(a) WHAT IS
THE MOLE CONCEPT? and WHAT IS ONE
MOLE OF A SUBSTANCE?
Summary of main points about defining and using the
mole concept
For the purposes of calculations
chemical amounts are primarily measured in moles (but sometimes just
masses are ok).
The
symbol for the unit mole is mol and the mass of one mole of a substance in grams
is numerically equal to its relative formula mass (can be atomic mass or
molecular mass  see below).
This might be an atomic mass of an
element e.g. Fe = 56, so 1 mol = 56g
or the molecular mass of an element e.g.
O = 16, so 1 mole of oxygen molecules O_{2} = 2 x 16 = 32g
or the molecular mass of a covalent
compound e.g. C = 12, H = 1, so 1 mol of propane C_{3}H_{8}
= (3 x 12) + 8 = 44g
or formula mass of an ionic compound e.g.
K = 39, O = 16, so 1 mol of potassium oxide K_{2}O = 39 + 2 x 16) =
71g
In the latter you can consider
1 mole of K_{2}O made up of 2 mol of potassium ions and 1 mol
oxide ion.
One mole of a substance contains the same
number of the stated particles, atoms, molecules
or ions as one mole of any other substance. So all the three described above
have the same number of defined particles.
The number of atoms, molecules or ions in a
mole of a given substance is the Avogadro
constant.
The value of the Avogadro constant
(denoted by N_{A}) is
6.02 x 10^{23} per mole of particles of whatever you have
specifically defined e.g.
hydrogen atoms H, hydrogen
molecules H_{2}, methane molecules CH_{4},
ionic lattice NaCl
sulfate ion SO_{4}^{2},
sodium ion Na^{+}, ammonium ion NH_{4}^{+}
etc.
So the previous four examples described above
equate to:
6.02 x 10^{23} Fe atoms, 6.02 x 10^{23}
O_{2}
molecules, C_{3}H_{8} molecules
and 6.02 x 10^{23} 'K_{2}O's.
The latter consists of 2 x 6.02 x
10^{23} = 12.04 x 10^{23} = 1.204 x 10^{24}
potassium ions
and 6.02 x 10^{23} oxide
ions. You must be versatile in your mol thinking!
You
should understand and how to use measurement of the amounts in moles
and apply this knowledge to calculations involving atoms, molecules,
ions, electrons, formulae and equations.
This is a lot to take in, so
I've described in details lots of examples explained below on how to
use the 'mol'
TOP OF PAGE
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
calculations!
The theoretical basis is
explained in section (b).
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 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
calculation e.g.
Al metal
element atom, H_{2}O covalent molecule, an
element O_{2} molecule, Na^{+}Cl^{}
ionic compound or just any compound formula like CuSO_{4}
etc. etc.
This specificity
cannot be overemphasised and you should understand that the
measurement of amounts in moles can apply to atoms, molecules,
ions, electrons, formulae and equations !!!
M_{r} 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 ...
Lots of little examples of the
interconversion of mass and moles knowing the formula mass are 'jotted' down
below
1 mole of carbon atoms = 12 g, 1 mole
of carbon dioxide (12 + 2 x 16) = 44, 1 mole CO_{2} = 44 g
the number of C atoms = the
