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AS Chemistry OCR Your notes

2.2 Amount of Substance

Contents
2.2.1 Amount of Substance
2.2.2 Determining Formulae
2.2.3 Reaction Calculations
2.2.4 T he Ideal Gas Equation
2.2.5 Percentage Yield & Atom Economy

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2.2.1 Amount of Substance

Your notes
The Mole & the Avogadro Constant
Amount of substance is the name given when counting the number of particles in a substance
Amount of substance is often seen in calculations using the letter / symbol n
The units for amount of substance are moles / mol
Amount of substance links to the Avogadro constant, NA, which is the number of particles
equivalent to the relative atomic, molecular or formula mass of a substance
The Avogadro constant applies to atoms, molecules, ions and electrons
The value of NA is 6.02 x 10 23 g mol-1
The mass of a substance with this number of particles is called a mole (mol)
This can be called the molar mass
This is the mass of substance that contains the same number of fundamental units as exactly
12.00 g of carbon-12
The amount / number of moles of a substance, n, the mass of the substance, m, and the molar
mass, M, are linked by the equation:
mass , m
n=
Molar mass , M
The molar gas volume is the volume occupied by one mole of any gas, at room temperature and
pressure
The molar volume is equal to 24 dm3
One mole of any element is equal to the relative atomic mass of the element, in grams
For example, if you had one mole of carbon in your hand, you would be holding 6.02 x 1023
atoms of carbon which have a mass of 12.0 g
If you were holding one mole of water, you would be holding 6.02 x 1023 molecules of water
which have a mass of 18.0 g (2 hydrogen + 1 oxygen = (2 x 1.0) + 16.0 = 18.0 g)

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Worked example
Your notes
Molar mass and molar gas volume

1. Calculate the molar mass of:

a. Carbon dioxide, CO 2
b. Magnesium nitrate, Mg(NO 3)2
(Ar data: C = 12.0, O = 16.0, Mg = 24.3, N = 14.0)

2. Calculate the number of moles of each gas:

a. 36.0 dm3 of carbon monoxide, CO
b. 9.6 dm3 of chlorine, Cl2

Answers
Answer 1:
The molar mass is the mass of one mole of any substance
a. Carbon dioxide, CO 2 = 12.0 + (16.0 x 2) = 44.0 g
b. Magnesium nitrate, Mg(NO 3)2 = 24.3 + (14.0 x 2) + (16.0 x 3 x 2) = 148.3 g
Answer 2:
One mole of any gas occupies 24.0 dm3
a. 36.0 dm3 is 1.5 times the molar gas volume of 24.0 dm3, therefore, there are 1.5 moles
of carbon monoxide, CO
b. 9.6 dm3 is 0.4 times the molar gas volume of 24.0 dm3, therefore, there are 0.4 moles
of chlorine, Cl2

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2.2.2 Determining Formulae

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Defining Empirical & Molecular Formulae
The molecular formula shows the number and type of each atom in a molecule
E.g. the molecular formula of ethanoic acid is C 2 H4O 2
The empirical formula shows the simplest whole number ratio of the elements present in one
molecule of the compound
E.g. the empirical formula of ethanoic acid is CH2 O

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Worked example
Your notes
Deducing molecular & empirical formulae
Deduce the molecular and empirical formula of the following compounds:

Answer

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Your notes

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Calculating Empirical & Molecular Formulae

Empirical formula Your notes
Empirical formula is the simplest whole number ratio of the elements present in one molecule
or formula unit of the compound
It is calculated from knowledge of the ratio of masses of each element in the compound
The empirical formula can be found by determining the mass of each element present in a sample
of the compound
It can also be deduced from data that gives the percentage compositions by mass of the
elements in a compound

Worked example
Empirical formula from mass

Determine the empirical formula of a compound that contains 10 g of hydrogen and 80 g of

oxygen.

The above example shows how to calculate empirical formula from the mass of each element
present in the compound

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The example below shows how to calculate the empirical formula from percentage composition

Your notes
Worked example
Empirical formula from %

Determine the empirical formula of a compound that contains 85.7% carbon and 14.3%
hydrogen.

Molecular formula
The molecular formula gives the exact numbers of atoms of each element present in the formula
of the compound
The molecular formula can be found by dividing the relative molecular mass of the molecular
formula by the relative formula mass of the empirical formula
Multiply the number of each element present in the empirical formula by this number to find the
molecular formula

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Worked example
Your notes
Calculating molecular formula

The empirical formula of X is C 4H10 S and the relative molecular mass of X is 180.2
What is the molecular formula of X?

