SCHEME OF WORK
WEEK TOPIC
1 Introduction to Chemistry
2 Particulate Nature of Matter
3 Particulate Nature of Matter
4 Symbols, Formulae and Equations
5 Symbols, Formulae and Equations
6 Chemical Combination
7 Chemical Combination
8 Chemical industries
9/10 Revision.
REFERENCE. TEXTS:
1. Comprehensive certificate chemistry for senior secondary schools by G N C Ohia.et al
2. Chemistry for Senior Secondary Schools 1 by Magbagbeola O, et al; Melrose Books and Publishers.
3. New school chemistry for senior secondary schools by Osei yaw Ababio
4. Revised edition understanding chemistry for schools and colleges by Godwin O. Ojokuku.
1ST TERM
WEEK 1
TOPIC: Introduction to Chemistry
CONTENT: 1. Meaning of chemistry
2. Career prospects tied to chemistry
3. Application (i) Hospital (ii) Military (iii) Teaching
(iv) Chemical and petrochemical industries.
4. Adverse effects of chemicals, drug abuse, poisoning,
Corrosion and pollution.
5. Scientific methods.
SUB-TOPIC 1: MEANING OF CHEMISTRY
Chemistry is one of the three main branches of pure science, the other two being physics and biology. Chemistry which probes into the principles governing the changes that matter undergoes also deals with the composition, properties uses of matter. Some of the chemical changes which matter undergoes include; lighting a match, cooking, burning fire wood,making palm wine, rusting of nails, rotting of leaves. Chemical changes are otherwise known as chemical reactions. The knowledge of chemistry helps us to subject some matter to chemical processes thereby producing some materials for our every day today use. Such materials include: soaps, detergents, hair cream, perfumes, oil, margarine and plastics among others.
Career prospects tied to chemistry
Career prospects tied to chemistry simply mean the job opportunities that are available for the students with knowledge of chemistry. Such students can be employed with private and public sectors which includes: Teaching service, health service, food processing, petroleum and petrochemical industries, manufacturing industry, extractive industry, Agriculture and Forestry.
(i) Teaching service: Concerns those who teach in primary, secondary schools, colleges of education and universities and even the laboratory assistants in schools and universities.
(ii) Health service: Involves pharmacists, biochemists, chemists, nutritionists, dieticians, doctors, nurses, medical assistants, laboratory assistants and dispensers.
(iii) Food processing: Food processing involves food technologists and research chemists.
(iv) Petroleum and petrochemical industries –Involves application of the following people; chemists research chemists, chemical engineers and laboratory assistants.
(v) Extractive industry- Involves chemists, mining engineers and geologists.
(vi) Manufacturing Industry: This involves research chemists and chemical engineers in the wide variety of manufacturing industries such as iron and steel works and cement factories.
(vii) Agriculture-Involves agricultural scientists, chemists, biochemists and physiologists engaged in research to improve the quality and yield of crops and livestock, and to advise farmers.
(viii) Forestry: Scientists engaged in research to preserve and improve forests and forestry products.
EVALUATION
1. Define the term chemistry.
2. Mention five changes that matter undergoes.
3. Give the uses of chemistry in our day to day life.
4. List at least five career opportunities in chemistry
5. Explain any three of the career opportunities mentioned above.
SUB TOPIC 2: APPLICATION/ USES OF CHEMISTRY
The knowledge of chemistry can be apply in the following areas; namely
1. Hospital-The knowledge of chemistry makes it possible for people to involve in chemical research and technology which lead to production of medicine that we use today.
2. Military: The duty of the military is defense, to defend the territorial integrity of a nation or state. Military cannot effectively do this without ammunition. Chemistry contributes to the discovery and description of the theoretical bases for the behavior of chemical substances such as explosives used by the military. The gun powder used in the earliest guns was made by mixing sulphur, charcoal and potassium trioxonitrates(v), compounded by early chemists. The manufacture of smokeless powder was based upon gun cotton, which is made from cotton fibers soaked in a strong mixture of HNO3 and H2SO4.
3. Teaching- chemistry teachers and lecturers in secondary schools, polytechnics, colleges of education and universities.
4. Chemical and petrochemical industries:Application chemists, research chemists, chemical engineers and laboratory assistants.
5. Space science: chemistry is not out in space exploration. In our efforts to gain more knowledge of the other planets and outer space around us, special rockets called ‘space rocket’ are sent into space. The first rocket was sent into space on October 4, 1957 by Russia. In July, 1969, Apollo II astronauts Neil Armstrong and Edwin Aldrin landed on the moon. These are made possible by science and technology.
6. AGRICULTURE: Agricultural scientist, chemists, biochemists and physiologists engaged in research to improve the quality and yield of crops and livestock, and to advise farmers.
EVALUATION
1. Enumerate and explain five application of chemistry you know.
SUB- TOPIC 3- ADVERSE EFFECTS OF CHEMICALS AND SCIENTIFIC METHOD.
The adverse effects of existence of chemistry
The existence of chemistry brought about the existence of chemicals. The adverse effects of chemicals include; drug abuse, poisoning, corrosion and pollution.
(a) DRUG ABUSE-simply involves wrong usage of drugs. Some of these drugs include heroin, cocaine and morphine which are used as addictive. Unscrupulous people produce and sell them at huge profits. Drug addiction is a major problem in our society, especially among young people. Many countries have imposed strict laws to control pollution and drug abuse. However, the most effective control measure is education. We must use what we learnt to improve our life and to control these abuses.
(b) POISONING: This is where chemicals are used to poison the food we eat. This happens when the chemicals used as addictive probably as preservative are added more than required or expired in the food stuff where it was added, then instead of the food stuff bringing health to our body, it turns to poison.
(c) CORROSION: Corrosion of iron can also be called rusting and requires the presence of water and oxygen. Rusting can also be regarded as the slow deterioration of iron to iron (iii) oxide. This iron (iii) oxide is permeable to both air and water and cannot protect the iron from further corrosion of iron is given below.
Fe(s) Fe2+ (aq) + 2e- (1)
O2(g) + 2H2O(L) + 4e 4OH-(aq) (2)
Fe2+(aq) + 2OH-(aq) Fe(OH)2(s) (3)
This rusting can be prevented by four methods.
(i) Application of protective coating.
(ii) Application of sacrificial metal.
(iii) Alloying.
(iv) Cathodic protection.
(d) POLLUTION: Chemical industries through the action of production pollute our environment as the smoke enters into the air, and dirts of different kinds enter into the water thereby polluting the entire environment. Specifically chemical wastes from factories and oil refineries and radioactive wastes from nuclear plants pollute our environment. Oil spillage, exhaust from motor vehicles, pesticides, fertilizers and acid rain have made our environment unclean and endangered plant and animal life. Human health is also being threatened by environmental pollution. Presently, chemists are trying to come up with a fuel that will reduce the air pollution problem. They are also modifying chemical processes to recycle chemical wastes or change them to harmless products which can be safely discharged into the surrounding.
SCIENTIFIC METHOD
This is the method the scientist used to produce different materials that exists as a result of chemistry. In the light of this, the scientists use their senses to observe what is happening around them. From a given set of observations, they see a certain pattern. This often leads to a problem which they try to solve. They put forward a reasonable explanation or hypothesis and carry out appropriate experiments to test it. Then, they carefully record their observations and the results of their experiments.
If the experiments support the hypothesis, they carry out further investigations. They discuss the hypothesis and results with other scientists in the field so that the hypothesis can be further tested. When a hypothesis has been tested and found to be correct within the limits of available evidence. It becomes a theory. A scientific law or principle is established only after the theory has been extensively tested and proven true without any exception. If the experiments give negative results, then the scientist goes back to his hypothesis and either modifies it or puts forward a new hypothesis. This way of studying a problem is known as the scientific method. It is the very foundation of all scientific discoveries.
EVALUATION
1. Mention five adverse effects of existence of chemistry in the world.
2. What do you understand by hypothesis?
3. Differentiate the terms hypothesis and theory
4. Explain fully what you understand by scientific method?
GENERAL EVALUATION
OBJECTIVE TEST:
1. Chemistry is defined as
A. a branch of knowledge which produces chemicals
B. the branch of science which deals with changes in matter
C. the oldest branch science.
D. the branch of science which makes physics and biology clearer.
2. Scientific approach to discoveries follows the order which includes
A: Observation, hypothesis, and results
B: experiments, hypothesis and results
C.further experiments and problems solving.
D. theory, negative and positive results and experiments.
3. Chemical hypothesis is different from chemical law in that
A: hypothesis is a reasonable explanation to observations made while law is a statement from a scientist.
B: hypothesis is a reasonable explanation to observations while law is a statement which confirms the hypothesis after extensive tests.
C.hypothesis is not reasonable while law is reasonable.
D. none of the above.
4. Chemical changes around us includes all except;
A: rusting of iron nails
B: sieving
C: fading of coloured cloth
D. decomposing of green leaves in a compost
5. One of these professions has no need for chemistry
A: Miners
B: Engineers
C: Philosophers
D: Geologists
ESSAY QUESTIONS
1. Give five reasons why chemistry is important in your life.
2. Explain in detail two of the reasons given above.
3. List three adverse effect of existence of chemistry in this world.
4. Explain two of those adverse effects mentioned above.
5. Explain the term scientific method full.
WEEKEND ASSIGNMENT:
1.New school chemistry for senior secondary schools by Osei Yaw Ababio revised by L.E.S. Akpanisi Herbert Igwe. Pages 2-5 .
PRE- READING ASSIGNMENT
Read about physical and chemical changes and differentiate them.
WEEKEND ACTIVITY
Explain fully what you understand by the following terms.
Atoms, Molecules, Constituents of atoms.
REFERENCE:
1. New school chemistry for senior secondary schools by Osei Yaw Ababio revised by L.E.S. Akpanisi Herbert Igwe.
CONTENT: 1. Meaning of chemistry
2. Career prospects tied to chemistry
3. Application (i) Hospital (ii) Military (iii) Teaching
(iv) Chemical and petrochemical industries.