number of CO_{2} molecules = 6.0 x 10^{23} specified
particles
molecular mass M_{r} 18 for the water molecule
H_{2}O,
17 for the ammonia molecule NH_{3}
so 1 mole of water is 18g,
0.333 mole = 0.333 x 18 = 6g
for ammonia 1 mol = 17g, 34g =
34/17 = 2 mols ammonia
M_{r} = 16 for the methane molecule
CH_{4}
and 180 for the glucose sugar molecule C_{6}H_{12}O_{6}
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 N_{2} molecules (M_{r} = 28),
molar mass = 28g,
0.25 moles = 0.25 x 28 = 7.0g
Relative atomic mass of iron A_{r}
= 56, 7g = 7/56 = 0.125 mol Fe (Relative
atomic mass explained)
The mass of 1 mole of sodium chloride
NaCl? M_{r}(NaCl) = 23 + 35.5 = 58.5 g
The mass of 1 mole of calcium
carbonate? M_{r}(CaCO_{3}) = 40 + 12 + (3 x 16) = 100
g
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 +
35.5)
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) (NH_{4})_{2}SO_{4} or (NH_{4}^{+})_{2}(SO_{4}^{2})
= 130g
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 Al_{2}O_{3}
= (2 x 27) + (3 x 16) = 102 g
BUT being ionic the formula can
be written as (Al^{3+})_{2}(O^{2})_{3},
1 mole of Al_{2}O_{3} consists of 2 moles of
aluminium ions and 3 moles of oxide ions
The mole concept can be extended to
reading equations in terms of moles e.g.
MgCO_{3(s)} + 2HCl_{(aq)}
===> MgCl_{2(aq)} + H_{2}O_{(l)} + CO_{2(g)}
1 mol of magnesium carbonate reacts with 2 mol of
hydrochloric acid to give 1 mol of magnesium chloride plus 1 mol of
water and 1 mol of carbon dioxide
CH_{4(g)} +
2O_{2(g)} ==> CO_{2(g)} + 2H_{2}O_{(l)}
1 mol of methane requires 2 mol of oxygen for
complete combustion to 1 mol of carbon dioxide and 2 mol of water
2Al_{(s)} + 6HCl_{(aq)}
===> 2AlCl_{3(aq) }+_{ }3H_{2(g)}
2 mol of aluminium reacts with 6 mol of
hydrochloric acid to form 2 mol of aluminium chloride plus 3 mol of
hydrogen gas
2C_{2}H_{6(g)} + 7O_{2(g)}
==> 4CO_{2(g)} + 6H_{2}O_{(l)}
2 mol of ethane requires 7 mol of oxygen for
complete combustion giving 4 mol of carbon dioxide and 6 mol of water
but you can also say: C_{2}H_{6(g)} +
3½O_{2(g)}
==> 2CO_{2(g)} + 3H_{2}O_{(l)}
in other words, 1 mol of ethane requires 3½ mol of oxygen for
complete combustion giving 2 mol of carbon dioxide and 3 mol of water,
and this example illustrates the importance
of thinking about balanced equations in terms of the
mole ratio of reactants and products.
This reading of equations in terms of
moles of reactants and products is important in calculations AND the process can
be reversed, in other words, given the masses of reactants and products, you can
convert them to moles and work out the balanced symbol equation.
There is a periodic table of atomic masses near the end of the page for doing
mole calculations
(b) EXAMPLES OF MOLE CALCULATIONS
A variety
to illustrate the introduction section (a)
Part 1 of problem solving and working through calculation
questions on mole ratios and equations
REMINDER
For a substance 'Z' i.e. a
specifically defined chemical species
(cover over the one you want and what's left is what you
need to do)
(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). 