(Ar data: C = 12.0, H = 1.0, S = 32.1)

Answer
Step 1: Calculate relative mass of the empirical formula
Relative empirical mass = (C x 4) + (H x 10) + (S x 1)
Relative empirical mass = (12.0 x 4) + (1.0 x 10) + (32.1 x 1)
Relative empirical mass = 90.1
Step 2: Divide relative molecular mass of X by relative empirical mass
Ratio between Mr of X and the Mr of the empirical formula = 180.2 / 90.1
Ratio between Mr of X and the Mr of the empirical formula = 2
Step 3: Multiply each number of elements by 2
(C 4 x 2) + (H10 x 2) + (S x 2)
(C 8) + (H20 ) + (S2 )
Molecular formula of X is C 8H20 S2

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Hydrated salts & Water of Crystallisation

Water of crystallisation is when some compounds can form crystals which have water as part Your notes
of their structure
A compound that contains water of crystallisation is called a hydrated compound
The water of crystallisation is separated from the main formula by a dot when writing the chemical
formula of hydrated compounds
E.g. hydrated copper(II) sulfate is CuSO 4∙5H2 O
A compound which doesn’t contain water of crystallisation is called an anhydrous compound
E.g. anhydrous copper(II) sulfate is CuSO 4
A compound can be hydrated to different degrees
E.g. cobalt(II) chloride can be hydrated by six or two water molecules
CoCl2 ∙6H2 O or CoCl2 ∙2H2 O
The conversion of anhydrous compounds to hydrated compounds is reversible by heating the
hydrated salt:
Hydrated: CuSO 4•5H2 O ⇌ CuSO 4 + 5H2 O :Anhydrous

The degree of hydration can be calculated from experimental results:

The mass of the hydrated salt must be measured before heating
The salt is then heated until it reaches a constant mass
The two mass values can be used to calculate the number of moles of water in the hydrated
salt - known as the water of crystallisation

Worked example
Calculating water of crystallisation

10.0 g of hydrated copper sulfate are heated to a constant mass of 5.59 g. Calculate the formula
of the original hydrated copper sulfate.

(Mr data: CuSO 4 = 159.6, H2 O = 18.0)

Answer

List the components CuSO 4 H2 O

Note the mass of each component 5.59 g 10 - 5.59 = 4.41 g

Divide the component mass by the

components Mr

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5 . 59 4 . 41
= 0.035 = 0.245
159 . 6 18. 0 Your notes
Divide by the lowest figure to obtain the 0 . 035 0 . 245
ratio =1 =7
0 . 035 0 . 035
Hydrated salt formula CuSO 4•7H2 O

Exam Tip
A water of crystallisation calculation can be completed in a similar fashion to an empirical formula
calculation
Instead of elements, you start with the salt and water
Instead of dividing by atomic masses, you divide by molecular / formula masses
The rest of the calculation works the same way as the empirical formula calculation

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2.2.3 Reaction Calculations

Your notes
Mass Calculations
The number of moles of a substance can be found by using the following equation:
mass of a substance (g )
number of moles ( mol ) =
molar mass (g mol −1)
It is important to be clear about the type of particle you are referring to when dealing with moles
E.g. one mole of CaF2 contains one mole of CaF2 formula units, but one mole of Ca2+ and two
moles of F- ions
Reacting masses
The masses of reactants are useful to determine how much of the reactants exactly react with
each other to prevent waste
To calculate the reacting masses, the balanced chemical equation is required
This equation shows the ratio of moles of all the reactants and products, also called the
stoichiometry, of the equation
To find the mass of products formed in a reaction the following pieces of information are
needed:
The mass of the reactants
The molar mass of the reactants
The balanced equation

Worked example
Mass calculation using moles

Calculate the mass of magnesium oxide that can be made by completely burning 6 g of
magnesium in oxygen.
magnesium (s) + oxygen (g) → magnesium oxide (s)

Answer

Step 1: The balanced symbol equation is:

2Mg (s) + O 2 (g) → 2MgO (s)

Step 2: The relative formula masses are:

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Mg = 24.3, O 2 = 32.0, MgO = 40.3

Step 3: Calculate the moles of magnesium used in the reaction: Your notes
6. 0 g
number of moles = = 0.25 moles
24 . 3 g mol −1
Step 4: Find the ratio of magnesium to magnesium oxide using the balanced chemical
equation

Therefore, 0.25 mol of MgO is formed

Step 5: Find the mass of magnesium oxide
Mass = mol x Mr
Mass = 0.25 mol x 40 g mol-1
Mass = 10 g