4. Adverse effects of chemicals, drug abuse, poisoning,
Corrosion and pollution.
5. Scientific methods.
SUB-TOPIC 1: MEANING OF CHEMISTRY
Chemistry is one of the three main branches of pure science, the other two being physics and biology. Chemistry which probes into the principles governing the changes that matter undergoes also deals with the composition, properties uses of matter. Some of the chemical changes which matter undergoes include; lighting a match, cooking, burning fire wood,making palm wine, rusting of nails, rotting of leaves. Chemical changes are otherwise known as chemical reactions. The knowledge of chemistry helps us to subject some matter to chemical processes thereby producing some materials for our every day today use. Such materials include: soaps, detergents, hair cream, perfumes, oil, margarine and plastics among others.
Career prospects tied to chemistry
Career prospects tied to chemistry simply mean the job opportunities that are available for the students with knowledge of chemistry. Such students can be employed with private and public sectors which includes: Teaching service, health service, food processing, petroleum and petrochemical industries, manufacturing industry, extractive industry, Agriculture and Forestry.
(i) Teaching service: Concerns those who teach in primary, secondary schools, colleges of education and universities and even the laboratory assistants in schools and universities.
(ii) Health service: Involves pharmacists, biochemists, chemists, nutritionists, dieticians, doctors, nurses, medical assistants, laboratory assistants and dispensers.
(iii) Food processing: Food processing involves food technologists and research chemists.
(iv) Petroleum and petrochemical industries –Involves application of the following people; chemists research chemists, chemical engineers and laboratory assistants.
(v) Extractive industry- Involves chemists, mining engineers and geologists.
(vi) Manufacturing Industry: This involves research chemists and chemical engineers in the wide variety of manufacturing industries such as iron and steel works and cement factories.
(vii) Agriculture-Involves agricultural scientists, chemists, biochemists and physiologists engaged in research to improve the quality and yield of crops and livestock, and to advise farmers.
(viii) Forestry: Scientists engaged in research to preserve and improve forests and forestry products.
EVALUATION
1. Define the term chemistry.
2. Mention five changes that matter undergoes.
3. Give the uses of chemistry in our day to day life.
4. List at least five career opportunities in chemistry
5. Explain any three of the career opportunities mentioned above.
SUB TOPIC 2: APPLICATION/ USES OF CHEMISTRY
The knowledge of chemistry can be apply in the following areas; namely
1. Hospital-The knowledge of chemistry makes it possible for people to involve in chemical research and technology which lead to production of medicine that we use today.
2. Military: The duty of the military is defense, to defend the territorial integrity of a nation or state. Military cannot effectively do this without ammunition. Chemistry contributes to the discovery and description of the theoretical bases for the behavior of chemical substances such as explosives used by the military. The gun powder used in the earliest guns was made by mixing sulphur, charcoal and potassium trioxonitrates(v), compounded by early chemists. The manufacture of smokeless powder was based upon gun cotton, which is made from cotton fibers soaked in a strong mixture of HNO3 and H2SO4.
3. Teaching- chemistry teachers and lecturers in secondary schools, polytechnics, colleges of education and universities.
4. Chemical and petrochemical industries:Application chemists, research chemists, chemical engineers and laboratory assistants.
5. Space science: chemistry is not out in space exploration. In our efforts to gain more knowledge of the other planets and outer space around us, special rockets called ‘space rocket’ are sent into space. The first rocket was sent into space on October 4, 1957 by Russia. In July, 1969, Apollo II astronauts Neil Armstrong and Edwin Aldrin landed on the moon. These are made possible by science and technology.
6. AGRICULTURE: Agricultural scientist, chemists, biochemists and physiologists engaged in research to improve the quality and yield of crops and livestock, and to advise farmers.
EVALUATION
1. Enumerate and explain five application of chemistry you know.
SUB- TOPIC 3- ADVERSE EFFECTS OF CHEMICALS AND SCIENTIFIC METHOD.
The adverse effects of existence of chemistry
The existence of chemistry brought about the existence of chemicals. The adverse effects of chemicals include; drug abuse, poisoning, corrosion and pollution.
(a) DRUG ABUSE-simply involves wrong usage of drugs. Some of these drugs include heroin, cocaine and morphine which are used as addictive. Unscrupulous people produce and sell them at huge profits. Drug addiction is a major problem in our society, especially among young people. Many countries have imposed strict laws to control pollution and drug abuse. However, the most effective control measure is education. We must use what we learnt to improve our life and to control these abuses.
(b) POISONING: This is where chemicals are used to poison the food we eat. This happens when the chemicals used as addictive probably as preservative are added more than required or expired in the food stuff where it was added, then instead of the food stuff bringing health to our body, it turns to poison.
(c) CORROSION: Corrosion of iron can also be called rusting and requires the presence of water and oxygen. Rusting can also be regarded as the slow deterioration of iron to iron (iii) oxide. This iron (iii) oxide is permeable to both air and water and cannot protect the iron from further corrosion of iron is given below.
Fe(s) Fe2+ (aq) + 2e- (1)
O2(g) + 2H2O(L) + 4e 4OH-(aq) (2)
Fe2+(aq) + 2OH-(aq) Fe(OH)2(s) (3)
This rusting can be prevented by four methods.
(i) Application of protective coating.
(ii) Application of sacrificial metal.
(iii) Alloying.
(iv) Cathodic protection.
(d) POLLUTION: Chemical industries through the action of production pollute our environment as the smoke enters into the air, and dirts of different kinds enter into the water thereby polluting the entire environment. Specifically chemical wastes from factories and oil refineries and radioactive wastes from nuclear plants pollute our environment. Oil spillage, exhaust from motor vehicles, pesticides, fertilizers and acid rain have made our environment unclean and endangered plant and animal life. Human health is also being threatened by environmental pollution. Presently, chemists are trying to come up with a fuel that will reduce the air pollution problem. They are also modifying chemical processes to recycle chemical wastes or change them to harmless products which can be safely discharged into the surrounding.
SCIENTIFIC METHOD
This is the method the scientist used to produce different materials that exists as a result of chemistry. In the light of this, the scientists use their senses to observe what is happening around them. From a given set of observations, they see a certain pattern. This often leads to a problem which they try to solve. They put forward a reasonable explanation or hypothesis and carry out appropriate experiments to test it. Then, they carefully record their observations and the results of their experiments.
If the experiments support the hypothesis, they carry out further investigations. They discuss the hypothesis and results with other scientists in the field so that the hypothesis can be further tested. When a hypothesis has been tested and found to be correct within the limits of available evidence. It becomes a theory. A scientific law or principle is established only after the theory has been extensively tested and proven true without any exception. If the experiments give negative results, then the scientist goes back to his hypothesis and either modifies it or puts forward a new hypothesis. This way of studying a problem is known as the scientific method. It is the very foundation of all scientific discoveries.
EVALUATION
1. Mention five adverse effects of existence of chemistry in the world.
2. What do you understand by hypothesis?
3. Differentiate the terms hypothesis and theory
4. Explain fully what you understand by scientific method?
GENERAL EVALUATION
OBJECTIVE TEST:
1. Chemistry is defined as
A. a branch of knowledge which produces chemicals
B. the branch of science which deals with changes in matter
C. the oldest branch science.
D. the branch of science which makes physics and biology clearer.
2. Scientific approach to discoveries follows the order which includes
A: Observation, hypothesis, and results
B: experiments, hypothesis and results
C.further experiments and problems solving.
D. theory, negative and positive results and experiments.
3. Chemical hypothesis is different from chemical law in that
A: hypothesis is a reasonable explanation to observations made while law is a statement from a scientist.
B: hypothesis is a reasonable explanation to observations while law is a statement which confirms the hypothesis after extensive tests.
C.hypothesis is not reasonable while law is reasonable.
D. none of the above.
4. Chemical changes around us includes all except;
A: rusting of iron nails
B: sieving
C: fading of coloured cloth
D. decomposing of green leaves in a compost
5. One of these professions has no need for chemistry
A: Miners
B: Engineers
C: Philosophers
D: Geologists
ESSAY QUESTIONS
1. Give five reasons why chemistry is important in your life.
2. Explain in detail two of the reasons given above.
3. List three adverse effect of existence of chemistry in this world.
4. Explain two of those adverse effects mentioned above.
5. Explain the term scientific method full.
WEEKEND ASSIGNMENT:
1.New school chemistry for senior secondary schools by Osei Yaw Ababio revised by L.E.S. Akpanisi Herbert Igwe. Pages 2-5 .
PRE- READING ASSIGNMENT
Read about physical and chemical changes and differentiate them.
WEEKEND ACTIVITY
Explain fully what you understand by the following terms.
Atoms, Molecules, Constituents of atoms.
REFERENCE:
1. New school chemistry for senior secondary schools by Osei Yaw Ababio revised by L.E.S. Akpanisi Herbert Igwe.
WEEK 2
Topic: PARTICULATE NATURE OF MATTER
SUB- TOPICS:
(a) Physical and chemical changes
(b) Atoms, molecules, ions and atomicity.
(c) Dalton’s atomic theory
(d) Constituents of atoms, protons, neutrons and electrons.
PHYSICAL AND CHEMICAL CHANGES
PHYSICAL CHANGES:
DEFINITION: A Physical change is one which is easily reversed and in which no new substances are formed.
Examples:
i. Melting of solids to liquids.
ii. Freezing of liquids to solids.
iii. Vaporization of liquids to gases.
iv. Liquefaction of gases to liquids.
v. Sublimation of solids to vapour.
CHEMICAL CHANGE:
DEFINITION: A chemical change is one which is not easily reversed andin which a new substance is formed.
Examples:
i. Burning of substances.
ii. Dissolution of metals and limestone in acids.
iii. Chemical decomposition – like digestion of food.
iv. Rusting of iron.
v. Charring of sugar.
vi. Dissolution of metals in acids.
DIFFERENCES BETWEEN PHYSICAL AND CHEMICAL CHANGE
PHYSICAL CHANGE CHEMICAL CHANGE
1. Easily reversible Not easily reversible
2. No new substance is formed New substance is formed
3. No change is mass. There is a noticeable change in mass.
4. Not accompanied by heat usually accompanied by heat change.
Change.
