|
See also on other pages
14.1
% purity of a product and assay calculations
14.2b atom economy calculations
*
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
14.2a Percentage yield of the product of a reaction
Even though no atoms are gained or lost
in a chemical reaction (law of conservation of mass), unfortunately it is
never possible to obtain the calculated amount of a product (i.e. 100%
yield) because:
(i) the reaction may not go to completion because it may be
reversible,
(ii) some of the product may be lost when it is separated from the
reaction mixture (eg via crystallisation or distillation),
(iii) some of
the reactants may react in ways different to the expected reaction (side
reactions to give bye-products which may be unwanted or useful).
The amount of a product you actually obtain is known as the
yield.
When compared with the maximum theoretical amount calculated as a
percentage, it is called the percentage yield.
You should be able to
calculate the percentage yield of a product from the actual yield of a
reaction or calculate the theoretical mass of a product from a given mass of
reactant given the balanced equation for the reaction.
In industrial processes of the chemical
industry, the reactions with highest yield are going to be the most
economic and produce the least waste that has to be dealt with.
-
YIELD The actual
yield is the mass of useful product you get from a chemical reaction and
this actual yield can be compared with the maximum theoretical yield if
everything could be done perfectly, which you can't!
-
The % yield of
a reaction is defined as the percentage of the product obtained compared to the
theoretical maximum (predicted) yield calculated from the balanced equation.
-
You get the predicted maximum
theoretical yield from a
reacting mass calculation (see examples further down).
-
The comparison of the actual
yield and the theoretical maximum yield can be expressed as the percentage
yield.
| |
ACTUAL YIELD (e.g. in grams,
kg, tonnes) |
|
PERCENTAGE YIELD
= |
100 x
--------------------------------------------------------------------------------------------- |
| |
PREDICTED theoretical YIELD (same mass units as above) |
-
In carrying out a chemical
preparation, the aim is to
work carefully and recover as much of the desired reaction product as you
can, and as
pure as is possible and practicable.
-
Despite the law of conservation
of mass, i.e. no atoms lost or gained, in real chemical preparations things
cannot work out completely according to chemical theory, often for quite
simple, physical or sometimes chemical reasons,
-
and it doesn't matter if its a
small scale school laboratory preparation or a large scale industrial
manufacturing process, in reality. the percent yield is never 100%.
-
You may need to use the rearranged
equation i.e. change of equation subject ...
-
e.g. 100 x Actual Yield = % Yield X
Theoretical Yield
-
so, Actual Yield (mass) = %
Yield X Theoretical Yield (mass) / 100
-
In industry, you want the
highest possible yield to reduce costs and have less uneconomic waste.
-
REASONS why you never get 100 percent yield of the desired product in
chemical reaction preparations
-
LOSSES So, in any chemical process,
it is almost impossible to get 100% of the product because of many
reasons: Four reasons why you do not get a 100% yield in a chemical
reaction are described and explained below.
-
The reaction
might not be 100% due to an equilibrium
-
The reaction may not be
completed because it is reversible reaction and an
equilibrium is established (note the sign
in the equation below.
-
Both reactants and products co-exist in the same
reaction mixtures (solutions or gases) i.e. the reaction can never go to
completion.
- In the
Haber Synthesis
of ammonia the conversion of hydrogen and nitrogen to ammonia is
only about 6-15% depending on the reactor temperature and pressure.
- N2(g) + 3H2(g)
2NH3(g)
- BUT, it isn't all bad news, the ammonia is quite
easily condensed out from the reaction mixture and the nitrogen and
hydrogen gases are recycled through the reactor, so there is very little waste.
- -
-
You always get losses of
the desired product when separating out the product
-
There are always losses when
the product is separated from the reaction mixture by filtration,
distillation, crystallisation or whatever method is required e.g.
-
Bits of solid or
droplets are left behind on the sides of the apparatus or reactor
vessel e.g. in the reaction flask, filter funnel and paper when
recovering the product from the reaction mixture or transferring a
liquid from one container to another.
-
Small amounts of liquid will be left in distillation units or
solid particles on the surface of filtration
units.