Set 1 of mole calculations  using the simple mole formula
(see triangle relating mass, moles and relative formula
mass)
Set 2 of mole calculations  relating a formula, balanced equations to
the mole concept and mole ratios
Part 2 of problem solving and working
through calculation questions on mole ratios and equations

Mole
calculation Example
7.2.1

Mole
calculation Example
7.2.2

Mole
calculation Example
7.2.3

Mole
calculation Example
7.2.4

Mole
calculation Example
7.2.5

Mole
calculation Example
7.2.6

Mole
calculation Example
7.2.7

Mole
calculation Example
7.2.8

Mole
calculation Example
7.2.9

Mole
calculation Example
7.2.10

Mole ratio calculation
Example 7.2.11

High quality magnetite ore
contains mainly the compound diiron(II)iron(III) oxide, Fe_{3}O_{4}.
('triiron tetroxide')

Info.: Atomic mass A_{r}:
Fe = 55.8, C = 12.0, O = 16.0; 1 tonne ≡
1000 kg ≡
10^{6} g (which you should know!)

One of several
reduction reactions using coke (C) in a blast furnace is:

Fe_{3}O_{4}
+ 4CO ====> 3Fe + 4CO_{2}

(a) Theoretically, what
quantity of high grade magnetite ore (in tonne) is needed to
make 100 tonnes of iron for steel making?

From the equation 1 mol of Fe_{3}O_{4}
gives 3 mol of Fe, a ratio of 1 : 3

100 tonne of Fe = 100 x 10^{6}
g,

since: mol = mass/Ar, mol
iron = 100 x 10^{6} / 55.8 = 1.792 x 10^{6} mol

Since the mol ratio in the
equation is 1 : 3, you need to divide mol Fe/3

So, mol Fe_{3}O_{4}
= 1.792 x 10^{6} / 3 = 0.597 x 10^{6}

mass = mol x Mr, r(Fe_{3}O_{4})
= (3 x 55.8) + (4 x 16.0) = 231.4

mass Fe_{3}O_{4}
required = 0.597 x 10^{6} x 231.4 = 138.146 x 10^{6}
g

138.146 x 10^{6} /
106 = 138 tonnes
(3sf)


(b) For the same
amount of iron production (100 tonne) what is the minimum
quantity of coke (in kg) required to reduce the quantity of ore
calculated in (a)?

4 mol CO
≡
4 mol C ==> 3 mol Fe

This
gives a mole ratio of 4 : 3 for C : Fe

We need
to scale up moles of Fe by a factor of 4/3 get the moles of
carbon.

From
(a) mol Fe = 1.792 x 106

Therefore mol C required = 1.792 x 10^{6} x 4/3 = 2.389
x 10^{6}

mass =
mol x Ar = 2.389 x 10^{6} x 12 = 28.67 x 10^{6}
g

mass =
28.67 x 10^{6} / 1000 = 28.67 x 10^{3} = 2.867 x
10^{4} =
2.87 x 10^{4}
kg (3 sf)


In all honesty, I find it
easier to solve this problem using simple reacting mass ratio
calculation (see
section 6.)

However, you are expected to
interpret questions on equations using moles and mass units of
g, kg and tonne.


