ⏐
Cs Silvery 2.8.18.8.1 0.25 1.53 Conductor 39 Incr⏐eases
Fr Silvery 2.8.18.18.8.1 0.26 1.87 Conductor 29 ⏐
2.8.18.32.18.8.1 0.29 27 ⏐
– Conductor ↓
4.1
To study the reaction of alkali metals with oxygen
Problem statement Variables
How do lithium, sodium and potassium differ in
reactivity with oxygen? (a) Manipulated variable : The alkali metals used
Hypothesis (b) Responding variable : Reactivity of the metals
The alkali metals show similar chemical properties in
their reactions with oxygen but the reactivity of the with oxygen
alkali metals with oxygen increases down the group
(in the order from lithium, sodium to potassium). (c) Constant variable : Excess supply of oxygen
and the size of the metal
Materials piece used
Small pieces of Li, Na and K metals, oxygen gas,
filter paper and phenolphthalein indicator.
Periodic Table of Elements 76
Apparatus 3 The lithium metal is heated until its starts to burn.
The spoon is then put into a gas jar containing
Pen knife, tongs, gas jar with cover, gas jar spoon and oxygen gas.
Bunsen burner.
4 The observation is recorded.
Safety precautions 5 After the reaction has stopped, about 20 cm3
Alkali metals, especially sodium and potassium, are
very reactive. Therefore we have to of water is added to the gas jar. The gas jar is
(a) use small pieces of each metal. shaken so that the product of the combustion is
(b) use goggles while carrying out the experiment. dissolved in the water. The solution is tested with
(c) ensure that we do not handle the metal with our a few drops of phenolphthalein indicator. The
observation is recorded.
bare hands. 6 The experiment is repeated using sodium and
potassium metals.
Procedure 4
1 A piece of lithium metal is removed from the Figure 4.5 Reaction of Group 1 metals with
bottle with tongs. A small piece of the metal is oxygen
cut using a pen knife. A piece of filter paper is
used to absorb the paraffin oil from the piece of
metal.
2 The lithium metal is then transferred onto a gas
jar spoon using tongs.
Results
Metal Observation
Lithium The lithium metal burns with a red flame forming a white metal oxide. The metal oxide
dissolves in water producing a solution which turns phenolphthalein indicator red.
Sodium The sodium metal burns with a bright yellow flame forming a white metal oxide. The metal
oxide dissolves in water producing a solution which turns phenolphthalein indicator red.
Potassium The potassium metal burns with a very bright purplish flame forming a white metal oxide.
The metal oxide dissolves in water producing a solution which turns phenolphthalein indicator
red.
Discussion
1 Alkali metals burn in oxygen to form white The presence of hydroxide ions, OH– causes the
metal oxides solution to be alkaline.
SPM 4Li(s) + O2(g) → 2Li2O(s) ⎯⎯→ Reactivity Conclusion
increases down All alkali metals burn in oxygen gas to produce
’11/P1 the group white metal oxides. Group 1 elements show similar
chemical properties. Their reactivity increases down
4Na(s) + O2(g) → 2Na2O(s) the group from lithium, sodium to potassium.
4K(s) + O2(g) → 2K2O(s)
4M(s) + O2(g) → 2M2O(s),
2 These metal oxides dissolve in water to form where M = Group 1 element.
alkaline solutions which turn phenolphthalein
indicator red.
Li2O(s) + 2H2O(l) → 2LiOH(aq)
Na2O(s) + 2H2O(l) → 2NaOH(aq)
K2O(s) + 2H2O(l) → 2KOH(aq)
77 Periodic Table of Elements
Periodic Table of Elements
Periodic Table of Elements
Periodic Table of Elements
4
Periodic Table of Elements 78
4.2 SPM
To study the reaction of alkali metals with water ’05/P2
Problem statement ’08/P3
How do lithium, sodium and potassium differ in ’09/P1
reactivity with water? ’10/P3
Hypothesis Figure 4.8 Reaction of
The alkali metals show similar chemical properties potassium with water
in their reactions with water but the reactivity of
alkali metals with water increases down the group Results
(in the order from lithium, sodium to potassium).
Metal Observation Experiment 4.2 4
Variables Lithium
(a) Manipulated variable : The alkali metals used Sodium The lithium metal moves slowly on
(b) Responding variable : Reactivity of the metal the surface of the water (Figure 4.6).
Potassium A colourless solution is produced. This
with water solution turns red litmus paper blue.
(c) Constant variable : Size of alkali metal and
The sodium metal moves at a fast
the temperature of water speed on the surface of the water with
a ‘hissing’ sound. It is ignited during
Materials the reaction and burns with a yellow
Small pieces of lithium, sodium and potassium flame (Figure 4.7). A colourless
metals, basin filled with water, filter paper and red solution is obtained. This solution
litmus paper. turns red litmus paper blue.
Apparatus The potassium metal moves at a very
Pen knife and tongs. fast speed on the surface of the water.
It is ignited during the reaction and
Procedure burns with a purple flame with a
1 A piece of lithium metal is removed from the ‘pop’ sound (Figure 4.8). A colourless
solution is obtained. This solution
bottle. A small piece of the metal is cut using turns red litmus paper blue.
a pen knife. A piece of filter paper is used to
absorb the paraffin oil from the piece of metal. Discussion
2 The lithium metal is then dropped into a basin of
water carefully using a pair of tongs. 1 Alkali metals are reactive with water and can
3 The observation is recorded. SPM displace hydrogen from water. They react with
4 The solution formed in the basin is tested with a ’11/P2 water to produce hydrogen gas and aqueous
piece of red litmus paper.