-
In fact, whenever you have
to manipulate or transfer the product in some way, there are bound to be
residual losses somewhere in the laboratory apparatus or a full-scale
chemical plant.
-
If the product is a volatile
liquid, there will be losses due to evaporation.
-
You cannot avoid losing
traces of product in all stages of the manufacturing process.
-
See
Four techniques used in a particular and
separation and purification procedure,
-
DISTILLATION and
Separating
funnel, solvent extraction, centrifuging
-
where these sort of
losses may be encountered.
-
-
-
Some of the reactants
may react in another way to give a different product
-
Other reactions might take
place to the one you want
(so-called by-products).
-
By-products are very common
in organic chemistry due to different, but con-current, reactions.
-
(i) A + B ==> C + D
-
(ii) A + B ==> E + F
-
A con-current
reaction, maybe just involving a few % of the reactants to give the
minor, and often undesirable, by-products of E + F.
-
Sometimes the by-products can be separated as a useful product and sold to help the
economics of the chemical process overall.
-
-
-
The original reactants might
not be pure.
-
You can't make the desired
product from the wrong chemicals - unwanted chemicals in your
initial reactants.
-
When you measure out the
reactants, you are not taking into account any impurities, so
your yield will be automatically reduced because of the wrong
chemical has been added to the reaction vessel.
-
The aim is to
work as carefully as possible and recover as much of the desired reaction product, and as
pure as is possible and practicable
- If a chemical reaction gives a
low percentage yield of useful product, research would be undertaken to find
ways the reaction can been improved to increase the yield of useful product.
- It might be possible to find
another synthetic route to produce a particular chemical that gives a higher
percentage yield and less waste (see also
Atom Economy).
-
% yield
= actual amount of desired chemical obtained x 100 /
maximum theoretical amount that could be formed
-
If the reaction doesn't work the
yield is zero or 0%.
-
If the reaction works perfectly
and you obtain all the product, the yield is 100%, BUT this never happens in
reality (as already discussed above).
-
The theoretical yield can be
calculated from the balanced equation by doing a
reacting mass ratio calculation.
-
Yield and industrial production
-
The higher the yield of a
reaction, the more economic is the process.
-
There will be less waste to deal
with and dispose of, which involves extra cost.
-
Waste is of no commercial value
and may be harmful to people and the environment.
-
High yields means less energy
used, saving money.
-
Research chemists in the
chemical industry are always looking for the most efficient (cost effective)
of making a particular product and the main criteria being ...
-
A high yield reaction,
particularly if the raw materials are expensive, and the resource may not be
infinite!
-
that goes as fast as possible -
a good economic rate,
-
all of the products are
commercially viable, so can even by-products be converted to some saleable
chemical as well as the main desired product.
EXAMPLES OF YIELD CALCULATIONS
Note: If there is only one product in
the reaction, the atom economy is theoretically 100%
(i) Haber synthesis of ammonia
N2(g) + 3H2(g)
===> 2NH3(g)
(ii) The addition of bromine to an alkene
CH2=CH2 +
Br2 ===> BrCH2-CH2Br
(iii) Burning a metal to form its oxide
Mg(s) + O2(g)
===> 2MgO
BUT remember, this is NOT the yield, there are
always losses!
You should understand the %
yield calculation must be based on the reactant NOT in excess.
Examples (1) to (6) do NOT involve moles.
From
example (7) onwards moles are involved and are more suitable for
advanced level chemistry students
BASIC
CALCULATIONS (1) to (5),
but you must be able to do reacting mass
calculations
-
% yield calculation Example 14.2a
(1) (reduction of copper oxide with hydrogen)
- Copper(II) oxide can be reduced to copper by
heating the oxide carefully in a stream of oxygen.
- 5.0 g of copper(II) oxide where heated in oxygen
until there appeared to be no more change in colour.
- Any unreacted copper(II) oxide was dissolved in
dilute sulfuric acid and the copper filtered off, dried and weighed.
