Introducing 'moles' and their the connection with mass and formula
mass
The mole is most simply expressed as the relative 'formula 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 molar mass is Mr g mol-1 i.e. the molar mass in grams
Mr is 'shorthand' for relative formula mass or molecular mass in amu (atomic mass units).
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
e.g. molecular mass 18 for the water molecule H2O, 17 for the ammonia molecule NH3
16 for the methane molecule CH4 and 180 for the glucose sugar molecule C6H12O6
(atomic masses for these examples H = 1, O = 16, N = 14, C = 12, S = 32, Na = 23, Cl = 35.5)
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. molar mass of ionic sodium chloride NaCl or Na+Cl- is 58.5g
the molar mass of ammonium sulfate (ionic salt) (NH4)2SO4 or (NH4+)2(SO42-) = 130g
or 18g for the molar mass of water H2O, and
17g is the molar mass of the ammonia molecule NH3 etc. as for molecular mass
BUT in most cases either term for Mr is ok to use, and if in doubt, just call it Mr
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 there are that many atoms in 12g of carbon (C = 12)
or that many molecules in 18g water (H2O = 1+1+16 = 18, H = 1; O = 16) *.
* This is about 18cm3, so picture this number of molecules in a nearly full 20cm 3
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 mole!
= 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!
However, 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. 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. Important
Note. Relative atomic mass is just a number based on the carbon-12 relative atomic mass
scale. Molar mass is a term used to describe the mass of one mole i.e. the relative
atomic/formula/molecular mass in grams (g).
Examples:


Example 7.1.1
o
Consider the formation of 1 mole of ammonia, NH3,
o
consists of 1 mole of nitrogen atoms combined with 3 moles of hydrogen
atoms.
o
Or you could say 2 moles of ammonia is formed from 1 mole of nitrogen
molecules (N2) combining with 3 moles of hydrogen molecules (H2).
o
N2(g) + 3H2(g) ==> 2NH3(g)
o
You can then think in any ratio you want e.g. 0.05 mol nitrogen combines
with 0.15 mol hydrogen to form 0.10 mol of ammonia.
o
So, you can calculate using any ratio on the basis of the 1:3:2 ratio (or 1:3 ==>
2) of the reactants and products in the balanced symbol equation.
Example 7.1.2
o
Consider the formation of 1 or 2 moles of aluminium oxide,
o
Al2O3, consists of 2 moles of aluminium atoms combined with 3 moles of
oxygen atoms (or 1.5 moles of O2 molecules) to form 1 mole of aluminium
oxide.

o


2Al(s) + 3/2O2(g) ==> Al2O3(s)
To avoid fractions in equations you can say 4 moles of aluminium atoms
combine with exactly 3 moles of oxygen molecules to form 2 moles of
aluminium.

4Al(s) + 3O2(g) ==> 2Al2O3(s)

So the simplest whole number reacting molar ratio is 4:3:2 (or 4:3 ==>
2)
Example 7.1.3
o
How do you go from a reacting mole ratio to reacting mass ratio?
o
You read the equation in relative numbers of moles and convert the moles
into mass.

mass moles x formula mass - see triangle below

e.g. the formation of copper(II) chloride from copper(II) oxide and
hydrochloric acid.

CuO(s) + 2HCl(aq) ==> CuCl2(aq) + H2O(l)

1 mole + 2 moles ==> 1 mole + 1 mole

Atomic masses: Cu = 64, O = 16, H = 1, Cl =35.5

Therefore ...

(64 + 16)g CuO + 2x(1 + 35.5)g HCl ==> (64 + 2 x 35.5) CuCl2 + (2 x 1 +
16)g H2O

80g CuO + 73g HCl ==> 135g CuCl2 + 18g H2O

So, from a mole ratio of 1:2 ==> 1:1 you get a mass ratio of 80:73 ==>
135:18
Example 7.1.4
o
This can be useful for calculating the quantities of chemicals you need for e.g.
the chemical preparation of a compound.
o
Using the concept of mole ratio and the exemplar reactions above ...
o
(a) Calculate how many grams of copper(II) oxide you need to dissolve in
hydrochloric acid to make 0.25 moles of copper(II) chloride?
o
From the equation, 1 mole of copper oxide makes 1 mole of copper chloride,
o
therefore you need 0.25 moles of CuO
o
since mass = molar mass x formula mass
o
you need 0.25 x 80 = 20g of CuO
o
(b) What mass of aluminium metal do you need to make 0.1 moles of
aluminium oxide?
o
4Al(s) + 3O2(g) ==> 2Al2O3(s) and the atomic mass of aluminium is 27
o
4 moles of aluminium makes 2 moles of aluminium oxide, (ratio 4:2 or 2:1)
o
therefore 0.2 moles of aluminium metal makes 0.1 moles of aluminium oxide
(keeping the ratio of 2:1)
o
mass of aluminium metal needed = 0.2 x 27 = 5.4g of Al
o
Note that you can pick out the ratio you need to solve a problem - you
DON'T need all the numbers of the full molar ratio, all you do is pick out the
relevant ratio!
For calculation purposes learn the following formula for 'Z' and use a triangle if necessary.
For a substance 'Z'
(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).


