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Counting atoms and molecules
When conducting a chemical reaction, it is often important
to mix reactants in the correct proportions. This prevents
contamination of the products by wasted reactants.
However, atoms are very
small and impossible to count
out. In order to estimate the
number of atoms in a sample
of an element, it is necessary
to find their mass.
The mass of an atom is
quantified in terms of
relative atomic mass.
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Relative atomic mass
The relative atomic mass (Ar) of an element is the mass of
one of its atoms relative to 1/12 the mass of one atom of
carbon-12.
relative atomic mass
average mass of an atom × 12
=
(Ar)
mass of one atom of carbon-12
Most elements have more
than one isotope. The Ar of
the element is the average
mass of the isotopes, taking
into account the abundance of
each isotope. This is why the
Ar of an element is frequently
not a whole number.
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Relative molecular mass
The relative molecular mass (Mr) of a covalent substance
is the mass of one molecule relative to 1/12 the mass of
one atom of carbon-12.
Mr can be calculated by adding together the masses of
each of the atoms in a molecule.
Example: what is the Mr of H2SO4?
1. Count number of atoms
(2 × H) + (1 × S) + (4 × O)
2. Substitute the Ar values
(2 × 1.0) + (1 × 32.1) +
(4 × 16.0)
3. Add the values together
2.0 + 32.1 + 64.0 = 98.1
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Relative formula mass
The equivalent of relative molecular mass for an ionic
substance is the relative formula mass.
This is the mass of a formula unit relative to 1/12 the mass
of one atom of carbon-12. It is calculated in the same way
as relative molecular mass, and is represented by the same
symbol, Mr.
Example: what is the Mr of CaCl2?
1. Count number of atoms
(1 × Ca) + (2 × Cl)
2. Substitute the Ar values
(1 × 40.1) + (2 × 35.5)
3. Add the values together
40.1 + 71.0 = 111.1
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Calculating relative formula mass
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Moles and Avogadro's number
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Moles, mass and Ar / Mr
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Moles, mass and Mr calculations
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Avogadro’s law
In 1811 the Italian scientist Amedeo Avogadro developed a
theory about the volume of gases.
Avogadro’s law:
Equal volumes of different gases at the same pressure
and temperature will contain equal numbers of particles.
For example, if there are 2 moles of O2 in 50 cm3 of oxygen
gas, then there will be 2 moles of N2 in 50 cm3 of nitrogen
gas and 2 moles of CO2 in 50 cm3 of carbon dioxide gas at
the same temperature and pressure.
Using this principle, the volume that a gas occupies will
depend on the number of moles of the gas.
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Molar volumes of gases
If the temperature and pressure are fixed at convenient
standard values, the molar volume of a gas can be
determined.
Standard temperature is 273 K and pressure is 100 kPa.
At standard temperature and pressure, 1 mole of any gas
occupies a volume of 22.7 dm3. This is the molar volume.
Example: what volume does 5 moles of CO2 occupy?
volume occupied = no. moles × molar volume
= 5 × 22.7
= 113.5 dm3
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Ideal gas equation
How is the number of moles in a gas at other temperatures
and pressures calculated?
The ideal gas equation relates pressure, volume, number
of moles and temperature for a gas.
pV = nRT
p = pressure in Pa
n = number of moles
V = volume in m3
R = gas constant: 8.31JK-1 mol-1
T = temperature in Kelvin
A gas that obeys this law under all conditions is called
an ideal gas.
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Ideal gas equation: converting units
It is very important when using the ideal gas equation that
the values are in the correct units.
The units of pressure, volume or temperature often need
to be converted before using the formula.
Pressure
to convert kPa to Pa:
× 1000
Volume
to convert dm3 to m3:
to convert cm3 to m3:
÷ 1000 (103)
÷ 1 000 000 (106)
Temperature
to convert °C to Kelvin: + 273
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Calculating the Mr of gases
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Using the ideal gas equation
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Ideal gas calculations
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Types of formulae
The empirical formula of a compound shows the
relative numbers of atoms of each element present,
using the smallest whole numbers of atoms.
For example, the empirical formula of hydrogen peroxide
is HO – the ratio of hydrogen to oxygen is 1:1.
The molecular formula of a compound gives the actual
numbers of atoms of each element in a molecule.
The molecular formula of hydrogen peroxide is H2O2 – there
are two atoms of hydrogen and two atoms of oxygen in each
molecule.
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Determining empirical formulae
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Percentage by mass
Elemental analysis is an analytical technique used to
determine the percentage by mass of certain elements
present in a compound.
To work out the empirical
formula, the total mass of
the compound is
assumed to be 100 g, and
each percentage is turned
into a mass in grams.
If necessary, the mass of any elements not given by
elemental analysis is calculated. The empirical formula of
the compound can then be calculated as normal.
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Calculating empirical formulae
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Calculating molecular formulae
The molecular formula can be found by dividing the Mr by
the relative mass of the empirical formula.
Example: What is the molecular formula of hydrogen
peroxide given that its empirical formula is HO and
the Mr is 34?
1. Determine relative mass of empirical formula:
empirical formula mass = H + O = 1.0 + 16.0 = 17
2. Divide Mr by mass of empirical formula to get a multiple:
relative molecular mass = 34 = 2
multiple =
17
mass of empirical formula
3. Multiply empirical formula by multiple:
HO × 2 = H2O2
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Formulae calculations
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Glossary
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What’s the keyword?
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Multiple-choice quiz
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Balancing equations
An important principle in chemical reactions is that matter
cannot be created or destroyed. It is important that symbol
equations are balanced.
A balanced equation has the same number of each type of
atom on each side of the equation.
Unbalanced:
Na
+
Cl2
1 sodium 2 chlorine
Balanced:
2Na
+
Cl2
2 sodium 2 chlorine

