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
14.2b atom economy calculations
*
14.3 dilution of solutions calculations
*
14.4
water of crystallisation
calculations
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
14.5 (a)
Calculation of quantities required for a chemical reaction
and a brief mention of
% yield
and
atom economy
(a) In a chemical preparation reaction
involving two reactants, it is common to use an excess of one of the reactants
to ensure that all of the other reactant is used but this still requires you to
be able to calculate the quantities of chemicals needed.
(b) The reactant chemical that is
completely used up is called the limiting reactant because it limits the amount
of products after it has all been used up. Therefore you need to be able to explain the effect of a limiting
quantity of a reactant on the maximum amount of products it is theoretically possible to obtain in
terms of amounts in moles or masses in grams.
BUT, don't forget, you still
never get in reality 100% yield.
Calculations of quantities of chemicals required to do a preparation

Example 1. How much iron and
sulfur do you need to heat together to make 20.0 g of iron sulfide

Atomic masses: Fe = 56
and S = 32

Balanced equation: Fe + S ==>
FeS

I've set out the solution to the problem
in the form of a 'logic' table.

This is essentially a reacting mass
calculation 
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product 
comments 
Fe + S 
==> 
FeS 
You need the balanced chemical equation
and in this case its very simple. 
56g + 32g 
==> 
88g 
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses 
56/88 = 0.6364g + 32/88 = 0.3636 
==> 
88g/88 = 1.00g 
divide by 88 to scale down
to 1g of FeS product 
20 x 0.6364 = 12.7g + 20 x
0.3636 = 7.3g 
==> 
20 x 100 = 20.0g 
then scale by a factor of 20 (for
20g of FeS) 
Therefore to make 20g of iron
sulfide you need 12.7g of iron and 7.3g of sulfur. 



Example 2. How much iron do
you need to make 100.0g of iron(III) chloride by passing excess chlorine
gas over heated iron filings?

Atomic masses: Fe = 56
and Cl = 35.5, formula mass of FeCl_{3} = 162.5 (56 +
3x35.5)

Balanced equation: 2Fe(s) +
3Cl_{2}(g) ==> 2FeCl_{3}(s)

Again, I've set out the solution to the
problem in the form of a 'logic' table not using moles.

Be aware that to solve this sort of
problem, you only need to pick out the relevant ration.

This is essentially a reacting mass
calculation 
Pick out the particular
ratio you need to solve the reacting mass problem
e.g. the particular reactant and product 
comments 
2Fe 
==> 
2FeCl_{3} 
You need the balanced chemical equation
BUT only these bits of the
equation are needed to solve the problem, you can ignore
the rest of the equation 
2 x 56 = 112g 
==> 
2 x 162.5 = 325g 
basic reacting mass ratio
from the balanced symbol equation and the relevant
atomic or formula masses 
112/325 = 0.3446g 
==> 
325/325 = 1.00g 
divide by 325 to scale down
to 1g of FeCl_{3} product 
100 x 0.3446 = 34.46g 
==> 
100 x 1 = 100.00g 
then scale by a factor of 100 to
make 100g of FeCl_{3} 
Therefore to make 100g of iron(III)
chloride you need 34.5g of iron (3sf) 

You
can do this calculation using the mole concept as follows:

From the equation 2 moles of iron makes 2
moles of iron(III) chloride.

Therefore 1 mole Fe makes 1 mol FeCl_{3}

Atomic mass Fe = 56, Cl = 25.5. Formula
mass of FeCl_{3} = 162.5

Using the formula on the right twice ...

mol FeCl_{3} required = 100/162.5
= 0.615 mol

Therefore 0.6154 mol of iron is required.

mass Fe required = 0.6154 x 56 =
34.5 g



Example 3 A more complex
example based on a salt preparation

Suppose you want to make
50g of
the blue hydrated copper(II) sulphate crystals

Calculation 3(a) based on the mass of copper oxide

The blue crystals contain water of
crystallisation, which must be taken into account in doing the calculation.

The
preparation is briefly described in the GCSE Acids, Bases and Salts Notes.

Your reactants are dilute sulphuric
acid and the black solid copper(II) oxide.

You can use
copper(II) carbonate, but this is not a pure simple compound and the
predictive nature of the calculations will not be as good.

copper(II) oxide +
sulphuric acid ==> copper(II) sulphate
+
water

(i) CuO_{(s)} +
H_{2}SO_{4(aq)} ==> CuSO_{4(aq) }+
H_{2}O_{(l)}

on crystallisation you get the blue
hydrated crystals of formula
CuSO_{4}.5H_{2}O

so strictly speaking, after
evaporationcrystallisation the overall equation is

(ii) CuO_{(s)}
+ H_{2}SO_{4(aq)} + 4H_{2}O_{(l)}
==>
CuSO_{4}.5H_{2}O_{(s)}
(formula of the blue crystals)
So, how much copper(II)
oxide is needed?