- Using the atomic masses: Cu = 63.5, H = 1,
O =16
- Formula masses: CuO = 79.5, H2 =
2, Cu = 63.5, H2O = 18
- (a) Calculate the maximum quantity of copper you
could obtain.
- The relevant ratio is CuO ==> Cu
- which is 79.5 ==> 63.5
- Therefore scaling down to 5 g of the oxide
...
- the maximum yield of Cu = 5 x 63.5/79.5 =
3.99 g (2 dp, 3 sf)
- -
- (b) If 3.91 g of copper was actually obtained,
calculate the percentage yield of the reaction.
- % yield = 100 x actual yield / theoretical
maximum yield
- % yield = 100 x 3.91 / 3.99
-
% yield of Cu = 98.0% (3 sf, 1
dp)
- -
-
% yield calculation Example 14.2a
(2) (dissolving magnesium in
hydrochloric acid)
-
Magnesium metal
dissolves in hydrochloric acid to form magnesium chloride.
-
Mg(s) + 2HCl(aq)
==> MgCl2(aq)
+ H2(g)
-
Atomic masses : Mg = 24
and Cl = 35.5, and formula mass MgCl2 = 24 + (2 x 35.5) = 95
-
(a) What is the maximum theoretical mass of
anhydrous magnesium chloride which can be made from
12g of magnesium?
-
Reacting mass ratio
calculation from the balanced equation:
-
1 Mg ==> 1 MgCl2,
so 24g ==> 95g or 12g ==>
47.5g MgCl2
-
(b) If only 47.0g of
purified anhydrous magnesium chloride was obtained after crystallising the salt from
the solution and heating it to drive off the water of crystallisation, what is the % yield
from the salt preparation?
-
% yield = actual amount
obtained x 100 / maximum theoretical amount possible
-
% yield = 47.0 x
100 / 47.5 =
98.9% (to 1dp, 3sf)
-
More examples of % yield and atom economy calculations in section 6.
-
-
-
% yield calculation Example 14.2a
(3) (making iron sulfide from
iron and sulfur)
-
2.8g of iron was heated
with excess sulphur to form iron(II) sulphide.
-
Fe + S ==> FeS
-
The excess sulphur was
dissolved in a solvent and the iron sulphide filtered off, washed with clean
solvent and dried.
-
If 4.1g of purified iron
sulphide was finally obtained, what was the % yield of the reaction?
-
1st a reacting mass
calculation of the maximum amount of FeS that can be formed:
-
Relative atomic/formula
masses: Fe =56, FeS = 56 + 32 = 88
-
This means 56g Fe ==>
88g FeS, or by ratio, 2.8g Fe ==> 4.4g FeS
-
because 2.8 is 1/20th of
56, so theoretically you can get 1/20th of 88g of FeS or 4.4g.
-
2nd the % yield
calculation itself.
-
% yield = actual amount
obtained x 100 / maximum theoretical amount possible
-
% yield = 4.1 x
100 / 4.4 = 93.2% (to 1dp,
3sf)
-
More examples of % yield and atom economy calculations in section 6.
-
-
-
% yield calculation Example
14.2a (4) (reduction
of haematite iron ore in a blast furnace)
-
(a) Theoretically how
much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe2O3
in a blast furnace?
-
If we assume the
iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation
is:
-
Fe2O3(s)
+ 3CO(g) ==> 2Fe(l) + 3CO2(g)
-
(atomic masses: Fe = 56,
O = 16)
-
For every Fe2O3
==> 2Fe
can be extracted, formula mass of ore = (2 x 56) + (3 x 56) = 160
-
Therefore reacting mass
ratio is: 160 ==> 112 (from 2 x 56)
-
so, solving the ratio,
1000 ==> 112 / 160 =
700 tonne copper = max. can be extracted
-
% yield calculation Example 14.2a
(5) (burning magnesium in
air/oxygen)
-
Given the atomic masses:
Mg = 24 and O = 16,
-
and the reaction between
magnesium to form magnesium oxide is given by the symbol equation
-
2Mg(s) + O2(g)
==> 2MgO(s)
-
(a) What mass of
magnesium oxide can be made from 1.00g of magnesium?
