Notes chapter 1 SK015

TOPIC 1: MATTER

1.1 ATOMS & MOLECULES C1 C2 C3 C4
Learning outcomes √
(a) Write isotopic notation.
(b) Interpret mass spectrum. √√
(c ) Calculate the average relative mass of an element given the relatives
√
abundances of Isotopes or a mass spectrum.

Example?

Matter Substance Compound Molecule Atom Subatomic
s particle

Element

Mass number Atom: Sub-particle
Smallest unit of an element
Neutron: Electron:
Uncharged subatomic particle Atom Diagram A subatomic particle
found in nucleus of an atom and that made up an
has same mass as proton atom and carry
negative charge. It
Proton: moves rapidly
A subatomic particle found in the around nucleus of an
nucleus of atom and has positive atom
charge

SUBATOMIC PARTICLE?

Particle / symbol Mass (gram) Charge(Coulomb) Relative Charge
Electron / e 9.1 x 10-28 -1.6 x 10-19 -1
Proton / p 1.67 x 10-24 +1.6 x 10-19 +1
Neutron / n 1.67 x 10-24 0 0

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LO 1.1a - Write isotopic notation (C1) TOPIC 1: MATTER

Let see this story… Situation 3:

Situation 1: Situation 2:

What is the relationship????
They are isotope.

Definition of ISOTOPE:

Isotopes are 2 or more atoms of the same element with the same number of protons in the
nucleus but different number of neutrons.

Properties:

Have same chemical properties (same number of electrons determines the similar chemical
properties) but different physical properties (different number of neutrons determines the
physical properties such as boiling, melting, density etc)

Example 1:

Protium Deuterium Tritium

Isotopes of hydrogen 1 H 2 H 3 H Notes
Proton number 1 1 1
Isotopes of an element
Number of neutrons 1 1 1 have the same,

0 1 2 • proton number
• number of electrons
Isotopes of uranium Uranium-235 Uranium-238
Proton number U235 U238 in a neutral atom

92 92 Isotopes of an element
have different,
92 92
• nucleon number
• relative isotopic

mass

Number of neutrons 143 146 ISOTOPES

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TOPIC 1: MATTER

C1 HOW TO WRITE ISOTOPIC NOTATION?

Isotopic notation is a symbol used to designate a particular atom of an element.

** *ALSO KNOWN as nuclide symbol or atomic notation
Charge of ion:
Nucleon Number @ Mass Number :
charge No. of proton –
Total number of proton and neutron
present in nucleus of an atom A no. of electron

Nucleon no. = Proton no. + Atomic Symbol of Element:
Number of Neutron
Exercise : ZX According to the periodic table

Proton Number @ Atomic Number:

Total number of proton in nucleus of an atom

MOLECULE Molecule: Polyatomic Molecule:

Diatomic Molecule: 2 or more atoms Contains more than two
Contains only two form chemical atoms
bonds with each
atoms Eg: O3, H2O, C6H12O6
Eg: H2, O2, CO other
IONS

Cation (+ve) : Ion: Anion (-ve):

A positive charge An atom or group of A negative charge ion
ion formed when a atoms that loses or formed when a
neutral atom loses
gains electron neutral atom gains an
an electron(s) electron (s)

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TOPIC 1: MATTER

Example 1:

202 Hg 2+
80
2

 the right superscript shows that the total charge on the ion is 2+

 the right subscript shows that the ion is formed from two mercury atoms.

 the left superscript: nucleon number of mercury, A = 202

 the left subscript: proton number of mercury, Z = 80
the number of neutrons in a nucleus of mercury = 202 – 80 = 122
Example 2:

Write the appropriate notation for each of the following species

Species Proton Neutron Electron Isotope Notation
A 2 2 2
B 1 2 0
C 1 1 1
D 7 7 10

Example 3:

For the following nuclides:

136T , 14 R and 15 Z
6
7

a) write the nuclides that are isotopes.

136T and 14 R (reason: same proton number but different nucleon number)
6

b) state the atoms that have the same number of neutrons.

14 R (no. neutrons = 14 − 6 = 8) and 15 Z (no. neutrons =15 − 7 = 8)
6 7

Exercise 1:

1. Write The Isotope Notation of Each

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TOPIC 1: MATTER

2. The appropriate notation for each of the following species

Species Z A No. n No. p No. e Charge on the Isotope
M 24 species Notation

28 3+

Q 13 14 13 0

F 19 9 10

T 10 8 10

3. What is isotope? Write the isotope notations for the three isotopes of carbon;
carbon-12, carbon-13, carbon-14 and state the number of neutron for each
isotope.

4. Compare and contrast between the isotopes of an element by referring to Ne-20,
Ne-21 and Ne-22

5. By referring to the periodic table, determine the subatomic particles for each of
elements:
a) Carbon-14
b) Sodium-23
c) Cobalt-60
d) Phosphorus-32

6. By referring to the periodic table, complete the table below:

Symbol No. of proton No. of electron No. of neutron Charge Isotope notation
-
28000Hg -
-
C63 u -

29

O17 2−

8

2579Co3+

K+

Br-

Mg2+

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TOPIC 1: MATTER

LO 1.1a - Interpret mass spectrum. (C2 & C4)

TERMS DEFINITION / DESCRIPTION

➢ Masses of atoms expressed relative to mass of Carbon-12 atom

CARBON-12 SCALE ➢ What does it imply?
Mass 1 atom 12 C = 12 a.m.u , thus, 1/12 mass of 12 C = 1 a.m.u

ATOMIC MASS ➢ Mass of an atom in atomic mass unit
➢ Unit : a.m.u @ u

AVERAGE ATOMIC ➢ The weighted average of all naturally occurring isotopes of the element.
MASS, Ar ➢ Unit: a.m.u @ u

➢ The average atomic mass of one atom when compared to mass of 1/12
carbon-12

RELATIVE ATOMIC ➢ Formula: Relative Atomic Mass,
MASS, R.A.M
RAM = average atomic mass of one atom (a.m.u)
1/12 x mass of one atom of 12C (12.000 a.m.u)

➢ Unit : No unit

RELATIVE ➢ The average molecular mass of one molecule when compared to mass
MOLECULAR MASS, of 1/12 carbon-12

R.M.M ➢ Formula:

Relative Molecular Mass,

RMM = average molecular mass of one molecule (a.m.u)
1/12 x mass of one atom of 12C (12.000 a.m.u)

➢ Unit : No unit

ATOMIC MASS RELATIVE ATOMIC MASS (Ar)
Has unit (a.m.u @ u) No unit
Example: Example:

Sodium, Na Sodium, Na
Atomic mass: 23.0 a.m.u Relative atomic mass: 23.0

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TOPIC 1: MATTER

Let discuss this diagram…

Abundance :
Fraction or percentage (%)
of each isotope in a sample
of element.

HOW TO DETERMINE ISOTOPIC MASSES AND ABUNDANCE?

