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Section 1: Principles of Chemistry - IGCSE Chemistry

Chemistry > Section 1: Principles of Chemistry

a) States of Matter

1.1 understand the arrangement, movement and energy of the particles in each of the three states of matter: solid, liquid and gas











Cannot move, vibrate

Particles can move

Particles can move



throughout the liquid



slight past each other


Energy of Particles

Particles have least

Particles have more

The particles have the


kinetic energy

kinetic energy than

most kinetic energy




Distance between

Closely packed

Not closely paced

Far apart




3D structure

Takes the shape of the

No fixed shape





1.2 understand how the interconversions of solids, liquids and gases are achieved and recall the names used for these interconversions



  • Melting: The process of converting from solid to liquid due to increase of temperature.
  • Melting point: The temperature at which a solid starts to melt. Ice melts at 0 degree C.
  • Boiling: The process of converting from liquid to gas due to increase of temperature.
  • Boiling point: The temperature at which liquid starts to boil. Water boils at 100 degree C.
  • Condensation: The process by which a gas turns to a liquid, this process is called condensation.
  • Sublimation: The process by which a solid directly turns into a gas without melting.
  • Solidification: The process at which gas directly turns to solid.
  • Vaporisation: A process by which liquid turns to a gas at its boiling point. (same as boiling)
  • Evaporation: A process by which a liquid turns to a gas below its boiling point.
  • Volatile: The liquids which evaporates at room temperature that liquids are called volatile.


1.3 explain the changes in arrangement, movement and energy of particles during these interconversions.


  • Solid to liquid: When a solid is heated the particles absorb energy and starts to vibrate faster about their fixed position. When the temperature is high enough, the vibration of the particles become sufficient to overcome the attraction between them. The particles begin to break away from their fixed position. The particles can now move slight past each other. The solid is formed liquid.
  • Liquid to solid: When a liquid is cooled, the energy is given out by the particles and begins to move slowly. When the temperature is low enough, the particles no longer have the energy to slide over each other. The particles start to settle into a fixed position.
  • Liquid to Gas: When a liquid is heated, the particles move fast which break all the forces of attraction in liquid. Bubbles of gaseous particles are formed throughout the whole liquid.
  • Gas to liquid: When a gas is cooled the particles eventually move slowly enough that attraction between them hold the particles as liquids. The gas is now condensed to liquid.


b) Atoms


1.4 describe and explain experiments to investigate the small size of particles and their movement including:


  • Dilution of coloured solutions
  • Diffusion experiments

Experiment on dilution

In a beaker of water, a deeply colored substance (Potassium mangnete) is kept. After 10-20 min, the whole solution turned purple. That's because Potassium Mangnete is diffused in the solution.If is it was replaced with heavier or bigger particles, it will take long time to diffuse and often it wont.

Diffusion experiment-1

When the glass lid is removed, the brome rises to the top because bromine is made up of small particles which occupies spaces inside both jars by diffusion.

Diffussion experiment-2

From the cotton wool which is soaked in aqueous ammonia, ammonia gas is produced. And from the cotton wool which is soaked in hydrochloric acid, hydrogen chloride gas is produced. Both gases move to each other inside the glass and meet near the cotton wool which is soaked in hydrochloric acid. The gases forms a white smoke of ammonium chloride.

NH3 (g) + HCl(g) --> NH4Cl(s)

This shows that ammonia gas move faster than hydrogen chloride gas in the same length of time. Because ammonia is lighter than hydrogen chloride gas. So ammonia diffuses faster.

The factors that affects the diffusion process are:-

(i) Molecular Mass(Mr)

(ii) Temperature

The molecules with low molecular mass diffuse faster than the higher molecular masses.

Increase in temperature increases the diffusion rate.


1.5 understand the terms atom and molecule






The smallest particle of an element is called atom.

The smallest particles of an element or compound which


can exist independent is called molecules.


Atoms consists of electron, proton and neutron.

Molecule are made up of different types of atom.


Atom can take part in chemical reactions directly.

Molecules cannot take part in chemical reaction without


breaking up into atoms.


There are limited types of atom in the earth.

There are unlimited types of molecules.



1.6 understand the differences between elements, compounds and mixtures



Difference between element and compound





Element is a pure substance that cannot be split into

A compound is a pure substance which contain two or


simplier substances by chemical process or electricity.

more different types of element.


