IGCSE Chemistry
Topic 4: Stoichiometry
1) Chemical Formula
2) Chemical Equations
3) Relative Molecular/Formula mass (Mr)
4) Moles (3 laws)
5) Empirical formula & Molecular Formula
1) Chemical Formula
Valency or combining power:
 The valency of an element is the number of electrons lost, gained or shared to from
compounds. The valency of an element depends on the number of valence electrons.
 In covalent molecules; the valency gives the number of covalent bonds which the
atoms can form.
 In ionic compounds; the valency gives the charge on the ions of the element.
 The following table shows the valencies & charges of common elements used in
chemistry:
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Another way to memorize Valency
 Silver – Ag – is mainly monovalent (valency = +1)
 Lead – Pb – is in group 4, but it exists mainly as divalent (valency = -2)
 +ve sign indicate losing e-, -ve sign indicate gaining e-, no sign indicate sharing e-
Polyatomic ions (formerly known as Radicals):
Groups of atoms that are combined together covalently & carry a charge, behave like ions.
They act as one unit in chemical reactions.
Valency
Radical name & Symbol
Charge
Hydroxide
OH-
-1
Nitrate
NO3-
-1
Nitrite
NO2-
-1
Hdrogen carbonate
HCO3-
-1
Ammonium
NH4+
+1
Carbonate
CO32-
-2
Sulfate
SO42-
-2
Sulphite
SO32-
-2
Phosphate
PO4-3
-3
Monovalent
Divalent
Trivalent
Names of some covalent compounds to be learned:
Methane
CH4
Carbon dioxide
CO2
Nitrogen dioxide
NO2
Ammonia
NH3
Carbon monoxide
CO
Nitrogen monoxide
NO
Water
H2O
Sulfur dioxide
SO2
N2O4
Hydrogen peroxide
H2O2
Sulfur trioxide
SO3
SF6
Number of atoms in covalent compounds: mono = 1 … di=2 … tri = 3 … tetra = 4 … penta = 5 … hexa = 6
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Steps of writing a Chemical formula:
1.
2.
3.
4.
Write the symbol of elements (metals on the left, non-metals or radicals on the right).
Write the valency or charge of each element above its symbol, simplify if needed (?)
Cross valencies
Radical (polyatomic ions) must be placed between brackets "( )" if they acquire a
valency of more than their own after crossing (?)
 Compounds ending with (-ide) are made of 2 elements – except hydroxide (OH-)