Technique used to determine isotopic Description:
masses & abundance is
✓ Technique used to determined relative atomic
MASS SPECTROMETRY mass and the relative abundance of isotopes

✓ Different atoms / isotopes / molecules can be
identified by their characteristic pattern of
lines (peak)

WHAT DEVICE USED TO DETERMINE RELATIVE ATOMIC MASS AND THE RELATIVE
ABUNDANCE OF ISOTOPES?

A device called MASS SPECTROMETER. Description:

✓ A device used to produce mass spectrum from
unknown/known sample

✓ Function of this mass spectrometer is used to
determine

i. atomic mass of an element
ii. molecular mass of a compound
iii. type of isotopes, the abundance and its

relative isotopic mass

iv. structure of compounds in an unknown

sample

Mass Spectrometer

C2 & C4

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TOPIC 1: MATTER

HOW TO DETERMINE RESULT FROM MASS SPECTROMETER?

Result data from detector called as Description:
MASS SPECTRUM.
✓ A plot of intensity vs mass (more correctly: mass-to-
Different atoms / isotopes / charge ratio) of a sample.
molecules can be identified by their
characteristic pattern of lines ✓ The plot represents the distribution of ions by mass.
(peak)

The plot represent distribution of ions ✓ The horizontal axis shows the m/e ratio or nucleon
by mass. number or isotopic mass or relative atomic mass of
y-axis : Abundance (% or relative) the ions entering the detector.
x-axis : Isotopic mass (a.m.u)
✓ The vertical axis shows the abundance or detector
current or relative abundance or ion intensity or
percentage abundance of the ions.

✓ The height is proportional to the amount of each
isotope present.

✓ The number of line represent the number of isotope
present in the sample.

*Information from a mass spectrum of an element

i.the isotopes which are present in the element

ii.the relative isotopic mass of each isotope

iii.the abundance of each isotope

Thus, the relative atomic mass of the element can be
determined

HOW TO ANALYSE MASS SPECTRUM?

Example: The mass spectrum of Mg • The height of each line is represent abundance of each
isotope

• How many isotopes of Mg? List

• According to mass spectrum, which one of three
isotopes is

i) Most abundance:
ii) Least abundance:

Exercise :

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TOPIC 1: MATTER

1. A sample of carbon dioxide gas that composes of isotopes 12C, 13C, 16O and 18O is
analysed using the mass spectrometer. How many peaks will be observed in the
mass spectrum?

2. Neon has three isotopes Ne (20), Ne (21) and Ne (22), each has % abundances
91.0%, 0.2% and 8.8% respectively. Draw the mass spectrum of Neon.

LO 1.1a - Calculate the average relative mass of an element given the relatives abundances of
Isotopes or a mass spectrum. (C4)

C4 HOW TO CALCULATE RELATIVE ATOMIC MASS FROM MASS SPECTRUM?

Step 1: Determine the average atomic mass.

Formula :
Average atomic mass = ⅀( ) @ ⅀( )
⅀ ⅀

Qi = relative abundance / percentage abundance of an isotope of element
mi = relative isotopic mass of the element

Step 2: Compare to mass of 1/12 of carbon-12

Formula :
Relative Atomic Mass,
RAM = average atomic mass of one atom (a.m.u)
1/12 x mass of one atom of 12C (12.000 a.m.u)

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Example 1: TOPIC 1: MATTER
Relative abundance
Calculate Isotopes

63 Mass Spectrum of Magnesium

9.1

8.1

0 24 25 26 m/e

The mass spectrum of magnesium shows that naturally occurring magnesium
consists of three isotopes: 24Mg, 25Mg and 26Mg.

The height of each line is proportional to the abundance of each isotope. In this
example, 24Mg is the most abundance of the three isotopes.

Average atomic mass = ⅀( )
⅀
Average atomic = (24 a.m.u. x 63) + (25 a.m.u. x 8.1) + (26 a.m.u x 9.1)

mass of Mg (63 + 8.1 + 9.1)

= 24.33 a.m.u

Example 2:

The ratio of relative abundance of naturally occurring of copper isotopes is as follow:
63Cu / 65Cu = 2.333

Based on the carbon-12 scale, the relative atomic mass of 63Cu=62.9396 and
65Cu=64.9278. Calculate the average atomic mass of copper.

Solution:

63Cu = 2.333

65Cu 1

Let’s assume

The abundance of Cu-63 = 2.333 and the abundance of Cu-65 = 1

Average atomic mass = ⅀( )
⅀

Average atomic mass Cu = (62.9396 x 2.333) + (64.9278 x 1)
(2.333 + 1)

= 63.54 a.m.u

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TOPIC 1: MATTER

Example 3:

The atomic masses of 6 Li and 7 Li are 6.0151 amu and 7.0160 amu respectively. What is
3 3

the relative abundance of each isotope if the relative atomic mass of lithium is 6.941 amu?

Solution:
Let the abundance of 6Li = w and the abundance of 7Li = 1-w

Average atomic mass = ⅀( )
⅀

6.941 = (6.0151 x (w)) + (7.0160 x (1-w))
w + (1-w)

w = 0.075

∴ Thus, the abundance of 6Li = 0.075 The abundance of 7Li = 1 – 0.075
= 0.925

Exercise :
1. Fig 1.3 shows the mass spectrum of the element rubidium, Rb.

a) What isotopes are present in Rb?
b) What is the percentage abundance of each isotope?
c) Calculate the relative atomic mass of Rb.

2. The mass spectrum of neon, Ne consists of three lines corresponding to m/e ratio of 20, 21
and 22 with relative intensities of 91.0, 0.2 and 8.8, respectively. Calculate the relative
atomic mass of Ne.

3. Naturally occurring iridium, Ir is composed of 2 isotopes 191Ir and 193 Ir in the ratio of

5:8. The relative isotopic mass of 191Ir and 193 Ir are 191.021 and 193.025

respectively. Calculate the relative atomic mass of iridium. Ans:(192.254)

Extra exercise:

1. The ratio of relative abundance of naturally occurring of chlorine isotopes as follow:

35 = 3.127. The relative atomic mass of 35Cl = 34.9689 and 37Cl = 36.9659. Calculate

37

relative atomic mass of Cl.

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TOPIC 1: MATTER

2. The relative isotopic mass of 6Li and 7Li are 6.01 and 7.02 respectively. What is the
percentage abundance of each isotope if the relative atomic mass of Li is 6.94?

3. Silver, Ag has 46 unknown isotopes, but only two occur naturally, 107Ag and 109Ag.
Given the following mass spectrometer data, calculate the abundance of each isotope
if relative atomic mass of Ag is 107.87?

Isotope Relative atomic mass
107Ag 106.90509
109Ag 108.9476

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TOPIC 1: MATTER

1.2 MOLE CONCEPT
INTRODUCTION

A pair : 2 things
A dozen : 12 things
A mole : 602, 000, 000, 000, 000, 000, 000, 000 things

(602 hexillion)
A mole = 6.02 x 1023 particles

Conclusion:
*Mole: unit of measurement for the amount of
substance or chemical amount.