The smallest particle of an element is atom.

The smallest particles of an compound is molecule.


there are limited types of element

There are unlimited types of compound.


The elements are expressed by using chemical symbol.

The compounds are expressed by using chemical formula.


Difference between compound and molecules





Compounds are made up of two or more elements

A gathering of two or more substances without chemical


chemically combined.



To make a compound, chemical reactions are needed.

To make a mixture chemical reactions are not needed.


Compounds cannot be separated by physical method of

Mixture can be separated by physical method of





In compounds there must be a fixed proportion of substance.

There is no fixed proportion of substance in mixture.
To make compounds energy change take place. To make mixture no energy change take place.





1.7 describe experimental techniques for the separation of mixtures, including simple distillation, fractional distillation, filtration, crystallisation and paper chromatography


i) Filtration:

A suspension or insoluble solid-liquid mixture can be separated by this process.

The suspended mixture is poured into the filter funnel. The filter paper has tiny holes, which allows the liquid molecules to pass through. But the solid particles get trapped on the filter paper as residue. The liquid is collected below the funnel as filtrate.


  • Precipitate from a reaction mixture can be separated by filtration.


Soluble solid can be separated from a solution by distillation process.

The solution is heated in a round flask. The liquid solvent evaporates and pass through the condenser. While passing, it condenses and drops of liquids are poured in an beaker. The collected liquid is called distillate.


Salt water solution can be separated.

Fractional Distillation:

Miscible liquid with different boiling points can be separated by fractional distillation.

Ethanol water can be separated by this process. The boiling point of ethanol is 78 degree Celsius and boiling point of water is 100 degree Celsius. The temperature of the mixture increases as it is heated. In 78 degree ethanol distills over. The temperature remains constant until all the ethanol has distilled out of the round bottom flask. Then temperature increases until 100 degree Celsius. At this temperature, water starts to boil off. This time heating should be stopped. The ethanol and the water is now separated.


  • The fractions can be separated from the crude oil by this process.
  • Nitrogen and oxygen can be separated from liquid air.
  • Ethanol can be separated from fermented mixture.


Soluble solid can be separated from a solution by crystallisation.

Heat the salt-solution to make it saturated.

Test the solution whether it is saturated or not by dipping a clean dry cold glass rod into the solution. If the solution is saturated, crystals will form in the glass rod.

Filter the solution to collect the crystals. Then wash the crystals with a liquid cold distilled water to remove impurity. Dry the crystals, keeping it in between few sheets of filter paper.

Paper chromatography:

A colored mixture with different solubility can be separated by chromatography.

Follow the steps:

  • A small step of ink is placed in the center of a filter paper.
  • When the drop has dried up another drop is added in exactly the same spot and this is allowed to dry.
  • Ethanol which is a solvent is slowly added drop by drop onto the spot
  • The addition of ethanol causes the spot of ink to slowly spread out into different coloured rings.

Ascending chromatography:

To separate the dyes in a mixture using ascending chromatography, we can set up the apparatus as shown below:


1. Draw a pencil line near the end of a chromatography paper.

2. Hang the paper inside a glass tank containing a solvent.

3. The paper will absorb the solvent and will rise up along with colors.

4. Those with higher solubility will spread further than others. Thus the colors will be separated.


1.8 explain how information from chromatograms can be used to identify the composition of a mixture.


Chromatography paper is taken. A sample(unknown) is placed along the pencil line. Other known elements are placed

side by the sample in the same line. The sheet of chromatography is coiled into a cylinder and secured with adhesive. It is put in a beaker containing a suitable solvent. The following chromatogram is obtained.

Identical dyes produce spots at the same height in the same color when the same solvent is used. In the diagram above the sample contains linileic acid.

c) Atomic Structure


1.9 understand that atoms consist of a central nucleus, composed of protons and neutrons, surrounded by electrons, orbiting in shells



Atoms are the smallest particle of an element which consists of a nucleus and electron shell. The nucleus is composed of protons and neutrons and the electrons orbit round the nucleus in different shells. A model diagram for an atom is shown below:

Electron Shells: The electrons move round the nucleus in different energy levels. This energy levels are called electron shells. In first shell two electron can stay. Second shell can hold upto eight electrons. Third shell can hold upto 8 electrons.


