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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 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
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. (see also section 6. for reacting masses not using
moles)
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
-
Consider the
formation of 1 mole of ammonia, NH3,
-
consists of
1 mole of nitrogen atoms combined with 3 moles of hydrogen atoms.
-
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).
-
N2(g)
+ 3H2(g) ==> 2NH3(g)
-
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.
-
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
-
Consider the
formation of 1 or 2 moles of aluminium oxide,
-
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.
-
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.
-
Example 7.1.3
-
Example 7.1.4
-
This can be useful
for calculating the quantities of chemicals you need for e.g.
the chemical preparation of a compound.
-
Using the concept of
mole ratio and the exemplar reactions above ...
-
(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?
-
From the equation, 1
mole of copper oxide makes 1 mole of copper chloride,
-
therefore you need
0.25 moles of CuO
-
since mass = molar
mass x formula mass
-
you need 0.25 x 80 =
20g of CuO
-
(b) What mass
of aluminium metal do you need to make 0.1 moles of aluminium
oxide?
-
4Al(s) +
3O2(g) ==> 2Al2O3(s)
and the atomic mass of aluminium is 27
-
4 moles of aluminium
makes 2 moles of aluminium oxide, (ratio 4:2 or 2:1)
-
therefore 0.2 moles
of aluminium metal makes 0.1 moles of aluminium oxide (keeping
the ratio of 2:1)
-
mass of aluminium
metal needed = 0.2 x 27 = 5.4g of Al
-
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?
-
Example
7.2.2: How many moles of calcium ions and
chloride ions in 2.5 moles of calcium chloride?
-
Example
7.2.3: How many moles of lead and oxygen
atoms are needed to make 5 moles of lead dioxide?
-
1 mole of PbO2 contains 1 mole of lead combined with
2 moles of oxygen atoms (or 1 mole of oxygen molecules O2).
-
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?
-
Example
7.2.5: How many moles of chlorine gas
in 6.5g? Ar(Cl) = 35.5)
-
chlorine consists of Cl2 molecules, so
Mr = 2 x 35.5 = 71
-
moles chlorine = mass / Mr = 6.5 / 71
= 0.0944 mol
-
Example
7.2.6: How many moles of iron in 20g?
(Fe = 56)
-
Example
7.2.7: How many grams of propane C3H8
are there in 0.21 moles of it? (C = 12, H = 1)
-
Example
7.2.8: 0.25 moles of molecule X was
found to have a mass of 28g. Calculate its molecular mass.
-
Example
7.2.9: What mass and moles of magnesium
chloride is formed when 5g of magnesium oxide is dissolved in excess
hydrochloric acid?
-
reaction equation: MgO + 2HCl ==> MgCl2
+ H2O
-
means 1 mole magnesium oxide forms 1 mole of
magnesium chloride (1 : 1 molar ratio)
-
formula mass MgCl2 = 24+(2x35.5) =
95,
-
MgO = 24+16 = 40, 1 mole MgO = 40g, so 5g MgO =
5/40 = 0.125 mol
-
which means 0.125 mol MgO forms 0.125 mol MgCl2,
-
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?
-
reaction equation: Na2CO3
+ 2HCl ==> 2NaCl + H2O + CO2
-
means 1 mole sodium
carbonate gives 2 moles of sodium chloride (1:2 ratio
in equation)
-
moles Na2CO3
= 21.2/106 = 0.2 mole
-
therefore 2 x 0.2 = 0.4
mol of NaCl formed.
-
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 ?
-
formula mass of water =
(2 x 1) + 16 = 18
-
every mole of a
substance contains 6 x 1023 particles of 'it' (the
Avogadro Constant).
-
moles water = 1 / 18 =
0.0556
-
molecules of water
= 0.0556 x 6 x 1023 = 3.34 x 1022
-
Since water has a
density of 1g/cm3, it means in every cm3 or ml there are
-
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 ?
-
0.001g of iron = 0.001 /
56 = 0.00001786 mol
-
atoms of iron in the
nail = 0.00001786 x 6 x 1023 = 1.07 x 1019 actual Fe atoms
-
(10.7 million million
million atoms!)
-
Example 7.3.2: (a) How
many particles of 'Al2O3' in 51g of aluminium oxide?
-
Atomic masses: Al =27, O
= 16, f. mass Al2O3 = (2x27) + (3x16) = 102
-
moles 'Al2O3'
= 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
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 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 AS(basic)
[mam]
Type in answer
GCSE/IGCSE H
only or AS(basic) or
multiple choice
GCSE/IGCSE H
only or AS(basic)
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)
(this page)
-
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 Gay-Lussac's Law (ratio of gaseous
reactants-products)
-
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
-
How to
do volumetric titration calculations e.g. acid-alkali titrations
(and diagrams of apparatus)
-
Electrolysis products calculations (negative cathode and positive anode products)
-
Other calculations
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, 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 & half-life calculations including
dating materials
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