MOLE Amount of substances that contains the same number of particles (atoms,
molecules, formula units, ions) as there are atoms in exactly in 12 g of C–12

AVOGADRO’S CONSTANT
The number of particles in one mole of substance that
contains 6.02 x 1023 atoms.

1 mole = 6.02 x 1023 particles

i. Conversion between moles and no of atoms & molecules

Remember : 1 mole of any element ≡ 6.02 x 1023 particles
≡ 6.02 x 1023 atoms
≡ 6.02 x 1023 molecules

Example 1:

1 mole of students in KMS =
2 mole of students in KMS =

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TOPIC 1: MATTER

Example 2:
1 mole of sulphur atoms ≡
How many atoms are in 5.5 moles of sulphur atoms?

How many moles in 4.6 x 1024 sulphur atoms?

Example 3:
1 mole of oxygen molecules ≡
How many oxygen atoms in 1 mole of oxygen molecules?

Example 4:
1 mole of H20 molecules ≡
How many oxygen atoms in 1 mole of H20 molecules?
How many hydrogen atoms in 1 mole of H20 molecules?
How many atoms in 1 mole of H20 molecules?

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TOPIC 1: MATTER

EXERCISE:
1. Calculate the number of H atom in 1 mole of NH3

2. Calculate the number of bromide ions in 2 moles of CaBr2

3. Calculate the no. of mole of CO2 which has 2.88x1016 molecules CO2

4. Calculate the no. of mole of oxygen atom in 2.88x1016 molecules CO2

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TOPIC 1: MATTER

ii. Conversion between moles and mass
Remember: 1 mole of any element ≡ mass of element (g)

Terms Definition / Description
Molar mass
• The mass of one mole of an any substance is referred to as the
molar mass (The mass of 6.02 x 1023 particles)

• Unit : gmol-1

The molar mass of any substance is numerically equal to its relative mass;
➢ relative atomic mass,
➢ relative molecular mass or
➢ relative formula mass.

RELATIONSHIP

Mole Particles Molar mass Atomic mass / Molecular mass
1 mole of C 12 gmol-1 12 a.m.u
6.02 x 1023 atom C 18 gmol-1 18 a.m.u
atom 57.5 gmol-1 57.5 a.m.u
6.02 x 1023 molecule
1 mole of H20 H20
molecule
6.02 x 1023 formula unit
1 mole of NaCl NaCl
formula unit

EXERCISE:
1. Sample has 14 g of N2 gas. Determine:

i) The number of mole
ii) Number of molecules
iii) Number of atoms

2. Calculate number of C, H and O atoms in 1.50 g of glucose, C6H12O6

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TOPIC 1: MATTER

iii. Conversion between moles and volume

Remember: 1 mole of any gas ≡ volume of gas at s.t.p or r.t.p (L)

Terms Definition / Description
Molar volume
• The volume occupied by one mole of any gas at s.t.p or r.t.p
• Unit : Lmol-1

Condition: At r.t.p @ room temperature Note:
At s.t.p Temperature : 25°C 1 dm3 = 1000 cm3
Temperature : 0°C Pressure : 1 atm
Pressure : 1 atm Molar volume : 24.0 Lmol-1 1 dm3 = 1L
Molar volume : 22.4 Lmol-1 1 cm3 = 1 mL
1 L = 1000 mL

RELATIONSHIP

Mole Particles (only gas) Molar volume (at s.t.p) Molar volume (at r.t.p)
6.02 x 1023 molecule O2 22.4 L mol-1 24.0 L mol-1
1 mole of O2
molecule 6.02 x 1023 molecule 22.4 L mol-1 24.0 L mol-1
CO2
1 mole of CO2
formula unit

EXERCISE:

1. A sample of CO2 has a volume of 56 cm3 at STP. Calculate:
i. The number of moles of gas molecules.
ii. The number of molecule.

iii. The number of oxygen atoms in the sample.

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TOPIC 1: MATTER

KEEP IN MIND!!!!

TRY THIS!! (MOLE)
1. Molecule Ca3(PO4)2. Determine

i) Number of Oxygen atoms
ii) Number of Ca2+ ions
iii) Number of PO43-

2. How many molecules of ethane, C2H6 are present in 0.334g of C2H6?

3. Iron (Fe), the main component of steel is the most important metal in industrial
society. How many Fe atoms are in 95.8 g of Fe?

4. X gas occupies 2.24 dm3 at STP. Calculate number of mole for X gas.

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TOPIC 1: MATTER

1.2 EMPIRICAL AND MOLECULAR FORMULA

Learning outcomes C1 C2 C3 C4

(a) Define the terms empirical and molecular formulae √

(b) Determine empirical and molecular formulae from mass composition √
or combustion data

(c ) Determine empirical and molecular formulae from experiment. √

LO 1.2a - Define the terms empirical and molecular formulae. (C1)

C1 TERM DEFINITION
Chemical formula that shows the simplest ratio of atoms of the
EMPIRICAL
FORMULA elements present in a compound.
MOLECULAR
FORMULA Chemical formula that shows the actual number of atoms of

each element in a molecule.

LO 1.2b - Determine empirical and molecular formulae from mass composition or
combustion data. (C3)

The relationship between empirical formula and molecular formula is:
Molecular formula = n (empirical formula)

To determine the value of n: n = Molar mass of molecular formula
Mass of empirical formula

C3

How to determine the empirical formula?
1. By using the masses of compounds
2. By using percentage of composition of a compound
3. By using the combustion data (elemental analysis data)

**Take note!!!
➢ Combustion of hydrocarbons, CxHy in excess oxygen produce CO2 and H2O
➢ Combustion of organic compound, CxHyOz in excess oxygen produce CO2 and H2O

Steps to answer: Empirical formula

Step 1: Find mass of each element
Step 2: Calculate no. of mole of each element
Step 3: Calculate simplest ratio
Step 4: Determine empirical formula
Step 5: Determine molecular formula

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TOPIC 1: MATTER

METHOD 1: By using the masses of compounds

EXAMPLE 1:

18.3 g sample of hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of

water only. Calculate its empirical formula.

Element Ca Cl H2O

Mass (g) 4.0 7.1 7.2

Number of moles (mol) 4.0 = 0.10 7.1 = 0.20 7.2 = 0.40
Simplest ratio of moles 40.0 35.5 18

0.10 = 1.0 0.20 = 2.0 0.40 = 4.0
0.10 0.10 0.10

1 2 4

Empirical Formula Empirical formula = CaCl2.4H2O

METHOD 2: By using percentage of composition of a compound

EXAMPLE 2:

Ascorbic acid (vitamin C) cures scurvy and may help to prevent the common cold. It is
composed of 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. The molar mass
of ascorbic acid is 176 g mol-1. Determine the empirical formula and molecular formula.