Element + Oxygen  Oxide
Element + Sulfur  Sulphide
Element + Nitrogen  Nitride
Element + Phosphorus  Phosphide
Element + Halogen  Halide
Element + Carbon  Carbide
Element + Hydrogen  Hydride
Compounds ending with (-ate) are made of 3 elements, one element is oxygen  ex: SO42Compounds ending with (-ite) are made of 3 elements, one element is oxygen but with less
number of oxygen than (-ate)  ex: SO32-
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2) Chemical Equations
 A chemical reaction occurs when substances react together to form new substances.
No atoms are created or destroyed in the process, they are just rearranged.
 Reacting substances are called reactants. New chemicals formed are called products.
Steps of writing symbol equation for a reaction:
1)
2)
3)
4)
Write the equation in words (word equation)
Write the equation again using correct symbols & formula for elements & compounds.
Balance the equation for each atom in turn. Make sure you do not change any formula.
Some chemical equations include state symbols after the formula. They show if the
substance is a solid (s), liquid (l), gas (g), or dissolved in water/aqueous (aq).
Example 1:
 Sodium reacts with chlorine to form sodium chloride
Sodium + Chlorine
Sodium chloride
Na + Cl2
NaCl
1 Na
1 Na
2 Cl
1 Cl
2 Na (s) + Cl2 (g)
2 NaCl (s)
2 Na
2 Na
2 Cl
2 Cl
2 sodium atoms react with 1 molecule of chlorine to form 2 sodium chloride compounds.
Example 2:
 Methane burns in air (oxygen) to form carbon dioxide & water
1 methane molecule reacts with 2 oxygen molecules to form 1 molecule of carbon dioxide & 2
molecules of water.
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Q: Write down the formula for each of these ionic compounds:
1) Potassium carbonate
……………………
8) Lead (II) nitrate
……………………
2) Zinc Oxide
……………………
9) Magnesium nitride
……………………
3) Iron (II) phosphate
……………………
10) Ammonium phosphate ……………………
4) Silver chloride
……………………
11) Iron (III) oxide
……………………
5) Aluminium hydroxide
…………………..
12) Lithium carbonate
……………………
6) Copper (II) bromide
……………………
13) Zinc sulphide
……………………
7) Barium sulfate
……………………
14) Calcium hydroxide
……………………
Q: Balance the following equations:
1) N2 + ___ H2
___ NH3
2) ___ Al + ___ O2
3) C + CO2
4) ___ Li+ ___ H2O
5) C2H5OH + ___O2
___ Al2O3
___ CO
___ LiOH + H2
___CO2 + ___ H2O
Q: Write the following word equation into symbol equations with proper balance:
1) Sodium + Copper II Sulfate  Sodium Sulfate + Copper
2) Zinc + Hydrochloric Acid  Zinc Chloride + Hydrogen
3) Iron III Oxide + Carbon Monixide  Iron + Carbon Dioxide
4) Magnesium Sulphide + Oxygen  Magnesium Oxide + Sulfur Dioxide
5) Iron III Chloride + Sodium Hydroxide  Iron III hydroxide + Sodium Chloride
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3) Relative Molecular/Formula Mass (Mr)
 Each element has its own relative atomic mass (Ar). This is the average mass of its
isotopes compared with the mass of a standard atom of carbon-12.
 Compounds have a relative molecular mass (or formula mass). It's the sum of the
relative atomic masses of the all the elements shown in the formula.
 Mr = RMM for covalent molecules while RFM for ionic compounds.
Examples:
Ionic: Sodium chloride, NaCl
1 atom of Na 1 X 23
= 23
Covalent: Ethane, C2H4
+
2 atoms of C 2 X 12 = 24
1 atom of Cl 1 X 35.5 = 35.5
4 atoms of H 4 X 1 = 4
Relative formula mass = 58. 5
Relative molecular mass = 28
+
Q1: Calculate the relative formula mass of the following:
1) CaCO3 : ……………………………………………………………….…………………………………………………….
2) C4H10O: ……………………………………………………………………………………………………………………..
3) KMnO4: ………………………………………….………………………………………………………………………….
4) (NH4)2SO4: ………………………………………………………..……………………………………………………….
Q2: Complete the following table:
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Percentage Mass:
The % of an element by mass =
𝐴𝑟 𝑜𝑓 𝑡𝑕𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ×𝑛𝑜 .𝑜𝑓 𝑖𝑡𝑠 𝑎𝑡𝑜𝑚𝑠
𝑀𝑟 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
X 100
In a compound
Examples:
1) The percentage mass of calcium in calcium carbonate, CaCO3
=
𝐶𝑎
𝐶𝑎+𝐶+3×0
× 100 =
40
40+12+48
× 100 = 40%
40% of the mass of calcium carbonate is calcium.
2) The percentage mass of nitrogen in ammonium nitrate, NH4NO3
=
2 ×𝑁
2×𝑁+4×𝐻+3×0
× 100 =
28
28+4+48
× 100 = 35%
35% of the mass of ammonium nitrate is nitrogen.
Reacting Masses:
Q1: When 16 g of methane reacts completely with an excess of steam, 6 g of hydrogen are
produced. Calculate the mass of methane required to produce 300 g of hydrogen.
16 g of CH4
6 g of H2
Simple cross multiplication
? g of CH4
300 g of H2
Mass of CH4 = 300 x 16 / 6 = 800 grams
Q2: The equation for the reaction between magnesium and dilute sulfuric acid is shown.
Mg + H2SO4
MgSO4 + H2 ; Where the Mr of MgSO4 is 120
Which mass of MgSO4 will be formed if 12 g of magnesium are reacted with H2SO4?
Eq: Mg + H2SO4
Mr: 24 + 98
MgSO4 + H2
120 + 2
Simple cross multiplication
24 grams of magnesium forms 120 grams of MgSO 4
Then, 12 grams of magnesium forms 60 grams of MgSO4
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IGCSE Chemistry
4) The Mole
 A mole (or molar mass) of any substance is: the relative atomic mass (Ar) – if it is an
element – or relative molecular mass (Mr) – If it is a compound – expressed in grams.
Calculations
 For Elements:
1) 1 mole of sulfur (32S)
2) 1 mole of calcium (40Ca)
= the relative atomic mass (Ar) in grams = 32 g
= the relative atomic mass (Ar) in grams = 40 g
 For compounds:
1) 1 mole of Water
2) 1 mole of CaCO4
= the relative molecular mass (Mr) of water in grams
= (2 X H) + O = 2 + 16 = 18 g
= the relative molecular mass (Mr) of calcium carbonate in grams
= Ca + C + (3 X O) = 40 + 12 + 48 = 100 g
1) What is the mass of one mole of each of the following:
a) Sodium (Na)
………………………………………
b) Copper (II) oxide (CuO)
……………………………………….
2) What is the mass of:
a) 0.2 mole of nitrogen (N2) ……………………………………..
b) 0.5 moles of carbon (C)
………………………………………
c) 4 moles of ammonia (NH3) ………………………………….…..
3) How many moles are in:
a) 2.4 g of magnesium (Mg) ……………………………………….
b) 36 g water of (H2O)?
……………………………………….
c) 50 g of calcium carbonate? ……………………………………….
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Avogadro's number:
It's it the number of particles (atoms or molecules) in 1 moles of any substance.
There are 6.02 x 1023 particles in 1 moles of any substance.
 1 mole of potassium (element) contains 6.023 x 1023 atoms of potassium.
 1 mole of carbon dioxide (compound) contains 6.023 x 1023 molecules of carbon dioxide.
Calculations involving moles
Example 1:
Calcium burns in oxygen to form calcium oxide. Calculate the mass of calcium needed to
prepare 28 grams of calcium oxide.
1st step: Write a balanced equation to find the moles ratio.
Calcium
+
Oxygen
2 Ca(s)
+
O2(g)
2
:
1
Moles ratio:
Calcium oxide
2 CaO(s)
:
2
2nd step: Calculate the moles in 28 grams of CaO




