Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Relative Atomic Mass & Relative Molecular Mass
 Relative atomic Mass (Ar) of an element
Ar
average mass of one atom of the element
= ---------------------------------------------------------------------mass of 1/12 of an atom of carbon-12
The values of the Ar for all the elements can be found in the Periodic Table.
e.g.
Q
Ar of carbon = 12;
Ar of Oxygen = 16;
What is the Ar of Calcium ? Ans [ 40.0 ] [40.0]
 Relative Molecular Mass (Mr) of a molecule
is the sum of the relative atomic masses, Ar of atoms in the molecule.
e.g. What is the relative molecular mass of ammonia ?
Mr of ammonia [NH3] = 14 + 3 x 1 = 17
Q
What is the Mr of carbon dioxide ?
Mr of carbon dioxide [
] = 44.0
 Relative Formula Mass (Mr) of an ionic compound
is the sum of the relative atomic masses, Ar of atoms in the compound.
e.g. What is the relative formula mass of magnesium sulfate ?
Relative formula mass of MgSO4 = 24 + 32 + 16x4 = 120
Q
What is the relative formula mass of calcium carbonate ?
Relative formula mass of calcium carbonate [
] = 100
Ar & Mr have NO UNITS.
By Mrs. Liew SG/Science Department /HCI
Page 1
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Composition
Sum of A r of all the atoms of the element
Percentage composition = ------------------------------------------------------------- x 100%
M r of the substance
e.g.
Calculate the percentage by mass of nitrogen in ammonium sulfate,
[(NH4)2SO4].
14x2
% by mass of Nitrogen = ----------------------------------- x 100% = 21.2%
(14+4) x 2 + 32 + 16x4
Q What is the percentage by mass of calcium in calcium carbonate ? [40.0%]
Sum of A r of all the atoms of the element
Composition = ------------------------------------------------------------- x mass of substance
By mass
M r of the substance
e.g. Calculate the mass of copper in 10.0 g of copper(II) oxide.
64
Mass of copper = ------------- x 10 = 8.00 g [express in 3 significant figures]
64 + 16
e.g. Calculate the mass of water in 12.5g of hydrated copper(II) sulfate.
[Hydrated copper (II) sulfate = CuSO4 .5H2O]
5 x [2x1 + 16]
Mass of water = -------------------------------------- x 12.5
64 + 32 + 16x4 + 5x 18
= 4.50 g [express in 3 significant figures]
Q What is the mass of silver in 20.0 g of silver nitrate, AgNO3 ? [12.7g]
By Mrs. Liew SG/Science Department /HCI
Page 2
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Mole Concept
 A mole is the amount of substance which contains 6.02 x 1023 particles.
(Particles : atoms, molecules, ions or electrons)
e.g.
One mole of carbon contains 6.02 x 1023 carbon atoms.
One mole of magnesium contains 6.02 x 1023 magnesium atoms.
One mole of oxygen molecules contains 6.02 x 1023 oxygen molecules.
One mole of sodium ion [Na+] contains 6.02 x 1023 sodium ions.
This number, 6.02 x 1023 is also called Avogadro’s number.
Mole is a Number
with NO UNITS
The symbol for mole is
“mol”.
e.g.
Determine the number of moles of aluminium atom in 1.80 x 1024 atoms of
Al.
Solution:
Since 1 mole = 6.02 x 1023
1.80 x 1024
1.80 x 1024 of aluminium atom contain 6.02 𝑥 1023 = 2.99 moles of aluminium atoms
e.g.
What is the number of magnesium atoms in 0.100 mole of magnesium ?
Solution:
Since 1 mole = 6.02 x 1023
0.100 mole of Mg = 0.100 x 6.02 x 1023 = 6.02 x 1022 magnesium atoms.