5 The experiment is repeated using small pieces of solutions of metal hydroxides.
sodium and potassium metals.
22LLiiO(s)H+(a2qH) 2+OH(l2)(g→) Reactivity
Figure 4.6 Reaction of 22NNaaO(s)H+(a2qH) 2+OH(l2)(g→) increases down
lithium with water 22KKO(sH) +(a2qH) +2OH(l2)(g→) the group
Figure 4.7 Reaction of 2 The metal hydroxides are alkaline and turn the
sodium with water colour of red litmus paper blue.
Conclusion
1 Elements of Group 1 have similar chemical
properties. All of them react with cold water to
produce an alkaline solution and hydrogen gas.
2M(s) + 2H2O(l) → 2MOH(aq) + H2(g),
where M = Group 1 element.
2 The reactivity increases down the group from
lithium, sodium to potassium.
79 Periodic Table of Elements
When the Group 1 elements react with water they give react vigorously with water forming hydrogen gas
off hydrogen gas. The heat generated by the chemical and alkali solutions. If the size of the metal is
reaction sets the hydrogen gas alight and it burns with quite large, it will ignite and start to burn giving
a coloured flame. Sodium burns with a yellow flame. out ‘pop’ sounds.
When carrying out an experiment on the reaction
Safety Precaution to be Taken When of alkali metals with water, we must ensure that
Handling Alkali Metals no flamm able organic solvents are nearby because
the fire from the burning alkali metals can spread
Alkali metals are extremely reactive. With the to the organic solvents.
exception of lithium, the rest of the alkali metals Therefore alkali metals must be kept in paraffin
oil in bottles to prevent them from reacting with
water.
Experiment 4.3 4.3 SPM
4
’09/P1
To study the reaction of alkali metals with halogens
Problem statement Results
How do lithium, sodium and potassium differ in Metal Observation
reactivity with chlorine gas? Lithium
Lithium burns slowly with a reddish
Hypothesis Sodium flame. A white solid is obtained.
The alkali metals show similar chemical properties Sodium burns brightly with a
in their reactions with chlorine but the reactivity of yellowish flame. A white solid is
alkali metals with chlorine increases down the group obtained.
(in the order from lithium, sodium and potassium).
Variables Potassium Potassium burns very brightly with
a purplish flame. A white solid is
(a) Manipulated variable : The alkali metals used obtained.
(b) Responding variable : Reactivity of the metal
Discussion
with chlorine 1 All alkali metals react with chlorine gas to
(c) Constant variable : Supply of chlorine gas and
form white metal chlorides
size of metal pieces used The reactivity of the metal increases down the
Materials group from Li, Na to K.
Small pieces of lithium, sodium and potassium
metals chlorine gas and filter paper.
Apparatus 2Li(s) + Cl2(g) → 2LiCl(s) ⎯⎯⎯→ Reactivity
2Na(s) + Cl2(g) → 2NaCl(s) increases down
Pen knife, tongs, gas jar with cover, gas jar spoon 2K(s) + Cl2(g) → 2KCl(s) the group
and Bunsen burner.
2 If the experiments are repeated using bromine gas,
Procedure the brown colour of bromine gas will disappear
and white metal bromides will be formed.
1 A small piece of lithium is cut and the paraffin
oil on it is blotted using filter paper. 2Li(s) + Br2(g) → 2LiBr(s) ⎯⎯→ Reactivity
2Na(s) + Br2(g) → 2NaBr(s) increases down
2 The lithium metal is then transferred onto a gas 2K(s) + Br2(g) → 2KBr(s) the group
jar spoon using tongs.
3 Similarly, Group 1 elements will react with
3 The lithium metal is heated until it starts to burn. iodine vapour to produce white metal iodides.
The spoon is then put into a gas jar containing The purple colour of iodine vapour will
chlorine gas. disappear in the reactions.
4 The observation is recorded.
5 The experiment is repeated using sodium and
potassium metals.
Periodic Table of Elements 80
4 Generally, all alkali metals react with halogens 2M(s) + Cl2(g) → 2MCl(s)
to produce metal halides. where M = Group 1 elements
2M(s) + X2(g) → 2MX(s) 2 The reactivity increases down the group from
where M = Group 1 elements, X = halogen. lithium, sodium to potassium.
Conclusion
1 Elements of Group 1 have similar chemical
properties. All react with chlorine to produce
white metal chlorides.
3 ’05 Comment
Element Electron arrangement 4
X 2.8.1
An atom of element X has three electron filled Y 2.8.8.1
shells. When element X reacts with chlorine, it forms
a compound with the formula XCl. Which of the
following is element X? The atomic radius of Y is larger than X. Hence Y
[Proton number of Li, 3; Na, 11; Mg, 12; K, 19] can donate its valence electron more easily than X.
A Lithium C Magnesium Thus Y is more reactive. (I is correct)
B Sodium D Potassium Both X and Y have one valence electron. Both are
Comment alkali metals which can conduct electricity.
Alkali metals react with chlorine to form metal
chloride with formula XCl. (II and III correct)
X is sodium because sodium has three electron
shells. Alkali metals react with water to form metal
Answer B hydroxides and hydrogen gas. (IV is incorrect)
Answer A
4 ’03 4.3
The table below shows the proton numbers of two 1 Imagine that a new element is discovered. It is
elements: named Pentium (Pn) and is below sodium metal
in Group 1 of the Periodic Table.