Mr of CaO = 40 + 16 = 56
No. of moles of CaO = Mass / Mr = 28 / 56 = 0.5
Since the moles ratio between CaO : Ca is 2 : 2 or 1 : 1
Then the no. of moles of CaO = No. of moles of Ca = 0.5
3rd step: Calculate the mass of 0.5 moles Ca

Mass of Ca = No. of moles x Ar = 0.5 x 40 = 20 grams
Example 2:
6 grams of magnesium reacts completely with 30 grams of excess Cl2. What is the
maximum yield (mass) of magnesium chloride that can be formed from this reaction?
What is the mass of Cl2 left unreacted?
1st step: Write a balanced equation to find the moles ratio
Magnesium
+
Chlorine
Mg(s)
+
Cl2(g)
1
:
1
Moles ratio:
Mr. Fady Youssuf
Magnesium Chloride
MgCl2(s)
:
45
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IGCSE Chemistry
2nd step: Calculate moles of Mg that completely reacted



No. of moles of Mg = Mass / Ar = 6/24 = 0.25
Since the moles ratio between Mg : MgCl2 is 1 : 1
Then the No. of moles of Mg = No. of moles of MgCl2 = 0.25
3rd step: Calculate mass of 0.25 moles of MgCl2

Mass of MgCl2 No. of moles x Mr = 0.25 x (24 + 71) = 23.75 grams
4th step: Calculate mass of Cl2 left unreacted



No. of moles of Cl2 reacted = 0.25
Mass of Cl2 reacted = moles x Ar = 0.25 x 71 = 17.75 grams
Mass of Cl2 left unreacted = 30 g – 17.75 g = 12.25 grams
Calculations involving gases – Molar gas volume

The volume of one mole of any gas is 24 dm3 (24 liters or 24000 cm3) at standard
room temperature and pressure (r.t.p.). This rule applies to all gases.
Example 1:
What volume does 0.25 moles of CO2 occupy at r.t.p?

