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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level
Online Chemical Calculations
14.
Other GCSE chemical calculations
The methods of how to calculate % purity?, how do you
calculate % yield? are all explained with worked out examples, What is the 'atom economy' of a chemical reaction?,
How do you calculate atom economy?, How do you do solution dilution
calculations, what method is use to measure water of crystallisation?,
how to you calculate water of
crystallisation in a salt?, How do you theoretically calculate quantities
of chemicals required for a chemical reaction from its symbol equation? Read on
.... !
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14.
Other GCSE chemical calculations
% purity of a product
* % reaction yield * atom economy *
dilution of solutions
water of crystallisation
calculation
* how
much of a reactant is needed?
14.1 Percentage purity of a chemical reaction product

Purity is very
important e.g. for analytical standards in laboratories or
pharmaceutical products where impurities could have dangerous side
effects in a drug or medicine. However in any chemical process it is
almost impossible to get 100.00% purity and so samples are always
analysed in industry to monitor the quality of the product.

% purity is the
percentage of the material which is the actually desired chemical in a sample
of it.

Example 14.1.1

A 12.00g sample of a
crystallised pharmaceutical product was found to contain 11.57g of the
active drug.

Calculate the % purity of the sample of the drug.

% purity = actual
amount of desired material x 100 / total amount of material

% purity =
11.57 x 100 / 12 = 96.4% (to 1dp)

Example 14.1.2

Sodium chloride was
prepared by neutralising sodium hydroxide solution with dilute
hydrochloric acid. The solution was evaporated to crystallise the salt.

The salt is required to be completely anhydrous, that is, not containing
any water.

The prepared salt
was analysed for water by heating a sample in an oven at 110^{o}C
to measure the evaporation of any residual water.

The following results
were obtained and from them calculate the % purity of the salt.

Mass of evaporating
dish empty = 51.32g.

Mass of impure salt
+ dish = 56.47g

Mass of dish + salt
after heating = 56.15g

Therefore the mass
of original salt = 56.47  51.32 = 5.15g

and the mass of pure
salt remaining = 56.15  51.32 = 4.83g

% salt purity
= 4.83 x 100 / 5.15 = 93.8% (to 1dp)

Advanced AS Q17 & Q20 examples
of puritytitration calculation based on an organic acid titrations


14.2a Percentage yield of the product of a reaction

The % yield of
a reaction is the percentage of the product obtained compared to the
theoretical maximum as calculated from the balanced equation.

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.

However in any chemical process
it is almost impossible to get 100% of the product because of several
reasons:

the reaction
might not be 100% completed because it is reversible reaction and an
equilibrium is established.

You always get losses of
product as it is separated from the reaction mixture by filtration,
distillation, crystallisation or whatever method is required.

Some of the reactants
may react in another way to give a different product to the one you want.

The aim is to
work carefully and recover as much of the desired reaction product, and as
pure as is possible and practicable

% yield
= actual amount of desired chemical obtained x 100 /
maximum theoretical amount formed

Example 14.2a.1

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 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 magnesium chloride was obtained after crystallising the salt from
the solution, what is the % yield of the salt preparation?

Example 14.2a.2

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.

2nd the % yield
calculation itself.

Example
14.2a.3

(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

(b) If in reality, only 670 tonne
of iron is produced what is the % yield of the overall
blast furnace process?

% yield = actual yield x
100 / theoretical yield

% yield = 670 x 100 /
700 = 95.7%

In other words, 4.3% of
the iron is lost in waste e.g. in the slag.

Example 14.2a.4

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 yield calculation

I've added more percentage yield
calculations to the
Reacting mass ratio calculations of reactants and products from equations
page
14.2b
The
Atom economy of a chemical reaction
The atom economy of a
reaction is a theoretical measure of the amount of starting materials that ends up
as 'desired' reaction product. 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. It can be defined numerically in words in
several ways, all of which amount to the same theoretical % number!
Atom
economy 
mass of atoms of desired
product 
mass of useful product 
total M_{r} of useful product 
 x 100
= 
 x 100
= 
 x 100 
total mass of reactant atoms 
mass of
all products 
total M_{r} of all products 
Example 14.2b.1
This is illustrated by using the
blast furnace reaction from example 14.2a.3 above.
Fe_{2}O_{3(s)}
+ 3CO_{(g)} ===> 2Fe_{(l)} + 3CO_{2(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 Fe_{2}O_{3}]
+ [84 of CO] ===> [112 of Fe] + [132 of CO_{2}]
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 112 + 132 = 244.
Therefore the atom economy = 112 x
100 / 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 from the law of conservation
of mass.
14.3
Dilution of solutions calculations
calculating
dilutions  volumes involved etc.