A 'nonmoles'
calculation first of all, involving a
reacting mass
calculation.

The crucial change overall is
CuO
==> CuSO_{4}.5H_{2}O (mass ratio 80
==> 250)

(Reminder: In reacting mass
calculations you can often ignore other reactant/product masses, pick out
the ratio you need)

Atomic masses: Cu = 64,
S
= 32, H = 1, O = 16

Formula masses are for:
CuO = 64 + 16 = 80, CuSO_{4} = 64 + 32 + (4 x 16) = 160

and CuSO_{4}.5H_{2}O
= 64 + 32 + (4 x 16) + [7 x (1+1+16)] = 250

The crucial reacting mass ratio
is: 80 ==> 250 since formula mole ratio is 1:1 in the equation.

Therefore, theoretically,
to make 50g of the crystals (^{1}/_{5}th of 250),

you need ^{1}/_{5}th of 80g of
copper(II) oxide,

and 80/5 = 16g, the mass of copper(II)
oxide required.



Calculating
the mass of copper oxide needed using moles.

Formula mass of CuSO_{4}.5H_{2}O
= 250, formula mass of CuO = 80 (details above)

50 g of CuSO_{4}.5H_{2}O
= 50/250 = 0.20 mol

So you need 0.20 mol of CuO (1 : 1
molar ratio in the balanced symbol equation)

mass = mol x formula mass
(see triangle on right!)

mass CuO needed = 0.20 x 80 = 16 g

The same answer as above, and in my
opinion, easier to manage using moles and its better to be able to do mole
calculations as well as 'nonmole' reacting mass calculations.



However, in reality,
things are not so simple because the method involves adding excess
copper(II) oxide to the dilute sulphuric acid. (see salt
preparation method (b) and see the note on 'limiting
reactant', the last section of this page.

So in practice you would
need to use more of the CuO to get anything like 50g of the salt crystals.



Calculation 3(b) based on the volume of dilute sulfuric
acid

There is another
way to calculate the quantities required based on the acid and
this is the better calculation in which you make the dilute sulfuric acid
the limiting reactant and then you can add excess copper oxide (i.e. more
than the 16 g of CuO calculated above..

How much dilute
sulphuric acid (of concentration 1 mol dm^{3}) is
required to make 50 g of CuSO_{4}.5H_{2}O?

Mol = mass in g / formula mass,

so moles of CuSO_{4}.5H_{2}O
required = 50/250 = 0.2 mol (see
basics of moles)

Therefore 0.2 mol of H_{2}SO_{4}
is required (^{1}/_{5}th mol), since the mole
ratio CuO : H_{2}SO_{4} is 1 : 1 reading the equation in
terms of moles of reactants and moles of products.

1dm^{3} (1000 cm^{3})
of a 1 mol dm^{3} solution of contains 1 mole (by
definition  see
molarity page)

Therefore 1/5th of 1 dm^{3}
is required to provide 1/5th of mole of the sulfuric acid.

so 200 cm^{3}
of 1 mol dm^{3} dilute sulphuric acid is required,

or using just 100 cm^{3}
of 2 mol dm^{3} (often labelled 2M on many laboratory reagent bottles)

You would then add
copper(II) oxide in small amounts until no more dissolves in the
warmhot acid and the excess black powder is
filtered off. There is no need to weigh out an exact amount of
copper oxide.

If you want just 25g of
copper sulfate crystals you would use 100 cm^{3} of 1 molar
sulfuric acid, or 50cm^{3} of 2 molar sulfuric acid.

BUT REMEMBER

(i) in
practice, you will NOT get a 100% yield, see calculation
below.

(ii) It would be normal to use
excess of the copper oxide, because it is easy to separate by
filtration the unreacted oxide to leave a neutral solution of the
salt, so you would use more than the 16 g of CuO calculated in part
3(a).

3(c) Suppose after carrying
out the preparation you finally crystallise 34g of pure the blue crystals of CuSO_{4}.5H_{2}O
after weighing the dry product. Losses are inevitable, but now
consider the two possible equations, the atom economy and a possible
yield in practice.

