|
See also
14.1
% purity of a product and assay calculations
14.2a
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
14.3 dilution of solutions calculations
*
14.4
water of crystallisation
calculations
14.5
how
much of a reactant is needed? calculation of quantities required, limiting
reactant quantities
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
14.2b
The
Atom economy of a chemical reaction
The atom economy (a measure of atom
utilisation or efficiency) is a measure of the amount of starting materials
that end up as useful products. It is important for sustainable development
and for good economic reasons to use reactions with high atom economy.
A chemical reaction may give, and often does, more
than one product, but of the mixture of products, perhaps only one of
them is the desired useful product.
The
percentage atom economy of a reaction is readily calculated using the
balanced equation for the reaction expressed in reacting masses.
You need to
be able to calculate the atom economy of a reaction to form a desired
product from the balanced equation and ...
... perhaps explain why a
particular reaction pathway is chosen to produce a specified product given
appropriate information such as atom economy, percent yield, rate of
reaction, equilibrium
position and usefulness of by-products.
The atom economy of a
reaction is a theoretical percentage measure of the amount of starting materials that ends up
as the 'desired' useful reaction products. Its sometimes referred to
as atom utilisation.
| |
MASS of desired USEFUL PRODUCT |
|
ATOM ECONOMY
= 100 x |
------------------------------------------------------------------------------- |
| |
TOTAL MASS of all REACTANTS or PRODUCTS |
In atom economy calculations you can say REACTANTS
or PRODUCTS because of the law of conservation of mass.
The greater the % atom economy of
a reaction, the more 'efficient' or 'economic' it is likely to be,
though this is a gross simplification when complex and costly chemical
synthesis are looked at.
Many reactions give more than one
product, and not all of them are useful, so it is useful to calculate
what % of the products is theoretically useful, and we call this the
atom economy of the reaction.
The reactions that only give one product,
have the maximum atom economy of 100% and these are the most economic
reactions e.g. synthesis of ammonia and reacting ethene with water to
make ethanol.
N2 + 3H2 ==> 2NH3
and CH2=CH2 +
H2O ===> CH3CH2OH
These are simple addition reactions
where two reactants give one product e.g. synthesis of ammonia from
nitrogen and hydrogen and the synthesis of ethanol from ethene and
water.
100 minus the atom economy gives you the % waste,
but reactions with only one product will always give the highest atom
economy
If a reaction gives more than one product it may be possible to use
and sell these other products, thereby making the
process more economic overall.
Whatever, waste products have to be
dealt with and disposed of in some safe way, and this costs money!
Reactions with a low atom economy are very
wasteful and use up resources at faster rate than high atom economy
reactions.
They are usually less sustainable in the long run and the cost of raw
materials will increase over time.
e.g. the process may involve non-renewable raw
materials which will increase in costs as reserves become depleted.
BUT take care on this 'low economy' point,
e.g. many products can be recycled and, as already mentioned, the
waste products may have some economic value after further separation
and processing, known as useful by-product.
The less waste there is, the higher
the atom economy, the less materials are wasted, less energy used, so
making the process more economic, 'greener' and more sustainable.
Quite simply, the larger the atom economy
of a reaction, the less waste products are produced and hopeful the
waste is easy and cheap to deal with, or even better, some use can be
found for these waste 'by-products'.
It can be defined numerically in words in
several ways, all of which amount to the same theoretical % number!
The chemical industry is continuously working
on chemical synthetic routes that have the highest atom economy and give
the highest % yield (see section 14.2a
% reaction yield)
If a chemical
reaction that has a low atom economy or gives a low yield of useful
product research would be undertaken to find ways the reaction can
been improved to increase the yield of useful product.
Note that there are other factors when considering industrial production.
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
The factors affecting the rate of a
reaction - is the chemical process fast enough to be economic?
Is the
reaction reversible? and
Does
it involve a chemical equilibrium? - can you alter reaction conditions to
increase the yield of the desired product?
CALCULATING
the ATOM ECONOMY of a chemical reaction
You can do the calculation in any mass
units you want, or non at all by simply using the atomic/formula
masses of the reactants and products as appropriate , and I suggest
you just think like that.