1.10 recall the relative mass and relative charge of a proton, neutron and electron


Substance particle

Relative mass

Relative charge


Formed in:


1 1840



Electron shells












1.11 understand the terms atomic number, mass number, isotopes and relative atomic mass (Ar)


Atomic number : the number of protons in an atom.

Mass number: The total number of protons and neutrons.

Isotopes : Isotopes are the atoms of the same element with different neutron number.

Relative Atomic Mass (Ar) : Relative atomic mass is the weighted average mass of the isotope of the element. It is measured by comparing with the mass of 1 12th of carbon-12.


1.12 calculate the relative atomic mass of an element from the relative abundances of its isotopes


To calculate it use the formula:

Ar = (% x mass number of the isotope) + (% x mass number of the second isotope) and so on...

The natural abundance for chlorine isotopes: 3517Cl is 75% and 3717Cl is 25%


Ar = {(75 x 35) 100} + {(25 x 37) 100}

= 26.5 +9.25

= 35.5

The natural abundance of boron isotopes: 115B is 80% and 105B is 20%

Ar = {(80 x 11) 100} + {(20 x 10) 100}

= 8.8+2

= 10.8


1.13 understand that the Periodic Table is an arrangement of elements in order of atomic number



Periodic Table is an arrangement of elements in order of atomic numbers. It consists of 8 groups and 7 periods.


1.14 deduce the electronic configurations of the first 20 elements from their positions in the Periodic Table


Electronic configuration can be easily figured out by following few rules:

  • 1st shell must contain only two electrons
  • 2nd shell must contain only 8 electrons
  • 3rd shell must contain only 8 electrons
  • 4th shell must contain only 18 electrons

· Atomic number is equal to electrons in an “atom”


Na(Sodium) : 2,8,1

Ca(Calcium): 2,8,8,2


1.15 deduce the number of outer electrons in a main group element from its position in the Periodic Table.


The group number determines outer electron of an element.


Group I elements have 1 electrons in their outer shell

Group 7 elements have 7

Group 0 have 8(since 8 comes after 7)


d) Relative formula masses and molar volumes of gases


1.16 calculate relative formula masses (Mr) from relative atomic masses (Ar)


The total mass of the molecule calculated from the relative atomic masses of the atoms present in the compound is called relative formula mass.



=(1 x 2) + 16 =18


=(22 + 1 x 4) x 2 + 12 + (16 x 2) =96


1.17 understand the use of the term mole to represent the amount of substance


A mole is a certain amount of substance. It is a general term to describe an amount of atoms, ions, or molecules.


1.18 understand the term mole as the Avogadro number of particles (atoms, molecules, formulae, ions or electrons) in a substance


A mole is defined as the amount of substance with contains the Avogadro number of particles.

Avogadro Number = 6 x 1023

1.19 carry out mole calculations using relative atomic mass (Ar) and relative formula mass (Mr)

mole = mass / Mr or Ar

Mole = Volume / 24000cm 3

Mole = Number of Particles 6 x 10 23

Mole = Volume x Concentration


1.20 understand the term molar volume of a gas and use its values (24 dm3 and 24,000 cm3) at room temperature and pressure (rtp) in calculations.


Volume of one mole of any gas is molar volume. It is 24dm3 or 24000 cm3 at R.T.P.

e) Chemical formulae and chemical equations


1.21 write word equations and balanced chemical equations to represent the reactions studied in this specification


A chemical equation which is expressed by using word is called word equation.

Eg: Carbon + oxygen => carbon dioxide

A chemical equation which is expressed by using symbols is called chemical equation.

Eg: C(s) + O2 (g ) => CO2(g)


1.22 use the state symbols (s), (l), (g) and (aq) in chemical equations to represent solids, liquids, gases and aqueous solutions respectively


  • (s) for solids
  • (l) for liquids
  • (g) for gases
  • (aq) for aqueous solution


1.23 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation


Finding the n in BaCl2.nH2O

Mass of crucible = 30.00g Mass of crucible + barium chloride crystals, BaCl2.nH2O

= 32.44 g Mass of crucible + anhydrous barium chloride, BaCl2

= 32.08 g


Mass of Bacl2

= 32.08 - 30.00


= 208 g


Mass of water

= 32.44 - 32.08 =


0.36 g


Bacl 2


H 2 O


Combining masses (g)






Number of moles


2.08 / 208


0.36 / 18








Ratio of moles




Simplest (empirical) formula





1.24 calculate empirical and molecular formulae from experimental data







Combining masses(g)





Number of moles in atoms

85.7 / 12


14.3 / 1







Ratio of moles




Simplest (empirical) formula




Converting empirical formulae into molecular formulae

Relative formula mass of the compound was 56. (56 / 14 = 4)

So you need four lots of CH2 The molecular formula is C4H8.