EVALUATION:
1. Mention three familiar process/changes which you know to be:
(a) Physical changes (b) chemical changes.
2. List three differences between physical change and chemical change.
3. Classify each of the following as physical change or chemical change.
(a) Boiling of egg (b) Burning of kerosene (c) Melting of wax (d) Rusting of iron(e) Digestion of glucose.(f) Dissolving iron in an acid.
4. State with a reason in each case whether each of the following is physical or chemical change.
(a) Dissolving common salt in water.
(b) Burning of petrol.
(c) Digestion of glucose.
ATOMS, MOLECULES AND IONS
DEFINITION: An atom is the smallest particle of an element which can take part in a chemical reaction.
The atom is the basic (fundamental) unit of elements and they are the units which are concerned in chemical reactions.
An atom of Symbol
Oxygen O
Chlorine Cl
Silver Ag
Carbon C
MOLECULES
DEFINITION: A molecule is the smallest particle of a substance that normally exist alone and still retain the chemical properties of that substance be it an element or a compound.
Some molecules can exist independently as single atoms e.g. He, Na, Ca, Mg etc.
Some molecules may be made up of atoms of the same element e.g. a molecule of hydrogen is H2, that of chlorine is Cl2, Oxygen is O2, phosphorus is P4, Sulphur is S8.
Some molecules may be made up of different elements e.g. a molecule of water is H20, Methane is CH4, ammonia is NH3, carbon (iv) oxide is CO2 etc.
ATOMICITY: Atomicity is the number of atoms in each molecule of an element.
ATOMICITY OF SOME MOLECULAR ELEMENT
Element Formula of molecule Atomicity
Hydrogen H2 2
Oxygen O2 2
Ozone O3 3
Phosphorus P4 4
Sulphur S8 8
Chlorine Cl2 2
Neon Ne 1
Argon Ar 1
IONS
DEFINITION:
An ion is any atom or group of atoms which possess an electric charge.
Some substances are not built of atoms or molecules but are made up of charged particles called ions.
There are two types of ions. The positively charged ions or cations e.g. k+, Ca2+, Zn2+, Al3+ etc. The negatively charged ions or anions e.g. Cl-, N3-, O2-, S2- etc. Ions made up of group of atoms are known as radicals e.g.co32-(trioxocarbonate (IV)),NO32-(trioxonitrate (v)) etc
EVALUATION
Define (a) an atom (b) a molecule (c) an ion. What is the relationship between an atom and a molecule?
DALTONS ATOMIC THEORY
In 1808 John Dalton proposed the following Atomic theory.
1. All elements are made up of small indivisible particles called atoms.
2. Atoms can neither be created or destroyed
3. Atoms of the same element are alike in every aspect, and differ from atoms of all other elements.
4. When atoms combine with other atoms, they do so in simple ratios.
5. All chemical changes result from the combination or the separation of atoms.
CONSTITUENTS OF ATOMS: PROTONS, NEUTRONS AND ELECTRONS
STRUCTURE OF AN ATOM
Electron
Nucleus
An atom is made up of the three sub-particles known as protons, electrons and neutrons as shown in the above diagram. Their characteristics are summarized in the table below:
CHARACTERISTICS OF THE SUB-PARTICLES
SUB-PARTICLE LOCATION RELATIVE CHARGE RELATIVE MASS SYMBOL
Proton Nucleus + 1 p
Electron Outside nucleus _ 0.005 e-
Neutron Nucleus Zero 1 n
EVALUATION:
1. What are the three fundamental units of all matter? Give their relative mass and charges.
2. Describe their relative positions to one another in an atom.
3. Name the two main part of an atom?
4. Mention the three subatomic particles in an atom.
5. Which particles are found in each part of the atom?
6. List three sub atomic particles with their corresponding charges.
GENERAL EVALUATION
OBJECTIVE TEST:
(1) The simplest unit of water that retains its properties is called. (a) an atom (b) an element (c) an hydroxide (c) a molecule
(2) Of the basic particles that make up an atom the one with the smallest mass is? (a) a proton (b) a neutron (c) an x- particle (d) an electron
(3) The following are physical changes except: (A) melting of candle wax (b) Dissolving common salt in water (c) Freezing of water (d) Rotting of leaves.
(4) Which of the following is not a molecule of the same element. (a) O2 (b) P4 (c) S8 (d)CO2
(5) The atomicity of O3 is (a) 1 (b) 2 (c) 3 (d )4
ESSAY QUESTIONS
1. Write four examples in each case of (a) Physical change (b) Chemical change
2. Define the following giving two examples in each case. (a) Atom (b) Ion
3. Write the symbols for proton, electron and neutrons.
4. Define atomicity giving two examples.
5. Write four Dalton’s atomic theory.
WEEKEND ASSIGNMENT:
1. Read about modifications of Dalton’s atomic theory.
2. Read the topic ‘’composition of the atom on page 32 of Comprehensive Certificate Chemistry.
WEEK ACTIVITY:
In a cardboard sheet draw a clearly labelled structure of an atom. From page 45 of New School Chemistry by Ababio.
PRE-READING ASSISNMENT:
Read the topic ‘’Arrangement of electrons around the nucleus’’ from page 33 of Comprehensive Certificate Chemistry by Ojokuku and others.
SUB- TOPICS:
(a) Physical and chemical changes
(b) Atoms, molecules, ions and atomicity.
(c) Dalton’s atomic theory
(d) Constituents of atoms, protons, neutrons and electrons.
PHYSICAL AND CHEMICAL CHANGES
PHYSICAL CHANGES:
DEFINITION: A Physical change is one which is easily reversed and in which no new substances are formed.
Examples:
i. Melting of solids to liquids.
ii. Freezing of liquids to solids.
iii. Vaporization of liquids to gases.
iv. Liquefaction of gases to liquids.
v. Sublimation of solids to vapour.
CHEMICAL CHANGE:
DEFINITION: A chemical change is one which is not easily reversed andin which a new substance is formed.
Examples:
i. Burning of substances.
ii. Dissolution of metals and limestone in acids.
iii. Chemical decomposition – like digestion of food.
iv. Rusting of iron.
v. Charring of sugar.
vi. Dissolution of metals in acids.
DIFFERENCES BETWEEN PHYSICAL AND CHEMICAL CHANGE
PHYSICAL CHANGE CHEMICAL CHANGE
1. Easily reversible Not easily reversible
2. No new substance is formed New substance is formed
3. No change is mass. There is a noticeable change in mass.
4. Not accompanied by heat usually accompanied by heat change.
Change.
EVALUATION:
1. Mention three familiar process/changes which you know to be:
(a) Physical changes (b) chemical changes.
2. List three differences between physical change and chemical change.
3. Classify each of the following as physical change or chemical change.
(a) Boiling of egg (b) Burning of kerosene (c) Melting of wax (d) Rusting of iron(e) Digestion of glucose.(f) Dissolving iron in an acid.
4. State with a reason in each case whether each of the following is physical or chemical change.
(a) Dissolving common salt in water.
(b) Burning of petrol.
(c) Digestion of glucose.
ATOMS, MOLECULES AND IONS
DEFINITION: An atom is the smallest particle of an element which can take part in a chemical reaction.
The atom is the basic (fundamental) unit of elements and they are the units which are concerned in chemical reactions.
An atom of Symbol
Oxygen O
Chlorine Cl
Silver Ag
Carbon C
MOLECULES
DEFINITION: A molecule is the smallest particle of a substance that normally exist alone and still retain the chemical properties of that substance be it an element or a compound.
Some molecules can exist independently as single atoms e.g. He, Na, Ca, Mg etc.
Some molecules may be made up of atoms of the same element e.g. a molecule of hydrogen is H2, that of chlorine is Cl2, Oxygen is O2, phosphorus is P4, Sulphur is S8.
Some molecules may be made up of different elements e.g. a molecule of water is H20, Methane is CH4, ammonia is NH3, carbon (iv) oxide is CO2 etc.
ATOMICITY: Atomicity is the number of atoms in each molecule of an element.
ATOMICITY OF SOME MOLECULAR ELEMENT
Element Formula of molecule Atomicity
Hydrogen H2 2
Oxygen O2 2
Ozone O3 3
Phosphorus P4 4
Sulphur S8 8
Chlorine Cl2 2
Neon Ne 1
Argon Ar 1
IONS
DEFINITION:
An ion is any atom or group of atoms which possess an electric charge.
Some substances are not built of atoms or molecules but are made up of charged particles called ions.
There are two types of ions. The positively charged ions or cations e.g. k+, Ca2+, Zn2+, Al3+ etc. The negatively charged ions or anions e.g. Cl-, N3-, O2-, S2- etc. Ions made up of group of atoms are known as radicals e.g.co32-(trioxocarbonate (IV)),NO32-(trioxonitrate (v)) etc
EVALUATION
Define (a) an atom (b) a molecule (c) an ion. What is the relationship between an atom and a molecule?
DALTONS ATOMIC THEORY
In 1808 John Dalton proposed the following Atomic theory.
1. All elements are made up of small indivisible particles called atoms.
2. Atoms can neither be created or destroyed
3. Atoms of the same element are alike in every aspect, and differ from atoms of all other elements.
4. When atoms combine with other atoms, they do so in simple ratios.
5. All chemical changes result from the combination or the separation of atoms.
CONSTITUENTS OF ATOMS: PROTONS, NEUTRONS AND ELECTRONS
STRUCTURE OF AN ATOM
Electron
Nucleus
An atom is made up of the three sub-particles known as protons, electrons and neutrons as shown in the above diagram. Their characteristics are summarized in the table below:
CHARACTERISTICS OF THE SUB-PARTICLES
SUB-PARTICLE LOCATION RELATIVE CHARGE RELATIVE MASS SYMBOL
Proton Nucleus + 1 p
Electron Outside nucleus _ 0.005 e-
Neutron Nucleus Zero 1 n
EVALUATION:
1. What are the three fundamental units of all matter? Give their relative mass and charges.
2. Describe their relative positions to one another in an atom.
3. Name the two main part of an atom?
4. Mention the three subatomic particles in an atom.
5. Which particles are found in each part of the atom?
6. List three sub atomic particles with their corresponding charges.
GENERAL EVALUATION
OBJECTIVE TEST:
(1) The simplest unit of water that retains its properties is called. (a) an atom (b) an element (c) an hydroxide (c) a molecule
(2) Of the basic particles that make up an atom the one with the smallest mass is? (a) a proton (b) a neutron (c) an x- particle (d) an electron
(3) The following are physical changes except: (A) melting of candle wax (b) Dissolving common salt in water (c) Freezing of water (d) Rotting of leaves.