(i) CuO_{(s)} +
H_{2}SO_{4(aq)} ==> CuSO_{4(aq) }+
H_{2}O_{(l)}

(ii) CuO_{(s)}
+ H_{2}SO_{4(aq)} + 4H_{2}O_{(l)}
==>
CuSO_{4}.5H_{2}O_{(s)}
(formula of the blue crystals)
What is the 'atom
economy' of the preparation? (you need to refer to equations
(i) and (ii) above.

Atom economy = useful
theoretical products x 100/mass of all reactants

based on equation (i)
Atom economy = 100 x mass CuSO_{4} / (mass CuO + H_{2}SO_{4})

= 100 x 160 / (80 + 98)
= 16000/178 = 89.9%

based on equation (ii)
the atom economy is 100% if you include water as a 'reactant', can you see
why?

If you are asked to calculate the
atom economy of a reaction you will be given the equation.

What is the %
yield? i.e. comparing what you actually get with the maximum
possible, i.e. a 'reality check'!

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

and specific sections
14.2a
% reaction yield * 14.2b
atom economy



Example 4


14.5 (b)
Calculating limiting quantities of a reactant
Diagram 'borrowed' from one of my 'rates
of reaction pages'
The graph above illustrates the effect of a
'limiting reactant'. It might represent a graph when a gas is
evolved from e.g. (i) a metal reacting with an acid or (ii) a
carbonate dissolving in acid. The little vertical arrow indicates
when the reaction has stopped. At this point one of the reactants,
the limiting reactant, has all been used up in the reaction.
It might be the metal or the carbonate (all the solid dissolved in
excess acid) or it might be the acid, whose concentration falls to
zero in the presence of excess of the reactant solid.
You need to be able to logically deduce from given quantities of
reactants and the chemical equation, which reactant is the
limiting reactant?
In a preparation chemical reaction involving two
reactants, it is quite common to use an excess of one of the
reactants to ensure that all of the other reactant is used up.
The
reactant that is completely used up is called the limiting
reactant because it limits the amount of products that can be
formed i.e. when it is all used up in the reaction, but from the
amount of it, you can theoretically calculate the maximum yield of
product possible.
The amount of product you can make is directly
proportional to the quantity of limiting reactant used i.e. double
the mass of the limiting reactant you will double the mass of
product. Double the particles present, you double the product
particles!
You need to be able to explain the effect of a limiting
quantity of a reactant on the amount of products it is possible to
make e.g. in terms of amount of product in moles or masses in grams.
How to
deduce the limiting reactant and then calculate the maximum quantity
of the product
You
need know or be given:
the correctly balanced symbol equation,
The relevant relative atomic masses to work out
formula masses
The relationship between moles, mass and
atomic/formula mass
mol = mass / (A_{r} or M_{r})
and mass = mol x (A_{r} or M_{r})
(help triangle on the right, cover over what
you want and the rest is how to work it out!)
You can then compare the moles of reactants and
see which one (the excess reactant) outweighs the other (the limiting
reactant).
AND work out the maximum moles or mass of product
from the limiting reactant.
These calculations do assume 100% yield, but so in
reality (% reaction yield and
theoretical yield calculations)
Example 1a.
Deducing the limiting reactant and calculating the maximum quantity
of the product
Making iron sulfide
If 40 g of sulfur is heated with 60g of iron
filings to make iron sulfide, which is the limiting reactant? and
what is the maximum mass of iron sulfide that can be made?
The reaction is: Fe + S ===> FeS;
atomic masses: Fe = 56; S = 32
(i)
Calculating the limiting reactant
moles = mass in g / formula mass
moles S = 40 / 32 = 1.25 mol S
moles Fe = 60 / 56 = 1.07 mol Fe
Since one mole of iron reacts with one mole of
sulfur, there is insufficient iron to react with all of the sulfur used
in the experiment, therefore the iron is the limiting reactant.
(ii)
Calculating the maximum yield from the limiting reactant.
This part of the calculation MUST be based on
the limiting reactant (1.07 mol Fe)