The formula to calculate atom economy
can be written in several different ways and they are ALL equivalent to
each other because of the law of conservation of mass e.g.
| |
MASS of desired USEFUL PRODUCT |
|
ATOM ECONOMY
= 100 x |
--------------------------------------------------------------- |
| |
TOTAL MASS of all REACTANTS |
| |
MASS of desired USEFUL PRODUCT |
|
ATOM ECONOMY
= 100 x |
--------------------------------------------------------------- |
| |
TOTAL MASS of all PRODUCTS |
| |
TOTAL FORMULA MASSES of USEFUL PRODUCT |
|
ATOM ECONOMY
= 100 x |
----------------------------------------------------------------------------------- |
| |
TOTAL FORMULA MASSES of all REACTANTS |
| |
TOTAL FORMULA MASSES of USEFUL PRODUCT |
|
ATOM ECONOMY
= 100 x |
----------------------------------------------------------------------------------- |
| |
TOTAL FORMULA MASSES of all PRODUCTS |
THEY ALL GIVE THE SAME
ANSWER!
As long as you take into account ANY
balancing numbers in the equation, which itself should be balanced!
but they are theoretical values,
you don't get these results in real life chemistry -
see
% purity of a product &
% reaction yield
Examples of how to
work out atom economy calculations
Atom economy calculation Example 14.2b (1) See
Extraction of Iron and Steel Making for detailed
chemistry
This is illustrated by using the
blast furnace reaction from example 14.2a.3 above.
Fe2O3(s)
+ 3CO(g) ===> 2Fe(l) + 3CO2(g)
Using the atomic masses of Fe =
56, C = 12, O = 16, we can calculate the atom economy for extracting
iron.
the reaction equation can be expressed in terms of
theoretical reacting mass units
[(2 x 56) + (3 x 16)] + [3 x
(12 + 16)] ===> [2 x 56] + [3 x (12 + 16 + 16)]
[160 of Fe2O3]
+ [84 of CO] ===> [112 of Fe] + [132 of CO2]
so there are a total of 112 mass
units of the useful/desired product iron, Fe,
out of a total mass of reactants
or products of 160 + 84 = 112 + 132 = 244.
Therefore the
atom economy = 100 x 112 / 244 =
45.9%
Note: It doesn't matter whether you use the
total mass of reactants or the total mass products in the calculations,
they are the same due to the law of conservation
of mass
Atom economy calculation Example 14.2b (2) See
Ethanol Chemistry
The fermentation of sugar to make ethanol
('alcohol') and (b) converting ethanol to ethene
(a) glucose (sugar) == enzyme ==> ethanol + carbon
dioxide
C6H12O6(aq)
==> 2C2H5OH(aq) + 2CO2(g)
atomic masses: C = 12, H = 1
and O = 16
formula mass of glucose reactant = 180
(6x12 + 12x1 + 6x16)
formula mass of ethanol product = 46 (2x12 +
5x1 + 1x16 + 1x1)
relative mass of desired useful product in the
equation = 2 x
46 = 92
Atom economy = 100 x 92/180 =
51.1%
(b) ethanol === heat/catalyst ===> ethene +
water
It possible to use ethanol from fermentation
to produce ethene for plastics (polymers) manufacture instead of
relying on the cracking of crude oil fractions (e.g. a country like
Brazil with a huge agriculture system but no oil deposits.
CH3CH2OH ===>
CH2=CH2 + H2O
formula masses: ethanol = 46, ethene = 28,
water = 18
% atom economy = 100 x 28/46 =
60.9%
(c) ethene + water
===> ethanol
CH2=CH2 + H2O
====> CH3CH2OH
BUT, it is also possible to manufacture
ethanol by catalytically reacting steam with ethene from cracking
oil fractions.
Since this is a 'simple' addition reaction,
the
atom economy is 100%.
This is a much cleaner and efficient process
than fermentation which only has a 51% atom economy.
Atom economy calculation Example 14.2b (3)
All about making
hydrogen
See Ammonia Synthesis
In these three examples, Reactions 1 to 3:
Relative atomic masses used: C = 12.0, H = 1.0, O = 16.0
Hydrogen is used in synthesising ammonia and
making margarine, and
is made on a large scale from reacting methane with water
methane + water ==> hydrogen +
carbon monoxide
Reaction equation 1.: CH4(g) + H2O(g) ==>
3H2(g) + CO(g)
using formula masses gives the ratios gives
16 + 18 ==> (3
x 2) + 28
34 mass units of reactants ==>
6
mass units of useful product
Atom economy = 100 x 6 / 34 =
17.6%
BUT, the 82.4% of waste toxic carbon monoxide must be
dealt with in some way!