Therefore, the mass of magnesium oxide produced is 10 g

Stoichiometric relationships
The stoichiometry of a reaction can be found if the exact amounts of reactants and products
formed are known
The amounts can be found by using the following equation:
mass of a substance (g )
number of moles ( mol ) =
molar mass (g mol −1)
The gas volumes can be used to deduce the stoichiometry of a reaction

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E.g. in the combustion of 50 cm3 of propane reacting with 250 cm3 of oxygen, 150 cm3 of
carbon dioxide is formed suggesting that the ratio of propane : oxygen : carbon dioxide is 1 : 5
:3 Your notes
C 3 H8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H2 O (l)

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Volume Calculations
The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of Your notes
solution
The solute is the substance that dissolves in a solvent to form a solution
The solvent is often water

number of moles ( mol )

concentration ( mol dm −3) =
volume of solution (dm3)
A concentrated solution is a solution that has a high concentration of solute
A dilute solution is a solution with a low concentration of solute
When carrying out calculations involve concentrations in mol dm-3 the following points need to
be considered:
Change mass in grams to moles
Change cm3 to dm3
To calculate the mass of a substance present in solution of known concentration and volume:
Rearrange the concentration equation
number of moles (mol) = concentration (mol dm-3) x volume (dm3)
Multiply the moles of solute by its molar mass
mass of solute (g) = number of moles (mol) x molar mass (g mol-1)

Worked example
Calculating volume from concentration

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm-3 that is required to react
completely with 2.5 g of calcium carbonate

Answer

Step 1: Write the balanced symbol equation

CaCO 3 + 2HCl → CaCl2 + H2 O + CO 2
Step 2: Calculate the amount, in moles, of calcium carbonate that reacts

2.5 g
number of moles ( CaCO 3) = = 0 . 025 mol
100 g mol −1

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Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry
1 mol of CaCO 3 requires 2 mol of HCl Your notes
So 0.025 mol of CaCO 3 requires 0.05 mol of HCl
Step 4: Calculate the volume of HCl required

amount (mol ) 0 . 05 mol

volume of HCl (dm 3 ) = = = 0 . 05 dm 3
concentration (mol dm −3 ) 1 . 0 mol dm −3
Volume of hydrochloric acid = 0.05 dm3

Worked example
Neutralisation calculation

25.0 cm3 of 0.050 dm–3 sodium carbonate was completely neutralised by 20.0 cm3 of dilute
hydrochloric acid.
Calculate the concentration in mol dm–3 of the hydrochloric acid.

Answer
Step 1: Write the balanced symbol equation
Na2 CO 3 + 2HCl → 2NaCl + H2 O + CO 2
Step 2: Calculate the amount, in moles, of sodium carbonate reacted by rearranging the
equation for amount of substance (mol) and dividing the volume by 1000 to convert cm3 to
dm3
amount (Na2 CO 3 ) = 0.025 dm3 x 0.050 mol dm-3 = 0.00125 mol
Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry
1 mol of Na2 CO 3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2
Therefore 0.00125 moles of Na2 CO 3 react with 0.00250 moles of HCl
Step 4: Calculate the concentration, in mol dm-3, of hydrochloric acid

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amount (mol ) 0 .00250

concentration of HCl (mol dm −3 ) = = = 0 .125 mol dm −3
3
volume (dm ) 0 . 0200
Your notes
Volumes of gases
Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’ (also
called Avogadro’s hypothesis)
At room temperature (20 degrees Celsius) and pressure (1 atm) one mole of any gas has a volume
of 24.0 dm3
This molar gas volume of 24.0 dm3 mol-1 can be used to find:
The volume of a given mass or number of moles of gas:
volume of gas (dm3) = amount of gas (mol) x 24 dm3 mol-1
The mass or number of moles of a given volume of gas:

volume of gas (dm3)

amount of gas ( mol ) =
24. 0 (dm3 mol −1)

Worked example
Calculation volume of gas using excess & limiting reagents

Calculate the volume the following gases occupy:

1. Hydrogen (3 mol)
2. Carbon dioxide (0.25 mol)
3. Oxygen (5.4 mol)
4. Ammonia (0.02 mol)
Calculate the moles in the following volumes of gases:
1. Methane (225.6 dm3)
2. Carbon monoxide (7.2 dm3)
3. Sulfur dioxide (960 dm3)

Answer

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Your notes

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2.2.4 The Ideal Gas Equation