(4) Which of the following is not a molecule of the same element. (a) O2 (b) P4 (c) S8 (d)CO2
(5) The atomicity of O3 is (a) 1 (b) 2 (c) 3 (d )4
ESSAY QUESTIONS
1. Write four examples in each case of (a) Physical change (b) Chemical change
2. Define the following giving two examples in each case. (a) Atom (b) Ion
3. Write the symbols for proton, electron and neutrons.
4. Define atomicity giving two examples.
5. Write four Dalton’s atomic theory.
WEEKEND ASSIGNMENT:
1. Read about modifications of Dalton’s atomic theory.
2. Read the topic ‘’composition of the atom on page 32 of Comprehensive Certificate Chemistry.
WEEK ACTIVITY:
In a cardboard sheet draw a clearly labelled structure of an atom. From page 45 of New School Chemistry by Ababio.
PRE-READING ASSISNMENT:
Read the topic ‘’Arrangement of electrons around the nucleus’’ from page 33 of Comprehensive Certificate Chemistry by Ojokuku and others.
WEEK 3
TOPICS: PARTICULATE NATURE OF MATTER
SUB-TOPICS:
1. Arrangement of electrons around the nucleus.
2. Atomic number, mass number.
3. Relative atomic mass based on c-12 isotope.
4. Isotopy.
ARRANGEMENT OF ELECTRONS AROUND THE NUCLEUS
Electrons are found revolving around the nucleus of an atom in circular paths known as rings, orbits, energy levels or shells. Each shell contains electrons with similar energy.Those with the lowest energies being nearest to the nucleus.
3 (M shell)
Nucleus
2(L shell)
1 (K shell)
Thus, the arrangement of electrons in the atom according to energy is called ELECTRONIC CONFIGURATION. Letters and figures are associated with these orbits or shells as shown above. The maximum possible number of electrons that can be accommodated in a shell is given by the formula:
Nmax= 2n2. Where Nmax = Maximum no of electron. n= no, of shell.
Thus K- shell can contain 2 × 12 = 2 electrons.
L- Shell can contain 2 × 22 = 8 electrons.
M- Shell can contain 2 × 32= 18 electrons etc
The electron structures of the atoms of the first twenty elements are given in the table below.
Element Symbol Number of protons(or atomic number) Number of electrons and their distribution in the shells. K L M N
Hydrogen H 1 1
Helium He 2 2
Lithium Li 3 2 1
Beryllium Be 4 2 2
Boron B 5 2 3
Carbon C 6 2 4
Nitrogen N 7 2 5
Oxygen O 8 2 6
Fluorine F 9 2 7
Neon Ne 10 2 8
Sodium Na 11 2 8 1
Magnesium Mg 12 2 8 2
Aluminium Al 13 2 8 3
Silicon Si 14 2 8 4
Phosphorus P 15 2 8 5
Sulphur S 16 2 8 6
Chlorine Cl 17 2 8 7
Argon Ar 18 2 8 8
Potassium Ic 19 2 8 8 1
Calcium Ca 20 2 8 8 2
The electronic configurations of some elements are shown below:
Hydrogen, H (atomic number 1)
Electronk1
Nucleus
Helium, He(atomic number 2)
electrons k2
nucleus
Lithium, Li (atomic number 3).
K2 L1
Neon, Ne (atomic number 10)
K2 L8
2,8
Sodium, Na (atomic number 11)
K L M
2 8 1
2, 8, 1
Argon, Ar (atomic number 18 )
K L M
2 8 8
2,8,8
Potassium, K (atomic number 19) K L M N
2 8 8 1
2, 8, 8, 1
Calcium, Ca (atomic number 20)
K L M N
2 8 8 2
2,8,8,2
EVALUATION: Draw the electronic configuration of the following elements.
Carbon (b) Fluorine (c) Aluminium (d) Nitrogen
ATOMIC NUMBER AND MASS NUMBER.
ATOMIC NUMBER:
DEFINITION: Atomic number is the number of protons in an atom of an element.
The atomic number of an element is a whole number and is designated z. In a neutral atom the number of protons must be equal to the number of electrons(since protons are positively charged and electron are negatively charged.)
All the atoms of a particular element have the same number of protons in their nuclei (i.e. they have the same atomic number). NO two elements have the same number of protons in their atoms.
DEFINITION:
MASS NUMBER: The mass number is the sum of the protons and neutrons in an atom of an element.
Mass number is represented by the letter A.
Mass number A = Number of protons + number of neutrons.
i.e. A =p + n. Where p=protons, n= neutrons.
Or number of neutrons n =A – Z.
An atom of an element can be described by writing its symbol together with its atomic number and mass number.
(Mass number) A
Symbol of an atom of the element.
(Atomic number)Z
Examples: The atom of carbon, oxygen and sodium can be written as 126C, 168O and2311Na respectively.
EVALUATION:
(1) Define the following.
Atomic number (b) Mass number.
(2) Describe the atoms of the following elements using their symbol, atomic number and mass number.
(a) Phosphorus (b) Silicon (c) Calcium
ISOTOPY
Definition: Isotopy is a phenomenon whereby atoms of an element exhibit different mass number but have the same atomic number.
Mass spectrometric studies show that the atoms of most elements exist in more than one form. This is due to the difference in number of neutrons present in these atoms. Such atoms are known as isotopes. Isotope of an element is represented by the original symbol of the element with the mass number and atomic numbers. For example 126C, 136C, 146C represent atoms of the isotopes of carbon. For each atom, the number of neutrons can be obtained by finding the difference between the mass number A and the atomic number Z i.e. A – Z. Each isotope of an element has its own mass known as isotopic mass.
Isotopes of an element have slightly different physical properties because neutrons contribute only to the mass of an atom and not its chemical behaviour. But isotopes of an element exhibit the same chemical properties because the number of valence electrons in an atom of an element determines its chemical behaviour (properties) and since isotopes have the same number of valence electrons they will be chemically alike.
NOTE:
(i) An analysis of the chlorine isotopes.
Isotope 3517Cl Isotope 3717Cl
Mass number, A 35 37
Atomic number,Z 17 17
Number of protons 17 17
Number of electrons 17 17
Number of neutrons (A-Z) 35 – 17=18 37 – 17=20
Abundance in nature (%) 75 25
ISOTOPES OF THE SAME ELEMENTS.
Element Carbon Oxygen
ISOTOPES 126C 136C 168O 178O 188O
ABUNDANCE IN NATURE (%) 98.9 1.1 99.76 0.04 0.20
(ii) The names of the isotopic forms of hydrogen
11H --- Protium (or hydrogen)
21H--- Deutrium (or heavy hydrogen or D)
31H--- Tritium or T
(iii) The relative atomic mass, RAM of an element which exhibits isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element and they are not usually in whole numbers.
CALCULATION INVOLVING ISOTOPY.
WORKED EXAMPLE:
Determine the relative atomic mass of element X from the data below
ISOTOPE MASS %ABUNDANCE
24X 24 78.70
25x 25 10.13
26x 26 11.7
((24 ×78.70)+ (25 ×10.13)+(26 × 11.17)= X)/100
(1888.8+253.25+290.42=X)/100
(2432.47=X)/100
∴X=24.3247
≅24
An element X has two isotopes of 2010X and 2210X in the ratio 1:3. What is the relative atomic mass?
Add ratio of occurrence together.
1 + 3= 4
((20 ×1)+ (22 ×3)=X)/4
(20+66=X)/4
( 86=X)/4
X = 21.5
Isotopes of an element X have isotopic masses 65 and 63 respectively. If the relative atomic mass of X is 63.60. Find the relative abundance of each isotope of the element.
Let the relative abundance of element X be y and Z respectively.
Z +Y =100
∴ Z = 100 – y....... (i)
((65 ×y)+ (63 ×y)=63.60)/100
65y + 63z = 63.60 × 100
65y + 63z = 6360 ......(ii)
65y +63(100 – y) = 6360
65y – 63y + 6300 = 6360
2y = 60
Y = 60/2 = 30
Y= 30
Z = 100 – y= 100 – 30 = 70
Y = 30, Z = 70
The relative abundance of X = 30% 0f 65X and 70% of 63X
EVALUATION:
(a) How many neutrons are present on the isotopes of 4119X (45%) and 4019x(55%)
(b)Calculate the relative atomic mass of X.
RELATIVE ATOMIC MASSES BASED ON C-12 ISOTOPE
Definition: The Relative Atomic Mass of an element is the number of times the average mass of an atom of the element is heavier than one-twelfth of the mass of one atom of carbon -12
RAM of an element= (Average massofoneatomofX)/(1/2 massof 1 atomofcarbon-12)
Thus the atom of carbon -12 is adopted as the standard for defining the relative atomic mass of the other elements and is given a basic mass value of 12units.
The relative atomic mass of each element has been determined accurately with the aid of the mass spectrometer. This instrument measures the masses of the isotopes of the elements and their abundance and the relative atomic mass is calculated from the data.
Relative atomic masses of the first twenty elements in the periodic table
Element Atomic number Relative atomic mass
Hydrogen 1 1.008
Helium 2 4.0026
Lithium 3 6.939
Beryllium 4 9.0122
Boron 5 10.81
Carbon 6 12.011
Nitrogen 7 14.0067
Oxygen 8 15.9994
Fluorine 9 18.9884
Neon 10 20.183
Sodium 11 22.9898
Magnesium 12 24.312
Aluminum 13 26.9812
Silicon 14 28.086
phosphorus 15 30.9738
Sulphur 16 32.06
Chlorine 17 35.453
Argon 18 39.948
Potassium 19 39.102
Calcium 20 40.08
The relative atomic masses of the first twenty elements in the periodic table are given in the table below.
