See also
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
not always possible to obtain the calculated amount of a product (i.e. 100%
yield) because the reaction may not go to completion because it may be
reversible or some of the product may be lost when it is separated from the
reaction mixture (eg via crystallisation or distillation) or some of
the reactants may react in ways different to the expected reaction (side
reactions).
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 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, 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: Three reasons why you do not get a 100% yield in a chemical
reaction are explained and discussed.

The reaction
might not be 100% completed because it is reversible reaction and an
equilibrium is established (note thesign
in the equation below.

You always get losses of
the desired product as it 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 fullscale
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 to the one you want
(socalled byproducts).

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 byproducts be converted to some saleable
chemical as well as the main desired product.

% yield calculation Example 14.2a
(1)
 CuO + H_{2} ==> Cu +
H_{2}O
 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, H_{2} =
2, Cu = 63.5, H_{2}O = 18
 (a) Calculate the maximum quantity of copper you
could obtain.
 The relevant ratio is CuO ==> CuO
 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)

Magnesium metal
dissolves in hydrochloric acid to form magnesium chloride.

Mg_{(s)} + 2HCl_{(aq)}
==> MgCl_{2(aq)}
+ H_{2(g)}

Atomic masses : Mg = 24
and Cl = 35.5, and formula mass MgCl_{2} = 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 MgCl_{2},
so 24g ==> 95g or 12g ==> 47.5g MgCl_{2}

(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?

More examples of % yield and atom economy calculations in section 6.



% yield calculation Example 14.2a
(3)

2.8g of iron was heated
with excess sulphur to form iron 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.

More examples of % yield and atom economy calculations in section 6.



% yield calculation Example
14.2a (4)

(a) Theoretically how
much iron can be obtained from 1000 tonne of pure haematite ore, formula Fe_{2}O_{3}
in a blast furnace?

If we assume the
iron(III) oxide ore (haematite) is reduced by carbon monoxide, the equation
is:

Fe_{2}O_{3(s)}
+ 3CO_{(g)} ==> 2Fe_{(l)} + 3CO_{2(g)}

(atomic masses: Fe = 56,
O = 16)

For every Fe_{2}O_{3}
==> 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)

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)} + O_{2(g)}
==> 2MgO_{(s)}

(a) What mass of
magnesium oxide can be made from 1g 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

(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 ths of the product ...

... you need to use
100/80 ths of the reactants in the first place

therefore Mg needed =
18g x 100 / 80 = 22.5g Mg

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.



Below is an example of a more
advanced level percentage yield calculation

Rather than here, I've added more
% yield calculations to the
Reacting mass ratio calculations of reactants and products from
equations page.
TOP OF PAGE
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
TOP OF PAGE
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 GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)

How to do acidalkali
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 & halflife calculations including
dating materials
percentage % yield calculations
from chemical preparations Revision KS4 Science
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