Element C H O
Mass (%) 40.92 4.58 54.50
Number of moles
40.92 = 3.41 4.58 54.50 = 3.41
(mol) 12 16
Simplest ratio of 4.58 = 1.33
3.41 = 1 3.41 = 1
moles 3.41
3.41 3.41
1.33x3
1x3 4 1x3
3 3

Empirical formula = C3H4O3 ∴ Molecular formula = (C3H4O3)2
= C6H8O6
Molecular formula = (C3H4O3) n
n = molar mass
empirical formula mass
= 176 = 2
88

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TOPIC 1: MATTER

METHOD 3: By using the combustion data (elemental analysis data)

EXAMPLE 3:
1.00 g sample of organic compound A was burnt in excess of oxygen. The combustion
produced 2.52 g of CO2 and 0.443 g of H2O. Determine the empirical formula of the organic
compound A.

Solution: In 1 mol of CO2 there is 1 mol of C

m = 2.52 g CO x 12.0 g C = 0.688 g C
C 2 44.0 g CO

2

In 1 mol of H2O there are 2 moles of H atoms

m = 0.443 g H O x 2 g H = 0.0492 g H
H 2 18.0 g H O

2

the mass of oxygen = mass of sample – (mass of C + mass of H)
= 1.00 g – (0.688 g + 0.0492 g)
= 0.263 g O

Element C H O
Mass/g 0.688 0.0492 0.263

Amount/mol 0.688 = 0.0573 0.0492 = 0.0492 0.263 =
12.0 1.00 16.0
0.0164

Simplest ratio of relative 0.0573 = 3.49 0.0492 = 3.00 0.0164 = 1.00
amount 0.0164 0.0164 0.0164

3.49 x 2 3.00 x 2 1.00 x 2
7 6 2

∴ Empirical formula = C7H6O2

TRY THIS!! (EMPIRICAL FORMULA & MOLECULAR FORMULA)

Exercise 1:

1) The combustion of 0.146 g of compound B gave 0.374 g of carbon dioxide and 0.154
g of water. Assuming that B contains carbon, hydrogen and oxygen only, determine
its empirical formula.

2) An organic compound, X which contains only carbon, hydrogen and oxygen, has a
molar mass of about 85 g mol−1. When 0.43 g of X is burnt in excess oxygen, 1.10 g
of carbon dioxide and 0.45 g of water are formed.

a) What is the empirical formula of X?
b) What is the molecular formula of X?

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TOPIC 1: MATTER

3) A compound Y with chemical formula as shown below:
CH2=CHCOOCH3

a) Write the empirical formula and molecular formula of the compound.
b) Calculate the percentage composition of carbon in the compound X.

Ans:(55.8%)
4) A complete combustion of a hydrocarbon compound yields 2.64 g of carbon dioxide

and 0.540 g of water. What is the empirical formula of the hydrocarbon? If the
relative molecular mass of the hydrocarbon is 78, what is the molecular formula?
LO 1.2c - Determine empirical and molecular formulae from experiment (C3)

C3

*** Experiment 1

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TOPIC 1: MATTER

1.2 CONCENTRATION MEASUREMENT

Learning outcomes C1 C2 C3 C4
√
(d) Define each of the following concentration measurements:
i.Molarity (M) ii.Molality (m) iii.Mole fraction (X) √√
iv.Percentage by mass(%w/w) v.Percentage by volume (%v/v)

(e) Calculate each of the following concentration measurements:
i.Molarity (M) ii.Molality (m) iii.Mole fraction (X)
iv.Percentage by mass(%w/w) v.Percentage by volume (%v/v)

A cup of coffee
Is it a SOLUTION???

What is SOLUTION???

A CUP OF COFFEE COFFEE POWDER HOT WATER

+

TERMS DEFINITION
SOLUTE The substance present in a smaller amount in a solution
SOLVENT The substance present in a larger amount in a solution
A homogeneous mixture formed when an amount of solute dissolve
SOLUTION completely in a solvent

Density of solution , ρ =
mass of solution (g)

Volume of solution (mL)

* Unit: gmL-1

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TOPIC 1: MATTER

CONCENTRATION OF SOLUTION

C1 Terms Definition / Description
➢ Molarity of a solution is defined as number of moles of solute in 1 L solution
Molarity, M

➢ Formula: Molarity, M = Mole of solute (mol)
Volume of solution (L)

➢ Unit: mol L-1 @ mol dm-3 @ molar

Molality, m ➢ Molality of a solution is defined as number of moles of solute required per
kilogram of solvent in solution

➢ Formula: Molality, m = Mole of solute (mol)
Mass of solvent (kg)

➢ Unit: mol kg-1 @ molal

** Mass of solution = mass of solute + mass of solvent
** Volume of solution ≠ Volume of solvent

➢ Mole fraction of a solution is defined as ratio number of moles of one
component to the total number of moles of all components present.

➢ Formula:

Mole Fraction, X Mole fraction of component A, XA = Mole of A (mol)

Total mole of all component (mol)

➢ Unit: no unit

Percentage by ➢ Percentage by mass is defined as percentage of the mass of solute per mass of
mass, % w/w solution (also known as weight percent)

➢ Formula: Percentage by mass, % w/w = Mass of solute (g) X 100%
Mass of solution (g)

Percentage by ➢ Percentage by volume is defined as percentage of the volume of solute per
volume, % v/v volume of solution

➢ Formula: Percentage by volume, % v/v = Volume of solute (mL) X 100%
Volume of solution (mL)

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TOPIC 1: MATTER

C3 & C4 CALCULATION FOR EACH CONCENTRATION MEASUREMENT

a) MOLARITY

Example 1: Scan me: Molarity

A matriculation student prepared a solution by dissolving 0.586 g of sodium carbonate,
Na2CO3 in 250.0 cm3 of water. Calculate its molarity.

Solution:

Mol of solute = 0.586 = 5.5283 x 10-3 mol

106 /

Molarity, M = Mole of solute (mol)
Volume of solution (L)

= 5.5283 x 10-3 (mol)
250 / 1000 (L)

= 0.02211 mol L-1

Exercise 1:

1) What is the molarity of 85.0 mL ethanol solution that contains 1.77 g of ethanol,
C2H5OH? [Ar C = 12.0, H = 1.0, O = 16.0]
Ans:(0.453)

2) Calculate the molarity of a solution of 1.71 g sucrose (C12H22O11) dissolved in ½
litre of water. [Ar H = 1.0, C = 12.0, O = 16.0]
Ans:(0.0100)

3) How many grams of potassium dichromate, K2Cr2O7 required to prepare a
solution of 250 mL with 2.16 M?
Ans:(159)

4) Calculate the amount of moles of solute in 200 cm3 of ammonia solution, having
a concentration of 0.125 mol dm−3.
Ans:(0.0250)

Extra exercise:

1. Determine the molarity of a solution of 1.71 g sucrose, C12H22O11 dissolved in water
to form 0.5 L of solution.
[ Given : Mr H = 1.01, C = 12.00, O = 16.00 ]

Page | 25 |

TOPIC 1: MATTER

2. A solution is prepared by dissolving 16.0 g of calcium chloride, CaCl2 in 64.0 g of
water. The solution formed has a density of 1.18 g mL-1.
Calculate the molarity of CaCl2 solution.[ Given : Mr Ca = 40.1 , Cl = 35.5 ]

3. An aqueous ammonia solution with density of 0.898 g mL-1 contains 28% ammonia
by mass. Calculate the molarity of solution.
[ Given : Mr N = , H = 1.01 ]

Page | 26 |

TOPIC 1: MATTER

b) MOLALITY Scan me: Molality
Example 1:

Calculate the molal concentration of ethylene glycol (C2H6O2) solution containing 8.40 g
of ethylene glycol in 200 g of water. The molar mass of ethylene glycol is 62 g/mol.