EVALUATION: From the complete periodic table of elements write out the relative atomic masses of (A) Magnesium (b) Oxygen (c) Chlorine (d) Carbon
GENERAL EVALAUATION
OBJECTIVES TEST :
The maximum number of electrons that can be accommodated in the M-shell is. (a) 18 (b) 8 (c) 2 (d) 32
The atomic number of chlorine is (a) 17 (b) 18 (c) 20 (d) 7
Which of the following is an isotope of hydrogen?
(a)41H (b) 51H (c) 31H (d) 01H
4. Which of the following is the electronic configuration of carbon? (a) k L M (b) K L M (c) K L M (d) K L M
22 224 1 2 3 2 1 3
ESSAY QUESTIONS
(1) Chlorine exists in two isotopic mixtures. The first has 17protons and 18 neutrons while the second isotope has 17 protons and 20 neutrons. If the two isotopes are present in ratio 3:1 respectively, calculate the relative atomic mass of chlorine.
(2)Show the electron structure of the following.
(a) Calcium (b) Magnesium (c) Sodium (d) Oxygen
(3) If the numbers of charged and unchanged particles in the centre of an atom are 6 and 7 respectively, what is the mass number of the atom?
(4) Calculate the number of neutrons in
(a) 2311Na (b) 3717Cl
(5) (a)What is an ‘isotopy’?
(b) Explain briefly why the chemical properties of isotopes of an element are similar.
WEEKEND ASSIGNMENT: Read about the topic ‘’mass spectrometer’’
WEEK ACTIVITY:
(a) Draw a labelled structure of a mass spectrometer showing its basic features. (b) Write two features of a mass spectrometer.
PRE-READING ASSIGNMENT:
Read the topic ‘’symbols of elements’’ page 26 of New school Chemistry. By Osai Yaw Ababio.
SUB-TOPICS:
1. Arrangement of electrons around the nucleus.
2. Atomic number, mass number.
3. Relative atomic mass based on c-12 isotope.
4. Isotopy.
ARRANGEMENT OF ELECTRONS AROUND THE NUCLEUS
Electrons are found revolving around the nucleus of an atom in circular paths known as rings, orbits, energy levels or shells. Each shell contains electrons with similar energy.Those with the lowest energies being nearest to the nucleus.
3 (M shell)
Nucleus
2(L shell)
1 (K shell)
Thus, the arrangement of electrons in the atom according to energy is called ELECTRONIC CONFIGURATION. Letters and figures are associated with these orbits or shells as shown above. The maximum possible number of electrons that can be accommodated in a shell is given by the formula:
Nmax= 2n2. Where Nmax = Maximum no of electron. n= no, of shell.
Thus K- shell can contain 2 × 12 = 2 electrons.
L- Shell can contain 2 × 22 = 8 electrons.
M- Shell can contain 2 × 32= 18 electrons etc
The electron structures of the atoms of the first twenty elements are given in the table below.
Element Symbol Number of protons(or atomic number) Number of electrons and their distribution in the shells. K L M N
Hydrogen H 1 1
Helium He 2 2
Lithium Li 3 2 1
Beryllium Be 4 2 2
Boron B 5 2 3
Carbon C 6 2 4
Nitrogen N 7 2 5
Oxygen O 8 2 6
Fluorine F 9 2 7
Neon Ne 10 2 8
Sodium Na 11 2 8 1
Magnesium Mg 12 2 8 2
Aluminium Al 13 2 8 3
Silicon Si 14 2 8 4
Phosphorus P 15 2 8 5
Sulphur S 16 2 8 6
Chlorine Cl 17 2 8 7
Argon Ar 18 2 8 8
Potassium Ic 19 2 8 8 1
Calcium Ca 20 2 8 8 2
The electronic configurations of some elements are shown below:
Hydrogen, H (atomic number 1)
Electronk1
Nucleus
Helium, He(atomic number 2)
electrons k2
nucleus
Lithium, Li (atomic number 3).
K2 L1
Neon, Ne (atomic number 10)
K2 L8
2,8
Sodium, Na (atomic number 11)
K L M
2 8 1
2, 8, 1
Argon, Ar (atomic number 18 )
K L M
2 8 8
2,8,8
Potassium, K (atomic number 19) K L M N
2 8 8 1
2, 8, 8, 1
Calcium, Ca (atomic number 20)
K L M N
2 8 8 2
2,8,8,2
EVALUATION: Draw the electronic configuration of the following elements.
Carbon (b) Fluorine (c) Aluminium (d) Nitrogen
ATOMIC NUMBER AND MASS NUMBER.
ATOMIC NUMBER:
DEFINITION: Atomic number is the number of protons in an atom of an element.
The atomic number of an element is a whole number and is designated z. In a neutral atom the number of protons must be equal to the number of electrons(since protons are positively charged and electron are negatively charged.)
All the atoms of a particular element have the same number of protons in their nuclei (i.e. they have the same atomic number). NO two elements have the same number of protons in their atoms.
DEFINITION:
MASS NUMBER: The mass number is the sum of the protons and neutrons in an atom of an element.
Mass number is represented by the letter A.
Mass number A = Number of protons + number of neutrons.
i.e. A =p + n. Where p=protons, n= neutrons.
Or number of neutrons n =A – Z.
An atom of an element can be described by writing its symbol together with its atomic number and mass number.
(Mass number) A
Symbol of an atom of the element.
(Atomic number)Z
Examples: The atom of carbon, oxygen and sodium can be written as 126C, 168O and2311Na respectively.
EVALUATION:
(1) Define the following.
Atomic number (b) Mass number.
(2) Describe the atoms of the following elements using their symbol, atomic number and mass number.
(a) Phosphorus (b) Silicon (c) Calcium
ISOTOPY
Definition: Isotopy is a phenomenon whereby atoms of an element exhibit different mass number but have the same atomic number.
Mass spectrometric studies show that the atoms of most elements exist in more than one form. This is due to the difference in number of neutrons present in these atoms. Such atoms are known as isotopes. Isotope of an element is represented by the original symbol of the element with the mass number and atomic numbers. For example 126C, 136C, 146C represent atoms of the isotopes of carbon. For each atom, the number of neutrons can be obtained by finding the difference between the mass number A and the atomic number Z i.e. A – Z. Each isotope of an element has its own mass known as isotopic mass.
Isotopes of an element have slightly different physical properties because neutrons contribute only to the mass of an atom and not its chemical behaviour. But isotopes of an element exhibit the same chemical properties because the number of valence electrons in an atom of an element determines its chemical behaviour (properties) and since isotopes have the same number of valence electrons they will be chemically alike.
NOTE:
(i) An analysis of the chlorine isotopes.
Isotope 3517Cl Isotope 3717Cl
Mass number, A 35 37
Atomic number,Z 17 17
Number of protons 17 17
Number of electrons 17 17
Number of neutrons (A-Z) 35 – 17=18 37 – 17=20
Abundance in nature (%) 75 25
ISOTOPES OF THE SAME ELEMENTS.
Element Carbon Oxygen
ISOTOPES 126C 136C 168O 178O 188O
ABUNDANCE IN NATURE (%) 98.9 1.1 99.76 0.04 0.20
(ii) The names of the isotopic forms of hydrogen
11H --- Protium (or hydrogen)
21H--- Deutrium (or heavy hydrogen or D)
31H--- Tritium or T
(iii) The relative atomic mass, RAM of an element which exhibits isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element and they are not usually in whole numbers.
CALCULATION INVOLVING ISOTOPY.
WORKED EXAMPLE:
Determine the relative atomic mass of element X from the data below
ISOTOPE MASS %ABUNDANCE
24X 24 78.70
25x 25 10.13
26x 26 11.7
((24 ×78.70)+ (25 ×10.13)+(26 × 11.17)= X)/100
(1888.8+253.25+290.42=X)/100
(2432.47=X)/100
∴X=24.3247
≅24
An element X has two isotopes of 2010X and 2210X in the ratio 1:3. What is the relative atomic mass?
Add ratio of occurrence together.
1 + 3= 4
((20 ×1)+ (22 ×3)=X)/4
(20+66=X)/4
( 86=X)/4
X = 21.5
Isotopes of an element X have isotopic masses 65 and 63 respectively. If the relative atomic mass of X is 63.60. Find the relative abundance of each isotope of the element.
Let the relative abundance of element X be y and Z respectively.
Z +Y =100
∴ Z = 100 – y....... (i)
((65 ×y)+ (63 ×y)=63.60)/100
65y + 63z = 63.60 × 100
65y + 63z = 6360 ......(ii)
65y +63(100 – y) = 6360
65y – 63y + 6300 = 6360
2y = 60
Y = 60/2 = 30
Y= 30
Z = 100 – y= 100 – 30 = 70
Y = 30, Z = 70
The relative abundance of X = 30% 0f 65X and 70% of 63X
EVALUATION:
(a) How many neutrons are present on the isotopes of 4119X (45%) and 4019x(55%)
(b)Calculate the relative atomic mass of X.
RELATIVE ATOMIC MASSES BASED ON C-12 ISOTOPE
Definition: The Relative Atomic Mass of an element is the number of times the average mass of an atom of the element is heavier than one-twelfth of the mass of one atom of carbon -12
RAM of an element= (Average massofoneatomofX)/(1/2 massof 1 atomofcarbon-12)
Thus the atom of carbon -12 is adopted as the standard for defining the relative atomic mass of the other elements and is given a basic mass value of 12units.
The relative atomic mass of each element has been determined accurately with the aid of the mass spectrometer. This instrument measures the masses of the isotopes of the elements and their abundance and the relative atomic mass is calculated from the data.
