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Title: IB CHEMISTRY NOTES - Topic 1 Quantitative Chemistry
Description: Comprehensive notes on topic 1 of IB chem, with sample questions and diagrams. Includes concepts such as the mole concept, the various ways of calculating moles, gases, concentrations and more. Written by a 7 student.

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IB CHEMISTRY:
TOPIC 1 QUANTITATIVE CHEMISTRY NOTES
1
...
g
...
g
...

In a chemical change, new chemical substances are formed
o The atoms in the reactants are rearranged to form new products

1
...
02 x 1023 particles (atoms, molecules,
ions)
𝑁

In order to calculate number of particles: 𝑛 = π΄π‘£π‘œπ‘”π‘Žπ‘‘π‘Ÿπ‘œ ′𝑠 π‘›π‘’π‘šπ‘π‘’π‘Ÿ OR 𝑁 = 𝑛 βˆ—
π΄π‘£π‘œπ‘”π‘Žπ‘‘π‘Ÿπ‘œβ€² 𝑠 π‘›π‘’π‘šπ‘π‘’π‘Ÿ
o 𝑁: number of particles (atoms, molecules, ions)
o 𝑛: number of moles

Mole question:
Calculate the number of O2 molecules in 1
...
πŸ“π’Žπ’π’
π΄π‘£π‘œπ‘”π‘Žπ‘‘π‘Ÿπ‘œβ€² 𝑠 π‘›π‘’π‘šπ‘π‘’π‘Ÿ = πŸ”
...
𝟎 Γ— πŸπŸŽπŸπŸ‘

Stoichiometric relationships
β€’

β€’

A chemical formula shows the stoichiometric relationship between the individual atoms
that make up that molecule
...
Calculate the number of mol of oxygen in 0
...
πŸŽπŸ“
𝑡(𝑢) = 𝟎
...
Calculate the number of mol of SO42- ions in a 2
...
πŸ‘πŸ— Γ— πŸπŸŽβˆ’πŸ‘
𝒏(π‘Ίπ‘ΆπŸ’ πŸβˆ’ ) = 𝒏(π‘·π’ƒπ‘Ίπ‘ΆπŸ’ ) It is equal because the ratios are equal
𝒏(π‘Ίπ‘ΆπŸ’ πŸβˆ’ ) = 𝟐
...
2g of Nitric Oxide (NO)
π’Ž = 𝟏
...
𝟐
πŸ‘πŸŽ

= 𝟎
...
𝟏𝟎

%𝑲 = 𝟏𝟎𝟏
...
67%

%𝑢 =

πŸ‘ Γ— πŸπŸ”
...
𝟏𝟏

Γ— 𝟏𝟎𝟎 = πŸ’πŸ•
...
πŸ–πŸ”%

Empirical formula
β€’
β€’

Empirical formula: The formula of a compound that shows the lowest whole number
ratio of each type of atom
To calculate the empirical formula of compounds we:
1
...
Write each elements % composition or mass
3
...
Divide each ratio by the smallest ratio present to get a whole number ratio
5
...
Determine
empirical formula
π‘ͺ: 𝑯
πŸ•πŸ“
πŸπŸ“
:
𝟏𝟐
...
πŸ–
∢
πŸ”
...


Molecular formula
β€’
β€’
β€’
β€’

Molecular formula: The formula of a compound that shows the actual number of each
type of atom in the molecule
A molecular formula gives the actual number of different atoms covalently bonded in one
molecule
The molecular formula is always a whole multiple of the empirical formula
To work out molecular formula:
π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘šπ‘Žπ‘ π‘ 

o

First use the formula π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘Žπ‘Ÿ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž =

o

relative atomic masses of the empirical formula
Multiply all the elements in the empirical formula by the result to obtain the
molecular formula