It seems a very inefficient process BUT ...
(i) hydrogen has very small molecular mass and
75% of the molecules are the desired product
(ii) you can actually burn the carbon monoxide
as a fuel to provide energy for power generation.
(iii) methane is readily available from the
petrochemical industry and is much cleaner to work with than impure
coal (see reactions 2. and 3.), which must be converted to coke
first - an extra process.
So, a low atom economy doesn't always mean the
process is not economically viable or necessarily produces undue waste.
Hydrogen can be made from reacting water with coke
(a form of carbon made by roasting coal)
A second reaction used to manufacture hydrogen
carbon + water ===> hydrogen +
carbon monoxide
Reaction equation 2.: C(s) + H2O(g)
===> H2(g) + CO(g)
formula masses: 12 + 18 (= 30) ==> 28 +
2 (= 30)
% atom economy = 100 x 2 / 30 =
6.67 %
This is a much lower atom economy than
reaction, and only 50% of the molecules are useful product.
However, the carbon monoxide can be made to react
further with water to form more hydrogen
carbon monoxide + water ===>
hydrogen + carbon dioxide
Reaction equation 3.: CO(g) + H2O(g)
===> H2(g) + CO2(g)
formula masses: 28 + 18 (= 46)
===> 2 + 44 (= 46)
% atom economy = 100 x 2 / 46 =
4.35
%
This is an even lower atom economy than
reaction, and again, only 50% of the molecules are useful product.
However: (a) it does take care of poisonous
carbon monoxide, but there is also something else you can do ...
(b) You can combine these process, so
combining equations 2. and 3. we get ...
C(s) + H2O(g) ===>
H2(g) + CO(g) plus
CO(g) + H2O(g)
===> H2(g) + CO2(g)
gives
------------------------------------------------------------------
C(s) + 2H2O(l)
====> 2H2(g) + CO2(g)
------------------------------------------------------------------
formula masses: 12 + 2x18 (= 48)
===> 4 + 44 (= 48)
% atom economy =
100 x 4 / 48 =
8.33 %
This is a higher atom economy than
reactions 2 or 3, and 67% of the molecules are useful
product.
Only 50% of the molecular products are
the desired product in reactions 2. and 3.
Therefore employing a secondary process
improves the efficiency of this particular process for making
hydrogen.
It should be noted that carbon monoxide
is a water insoluble toxic gas and not easy to deal with.
As already mentioned, you can burn it as a
fuel for power generation or heating a reactor vessel.
BUT, carbon dioxide is not toxic (as long
as plenty of air is around!) and importantly, it readily
dissolves in sodium hydroxide solution in gas flow 'scrubbers'
to leave hydrogen as the only remaining gas.
This greatly reduces the cost of
purifying the hydrogen to use e.g. in the Haber Synthesis of
ammonia and the hydrogenation of unsaturated plant fats to
make spreadable margarine.
Also, the reaction between steam and
carbon monoxide is exothermic, this reduces the energy needs
of the overall process.
Making hydrogen using electrolysis
(I'm only interested in atom economy here -
see
other notes on
electrolysis for details)
Relative atomic masses: Na = 23.0, Cl =
35.5, H = 1.0, O = 16.0
Process A. Electrolysis of aqueous sodium
chloride solution (brine)
Overall change: 2NaCl(aq) + 2H2O(l)
====> 2NaOH(aq) + Cl2(g) +
H2(g)
total masses for reactants or products =
(2 x 58.5) + (2 x 18) =
mass of desired product 2
% atom economy
= 100 x 2 / = 153 =
1.31%
Very low atom economy and much waste to deal
with?, BUT, sodium hydroxide and chlorine are useful saleable
chemicals.
Process B. Electrolysis of acidified water
Overall change: 2H2O(l) ===>
2H2(g) + O2(g)
total masses for reactants or products = (2 x
18) = (4 + 32) = 36
mass of desired product 2
%
atom economy = 100 x 4 / = 36 =
11.1%
This has a much greater atom economy than
process B and no waste products, no pollution.