Relative atomic masses of the first twenty elements in the periodic table
Element Atomic number Relative atomic mass
Hydrogen 1 1.008
Helium 2 4.0026
Lithium 3 6.939
Beryllium 4 9.0122
Boron 5 10.81
Carbon 6 12.011
Nitrogen 7 14.0067
Oxygen 8 15.9994
Fluorine 9 18.9884
Neon 10 20.183
Sodium 11 22.9898
Magnesium 12 24.312
Aluminum 13 26.9812
Silicon 14 28.086
phosphorus 15 30.9738
Sulphur 16 32.06
Chlorine 17 35.453
Argon 18 39.948
Potassium 19 39.102
Calcium 20 40.08
The relative atomic masses of the first twenty elements in the periodic table are given in the table below.
EVALUATION: From the complete periodic table of elements write out the relative atomic masses of (A) Magnesium (b) Oxygen (c) Chlorine (d) Carbon
GENERAL EVALAUATION
OBJECTIVES TEST :
The maximum number of electrons that can be accommodated in the M-shell is. (a) 18 (b) 8 (c) 2 (d) 32
The atomic number of chlorine is (a) 17 (b) 18 (c) 20 (d) 7
Which of the following is an isotope of hydrogen?
(a)41H (b) 51H (c) 31H (d) 01H
4. Which of the following is the electronic configuration of carbon? (a) k L M (b) K L M (c) K L M (d) K L M
22 224 1 2 3 2 1 3
ESSAY QUESTIONS
(1) Chlorine exists in two isotopic mixtures. The first has 17protons and 18 neutrons while the second isotope has 17 protons and 20 neutrons. If the two isotopes are present in ratio 3:1 respectively, calculate the relative atomic mass of chlorine.
(2)Show the electron structure of the following.
(a) Calcium (b) Magnesium (c) Sodium (d) Oxygen
(3) If the numbers of charged and unchanged particles in the centre of an atom are 6 and 7 respectively, what is the mass number of the atom?
(4) Calculate the number of neutrons in
(a) 2311Na (b) 3717Cl
(5) (a)What is an ‘isotopy’?
(b) Explain briefly why the chemical properties of isotopes of an element are similar.
WEEKEND ASSIGNMENT: Read about the topic ‘’mass spectrometer’’
WEEK ACTIVITY:
(a) Draw a labelled structure of a mass spectrometer showing its basic features. (b) Write two features of a mass spectrometer.
PRE-READING ASSIGNMENT:
Read the topic ‘’symbols of elements’’ page 26 of New school Chemistry. By Osai Yaw Ababio.
WEEK 4
TOPIC: Symbols, Formulae and Equations.
CONTENT
Chemical symbols and their valences
Formulae and equations
Empirical formula and molecular formula
SUB- TOPIC
CHEMICAL SYMBOLS AND THEIR VALENCIES
Symbols are modern ways of representing atoms of elements by using abbreviations. The modern symbols were developed by Berzelius in 1814.
First, He used the first letter in the name of the element.
Examples
ELEMENT SYMBOL
Hydrogen H
Boron B
Carbon C
Nitrogen N
Oxygen O
Fluorine F
Phosphorus P
Sulphur S
Iodine I
The second principle uses the first two letters.
Examples:
ELEMENTS SYMBOL
Helium He
Lithium Li
Beryllium Be
Neon Ne
Aluminum Al
Silicon Si
Argon Ar
Calcium Ca
Bromine Br
Barium Ba
The third principle is when the first letter and another letter in the name are used.
Example:
ELEMENTS SYMBOL
Magnesium Mg
Chlorine Cl
Chromium Cr
Manganese Mn
Zinc Zn
The fourth principle is the elements that derived symbols from their Latin names.
Example:
ELEMENT LATIN NAME SYMBOL
Sodium Natrium Na
Potassium Kalium K
Iron Ferrum Fe
Copper Cuprum Cu
Silver Argentum Ag
Tin Stannum Sn
Gold Aurum Au
Mercury Hydrargyrum Hg
Lead Plumbum Pb
VALENCY
The valency of an element is the combining power of the element. It is defined as the number of atoms of hydrogen that will combine with or displace one atom of the element in chemical reactions. Some elements have more than one valency. For example.
1st twenty elements with their valencies show this character.
Atomic Number Element Symbol Valency
1 Hydrogen H 1
2 Helium He Nil
3 Lithium Li 2
4 Beryllium Be 2
5 Boron B 3
6 Carbon C 2 or 4
7 Nitrogen N 3 or 5
8 Oxygen O 2
9 Fluorine F 1
10 Neon Ne Nil
11 Sodium Na 1
12 Magnesium Mg 2
13 Aluminum Al 3
14 Silicon Si 2 or 4
15 Phosphorus P 3 or 5
16 Sulphur S 2, 4 or 6
17 Chlorine Cl 1
18 Argon Ar Nil
19 Potassium Ic 1
20 Calcium Ca 2
EVALUATION:
1. What do you understand by the word chemical symbol.
2. Define valency of an element.
READING ASSIGNMENT:
Find out the valency of the following element.
(i)Gold (ii) Silver (iii) Zinc and copper
SUB- TOPIC 2:FORMULAE AND EQUATIONS
The formulae for compounds contain the symbols for the different elements that are combines chemically to form the compound. Compounds are made up of molecules and the formula of a molecule of a compound is made of the symbols of the atoms of the element of the compound, combined in correct proportions or valencies.
For example: Consider Formulae of some compounds
Compounds Formulae
Oxygen molecule O2
Hydrogen molecule H2
Hydrogen chloride HCl
Potassium chloride KCl
Magnesium tetraoxosulphate (vi) MgSO4
Hydrogen sulphide H2S
Bromine Molecule Br2
Ozone O3
RADICALS: These are groups of atoms of different elements that come together and react as a unit. These radicals are charged, that is they either carry a positive or a negative charge. An acid radical is thus a small group or cluster of atoms carrying a negative charge that keeps its identity. These groups of atoms originate from the acids which have formed the salts. For examples
Radical Symbol Valency Oxidation No
Ammonium ion NH4+ 1 +1
Hydroxyl ion OH- 1 -1
Trioxocarbonate (iv) CO32- 2 -2
Tetraoxosulphate(vi) SO42- 2 -2
Trioxonitrate(v) NO3- 1 -1
Chemical reactions are represented in form of equations which show the reactants and products in any given chemical reaction. For example, the reaction of aqueous hydrogen chloride and aqueous sodium hydroxide is represented by the equation:
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
An equation must be balanced. A balanced equation contains the same number of atoms of the elements of the compounds on both sides of the equation.
The equation must also show the physical states of the reactants and the products i.e. whether in aqueous solution or gaseous or solid state.
Examples
Balance the equation below:
CaCO3(s) +HCl(aq) Cacl2(s) + H2O(l) + CO2(s)
CO2(g) + NaOH(aq) Na2CO3(aq) + H2O(l)
Answer
Balanced equation:
CaCO3(s) + 2HCl(aq) CaCl2 + H2O(l)
CO2(g) +2NaOH(aq) Na2CO3(aq) + H2O(l)
OXIDATION NUMBER
To be able to write correct chemical formulae for compounds a system of small whole numbers, related to the combining ratio of element has been developed on the basis of arbitrary rules. Such numbers are called oxidation numbers or oxidation states.
Four of the rules are:
The oxidation states of an uncombined element are assigned negative oxidation states, and more electro positive elements are assigned positive states .e.g. H2, O2, Fe, Mg etc.
In a compound, the more electro negative elements are assigned negative oxidation states, and more electropositive elements are assigned positive states. E.g. in NaCl, Na is more electropositive than Cl so we have. Na+ Cl-.
HCl H+ Cl-
In a compound the sum of the positive oxidation on state and the negative oxidation states is zero e.g.
NaCl Na+Cl-
(+1) + (-1) = zero
In an ion, the sum of the oxidation number is equal to the charge on the ion. E.g. Na+ = (+1)
EVALUATION
Define chemical formula
Write the formula of the following compound.
Hydrogen chloride
Sodium chloride
What is oxidation number?
SUB-TOPIC 3: EMPIRICAL AND MOLECULAR FORMULAE
The empirical and molecule formula have their bases formthe relative molecular mass, molar mass and percentage composition of a compound.
Relative molecular mass and molar mass.
If the formulae of a substance and the relative atomic Masses of each of the elements are known, then it is possible to determine the relative molecular mass of that substance.
The relative molecular mass of a compound is the sum of the masses of all the atoms present in one molecule of the compound .e.g.
For NaCl, the relative molecular mass= (23 +35.5) = 58.5
For ethanol = C2H5 OH (carbon=12, H=1, O =16)
The relative molecular mass of ethanol
= C2H5OH
(12×2) + (1×5) + (16) + (1)
24 + 5 + 16 +1 = 46
The relative molecular mass refers to the number of times a mole is heavier than one- twelfth the mass of one atom of carbon -12.
THE MOLAR MASS
This is the relative molecular mass expressed in grams. E.g. the molar mass of ethanol is 46g.
In 12g of carbon -12, there are 6 × 1023 atoms of carbon. This is one mole of carbon -12.
A mole of any substance is the amount of that substance which contains 6× 1023 particles of that substance e.g. One mole of ethanol has a mass of 46g and contains 6× 1023 ethanol molecules.
NOTE: The relative molecular mass has no units but the molar mass of any substance is expressed in grams.
Percentage composition of a compound.
To calculate the percentage position of ethanol whose molecular formula is C2H5OH, given that the relative atomic masses of carbon, hydrogen and oxygen are 12, 1, and 16 respectively?
First calculate the molar mass of
C2H5OH =(12×2) +(1×5) + (16)+ (1) = 46g
Then determine the masses of C H and O present;
Mass of carbon= 12×2=24g
Mass of hydrogen= 6×1 = 6g
Mass of oxygen = 16× 1 = 16g
Molar mass of C2H5OH= 46g
Therefore, percentage of C= 24/46 × 100 = 52.17%
Percentage of hydrogen=6/( 46) × 100 = 13.04%
Percentage of oxygen = 16/46 × 100 = 13.04%
Calculation of the chemical formula from percentage composition by mass.