No. of moles x 24 dm3 = 0.25 x 24 = 6 dm3
Example 2:
What is the volume of SO2 gas produced from burning 5 grams sulfur?
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1st step: Write a balanced equation to find the moles ratio
Moles ratio:
Sulfur
+
Oxygen
S(s)
+
O2(g)
1
:
1
Sulfur dioxide
SO2(g)
:
1
2nd step: Calculate the moles of SO2 produced



No. of moles of S = Mass / Ar = 5/32 = 0.156
Since the moles ratio between S : SO2 is 1 : 1
Then the No. of moles of S = No. of moles of SO2 = 0.156
3rd step: Calculate the volume of SO2 produced

No. of moles x 24 = 0.156 x 24 = 3.75 dm3
Example from past papers:
The equation for the complete combustion of butane is given below. Insert the two
missing volumes.
2C4H10(g) + 1302(g )
…………..
8CO2(g) + 10H2O(g)
……………
40
volume of gas/cm3
[2]
Example from past papers
Iron (II) sulphate decomposes when heated. Calculate the mass of iron (III) oxide formed
and the volume of sulphur trioxide produced when 10.0g of iron(III) sulphate was heated.
Mass of one mole of Fe2(SO4)3 is 400 g.
Fe2(SO4)3 (s)
Fe2O3 (s) + 3SO3 (g)
Number of moles of Fe2(SO4)3 =
Number of moles of Fe2O3 formed =
Mass of iron (III) oxide formed =
g
Number of moles of SO3 produced =
dm3
Volume of sulphur trioxide at r.t.p. =
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A 20 cm3 sample of butyne, C4H6, is burnt in 150 cm3 of oxygen. This is an excess of
oxygen.
2C4H6(g) + 11O2(g)
8CO2(g) + 6H2O (l)
(i) What volume of oxygen reacts?
………………………………………………………………………………………………………….. [1]
(ii) What volume of carbon dioxide is produced?
………………………………………………………………………………………………………….. [1]
(iii) What is the total volume of gases left at the end of the reaction?
………………………………………………………………………………………………………….. [1]
A compound contains only aluminium and carbon. 0.03 moles of this compound
reacted with excess water to form 0.12 moles of Al(OH)3 and 0.09 moles of CH4.
While a balanced equation for this reaction.
………………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………….………..
……………………………………………………………………………………………………………….……………………..
……………………………………………………………………………………………………………………………….. [2]
Calculations involving concentration of solutions
 Concentration of a solution is the amount of solute (in grams or moles) that is dissolved
in 1 dm3 (1 liter) of solution.
 Molar solution: A solution that contains 1 mole of solute in 1 dm3 of solution. The
concentration of solutions is expressed in g / dm3 or mol / dm3.
o If we want to prepare 1 mol/dm3 of NaCl solution
Dissolve 58.5 g of NaCl (1 mole) in water to give a total volume of 1 dm3
o
If we want to prepare 2 mol/dm3 of NaCl solution
Dissolve 2 x 58.5g (2 mole) in water to give a total volume of 1 dm3
o
If we want to prepare 0.1 mol/dm3 of NaCl solution
Dissolve 5.85g (0.1 mole) in water to give a total volume of 1 dm3
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Example:
What is the concentration of a solution when 11.7 grams of NaCl dissolves in 500 cm3?