-
2Mg ==> 2MgO
-
in terms of reacting
masses (2 x 24) ==> {2 x (24 +16)}
-
so 48g Mg ==> 80g MgO
(or 24g ==> 40g, its all the same)
-
therefore solving the
ratio
-
1g Mg ==> w g
MgO, using the ratio 48 : 80
-
w = 1 x 80 / 48
= 1.67g MgO
(3 sf)
-
(b) Suppose the % yield
in the reaction is 80%. That is only 80% of the magnesium oxide formed is
actually recovered as useful product. How much magnesium needs to be
burned to make 30g of magnesium oxide?
-
This is a bit tricky and
needs to done in two stages and can be set out in several ways.
-
Now 48g Mg ==> 80g MgO
(or any ratio mentioned above)
-
so y g Mg ==> 30g
MgO
-
y = 30 x 48 / 80
= 18g Mg
-
BUT you only get back
80% of the MgO formed,
-
so therefore you need to
take more of the magnesium than theoretically calculated above.
-
Therefore for practical
purposes you need to take, NOT 18g Mg, BUT ...
-
... since you only get
80/100 th's of the product ...
-
... you need to use
100/80 th's of the reactants in the first place
-
therefore Mg needed =
18g x 100 / 80 =
22.5g Mg (3sf)
-
CHECK: reacting mass
calculation + % yield calculation CHECK:
-
22.5 Mg ==> z MgO, z =
22.5 x 80 / 48 = 37.5g MgO,
-
but you only get 80% of
this,
-
so you actually get 37.5
x 80 / 100 = 30g
-
This means in
principle that if you only get x% yield ...
-
... you need to take
100/x quantities of reactants to compensate for the losses.
-
-
-
% yield calculation Example 14.2a
(5) (Haber synthesis of ammonia)
-
In laboratory experiments to
improve the catalyst used to synthesise, the best catalyst produced
a yield of 20.0% at a particular temperature and pressure.
-
In the industrial sized plant,
what yield of ammonia would be obtained after 8.00 tonne of nitrogen
passed through the reactor vessel?
MORE ADVANCED CALCULATIONS
from (6) onwards, but you must be familiar with
moles related to equations and
moles related to gas volumes,
otherwise you can't proceed to calculate a yield from the theoretical
100% yield.
REMEMBER: The % yield calculation
must be based on the reactant NOT in excess
(6) involves using the molar
volume
(7) and (8) involve reacting mass
calculations, though they can be solved using mole ratios (a simpler
way?)
(9) to (11) involve problem
solving using the mole ratio from the reaction equation.
Percentage yield calculation
using moles
Percentage yield calculation example (9):
Aspirin preparation
3.50 g of 2-hydroxybenzoic acid was reacted
with excess ethanoic anhydride to yield aspirin and ethanoic acid.
Simplified equation: HO-C6H4-COOH
+ (CH3CO)2O ===> CH3CO-C6H4-COOH
+ CH3COOH
After the final crystallisation and
drying procedures, 3.90 g of aspirin was obtained.
Calculate the percentage yield of
aspirin from the preparation.
(a) Calculate the molar masses of
2-hydroxybenzoic acid and aspirin
Mr of HO-C6H4-COOH
= 138 and Mr CH3CO-C6H4-COOH
= 180
that is 138 g/mol and
180 g/mol
(b) Calculate the moles of 2-hydroxybenoic
acid used in the preparation
mol acid = 3.50/138 = 0.02536
(c) Calculate the theoretical mass for 100%
yield
Note the % yield calculation must be based
on the reactant NOT in excess.
The mole ratio of reactant to product is
1:1
Therefore maximum yield of aspirin is
0.02536 mol.
Therefore the maximum mass of aspirin that
can be formed is:
mol x molar mass = 0.02536 x 180 =
4.565 g
(d) Calculate the % yield
% yield = 100
x actual yield / maximum theoretical yield = 100 x 3.90 / 4.565
= 85.4 %
Percentage yield calculation example (10):
Dehydration of an alcohol
Refluxing propan-2-ol with conc. sulfuric acid
yields propene gas with a typical yield of 70%.
