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 oxygenMost metals react with oxygen to form metallic oxides.

All the oxides formed are ionic compounds. Why?

Metal + Oxygen Metal oxide

Reactions of metals with oxygenFor the reaction of iron with oxygen:

Step 1: Write the word equation

Iron + oxygen Iron oxide

Step 2: Write the forumla

Fe + O2 Fe 2O 3

Where do the subscripts come from?

Reactions of metals and oxygen Step 3: Balance the equation

4Fe + 3O2 2Fe2O3

Step 4: Add states

4Fe (s) + 3O2 (g) 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 acidsFor the reaction of zinc and hydrochloric acid: Step 1: Write the word equation

Zinc + hydrochloric zinc + hydrogenacid chloride

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 featureA 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 SO42-

MgCl2 = ionic compound containing Mg2+ and Cl22-

Ionic equationsZn (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) Zn 2+

(aq)+ 2Cl-(aq)+ H2(g)

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+ + 2e-

Hydrogen ions have changed to hydrogen ions have changed to hydrogen molecules. Therefore each hydrogen ion must have gained one electron.

2H+ + 2e- H2

Net ionic equations 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 equationsIonic equation for the reaction of Al and HCl

Step 1: Write the word equation

aluminium + hydrochloric Al + hydrogenacid 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

productsReactants: 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 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) Mg2+(s) + 2e-

Reduction: O2(g) + 4e- 2O2-(s)

Relative atomic massBecause 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

IsotopesMost 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.

IsotopesTherefore 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 massRelative 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 massThe 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)

= 2 x 1 + 16

= 18

Relative formula massMany 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 moleChemists measure the amount of any substance in moles.

Mole: the quantity or amount of a substance that containsthe same number of particles as there are atoms in

exactly12 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 molesWhen 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 mass1 mole of a substance has a mass equal to its:

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 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

A number of particles(atoms, ions, molecules)6.022 x 1023

SummaryThere are therefore two ways of looking at a mole:

6.022 x 1023 = 1 mole of Ti = 48 grams of Ti

atoms of Ti

A mass (the relative atomic, molecular or formula mass in

grams)

MOLE

Moles and numbers of particlesRelationship 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 particlesUsing 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 massThe relationship between the number of moles

and mass of a substances is:

Number of moles (n) = mass (m) in gramsmass of 1 mole

(M)n = m

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

Summaryn = m

Mn = N

6.022 x 1023

Therefore:

N = m6.022 x 1023 M

Percentage compositionPercentage 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 compositionCalculate 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

SummaryTherefore:

% 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 formulaThe 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: C H OMass in 100g: 40 6.6 53.3

Step 2: Determine the moles of each element in 100 gramsn = m 40 6.6 53.3

M 12.01 1.008 16.00= 3.33 = 6.55 = 3.33

Mole ratio 1 : 2 : 1

Step 3: Empirical formula CH2O

Determining molecular formulaIf 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 = molecular

formulaRelative empirical formula mass empirical formula

Relative molecular mass = molecular formulaRelative empirical formula mass empirical formula

180.16 = molecular formula 30.03 CH2O

molecular formula = 180.16 x CH2O 30.03

molecular formula = 6 x CH2O

molecular formula = C6H12O6

Joseph Gay-LussacGay 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 ratiosy

1 mole N2 3 mole H2 2 mole NH3

1 volume 3 volume 2 volume

Step 3: calculate volume V (H2 ) = 3 x V(N2)

= 3 x 6.0L

= 18 L

Mass-mass calculationsChemical 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) + 3H2 (g) 2NH3 (g)

1 mole 3 mole 2 mole

28g + 6g 34g

Remember this!When carrying out mass-mass calculations

remember the following step by step method:

Mass of the

known

Moles of the

known

Moles of the

unknown

Moles of the

unknown

N = m/NFrom the equation N = m/N