1.25 calculate reacting masses using experimental data and chemical equations


Calculate the mass of magnesium oxide that can be made by completely burning 6g of magnesium in oxygen.

Equation for reaction: 2Mg + O2 => 2MgO

1. Calculate the amount, in moles, of magnesium reacted

Ar of Mg is 24

Amount of magnesium = (6 / 24) = 0.25mol

2. Calculate the amount of magnesium oxide formed

The equation tells us that 2 mol of Mg for 2 mol of MgO, hence the amount of MgO formed is the same as the amount of

Mg reacted.

Amount of MgO formed is 0.25 mol

3. Calculate the mass of MgO formed

Mr of MgO = (24 + 16) = 40

Mass of magnesium oxide = (0.25 x 40) = 10g


1.26 calculate percentage yield


The actual yield is the amount of product that's actually there to be used at the end of the manufacturing process. The predicted yield is the amount that might have been expected if nothing had got lost along the way.

In practice, some product will be lost during the process when purifying the product by filtration or evaporation or when transferring a liquid or when heating.

The percentage yield is a way of comparing the actual yield with the predicted yield. It's calculated using a formula:

Percentage yield = (actual yield x 100%) / predicted yield


1.27 carry out mole calculations using volumes and molar concentrations.


Mole = Volume x Concentration

f) Ionic Compounds


1.28 describe the formation of ions by the gain or loss of electrons


Ions are charged particles. Atoms become charged by losing or gaining electrons. Atoms that lose electron become positively charged and atoms that gain electrons become negatively charged.


1.29 understand oxidation as the loss of electrons and reduction as the gain of electrons


Oxidation means lose of electrons. If an atom loses electron, it is oxidized. Reduction means gain of electrons. If an atom gain electron, it is reduced.


1.30 recall the charges of common ions in this specification



There are only two types of charges:


  • Positive ion eg: Zn2+, Ag+, H+, NH4+
  • Negative ion eg: NO3+, OH-, HCO3-


1.31 deduce the charge of an ion from the electronic configuration of the atom from which the ion is formed


Here are the steps:

1. Find the number of electrons in the outer electron shell.

2. Find out if it is easy for the atom to gain electron or to donate electron. (in most cases atoms that have below four electrons, donate electrons and atoms that have above 4 electrons, receive electrons)

3. Atoms that gain electron becomes negative ion and atoms that donate electron forms positive ion.


1.32 explain, using dot and cross diagrams, the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7



1.33 understand ionic bonding as a strong electrostatic attraction between oppositely charged ions


When a metal metals donates electron, a non-metal non-metals receives it. The metal becomes positively charged and non-metal becomes negatively charged. As positive and negative charge attract each other, they form a electrostatic force between them. That’s how they form ionic compound.


1.34 understand that ionic compounds have high melting and boiling points because of strong electrostatic forces between oppositely charged ions


Melting and boiling point depends on the force that holds the particles. In case of ionic compounds, electrostatic force is very strong. So many strong forces are needed to break at once. That’s and why they have high melting and boiling point.


1.35 understand the relationship between ionic charge and the melting point and boiling point of an ionic compound


Ionic charge is directly proportional to the melting and boiling point in an ionic compound.


1.36 describe an ionic crystal as a giant three-dimensional lattice structure held together by the attraction between oppositely charged ions



1.37 draw a diagram to represent the positions of the ions in a crystal of sodium chloride.


g) Covalent Compounds


1.38 describe the formation of a covalent bond by the sharing of a pair of electrons between two atoms


When two or more non-metal share electrons to fulfill their outer electron shell, they form a bond called covalent bond.


1.39 understand covalent bonding as a strong attraction between the bonding pair of electrons and the nuclei of the atoms involved in the bond


When non-metals share electrons, they are attracted by the nucleus. Since nucleus is positively charged it attracts negatively charged electrons.


