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FORMULAE, EQUATIONS AND MOLES
Important terms
In order to use formulae and equations it is necessary to understand the meaning of important terms
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Ion

An atom or group of atoms possessing a negative charge

Element

A substance containing just one type of atom
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Molecular formulae

The actual numbers of each type of atom in a molecule of the substance
...

This is done using the mole
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If we take a mole of atoms of a particular element, the mass will be the Relative Atomic Mass for that
element
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023 x 10 23)
The mole is the amount of a substance that contains the same number of particles as there are atoms
in 12
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This number of atoms is 6
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Amount of substance = number of particles
Avogadro constant

Moles from mass
To find the number of moles of atoms we use the formula:
Moles =

mass of substance
relative atomic mass

If we take a mole of ionic lattice or molecules of a material, the mass will be the molar mass
To find the number of moles of molecules or compounds we can use the formula:
Moles =

mass
molar mass (RFM, Mr)

Examples
1
...
)
Molar mass of MgSO4 = 24 + 32 + (4x 16) = 120
Moles = 10 /120 = 0
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Calculate the mass of 0
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5 + (2 x 14) + (6x 16) = 178
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04 x 178
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5g

Formulae determination
An empirical formula is the simplest formula which shows the ratio of each type of atom present in a
compound
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The formula of a compound can be calculated from the mass of elements in it
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Step 2:
From the number of moles find the simplest ratio
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Example
A hydrocarbon consists of 14
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2g hydrogen
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C
14
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2 mol

H
1
...
2 mol

Ratio = 1
:
1
So the empirical formula is CH
If we try to work out the structure of this we can see that this is not the formula of the molecule
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If the molar mass of this compound is 78gmol-1, what is the molecular formula?
CH has a mass of 13
the number of these units is 78/13 = 6
so the formula is C6H6
A similar process is used to find formulae from percentage composition
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00% Carbon, 6
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33% Oxygen
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Calculate the molecular formula
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00 / 12
= 3
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67 / 1
= 6
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33 / 16
= 3
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(This is Avogadro’s Law)
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5 moles of oxygen
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5 volumes oxygen
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5 = 97
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At room temperature and 1Atm this is 24dm3 (24000cm3)
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4dm3)
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Moles = Volume (in dm3)
24
Example
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Moles = 480 / 24,000 = 0
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Concentration =

Moles
Volume (in dm3)

The concentration of a solution can be stated as the mass of solute per cubic decimeter of solution
(g/dm3) or the amount in moles of a solute present in 1dm3 of solution (mol/dm3)
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Examples
1
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4g MgSO4 in 500cm3 of solution
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4 / 120 = 0
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02mol / 0
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04moldm-3
2
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2 mol dm-3 solution
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2 x 0
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02mol
Mass = Moles x Formula mass
= 0
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8g

Parts per million
This is a way of expressing very dilute concentrations of substances (usually pollutants)
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Usually describes the concentration of something in water or soil
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or the volume of a gas pollutant in air
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Calculate the mass of gold in the 25cm3 sample of river water
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056 mg of gold
25 cm3 of water contains 0
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4 mg = 1
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Would you be at serious risk if you were exposed 380 cm3 of carbon monoxide in a 3500 dm3 room?
380 cm3 in 3500000 cm3 = 380 cm3 in 3
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5 = 108
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Whether using masses, volumes of solutions (of known concentration), or volumes of gases the same
general method can be used
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Step 3: From the number of moles of this substance calculate the mass / volume /
concentration of the desired substance
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84g of H2SO4
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84 / 98 = 0
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08
Step 3 – Mass of Iron(II)sulphate = moles x RFM = 0
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2g (3 sig
...
)
Example 2:
What mass of aluminium is needed to react with 7
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5 = 0
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So 1 mol of CuSO4 reacts with ⅔ mol of Al
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0439 mol CuSO4 reacts with 0
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0293 mol Al

Step 3 – Mass of Al = moles x RFM = 0
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790 g (3 sig
...
)
Example 3:
Calculate the volume carbon dioxide gas which will be produced from the combustion of 2
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C3H8 + 5O2
3CO2 + 4H2O
Step 1 - Moles of propane used = mass / RFM = 2
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05 mol
Step 2 - From equation 1mol propane forms 3mol carbon dioxide
So mole of carbon dioxide = 0
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15 x 24,000 = 3,600cm3

Titrations
A titration is a method of finding out how much of one material will react with how much of another
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A pipette-filler is added to the volumetric pipette
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Some of the solution is drawn into the pipette
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3
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4
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5
...

