The document provides information about chemical formulas and equations. It defines empirical and molecular formulas, and explains how to determine them through calculation of moles and mass ratios of elements in a compound. It also describes writing and balancing chemical equations, naming ionic compounds based on their constituent ions, and using formulas and equations to solve stoichiometric problems. Key topics covered include determining formulas from experimental data, relating formulas to molecular structure and mass, and representing chemical reactions systematically.

1. Chemistry
Chapter 3 Chemical
Formula & Equations
-Ms Cindy-

2. 2
Get Ready Your

3. What is chemical formula?
3
Representation of a chemical substance using letters for atoms and
subscript numbers to show the number of each type of atoms that are
present in the substance.

4. 4
What is chemical formula?
Molecular formula Empirical formula
Formula that show the
actual number of atoms of
each element present in
one molecule of the
compound.
Formula that shows the
simplest whole number
ratio of atoms of each
element present in the
compound.
Compound Molecular
formula
Simplest whole
number ratio
Empirical
formula
Water H20 H:O = 2:1 H20
Benzene C6H6 C:H = 1:1 CH
Vitamin C C6H8O6 C:H:O = 3:4:3 C3H4O3

5. 5
Determining empirical formula
Determining the mass
of each element in the
compound.
Converting the mass of
each element into the
number of moles of
atoms.
Determining the
simplest ratio of number
of moles of atoms of
each element.
1 2
3
𝒎𝒂𝒔𝒔(𝒈)
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀

6. 6
Determining empirical formula
Example
1.08g of aluminium powder combines completely with 0.96g of oxygen
to produce an oxide compound. What is the empirical formula of the
oxide? [Relative atomic mass: O, 16; Al, 27]
Element Aluminium, Al Oxygen, O
Mass (g) 1.08 0.96
Number of
moles of atoms
1.08
27
= 0.04
0.96
16
= 0.06
Ratio of
number moles
0.04
0.04
= 1
0.06
0.04
= 1.5
Simplest ratio
of number of
moles
2 3
Solution
Al2O3

7. 7
Determining empirical formula
Procedure:
1. A crucible with its lid is weighed.
2. Mg ribbon is cleaned with sand paper to remove its oxide
layer.
3. The ribbon is put into the crucible covered with lid, and
weighed.
4. When the burning is complete, the lid is removed and the
crucible is heated strongly for 1 to 2 minutes.
5. After cool down, the crucible, its lid and content are weighed
again.
Results:
Items Mass
Crucible + lid X
Crucible + lid + mg ribbon Y
Crucible + lid + magnesium oxide z
Element Mg O
Mass
Number of moles of atoms
Simplest ratio of number of
moles of atom
Empirical formula = MgpOq

8. 8
Check Point
1. A compound contains 3.06% of hydrogen, 31.63% of phosphorus and 65.31% of
oxygen. What is the empirical formula of the compound?
[RAM: O, 16; H, 1; P, 31]
Element Hydrogen, H Phosphorus, P Oxygen, O
Mass 3.06 31.63 65.31
Number of
moles of atoms
3.06
1
= 3.06
31.63
31
=1.02
65.31
16
=4.08
Ratio of
number moles
4.08
1.02
= 4
3.06
1.02
= 3
1.02
1.02
= 1
Simplest ratio
of number of
moles
3 1 4
H3PO4

9. 9
Determining molecular formula
• Molecular formula is a multiple of empirical formula, where n is
an integer.
Molecular formula = (Empirical formula)n
Compound Empirical Formula n Molecular formula
Water H20 1 (H20)1 = H20
Ethane CH3 2 (CH3) 2 = C2H6
Glucose CH2O 6 (CH2O) 6 = C6H12O6
• The determination of a molecular formula requires the following information:
 Its empirical formula
 Its relative molecular mass or its molar mass

10. 10
Determining molecular formula
Example
A compound has an empirical formula of CH2 and a relative
molecular mass of 70. What is the compound’s molecular formula?
[Relative atomic mass: H,1; C, 12]
Solution
Let the molecular formula of the compound to be (CH2)n.
(CH2)n = n[12 + 2(1)]
= 14n
So, 14n = 70
n =
70
14
= 5
Therefore, the molecular formula of the compound is (CH2) 5 , which is C5H10.