7.2.12 ???
Set 3 of mole calculations  deducing a balanced symbol equation from the
masses of reactants and products
Part s of problem solving and working through calculation
questions on mole ratios and equations
Since you can interpret equations in terms
of moles and masses of reactants and products, you can also work in the other
direction.
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: A_{r}
= atomic mass, M_{r} = 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
(a)
Convert the reacting masses given into moles
mol = mass / A_{r} or M_{r}
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 ==>
0.05 moles
scaling up to the simplest whole
number ratio is 2 : 1 ==> 2 moles (divide the 1st ratios by
0.025, always try dividing by the lowest number first)
(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
equation
2PbO + C ===>
2Pb + CO_{2}
Equation  mole calculation Example 7.3.2
to work out an equation from reacting masses using moles
(a) 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
(i) 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 (O_{2}) = 11.6
 8.4 = 3.2 g
(ii)
Calculate the moles of iron and oxygen that combined to give the iron oxide.
mol Fe = 8.4 / 56 = 0.15,
mol O_{2} = 3.2 / 32 = 0.10
Note: O is 16, BUT it is oxygen
molecules that are in air, formula mass O_{2} = 32
(iii) Convert the Fe : O2 ratio to the
simplest whole number ratio
Fe : O_{2} is 0.15 : 0.10,
diving by 0.05 gives a reacting mole ratio of 3 : 2 for Fe : O_{2}
(iv) We can therefore write the lefthand
side of the equation as
3Fe + 2O_{2}
===> ?
Since there was only one product of
the reaction, the empirical formula of this particular iron oxide
must be composed of three iron atoms and four atoms of oxygen, so the
full equation is
3Fe + 2O_{2}
===> Fe_{3}O_{4}
(b) It was found that 14.0 g of
lithium (Li) reacted with 16.0 g of oxygen (O_{2}) to form 30 g of
lithium oxide (Li_{2}O).
Convert
the masses to moles and deduce the balanced equation from the mole
ratios calculated.
Relative masses: Li = 7,
O = 16, O_{2} = 32, Li_{2}O
= (2 x 7) + 16 = 30
mol Li = 14/7 = 2.0,
mol O_{2} = 16/32 = 0.5, mol Li_{2}O
= 30/30 = 1.0
To get a set of whole numbers for the
ratios you just double everything giving the following ratios of 4Li
: 1 O_{2} : 2 Li_{2}O, so the balanced
equation is:
4Li_{(s)} +
O_{2(g)} ===> 2Li_{2}O_{(s)}
Example 7.3.3
to work out an equation from reacting masses using moles (typical GCSE
question)
It was found that 7.95 g copper oxide
(CuO) was reduced by 0.6 g of carbon to form 6.35 g of copper and releasing
2.2 g of carbon dioxide.
Convert the reactant and product masses
to moles and deduced the balanced equation.
Atomic masses: Cu = 63.5, C = 12,
O = 16 (I've set out the solution to the problem as a table of
'logic')
reactant => product 
? CuO 
? C 
==> 
? Cu 
? CO_{2} 
atomic/formula mass 
63.5 + 16 = 79.5 
12 
==> 
63.5 
44 
mass in g 
7.95 
0.6 
==> 
6.35 
2.2 
moles = mass/formula mass 
7.95/79.5 = 0.10 
0.6/12 = 0.05 
==> 
6.35/63.5 = 0.1 
2.2/44 = 0.05 
simplest whole number mol ratio 
0.10/0.05 = 2 mol 
0.05/0.05 = 1 mol 

