Transcript Metals

Metals
Metal Reactivity



Metals display a wide range of reactivity with other
substances, varying from very reactive to no reaction
at all.
Question: Why is knowing the reactivity of a metal
useful to us?
The other substances that most influence the choice
of metal for a particular purpose are oxygen, water
and acids.
The order of metal reactivity is called the activity
series of metals.
Reaction of metals with oxygen
Most metals react with oxygen to form metallic oxides.
All the oxides formed are ionic compounds. Why?
Metal + Oxygen
Metal oxide
Reactions of metals with oxygen
For the reaction of iron with oxygen:


Step 1: Write the word equation
Iron + oxygen
Step 2: Write the forumla
Fe + O2
Iron oxide
Fe 2O 3
Where do the subscripts
come from?
Reactions of metals and oxygen


Step 3: Balance the equation
4Fe + 3O2
Step 4: Add states
4Fe (s) + 3O2 (g)
2Fe2O3
2Fe2O3 (s)
Reaction of metals with water

Most metals undergo no change when placed in cold
water. Some exceptions to this are: lithium, potassium,
sodium, calcium. These react with cold water to form
hydrogen and a metal hydroxide.
Sodium + water

sodium hydroxide + hydrogen
This reaction involves the transfer of electrons from
sodium atoms to hydrogen atoms in the water.
Reaction of a metal and water
For the reaction of sodium with water:
 Step 1: Write the word equation
Sodium + water

sodium hydroxide + hydrogen
Step 2: Write the formula
Na + H2O
NaOH + H2
Reaction of a metal and water

Step 3: Balance the equation
2Na + 2H2O

2NaOH
+ H2
Step 2: Add states
2Na (s) + 2H2O (l)
2NaOH (aq) + H2 (g)
Reaction of metals with water

Some less reactive metals (Al, Zn, Fe) will NOT
react with cold water but will react with steam to
produce steam to produce hydrogen and a metal
oxide.
Zinc + steam
zinc oxide + hydrogen
Reaction of metals with acids

Some metals will react with acids to produce salt and
hydrogen.
Metal + acid

salt + hydrogen
During the reaction between a metal and an acid the
metal loses electrons and becomes positively charge
ions. Hydrogen ions from the acid gain electrons to
form hydrogen gas.
Reaction of metals and acids
For the reaction of zinc and hydrochloric acid:
 Step 1: Write the word equation
Zinc + hydrochloric
acid

zinc
+
chloride
hydrogen
Step 2: Write the formula
Zn + HCl
ZnCl2 + H2
Reaction of metals with acids

Step 3: Balance the equation
Zn + 2HCl

ZnCl2 + H2
Step 4: Add states
Zn (s) + 2HCl (aq)
ZnCl2 (aq) + H2 (g)
A common feature
A common feature of all reactions of metals with oxygen,
water and dilute acids is that atoms of the metals lose
electrons to become positive ions.
 Reactions with oxygen: ionic oxides are formed
MgO = ionic compound containing Mg 2+ and O 2
Reactions with water: ionic hydroxides are formed
LiOH = ionic compound containing Li+ and H-
A common feature

Reactions with acids: ionic metallic chlorides and
sulfates are formed
FeSo4 = ionic compound containing Fe2+ and SO42MgCl2 = ionic compound containing Mg2+ and Cl22-
Ionic equations
Zn (s) + 2HCl (aq)
ZnCl2 (aq) + H2 (g)

Species of the reactants are?
Zn atoms, H+ ions and Cl- ions

Species of the products are?
Zn 2+ ions, H2 molecules and Cl- ions
So we could write the complete ionic equation:
Zn(s) + 2H+(aq)+ 2Cl-(aq)
Zn 2+(aq) + 2Cl-(aq) + H2(g)
Zn(s) + 2H+(aq)+ 2Cl-(aq)
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there are two Cl- ions on the right and two on the left.
Zn atoms have changed to Zn 2+ ions. For this to
happen zinc atoms must have given up 2 electrons.
We can write this as:
Zn