11. 11
Check Point
(a) What is meant by empirical formula? [1 mark]
(b) A carbon compound contains 92.3% of carbon and 7.7% of hydrogen by
mass. The relative molecular mass of this compound is 78. Find the
molecular formula of this compound. [RAM: C = 12; H =1] [5 marks]
(c) Describe how you could determine the empirical formula of magnesium
oxide in the laboratory. Your description should include
- the procedure of the experiment
- tabulation of the data
- the calculation of the results
[10 marks]
Empirical formula is a chemical formula that shows the simplest ratio of the
number of atoms for each element in the compound.
Element C H
Mass
No. of moles
Ratio of moles
Empirical formula = CH
Relative molecular mass = (CH)n
(CH)n = 78
(12+1)n = 78
13n = 78
n = 6 Molecular formula: C6H6
92.3% 7.7%
92.3/12 = 7.7 7.7/1 =7.7
1 1

12. 12
Calculation involving empirical
and molecular formula
• Empirical formula and molecular formula can help us to solve
calculation problems related to the composition of compounds.
Example
What is the mass of zinc that is needed to combine with 0.5 mole of chlorine
atoms to form zinc chloride, ZnCl2 ? [Relative atomic mass: Zn, 65]
Solution
Number of moles of Cl = 0.5
mass?
Zn + Cl2  ZnCl2
0.5 mole
Therefore, number of moles of Zn = 0.5÷ 2
= 0.25
Number of moles =
𝒎𝒂𝒔𝒔
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
0.25 =
𝒎𝒂𝒔𝒔
𝟔𝟓
mass = 0.25 x 65
= 16.25g

13. 13
Calculation involving empirical
and molecular formula
The percentage composition based on mass can be calculated as follows:
Percentage of an element by mass in a compound = Mass of the element in 1
mole of the compound
Mass of 1 mole of the
compound
X 100%
Example
Calculate the percentage by mass of carbon in octane, C8H18 [RAM: H, 1; C, 12]
Solution
The mass of 1 mole of octane = 8(12) +18(1)
=114
The mass of carbon in 1 mole of octane = 8 x 12
= 96
So, the percentage by mass of carbon =
96
114
x 100% = 84.21%

14. 14
Chemical formula of ionic compound
Cation Anion+
Formula of ionic compound:
Sodium ion Na+
Magnesium ion Mg2+
Aluminium ion Al3+
Lead(IV) ion Pb4+
Chloride ion Cl-
Hydroxide ion OH-
Oxide ion O2-
Phosphate ion PO4
3-

15. 15
Chemical formula of ionic compound
List of Cations and Anions

16. 16
Chemical formula of ionic compound
• The formula of an ionic compound combines the formula of its cation
(positive ion) and the formula of its anion (negative ion).
• However, the formula of an ionic compound is neutral because the
total positive charges equal the total negative charges.
Example
Iron (III) chloride
Fe3+
Cl-
FeCl3

17. 17
Chemical formula of ionic compound
• When polyatomic ions such as SO4
2-, OH-, NO3
- , NH4
+ are involved,
brackets are used to show the number of ions in the formula.