This is an ideal process, especially if you
can use green energy e.g. from wind turbine, solar cell or
hydroelectric sources of electricity.
BUT, both processes have their place in the
chemical industry - from Process A we do need sodium hydroxide and
chlorine to manufacture other products e.g. soap from NaOH + plant
oil and PVC from ethene and chlorine.
Note 1:
These reactions contrast with the 100% atom economy of ammonia production
(for which much of the hydrogen is made),
because its an addition reaction with no extra waste
products
N2 + 3H2 ==> 2NH3
mass of reactants = mass of useful
products = 100% atom economy
Note 2:
In organic chemistry, these three types of reaction can never have a
100% atom economy because there are always at least two products -
desired useful product and often a waste product e.g.
(i)
condensation reaction - a small molecule formed in
joining two molecules to make a larger molecule
e.g. n HOOC-[][][][]-COOH
+ n H2N-[][][][]-NH2
==> -(-OC[][][][]-CONH-[][][][]-NH-)n-+
2n H2O
a schematic example of condensation
polymerisation
(ii)
elimination
reaction - a group of atoms eliminated from a molecule
e.g. CH3CH2OH ===>
CH2=CH2 + H2O
water eliminated from ethanol to form
ethene
(iii)
substitution reaction - one or more atoms substituted
by another atom or group of atoms
e.g. CH3CH2Br
+ NaOH ===> CH3CH2OH
+ NaBr
-Br group substituted by an -OH group
Atom economy calculation Example 14.2b (4)
See Ammonia Synthesis
You can use either (a) hydrogen or (b) a
hydrocarbon gas like methane to reduce the oxides of metals of low
reactivity to obtain the metal itself.
e.g. the reduction of copper(II) oxide.
Using the atomic masses: Cu = 63.5, H = 1,
O =16, C = 12
(a) CuO + H2 ==>
Cu + H2O
Calculate the atom economy of the reaction.
Formula masses: CuO = 79.5, H2
= 2, Cu = 63.5, H2O = 18
Total mass of reactants = total mass of
products 79.5 + 2 = 63.5 + 18 = 81.5
atom economy of desired product Cu
= 100 x 63.5/81.5 = 77.9%
(b) 4CuO + CH4 ==>
4Cu + 2H2O + CO2
Calculate the atom economy of the reaction.
Formula masses: CuO = 79.5, CH4
= 16, Cu = 63.5, H2O = 18, CO2
= 44
Total mass of reactants = total mass of
products
(4 x 79.5) + 16 = (4 x 63.5) + (2 x 18) + 44 =
334
atom economy of desired product Cu
= 100 x (4 x 63.5)/334 =
76.0%
(c) Which reaction has the higher atom economy?
BUT, would this be the preferential method used?
Reaction (a) has the higher atom economy, BUT,
hydrogen is probably more costly to produce than cheap methane gas
from crude oil. Therefore method (b) is probably more economic.
Rather than here, I've added more atom
economy calculations to the Reacting mass ratio calculations of reactants and products from
equations page.
See also
14.1
% purity of a product and assay calculations
14.2a
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
14.3 dilution of solutions calculations
*
14.4
water of crystallisation
calculations
14.5
how
much of a reactant is needed? calculation of quantities required, limiting
reactant quantities
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
TOP OF PAGE
Above is typical periodic table used in GCSE science-chemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
OTHER CALCULATION PAGES
-
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
-
Calculating relative
formula/molecular mass of a compound or element molecule
-
Law of Conservation of Mass and simple reacting mass calculations
-
Composition by percentage mass of elements
in a compound
-
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
-
Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination
-
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
-
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
-
Moles and the molar volume of a gas, Avogadro's Law
-
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants-products)
-
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
-
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
-
Electrolysis products calculations (negative cathode and positive anode products)
-
Other calculations
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
-
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
-
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
-
Radioactivity & half-life calculations including
dating materials
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Keywords: Quantitative chemistry calculations What is atom economy? Help on how to do atom
economy calculations. What is the formula to calculate the 'atom economy' of a chemical reaction?
Calculating the atom economy of a chemical reaction is explained on this
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