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Topic 1: Quantitative Chemistry
1.1 Introduction to the particulate nature of matter and chemical change
1.1.1 Atoms of different elements combine in fixed ratios to form compounds, which have different properties from their
component elements
1.1.2 Mixtures contain more than one element and/or compound that are not chemically bonded together and so retain their
individual properties
1.1.3 Mixtures are either homogeneous or heterogeneous
1.1.4 Deduction of chemical equations when reactants and products are specified
1.1.5 Application of the state symbols (s), (l), (g), and (aq) in equations
1.1.6 Explanation of observable changes in physical properties and temperatures during changes of state
Particle Nature of Matter
Matter is anything that takes up space
Matter can either refers to the particles (pure substances) or combination of a substances (mixtures):
Pure Substances
A pure substance has definite and constant composition
For a pure substance, from a particle perspective all particles will look and remain the same
From their definitions: All compounds are molecules, but not all molecules are compounds
When the elements are joined, the atoms lose their individual properties and have different properties from the elements
they are composed of
Mixtures
Mixture: A combination of pure substances
Mixtures contain more than one element and/or compound that are
not chemically bonded together, so retain their individual properties
Mixtures are either homogeneous or heterogeneous:
Homogeneous mixtures are the same mixture throughout
o They will have a uniform composition
Heterogeneous mixtures have a different mixture throughout
o They will have visibly different substances or phases
throughout, a non-uniform composition
Chemical Equation
Chemical Equation: Describes what happens during a chemical reaction
A chemical reaction will always have reactants and products as well as some special reaction conditions if required
Reactants are always on the left, and products are always on the right: Reatants → Products
Chemical equations usually use state symbols to identify the state of the products and reactants
Definitions
Element – Atoms all having the same number of protons
Molecule – Two or more elements chemically join together
Compounds – Two or more different elements chemically joined together in a fixed ratio
State Symbols
Reactants and products can be in one of four states
o (s): solid
o (l): liquid
o (g): gas
o (aq): aqueous solution (dissolved in a solvent)
The changes of state are to the left:
A heating curve is a graph showing the temperature of a
substance plotted against the amount of energy it has
absorbed
Note, during a state change there will be no increase or decrease in temperature
Adding temperature only increases the kinetic energy of the molecules, which will eventually break the bonds, then the
molecules will change state
It also takes a higher temperature to turn a solid to a liquid, and an even greater temperature to turn a liquid to a gas
Physical and Chemical Changes
In a physical change, no new substances are produced
o Example: The melting of ice is a physical change. It is being changed physically
In a chemical change, new chemical substances are formed
o The atoms in the reactants are rearranged to form new products. It is being changed chemically
1.2 The mole concept and Avogadro’s Constant
1.2.1 The mole is a fixed number of particles and refers to the amount, n , of substance
1.2.2 Masses of atoms are compared on a scale relative to
12
C and are expressed as relative atomic mass (A r
) and relative
formula/molecular mass (M r
)
1.2.3 Molar mass (M) has the units g mol
1.2.4 Calculation of the molar masses of atoms, ions, molecules and formula units
1.2.5 Solution of problems involving the relationships between the number of particles, the amount of substance in moles and
the mass in grams
1.2.6 Interconversion of the percentage composition by mass and the empirical formula and molar mass
1.2.7 Determination of the molecule formula of a compound from its empirical formula and molar mass
1.2.8 Obtaining and using experimental data for deriving empirical formulas from reactions involving mass changes
The Mole Concept
Masses of atoms are compared on a scale relative to
12
C and are expressed as relative atomic and molecular mass
The relative molecular mass (M r
) also called the molar mass can be calculated from its chemical formula using the relative
atomic masses (A r
) of the elements from the periodic table
Some elements will have a greater atomic mass than others despite their atomic number because they will either have a
greater proportion of heavier isotopes or they will have a greater number of neutrons
Relative atomic and molecular mass are relative therefore it has no units
Molar mass (M) has the units g mol
Relative molecular mass question:
Calculate the relative formula mass (molar mass) of Vitamin C: C
6
H
8
O
6
M
r
( Vitamin C )=
[
6 × A
r
(C )
]
[
8 × A
r
( H )
]
+[ 6 × A
r
(O) ]
Note the multiples because there is a # of atoms
M
r
( Vitamin C )=( 6 × 12 ) +( 8 × 1 )+( 6 × 16 )
M
r
( Vitamin C )= 176
Therefore M
r
Vitamin C
= 176 gmo l
− 1
Amount of moles
In order to calculate number of moles: n=
m
M
where
o n
: moles
o m
: mass
o M
: molar mass
Number of moles question:
Calculate the amount in mol of 1.2g of Nitric Oxide (NO)
m=1.2 M = 14 + 16 = 30
Therefore, since
n=
m
M
→ n=
=0.04 mol
Percentage Composition
The values of molar masses of elements in compounds can be used to calculate the % compositions of a compound once its
formula is known
This is given by the following equation:
% composition by mass of element=
molar mass of x
molar mass of the compound
Question: Determine the % composition by mass of each element in potassium nitrate (KNO 3
)
Definitions
Relative atomic mass (A r
) – The average mass of all isotopes of an element compared to
the mass of C 12
atom
Relative molecular/formula mass (M r
) – The mass of a molecule compared to
the mass of C 12
atom
%K =
× 100 = 38.67% %O=
3 ×16.