Your notes
Ideal Gas Equation & Calculations
Kinetic theory of gases
The kinetic theory of gases states that molecules in gases are constantly moving
The theory makes the following assumptions:
That gas molecules are moving very fast and randomly
That molecules hardly have any volume
That gas molecules do not attract or repel each other (no intermolecular forces)
No kinetic energy is lost when the gas molecules collide with each other (elastic collisions)
The temperature of the gas is related to the average kinetic energy of the molecules
Gases that follow the kinetic theory of gases are called ideal gases
However, in reality gases do not fit this description exactly but may come very close and are
called real gases
Ideal gases
The volume that an ideal gas occupies depends on:
Its pressure
Its temperature
When a gas is heated (at constant pressure) the particles gain more kinetic energy and undergo
more frequent collisions with the container wall
To keep the pressure constant, the molecules must get further apart and therefore the volume
increases
The volume is therefore directly proportional to the temperature (at constant pressure)

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Your notes

The volume of a gas increases upon heating to keep a constant pressure (a); volume is directly
proportional to the temperature (b)
Limitations of the ideal gas law
At very low temperatures and high pressures real gases do not obey the kinetic theory as under
these conditions:
Molecules are close to each other

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There are instantaneous dipole- induced dipole or permanent dipole- permanent dipole
forces between the molecules
These attractive forces pull the molecules away from the container wall Your notes
The volume of the molecules is not negligible
Real gases therefore do not obey the following kinetic theory assumptions at low temperatures
and high pressures:
There is z ero attraction between molecules (due to attractive forces, the pressure is lower
than expected for an ideal gas)
The volume of the gas molecules can be ignored (volume of the gas is smaller than expected
for an ideal gas)
Ideal gas equation
The ideal gas equation shows the relationship between pressure, volume, temperature and
number of moles of gas of an ideal gas:
PV = nRT
P = pressure (pascals, Pa)
V = volume (m3)
n = number of moles of gas (mol)
R = gas constant (8.314 J mol-1 K-1)
T = temperature (kelvin, K)

Worked example
Calculating the volume of a gas

Calculate the volume occupied by 0.781 mol of oxygen at a pressure of 220 kPa and a
temperature of 21 °C.

Answer
Step 1: Rearrange the ideal gas equation to find volume of gas

nRT
V=
P
Step 2: Calculate the volume the oxygen gas occupies

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p = 220 kPa = 220 000 Pa

n = 0.781 mol Your notes
R = 8.314 J mol-1 K-1
T = 21 oC = 294 K

0. 781 mol × 8. 314 J K−1 mol −1 × 294 K

V= = 0. 00867 m 3 = 8. 67 dm 3
220000 Pa

Worked example
Calculating the molar mass of a gas

A flask of volume 1000 cm3 contains 6.39 g of a gas. The pressure in the flask was 300 kPa and
the temperature was 23 °C.
Calculate the relative molecular mass of the gas.

Answer
Step 1: Rearrange the ideal gas equation to find the number of moles of gas

PV
n=
RT
Step 2: Calculate the number of moles of gas
p = 300 kPa = 300 000 Pa
V = 1000 cm3 = 1 dm3 = 0.001 m3
R = 8.314 J mol-1 K-1
T = 23 oC = 296 K

300000 Pa × 0. 001 m 3
n= = = 0. 1219 mol
8. 314 J K−1 mol −1 × 296 K

Step 3: Calculate the molar mass using the number of moles of gas

mass
n-
molar mass

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6. 39 g
Molar mass = = 52 . 42 g mol −1
0. 1219 mol
Your notes
Exam Tip
To calculate the temperature in Kelvin, add 273 to the Celsius temperature - e.g. 100 oC is 373
Kelvin

You must be able to rearrange the ideal gas equation to work out all parts of it

The units are incredibly important in this equation - make sure you know what units you should
use, and do the necessary conversions when doing your calculations!

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2.2.5 Percentage Yield & Atom Economy

Your notes
Percentage Yield Calculations
Percentage yield
In a lot of reactions, not all reactants react to form products which can be due to several factors:
Other reactions take place simultaneously
The reaction does not go to completion
Products are lost during separation and purification
The percentage yield shows how much of a particular product you get from the reactants
compared to the maximum theoretical amount that you can get:
actual yield
percentage yield = × 100
theoretical yield
The actual yield is the number of moles or mass of product obtained experimentally
The theoretical yield is the number of moles or mass obtained by a reacting mass calculation

Worked example
In an experiment to displace copper from copper(II) sulfate, 6.5 g of z inc was added to an
excess of copper(II) sulfate solution. The resulting copper was filtered off, washed and dried.
The mass of copper obtained was 4.8 g.
Calculate the percentage yield of copper.