Solution:

Mol of solute = 8.40 = 0.1355 mol

62 /

Molality, m = Mole of solute (mol)
Mass of solution (kg)

= 0.1355 (mol)
200 / 1000 (kg)

= 0.6775 mol kg-1

Exercise 1:

1) A mixture is prepared from 45.0 g of benzene, C6H6 and 80.0 g of toluene, C7H8.

Calculate the molarity of the solution. [Ar C=12; H=1] Ans: (7.21m)

2) A solution containing 121.8 g of Zn(NO3)2 per litre has a density of 1.107 g/mL.
Calculate its molal concentration. [Ar Zn=65; N=14; O=16] Ans: (0.654m)

3) What is the molal concentration of a solution prepared by dissolving 0.30 mol

of CuCl2 in 40.0 mol of water? Ans: (0.417m)

Extra exercise:
1. Calculate the molality of a solution containing 40.0 g of KCl in 190 g of water.

[ Given : Ar K = 39, Cl = 35.5 ]

Page | 27 |

TOPIC 1: MATTER

2. What is the molal concentration of a solution prepared by dissolving 0.30 mol of
CuCl2 in 40.0 mol of water. [ Given : Ar Ca = , Cl = ]

3. The density of solution containing 10% ethanol, C2H5OH by mass is 0.98 g mL-1.
Calculate:
i. The molality of solution
ii. Its molarity

Page | 28 |

TOPIC 1: MATTER

c) MOLE FRACTION

Exercise 1:

1) What is the mole fraction of CuCl2 in a solution prepared by dissolving 0.30

mol of CuCl2 in 40.0 mol of H2O? Ans:(0.0074)

2) A solution is prepared by mixing 55 g of toluene, C7H8 and 55 g of
bromobenzene, C6H5Br. What is the mole fraction of each component?
Ans:(0.63; 0.37)

3) A mixture containing benzene and toluene has 18.4 g of toluene and its
percentage composition is 30%. Calculate the number of moles of benzene in
this solution.

[Mr benzene = 78 and toluene = 92] Ans:(0.55)

Extra Exercise
1. What is the mole fraction of CuCl2 in a solution prepared by dissolving 0.30 mol of
CuCl2 in 40.0 mol of water. [ Given Mr H2O = 18.02 gmol-1 ]

2. A solution is prepared by mixing 55 g of toluene, C7H8 and 55 g of bromobenzene,
C6H5Br. What is the mole fraction of each component ?

[ Ar C = 12.01, H = 1.01, Br = 79.90 ]

Page | 29 |

TOPIC 1: MATTER

d) PERCENTAGE BY MASS

Example 1:

A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.3 g of water. What is the
percent by mass of KCl in this solution?

Percent by mass of KCl = 0.892 g x 100% = 1.61%
0.892 g + 54.2 g

Exercise 1:

1) Calculate the percent by mass of the solute in an aqueous solution containing

5.50 g of NaBr in 78.2 g of solution. Ans:(7.03)

2) Calculate the amount of water (in grams) that must be added to 5.00 g of urea,
(NH2)2CO in the preparation of a 16.2 percent by mass solution. Ans:(25.9)

3) How many grams of NaOH and water are needed to prepare 250.0 g of 1.00%

NaOH solution? Ans:(2.50; 248)

4) Hydrochloric acid can be purchased as a solution of 37% HCl. What mass of

this solution contains 7.5 g of HCl? Ans:(20.3)

Extra exercise
1. Hydrochloric acid can be purchased as a solution of 37% HCl. What is the mass of
this solution contains 7.5 g of HCl ?

2. Calculate the amount of water (in grams) that must be added to 5.00 g of urea in the
preparation of 16.2% by mass of solution.

Page | 30 |

TOPIC 1: MATTER

e) PERCENTAGE BY VOLUME
Example 1:
What mass of NaCl is needed to prepare 250 mL of 0.9% w/v solution?

0.9% w/v NaCl = mass of NaCl x 100%
250 mL solution

mass of NaCl = 2.25 g
Exercise 1:

Describe how to prepare a 10% w/v KOH solution.
Extra exercise:
1. A 200 mL of unknown perfume contains 28 mL of alcohol. What is the percentage
by volume of alcohol in this solution ?

2. Calculate the volume of antifreeze required to make 10 L of a solution of antifreeze
which is 40 % by volume.

TRY THIS!! (CONCENTRATION MEASUREMENT)
The density of 10.5 molal NaOH solutions is 1.33 g mL-1 at 20˚C. Determine
i) mole fraction of NaOH
ii) percentage by mass of NaOH
iii) molarity of the solution

Page | 31 |

TOPIC 1: MATTER

1.3 STOICHIOMETRY

Learning outcomes C1 C2 C3 C4
(a) Write and balance: √

i. Chemical equation by inspection method √

ii.Redox equation by ion-electron method

(b) Define limiting reactant and percentage yield √

(c) Perform stoichiometric calculations using mole concept including limiting √
reactant and percentage yields

LO 1.3 a - Write and balance:
i. Chemical equation by inspection method. (C2)

ii. Redox equation by ion-electron method. (C3)

BALANCING CHEMICAL EQUATION:
iii.
• A chemical equation shows a chemical reaction using symbols for the

Definition reactants and products.
• The formulae of the reactants are written on the left side of the equation

while the products are on the right.

• Example: xA + yB → zC + wD

Example Reactant Product

• The total number of atoms of each element is the same on both sides of a balanced
equation.

• x, y, z and w, = stoichiometric coefficients.

• Methods to balance an equation: i) Inspection Method
ii) Ion-electron Method

i) BALANCE EQUATION USING INSPECTION METHOD

C2 ➢ Follow the steps systematically so that equations become easier to balance.