× 100 =47.47 %
%N = 100 −38.67−47.47=13.86 %
Empirical formula
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:
- Write the elements present in the compound
- Write each elements % composition or mass
- Divide the % or mass by the relative atomic mass and calculate the ratio
- Divide each ratio by the smallest ratio above to get a whole number ratio
- Express as an empirical formula
Question: A compound consists of carbon 75% and hydrogen 25% by mass. Determine empirical formula
C : H
Molecular formula
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
A molecular formula can be found is the molar mass is known
Question: Work out the molecular formula of CH 2
(M r
= 70)
Empirical Formula: A
r
( C) + A
r
H
2
70 ÷ 14 = 5
C H
2
× 5 =C
5
H
10
Atom Economy
The atom economy of a chemical reaction is a measure of the amount of starting materials that become useful products
A high atom economy means that less waste is created and the reaction has a higher efficiency
To calculate:
Atomeconomy=
total mass of desired products
total mass of all products /reactants
× 100
1.3 Reacting masses and volumes
1.3.1 Reactants can either be limiting or excess
1.3.2 The experimental yield can be different from the theoretical yield
1.3.3 Avogadro’s law enables the mole ratio of reacting gases to be determined from volumes of the gases
1.3.4 The molar volume of an ideal gas is a constant at specified temperature and pressure
1.3.5 The molar concentration of a solution is determined by the amount of solute and the volume of solution
Percentage yield ( %) =
actual yield
t h eoretical yield
× 100
× 100 =74.68 %
Therefore, percentage yield if 74.68%
Theory of an ideal gas
The kinetic molecular theory is a model used to explain the behavior of gases. The essential ideas are:
o Gaseous particles are in continuous random motion, in straight lines not curved
o 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
Ideal Gas Equation
Ideal gas equation: PV =nRT where:
o P: Pressure in kilopascals (kPa) In IB convert to Pa
o V: Volume decimeters cubed (dm
3
) In IB convert to m
3
o n: Number of moles
o T: Temperature in kelvin
o R: 8.31 (Universal gas constant)
Question:
0.25 mol of nitrogen is placed in a flask of volume 5.0dm
3
at a temperature of 5°C. What is the pressure?
P=x , V =5.0 d m
3
, n=0.25 mol , T = 278 K , R=8.
Therefore: P
N
2
nRT
V
0.25 ×8.31 × 278
= 116 kPa
Combined Gas Equation
The three gas laws applied to a fixed mass of gas can be summarized:
o P ∝
V
at constant temperature
o V ∝ T
at constant pressure
o P ∝ T at constant volume
These three laws can combine to form the combined gas law:
P
1
V
1
T
1
P
2
V
2
T
2
Law Result Formula
Combined gas law PV
T
=k
P
1
V
1
T
1
P
2
V
2
T
2
Gay-Lussacs’ law P
T
=k
P
1
T
1
P
2
T
2
Boyles’ law PV =k P
1
V
1
=P
2
V
2
Charles’s law V
T
=k
V
1
T
1
V
2
T
2
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 27°C. It rises to an altitude of 15km
where the temperature is -30°C and the pressure 12kPa.
What is the volume of the balloon at this altitude
V
1
= 150 L V
2
=x
P
1
= 101 kPa P
2
= 12 kPa
T
1
= 27 + 273 K T
2
=− 30 + 273 K
Therefore: V
2
P
1
V
1
T
2
T
1
P
2
101 × 150 ×( 273 + 27 )
( 273 − 30 )× 12
= 1559 L
Real vs Ideal Gases:
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
The molar volume of an ideal gas is a constant at specified temperature and pressure
o Molar volume (V m
): 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 22.7dm
3
o Standard temperature and pressure (STP) conditions are at 273K and 100kPa
Avogadro’s law enables the mole ratio of reacting gases to be determined from volumes of the gases
n
CaC l
2
=c × v n
CaC l
2
=c × v
n
CaC l
2
=0.40 × 0.025 n
CaC l
2
=0.40 × 0.
n
CaC l
2
=0.010 mol n
CaC l
2
=0.060 mol
Therefore, by adding both mols together we get 0.070mol. Now we need to calculate concentration:
[
CaC l
2
]
n
v
=0.93 M
Dilution
Dilution: The process of adding more solvent to a solution
When a solution is diluted, the solute particles are more widely spread. There is a direct relationship between volume and
concentration
The dilution formula is then:
C
1
V
1
=C
2
V
2
Question:
Calculate the molarity of CalCl 2
in 200cm
3
of 0.40M CaCl 2
diluted to 400cm
3
of water
C
1
=0.40 M C
2
=x
V
1
= 200 c m
3
V
2
= 400 c m
3
Therefore : C
2
C
1
V
1
V
2
0.40 ×0.
=0.20 M