Answer:
Step 1: The balanced symbol equation is:
Zn (s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu (s)

Step 2: Calculate the amount of z inc reacted in moles

6.5 g
number of moles = = 0 . 10 mol
65. 4 g mol −1

Step 3: Calculate the maximum amount of copper that could be formed from the molar ratio:

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Since the ratio of Zn(s) to Cu(s) is 1:1 a maximum of 0.10 moles can be produced
Your notes
Step 4: Calculate the maximum mass of copper that could be formed (theoretical yield)

mass = mol x M
mass = 0.10 mol x 63.55 g mol-1
mass = 6.4 g (2 sig figs)

Step 5: Calculate the percentage yield of copper

4.8 g
percentage yield = × 100 = 75%
6.4 g

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Atom Economy Calculations

The atom economy of a reaction shows how many of the atoms used in the reaction become the Your notes
desired product
The rest of the atoms or mass is wasted
It is found directly from the balanced equation by calculating the Mr of the desired product

molecular mass of desired product

Atom economy = × 100
sum of molecular masses of ALL reactants
In addition reactions, the atom economy will always be 100% because all of the atoms are used
to make the desired product
Whenever there is only one product, the atom economy will always be 100%
For example, in the reaction between ethene and bromine:
CH2 =CH2 + Br2 → CH2 BrCH2 Br
The atom economy could also be calculated using mass, instead or Mr
In this case, you would divide the mass of the desired product formed by the total mass of all
reactants, and then multiply by 100
Questions about atom economy often asked in qualitative or quantitative terms

Worked example
Qualitative atom economy

Ethanol can be produced by various reactions, such as:

Hydration of ethene: C 2 H4 + H2 O → C 2 H5OH
Substitution of bromoethane: C 2 H5Br + NaOH → C 2 H5OH + NaBr

Explain which reaction has a higher atom economy.

Answer
Hydration of ethene has a higher atom economy (of 100%) because all of the reactants are
converted into products, whereas the substitution of bromoethane produces NaBr as a waste
product

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Worked example
Your notes
Quantitative atom economy
The blast furnace uses carbon monoxide to reduce iron(III) oxide to iron.
Fe 2 O 3 + 3CO → 2Fe + 3CO 2

Calculate the atom economy for this reaction, assuming that iron is the desired product.

(Ar / Mr data: Fe 2 O 3 = 159.6, CO = 28.0, Fe = 55.8, CO 2 = 44.0)

Answer
Step 1: Write the equation:

molecular mass of desired product

Atom economy = × 100
sum of molecular masses of ALL reactants
Step 2: Substitute values and evaluate:

2 × 55. 8
Atom economy = × 100 = 45. 8%
159 . 6 + (3 × 28. 0)

Exam Tip
Careful: Sometimes a question may ask you to show your working when calculating atom
economy.
In this case, even if it is an addition reaction and it is obvious that the atom economy is 100%, you
will still need to show your working.

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Benefits of High Atom Economy

Chemists use percentage yield as one possible measure of how efficient a reaction is Your notes
A high percentage yield suggests that a process is effective at converting reactants into
products
The estimated percentage yield for a single run of the Haber Process is around 15%
This is a compromise due to the cost and safety of the required conditions against the
overall rate of ammonia production
Any unreacted materials are also recycled so it is estimated that the percentage conversion
of all reactants to products is around 97%
Whilst a high percentage yield can be good for profits, it does not account for any waste
products
A reaction can have a high percentage yield but low atom economy which essentially means
that more waste products are produced
Atom economy is a measure of the percentage of reactants that become useful products and is
calculated by:
mass of desired product
Percentage atom economy = × 100
total mass of ALL reactants
Atom Economy and Green Chemistry
Chemists will often have several choices of reaching a target molecule and those choices need
to take into the principles of Green Chemistry

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Your notes

The twelve principles of green chemistry

By choosing a reaction pathway that has fewer steps, you can prevent waste and reduce energy
demands which is better for the environment
This also reduces production costs
One of the key ideas behind Green Chemistry is to find reaction pathways with high percentage
yield and high atom economy
By analysing the atom economy of each step, you can select reactions that give a higher atom
economy and / or select other reactions to reduce the number of steps involved in a reaction
pathway
For example, the synthesis of ibuprofen that was patented by Boots in the 1960's was a six-
step synthesis
Even if each step has an atom economy of 90%, a six-step synthesis would have an overall
atom economy of 53%
The modern production of ibuprofen is a three-step synthesis, which with the same
assumptions as before, gives an overall atom economy of 73%
Higher atom economy means that there is less waste produced
This can be considered environmentally friendly even though it may not influence the reaction
conditions
It also means that reactions are more sustainable and often use less natural / finite resource

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