Element Proton number (a) Predict three physical and three chemical
X 11 properties of pentium.
Y 19 (b) Which is more reactive, pentium or sodium?
Explain your answer.
Which of the following statements are true?
2 List the alkali metals in order of decreasing reactivity.
I Y is more reactive than X.
3 How do each of the following properties of alkali
II Both X and Y can conduct electricity. metals change as we go down the group?
(a) Melting point
III Both X and Y are in the same group of the (b) Density
(c) Hardness
Periodic Table. (d) Chemical reactivity
IV Both X and Y react with water to form metal 4 When a piece of burning sodium metal is placed in
a gas jar containing bromine gas, the brown colour
oxides and hydrogen gas. of bromine gas will disappear and white powder
A I, II and III only will be formed. Explain the above observation with
B I, II and IV only a suitable equation.
C II, III and IV only
D I, II, III and IV
81 Periodic Table of Elements
4.4 Group 17 Elements SPM
’08/P2
4 General Physical properties
• The elements in Group 17 are also
Physical properties of the halogens
known as the halogens. • All Group 17 elements are non-metals. Hence,
• The elements in Group 17 are fluorine
they are non-conductors of heat and electricity.
(F), chlorine (Cl), bromine (Br), • The atomic radius increases down the group.
iodine (I) and astatine (At).
• Halogens are very reactive elements The reason is that as the number of filled
and most of them exist naturally as electron shells increases down the group, the
halide salts. distance between the outermost electron shell
• The halogen molecules exist as diatomic and the nucleus increases.
molecules: F2, Cl2, Br2, I2 and At2. • The density increases down the group. This is
due to the increase in relative molecular mass.
Figure 4.9 The elements of Group 17 in the • Halogens have low boiling points. The forces of
Periodic Table attraction between the molecules are weak.
• The melting points and boiling points of the
Halothane (dCuHriCnlgBraCFm3)ajiosr used as a general halogens increase down the group. This is
anaesthetic operation. Can you because the molecular size increases down the
group. As the size increases, the van der Waals
name the halogens present in this compound? forces of attraction between the molecules
become stronger. More heat is required to
Answer Bromine, chlorine and fluorine overcome the forces of attraction and therefore
the melting points and boiling points increase.
The first two elements (fluorine and chlorine) are
gases at room temperature. Bromine is a liquid
whereas iodine and astatine are solids at room
temperature.
• The colour of the halogen becomes darker
down the group. Fluorine is a colourless gas;
chlorine is a yellowish-green gas; bromine is a
dark brown liquid and iodine is a black solid.
• All halogens have high electronegativities. They
are electronegative non-metals. Electronegativity
is a measurement of the tendency of an element
to attract electrons. The electronegativity
decreases down the group from fluorine to
iodine. As the atomic radius becomes larger
down the group, the force of attraction between
the nucleus and the electrons becomes weaker
and thus electronegativity decreases.
Chemical properties
Chemical properties of Group 17 elements
• The elements in Group 17 have the same chemical properties because each has seven valence
electrons.
• Group 17 elements react with
– water to produce acids,
– metals such as iron to produce metal halides,
– sodium hydroxide to produce salts and water.
Periodic Table of Elements 82
Reactivity SPM
’06/P2, ’08/P2
Reactivity of Group 17 elements X2 + 2e– → 2X–, where X = F, Cl, Br or I
• All Group 17 elements are very reactive. However, the
Cl2 + 2e– → 2Cl–
reactivity decreases down the group. 2.8.7 2.8.8
• The elements in Group 17 have seven valence electrons
each. During a chemical reaction, the atom of a Br2 + 2e– → 2Br–
Group 17 element will accept a valence electron to 2.8.18.7 2.8.18.8
form univalent negative ion to attain the stable octet
I2 + 2e– → 2I–
in its electron arrangement. 2.8.18.18.7 2.8.18.18.8
• The reactivity of Group 17 elements depends on
its ability to gain an electron. The atomic radius of Group 17 elements increases down the group.
Thus the force of attraction between the nucleus and the valence electrons become weaker. As a
result, the halogen lower in the group has a lower tendency to attract an electron to form a negative Experiment 4.4 4
ion. Therefore, the reactivity of halogens decreases down the group.
SPM
Table 4.7 Some physical properties of three halogens ’08/P2
Halogen Proton Electron Atomic Melting Boiling Physical Electro Colour
number arrangement radius point point state at room negativity
(nm) (°C) (°C) temperature
Chlorine 17 2.8.7 0.099 –101 –35 Gas 3.0 Yellowish-
green gas
Bromine 35 2.8.18.7 0.114 –7 58 Liquid
Iodine 53 2.8.18.18.7 0.133 114 183 Solid 2.8 Brown liquid
2.5 Black solid
4.4
To study the reactions of chlorine, bromine and iodine with water
Problem statement Materials
How do chlorine, bromine and iodine react with water? Chlorine gas, liquid bromine, iodine crystals and blue
litmus paper.
Hypothesis
The halogens show similar chemical properties when Apparatus
they react with water but the reactivity decreases Test tube, rubber stopper, test tube holder, delivery
down the group from chlorine to iodine. tube and teat pipette.
Variables Safety precautions
(a) Manipulated variable : Types of halogens used (a) Chlorine gas and bromine vapour are poisonous.