No. Steps
1. Write the unbalanced equation and the correct formulae for the reactants and products.
2. Balance metallic atoms. (M)
3. Balance non-metallic atoms. (N)
4. Balance oxygen atoms.(O)
5. Balance hydrogen atoms. (H)
6. Ensure that total number of atoms is the same on both sides of equation. (S)

Page | 32 |

TOPIC 1: MATTER

Example 1:
Balance the chemical equation below by applying inspection method.
1. NH3 + CuO → Cu + N2 + H2O

2. Al + H2SO4 → Al2(SO4)3 + H2

Exercise 1 :
Balance the chemical equation below by applying inspection method.
1. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
2. C6H6 + O2 → CO2 + H2O

TRY THIS!! (BALANCE EQ. USING INSPECTION METHOD)

Balance the chemical equation below:

1. Al + H2SO4 → Al2 (SO4)3 + H2

2. Fe (OH) 3 + H2SO4 → Fe2 (SO4)3 + H2O

3. C6H6 + O2 → CO2 + H2O

4. ClO2 + H2O → HClO3 + H2

5. N2H4 + H2O2 → HNO3 + H2

Page | 33 |

TOPIC 1: MATTER

ii) BALANCE EQUATION USING ION-ELECTRON METHOD
This method is used to balance redox reaction that involves oxidation and reduction.

Redox reaction: A reaction where both reduction and oxidation reaction occurs

Reduction Oxidation

a) Decrease in oxidation number a) Increase in oxidation number
b) The substance gains one or more electrons b) The substance loses one or more electrons
c) Act as an oxidising agent (oxidant) c) Act as a reducing agent (reductant)

Oxidation Number Number of electrons that an atom either gains or losses in order to form
a chemical bond with another atom

➢ How to determine oxidation number of an element in chemical formula?

By applying following rules:
1. In a free element, as an atom or a molecule the oxidation number is zero.

Example: Na=0 , Cl2=0 , Br2=0 , O2=0 , Mg=0

2. For monoatomic ion, the oxidation number is equal to the charge of the ion.

Example: Na+ = +1 Mg2+ = +2 Al3+ = +3 S2- =-2

3. Fluorine and other halogens always have oxidation number of -1 in its compound. Positive number
occurs when combine with oxygen.

Example: Oxidation number of F in NaF = -1
Oxidation number of Cl in HCl = -1
Oxidation number of Cl in Cl2O7 = +7

4. Hydrogen has an oxidation number of +1 in its compound except in metal hydrides which
hydrogen has an oxidation number of -1.

Example: Oxidation number of H in HCl = +1
Oxidation number of H in MgH2 = -1

5. Oxygen has an oxidation number of -2 in most of its compound.

Example: Oxidation number of O in MgO = -2 Oxidation number of O in H2O = -2

6. In neutral molecule, the sum of the oxidation number of all atoms that made up the molecule is
equal to zero.

Example: Oxidation number of H2O= 0 Oxidation number of KMnO4= 0

7. For polyatomic ions, the total oxidation number of all atoms that made up the polyatomic ion must
be equal to the net charge of the ion.

Example: Oxidation number of MnO4- = -1 Oxidation number of Cr2O72-= -2

Page | 34 |

TOPIC 1: MATTER

Example 1:

1. Assign (determine) the oxidation number of Cr in Cr2O72-
2Cr + 7(-2) = -2
2Cr = -2 + 14
Cr = 12/2
= +6

2. Assign the oxidation number of Mn in the following chemical compounds.
i. MnO2
Mn + 2(-2) = 0
Mn = +4

ii. MnO4-
Mn + 4(-2) = -1
Mn = -1 + (+8)
= +7

Exercise 1 :

1.Assign the oxidation number of S in SO42-.

2. Assign the oxidation number of Cl in the following chemical compounds.

i. KClO3 ii. Cl2O72-

TRY THIS!! (OXIDATION NUMBER)

1. Assign the oxidation number for Mn in the following chemical compounds.
i. MnO2
ii. MnO4-

2. Assign the oxidation number of the following elements:

i. U in UO22+
ii. C in C2O42-

Page | 35 |

TOPIC 1: MATTER

C3 ii) BALANCE EQUATION USING ION-ELECTRON METHOD

Redox reaction occurs in

Acidic Medium Basic Medium Balancing Chemical Equations
Using the Inspection Method

IN ACIDIC MEDIUM

No. Steps
1. Divide equation into 2 half equations based on same element (one involving

oxidation and the other reduction)
2. Balance each of half equation reaction:

i. Balance the element other than oxygen and hydrogen
ii. Balance the oxygen atom – by adding H2O
iii. Balance the hydrogen atom – by adding H+
iv. Balance the total charge for both equation – by adding electrons to the

side with greater positive charge
3. For overall equation, balance number of electron for each half equation. Multiply

each half equation by an integer so that number of electron lost equals to the
number gained
4. Add 2 half equations and simplify where possible by cancelling species
appearing on both sides of equation.
5. Check the equation to make sure that there are the same number of atoms of
each and same total charge on both sides.

IN BASIC MEDIUM

No. Steps
1. Follow all the 5 steps (step 1- step 5) on how to balance redox equation in acidic

medium
2. Calculate the total number of H+ in the equation
3. Add OH- to both sides of equation. The number of OH- added is equal to number

of H+ in the equation
4. Combined OH- added with H+ to form H2O

Page | 36 |

TOPIC 1: MATTER

Example 1:

Balance the chemical equation below in acidic medium :

1. Cr2O72- + Cl- → Cr3+ + Cl2

Step 1 : Divide the equation into two half reactions, one involving oxidation and the other
reduction.
Reduction : Cr2O72- → Cr3+
Oxidation: Cl- → Cl2

Step 2 :

i.Balance the element other than oxygen and hydrogen

Reduction : Cr2O72- → 2Cr3+
Oxidation: 2Cl- → Cl2

ii)balance the oxygen atom by adding H2O

Reduction : Cr2O72- → 2Cr3+ + 7H2O
Oxidation: 2Cl- → Cl2

iii)balance hydrogen atom by adding H+
Reduction : Cr2O72- + 14H+ → 2Cr3+ + 7H2O
Oxidation: 2Cl- → Cl2

iv)balance the charge by adding electrons to the side with the greater overall positive charge.

Reduction : Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
Oxidation: 2Cl- → Cl2 + 2e

Step 3 :

Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction
equals the number gained in the other.
Reduction : Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
Oxidation: { 2Cl- → Cl2 + 2e} x 3

Step 4 :

Add the two half-reactions and simplify where possible by canceling species appearing on both

sides of the equation.
Reduction : Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O

Oxidation : 6Cl- → 3Cl2 + 6e
Cr2O72- + 14H+ + 6Cl- → 2Cr3+ + 7H2O + 3Cl2

Step 5 :

Check the equation to make sure that there are the same number of atoms of each kind and the

same total charge on both sides.
Cr2O72- + 14H+ + 6Cl- → 2Cr3+ + 7H2O + 3Cl2

+6 +6

Page | 37 |

TOPIC 1: MATTER

Example 2:
Balance the chemical equation below in basic medium:
Cu + NO3 - → Cu2+ + NO

Step 1 : Divide the equation into two half reactions, one involving oxidation and the other reduction.
Oxidation : Cu → Cu2+
Reduction : NO3 - → NO