18. 18
Naming of chemical compounds
• Chemical compounds are named systematically according to the
guidelines given by the International Union of Pure and Applied
Chemistry (IUPAC).
Name of cation Name of anion
Magnesium sulphate
Chemical
formula
Name
CO Carbon monoxide
CO2 Carbon dioxide
SO3 Sulphur trioxide
PCl5 Phosphorus pentachloride
MgSO4
Chemical
formula
Name
Ca(OH) Calcium hydroxide
KCl Potassium chloride
FeCl3 Iron (III) chloride
NaBr Sodium bromide

19. 19
Check Point
Name each of the following ionic compound.
(a) Al(OH) 3
(b) FeSO4
(c) NH4Cl
(d) Ca(NO3) 2
(e) K2CO3
(f) ZnS
Aluminium hydroxide
Iron(II) sulphate
Ammonium chloride
Calcium nitrate
Potassium carbonate
Zinc sulphide

20. 20
Chemical Equations
• Describe a chemical reaction.
• Can be written in words or chemical formulae.
2 H2 (g) + O2 (g)  2 H2O (l)
reactants products
• The state symbols (s), (l) and (g) represent the solid, liquid and
gaseous states respectively. The symbol (aq) represents the
aqueous solution.

21. 21
Chemical Equations
• According to the law of conservation of mass, a matter can neither
be created nor destroyed. So, the numbers of atoms before and
after a chemical reaction are the same. Therefore, a chemical
equation must be balanced.
Example
Iron fillings react with copper(II) chloride solution to form iron (III) chloride and
copper metal.
Fe CuCl2+ FeCl3 + Cu3 22 3
(2mol) (3mol) (2 mol) (3 mol)

22. 22
Chemical Equations
Example
What is the mass of sodium needed to react with 0.5 mole of chlorine?
[Relative atomic mass: Na, 23]
Solution
Na + Cl2  NaCl22
0.5molemass ?
So, number of mole of Na = 1 mole
Number of moles =
𝑚𝑎𝑠𝑠
𝑅𝐴𝑀
1 =
𝑚𝑎𝑠𝑠
23
mass = 23 x 1
= 23g

23. 23
SPM Target
Question 1
Sodium thiosulphate has the formula of Na2S2O3. [RAM: Na, 23; S, 32; O, 16]
(a) What is the mass of sulphur in 0.2 mole of sodium thiosulphate?
(b) How many moles of oxygen atoms are present in 79g of sodium
thiosulphate?
Number of moles =
𝒎𝒂𝒔𝒔(𝒈)
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
0.2 =
𝒈
𝟐 𝟐𝟑 +𝟐(𝟑𝟐)+𝟑(𝟏𝟔)
0.2 =
𝒈
𝟏𝟓𝟖
g = 31.6
158  31.6g
2(32)  x g
2022.4 = 158 x
Let x be the mass of sulphur
X = 12.8g
Number of moles of sodium thiosulphate =
𝒎𝒂𝒔𝒔(𝒈)
𝑅𝐴𝑀 𝑜𝑟 𝑅𝑀𝑀
=
𝟕𝟗
𝟐 𝟐𝟑 +𝟐(𝟑𝟐)+𝟑(𝟏𝟔)
= 0.5
1 mol of Na2S2O3 can produce 3 atoms of
oxygen.
Therefore, number of moles of oxygen atoms
= 3 x 0.5
= 1.5 moles

24. SPM Target
24
Diagram shows the structural formula of vitamin C.
(a) What is the molecular formula of vitamin C?
(b) What is the empirical formula of vitamin C?
(c) State the molar mass of vitamin C.
(d) Calculate the percentage of oxygen by mass in a molecule of vitamin C.
Question 2
C6H8O6
C3H4O3
6(12) + 8(1) + 6(16) = 176g/mol
𝟔(𝟏𝟔)
𝟏𝟕𝟔
x 100% = 54.55%

25. 25
SPM Target
Equation for the reaction between potassium and oxygen:
K + O2  K2O
Question 3
24
What is the maximum mass of potassium oxide that is formed when 19.5g
potassium is burned completely in excess oxygen? [RAM: O, 16; K, 39]
19.5g Mass?
No.of moles of potassium = 19.5
39
=0.5mol
No.of moles of potassium oxide
= 0.5/2
=0.25mol
0.25 = mass
2(39) +16
Mass = 0.25 x 94
= 23.5g

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26

27. 27
Form 4: Chapter 6 Acid Bases and Salts
Form 5: Chapter 3 Oxidation and
Reduction
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