We can determine the simplest chemical formula of a compound, given its percentage composition e.g. If the formula for anhydrous disodium trioxocarbonate (iv) is not known, if its percentage composition by mass is known then it chemical formula could be calculated.
For example, the percentage composition of the compound was found to be Na=43.40%,C= 11.32% and O = 45.28%. This would mean that in every 100g of the compound, the masses of Na, C and O were 43.40g, 11.32g and 45.28g respectively.
∴The amount in moles of Na, C and O would be.
43.40/23 , 11.32/(12 ) and 45.28/(16 ) respectively.
Therefore, the amount in moles of Na =(43.40)/23 = 1.89.
Amount in mole of C= (11.30)/12=0.94.
Amount in mole of O= (45.28)/16=2.83.
Molar ratio of H: C: O is 1.89: 0.94: 2.83.
2: 1 : 3
Na: C: O= Na2C O3
The simplest formula is therefore, Na2Co3
EMPIRICAL FORMULAEAND MOLECULARFORMULAE
The empirical formula is the chemical formulae from percentage composition by mass. The molecular formula of a compound is a whole number multiple of its empirical formula.
Examples:
Find the empirical formula of a compound which contains 80% carbon and 20% hydrogen by mass.
[Relativeatomicmasses;(C=12,H=1)]
C H
Percentage by mass 80% 20%
Percentage by mass
80/12 20/1
6.6 20
Divided by the
Smaller number
6.6/(6.6 ) 20/6.6
1 3
C H3
The empirical formula= CH3
Find the empirical formula of a compound which on analysis yields the following as the reacting masses carbon= 2.0, hydrogen=0.34g, Oxygen = 2.67g. From your result find the molecular formula of the compound. If it’s relative molecular mass is 60. (C=12, H=1, O=16)
Solution:
Carbon Hydrogen Oxygen
2.0/(12 ) ( 0.34 )/1 2.67/16
0.17 0.34 0.17
Divide by
The smallest nunber
0.17/0.17 (0.34 )/(0.17 ) 0.17/0.17
1 2 1
Empirical formula of the compound is CH2O.
(CH2O)X= 60
(12 + (2×1) + 16)x =60
(12+2+16)x=60
(30)x= 60
X = 2
Molecular formula is therefore (CH2O)2
=C2H4O2 or CH3COOH
EVALUATION:
Calculate the relative molecular mass of each of the following
CaCO3 (b) Mg(NO3)2
Calculate the percentage by mass present in NaOH and MgO.
Calculate the empirical formula of 15.8% Al, 28.1% S, 56.1%O
GENERAL EVALUATION
ESSAY QUESTION
1. Differentiate between valency and oxidation number.
2. Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, deduce its molecular formula.
3. Balance this chemical equation
NaOH + H2SO4 Na2SO4 + H2O
PRE-READING:
Read Ababio, pg34-35.On chemical combinations
WEEKEND ASSIGNMENT:
Read comprehensive chemistry pg 19-20
CONTENT
Chemical symbols and their valences
Formulae and equations
Empirical formula and molecular formula
SUB- TOPIC
CHEMICAL SYMBOLS AND THEIR VALENCIES
Symbols are modern ways of representing atoms of elements by using abbreviations. The modern symbols were developed by Berzelius in 1814.
First, He used the first letter in the name of the element.
Examples
ELEMENT SYMBOL
Hydrogen H
Boron B
Carbon C
Nitrogen N
Oxygen O
Fluorine F
Phosphorus P
Sulphur S
Iodine I
The second principle uses the first two letters.
Examples:
ELEMENTS SYMBOL
Helium He
Lithium Li
Beryllium Be
Neon Ne
Aluminum Al
Silicon Si
Argon Ar
Calcium Ca
Bromine Br
Barium Ba
The third principle is when the first letter and another letter in the name are used.
Example:
ELEMENTS SYMBOL
Magnesium Mg
Chlorine Cl
Chromium Cr
Manganese Mn
Zinc Zn
The fourth principle is the elements that derived symbols from their Latin names.
Example:
ELEMENT LATIN NAME SYMBOL
Sodium Natrium Na
Potassium Kalium K
Iron Ferrum Fe
Copper Cuprum Cu
Silver Argentum Ag
Tin Stannum Sn
Gold Aurum Au
Mercury Hydrargyrum Hg
Lead Plumbum Pb
VALENCY
The valency of an element is the combining power of the element. It is defined as the number of atoms of hydrogen that will combine with or displace one atom of the element in chemical reactions. Some elements have more than one valency. For example.
1st twenty elements with their valencies show this character.
Atomic Number Element Symbol Valency
1 Hydrogen H 1
2 Helium He Nil
3 Lithium Li 2
4 Beryllium Be 2
5 Boron B 3
6 Carbon C 2 or 4
7 Nitrogen N 3 or 5
8 Oxygen O 2
9 Fluorine F 1
10 Neon Ne Nil
11 Sodium Na 1
12 Magnesium Mg 2
13 Aluminum Al 3
14 Silicon Si 2 or 4
15 Phosphorus P 3 or 5
16 Sulphur S 2, 4 or 6
17 Chlorine Cl 1
18 Argon Ar Nil
19 Potassium Ic 1
20 Calcium Ca 2
EVALUATION:
1. What do you understand by the word chemical symbol.
2. Define valency of an element.
READING ASSIGNMENT:
Find out the valency of the following element.
(i)Gold (ii) Silver (iii) Zinc and copper
SUB- TOPIC 2:FORMULAE AND EQUATIONS
The formulae for compounds contain the symbols for the different elements that are combines chemically to form the compound. Compounds are made up of molecules and the formula of a molecule of a compound is made of the symbols of the atoms of the element of the compound, combined in correct proportions or valencies.
For example: Consider Formulae of some compounds
Compounds Formulae
Oxygen molecule O2
Hydrogen molecule H2
Hydrogen chloride HCl
Potassium chloride KCl
Magnesium tetraoxosulphate (vi) MgSO4
Hydrogen sulphide H2S
Bromine Molecule Br2
Ozone O3
RADICALS: These are groups of atoms of different elements that come together and react as a unit. These radicals are charged, that is they either carry a positive or a negative charge. An acid radical is thus a small group or cluster of atoms carrying a negative charge that keeps its identity. These groups of atoms originate from the acids which have formed the salts. For examples
Radical Symbol Valency Oxidation No
Ammonium ion NH4+ 1 +1
Hydroxyl ion OH- 1 -1
Trioxocarbonate (iv) CO32- 2 -2
Tetraoxosulphate(vi) SO42- 2 -2
Trioxonitrate(v) NO3- 1 -1
Chemical reactions are represented in form of equations which show the reactants and products in any given chemical reaction. For example, the reaction of aqueous hydrogen chloride and aqueous sodium hydroxide is represented by the equation:
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
An equation must be balanced. A balanced equation contains the same number of atoms of the elements of the compounds on both sides of the equation.
The equation must also show the physical states of the reactants and the products i.e. whether in aqueous solution or gaseous or solid state.
Examples
Balance the equation below:
CaCO3(s) +HCl(aq) Cacl2(s) + H2O(l) + CO2(s)
CO2(g) + NaOH(aq) Na2CO3(aq) + H2O(l)
Answer
Balanced equation:
CaCO3(s) + 2HCl(aq) CaCl2 + H2O(l)
CO2(g) +2NaOH(aq) Na2CO3(aq) + H2O(l)
OXIDATION NUMBER
To be able to write correct chemical formulae for compounds a system of small whole numbers, related to the combining ratio of element has been developed on the basis of arbitrary rules. Such numbers are called oxidation numbers or oxidation states.
Four of the rules are:
The oxidation states of an uncombined element are assigned negative oxidation states, and more electro positive elements are assigned positive states .e.g. H2, O2, Fe, Mg etc.
In a compound, the more electro negative elements are assigned negative oxidation states, and more electropositive elements are assigned positive states. E.g. in NaCl, Na is more electropositive than Cl so we have. Na+ Cl-.
HCl H+ Cl-
In a compound the sum of the positive oxidation on state and the negative oxidation states is zero e.g.
NaCl Na+Cl-
(+1) + (-1) = zero
In an ion, the sum of the oxidation number is equal to the charge on the ion. E.g. Na+ = (+1)
EVALUATION
Define chemical formula
Write the formula of the following compound.
Hydrogen chloride
Sodium chloride
What is oxidation number?
SUB-TOPIC 3: EMPIRICAL AND MOLECULAR FORMULAE
The empirical and molecule formula have their bases formthe relative molecular mass, molar mass and percentage composition of a compound.
Relative molecular mass and molar mass.
If the formulae of a substance and the relative atomic Masses of each of the elements are known, then it is possible to determine the relative molecular mass of that substance.
The relative molecular mass of a compound is the sum of the masses of all the atoms present in one molecule of the compound .e.g.
For NaCl, the relative molecular mass= (23 +35.5) = 58.5
For ethanol = C2H5 OH (carbon=12, H=1, O =16)
The relative molecular mass of ethanol
= C2H5OH
(12×2) + (1×5) + (16) + (1)
24 + 5 + 16 +1 = 46
The relative molecular mass refers to the number of times a mole is heavier than one- twelfth the mass of one atom of carbon -12.
THE MOLAR MASS
This is the relative molecular mass expressed in grams. E.g. the molar mass of ethanol is 46g.
In 12g of carbon -12, there are 6 × 1023 atoms of carbon. This is one mole of carbon -12.
A mole of any substance is the amount of that substance which contains 6× 1023 particles of that substance e.g. One mole of ethanol has a mass of 46g and contains 6× 1023 ethanol molecules.
NOTE: The relative molecular mass has no units but the molar mass of any substance is expressed in grams.
Percentage composition of a compound.
To calculate the percentage position of ethanol whose molecular formula is C2H5OH, given that the relative atomic masses of carbon, hydrogen and oxygen are 12, 1, and 16 respectively?