Formula mass iron sulfide FeS = 56 + 32 = 88
From the reaction equation, reading in molar
quantities, one mole of iron forms one mole of iron sulfide.
mass FeS = moles of FeS x formula mass FeS
mass FeS = 1.07 x 88 = 94.2 g
Example 1b.
Deducing the limiting reactant and calculating the maximum quantity
of the product
Making oxides by heating a metal in air
In this case the metal is automatically the
limiting reactant if it is fully exposed to excess of air,
effectively excess oxygen.
In these examples I have omitted the experiment
details and details of weighings.
Atomic masses for these three examples: Al = 27,
Mg = 24, Fe = 56, O = 16
Example (i)
Heating magnesium in air in
a crucible
2Mg + O_{2} ===>
2MgO
If
you burn 3.0 g of magnesium what is the maximum mass of magnesium oxide
that can be formed?
mol Mg = 3/24 = 0.125 mol
In the equation 1 mol of Mg gives 1 mol of MgO
(formula mass = 24 + 16 = 40)
Therefore maximum mass of MgO product = 0.125 x 40
= 5.0 g MgO
You can also deduce that 2.0 g of oxygen was
consumed in the reaction.
Example (ii)
Heating
aluminium air in a crucible
2Al + 3O_{2} ===>
2Al_{2}O_{3}
What mass of aluminium do you need to make 20.0 g
of aluminium oxide?
Formula mass of Al_{2}O_{3} = (2 x
27) + (3 x 16) = 102
mol
Al_{2}O_{3} required = 20.0/102 = 0.196 mol
In the equation 2 mol of Al gives 2 mol Al_{2}O_{3}
This is a 1 : 1 ration so mol Al = mol Al_{2}O_{3}
Therefore mass of Al needed = 0.196 x 27 =
5.29 g (3 sf, 2 dp)
Example (iii) Heating iron in a stream of steam
When excess steam is passed over iron at
450^{o}C the oxide formed has the formula Fe_{3}O_{4}.
This means the iron is the limiting reactant.
The equation is: 3Fe + 2O_{2}
===> Fe_{3}O_{4}
If 14.0g of iron is heated in excess steam what
mass of the iron oxide is formed?
Atomic
mass Fe = 56, formula mass of Fe_{3}O_{4} = {(3 x 56) +
(4 x 16)} = 168 + 64 = 232
mol Fe = 14.0 / 56 = 0.25 mol
Now according to the equation above 3 mol of Fe
gives 1 mol of Fe_{3}O_{4}
Therefore mol Fe_{3}O_{4} = mol Fe
/ 3 = 0.25 / 3 = 0.08333
Therefore mass of Fe_{3}O_{4} =
0.08333 x 232 = 19.3 g (3 sf, 1 dp)
You can work the calculation the other way and
work out the amount of iron needed to make a given mass of Fe_{3}O_{4}.
e.g. How much iron is needed to make 50.0
g of Fe_{3}O_{4}?
From the equation you need 3 mol of
iron to make 1 mol of Fe_{3}O_{4}
mol
of Fe_{3}O_{4} = 50.0 / 232 = 0.2155
mol Fe needed = 3 x mol Fe_{3}O_{4}
mol Fe needed = 3 x 0.2155 = 0.6465
mass of Fe required = 0.6465 x 56 =
36.2 g
Example 2.
Deducing the limiting reactant and calculating the maximum quantity
of the product
Making copper sulfate from copper
oxide and sulfuric acid
If
10 g of copper oxide (CuO) is added to 50 cm^{3} of sulfuric
acid of concentration 2.0 mol/dm^{3}, deduce:
(a) which is the
limiting reactant?
(a) What is the maximum amount of copper sulfate (as CuSO_{4})
you could theoretically make?
The reaction is: CuO(s) + H_{2}SO_{4}(aq)
===> CuSO_{4}(aq) + H_{2}O(l)
Atomic masses: Cu = 64; O = 16; H = 1; S = 32 (but
you might not need all of these!)
(a) Deducing the limiting reactant
Formula
mass copper sulfate = 64 + 16 = 80
moles = mass in g / formula mass
mol CuO = 10 / 80 = 0.125 mol
mol H_{2}SO_{4} = molarity
x
volume (in dm^{3}) (dm^{3} = cm^{3}
/ 1000)
mol H_{2}SO_{4} = 2.0 x 50
/ 1000 = 0.10 mol
From the equation, reading in molar quantities,
one mole of copper oxide reacts with one mole of sulfuric acid to make one mole of copper sulfate.
Clearly the copper oxide is in excess, so
the sulfuric acid is the limiting reactant, because you would
need 0.125 moles of the acid (only got 0.10 mol) to react with all 10 g
(0.125 mol) of the copper oxide.
In the actual
preparation of the copper
sulfate salt crystals you would filter off the excess copper oxide,
so this is a viable recipe for making copper(II) sulfate crystals..
(b)
Calculating the theoretical maximum yield of copper sulfate from the
limiting reactant
This part of the calculation MUST be based on
the limiting reactant (0.10 mol H_{2}SO_{4})
From the equation, reading in molar quantities:
one mole of sulfuric acid makes one mole of copper sulfate (0.1
mol).