Moles of NaCl = Mass/Mr = 11.7/58.5 = 0.2
Conc. Of NaCl solution = 0.2/500 x 1000 = 0.4 mol / dm3
Example from past papers
In an experiment, 25.0 cm3 of aqueous sodium hydroxide, 0.4 mol/dm3, was neutralized
by 20.0 cm3 of aqueous oxalic acid, H2C2O4.
2NaOH + H2C2O4
Na2C2O4 + 2H2O
Calculate the concentration of the oxalic acid mol/dm3.
(i) Calculate the number of moles of NaOH in 25.0 cm3 of 0.4 mol/dm3 solution.
………………………………………………………………………………………………………………………. [1]
(ii) Use your answer to (l) and the mole ratio in the equation to find out the number
of moles of H2C2O4 in 20 cm3 of solution.
………………………………………………………………………………………………………………………. [1]
(iii) Calculate the concentration, mol/dm3, of the aqueous oxalic acid.
………………………………………………………………………………………………………………………. [2]
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Percentage yield & Percentage purity
Percentage Yield
 Some reactions do not to completion Not all of the reactants are converted to
products.
 The theoretical yield is the maximum (calculated) yield expected from the balanced
chemical equation.
 The actual yield is the true yield that is produced from the reaction in a real
experiment.
 Yield is the mass (in grams)
Percentage yield =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 (𝑔𝑖𝑣𝑒𝑛 )
𝑡𝑕𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 (𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 )
× 100
Example: On heating 12.4 g of calcium carbonate, 7 grams of calcium oxide are formed.
Knowing that the calculated amount is 8 grams; what is the percentage yield of the reaction?
…………………………………………………………………………………………….………………………………………………
Example from past papers
Strontium chloride-6-water can be made from the insoluble compound, strontium
carbonate, by the following reactions.
SrCO3(s) + 2HCl(aq)  SrCl2(aq) + CO2(g) + H2O(l)
SrCl2(aq) + 6H2O(l)  SrCl2.6H2O(s)
In the above experiment, 50.0 cm3 of hydrochloric acid of concentration 2.0 mol/dm3
was used. 6.4 g of SrCl2.6H2O was made.
Calculate the percentage yield.
number of moles of HCl used = ……………………….
number of moles of SrCl2.6H2O which could be formed = ……………………..
mass of one mole of SrCl2.6H2O is 267 g
theoretical yield of SrCl2.6H2O = ……………………… g
percentage yield = ………………………. %
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Percentage Purity
Percentage purity =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑢𝑟𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑖𝑚𝑝𝑢𝑟𝑒 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
× 100
Limiting reactant & Excess reactant



No matter how many tires there are, if there are only 8 car bodies  only 8 cars
can be made. Likewise with chemistry, if there is only a certain amount of one
reactant available for a reaction, the reaction must stop when that reactant is
consumed whether or not the other reactant has been used up.
Limiting Reactant: is the reactant that limits the amount of product that can be
formed. The reaction will stop when all of the limiting reactant is consumed.
Limiting reactant is always in a smaller portion than the other reactant.
Excess Reactant: is the reactant that still remains present in the reaction mixture
when a reaction stops. The excess reactant remains because there is nothing with
which it can react. (limiting reactant is consumed).
Example from past papers
0.07 moles of silicon reacts with 25 g of bromine.
Si + 2Br2
SiBr4
(i) Which one is the limiting reagent? Explain your choice.
………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………….…………. [3]
(ii) How many moles of SiBr4 are formed?
………………………………………………………………………………………………………..………………… [1]
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5) Empirical Formula & Molecular Formula

Empirical formula shows the simplest ratio of the number of atoms of the different
elements present in it.
Molecular formula shows the actual number of atoms of the different elements
present in it.
Ionic
Covalent

Compound
Molecular formula
Empirical formula
Ethane
C2H6
CH3
Benzene
C6H6
CH
Ethanoic acid (CH3COOH)
C2H4O2
CH2O
Methane
CH4
CH4
Glucose
C6H12O6
CH2O
Magnesium chloride
MgCl2
MgCl2
Sodium phosphide
Na3P
Na3P
Example 1 (Covalent compound)
Oxalic acid was found to contain 26.7% carbon, 2.2% hydrogen by mass and the rest is
oxygen. The Mr of oxalic acid is 90. Find the empirical formula and the molecular formula.
A) Find the empirical formula by following the steps in the table:
C
H
O
26.7
2.2
100-28.9=71.1
12
1
16
26.7/12=2.2
2.2/1=2.2
71.1/16=4.4
1
1
2
1- Mass (%)  always given
2- Relative atomic mass (Ar)  from P.T
3- Moles  Step 1 + Step 2
4- Ratio (dividing) by the smallest value)
Empirical formula
CHO2
B) Find the molecular formula by following the steps in the table:
1- Mr of empirical formula
2-
𝐌𝐫 𝐨𝐟 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐟𝐨𝐫𝐦𝐮𝐥𝐚
𝐌𝐫 𝐨𝐟 𝐞𝐦𝐩𝐢𝐫𝐢𝐜𝐚𝐥 𝐟𝐨𝐫𝐦𝐮𝐥𝐚
CHO2 = 12 + 1 + (2 x 16)
𝟗𝟎 (𝐠𝐢𝐯𝐞𝐧)
𝟒𝟓 (𝐜𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞𝐝)
3- Multiply each element in the
empirical formula by 2
= 45
Molecular formula
N=2
is twice the
empirical formula
2 x CHO2 = C2H2O4
n = shows how many times the empirical formula is multiplied / repeated.
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Example 2 (Ionic Compound)
A compound contains 72% magnesium & 28% nitrogen. What is the empirical formula?
Mg
N
1- Mass (%)
72
28
2- Relative atomic mass (Ar)
24
14
72 / 24 = 3
28 / 14 = 2
3
2
3- Moles
4- Ratio (dividing by the smallest value)
Empirical formula
Mg3N2
N.B: In ionic compounds, the empirical formula is the same as the molecular formula
Example from past papers
Maleic acid is an unsaturated acid. 5.8g of this acid contained 2.4g of carbon, 0.2g of
hydrogen and 3.2g of oxygen.
(i) How do you know that the acid contained only carbon, hydrogen and oxygen?
……………………………………………………………………………………………………………..……………………
……………………………………………………………………………………………………………………………. [1]
(ii) Calculate the empirical formula of maleic acid.
Number of moles of carbon atoms = …………………………………..
Number of moles of hydrogen atoms = ………………………………
Number of moles of oxygen atoms = …………………………………..
The empirical formula is ………………………………………………………….…………….……………. [3]
(iii) The mass of one mole of maleic acid is 116g. What is its molecular formula?
………………………………………………………………………...…………………………………….…………… [2]
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Types of Chemical Reactions