𝐸𝐹𝑅

, where 𝐸𝐹𝑅 is the sum of

Question: Work out the molecular formula of CH2 (Mr = 70)
π‘¬π’Žπ’‘π’Šπ’“π’Šπ’„π’‚π’ π‘­π’π’“π’Žπ’–π’π’‚: 𝑨𝒓 (π‘ͺ) + 𝑨𝒓 (π‘―πŸ ) = 𝟏𝟐 + 𝟐 = πŸπŸ’
πŸ•πŸŽ Γ· πŸπŸ’ = πŸ“
π‘ͺπ‘―πŸ Γ— πŸ“ = π‘ͺπŸ“ π‘―πŸπŸŽ

1
...
It is
formed by burning sulfur in the atmosphere of fluorine
...
Which will be the limiting reagent?
S + 3F2 β†’ SF6
𝒏(𝒔) Γ· 𝟏 = πŸ’ π’Žπ’π’ Γ· 𝟏 = πŸ’

Divided by 1 because coefficient is one

𝒏(π‘­πŸ ) Γ· πŸ‘ = 𝟐𝟎 π’Žπ’π’ Γ· πŸ‘ = πŸ”
...

A percentage yield is the amount of product produced experimentally compared to the
theoretical amount
In order to calculate percentage yield:
π‘Žπ‘π‘‘π‘’π‘Žπ‘™ 𝑦𝑖𝑒𝑙𝑑
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ 𝑦𝑖𝑒𝑙𝑑 (%) =
Γ— 100
π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘’π‘‘π‘–π‘π‘Žπ‘™ 𝑦𝑖𝑒𝑙𝑑

Question – percentage yield:
10
...
95g of bromine
...
95
...
00g
...

π‘·π’†π’“π’„π’†π’π’•π’‚π’ˆπ’† π’šπ’Šπ’†π’π’… (%) =

𝒂𝒄𝒕𝒖𝒂𝒍 π’šπ’Šπ’†π’π’…
Γ— 𝟏𝟎𝟎
π’•π’‰π’†π’π’“π’†π’•π’Šπ’„π’‚π’ π’šπ’Šπ’†π’π’…

πŸ“πŸŽ
Γ— 𝟏𝟎𝟎 = πŸ•πŸ’
...
πŸ—πŸ“
Therefore, percentage yield if 74
...
The
essential ideas are:
o Gaseous particles are in continuous random motion
o They exhibit perfect elastic collision
o Average kinetic energy is directly proportional to temperature
o Volume of gas is negligible
o No intermolecular forces (no attraction between particles)
Note that no gas is perfectly ideal – but we use ideal gas theory to approximate
...
31 (Universal gas constant)

Question:
0
...
0dm 3 at a temperature of 5Β°C
...
πŸ‘πŸ
Therefore: 𝑷(π‘΅πŸ ) =

𝒏𝑹𝑻
𝑽

=

𝟎
...
𝟎

= πŸπŸπŸ”π’Œπ‘·π’‚

Gas laws
β€’

β€’

The three gas laws applied to a fixed mass of gas can be summarized:
1

at constant temperature (Boyle’s law)

o

π‘ƒβˆ

o
o

𝑉 ∝ 𝑇 at constant pressure (Charles’s law)
𝑃 ∝ 𝑇 at constant volume (Gay-Lussacs’ law)

𝑉

These three laws can combine to form the combined gas law:

𝑃1 𝑉1
𝑇1

=

Law

Result

Formula

Combined gas law

𝑃𝑉
=π‘˜
𝑇

𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2

Gay-Lussacs’ law

𝑃
=π‘˜
𝑇

𝑃1 𝑃2
=
𝑇1 𝑇2

Boyles’ law

𝑃𝑉 = π‘˜

𝑃1 𝑉1 = 𝑃2 𝑉2

𝑃2𝑉2
𝑇2

Charles’s law

β€’

𝑉
=π‘˜
𝑇

𝑉1 𝑉2
=
𝑇1 𝑇2

An ideal gas will have the greatest volume at a high temperature and low pressure

Question:
A balloon has a volume of 150L at a pressure of 101kPa and a temperature of 27Β°C
...