Formula mass copper sulfate = 64 + 32 + (4 x 16) =
160
mass = moles x formula mass = 0.10 x 160 =
16g of CuSO_{4}
Strictly speaking the actually product is the
'blue' hydrated copper sulfate crystals, formula CuSO_{4}.5H_{2}O
This has a formula mass of 160 + (5 x 18) = 250
Therefore the maximum possible mass of 'blue'
crystals is 0.1 x 250 = 25g of CuSO_{4}.5H_{2}O
Example 3.
Deducing the limiting reactant and calculating the maximum quantity of the
product
Making hydrogen from
magnesium and hydrochloric acid
Suppose we take a strip of magnesium weighing
0.36g and dissolve it in 50 cm^{3} of hydrochloric acid of
molarity 1.0 mol/dm^{3}..
The equation is: Mg(s) + 2HCl(aq)
===> MgCl_{2}(aq) + H_{2}(g)
and A_{r}(Mg) = 24
(a) Which is the limiting reactant?
To calculate the limiting reactant we need to
compare the number of moles of each reactant, AND then the relative
reaction mole ratio to see which one is in excess  take care, you must
not automatically just compare the actual number of moles of each
reactant, you must consult the equation too!
mol Mg = mass Mg / A_{r}(Mg) = 0.36 / 24 =
0.015 mol Mg
moles = molarity x volume (in dm^{3})
mol HCl = [molarity HCl] x [volume of HCl(aq) in
dm^{3}] (dm^{3} = cm^{3} / 1000)
mol HCl = 1.0 x (50 / 1000) = 0.05 mol HCl
According to the equation each mole of Mg requires
two moles of HCl to complete the reaction.
Therefore 0.015 mol Mg requires 0.030 mol of HCl
to react completely.
Since the moles of HCl required (0.03) is exceeded
by the moles of HCl used (0.05),
the magnesium must be the limiting
reactant
(b)
What is the maximum volume of hydrogen gas you can make in this
experiment from the limiting reactant?
This part of the calculation MUST be based on
the limiting reactant (0.015 mol Mg)
Volume of 1 mol of gas = 24 dm^{3} at RTP
(room temperature and pressure)
From the equation, reading in molar quantities,
each mole of magnesium gives one mole of hydrogen gas,
so mol Mg used = mol H_{2} formed = 0.015
(limiting moles)
volume H_{2} = mol H_{2} x 24
= 0.015 x 24 = 0.36 dm^{3}
(360 cm^{3})
Example 4.
Deducing the limiting reactant and calculating the maximum quantity of the
product
Preparing carbon
dioxide from limestone and acid
2g
of calcium carbonate granules (limestone) was dissolved in 20 cm^{3}
of hydrochloric acid of concentration 0.50 mol/dm^{3}.
The reaction is: CaCO_{3}(s) +
2HCl(aq) ===> CaCl_{2}(aq) + H_{2}O(l)
+ CO_{2}(g)
Atomic masses: Ca = 40; C = 12; O = 16; H = 1, Cl
= 35.5; but you might not need all these A_{r} values!
(a) Which is the limiting reactant?
Formula
mass CaCO_{3} = 40 + 12 + (3 x 16) = 100
mol CaCO_{3} = 2 / 100 = 0.02
mol HCl = [molarity HCl] x [volume of HCl(aq) in
dm^{3}] (dm^{3} = cm^{3} / 1000)
mol HCl = 0.5 x 20 / 1000 = 0.01 mol
From the equation one mole of calcium carbonate
requires two moles of hydrochloric acid to complete the reaction.
Therefore the 0.02 mol of CaCO_{3} needs 0.04 mol of HCl to
react completely.
Since only 0.01 mol of HCl was used in the
mixture, clearly the calcium carbonate is in great excess and the
hydrochloric acid is the limiting reactant.
Note: At the end of the reaction, indicated by no
more 'fizzing', you would see white unreacted granules of limestone at
the bottom of the flask.
(b)
What is the maximum volume of carbon dioxide gas that can be
obtained using this mixture from the limiting reactant?
This part of the calculation MUST be based on
the limiting reactant (0.01 mol HCl)
From the equation, reading in molar quantities,
two moles of hydrochloric acid produces one mole of carbon dioxide (so
the moles of HCl will only make half the amount of moles of CO_{2}).
Therefore mole CO_{2} = mol HCl / 2 = 0.01
/ 2 = 0.005
volume CO_{2} = mol CO_{2} x 24 dm^{3}
= 0.005 x 24 = 0.12 dm^{3}
(or 0.12 x 100 = 120 cm^{3})
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
14.2b atom economy calculations
*
14.3 dilution of solutions calculations
*
14.4
water of crystallisation
calculations
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
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
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
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