Chemical reactions can be classified according to 1) type, 2) heat change (endo or exo) or
3) gain or loss of oxygen (Redox)
To know more about chemical reactions types, chemists classified them into five major
categories according to the rearrangement of atoms during the reaction.
(1) Synthesis reaction (direct combination)

Takes place between two or more simple substances (reactants) to form one product.
𝒉𝒆𝒂𝒕
Fe + S

FeS
Photosynthesis is an essential process of life. It's a type of synthesis reaction; it needs
sunlight & chlorophyll to occur (endothermic reaction).
Carbon dioxide + Water
6 CO2 + 6 H2O
𝑺𝒖𝒏𝒍𝒊𝒈𝒉𝒕
𝑪𝒉𝒍𝒐𝒓𝒐𝒑𝒉𝒚𝒍𝒍
Glucose + Oxygen
C6H12O6 + 6 O2
(2) Decomposition reaction (opposite of synthesis)


Takes place when one reactant breaks down to give two or more simpler products.
Decomposition may take place by adding a catalyst (catalytic decomposition), heating
(thermal decomposition), light (photo-degradation) or by electrolysis (electric
decomposition).
2 H2O2
CaCO3
2AgCl
𝑴𝒏𝑶𝟐
2 H2O + O2
𝒉𝒆𝒂𝒕
𝒍𝒊𝒈𝒉𝒕
(Catalytic decomposition)
CaO + CO2
(Thermal decomposition)
2 Ag + Cl2
2 NaCl (l)
Mr. Fady Youssuf
(Photochemical degradation)
2Na (s) + Cl2 (g)
(Electrolysis)
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N.B: Silver bromide (cream) & Silver iodide (yellow)also photo-degrade  topic 10
(3) Displacement reactions
A) Single displacement reaction

Takes place when a more reactive element displaces a less reactive element from its
aqueous salt solution. Elements are arranged according to their reactivity in the
"Electrochemical / Reactivity series".
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B) Double displacement reaction
i) Neutralization (involves an acid)




Takes place between an acid and a base to form a salt  topic 7
Acid: Substance that dissolves in water to give H+ in water (proton donor)
Base: Substance (metal oxide & metal hydroxides) that neutralizes an acid, accepting
give H+ from acid to produce salt and water (proton acceptor)
Alkali: Soluble base that dissolves in water to give OH- in water.
HCL + NaOH
NaCl + H2O
H2SO4 + ZnO
ZnSO4 + H2O
2 HBr + Ca(OH)2
CaBr2 + 2 H2O
ii) Precipitation (product is insoluble in water).