Example 7.2.1: How many moles of potassium ions and bromide ions in 0.25 moles
of potassium bromide?
o
1 mole of KBr contains 1 mole of potassium ions (K+) and 1 mole of bromide
ions (Br-).
o
So there will be 0.25 moles of each ion.
Example 7.2.2: How many moles of calcium ions and chloride ions in 2.5 moles of
calcium chloride?




o
1 mole of CaCl2 consists of 1 mole of calcium ions (Ca2+) and 2 moles of
chloride ion (Cl-).
o
So there will be 2.5 x 1 = 2.5 moles of calcium ions and 2.5 x 2 = 5 moles
chloride ions.
Example 7.2.3: How many moles of lead and oxygen atoms are needed to make 5
moles of lead dioxide?
o
1 mole of PbO2 contains 1 mole of lead combined with 2 moles of oxygen
atoms (or 1 mole of oxygen molecules O2).
o
So 1 x 5 = 5 mol of lead atoms and 2 x 5 = 10 mol of oxygen atoms (or 5 mol
oxygen molecules) are needed.
Example 7.2.4: How many moles of aluminium ions and sulphate ions in 2 moles of
aluminium sulphate?
o
1 mole of Al2(SO4)3 contains 2 moles of aluminium ions (Al3+) and 3 moles of
sulphate ion (SO42-).
o
So there will be 2 x 2 = 4 mol aluminium ions and 2 x 3 = 6 mol of sulphate
ion.
Example 7.2.5: How many moles of chlorine gas in 6.5g? Ar(Cl) = 35.5)
o
chlorine consists of Cl2 molecules, so Mr = 2 x 35.5 = 71
o
moles chlorine = mass / Mr = 6.5 / 71 = 0.0944 mol
Example 7.2.6: How many moles of iron in 20g? (Fe = 56)
o

Example 7.2.7: How many grams of propane C3H8 are there in 0.21 moles of it? (C =
12, H = 1)
o

Mr of propane = (3 x 12) + (1 x 8) = 44, so g propane = moles x M r = 0.21 x 44
= 9.24g
Example 7.2.8: 0.25 moles of molecule X was found to have a mass of 28g.
Calculate its molecular mass.
o

iron consists of Fe atoms, so moles iron = mass/Ar = 20/56 = 0.357 mol Fe
Mr = mass X / moles of X = 28 / 0.25 = 112
Example 7.2.9: What mass and moles of magnesium chloride is formed when 5g of
magnesium oxide is dissolved in excess hydrochloric acid?

o
reaction equation: MgO + 2HCl ==> MgCl2 + H2O
o
means 1 mole magnesium oxide forms 1 mole of magnesium chloride (1 : 1
molar ratio)
o
formula mass MgCl2 = 24+(2x35.5) = 95,
o
MgO = 24+16 = 40, 1 mole MgO = 40g, so 5g MgO = 5/40 = 0.125 mol
o
which means 0.125 mol MgO forms 0.125 mol MgCl2,
o
Mass = moles x formula mass = 0.125 x 95 = 11.9g MgCl2
Example 7.2.10: What mass and moles of sodium chloride is formed when 21.2g of
sodium carbonate is reacted with excess dilute hydrochloric acid?
o
reaction equation: Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
o
means 1 mole sodium carbonate gives 2 moles of sodium chloride (1:2 ratio
in equation)

Formula mass of Na2CO3 = (2x23) + 12 + (3x16) = 106

Formula mass of NaCl = 23 + 35.5 = 58.5
o
moles Na2CO3 = 21.2/106 = 0.2 mole
o
therefore 2 x 0.2 = 0.4 mol of NaCl formed.
o
mass of NaCl formed = moles x formula mass = 0.4 x 58.5 = 23.4g NaCl
Using the Avogadro Constant, you can actually calculate the number of particles in known
quantity of material.

Example 7.3.1: How many water molecules are there in 1g of water, H2O ?
o
formula mass of water = (2 x 1) + 16 = 18
o
every mole of a substance contains 6 x 1023 particles of 'it' (the Avogadro
Constant).
o
moles water = 1 / 18 = 0.0556
o
molecules of water = 0.0556 x 6 x 1023 = 3.34 x 1022
o
Since water has a density of 1g/cm3, it means in every cm3 or ml there are
o
33 400 000 000 000 000 000 000 individual H2O molecules or particles!


Example 7.3.2: How many atoms of iron (Fe = 56) are there in an iron filing of mass
0.001g ?
o
0.001g of iron = 0.001 / 56 = 0.00001786 mol
o
atoms of iron in the nail = 0.00001786 x 6 x 1023 = 1.07 x 1019 actual Fe
atoms
o
(10.7 million million million atoms!)
Example 7.3.2: (a) How many particles of 'Al2O3' in 51g of aluminium oxide?
o
Atomic masses: Al =27, O = 16, f. mass Al2O3 = (2x27) + (3x16) = 102
o
moles 'Al2O3' = 51/102 = 0.5 mol
o
Number of 'Al2O3' particles = 0.5 x 6 x 1023 = 3 x 1023
o
(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.
o
For every Al2O3 there are two Al3+ and three O2- ions.
o
So in 51g of Al2O3 there are ...
o
0.5 x 2 x 6 x 1023 = 6 x 1023 Al3+ ions, and
o
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)

You can have a mole of whatever you want in terms of chemical species e.g.

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 ...
o

2 x 2.5 = 5 moles of aluminium ions (Al3+) and 3 x 2.5 = 7.5 mol of oxide ions
(O2-)
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 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 24dm3 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.
Calculate ....
(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 cm 3 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.