Zn 2+(aq)+ 2Cl-(aq)+ H2(g)
Zn 2+ + 2e-
Hydrogen ions have changed to hydrogen ions have
changed to hydrogen molecules. Therefore each
hydrogen ion must have gained one electron.
2H+ + 2eH2
Net ionic equations
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
Notice that this reaction is really between Zn and H
because they are the only species that undergo a
change.
A net ionic equation only shows the ionic species
that undergo a CHANGE in the reaction.
Zn (s) + 2H+ (aq)
Zn 2+ (aq) + H2 (g)
Spectator ions

Note that in the reaction:
Zn(s) + 2H+(aq)+ 2Cl-(aq)
Zn 2+(aq) + 2Cl-(aq) + H2(g)
the Cl- ions do not undergo a chemical change: there are
two Cl- ions on the right and two Cl- ions on the left. Ions
that do not undergo a chemical change during the reaction
are called spectator ions
Working out net ionic equations
Ionic equation for the reaction of Al and HCl

Step 1: Write the word equation
aluminium + hydrochloric
acid

Al + hydrogen
chloride
Step 2: Write the formula
Al + HCl
AlCl3 + H2
Working out net ionic equations

Step 3: Balance the equation
2Al + 6HCl

2AlCl3 + 3H2
Step 4: Add states
2Al (s) + 6HCl (aq)
2AlCl3 (aq) + 3H2 (g)
Working out net ionic equaitons

Step 5: Determine the species of the reactants and
products
Reactants: Al atoms, H+ ions and Cl- ions
Products: Al 3+ atoms, H2 molecules and Cl- ions

Step 6: write the complete ionic equation
2Al (s) + 6H+ (aq) + 6Cl- (aq)
2Al 3+ (aq) + 6Cl- (aq) + 3H2 (g)
Complete ionic equations

Step 7: Write the net ionic equation
2Al (s) + 6H+ (aq)
2Al 3+ (aq) + 3H2 (g)
Ionisation energy

The reactivity of a metal is related to the ease with
which it loses valence electrons to form ions.
Ionisation energy is a measure of the energy needed
to remove the most loosely bound electron from an
atom in the gaseous state.

In general reactive metals have low ionisation
energies and less reactive metals have high
ionisation energies
Oxidation – reduction reactions
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
Reactions which involve the transfer of electrons are
called oxidation-reduction reactions.
When an atom LOSES one or more electrons we say
it has been oxidised.
When an atom GAINS one or more electrons we say
it has been reduced.
OXIDATION = LOSS OF ELECTRONS
REDUCTION = GAIN OF ELECTRONS
Oxidation – reduction reactions

In normal chemical reactions there can be no overall
loss or gain of electrons. Hence oxidation and
reduction occur simultaneously.

We call these reactions REDOX reactions.

Half equations can be used to describe the oxidation
and reduction processes separately in terms of
electrons lost or gained.
Half equations

