(a) WHAT IS
THE MOLE CONCEPT? and WHAT IS ONE
MOLE OF A SUBSTANCE?
The 2nd part of the heading is 'easy', the first
part is a bit more 'abstract' to get your head round!
The mole concept is an invaluable way of solving
many quantitative problems in chemistry!
Its a very important way of doing chemical
The theoretical basis is
explained in section (b).
The mole is most simply expressed as the
mass in g' or the 'molecular mass in g' of the defined chemical 'species', and that is how it is
used in most chemical calculations. The mass of one mole of a substance is
sometimes referred to as the molar mass.
The atomic/formula mass in grams = one
mole of the defined substance.
If your are dealing with
individual atoms, one mole of equals
the relative atomic mass in grams.
This can be expressed as a
simple formula ...
moles of species = (actual mass of species in g)
/ (atomic/formula mass of species)
therefore (using triangle on
right if necessary)
mass of species in g =
moles species x atomic/formula mass of species
atomic/formula mass of
species = mass of species in g / moles of species
Note these equations are
for either an element or a compound,
but, whatever, you must
clearly define the chemical species you mean for any mole
element atom, H2O covalent molecule, an
element O2 molecule, Na+Cl-
ionic compound or just any compound formula like CuSO4
cannot be overemphasised and you should understand that the
measurement of amounts in moles can apply to atoms, molecules,
ions, electrons, formulae and equations !!!
Mr is 'shorthand' for relative formula mass or
molecular mass in amu (atomic mass units, u or dalton Da) and you must be able to work these
out correctly from a given formula (Calculating relative
formula/molecular mass of a compound or element molecule).
The term relative molecular mass (sum of the atomic
masses of the atoms in a single molecule of the substance) is usually
applied to definite molecular species.
Using the following atomic
masses: H = 1, O = 16, N = 14, C = 12, Na = 23, Al = 27, Cl = 35.5, S = 32,
Ca = 40
and the three formulae above
relating moles, mass and formula mass ...
1 mole of carbon atoms = 12 g, 1 mole
of carbon dioxide (12 + 2 x 16) = 44, 1 mole CO2 = 44 g
the number of C atoms = the
number of CO2 molecules = 6.0 x 1023 specified
molecular mass Mr 18 for the water molecule
17 for the ammonia molecule NH3
so 1 mole of water is 18g,
0.333 mole = 0.333 x 18 = 6g
for ammonia 1mol = 17g, 34g =
34/17 = 2 mols ammonia
Mr = 16 for the methane molecule
and 180 for the glucose sugar molecule C6H12O6
so 0.5 mol methane = 0.5
x 16 8g, 72g
= 72/16 = 4.5 mols methane
for glucose 18g = 18/180
= 0.10 moles, 0.05 mole = 0.05 x 180 = 9.0g glucose
The element nitrogen
consists of N2 molecules (Mr = 28),
molar mass = 28g,
0.25 moles = 0.25 x 28 = 7.0g
Relative atomic mass of iron Ar
= 56, 7g = 7/56 = 0.125 mol Fe (Relative
atomic mass explained)
The mass of 1 mole of sodium chloride
NaCl? Mr(NaCl) = 23 + 35.5 = 58.5 g
The mass of 1 mole of calcium
carbonate? Mr(CaCO3) = 40 + 12 + (3 x 16) = 100
So, these calculations are
quite simple, but they are often just one part of solving a more complex
problem involving moles!
The term relative formula mass (sum of the atomic
masses of the atoms in a specified formula) can be used for ANY specified
formula of ANY chemical substance, though it is most often applied to ionic substances.
e.g. mass of
1 mole of ionic sodium chloride
NaCl or Na+Cl- is 58.5g (from 23 +
BUT, each mole of NaCl consists
of 1 mole of sodium ions and 1 mole of chloride ions
mass of 1 mole of ammonium sulfate (ionic
salt) (NH4)2SO4 or (NH4+)2(SO42-)
BUT 1 mole of the salt consists
of 2 moles of ammonium ions and 1 mole of sulfate ion
mass of 1 mole of aluminium oxide Al2O3
= (2 x 27) + (3 x 16) = 102 g
BUT being ionic, (Al3+)2(O2-)3,
1 mole of Al2O3 consists of 2 moles of
aluminium ions and 3 moles of oxide ions
There is a periodic table of atomic masses near the end of the page for doing
(b) EXAMPLES OF MOLE CALCULATIONS
to illustrate the introduction section (a)
REMINDER For a substance 'Z' i.e. a
specifically defined chemical species
(1) mole of Z = g of
Z / atomic
or formula mass of Z,
(2) or g of Z = mole of
Z x atomic or formula mass of Z
(3) or atomic or formula mass of Z
= g of Z /
mole of Z
where Z represents atoms, molecules or formula of
the particular element or compound defined in the question and all masses quoted
in grams (g).
