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 14.
Other GCSE chemical calculations
* % purity of a product
* % reaction yield * atom economy *
dilution of solutions *
* water of crystallisation
* how
much of a reactant is needed? *
14.1 Percentage purity
-
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 12g 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 110oC
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)
-
AS example
of purity-titration calculation based on an aspirin titration. This sort of analysis is called an 'assay'.
14.2a Percentage yield
-
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)
==> 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 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 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 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
-
(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) + O2(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.
-
-
14.2b
Atom economy
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 desired
product |
mass of useful product |
total Mr of useful product |
|
----------- -------------------------- x 100
= |
------- --------------------- x 100
= |
----------------------------------- x 100 |
|
total mass of reactant atoms |
mass of
all products |
total Mr of all products |
Example 14.2b.1
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 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.
See also
Example 6.6 in chemical calculations
section 6
14.3
Dilution calculations
-
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/dm3. How would you prepare 100 cm3 of
a 0.1 mol/dm3 solution to do a titration of an acid?
-
The required
concentration is 1/10th of the original solution.
-
To make 1dm3 (1000 cm3) of
the diluted solution you would take 100 cm3 of the original solution and mix
with 900 cm3 of water.
-
The total volume is 1dm3 but only 1/10th as much sodium
hydroxide in this diluted solution, so the concentration is 1/10th, 0.1
mol/dm3.
-
To make only 100 cm3 of the diluted solution you would dilute 10cm3
by mixing it with 90 cm3 of water.
-
How to do this in practice is described at
the end of Example 14.3.2 below and a variety of accurate/inaccurate apparatus
is illustrated above.
-
Example 14.3.2
-
Given a stock solution of sodium chloride
of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?
-
The
required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0
mol/dm3.
-
To make 1dm3 (1000 cm3) of a 0.5 mol/dm3 solution you would take 250
cm3 of the stock solution and add 750 cm3 of water.
-
Therefore to make only 250
cm3 of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm3 of
the stock solution plus 187.5 cm3 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 cm3 graduated-volumetric
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 cm3 of a 2.0 x 10-2 mol dm-3
(0.02M)
solution of paracetamol (C8H9NO2).
(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): Mr(paracetamol)
= (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151
-
1000 cm3 of
1.0 molar solution needs 151g
-
1000 cm3 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
cm3 of the solution you need 3.02 x 250/1000 = 0.755 g
method (ii): Mr(paracetamol)
= 151
-
moles = molarity x
volume in dm3
-
mol paracetamol required
= 2.0 x 10-2 x 250/1000 = 5.0 x 10-3 (0.005)
-
mass = mol x Mr
= 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
100cm3 volumetric flask to make 100 cm3 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 cm3
(1/4 of 100) of the 2.0 x 10-2 molar
solution is added to the 100 cm3 volumetric flask prior to
making it up to 100 cm3 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
14.4 Water of crystallisation
14.5
Calculation of quantities required for a reaction, % yield and atom economy
Copyright Dr W P Brown 2000-2010
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