By Mrs. Liew SG/Science Department /HCI
Page 3
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
Questions:
(a) How many moles of carbon atoms are contained in one mole of
methane gas, CH4 ? [1.00 mole]
(b) How many moles of hydrogen atoms are contained in one mole of
methane gas, CH4 ? [2.40x 1024]
(c) How many moles of hydrogen atom is present in 1 mole of hydrogen
chloride gas ? [1.00 mole]
** (d) How many moles of hydrogen ion is present in 1 mole of sulfuric acid?
[2 moles]
**
(e) Will one mole of oxygen and one mole of nitrogen have equal mass ?
[No, please explain…. ]
(f) Will one mole of oxygen and one mole of nitrogen have equal no of
molecules ? {Yes, please explain…}
(g) Will one mole of oxygen and one mole of nitrogen have equal “relative
molecular mass” ? [No, please explain….]
Number of particles = number of moles x Avogadro’s number
By Mrs. Liew SG/Science Department /HCI
Page 4
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Molar Mass : The mass of one mole of any substance.
Molar mass of an element
= One mole of an element
= Ar of the element in grams
Example:
1 mole of carbon atom has a mass of 12 g
(i.e. molar mass of carbon atom = 12g)
Ar of carbon = 12
(without unit!)
Example:
What is the mass of 3 moles of carbon atoms ?
Ans:
Mass of 3 moles of carbon atoms
= no. of moles of carbon atoms x molar mass of carbon atoms
= 3 x 12g
= 36.0 g
Mass of the substance in grams
No. of moles of a substance = ------------------------------------------------Molar mass of the substance
i.e.
mass
no of moles = ------------------molar mass
Example:
1 What is the number of hydrogen atoms in 4.80 g of methane ?
Step 1: calculate the no of moles of methane
Step 2: calculate the no of moles of hydrogen atoms
Step 3: calculate the no of hydrogen atoms using Avogadro’s number
By Mrs. Liew SG/Science Department /HCI
Page 5
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
Step 1:
Mass of methane
4.80
No of moles of methane =---------------------------------- = -------------- = 0.300
Molar mass of methane
12 + 4x1
Step 2:
1 mole of CH4 contains 4 moles of hydrogen atoms
Hence
No of moles of hydrogen atom = 0.300 x 4 = 1.20 moles of hydrogen atoms.
Step 3:
No of hydrogen atoms = no of moles of hydrogen atoms x 6.02 x 10 23
= 1.20 x 6.02 x1023
= 7.22 x 1023
Questions:
1 Calculate the mass of the following:
(a) 3 moles of Helium atoms
Mass of He atoms
No of moles of He atoms = -------------------------------Molar mass of He
Mass of He atoms = 3 x molar mass of He = 3 x 4 = 12.0 g
(b) 4 moles of nitrogen atoms [56.0g]
(c) 4 moles of nitrogen gas. [nitrogen gas exist as diatomic molecules]
[112g]
(d) 1 mole of chlorine molecules. [71.0g]
By Mrs. Liew SG/Science Department /HCI
Page 6
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
(e) 12.0 x 1023 magnesium atoms.[48.0g]
(f) 1.50 x 1023 copper atoms. [16.0g]
Answers: 1 (a) 12.0g (b) 56.0g (c) 112g (d) 71.0g (e) 48.0g (f) 16.0g
2
(a) How many moles of sulfuric acid are contained in 4.90 g of H2SO4 ?
Mass of H2SO4
4.90
No of moles of H2SO4 = ---------------------------------- = --------------------------Molar mass of H2SO4
2x1 + 32 + 16 x 4
= 0.0500
(a) How many moles of calcium carbonate are contained in 20.0 g of calcium
carbonate ? [ans: 0.200]
**
3 What is the number of chlorine molecules in 35.5 g of chlorine gas ?
[ans: 3.00 x 1023 ]
By Mrs. Liew SG/Science Department /HCI
Page 7
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Calculating the Empirical Formula & Molecular formula of
a compound
 Empirical Formula: Simplest formula showing the ratio of atoms in the
compound.
 Molecular Formula: Real formula showing all the atoms in the molecule.