For calculation purposes learn the following
formula for 'Z' and use a triangle if necessary.
Since you can interpret equations in terms
of moles and masses of reactants and products, you can also work in the other
e.g. if you know reactant and product
masses, you can convert them to moles and from the simplest whole number ratio
work out a balanced chemical equation.
Abbreviations used: Ar
= atomic mass, Mr = formula mass
Equation - mole calculation Example 7.3.1
to work out an equation from reacting masses using moles
It was found that 11.15 g of lead(II)
oxide PbO reacted with 0.30 g of carbon to produce 10.35 g of lead
Given the atomic masses: Pb = 207,
O = 16, C = 12, formula mass PbO = 207 + 16 = 223
Convert the reacting masses given into moles
mol = mass / Ar or Mr
mol PbO = 11.15 / 223 = 0.05,
mol C = 0.30 / 12 = 0.025, mol Pb = 10.35 / 207 = 0.05
(b) Convert the mole ratio from (a) into
the simplest whole number (integer) ratio
PbO : C ==> Pb is 0.05 : 0.025 ==>
scaling up to the simplest whole
number ratio is 2 : 1 ==> 2 moles (divide the 1st ratios by
(c) Deduce the balanced equation for the
reduction of lead(II) oxide to lead using carbon
From the whole number ratio above we
can then write
2PbO + C ===> 2Pb
BUT, the oxygen must have combined
with the carbon to give carbon dioxide because there are two atoms of
oxygen in the equation to one atom of carbon, so we can write the full
2PbO + C ===>
2Pb + CO2
Equation - mole calculation Example 7.3.2
to work out an equation from reacting masses using moles
8.4 g of iron was heated in air to form
an oxide, until there was no longer any gain in weight. The final mass of
the iron oxide was 11.6 g. There was only one product of the reaction.
Atomic masses: Fe = 56, O = 16
(a) Calculate the mass of oxygen that
combined with the iron.
mass of oxygen used = mass of iron
oxide - mass of iron that reacted
mass of oxygen (O2) = 11.6
- 8.4 = 3.2 g
Calculate the moles of iron and oxygen that combined to give the iron oxide.
mol Fe = 8.4 / 56 = 0.15,
mol O2 = 3.2 / 32 = 0.10
Note: O is 16, BUT it is oxygen
molecules that are in air, formula mass O2 = 32
(c) Convert the Fe : O2 ratio to the
simplest whole number ratio
Fe : O2 is 0.15 : 0.10,
diving by 0.05 gives a reacting mole ratio of 3 : 2 for Fe : O2
(d) We can therefore write the left-hand
side of the equation as
3Fe + 2O2
Since there was only one product of
the reaction, the empirical formula of this particular iron oxide
must be composed of three iron atoms to four atoms of oxygen, so the
full equation is
3Fe + 2O2
Equation - mole calculation Example 7.3.3
to work out an equation from reacting masses using moles
If you don't need to know about the Avogadro
Constant, you can skip this section.
One mole of a substance contains exactly the same number of
the stated particles, atoms, molecules or ions as one mole of ANY other
substance (must be specifically defined e.g.
iron atoms Fe, water molecules H2O,
glucose molecules C6H12O6, hydroxide ion OH-,
iron(III) ion Fe3+, electron e-
The number of atoms, molecules or ions in a mole of a given
substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x
1023 per mole. You should understand that the measurement of amounts
in moles can apply to atoms, molecules, ions, electrons, formulae and
So, every mole of any substance contains the
same number of the defined species.
The actual particle number is known and
is called the Avogadro Constant, denoted NA).
It is equal to 6.023 x 1023
'defined species' per mole i.e. 6.023 x 1023 mol-1
This means in a mole of any defined species e.g.
atom, molecule, ion etc. there are the Avogadro constant number of them.
e.g. this means there are that many atoms in 12g of carbon (C = 12)
= 6.023 x 1023
atoms of carbon
that many molecules in 18g water (H2O = 1+1+16 = 18, H = 1; O =
16) * = 6.023 x 1023
molecules of water
This is about 18cm3, so picture this
number of molecules in a nearly full 20cm3 measuring cylinder or a
100ml beaker less than 1/5th full!
The Avogadro number is 6.023 x
1023 = 602 300 000 000 000 000 000 000 atoms or molecules per
= six hundred and two
thousand and three hundred million million million 'particles' per mole !
A thimble full of water is about
1cm3, 1 mole of water = 18g and ~ 18cm3 because the density of
water is ~1.0 g/cm3
Therefore in a thimble full of
water there are ~6.023 x 1023/18 = ~3.3 x 1022 = 33 000
000 000 000 000 000 000 molecules!