6
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Some of the liquid will remain in the tip and this should be left as the
pipette is calibrated to allow for this
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A suitable indicator should be added to the conical flask
8
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9
...

10
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Add the solution from the burette until the indicator changes
colour
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This is the rough reading
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Discard the contents of the flask and rinse it with tap water and then distilled water
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Repeat the process, adding the solution from the burette fairly slowly with continual stirring
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When one drop changes the colour of the indicator, allow the solution to drain down the
sides of the burette before taking the reading
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Accurate burette readings should be recorded to two decimal places, the second decimal
place being 0 or 5
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Readings should continue to be taken in this way until one rough and two accurate readings
that are within 0
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15
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Titration calculations
Use the same three step method as before
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45 cm3 of 0
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0cm3 of sulphuric acid
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2NaOH + H2SO4  Na2SO4 + 2H2O
Step 1 - Moles of NaOH = 23
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2 = 0
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00469 / 2 = 0
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002345 / 0
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0938 mol dm-3 (3 sig
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)

Percentage Yield
Chemical reactions are carried out in the laboratory and by industry to make new materials
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The efficiency of the conversion process is measured using percentage yield or atom economy
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00g of the pure acid, is
reacted with an excess of copper oxide
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37g of the hydrated copper sulphate, CuSO4
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Calculate the % yield
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00 / 98 = 0
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0204mol
From equation 1mol sulphuric acid forms 1mol copper sulphate
So moles of copper sulphate formed = 0
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5 + 32 + 64 + 90 = 249
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5 x 0
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09g
% Yield = Actual yield / theoretical x 100 = 4
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09 x 100 = 85
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Molar mass of useful product
% atom economy=

x 100

Total molar mass of starting materials
Taking the laboratory preparation of copper(II) sulphate from sulphuric acid and copper(II) oxide as an
example
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5 + 16 = 79
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5 + 32 + 64 = 159
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5
159
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9%

The atom economy for the production of ethanol from ethene and steam is shown below
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5
C2H4 + H2O  C2H5OH
% atom economy = 46 / 46 x 100 = 100% (there are no unwanted products)
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Equations can be written to explain simple test tube reactions such as displacement and precipitation
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74g of manganese(IV) oxide is added to an excess of concentrated hydrochloric acid and warmed
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The mass of chlorine given off was 1
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Another
product of the reaction was purified and analysed
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10g of manganese and 1
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(a) Determine the formula of the salt
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(c) Calculate the number of moles of chlorine and salt produced in the reaction
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(a)

Mn
1
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02 mol
Ratio = 1

Cl
1
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5
=0
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74 / 87 = 0
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42 / 71 = 0
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52 / 126 = 0
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02

0
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02

1

1

1

Mole ratio

Balancing for Cl:

MnCl2 + Cl2

MnO2 + 4HCl  MnCl2 + Cl2

+

?

+

?

Balancing for H & O: MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O

Example 2
For the following reactions, write chemical and ionic equations using state signs
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Na2CO3(aq) + 2HNO3(aq)  2NaNO3(aq) + H2O(l) + CO2(g)
CO32-(aq) + 2H+(aq)



H2O(l) + CO2(g)

(b) When iron filings are added to copper sulphate solution, the blue solution turns green and the
grey solid is replaced by a red-brown solid
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Na2CrO4(aq) + 2AgNO3(aq)  2NaNO3(aq) + Ag2CrO4(s)
CrO42-(aq) + 2Ag+



Ag2CrO4(s)



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