Reaction between magnesium and oxygen
Oxidation:
Mg (s)
Reduction:
O2(g) + 4e-
Mg2+(s) + 2e2O2-(s)
Relative atomic mass
Because atoms are so small it is difficult to measure
their actual individual masses.
Relative atomic mass is NOT the mass of an atom of that
element. It is just a relative mass – relative to the mass of
a carbon atom. It is not a mass at all – just a number with
no units.
Example: a titanium atom is 4x the mass of a carbon atom
so relative atomic mass of Ti is 48
Isotopes
Most elements in nature consist of several isotopes with
slightly different masses. This is because isotopes have a
different number of neutrons in the nucleus.
Example:
75% of Cl atoms have 18 neutrons
25% of Cl atoms have 20 neutrons
There are two isotopes of Cl one with an atomic mass of
35 (Cl-35) and one with an atomic mass of 38 (Cl-37).
Remember all Cl atoms have 17 protons.
Isotopes
Therefore strictly speaking when we say the ‘mass of an
atom’ we actually mean the ‘average mass of the atoms
in the naturally occurring element’.
Average mass = 75 x 35 + 25 x 37 = 35.5
100
Note: average mass is closest to the atomic mass of the
most abundant isotope
Relative molecular mass
Relative atomic mass (Ar) is used to describe the mass
of atoms.
Relative molecular mass (Mr) is used to describe the
mass of molecules.
(Mr) = the mass of a molecule of a substance or
compound relative to the mass of an atom of the
carbon-12 isotope taken exactly as 12
Relative molecular mass
The relative molecular mass of a substance is found by
adding the relative atomic masses of the constituent
elements.
Example: molecular mass of H2O
= 2 x Ar(H) + 1 x Ar(O)
= 2x1
+
16
= 18
Relative formula mass
Many compounds, particularly ionic compounds (eg: NaCl)
exist as an array of ions or atoms bound to each other but
with no recognisable molecules. The formula NaCl instead
tells us that throughout a sample of NaCl sodium and
chlorine atoms are present in the ratio 1:1. Because ionic
compounds do not contain molecules the sum of the
relative atomic masses of the atoms in the formula is called
the relative formula mass (still given the symbol Mr).
The mole
Chemists measure the amount of any substance in moles.
Mole: the quantity or amount of a substance that contains
the same number of particles as there are atoms in exactly
12 grams of carbon-12
Avogadro’s number (NA): the number of atoms in
exactly 12 grams of carbon-12 = 6.022 x 1023
Therefore one mole of a substance contains 6.022 x 1023
particles of that substance.
Working with moles
When working in moles the particle or units being
counted should be stated as atoms, molecules or ions.
Eg:
1 mole Fe = 6.022 x 1023 atoms of Fe
1 mole of Pb = 6.022 x 1023 atoms of Pb
1 mole of H2SO4 = 6.022 x 1023 molecules of H2SO
1 mole N2 gas = 6.022 x 1023 molecules of nitrogen
Molar mass
6.022 x 1023 atoms of carbon has a mass of 12 grams
Since all relative atomic masses are measured against the
standard carbon 12 it follows that the atomic mass in
grams of an element (or the formula mass in grams of
any compound) is one mole of that substance and this
one mole contains avogadro’s number of particles.
Consider:
 Titanium (atomic mass 48)
 1 mole of titanium has 6.022 x 1023 atoms (each of
which have a mass 4x that of carbon)
 Therefore mass of 1 mole of titanium = 4x12 =48
1 mole of titanium has a mass equal to its relative
atomic mass.
Molar mass
1 mole of a substance has a mass equal to its:
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
relative atomic mass (expressed in grams)
relative molecular mass (expressed in grams)
relative formula mass (expressed in grams)
This is called the molar mass (M) of a substance. Molar
mass has units – usually grams per mole (g mol -1).
Calculating molar mass
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
Write down the symbol or formula of the substance
Add up the relative atomic masses of the elements
involved
This is the symbol mass or formula mass
In grams this is one mole of the substance
This is made of avogadro’s number of particles
Summary
There are therefore two ways of looking at a mole:
A number of particles
(atoms, ions,
molecules)
6.022 x 1023
6.022 x 1023 =
atoms of Ti
MOLE
1 mole of Ti
A mass (the relative
atomic, molecular
or formula mass in
grams)
= 48 grams of Ti
Moles and numbers of particles
Relationship between moles and number of particles (eg:
atoms, molecules):
Number of moles (n) = number of particles (N)
number of particles in one mole (NA)
n=
N
6.022 x 1023
Moles and numbers of particles
Using this formula it is possible to calculate:

Number of moles of a substance from the number of
particles or basic units of a substance