= thirty three thousand
million million million molecules of water!
So, just think how many
molecules of water are in your body!
AND just think how
useful the 'mole' is, to make life 'simple' in calculations! (well
The real importance of the mole is that
it allows you to compare ratios of the relative amounts of reactants and products, or the element
composition of a compound, at the atomic and molecular level.
1 mole of any defined chemical
species has an identical number of that species, and that number is the Avogadro
If you have
a mole ratio for A : B of 1 : 3, it means 1 particle of A to 3 particles of B
irrespective of the atomic or formula masses of A and B.
It also means that you can read
equations in terms of a mole ratio (see section (b) questions)
e.g. 2NaOH + H2SO4
==> Na2SO4 + 2H2O
can be read as 2 moles of sodium
hydroxide neutralises 1 mole of sulfuric acid to form 1 mole of the salt
sodium sulfate and 2 moles of water,
BUT the equation can be read in
terms of any molar quantities, as long as you keep the ratios the same!
e.g. by only taking 1/20th
of a mole of sodium hydroxide you can deduce (yes predict!)
0.05 moles of NaOH reacts with
0.025 moles of H2SO4 to form 0.025 moles of Na2SO4
and 0.05 moles of H2O
What more, since you can convert
moles to mass, you can do deduce the mass of product formed or the mass of
Also, since you can go from mass
to moles, you can deduce equations from measuring reacting masses.
(see also section 6. for reacting masses not using
advanced calculations using the Avogadro Constant
This is for more advanced
students and illustrates the concepts in the introduction
you can actually calculate the
number of particles in known quantity of material !
constant Mole calculation Example 7.3.1
Avogadro number calculation Example 7.3.2
Avogadro constant calculation Example 7.3.2
many particles of 'Al2O3' are there in 51g of aluminium oxide?
Atomic masses: Al =27, O
= 16, f. mass Al2O3 = (2x27) + (3x16) = 102
= 51/102 = 0.5 mol
Number of 'Al2O3'
particles = 0.5 x 6 x 1023 = 3 x 1023
(b) Aluminium oxide
is an ionic compound. Calculate the number of individual aluminium ions (Al3+)
and oxide ions (O2-) in the same 51g of the substance.
For every Al2O3
there are two Al3+ and three O2- ions.
So in 51g of Al2O3
there are ...
0.5 x 2 x 6 x 1023
= 6 x 1023 Al3+ ions, and
0.5 x 3 x 6 x 1023
= 9 x 1023 O2- ions.
More advanced use of the mole and Avogadro Number concepts
(for advanced level students only)
In terms of electric
charge, 1 Faraday = 96500 C (coulombs) = 6 x 1023 electrons
If you have 2.5 moles of
the ionic aluminium oxide (Al2O3) you have ...
When you write ANY
balanced chemical equation, the balancing numbers, including the un-written
1, are the reacting molar ratio of reactants and products.
Extra Advanced Level Chemistry Questions - more suitable for Advanced AS-A2 students
which can be completely tackled after ALSO studying
section 9 on the molar volume of gases and
ANSWERS to QA7.1
This question involves using the mole concept and the Avogadro Constant in a variety of situations.
The Avogadro Constant = 6.02 x 1023 mol-1
The molar volume for gases is 24.0dm3 at 298K/101.3kPa.
Atomic masses: Al = 27, O = 16, H =
1, Cl = 35.5, Ne = 20, Na = 23, Mg = 24.3, C = 12
Where appropriate assume the temperature is 298K and
the pressure 101.3kPa.
(a) how many oxide ions in 2g of aluminium oxide?
(b) how many molecules in 3g of hydrogen?
(c) how many molecules in 1.2 cm3 of oxygen?
(d) how many molecules of chlorine in 3g?
(e) how many individual particles in 10g of neon?
(f) the volume of hydrogen formed when 0.2g of sodium reacts with water.
(g) the volume of hydrogen formed when 2g of magnesium reacts with excess acid.
(h) the volume of carbon dioxide formed when the following react with excess acid
(1) 0.76g of sodium carbonate
(2) 0.76g sodium hydrogencarbonate
(i) the volume of hydrogen formed when excess zinc is added to 50 cm3 of hydrochloric acid, concentration 0.2 mol dm-3.
(j) the volume of carbon dioxide formed when excess calcium carbonate is added to 75 cm3 of 0.05 mol dm-3 hydrochloric acid.
ANSWERS to QA7.1
Self-assessment Quizzes for GCSE or A Level (the basics) on mole
Type in answer
Above is typical periodic table used in GCSE science-chemistry specifications in
doing mole chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
OTHER CALCULATION PAGES
What is relative atomic mass?,
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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
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Empirical formula and formula mass of a compound from reacting masses
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Reacting mass ratio calculations of reactants and products
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Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
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