 Finding Empirical Formula:
Example:
30 g of aluminium sulfide contains 19.2g of sulfur. Find the
empirical formula of the compound.
Sol:
Type of atom
Al
S
30-19.2=10.8
19.2
27
32
10.8/27=0.4
19.2/32=0.6
Mole ratio
1
1.5
Mole ratio
(change to whole
no.)
2
3
Mass/g
Molar mass, g/mol
No of moles
Hence, the empirical formula for aluminium sulfide is __Al2S3.
THINK: What if the calculated mole
ratio is 1: 1.3 ?
By Mrs. Liew SG/Science Department /HCI
Page 8
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
Question:
1. CFCs are compounds that contain only carbon, chlorine and fluorine.
They are atmospheric pollutants and destroy ozone in the upper atmosphere.
“CFC11” has the following composition, by mass:
C, 8.7%; F, 13.8%;
Cl, 77.5%
Calculate the empirical formula of CFC11.
[ans: CFCl3]
Type of atom
C
F
Cl
Mass/g
(in 100g)
Molar mass,
g/mol
No of moles
8.7
13.8
77.5
12
19
35.5
8.7/12
13.8/19=0.726
77.5/35.5=2.18
1
3
=0.725
Mole ratio
1
Mole ratio
(change to
whole no.)
Hence the empirical formula of “CFC11” is _________.
2 An oxide of titanium contains 60% by mass of titanium (Ti).
What is the empirical formula for this oxide ? [ans: TiO2 ]
By Mrs. Liew SG/Science Department /HCI
Page 9
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Finding Molecular Formula:
Molecular Formula = n x Empirical Formula
Example:
Lithium forms a compound with composition 8.00% lithium, 36.8% sulfur and 55.2%
oxygen.
(a) Find the empirical formula of this compound.
(b) The relative molecular mass of the compound is 174. Find the molecular formula
of the compound.
Sol:
(a)
Li
S
O
Mass [in 100g]
8
36.8
55.2
Molar mass
7
32
16
8/7 =1.14
36.8/32 = 1.15
55.2/16 = 3.45
1
1
3
Mols
Mole ratio
Hence empirical formula = LiSO3
(b)
Molecular formula = n x empirical formula
174 = n x Mr of empirical formula = n x [ 7 + 32 + 16x3] = n x 87
n = 174/87 = 2
Hence the molecular formula is (Li2SO3 )x 2 = Li2S2O6
By Mrs. Liew SG/Science Department /HCI
Page 10
Hwa Chong Institution

CHEMISTRY
Sec 3 IP
Molar Volume of gases
Avogadro’s Law states that equal volume of all gases at the same temperature
and pressure contain the same number of particles.
Molar gas volume
= 24 dm3
One mole of any gas occupies 24 dm3 (24000 cm3) at room
temperature and pressure (r.t.p.)
Example:
A sample of carbon monoxide gas, CO, has a volume of 6.00 dm3 at r.t.p,
(a) How many moles of gas are there in the sample ?
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝑪𝑶
No of moles of CO = 𝒎𝒐𝒍𝒂𝒓 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝑪𝑶
no. of moles of CO = 6/24 = 0.250
(b) If the number of particles in one mole is 6.02 x 1023, how many molecules
are there in the sample ?
no of CO molecules = 0.25 x 6.02 x 1023 = 1.51 x 1023
(c) What is the mass of the sample of CO gas?
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝑶
No of moles of CO = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝑶
mass of CO = 0.25 x (12+16) = 7.00g
By Mrs. Liew SG/Science Department /HCI
Page 11
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Calculation from Equations – Reacting masses & volumes.
Stoichiometry :
The relationship between the amounts (measured in moles) of reactants and products
involved in a chemical reaction is known as the stoichiometry of the reaction.
Example :
A cook puts 4.2 g of sodium hydrogen carbonate into a cake mixture.
2 NaHCO3 (s) →
Na2CO3 (s) + CO2 (g) + H2O (l)
(i) Calculate the mass of the solid obtained.