Number of particles or basic units of a substance
(atoms, formula units) from the number of moles
Moles and mass
The relationship between the number of moles
and mass
of a substances is:
Number of moles (n) =
n= m
M
mass (m) in grams
mass of 1 mole (M)
Using this formula it is possible to calculate:

Number of moles of any substance in a given mass

Mass of a substance in a given number of moles
Summary
n=
m
M
n=
N
6.022 x 1023
Therefore:
N
6.022 x 1023
=
m
M
Percentage composition
Percentage composition of a compound is simply
the percentage by mass of each element present
in the compound.
To determine percentage composition you need two
things :
 formula of the compound
 Relative atomic masses of the elements present
Steps in determining percentage
composition
Calculate the percentage composition of iron in Fe2O3
Step 1:
determine mass of one mole of Fe2O3
2 x 55.9 + 3 x 16 = 159.8 grams
Step 2:
One mole of Fe2O3 contains 2 moles of Fe.
2 moles of Fe : 2 x 55.9 = 111.8 grams
Step 3:
% composition of Fe = 111.8 x 100 = 70%
159.8
Summary
Therefore:
% A in a compound
= mass of A in 1 mole of the compound x 100
mass of 1 mole of the compound
Molecular v’s empirical formula

Molecular formula: specifies the actual number of
atoms of each element in a molecule
hydrogen peroxide: H2O2

Empirical formula: specifies the simplest whole
number ratio of each element
hydrogen peroxide: HO
Determining empirical formula
The empirical formula of a substance can be established
by first determining the percentage composition of a
substance by chemical analysis.
Ex: Calculate the empirical formula of glucose with a
chemical composition of 40% carbon, 6.6% hydrogen
and 53.3% chlorine

Step 1: List the elements and the mass of each element
in 100grams of freon-12
Element:
Mass in 100g:

H
6.6
O
53.3
Step 2: Determine the moles of each element in 100
grams
n=m
40
6.6
53.3
M
Mole ratio

C
40
12.01
= 3.33
1
:
Step 3: Empirical formula CH2O
1.008
= 6.55
2
:
16.00
= 3.33
1
Determining molecular formula
If the molecular mass of glucose =180.16 what is its
molecular formula?
Step 1: Determine the relative empirical formula mass
CH2O = 1 x12.01 + 2 x 1.008 + 1 x 16.00 = 30.03
Step 2:
Relative molecular mass
=
Relative empirical formula mass
molecular formula
empirical formula
Relative molecular mass
=
Relative empirical formula mass
180.16
30.03
=
molecular formula =
molecular formula
empirical formula
molecular formula
CH2O
molecular formula =
180.16 x CH2O
30.03
6 x CH2O
molecular formula =
C6H12O6
Joseph Gay-Lussac
Gay Lussac’s law of combining gas volumes states that
when two gaseous elements combine:
‘the ratios of the volumes of gases involved, if measured
at the same temperature and pressure, are expressed by
small whole numbers’
Example: calculate the volume of hydrogen that will
combine with 6 L of nitrogen to form ammonia

Step 1: write the balanced equation for the reaction
N2 (g)
+
3H2 (g)
2NH3 (g)

Step 2: determine the ratios
y
1 mole N2
1 volume

3 mole H2
3 volume
Step 3: calculate volume
2 mole NH3
2 volume
V (H2 )
= 3 x V(N2)
= 3 x 6.0L
= 18 L
Mass-mass calculations
Chemical equations show the number of moles of reactants
and products in a chemical reaction. They can also be used
to determine the relationship between the masses of the
reactants and products:
N2 (g)
1 mole
28g
+
+
3H2 (g)
3 mole
6g
2NH3 (g)
2 mole
34g
Remember this!
When carrying out mass-mass calculations
remember the following step by step method:
N = m/N
Mass of
the
known
From the
equation
Moles of
the
known
N = m/N
Moles of
the
unknown
Moles of
the
unknown