(b) Responding variable : The rate at which the The experiments should be carried out in a fume
cupboard.
halogen dissolves in water
and products of reactions (b) The chlorine gas and bromine vapour irritate the
(c) Constant variable : Temperature of water eyes. So goggles should be worn while carrying
out the experiments.
Figure 4.10 Reaction
of halogen with water
(a) (b) (c)
83 Periodic Table of Elements
Procedure 2 The test tube is tightly closed with a rubber
stopper and then shaken.
(A) Reaction of chlorine with water
3 The solution produced is tested with blue litmus
1 Chlorine gas is passed into a test tube containing paper.
water.
(C) Reaction of iodine with water
2 The solution produced is tested with blue litmus 1 Some iodine crystals are added to some water in
paper.
a test tube.
(B) Reaction of bromine with water 2 The test tube is tightly closed with a rubber
1 A few drops of liquid bromine are added to some stopper and then shaken.
water in a test tube. 3 The solution produced is tested with blue litmus
Results SPM paper.
Halogen
4 ’11/P1
Observation
Solubility Effect on litmus paper
Chlorine Dissolves quickly in the water to form a light The solution first turns the blue litmus paper
yellowish solution red, then it quickly decolourises it.
Bromine Dissolves slowly in water to form a brown The solution first turns the blue litmus paper
solution red. The red colour of the litmus takes a
longer time to be decolourised.
Iodine A little of the iodine crystals dissolves The solution turns the litmus paper from
slightly in water to form a pale brown blue to red. The red litmus paper is not
solution decolourised.
Discussion Hypobromous(I) acid is a weak bleaching agent
and takes a longer time to decolourise the red
1 Chlorine, bromine and iodine dissolve in water colour of litmus paper.
to form acidic solutions which turn blue litmus
paper red. The solubility of halogens decreases 4 Iodine is only very slightly soluble in water. It
from chlorine to iodine. forms hydroiodic acid and hypoiodous(I) acid.
2 Chlorine dissolves in water to form hydrochloric I2(s) + H2O(l) → HI(aq) + HOI(aq)
acid and hypochlorous(I) acid. hydroiodic hypoiodous(I)
Cl2(g) + H2O(l) → HCl(aq) + HOCl(aq) acid acid
hydrochloric hypochlorous(I)
acid acid Hydroiodic acid, HI is an acid and it turns blue
litmus red.
Hydrochloric acid, HCl is an acid and it turns
blue litmus red. Hypoiodous(I) acid has a very weak bleaching
property.
Hypochlorous(I) acid, HOCl is a strong bleaching
agent. It decolourises the red colour of litmus Conclusion
paper quickly.
1 Chlorine, bromine and iodine show similar
3 Bromine dissolves slowly in water to form chemical properties. They dissolve in water to
hydrobromic acid and hypobromous(I) acid. form acidic solutions.
Br2(l) + H2O(l) → HBr(aq) + HOBr(aq) 2 The solubility of halogens in water decreases
hydrobromic hypobromous(I) down the group.
acid acid 3 Aqueous chlorine and bromine solutions have
bleaching properties.
Hydrobromic acid, HBr is an acid and it turns
blue litmus red. Aqueous iodine solution does not bleach the
colour of litmus paper.
Periodic Table of Elements 84
4.5
To study the reactions of halogens with aqueous sodium hydroxide solution SPM
’06/P2
Problem statement Results
How do chlorine, bromine and iodine react with Halogen Observation Experiment 4.5 4
aqueous sodium hydroxide solution? Chlorine
Bromine The greenish chlorine gas dissolves
Hypothesis quickly in NaOH solution to form a
The halogens show similar chemical properties when Iodine colourless solution.
they react with sodium hydroxide solution but the
reactivity decreases down the group from chlorine to The brownish liquid bromine
iodine. dissolves steadily in NaOH
solution to form a colourless
Variables solution.
(a) Manipulated variable : Types of halogens used
(b) Responding variable : The products of the The dark iodine crystal dissolves
slowly in NaOH solution to form a
reactions colourless solution.
(c) Constant variable : Concentration of sodium
Discussion
hydroxide solution
1 Chlorine gas reacts rapidly with sodium
Materials
Chlorine gas, liquid bromine, iodine crystals and hydroxide solution to produce sodium chloride
sodium hydroxide solution.
salt, sodium chlorate(I) salt and water.
Apparatus
Test tube, rubber stopper, test tube holder and teat Cl2(g) + 2NaOH(aq) → H2O(l)
pipette. NaOCl(aq) + NaCl(aq) +
sodium sodium
Procedure chlorate(I) chloride
(A) Reaction of chlorine with aqueous sodium
2 Bromine reacts moderately fast with sodium
hydroxide solution
1 Chlorine gas is bubbled into aqueous sodium hydroxide solution to produce sodium bromide
hydroxide solution. salt, sodium bromate(I) salt and water.
2 The colour change of chlorine is recorded.
Br2(l) + 2NaOH(aq) →
NaOBr(aq) + NaBr(aq) + H2O(l)
sodium bromate(I) sodium bromide
3 Solid iodine reacts slowly with sodium
hydroxide solution to produce the salts sodium
iodide, sodium iodate(I) and water.
(B) Reaction of bromine with aqueous sodium I2(s) + 2NaOH(aq) →
hydroxide solution NaOI(aq) + NaI(aq) + H2O(l)
1 Two drops of liquid bromine are added to sodium iodate(I) sodium iodide
aqueous sodium hydroxide solution using a teat
pipette. [Sodium chlorate(I), sodium bromate(I), sodium
iodate(I) are also called sodium hypochlorite, sodium
2 The test tube is tightly closed with a rubber hypobromite and sodium hypoiodite respectively]
stopper and the mixture is shaken.