Step 2 :
i.Balance the element other than oxygen and hydrogen
Oxidation : Cu → Cu2+ unchange
Reduction : NO3 - → NO unchange

ii)balance the oxygen atom by adding H2O
Oxidation : Cu → Cu2+ unchange
Reduction : NO3 - → NO + 2 H2O

iii)balance hydrogen atom by adding H+
Oxidation : Cu → Cu2+ unchange
Reduction : NO3 - + 4H+ → NO + 2 H2O

iv)balance the charge by adding electrons to the side with the greater overall positive charge.
Oxidation : Cu → Cu2+ + 2e
Reduction : NO3 - + 4H+ + 3e → NO + 2H2O

Step 3 :

Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals
the number gained in the other.
Oxidation : {Cu → Cu2+ + 2e } x 3
Reduction : { NO3 - + 4H+ + 3e → NO + 2H2O} x 2

Step 4 :

Add the two half-reactions and simplify where possible by canceling species appearing on both

sides of the equation.
Oxidation : 3Cu → 3Cu2+ + 6e
Reduction : 2 NO3 - + 8H+ + 6e → 2NO + 4H2O
3Cu + 2 NO3 - + 8H+ → 3Cu2+ + 2NO + 4H2O

Step 5 :

Check the equation to make sure that there are the same number of atoms of each kind and the

same total charge on both sides.
3Cu + 2 NO3 - + 8H+ → 3Cu2+ + 2NO + 4H2O

+6 +6
Step 6 :
Neutralize H+ by adding the same number of OH- to the both side of equations

3Cu + 2 NO3 - + 8H++ 8OH- → 3Cu2+ + 2NO + 4H2O + 8OH-
+ 8H2O

Equation: 3Cu + 2 NO3 - + 4H2O → 3Cu2+ + 2NO + 8OH-

Page | 38 |

TOPIC 1: MATTER

Example:
1. Balance following equation: C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O

2. Balance following equation: Zn + SO42- + H+ → Zn2+ + SO2 + H2O

Page | 39 |

TOPIC 1: MATTER

Example:
3. Balance following equation in basic medium : Cr(OH)3 (aq) + IO3- (aq) ⎯→ CrO32- (aq) + I- (aq)

4. Balance following equation in basic medium : Cl2 (aq) ⎯→ ClO4–(aq) + Cl–(aq)

Page | 40 |

TOPIC 1: MATTER

TRY THIS!!

Exercise 1:
Balance the following chemical equations by applying the ion-electron method.

1. MnO − + C2O 2− + H+ → Mn2+ + CO2 + H2O
4 4

2. Cr2O 2 − + Fe2+ + H+ → Cr3+ + Fe3+ + H2O
7

3. MnO − + SO2 + H+ → SO 2− + Mn2+ + H2O
4 4

4. Cr(OH)3 + IO3− + OH− → CrO32− + I− + H2O

5. ClO− + S2O32− → Cl− + SO42− (basic medium)
6. Cl2 → ClO3− + Cl− (basic medium)

7. NO2 → NO3− + NO
8. P + Cu2+ → Cu3P + H3PO3

9. H2SO3 + MnO4- → SO42- + Mn2+ + H2O + H+

10. Br- (aq) + MnO4- (aq) ⎯→ BrO3- (aq) + MnO2 (s)

Page | 41 |

TOPIC 1: MATTER

STOICHIOMETRY
➢ A coefficient that shows relative quantities of reactants and products in a chemical reaction.
➢ Stoichiometry can be used to calculate the quantity of species we are interested in during a

reaction. Thus, it must have a balanced chemical equation for each chemical reaction.
Example: CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)

1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CaCl2, 1 mole of CO2 and 1 mole of
H2O.

WHAT ARE THE OTHER CHEMISTRY CONCEPTS INVOLVED IN STOICHIOMETRY OF
REACTION IN SOLUTIONS?

1. Preparing a solution

2. Dilution of concentrated solution

No. of moles of No. of moles of
solute before = solute after

dilution (initial), ni dilution (final), nf
MiVi = MfVf

Page | 42 |

TOPIC 1: MATTER

3. Titration of solution (redox or acid-base titration)

From general equation:
aA + bB → cC + dD

=



MA = Molarity of acid MB = Molarity of base
VA = Volume of acid VB = Volume of base

Example::
1. Zn (s) + 2HCl (l) → ZnCl2 (s) + H2 (g)

How many moles of hydrochloric acid, HCl do we need to react with 0.5 moles of zinc?
Solution:

2. C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)
How many moles of H2O will be formed when 0.25 moles of C2H5OH burns in oxygen?

Solution:

Exercise:
1. 3HCl + Al(OH)3 → AlCl3 + 3H2O(l)

a) How many grams of aluminium hydroxide will react with 5.3 moles of HCl?
b) What quantity of HCl in grams, can a tablet with 0.750 g of Al(OH)3 consume?

2. A 16.50 mL 0.1327 M KMnO4 solution is needed to oxidize 20.00mL of a FeSO4 solution in an
acidic medium. What is the concentration of the FeSO4 solution? The net ionic equation is:
Fe 2+ + MnO4- → Mn2+ + Fe 3+

Page | 43 |

TOPIC 1: MATTER

LO 1.3 b - Define limiting reactant and percentage yield (C1)
Analogy. Let see this story..

Limiting reactant

C1 Limiting Reactant
Definition
Terms
Limiting reactant The reactant that completely consumed at the end of the reaction and its
quantity limit the amount of products formed.
Terms
theoretical yield Reaction yield and percentage yield

actual yield Definition
Excess reactant The amount of product that would result if the entire limiting reagent reacted.
Percentage yield
**is the maximum obtainable yield, predicted by the balanced equation
according to the number of mole of limiting reactant.

The amount of product actually obtained from a reaction.
** commonly stated or given yield in questions

The reactant that is left over (not completely consumed) in a reaction when

all the limiting reagent is used up

The ratio of the actual yield (obtained from exp.) to the theoretical yield

multiply by 100%.

Formula: Percentage yield= actual yield x 100%

theoretical yield

Page | 44 |

TOPIC 1: MATTER

LO 1.3 c - Perform stoichiometric calculations using mole concept including limiting
reactant and percentage yields (C4)

C4 Steps in Calculating Limiting Reactant

Step 1: Write a balanced equation for the reaction (if equation is not given by the question)

Step 2: Calculate no. of mole for each of the reactant (if masses are given)

Step 3: Write the stoichiometry equation
Step 4: Compare no. of mole given with the no. of mole from stoichiometry (needed)

Conclusion: Find limiting reactant (no. of mole needed is less than no. of mole given) and
make a conclusion

Take note!!!! Always using no. of mole of limiting reactant (mole given) to
calculate any amount (mole, mass, volume) of substances
(reactant @ product)

Example 1: A Comprehensive Stoichiometry
Problem: Mole Ratio, Limiting Reactant,

Given: 2Al + Fe2O3 ⎯→ Al2O3 + 2Fe Percent Yield

In one process 124 g of Al are reacted with 601 g of Fe2O3. Identify the limiting

reagent.