0.1/0.05 = 2 mol 
0.05/0.05 = 1 mol 
Therefore the equation must be 
2CuO_{(s)} +
C_{(s)} ===> 2Cu_{(s)} + CO_{2(g)} 
Notes:
(i) Always try dividing by the lowest
ratio number first to try to deduce the whole number ratio to deduce the
balanced equation
(ii) If you have done the calculation
correctly the equation should be balanced, so always do an atom count!
(ii) For GCSE students the numbers
are often 'perfect' but A level students may to make a reasonable
judgement from ratios like 2.02 : 0.99 ==> 2.01 : 0.97 etc.
(iv) Don't forget that state symbols
might be required, especially at A Level.
Example 7.3.4
to work out an equation from reacting masses using moles (more an A level
question?)
7.75 g of solid phosphorus consumed 10 g
of oxygen gas to form 17.75 g of a solid phosphorus oxide.
Convert the reactant and product masses
to moles and deduced the balanced equation.
Atomic masses: P = 31, O = 16
(a) convert the masses of phosphorus and
oxygen to moles of atoms and deduce the empirical formula of the phosphorus
oxide. Remember empirical formula is based on the simplest whole number
ratio expressed in the formula.
7.75/31 = 0.25 mol of P atoms,
10/16 = 0.625 mol O atoms
(this is a slightly awkward ratio to
deal with, so it needs a bit of trial and error patience to solve it!)
Dividing by the smallest number gives
0.25/0.25 : 0.625/0.25
gives 1 P : 2.5 O, then just
multiplying by 2 gives 2 P : 5 O
So the empirical formula is P_{2}O_{5}
(b) From experiments it was further
deduced that the molecular mass of the phosphorus oxide was 284.
(i) Deduce the correct molecular
formula of the phosphorus oxide
The empirical formula mass of P_{2}O_{5}
= (31 x 2) + (16 x 5) = 142
Since 284/142 = 2, means the true
molecular formula = 2 x P_{2}O_{5} = P_{4}O_{10}
(ii) From the mole ratios and your
answer to (i) deduce the correctly balanced symbol equation.
From your answer in (a) you need a
ratio of 2 : 5 for P : O, but you must allow for the formation of P_{4}O_{10}
molecules and oxygen exists as diatomic molecules (O_{2}).
Therefore to make a P_{4}O_{10}
you need 4 P atoms and 10 O atoms BUT as 5 O_{2} molecules
So the equation is: 4P_{(s)}
+ 5O_{2(g)} ===> P_{4}O_{10(s)}
(don't forget that state symbols
might be required)
Example 7.3.5
to work out an equation from reacting masses using moles (very much an A
level question)
Iodine monofluoride gas (IF) is very
unstable and above 0^{o}C it decomposes into solid iodine (I_{2})
and liquid iodine pentafluoride (IF_{5}).
It was found that 14.6 g of iodine
monofluoride decomposed to form 10.16 g of iodine and 4.44 g of iodine
pentafluoride.
Atomic masses: I = 127, F = 19
(a) Calculate the formula masses of the
reactants and products.
IF = 127 + 19 = 146,
I_{2} = 2 x 127 = 254,
IF_{5} = 127 + (5 x 19) = 222
(b) Convert the reactant and product
masses to mol
mol IF = 14.6/146 = 0.1,
mol I_{2} = 20.32/254 = 0.04, mol IF_{5}
= 4.44/222 = 0.02
(c) From the mole ratios of reactants and
products deduce the balanced symbol equation for the decomposition of iodine
monofluoride.
Initial mol ratio: IF : I_{2}
: IF_{5} is 0.1 : 0.04 : 0.02,
dividing by 0.02 gives a whole number
ratio of 0.1/0.02 = 5 (IF), 0.04/0.02 = 2 (I_{2}),
0.02/0.02 = 1 (IF_{5})
Therefore the equation is: 5IF_{(g)}
===> 2I_{2(s)} + IF_{5(l)}
(c) THE
AVOGADRO CONSTANT
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 H_{2}O,
glucose molecules C_{6}H_{12}O_{6}, hydroxide ion OH^{},
iron(III) ion Fe^{3+}, 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
10^{23} per mole. You should understand that the measurement of amounts
in moles can apply to atoms, molecules, ions, electrons, formulae and
interpreting equations
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 N_{A}).
It is equal to 6.023 x 10^{23}
'defined species' per mole i.e. 6.023 x 10^{23} 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 10^{23}
atoms of carbon
or
that many molecules in 18g water (H_{2}O = 1+1+16 = 18, H = 1; O =
16) * = 6.023 x 10^{23}
molecules of water
*
This is about 18cm^{3}, so picture this
number of molecules in a nearly full 20cm^{3} measuring cylinder or a
100ml beaker less than ^{1}/_{5}th full!
The Avogadro number is 6.023 x
10^{23} = 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
1cm^{3}, 1 mole of water = 18g and ~ 18cm^{3} because the density of
water is ~1.0 g/cm^{3}
Therefore in a thimble full of
water there are ~6.023 x 10^{23}/18 = ~3.3 x 10^{22} = 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
sort of!)
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
Number.
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 + H_{2}SO_{4}
==> Na_{2}SO_{4} + 2H_{2}O
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}/_{20}th
of a mole of sodium hydroxide you can deduce (yes predict!)
0.05 moles of NaOH reacts with
0.025 moles of H_{2}SO_{4} to form 0.025 moles of Na_{2}SO_{4}
and 0.05 moles of H_{2}O
What more, since you can convert
moles to mass, you can do deduce the mass of product formed or the mass of
reactants need.
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
moles)
(d) More
advanced calculations using the Avogadro Constant
This is for more advanced
students and illustrates the concepts in the introduction
section (b)
where
you can actually calculate the
number of particles in known quantity of material !