Conclusion
3 The colour change of bromine is recorded.
1 Chlorine, bromine and iodine react with
(C) Reaction of iodine with aqueous sodium sodium hydroxide solution to form two types
hydroxide solution of salts and water.
1 Some iodine crystals are added to aqueous X2(g) + 2NaOH(aq) → NaX(aq) + NaOX(aq) +
sodium hydroxide solution. H2O(l), where X = Cl, Br, I
2 The test tube is tightly closed with a rubber 2 The reactivity of halogens with sodium
stopper and the mixture is shaken. hydroxide solution decreases down the group
from chlorine to iodine.
3 The colour change of iodine crystal is recorded.
85 Periodic Table of Elements
4.6
To study the reactions of halogens with iron SPM ( B) Reaction of bromine gas with iron wool
’06/P2
Problem statement
How do chlorine, bromine and iodine react with
iron?
Experiment 4.6 Hypothesis Figure 4.12 Reaction of
4 The halogens show similar chemical properties when bromine with iron
they react with iron but the reactivity decreases down wool
the group from chlorine to iodine.
1 A small roll of iron wool is placed in the middle
Variables of a combustion tube and is heated strongly.
(a) Manipulated variable : Types of halogen used
(b) Responding variable : Products of reactions and 2 The liquid bromine is warmed up by using a
Bunsen burner.
rate of the reactions
(c) Constant variable : Iron wool 3 The bromine is vaporised and bromine gas
passed through the heated iron wool.
Materials
Chlorine gas, liquid bromine, iodine crystals, soda 4 The excess bromine gas is absorbed by the soda
lime, potassium manganate(VII), concentrated lime.
hydroc hloric acid and iron wool.
(C) Reaction of iodine with iron wool
Apparatus
Combustion tubes, Bunsen burner, retort stand and
clamp, conical flask and thistle funnel.
Procedure Figure 4.13 Reaction of iodine
with iron wool
(A) Reaction of chlorine gas with iron wool
1 A small roll of iron wool is placed in the middle 1 A few crystals of iodine are placed in a boiling tube.
2 A small roll of iron wool is then placed in the
of a combustion tube. The iron wool is then
heated strongly. middle of a combustion tube.
2 Chlorine gas is prepared in the laboratory 3 The iron wool is heated strongly first, followed by
by adding concentrated hydrochloric acid to
potassium manganate(VII). the iodine crystals (sublimation will take place).
3 The chlorine gas produced is allowed to pass 4 The iodine vapour produced is allowed to pass
through the heated iron wool.
4 The excess chlorine gas is absorbed by the soda through the hot iron wool.
lime.
Results
Halogen Observation
Figure 4.11 Reaction of chlorine with Chlorine Hot iron wool glows brightly when
iron wool Bromine chlorine gas is passed over it. A brown
Iodine solid is formed.
Hot iron wool glows moderately
bright when bromine gas is passed
over it. A brown solid is formed.
Hot iron wool glows dimly when
iodine vapour is passed over it. A
brown solid is formed.
Periodic Table of Elements 86
Discussion 2KMnO4(s) + 16HCl(aq) →
2KCl(aq) + 2MnCl2(aq) + 5Cl2(g) + 8H2O(l)
1 The halogens react with hot iron wool to form
iron(III) halides which are brown in colour but 3 Soda lime is a mixture of calcium hydroxide
the reactivity of the halogen decreases down and sodium hydroxide. It is used to absorb the
the group. excess halogen gas. The excess chlorine and
bromine gas have to be absorbed because they
2Fe(s) + 3Cl2(g) → 2FeCl3(s)⎯⎯⎯⎯⎯⎯⎯→ reactivity are poisonous.
iron(III) chloride decreases
2Fe(s) + 3Br2(g) → 2FeBr3(s) down the Conclusion
iron(III) bromide group
2Fe(s) + 3I2(g) → 2FeI3(s) 1 Chlorine, bromine and iodine show the same
iron(III) iodide chemical properties when they react with iron
wool, producing brown iron(III) halides.
4
2 The reactivity of the halogen decreases down the
2 The reaction between concentrated hydrochloric group from chlorine to iodine.
acid and potassium manganate(VII) produces
chlorine gas:
Safety Precaution to be Taken When X and Y reacts with water to form two kinds of
Handling Group 17 Elements
acids. (III is correct)
1 Fluorine, chlorine and bromine gases are very
poisonous. In fact chlorine was used in the X2 + H2O → HX + HOX
First World War to kill people. Answer C
2 Hence the experiments which involve the use 4.4
of these gases should be carried out in a fume
cupboard. 1 Explain why a solution of chlorine is (a) acidic (b)
able to bleach things.
5 ’05
2 Aqueous bromine and iodine solutions are both
X brown.
Y (a) How do you differentiate between the two
solutions?
Which of the following statements below are true (b) Carry out an experiment to show that bromine
concerning the elements X and Y in the above is more reactive than iodine.
Periodic Table?
I Y is more reactive than X. 3 Iodine is an element below chlorine in Group 17
II X is more electronegative than Y. of the Periodic Table.
III They react with water to produce two kinds of (a) Does iodine show similar chemical properties
as chlorine? Explain your answer.
acids. [Proton number of iodine is 53 and proton
IV Both form univalent negative ions of charge number of chlorine is 17]
(b) How does the
–1 when reacted with sodium.