It is important to study
Calculations Part 11
Introducing
Molarity, volumes
and the concentration of solutions before tackling this section and note
the triangle of relationships you need to know!

It is important to know how to
accurately dilute a more concentrated solution to a specified solution of
lower concentration. It involves a bit of logic using ratios of volumes. The
diagram above illustrates some of the apparatus that might be used when
dealing with solutions.

Example 14.3.1

A purchased standard solution of sodium
hydroxide had a concentration of 1.0 mol/dm^{3}. How would you prepare 100 cm^{3} of
a 0.1 mol/dm^{3} solution to do a titration of an acid?

The required
concentration is 1/10th of the original solution.

To make 1dm^{3} (1000 cm^{3}) of
the diluted solution you would take 100 cm^{3} of the original solution and mix
with 900 cm^{3} of water.

The total volume is 1dm^{3} but only 1/10th as much sodium
hydroxide in this diluted solution, so the concentration is 1/10th, 0.1
mol/dm^{3}.

To make only 100 cm^{3} of the diluted solution you would dilute 10cm^{3}
by mixing it with 90 cm^{3} of water.

How to do this in practice is described at
the end of Example 14.3.2 below and a variety of accurate/'less inaccurate' apparatus
is illustrated above.

Example 14.3.2

Given a stock solution of sodium chloride
of 2.0 mol/dm^{3}, how would you prepare 250cm^{3} of a 0.5 mol/dm^{3} solution?

The
required 0.5 mol/dm^{3} concentration is 1/4 of the original concentration of 2.0
mol/dm^{3}.

To make 1dm^{3} (1000 cm^{3}) of a 0.5 mol/dm3 solution you would take 250
cm^{3} of the stock solution and add 750 cm^{3} of water.

Therefore to make only 250
cm^{3} of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm^{3} of
the stock solution plus 187.5 cm^{3} of pure water.

This can be done, but rather
inaccurately, using measuring cylinders and stirring to mix the two liquids in
a beaker.

It can be done much more accurately by using a burette or a pipette
to measure out the stock solution directly into a 250 cm^{3} graduatedvolumetric
flask.

Topping up the flask to the calibration mark (meniscus should rest on
it). Then putting on the stopper and thoroughly mixing it by carefully shaking
the flask holding the stopper on at the same time!

Example
14.3.3

In
the analytical laboratory of a pharmaceutical company a laboratory assistant
was asked to make 250 cm^{3} of a 2.0 x 10^{2} mol dm^{3}
(0.02M)
solution of paracetamol (C_{8}H_{9}NO_{2}).

(a) How much
paracetamol should the laboratory assistant weigh out to make up the
solution? Atomic masses: C = 12, H = 1, N = 14, O = 16

method (i): M_{r}(paracetamol)
= (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151

1000 cm^{3} of
1.0 molar solution needs 151g

1000 cm^{3} of
2.0 x 10^{2} molar solution needs 151 x 2.0 x 10^{2}/1 =
3.02g

(this is just scaling
down the ratio from 151g : 1.0 molar)

Therefore to make 250
cm^{3} of the solution you need 3.02 x 250/1000 = 0.755 g

method (ii): M_{r}(paracetamol)
= 151

moles = molarity x
volume in dm^{3}

mol paracetamol required
= 2.0 x 10^{2} x 250/1000 = 5.0 x 10^{3} (0.005)

mass = mol x M_{r}
= 5.0 x 10^{3} x 151 = 0.755 g

(b) Using the 2.0 x
10^{2} molar stock solution, what volume of it should be added to a
100cm^{3} volumetric flask to make 100 cm^{3} of a 5.0 x 10^{3}
mol dm^{3} (0.005M) solution?

The ratio of the two
molarities is stock/diluted = 2.0 x 10^{2}/5.0 x 10^{3} =
4.0 or a dilution factor of 1/4 (0.02/0.005).