Takes place when two solutions soluble salts react to produce an insoluble salt
(precipitate)
Pb(NO3)2 (aq) + 2 Kl (aq)

Pbl2 (s) + 2 KNO3 (aq)
insoluble salt (ppt)
Precipitation reaction can also take place when a gas is bubbled into a solution.
Limewater test for carbon dioxide is a type of precipitation reaction.
CO2 (g) + Ca(OH)2 (aq)
CaCO3 (s) + H2O (l)
milky suspension
(4) Combustion


Combustion is an exothermic reaction between a fuel & oxygen accompanied by
releasing of heat energy. An example of this kind of reaction is the burning of the
hydrocarbon; like methane gas (natural gas).
Incomplete combustion of methane gas – due to insufficient oxygen or a poor
ventilated room can lead to the formation of carbon monoxide which is poisonous.
CH4 (g) + 2 O2 (g)
2 CH4 (g) + 3 O2 (g)
CO2 (g) + 2 H2O (g)
Complete combustion
2 CO (g) + 4 H2O (g)
Incomplete combustion
N.B:


Combustion is both an oxidation & exothermic reaction.
Fuels are mainly carbon, hydrogen, hydrocarbon & alcohols  topic 9
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Redox reactions
A) Redox in terms of oxygen/hydrogen loss or gain
Oxidation reaction
Reduction reaction
Gaining of Oxygen
Removal of Oxygen
Removal of Hydrogen
Gaining of Hydrogen
A substance is said to be oxidized when it
gains oxygen/loses hydrogen during
reaction
A substance is said to be reduced when it
loses oxygen/gains hydrogen during
reaction
A substance which is oxidized is called
reducing agent (reductant)
A substance which is reduced is called
oxidizing agent (oxidant)
Examples for redox reactions:
Oxidixing agent (Oxidant):

Substance that will give/add oxygen to neighbouring reactants & gets reduced itself.

Examples: Oxygen (O2), Ozone (O3), Hydrogen peroxide (H2O2), potassium chromate VI
(K2CrO4), potassium dichromate VI (K2Cr2O7), potassium manganate VII (KMnO4).
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Reducing agent (Reductant):


Substance that will take/remove oxygen from neighbouring reactants & gets oxidized
itself.
Examples: Carbon, Carbon monoxide, Hydrogen, Reactive metals (Na, K, Ca), Potassium
iodide (Kl) & Ethanol.
B) Redox in terms of electron transfer & oxidation state
Oxidation reaction
Reduction reaction
Oxidation is loss of electrons (OIL)
Reduction is gain of electrons (RIG)
increase in oxidation state
decrease in oxidation state
From the above definitions we can deduce that
Reducing agents lose/give electrons.
Oxidizing agents gain/accept electrons.
Example focusing on electron transfer:



Word equation:
Magnesium + Sulfer
Symbol equation:
Mg + S
Ionic equation:
Mg + S
Magnesium Sulphide
MgS
Mg2+ + S2-
During this reaction, two electrons are transferred from magnesium to sulfur.
Magnesium ion and sulphide ion are formed. The magnesium is oxidized and the sulfur
is reduced
Mg
Mg2+ + 2e- (Magnesium loses 2 electrons) (Magnesium is the
reducing agent)
S + 2eS2- (Sulfur gain 2 electrons) (Sulfur is the oxidizing agent)
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Oxidation state / Oxidation number:








It's the number given to an element to show whether it has been oxidized or reduced.
It's the charge on its ion. Some transition elements have more than one oxidation
state.
The oxidation state is the roman number written after the element in a compound.
The oxidation state of an uncombined element – not in a compound – is always zero.
The oxidation state of iron in iron (II) oxide is 2 and in iron (III) oxide is 3.
The oxidation state of manganese in manganese (IV) oxide is 4 and in potassium
manganate (VII) is 7.
The oxidation state of chromium in chromium (III) oxide is 3 and in potassium
dichromate (VI) is 6.
Oxidation involves an increase in oxidation state
Fe2+
Fe3+
Reduction involves a decrease in oxidation state
Fe3+
Fe2+
Q: Indicate on the following equation which reactant is the oxidant
2 V3+ + Zn
2 V2+ + Zn2+
Q: Which change in the following equation is oxidation? Explain your choice.
V3+ + Fe3+
Mr. Fady Youssuf
V4+ + Fe2+
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Oxidation reduction summary
Oxidation
Reduction
Oxygen
Addition of Oxygen
Removal of Oxygen
Hydrogen
Removal of Hydrogen
Addition of Hydrogen
Electrons
Loss of electrons (OIL)
Gain of electrons (RIG)
Oxidation state
Increase in OS
Decrease in OS
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