A I, II and III only (i) density
B I, III and IV only (ii) melting point of iodine compare to
C II, III and IV only
D I, II, III and IV chlorine?
(c) Write equations for the reaction of iodine with
Comment (i) aqueous sodium hydroxide solution
(ii) iron wool
X has a smaller atomic radius. Thus it has a higher
4 Explain why the reactivity of Group 17 elements
tendency to accept an electron to form a univalent decreases down the group.
negative ion of charge –1. Thus X is more reactive 5 Name five compounds containing halogens and
state their uses.
and more electronegative. (I is incorrect)
87 Periodic Table of Elements
4 Modern Periodic Table
(a) Elements are arranged in order of increasing proton numbers.
(b) The vertical column is known as Group whereas the horizontal rows are called Periods.
(c) The number of valence electrons in the element corresponds to the group the element is in.
(d) The number of filled electron shells of an element corresponds to the period of the element.
Group 18 elements
(a) Group 18 elements are inert.
(b) They have attained the duplet or octet electronic configuration.
(c) Thus they do not need to share, donate or receive electrons from other elements.
Group 1 elements (Alkali metals)
(a) As we go down Group 1,
(i) melting point decreases,
(ii) density increases,
(iii) reactivity and electropositivity increases.
(b) Chemical properties of Group 1 elements:
(i) Alkali metals react with chlorine to form metal halide salts.
2M(s) + Cl2(g) → 2MCl(s) (M = Li, Na, K, …)
(ii) Alkali metals react with water to form metal hydroxides and hydrogen gas.
2M(s) + 2H2O(l) → 2MOH(aq) + H2(g) (M = Li, Na, K, …)
(iii) Alkali metals burn in air to form metal oxides which are soluble in water.
4M(s) + O2(g) → 2M2O(s) (M = Li, Na, K, …)
M2O(s) + H2O(l) → 2MOH(aq)
Group 17 elements (Halogen)
(a) As we go down group 17,
(i) melting point and density increases,
(ii) reactivity and electronegativity decreases.
(b) Chemical properties of Group 17 elements:
(i) Halogens dissolve in water to form two types of acids.
X2(g) + H2O(l) → HX(aq) + HOX(aq) (X = F, Cl, Br, …)
(ii) Halogens react with iron to form brown iron(III) halide salts.
2Fe(s) + 3X2(g) → 2FeX3(s) (X = F, Cl, Br, …)
( iii) Halogens react with sodium hydroxide to form two types of salts and water.
2NaOH(aq) + X2(g) → NaX(aq) + NaOX(aq) + H2O(l) (X = F, Cl, Br, …)
Periodic Table of Elements 88
4.5 Elements in a Period period, other than Period 3, based on the changes
in the properties of elements in Period 3.
Elements in Period 3 4 Table 4.8 shows the trends across Period 3.
1 The elements in Period 3 are sodium (Na), Figure 4.14 The elements in Period 3 of the Periodic 4
magnesium (Mg), aluminium (Al), silicon Table
(Si), phosphorus (P), sulphur (S), chlorine
(Cl), and argon (Ar).
2 The study of the elements in Period 3 will show
a gradual change of physical and chemical
properties across the period from left to right.
3 Period 3 is a typical period. Thus we can predict
the trend of changes in properties across a
Table 4.8 The trends across Period 3 SPM
Group 1 2 13 14 15 16 17 ’11/P1
S Cl
Element Na Mg Al Si P 16 17 18
Ar
Proton number 11 12 13 14 15 2.8.6 2.8.7 18
Electron 2.8.1 2.8.2 2.8.3 2.8.4 2.8.5 2.8.8
arrangement
Number of valence 1 23 45678
electrons
Atomic radius 0.156 0.136 0.125 0.117 0.111 0.104 0.099 0.094
(nm)
Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 —
649 660
Melting point (°C) 98 1107 2467 1410 590 119 –101 –189
Metal Metal
Boiling point (°C) 883 MgO Al2O3 2355 Ignites 445 –35 –186
Basic Amphoteric
Nature of elements Metal Metalloid Non-metal Non-metal Non-metal Non-metal
Formula of oxide Na2O SiO2 P4O10 SO2 Cl2O7 None
Character of oxide Basic Acidic Acidic Acidic Acidic —
Trends of Changes across Period 3
SPM
’05/P2 Atomic Radius
’09/P2 Valence Electrons
The atomic radius decreases across the period. The number of valence electrons increases
• All the elements in Period 3 have three filled across the period.
• As the proton number increases, the
electron shells but the proton number increases
number of electrons increases.
by one unit across the period. • The number of valence electrons
• As a result, the increase in the number of protons increases by 1 from one element to the
increases the electrostatic force between the next across the period.
nucleus and the valence electrons.
• The valence electrons are pulled closer to the
nucleus, causing the atomic radius to decrease.
89 Periodic Table of Elements
Electronegativity Melting & Boiling Points
The electronegativity increases across the The melting points and boiling points
period. Electronegativity is a measurement of the
tendency of an atom to attract electrons. of the elements increase from the left
• The atomic radius decreases across the period.
• The proton number increases across the of the period to the middle of the
period. period and then decrease again.