What is the volume of the balloon at this altitude
π‘½πŸ = πŸπŸ“πŸŽπ‘³

π‘½πŸ = 𝒙

π‘·πŸ = πŸπŸŽπŸπ’Œπ‘·π’‚
π‘»πŸ = πŸπŸ• + πŸπŸ•πŸ‘π‘²
Therefore: π‘½πŸ =

π‘·πŸ = πŸπŸπ’Œπ‘·π’‚
π‘»πŸ = βˆ’πŸ‘πŸŽ + πŸπŸ•πŸ‘π‘²
π‘·πŸπ‘½πŸ π‘»πŸ
π‘»πŸ 𝑷 𝟐

=

πŸπŸŽπŸΓ—πŸπŸ“πŸŽΓ—(πŸπŸ•πŸ‘+πŸπŸ•)
(πŸπŸ•πŸ‘βˆ’πŸ‘πŸŽ)Γ—πŸπŸ

= πŸπŸ“πŸ“πŸ—π‘³

Real vs Ideal Gases:
β€’

A gas behaves more like an ideal gas at a high temperature and lower pressure:
o High temperature: The potential energy due to intermolecular forces becomes
less significant compared with the particles kinetic energy
o Low pressure: The size of the molecules becomes less significant compared to
the empty space between them
Real Gases

Ideal Gases

Gas particles have volume

Gas particles do not have
volume

Particles have attractive
forces

No attractive forces between
particles

Molar Volume
β€’

β€’
β€’

The molar volume of an ideal gas is a constant at specified temperature and pressure
o Molar volume (Vm): The volume occupied by one mole of a substance (chemical
element or chemical compound) at a given temperature and pressure
Avogadro’s law states 1 mol of any gas at STP will occupy 22
...
g of oxygen gas (O 2 at STP)
𝒏=

π’Ž
πŸπŸ”
=
= 𝟎
...
πŸ’

MOLAR CONCENTRATIONS
Definitions
Solute – The smallest component in a solution (what is being dissolved)
Solvent – The largest component of a solution (what is it being dissolved in)
Solution – The solute and solvent combined (A homologous mixture)
Concentration – A measure of solute (mol) per solution (dm-3)

π’Žπ’π’π’† 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

𝒏

Concentration can be calculated by: π’„π’π’π’„π’†π’π’•π’“π’‚π’•π’Šπ’π’ = π’—π’π’π’–π’Žπ’† 𝒐𝒇 π’”π’π’π’–π’•π’Šπ’π’ = 𝑽

β€’

Question:
What is the concentration of sodium chloride in a saline solution if 200cm 3 of the solution
contains 0
...
𝟎𝟏𝟎
𝟐𝟎𝟎/𝟏𝟎𝟎𝟎

= 𝟎
...
40M CaCl2 added
to 50cm3 of 1
...
πŸŽπŸπŸ“

𝒏(π‘ͺ𝒂π‘ͺπ’πŸ ) = 𝟎
...
πŸŽπŸπŸŽπ’Žπ’π’

𝒏(π‘ͺ𝒂π‘ͺπ’πŸ ) = 𝟎
...
070mol
...
πŸŽπŸ•πŸŽ
=
= 𝟎
...
πŸŽπŸ•πŸ“

Dilution
β€’
β€’
β€’

Dilution: The process of adding more solvent to a solution
When a solution is diluted, the solute particles are more widely spread
...
40M CaCl2 diluted to 400cm3 of water
π‘ͺ𝟏 = 𝟎
...
𝟐𝟎𝟎
𝟎
...
πŸπŸŽπ‘΄

π‘½πŸ = πŸ’πŸŽπŸŽπ’„π’ŽπŸ‘


Title: IB CHEMISTRY NOTES - Topic 1 Quantitative Chemistry
Description: Comprehensive notes on topic 1 of IB chem, with sample questions and diagrams. Includes concepts such as the mole concept, the various ways of calculating moles, gases, concentrations and more. Written by a 7 student.