Step 1:
No of moles of NaHCO3 = 4.2/(23+1+12+16x3) = 0.0500
Step 2:
From the equation,
Mole ratio of NaHCO3 : Na2CO3 = 2:1
[2 moles of NaHCO3 produces 1 mole of Na2CO3]
No of moles of Na2CO3 =
0.05
2
= 0.0250
Step 3:
Since
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂 𝑪𝑶
No of moles of Na2CO3 =𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇𝟐𝑵𝒂 𝟑𝑪𝑶
𝟐
𝟑
Hence mass of Na2CO3 = 0.025 x (23x2+12+16x3) = 2.65g [in 3 sf]
(ii) Calculate the volume of carbon dioxide given off when the mixture is heated
at room temperature and pressure.
From the equation,
Mole ratio of NaHCO3 : CO2 = 2:1
No of moles of CO2 =
Since
0.05
2
= 0.0250
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝑪𝑶
No of moles of CO2 =𝒎𝒐𝒍𝒂𝒓 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇𝟐𝑪𝑶
𝟐
Volume of CO2 = 0.025 x 24.0 = 0.600 dm3 = 600 cm3 [in 3sf]
By Mrs. Liew SG/Science Department /HCI
Page 12
Hwa Chong Institution

CHEMISTRY
Calculation involving Limiting Reactant
Sec 3 IP
(It limits or determine the
amount of products formed)
Reactant that is
completely consumed in a
reaction.
Example
A mixture containing 8g of iron filings and sulfur is heated to form iron(II) sulfide.
(a) Write the chemical equation to form iron(II) sulfide.
Fe + S → FeS
(b) One of the elements is a limiting reactant. Find out which is the element.
No of moles of Fe = 8/56 = 0.143
No of moles of S = 8/32 = 0.250
From the equation,
1 mole of Fe react with 1 mole of S,
0.143 mole of Fe will react with 0.143 mole of S.
Hence Sulfur is in excess and iron is the limiting reactant.
(c) Calculate the mass of iron(II) sulfide formed.
From the equation,
1 mole of Fe produces 1 mole of FeS,
Hence 0.143 mole of Fe will produce 0.143 mole of FeS.
Since
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑭𝒆𝑺
No of moles of FeS = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑭𝒆𝑺
Mass of FeS = 0.143 x (56+32) = 12.6g
By Mrs. Liew SG/Science Department /HCI
Page 13
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Calculation involving Percentage Yield
Experimental yield of product
Percentage Yield = ------------------------------------------- x 100%
Theoretical yield of product
Example
Millions of tonnes of urea are produced annually for use as fertilizers. The raw materials
used to make urea are ammonia and carbon dioxide.
2 NH3 (g) + CO2 (g) →
CO(NH2) 2 (s) + H2O (l)
If 48.8g of urea are produced when 1 mole of carbon dioxide reacts with excess
ammonia, what is the percentage yield of urea in this reaction ?
Sol:
From the equation,
1 mole of CO2 produces 1 mole of CO(NH2) 2
Since
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝑶(𝑵𝑯 )
𝟐 𝟐
no of moles CO(NH2)2 = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝑶(𝑵𝑯
𝟐 )𝟐
mass of CO(NH2) 2 produced = 1x (12+16+14x2+2x2) = 60.0g
Percentage yield of CO(NH2) 2 =
𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
x 100%
48.8
= 60.0 x 100%
= 81.3 %
By Mrs. Liew SG/Science Department /HCI
Page 14
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Calculation involving Percentage Purity
Mass of pure substance
Percentage Purity = ------------------------------------ x 100%
Mass of sample
Example:
Household bleach, hydrogen peroxide, decomposes to form water and oxygen.
2H2O2 (aq) → 2H2O (l) + O2 (g)
68g of hydrogen peroxide decomposed to give 1.2 dm3 of oxygen.
What was the percentage purity of the hydrogen peroxide?