If 5.0 g of propan-2-ol is refluxed with conc.
sulfuric acid, what volume of propene is formed?
(Gas volume measured at 298 K and
101 kPa and molar volume 24 dm3/24000 cm3)
Equation: CH3CH(OH)CH3
===> CH3CH=CH2 + H2O
(a) Calculate the molecular mass of
propan-2-ol and moles of it used in the preparation
Mr = CH3CH(OH)CH3
= 60, therefore moles of propan-2-ol used = 5.0 / 60 =
0.08333 mol
(b) Calculate the maximum volume of propene
that can be formed
The mole ratio for reactant : product from
the equation is 1 : 1,
therefore the maximum theoretical yield of
propene is 0.08333 mol,
max. volume = mol x molar volume = 0.08333
x 24000 = 2000 cm3.
(c) Calculate the actual volume of propene gas
that will be formed in the preparation.
therefore for a 70% yield,
volume propene formed
is 2000 x 70 / 100 =
1400 cm3
Percentage yield calculation example (11):
Bromination of phenol
Phenol readily reacts with an aqueous solution
of bromine to give a white precipitate of 2,4,6-tribromophenol.
When 6.00 g of phenol was reacted with
excess bromine water, and, after filtration, recrystallisation in
ethanol and drying, 19.5 g of 2,4,6-tribromophenol crystals
were obtained.
Calculate the % yield of product.
(a) Calculate the molar masses of phenol and
2,4,6-tribromophenol.
Mr C6H5OH
= 94 and Mr Br3C6H2OH
= 331
Molar mass of 94 g/mol and
331 g/mol
(c) Calculate moles of phenol used in the
preparation
mol phenol = 6.0 / 94 = 0.06383
(b) Calculate the maximum theoretical yield of
product.
From the equation reactant : product
mole ratio i 1 : !.
Therefore maximum moles of product 0.06383
Max. mole product = mol x formula mass =
0.6383 x 331 = 21.13 g
(c) Calculate the
% yield of the
2,4,6-tribromophenol.
% yield = 100 x actual yield / theoretical
maximum yield = 100 x 19.5 / 21.13 =
92.3%
Percentage yield calculation example (12):
An esterification reaction
41.0 g of ethanoic acid
was refluxed with 55.0 g of butan-1-ol and an acid catalyst
to synthesise the ester butyl ethanoate. The equation is:
CH3COOH +
CH3CH2CH2CH2OH
===> CH3COOCH2CH2CH2CH3
+ H2O
After separation of the product
and fractional distillation, 55.0 g of the ester was
obtained.
Calculate the % yield, but on
which quantity is the calculation based?
(a) Calculate the molar masses of
reactants and organic product. (C= 12, H =1, O = 16)
Mr CH3COOH
= 60; Mr CH3CH2CH2CH2OH
= 74; Mr CH3COOCH2CH2CH2CH3
= 116
(b) Calculate the moles of each
reactant
mol CH3COOH =
41/60 = 0.683; mol CH3CH2CH2CH2OH
= 55/74 = 0.743
(c) Calculate the mol of product,
and from (b) deduce what the % yield calculation is based on.
mol product = 55/116 =
0.474
From (b) the alcohol is in
excess, so the % yield calculation must be based on the acid.
Theoretical moles of product
= mol of acid = 0.683
Therefore % yield
= 0.474 x 100 / 0.683 =
69%
(13)
I've added some
% yield calculations to the
Reacting mass ratio calculations of reactants and products from
equations page, and see also
Chemical economics of processes
and sociological and environmental issues etc.
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
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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
percentage % yield calculations
from chemical preparations Revision KS4 Science
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Quantitative chemistry calculations Help for problem solving in doing % yield calculations. How do you
calculate % yield? How to calculate the percentage yield of a chemical
reaction is explained with worked out examples. Explaining what we mean
by % yield calculations
based on theoretical yield versus actual yield. Online practice exam
chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE
CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB
courses. These revision notes and practice questions on how to do
percentage yield chemical calculations and worked examples should prove
useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science
courses.
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