Therefore 25 cm^{3}
(^{1}/_{4} of 100) of the 2.0 x 10^{2} molar
solution is added to the 100 cm^{3} volumetric flask prior to
making it up to 100 cm^{3} with pure water to give the 5.0 x 10^{3}
mol dm^{3} (0.005M) solution.

There are more questions
involving molarity in
section 7.
introducing molarity and
section 12. on
dilution

Example 14.3.4

You are given a stock
solution of concentrated ammonia with a concentration of 17.9 mol dm^{3}
(conc. ammonia! ~18M!)

(a) What volume of the conc.
ammonia is needed to make up 1dm^{3} of 1.0 molar ammonia solution?

Method (i) using simple ratio
argument.

The conc. ammonia must be
diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.

Therefore you need 1.0/17.9 x
1000 cm^{3} = 55.9 cm^{3} of the conc.
ammonia.

If the 55.9 cm^{3} of
conc. ammonia is diluted to 1000 cm^{3} (1 dm^{3}) you will
have a 1.0 mol dm^{3} (1M) solution.

Method (ii) using molar
concentration equation  a much better method that suits any kind
of dilution calculation involving molarity.

molarity = mol / volume (dm^{3}),
therefore mol = molarity x volume in dm^{3}

Therefore you need 1.0 x 1.0
= 1.0 moles of ammonia to make 1 dm^{3} of 1.5M dilute ammonia.

Volume = mol / molarity

Volume of conc. ammonia needed =
1.0 / 17.9 = 0.0559 dm^{3} (55.9
cm^{3}) of the conc. ammonia is required,

and, if this is diluted to 1 dm^{3},
it will give you a 1.0 mol dm^{3} dilute ammonia solution.

(b) What volume of conc.
ammonia is needed to make 5 dm^{3} of a 1.5 molar solution?

molarity = mol / volume (dm^{3}),
therefore mol = molarity x volume in dm^{3}

Therefore you need 1.5 x 5 =
7.5 moles of ammonia to make 5 dm^{3} of 1.5M dilute ammonia.

Volume (of conc. ammonia
needed) = mol / molarity

Volume of conc. ammonia needed =
7.5 / 17.9 = 0.419 dm^{3} (419 cm^{3})
of the conc. ammonia is required,

and, if this is diluted to 5 dm^{3},
it will give you a 1.5 mol dm^{3} dilute ammonia solution.


14.4 Water of crystallisation
in a crystallised salt

Example 14.4.1:
Calculate the % of water in hydrated magnesium sulphate MgSO_{4}.7H_{2}O
salt crystals

Relative atomic masses:
Mg = 24, S = 32, O = 16 and H = 1

relative formula mass =
24 + 32 + (4 x 16) + [7 x (1 + 1 + 16)] = 246

7 x 18 = 126 is the mass
of water

so % water = 126
x 100 / 246 = 51.2%

Example 14.4.2 The %
water of crystallisation and the formula of the salt are calculated as
follows:

Suppose 6.25g of blue
hydrated copper(II) sulphate, CuSO_{4}.xH_{2}O, (x unknown)
was gently heated in a crucible until the mass remaining was 4.00g. This is
the white anhydrous copper(II) sulphate.

The mass of anhydrous
salt = 4.00g, mass of water (of crystallisation) driven off = 6.254.00 =
2.25g

The % water of
crystallisation in the crystals is 2.25 x 100 / 6.25 = 36%

[ Ar's Cu=64, S=32,
O=16, H=1 ]

The mass ratio of CuSO_{4}
: H_{2}O is 4.00 : 2.25

To convert from mass
ratio to mole ratio, you divide by the molecular mass of each 'species'

CuSO_{4} = 64 +
32 + (4x18) = 160 and H_{2}O = 1+1+16 = 18

The mole ratio of CuSO_{4}
: H_{2}O is 4.00/160 : 2.25/18

which is 0.025 : 0.125
or 1 : 5, so the formula of the hydrated salt is CuSO_{4}.5H_{2}O
14.5
Calculation of quantities required for a chemical reaction, % yield and atom economy
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 volumetric titration calculations e.g. acidalkali titrations
(and diagrams of apparatus)

Electrolysis products calculations (negative cathode and positive anode products)

Other calculations
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions, water of crystallisation, quantity of reactants
required, atom economy
(this page)

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
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