• The increase in the number of protons (positive
• Sodium, magnesium and
charge in the nucleus) and the decrease in the
distance between the nucleus and the outermost aluminium are metals with strong
electron shell across the period cause an increase
in the force of attraction of the nucleus. metallic bonds between the metal
The atoms will have a higher tendency to attract
electrons. Therefore electronegativity increases. atoms. Hence they have high melting
• Elements on the left side of the period tend to
lose electrons to form positive ions. Elements and boiling points. The strength of
on the right side of the period tend to gain
electrons to form negative ions. the metallic bonds increase with the
• Thus the elements on the left side of the period
4 (such as Na) are electropositive while the increase in the number of valence
elements on the right side of the period (such
as Cl) are electronegative. electrons in the order: Na < Mg < Al.
• Silicon has very high melting and
boiling points. It has strong covalent
bonds between atoms forming a
3-dimensional gigantic network.
• Phosphorus, sulphur, chlorine and
argon are non-metals with weak van
der Waals forces of attraction between
molecules. They are lowest at the right
with chlorine and argon existing as
gases at room temperature.
Nature of Metals
As the electronegativity of the elements increases, the elements change from metals to metalloid and
finally to non-metals across the period.
• The elements on the left of the period are metals (Na, Mg and Al).
• Silicon has some metallic and some non-metallic properties. It is called a metalloid or semi-metal.
• The elements on the right of the period are non-metals (P, S, Cl and Ar).
Nature of Oxides SPM
’08/P1, ’10/P1
The oxides of the elements change from basic to • Acidic oxides react with alkali to form salts and
amphoteric and then to acidic across the period. water.
• Elements on the left of the period, which are For example, sulphur trioxide reacts with
sodium hydroxide to form sodium sulphate
metals, form metal oxides. The metal oxides salt and water:
are usually basic oxides.
• Basic oxides react with acids to form salts and SO3(g) + 2NaOH(aq) → Na2SO4(aq) + H2O(l)
water.
For example, magnesium oxide reacts with • An amphoteric oxide can react with both acids
sulphuric acid to form a salt and water: and bases to form salt and water. Aluminium
oxide is an example of an amphoteric oxide.
MgO(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) Aluminium oxide can react with both acids and
alkalis to form salts and water.
• Elements on the right of the period are non-
metals. Non-metallic oxides are acidic oxides. • Argon as an inert gas, does not form oxide.
Periodic Table of Elements 90
4.7
To determine the properties of the oxides of elements in Period 3
Problem statement
How do the properties of the oxides of elements in
Period 3 change across the period?
Hypothesis
The oxides change from basic to amphoteric and
then to acidic across Period 3.
Variables Figure 4.15 Reaction of oxides of Period 3 with Experiment 4.7 4
(a) an acid, (b) an alkali
( a) Manipulated variable : Oxides of Period 3
( b) Responding variable : Reaction with acid or 3 Two drops of universal indicator are added and
the pH of the solution is noted.
alkali
( c) Constant variable : Concentrations of sodium 4 The experiment is repeated with magnesium oxide,
aluminium oxide, silicon(IV) oxide, phosphorus(V)
hydroxide and nitric acid oxide, sulphur dioxide and dichlorine heptoxide
solutions respectively in place of sodium oxide.
Materials (B) Reaction of the oxides of Period 3 elements
with 2 mol dm–3 nitric acid and 2 mol dm–3
Sodium oxide, magnesium oxide, aluminium oxide, sodium hydroxide solutions
silicon(IV) oxide, phosphorus(V) oxide, sulphur
dioxide, dichlorine heptoxide, nitric acid solution 1 A little sodium oxide powder is put into two
and sodium hydroxide solution of 2 mol dm–3, separate test tubes.
distilled water and universal indicator.
2 5 cm3 of nitric acid and 5 cm3 of sodium hydroxide
Apparatus are added separately to the contents in each test tube.
Test tube, test tube holder, Bunsen burner, rubber 3 The contents in each test tube are heated slowly
stopper and glass rod. while being stirred with glass rods.
Procedure 4 The solubility of sodium oxide in the two
(A) Reaction of the oxides of Period 3 elements solutions is recorded.
with water 5 The experiment is repeated with magnesium oxide,
1 A little sodium oxide powder is added to some aluminium oxide, silicon(IV) oxide, phosphorus(V)
oxide, sulphur dioxide and dichlorine heptoxide
distilled water in a test tube. respectively in place of sodium oxide.
2 The test tube is tightly closed with a rubber
stopper and the contents are shaken.
Results
Experiment A
Oxide Observation Inference
Na2O
MgO Solubility in water pH value of solution Solution obtained is a strong alkali. Sodium
oxide is basic. Sodium shows metallic properties.
Al2O3 Dissolves in water to form 13 – 14 Solution obtained is a weak alkali.
SiO2 a colourless solution 8–9 Magnesium oxide is basic.
Magnesium shows metallic properties.
Slightly soluble in water 7 pH measured is pH of water as the oxide is
to form a colourless insoluble in water.
solution pH measured is pH of water as the oxide is
insoluble in water.
Insoluble in water
Insoluble in water 7
91 Periodic Table of Elements
Oxide Observation Inference
P4O10
SO2 Solubility in water pH value of solution Solution obtained is acidic. P4O10 is acidic.
Cl2O7 Phosphorus shows non-metallic properties.
Dissolves in water to 2–3 Solution obtained is acidic. SO2 is acidic.
form a colourless solution 2–3 Sulphur shows non-metallic properties.