1.40 explain, using dot and cross diagrams, the formation of covalent compounds by electron sharing for the following substances:


1. Hydrogen

2. chlorine

3. hydrogen chloride

4. water

5. methane

6. ammonia

7. oxygen

8. nitrogen

9. carbon dioxide

10. ethane

11. ethene


1.41 understand that substances with simple molecular structures are gases or liquids, or solids with low melting points


In covalent bonds, the intermolecular forces are very weak. Because molecules are electrically neutral unlike ions. They have charges in proton and neutron which are very weak.


1.42 explain why substances with simple molecular structures have low melting and boiling points in terms of the relatively weak forces between the molecules


Simple molecules have low melting and boiling points because melting and boiling points depends on the electrostatic force. Since in covalent compounds no ions are formed, the electrostatic charge is very weak and they have low melting and boiling points.


1.43 explain the high melting and boiling points of substances with giant covalent structures in terms of the breaking of many strong covalent bonds


Giant covalent bond have strong intermolecular attraction. It need a lot of energy to separate them that is they have high melting and boiling point.


1.44 draw diagrams representing the positions of the atoms in diamond and graphite


(a)Diamond: Diamond is one of the allotropes of carbon. Here each carbon atom is joined to four other carbon atoms by strong covalent bonds.

(b) Graphite: Graphite is another allotropes of carbon. Within each layer, each carbon atom forms strong covalent bonds with three other carbon atoms. This forms of six carbon atoms that are joined to form two-dimensional flat layers.

Therefore, each layer is a giant molecule. The layers of carbon atoms, which on top of each other.


1.45 explain how the uses of diamond and graphite depend on their structures, limited to graphite as a lubricant and diamond in cutting.






  • Hard
  • Soft
  • Very high melting and boiling points
  • Very high melting and boiling points
  • Non-conductor of electricity
  • Conductor of electricity


  • As gemstones
  • In pencils
  • As tips of cutting, grinding and
  • As a dry lubricant
  • polishing tools
  • As inert electrodes

h) Metallic Crystals


1.46 understand that a metal can be described as a giant structure of positive ions surrounded by a sea of delocalised electrons


Metal atoms are held strongly to each other by metallic bonding. In the metal lattice, the atoms lose their valence electrons and become positively charged. The valence electrons no longer belong to any metal atom and are said to be delocalised. They move freely between the metal ions like sea of electrons. Hence, this lattice structure is described as a

lattice of positive ions surroundedWecanthereforedefinebyametallic“seabondastheofforce mob of attraction between positive metal ions and sea of delocalised electrons.


1.47 explain the electrical conductivity and malleability of a metal in terms of its structure and bonding.


Electrical conductivity : Metals are good at conductors because of the free electron diffusion. When a metal is used in an electrical circuit, electrons entering one end of the metal cause another similar electron to displace from the other end. That’s why they are very good conductors.

Malleability and ductility: In metallic bonding, the valence electrons do not belong to any particular metal atom. If sufficient force is applied to the metal, one layer of atom can slide over another without disrupting the metallic bonding. As a result metallic bonds are strong but flexible. Therefore, they can be hammered into different shapes without breaking.

i) Electrolysis


1.48 understand that an electric current is a flow of electrons or ions


Electric current is the flow of electrons.


1.49 understand why covalent compounds do not conduct electricity


In covalent compounds there are no free electrons to move. That's why they don't conduct electricity.


1.50 understand why ionic compounds conduct electricity only when molten or in solution


Ionic compounds normally don't have free moving electrons. But when they are molten or dissolved in solution they form ions, which allow them to conduct electricity.


1.51 describe experiments to distinguish between electrolytes and non- electrolytes


An electrolyte is a substance that conducts electricity.


Take a led bulb and attach the wire at the end of the substance. Now supply electricity from the other end. If the led bulb glows, that substance is an electrolyte or else its non-electrolyte.


1.52 understand that electrolysis involves the formation of new substances when ionic compounds conduct electricity


Electrolysis is the process of using electricity to break down or decompose a compound. The compound is usually dissolved in water or molten.


1.53 describe experiments to investigate electrolysis, using inert electrodes, of molten salts such as lead(II) bromide and predict the products


Lead bromide is heated until it becomes molten. Then it forms two ions: Pb2+ & Br-.