Solution:
Stoichiometric equivalency: 2 moles Al  1 mole Fe2O3

Mole Al = 124g /27gmol-1 = 4.5926 mol

Mole Fe2O3 = 601g/159.8gmol = 3.7610mol

2 moles Al  1 mole Fe2O3

4.5926 moles Al ≡ 4.5926 1 2 3 = 2.2963 mol of Fe2O3
2

if we used all the number 4.5926 mol of Al we only need 2.2963 mol of Fe2O3 to
react

Page | 45 |

TOPIC 1: MATTER

since the given mol of Fe2O3 in this reaction is 3.7610 mol the number of mole of
Fe2O3 is excess

 Thus the limiting reactant is Al

The excess reactant is Fe2O3

Example 2:
Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide:
2NH3(g) + CO2(g) ⎯→ (NH2)2CO(aq) + H2O(l)
In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2. The actual
mass of
a) Which of the two reactants is the limiting reagent?
b) Calculate the mass of (NH2)2CO formed.
c) How much of the excess reagent (in grams) is left at the end of the reaction?
d) Calculate the percentage yield for this reaction if the actual mass of (NH2)2CO
obtained is 955.4 g.

Solution:
a) Since we cannot tell by inspection which of the two reactants is the limiting
reagent, we have to proceed by first converting their masses into numbers of
moles.

Moles of NH3 = 637.2 g NH3 x 1 mol NH3 = 37.4823mol NH3
17.0 g NH3

Moles of CO2 = 1142 g CO2 x 1 mol CO2 = 25.9545 mol CO2
44.0 g CO2

The balanced equation shows that 2 mol NH3  1 mol CO2;
therefore,
the number of moles of NH3 needed to react with 25.9545 mol CO2 is given
by

25.9545 mol CO2 x 2 mol NH 3 = 51.909 mol NH3

1mol CO 2

Since there are only 37.4823 moles of NH3 present, not enough to react
completely with the CO2,
NH3 must be the limiting reagent and CO2 the excess reagent.

b) To calculate the maximum yield of (NH2)2CO formed, we must calculate
based on the stoichiometric proportions (the proportions indicated in the
balanced equation) between the product and the limiting reagent, NH3.

The balanced equation shows that

2 mol NH3 ≡ 1 mol (NH2)2CO
37.4823 mol NH3 ≡ 37.4823 mol NH3 x 1 mol (NH2 )2 CO

2 mol NH3

Mol of (NH2)2CO = 18.7412mol

Page | 46 |

TOPIC 1: MATTER

Mass of (NH2)2CO = 18.7412 mol x 60.0 g (NH2 )2 CO
1 mol (NH2 )2 CO

= 1124.47 g (NH2)2CO

c) Since NH3 is the limiting reagent,
CO2 must be the reactants left over after the reaction is finished.
The difference between the amount of CO2 available and the amount
consumed is the amount left over.

Mole of CO2 used,

2 mol NH3 ≡ 1 mol CO2

37.4823mol NH3 ≡ 37.4823mol NH3 1 2
2 3

= 18.7412 CO2

The number of moles of CO2 (the excess reagent) left over is
25.9545 mol CO2 − 18.7412 mol CO2 = 7.2133 mol CO2
mass of CO2 left over = 7.2133 mol CO2 x 44.0 g CO2

1 mol CO2
= 317.39 g CO2
(d)

Percentage yield= actual yield x 100%
theoretical yield

= 955.4 g / 1124.47 g x 100%
= 84.96 %

Exercise:
1) The reaction between aluminium and iron (III) oxide can generate temperatures

approaching 3000C and is used in welding metals:

2Al + Fe2O3 ⎯→ Al2O3 + 2Fe
In one process 124 g of Al are reacted with 601 g of Fe2O3.
a) Identify the limiting reagent.
b) Calculate the mass (in grams) of Al2O3 formed.
c) How much of the excess reagent (in grams) is left over at the end of the reaction?

2) Aluminium, Al reacts with sulphuric acid, H2SO4, which is the acid in automobile

batteries, according to the equation: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
If 20.0 g of Al is put into a solution containing 115 g of H2SO4,
a) Which reactant will be used up first? (Ans:Al)
b) Calculate the number of moles of Al2(SO4)3 will be produced? (Ans:1.11
mol)

Page | 47 |

TOPIC 1: MATTER

c) How many grams of Al2(SO4)3 can be formed? (Ans:127
g)
d) Calculate the mass (in grams) of the excess reagent remaining at the end of the
reaction.

(Ans: 6 g)
3) Industrially, vanadium metal, which is used in steel alloys, can be obtained by reacting
vanadium(V) oxide with calcium at high temperatures:

5Ca + V2O5 → 5CaO + 2V
In one process 1.54 x 103 g of V2O5 react with 1.96 x 103 g of Ca.
a) Calculate the theoretical yield of V. (Ans:863 g)

b) Calculate the percent yield if 803 g of V are obtained. (Ans:93.0 %)

4) Ethylene, C2H4 an important industrial organic chemical, can be prepared by heating
hexane, C6H14 at 800C:
C6H14 ⎯→ C2H4 + other products
If the yield of ethylene production is 42.5%, what mass of hexane must be reacted to
product 481g of ethylene?
(Ans:3.47x103 g)

Extra exercise:
1. S + 3F2 → SF6
If 4 mol of S reacts with 10 mol of F2, which of the two reactants is the limiting reagent?

2. 10 g of Zn is added to a beaker containing 0.18 mole of hydrochloric acid to form zinc chloride
and hydrogen gas. Determine mass in gram for hydrogen gas produced.

Page | 48 |

TOPIC 1: MATTER

3. An aqueous solution of MgSO4 is added to an aqueous solution of BaCl2to form BaSO4 and MgCl2. If
1.75 g of MgSO4 and 2.75 g of BaCl2 are used, determine
i) determine the limiting reactant.
ii) calculate the mass of BaSO4 produced.
iii) mass of excess reactant left

4. Consider the reaction:
3CCl4 + 2SbF3 → 3CCl2F2 + 2SbCl3

In an experiment, 14.6g of SbF3 was allowed to react in excess CCl4. 8.62g of CCl2F2 was obtained at
the end of the experiment.
a) Determine the theoretical yield for CCl2F2 in grams?
b) What was the percentage yield of CCl2F2?

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TOPIC 1: MATTER

5. Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide:
2NH3(g) + CO2(g) → (NH2)2CO(aq) + H2O(l)

637.2g of NH3 react with 1142g of CO2. Determine:
a) the limiting reactant
b) the mass of (NH2)2CO formed
c) the mass of excess reactant left.

6. Adipic acid, H2C6H8O4 is made by a reaction between cyclohexane and oxygen. The equation for the
reaction is:
2C6H12 (l) + 5O2 (g) → 2H2C6H8O4 (l) + 2H2O (l)
Assume that cyclohexane is the limiting reactant and the reaction starts with 45.0g of cyclohexane,
determine the theoretical yield of adipic acid. If the actual yield of adipic acid is 63.5g, calculate the
percentage yield of adipic acid.

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