Solution obtained is a strong acid. Cl2O7 is
Dissolves in water to 1 acidic. Chlorine shows non-metallic properties.
form a colourless solution
Dissolves in water to
form a colourless solution
Experiment B
4 Oxide Reaction with 2 mol Reaction with 2 mol dm–3 Inference
Na2O dm–3 NaOH HNO3
MgO Sodium oxide is basic because it reacts with the
Does not react Reacts with nitric acid to acid. Sodium shows metallic properties.
Al2O3 form a colourless solution Magnesium oxide is basic because it reacts with
Magnesium oxide the acid. Magnesium shows metallic properties.
SiO2 does not dissolve Dissolves in nitric acid
P4O10 forming a colourless Aluminium oxide is amphoteric because it
Dissolves in sodium solution reacts with both acid and alkali.
hydroxide forming a
colourless solution Dissolves in nitric acid Silicon(IV) oxide is acidic because it reacts with
Reacts with sodium forming a colourless an alkali. Silicon shows non-metallic properties.
hydroxide solution solution Phosphorus(V) oxide is acidic because it reacts
Reacts with sodium with an alkali. Phosphorus shows non-metallic
hydroxide solution Does not dissolve in nitric properties.
acid Sulphur dioxide is acidic because it reacts
with an alkali. Sulphur shows non-metallic
Does not react with nitric properties.
acid Dichlorine heptoxide is acidic because it reacts
with an alkali. Chlorine shows non-metallic
SO2 Reacts with sodium Does not react with nitric properties.
hydroxide solution acid
Cl2O7 Reacts with sodium Does not react with nitric
hydroxide solution acid
Discussion SiO2(s) + 2NaOH(aq) → Na2SiO3(aq) + H2O(l)
1 (a) Metallic oxides are basic. Metal oxides react P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l)
with acids to form salts and water.
SO2(g) + 2NaOH(aq) → Na2SO3(aq) + H2O(l)
Na2O(s) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l)
Conclusion
MgO(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2O(l)
(b) Aluminium oxide is amphoteric because it 1 Across Period 3 from left to right,
SPM can react with both acids and alkalis. (a) the element changes from being a metal, to a
metalloid and a non-metal.
’05/P2 (b) the oxides change from being basic to
amphoteric and acidic.
Al2O3(s) + 6HNO3(aq) →
2Al(NO3)3(aq) + 3H2O(l) 2 Silicon is classified as a metalloid because it is
a very weak conductor of electricity. However,
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → its oxide is acidic.
2NaAl(OH)4(aq) (sodium aluminate)
3 Aluminium is classified as a metal because it is
Other examples of amphoteric oxides are a very good conductor of electricity. However, its
lead(II) oxide and tin(II) oxide. oxide is amphoteric. Aluminium oxide shows
both metallic and non-metallic properties.
(c) Non-metallic oxides are acidic. Non-metallic
oxides react with alkalis to form salts and water.
Periodic Table of Elements 92
Uses of Semi-metals (or Metalloids) Do you know that Microchip wafer
a silicon wafer can
1 A semi-metal or metalloid is an element with contain hundreds
properties intermediate between those of of microchips? Each
metals and non-metals. microchip itself
contains hundreds of
2 For example, silicon is a non-metal and is a electronic components.
very poor conductor of electricity. However, Our earth contains
the conductivity increases with temperature. It about 27.7% silicon
becomes a good conductor of electricity at and a large portion is
high temperatures. found in sand.
3 This type of substance is known as a 4
semi-metal. Examples of semi-metals are
silicon, germanium, boron, antimony, 4.6 Transition Elements
and arsenic. These elements are important
industrial materials and are used to make Figure 4.16 The transition elements in the Periodic
semiconductors. Table
4 Adding of foreign elements (called doping) 1 The transition elements are elements in a
can increase the conductivity of semi-metals. block located in-between Group 2 and Group
(a) If silicon is doped with Group 13 elements 13 in the Periodic Table.
such as boron, a p-type semiconductor is
produced. 2 There are 10 elements in each series. The
(b) If it is doped with Group 15 elements first series is in Period 4 and consists of
such as arsenic or antimony, a n-type the elements: scandium(Sc), titanium(Ti),
semiconductor is produced. vanadium(V), chromium(Cr), manganese
(Mn), iron(Fe), cobalt(Co), nickel(Ni),
5 Semiconductors are very important in the copper(Cu) and zinc(Zn).
microelectronic industry and are used to
make transistors, diodes, rectifiers, thermistors 3 Table 4.9 shows some physical properties of
and microprocessors. Hundreds of these the transition elements in the first series in
electronic components can be built onto a Period 4.
crystal of silicon to make a microchip.
4 The transition elements are all metals with
4.5 the following physical properties:
(a) High density
1 Predict the changes in the properties of the (b) High hardness
elements across Period 3 in the Periodic Table. (c) High electrical conductivity
(a) Atomic radius (d) High tensile strength
(b) Electronegativity (e) Silvery surface
(c) Acidic-base property of the oxides (f) Ductile and malleable
(g) High melting point
2 The proton numbers of element X and Y are 3 (h) High boiling point
and 9 respectively.
(a) To which period of the Periodic Table do X 5 The atomic radius and electronegativity of the
and Y belong? transition elements are almost the same.
(b) Which is more electronegative? Explain your
answer.
3 Write the formula of the oxides of the elements in
Period 3. State whether the oxide is basic, acidic
or amphoteric.
4 Silicon is a semi-metal. State one difference
between
(a) silicon and sulphur,
(b) silicon and iron.
93 Periodic Table of Elements
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