Avogadro
constant Mole calculation Example 7.4.1

Avogadro number calculation Example 7.4.2

Avogadro constant calculation Example 7.4.3

(a) How
many particles of 'Al_{2}O_{3}' are there in 51g of aluminium oxide?

Atomic masses: Al =27, O
= 16, f. mass Al_{2}O_{3} = (2x27) + (3x16) = 102

moles 'Al_{2}O_{3}'
= 51/102 = 0.5 mol

Number of 'Al_{2}O_{3}'
particles = 0.5 x 6 x 10^{23} = 3 x 10^{23}


(b) Aluminium oxide
is an ionic compound. Calculate the number of individual aluminium ions (Al^{3+})
and oxide ions (O^{2}) in the same 51g of the substance.

For every Al_{2}O_{3}
there are two Al^{3+} and three O^{2} ions.

So in 51g of Al_{2}O_{3}
there are ...

0.5 x 2 x 6 x 10^{23}
= 6 x 10^{23} Al^{3+} ions, and

0.5 x 3 x 6 x 10^{23}
= 9 x 10^{23} O^{2} ions.

(e)
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.

In terms of electric
charge, 1 Faraday = 96500 C (coulombs) = 6 x 10^{23} electrons

If you have 2.5 moles of
the ionic aluminium oxide (Al_{2}O_{3}) you have ...

When you write ANY
balanced chemical equation, the balancing numbers, including the unwritten
1, are the reacting molar ratio of reactants and products.

This also applies to half equations e.g.
electrolysis examples ...

Al^{3+} + 3e^{}
===> Al, is read as 1 mol of aluminium ions is reduced by 3 mol of
electrons to 1 mol of aluminium atoms.

2Cl^{}
==> Cl_{2} + 2e^{},
is read as 2 mol chloride are oxidised to release 1 mol of chlorine gas and
2 mol of electrons.

See separate page for
Electrolysis products calculations (negative cathode and positive anode
products)
Extra Advanced Level Chemistry Questions  more suitable for Advanced ASA2 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 10^{23} mol^{1}
The molar volume for gases is 24.0dm^{3} 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 cm^{3} 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 cm^{3} of 0.05 mol dm^{3} hydrochloric acid.
ANSWERS to QA7.1
Selfassessment Quizzes for GCSE or A Level (the basics) on mole
calculations
Type in answer
QUIZ or
multiple choice
QUIZ
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing mole 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)
(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 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
introduction to mole calculations Revision KS4 Science
revising introduction to mole calculations Additional
Science Triple Award Science Separate Sciences Courses aid to introduction to
mole calculations textbook revision
GCSE/IGCSE/O level Chemistry introduction to mole calculations Information Study Notes for revising for AQA GCSE
Science introduction to mole calculations, Edexcel GCSE Science/IGCSE Chemistry
introduction to mole calculations & OCR 21st Century Science, OCR Gateway
Science introduction to mole calculations WJEC gcse science chemistry
introduction to mole calculations CEA/CEA gcse science chemistry O
Level Chemistry (revise courses equal to US grade 8, grade 9 grade 10
introduction to mole calculations) A level
Revision notes for GCE Advanced Subsidiary Level introduction to mole
calculations AS Advanced Level A2 IB Revising introduction to mole calculations
AQA GCE Chemistry OCR GCE Chemistry introduction to mole calculations Edexcel GCE Chemistry Salters Chemistry
introduction to mole calculations CIE
Chemistry introduction to mole calculations, WJEC GCE AS A2 Chemistry
introduction to mole calculations, CCEA/CEA GCE AS A2 Chemistry revising
introduction to mole calculations courses for preuniversity students (equal to
US grade 11 and grade 12 and AP Honours/honors level introduction to mole
calculations revision guide to introduction to mole calculations how to work out
a balanced symbol equation using reacting masses and moles of products.
















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