NaCl
1 sodium 1 chlorine

2NaCl
2 sodium 2 chlorine
This shows that two moles of sodium react with one mole
of chlorine to make two moles of sodium chloride.
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Balancing unfamiliar equations
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Balancing ionic equations
Equations containing ions should have the same overall
charge on each side in order to be balanced.
This can be achieved by balancing the equation in the
normal way:
Unbalanced:
Ca2+
+
Cl-
2 calcium 1 chloride
+1 charge
Balanced: Ca2+
+
2Cl-
2 calcium 2 chloride
no charge
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→
CaCl2
2 calcium 2 chloride
no charge
→
CaCl2
2 calcium 2 chloride
no charge
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Balancing ionic equations problems
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State symbols
State symbols are letters that are added to a formula to
indicate what state each reactant and product is in.
The four state symbols are:
s
solid
l
liquid
g
gas
aq
aqueous
These are added after the formula in brackets and subscript.
For example:
2H2(g)
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+
O2(g) 
2H2O(g)
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Adding state symbols
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Reacting masses
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Calculating reacting masses
To calculate the mass of a product given the mass of a
reactant, use the following steps:
1. Calculate no. moles of reactant:
no. moles = mass / Mr
2. Determine mole ratio of reactant to product:
ensure the equation is balanced
3. Calculate no. moles of product:
use the mole ratio
4. Calculate mass of product:
mass = moles × Mr
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Reacting masses example
What mass of sodium chloride is produced if 2.30 g of
sodium is burnt in excess chlorine?
1. Calculate
no. moles of Na:
no. moles = mass / Mr
= 2.30 / 23.0
= 0.100
2. Determine mole
2Na + Cl2  2NaCl
ratio of Na to NaCl: ratio = 2:2
= 1:1
3. Calculate
0.100 moles Na = 0.100 moles NaCl
no. moles of NaCl:
4. Calculate
mass = moles × Mr
= 0.100 × 58.5
mass of NaCl:
= 5.85 g
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Reacting masses calculations
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More reacting masses calculations
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What is concentration?
The concentration of a solution is a measure of how much
solute is dissolved per unit of solvent.
concentration = amount of solute / volume of solvent

amount of solute is measured in moles

volume of solvent is measured in dm3

concentration is measured in mol dm-3.
Volumes are often expressed in cm3, so a more useful
equation includes a conversion from cm3 to dm3.
concentration = (no. moles × 1000) / volume
mol dm3
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cm3
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Concentration, moles and volume
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Concentration calculations
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Standard solutions
A standard solution is a solution of known concentration.
Standard solutions are made by
dissolving an accurately weighed mass
of solid in a known volume of solvent
using a volumetric flask.
The volumetric flask has a thin neck,
which is marked with a line so it can be
filled accurately to the correct capacity.
The standard solution can then be used to find the
concentration of a second solution with which it reacts.
This is known as volumetric analysis or titration.
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Preparing standard solutions
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What is a titration?
A titration is a procedure used to identify the concentration
of a solution by reacting it with a solution of known
concentration and measuring the volume required for a
complete reaction.
The number of moles in the standard
solution is calculated. Using a
balanced equation for the reaction,
the number of moles in the solution
of unknown concentration can also
be calculated.
Once the number of moles for the
solution is known, the concentration
can be easily calculated.
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Carrying out a titration
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Titration calculations examples
What is the concentration of an NaOH solution if 25.0 cm3
is neutralized by 23.4 cm3 0.998 mol dm-3 HCl solution?
1. Calculate no.
moles HCl:
moles = (conc. × volume) / 1000
= (0.998 × 23.4) / 1000
= 0.0234
2. Determine ratio NaOH + HCl  NaCl + H2O
of NaOH to HCl: ratio NaOH:NaCl = 1:1
3. Calculate no.
0.0234 moles HCl = 0.0234 moles NaOH
moles of NaOH:
4. Calculate conc.
of NaOH:
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conc. = (moles × 1000) / volume
= (0.0234 × 1000) / 25.0
= 0.936 mol dm-3
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Titration calculations
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More titration calculations
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What are the different types of yield?
The percentage yield of a chemical reaction shows
how much product was actually made compared with
the amount of product that was expected.
To calculate the percentage yield, the theoretical yield
and the actual yield must be calculated.
The theoretical yield is the maximum mass of
product expected from the reaction, calculated
using reacting masses.
The actual yield is the mass of product that is
actually obtained from the real chemical reaction.
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Calculating yield
The percentage yield of a reaction can be calculated using
the following equation:
percentage yield = (actual yield × 100) / theoretical yield
Example: What is the percentage yield of a reaction
where the theoretical yield was 75 kg but the actual
yield was 68 kg?
percentage yield = (actual yield × 100) / theoretical yield
= (68 × 100) / 75
= 90.7%
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What is atom economy?
Atom economy is another measure of the efficiency of a
chemical reaction. It is the mass of reactants that end up as
the desired product – this is calculated as a percentage.
This concept is useful to chemical industry, because it
takes into account the atoms that end up in unwanted
waste products as well as the yield of the reaction.
This means a process that produces several worthless
by-products could have a high yield but a low atom economy.
Reactions with a high atom economy tend to be more
environmentally friendly as they tend to produce less waste,
use fewer raw materials and use less energy.
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Calculating atom economy
mass of desired products × 100
atom economy =
total mass of reactants
Example: What is the atom economy of a reaction
where the actual yield was 25 000 tonnes but the mass
of the reactants was 30 000 tonnes?
mass of desired products × 100
atom economy =
total mass of reactants
25 000 × 100
=
30 000
= 83.3%
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Yield and atom economy calculations
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Glossary
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What’s the keyword?
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Multiple-choice quiz
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