Sol:
Since
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝑶
no of moles of O2 =𝒎𝒐𝒍𝒂𝒓 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇𝟐 𝑶 =
𝟐
1.2
24
= 0.0500
From the equation,
Mole ratio of H2O2 : O2 = 2:1
No of moles of H2O2 = 2 x 0.05 = 0.100
Since
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑯 𝑶
no of moles of H2O2 = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇𝟐 𝑯𝟐 𝑶
𝟐 𝟐
Mass of H2O2 = 0.100 x (2x1+16x2) = 3.40 g
% purity of H2O2 =
=
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑯𝟐 𝑶𝟐
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆
3.40
68
x 100%
x 100%
= 5.00 %
By Mrs. Liew SG/Science Department /HCI
Page 15
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 The concentration of a solution
Concentration (g/dm3)
=
Mass of solute in grams
Volume of solution in dm3
Concentration in mol/dm3
=
No. of moles of solute
Volume of solution in dm3
=
Concentration (g/dm3)
Molar mass of solute
OR Molarity
Concentration (mol/dm3)
Note: 1 dm3 = 1000 cm3 = 1000 ml
Example:
A patient took some tablets containing magnesium oxide to relieve his gastric pain.
Each tablet contains 3.0 g of magnesium oxide. The pain was caused by an excess
amount of hydrochloric acid in his stomach. If his stomach contains the equivalent of
100 cm3 of excess hydrochloric acid of concentration 3.00 mol/dm3.
How many tablets should he take to relieve the pain ?
Sol:
MgO (s) + 2 HCl (aq) →
Since
MgCl2 (aq) + H2O (l)
No of moles = Molarity x volume
No of moles of HCl = 3.00 x (100/1000) = 0.300
From the equation, Mole ratio of MgO : HCl = 1 : 2
No of moles of MgO = 0.3/2 = 0.150
Mass of MgO needed = no of moles of MgO x molar mass of MgO
= 0.150 x (24+16)
= 6.00g
No of MgO tablet needed = 6/3 = 2
By Mrs. Liew SG/Science Department /HCI
Page 16
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
 Volumetric Analysis
Titration
experiments involve the measurement of volumes of solutions.
This technique is also called volumetric analysis.
In volumetric analysis, a standard solution is used to determine the
concentration of an unknown solution.
A standard solution is a solution of known concentration.
Example 1
A household ammonia solution was analysed to determine its ammonia content.
25.0cm3 of the ammonia solution required 21.9 cm3 of 0.110 mol/dm3 sulphuric acid to
achieve the end-point titration. Calculate the concentration of ammonia, in g/dm 3, in the
household ammonia solution.
Sol:
H2SO4(aq) + 2NH3(aq) → (NH4)2SO4 (aq)
From the equation,
No of moles of H2SO4
No of moles of NH3
(MV) of H2SO4
=
(MV) of NH3
Molarity of NH3
=
1
2
0.11 x 21.9/1000
M of NH3 x 25.0/1000
=
2 x 0.11 x 21.9
25.0
=
0.1927 mol/dm3
=
1
2
Since
Concentration of NH3 solution
= Molarity of NH3 solution x molar mass of NH3
Hence
Concentration of NH3 solution
= 0.197 x (14 + 3 x 1)
= 3.35 g/dm3
By Mrs. Liew SG/Science Department /HCI
Page 17
Hwa Chong Institution
CHEMISTRY
Sec 3 IP
Example 2
16.6g of a metal carbonate, M2CO3, was made up to 1000cm3 of aqueous solution. 25.0
cm3 of this solution required 30.0 cm3 of 0.200 mol/dm3 HCl for complete reaction.
(a) Write the equation for the reaction between M2CO3 and HCl.
(b) Calculate the number of moles of HCl used in this reaction.
(c) Calculate the number of moles of M2CO3 present in
(i) 25.0 cm3 of solution.
(ii) 1 dm3 of solution.
(d) Calculate
(i) the relative atomic mass of M.
(ii) the relative formula mass of M2CO3.
(e) Identify the metal M.
 THE END 
By Mrs. Liew SG/Science Department /HCI
Page 18