Electricity is provided. In the anode bubble of bromine gas is formed. Bromine ion loosed one electron and became bromine gas.

Br- ==> Br2 + e-

In the cathode lead received electrons and formed lead metal.

Pb 2+ + 2e - ==> Pb


1.54 describe experiments to investigate electrolysis, using inert electrodes, of aqueous solutions such as sodium chloride, copper(II) sulfate and dilute sulfuric acid and predict the products


Electrolysis of Sodium Chloride solution:

In an aqueous solution of sodium chloride contains four different types of ions. They are:

  • Ions from sodium chloride –Na+(aq) and Cl- (aq)
  • Ions from water –H+(aq) and OH- (aq)

When dilute sodium chloride solution is electrolysed using inert electrodes, the Na+ and H+ ions are attracted to the cathode. The Cl- and OH- ions are attracted to the anode.

At the cathode : The Na+ and H+ are attracted to the platinium cathode. H+ ions gain electrons from the cathode to form hydrogen gas.

2 H+(aq) + 2e- ==> H2(g)

Na+ remains in the solution.

At the anode: Cl- and OH- ions are attracted to the platinum anode. Cl- ions give up electrons to the anode to form chlorine gas.

Cl- (aq) ==> Cl2(g) + 2e-

OH- remains in the solution.

Electrolysis of Aqueous Copper (II) Sulphate Using Inert Electrodes

During electrolysis, the cathode is coated with a layer of reddish-brown solid copper. The blue colour of the solution fades gradually as more copper is deposited. The resulting electrolyte also becomes increasingly acidic.

An aqueous solution of copper (II) sulphate contains four types of ions:

Ions from copper(II) sulphate - Cu2+ and SO42-

Ions from water - H+ and OH-

At the anode: OH- ions and SO42- ions are attracted to the anode. OH- ions give up electrons more readily than SO42- ions.

4OH-(aq) ==> 2H2O(l) + O2(g) + 4e-

The SO42- ions remain in solution.

At the cathode: H+ ions and Cu2+ ions are attracted to the cathode. Copper is lower than hydrogen in the reactivity series. Cu2+ ions accept electrons more readily than H+ ions. As a result, Cu2+ ions are preferentially discharged as copper metal (atoms).

Cu2+(aq) + 2e- ==> Cu(s)

The H+ ions remain in the solution.

Electrolysis of dilute sulphuric acid

During electrolysis, the cathode is coated with a layer of reddish-brown solid copper. The blue colour of the solution fades gradually as more copper is deposited. The resulting electrolyte also becomes increasingly acidic.

An aqueous solution of copper (II) sulphate contains four types of ions:

Ions from copper (II) sulphate: Cu2+ and SO42-

Ions from water: H+ and OH-

At the anode:

OH- ions and SO42- ions are attracted to the anode. OH- ions give up electrons more readily than SO 42- ions. Consequently, OH- ions are preferentially discharged to give oxygen gas.

4OH - (aq) ==> 2H 2 O (l) + O 2 (g) + 4e -

The SO42 ions remain in solution.

At the cathode:

H+ ions and Cu2+ ions are attracted to the cathode. Copper is lower than hydrogen in the reactivity series. Cu2+ ions accept electrons more readily than H+ ions. As a result, Cu2+ ions are preferentially discharged as copper metal (atoms).

Cu 2+ (aq) + 2e - ==> Cu (s)

The H+ ions remain in solution.


1.55 write ionic half-equations representing the reactions at the electrodes during electrolysis



See 1.54



1.56 recall that one faraday represents one mole of electrons


One Faraday is 96500 coulombs. That is the amount of coulombs in one mole of electrons.


1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions.


What mass of copper is deposited on the cathode during the electrolysis of copper(II) sulphate solution if 0.15 amps flows for 10 minutes?

The electrode equation is: Cu2+(aq) +2e- Cu(s)

(RAM: Cu = 64. 1 faraday = 96000 coulombs.)

Number of coulombs = amps x time in seconds = 0.15 x 10 x 60 = 90

Now working from the equation:

2 moles of electrons give 1 mole of copper, Cu 2 x 96000 coulombs give 64 g of copper

192,000 coulombs give 64 g of copper

90 coulombs give 90 / 192000 x 64 = 0.03