First calculate the molar mass of
C2H5OH =(12×2) +(1×5) + (16)+ (1) = 46g
Then determine the masses of C H and O present;
Mass of carbon= 12×2=24g
Mass of hydrogen= 6×1 = 6g
Mass of oxygen = 16× 1 = 16g
Molar mass of C2H5OH= 46g
Therefore, percentage of C= 24/46 × 100 = 52.17%
Percentage of hydrogen=6/( 46) × 100 = 13.04%
Percentage of oxygen = 16/46 × 100 = 13.04%
Calculation of the chemical formula from percentage composition by mass.
We can determine the simplest chemical formula of a compound, given its percentage composition e.g. If the formula for anhydrous disodium trioxocarbonate (iv) is not known, if its percentage composition by mass is known then it chemical formula could be calculated.
For example, the percentage composition of the compound was found to be Na=43.40%,C= 11.32% and O = 45.28%. This would mean that in every 100g of the compound, the masses of Na, C and O were 43.40g, 11.32g and 45.28g respectively.
∴The amount in moles of Na, C and O would be.
43.40/23 , 11.32/(12 ) and 45.28/(16 ) respectively.
Therefore, the amount in moles of Na =(43.40)/23 = 1.89.
Amount in mole of C= (11.30)/12=0.94.
Amount in mole of O= (45.28)/16=2.83.
Molar ratio of H: C: O is 1.89: 0.94: 2.83.
2: 1 : 3
Na: C: O= Na2C O3
The simplest formula is therefore, Na2Co3
EMPIRICAL FORMULAEAND MOLECULARFORMULAE
The empirical formula is the chemical formulae from percentage composition by mass. The molecular formula of a compound is a whole number multiple of its empirical formula.
Examples:
Find the empirical formula of a compound which contains 80% carbon and 20% hydrogen by mass.
[Relativeatomicmasses;(C=12,H=1)]
C H
Percentage by mass 80% 20%
Percentage by mass
80/12 20/1
6.6 20
Divided by the
Smaller number
6.6/(6.6 ) 20/6.6
1 3
C H3
The empirical formula= CH3
Find the empirical formula of a compound which on analysis yields the following as the reacting masses carbon= 2.0, hydrogen=0.34g, Oxygen = 2.67g. From your result find the molecular formula of the compound. If it’s relative molecular mass is 60. (C=12, H=1, O=16)
Solution:
Carbon Hydrogen Oxygen
2.0/(12 ) ( 0.34 )/1 2.67/16
0.17 0.34 0.17
Divide by
The smallest nunber
0.17/0.17 (0.34 )/(0.17 ) 0.17/0.17
1 2 1
Empirical formula of the compound is CH2O.
(CH2O)X= 60
(12 + (2×1) + 16)x =60
(12+2+16)x=60
(30)x= 60
X = 2
Molecular formula is therefore (CH2O)2
=C2H4O2 or CH3COOH
EVALUATION:
Calculate the relative molecular mass of each of the following
CaCO3 (b) Mg(NO3)2
Calculate the percentage by mass present in NaOH and MgO.
Calculate the empirical formula of 15.8% Al, 28.1% S, 56.1%O
GENERAL EVALUATION
ESSAY QUESTION
1. Differentiate between valency and oxidation number.
2. Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, deduce its molecular formula.
3. Balance this chemical equation
NaOH + H2SO4 Na2SO4 + H2O
PRE-READING:
Read Ababio, pg34-35.On chemical combinations
WEEKEND ASSIGNMENT:
Read comprehensive chemistry pg 19-20
WEEK 5
TOPIC: Symbols, formulae and equations
CONTENT: chemical laws
Law of conservation of matters
Law of constant composition
Law of multiple proportions.
SUB-TOPIC 1
CHEMICAL LAWS OF COMBINATIONS
There are four laws of chemical combination which describe the general features of a chemical change.
Law of conservation of mass: This law was established by Lavoisier, a French chemist. The law of conservation of mass states that matter is neither created nor destroyed during chemical reaction, but changes from one form to another.
Experiment to verify the law of conservation of matter (mass)
Theory:
The equation of the chemical reaction chosen for study is as follows;
Silver nitrate + sodium chloride Silver chloride + Sodium trioxonitrate (v)
(White precipitate)
Method:
Put some sodium chloride solution in a conical flask
Fill a small test tube with silver trioxonitrate (iv) solution of string, suspend it in a conical flask as shown below:
Insert the stopper and weight the whole apparatus on a balance, note the mass of the whole system.
Mix the two liquids by pulling the string attached to the bottom end of the small test tube.
Weigh the whole apparatus again.
Result: When the two reactants are mixed together, a white precipitate is formed indicating that a chemical reaction has taken place. The new substances formed are known as the products of the chemical reaction. The masses of the system taken before and after the reaction are found to be the same, indicating that the mass of the reactants equals that of the products.
CONCLUSION: Since there is no overall change in mass when the products are formed, we can infer that matter is neither created nor destroyed during the chemical reaction. The law is, hence valid.
EVALUATION:
Mention another compound that could be used instead of silvertrioxo-nitrate(v) with sodium chloride
State the law of conservation of mass/matter.
SUB-TOPIC 2
LAW OF DEFINITE PROPORTION OR LAW OF CONSTANT COMPOSITION
The second law of chemical combination which is supported by the Atomic theory was proposed by provost (1755-1826) known as the Law of definite proportions or constant composition.
The law of definite proportions states that all pure samples of a particular chemical compound contain similar elements combined in the same proportion by mass. It is based on the fact that when elements combine to form a given compound, they do so in fixed proportions by mass, so that all pure samples of that compound are identical in composition by mass. Water for example: chemical analyses showed that as long as it is pure, its composition is always in the ratio of one mole of oxygen to two moles of hydrogen. i.e. 32g of O to 4g of H. Irrespective of whether the water comes from river, sea, rain or anywhere.
Experiment to verify the law of definite proportion
Method: Prepare two samples of black copper (ii) oxide, each by a different method as given below:
Sample A: Place some coppers turning in a crucible and add some concentrated trioxonitrate (v) acid, a little at a time, until the copper dissolves completely. Evaporate the resulting green solution of copper II oxide trioxonitrate (v) to dryness; continue to heat the residue until it decomposes to give a black solid which is copper II oxide. Keep the black residue dry in desiccator.
Sample B:Place some copper (i) trioxocarbonate (iv) in a crucible and decompose it into copper (ii) oxide and carbon (iv) oxide store the residue in a desiccator.
ANALYSES:
Determine the amount of copper present in the two samples of copper oxide by reducing the oxide in a stream of hydrogen or carbon II oxide as follows.
Weigh two clean metal boats.
Add a reasonable amount of sample A to one and sample B to the other
Reweigh and determine the mass of each sample. Place the boats inside a hard glass tube as shown. Heat the samples stronglywhile passing a stream of dry hydrogen gas through the tube. After some time, a reddish- brown copper residue is left in each boat. Remove the flame, but continue passing the hydrogen as the copper residues cool down. This presents the re-oxidation of the hot copper residue by atmospheric oxygen. Any water formed during the reaction is absorbed by the fused calcium chloride in the adjacent U-tube.
Result:
Sample A B
Mass of copper II oxide 3.55g 3.02g
Mass of copper residue 2.81g 2.42g
Percentage of copper present in copper (ii) oxide 2.81/3.55 ×100 2.42/3.02 × 100
79.2% 80.1%
The percentage of copper residue in the two samples in approximately 80.0, irrespective of the method of preparation of the copper(II) oxide samples.
CONCLUSION: In the pure copper(II) oxide copper and oxygen are always present in a definite proportion by mass of approximately 4 to 1 i.e.
Copper (II)oxide = copper + oxygen
100% 80% 20%
Ratio 4 : 1
EVALUATION:
1. State law of definite proportion.
SUB-TOPIC 3
LAW OF MULTIPLE PROPORTIONS
This law states that if two elements combine to form more than one compound, the masses of one of the elements which separately combine with a fixed mass of the other element are in simple ratio.
VERIFICATION OF THE LAW OF MULTIPLE PROPORTIONS
Some elements form more than one compound, depending on the conditions of the reaction and the valency copper forms. Copper (I) and copper(II) with oxygen. Also in an insufficient supply of air, carbon burns to form carbon(II) oxide and when the supply of air is sufficient, carbon(iv) oxide is obtained.
The sample of the copper (I) oxide and copper(II) are placed in porcelain, boats and placed in a combination tube as in the diagram below.
A current of dry hydrogen is passed through the combustion tube until the oxides are reduced to metallic coppers. They are now cooled and weighed and the masses of copper and oxygen are determined in the two samples.
Calculations
Sample A Sample B
(i)Mass of porcelain boat 4.55g 5.38g
(ii) Mass of porcelain boat + copper oxide 6.44g 8.21g
(iii) Mass of copper oxide 1.89g 2.83g
(iv) Mass of porcelain boat + copper 6.05g 7.90g
(v) Mass of copper (iv) – (i) 1.50g 2.52g
(vi) Mass of oxygen (iii) –(v) 0.39g 0.31g
For example A1.50g of copper combines with 0.39 of oxygen.
∴ 100g of copper combines with 0.39/1.50 × 100 = 26g
For sample (b) 2.52g of copper combines with 0.31g of oxygen
∴ 100g of copper combines with 0.31/2.52× 100= 12.3g
From these calculations, the masses of oxygen (26g and 12.3g) which combine with a fixed mass (100g) of copper are in simple ratio 2:1
LAW OF RECIPROCAL PROPORTION
This is the fourth law of chemical combination. This law states that the masses of several elements,A,B,C,which combine separately with a fixed mass of another element,D,are the same as ,or simple multiples of ,the masses in which A,B,C,themselves combine with one another. For example C, H, O (12, 1, 16) respectively. Carbon and hydrogen combine to form methane (CH4). Carbon and oxygen combine to form carbon (iv) oxide, (CO2) and hydrogen and oxygen combine to form water (H2O).
In water,
H2+ O H2O
2moles + 1 mole
2g 16g H2O
Ratio= 2g: 16g or 4g : 32g H2O
This is the prediction of the law of reciprocal proportions. For example.
23g of calcium trioxocarbonate (iv) on heating decomposes to give calcium oxide (CaO) and carbon (iv) oxide. Calculate the masses of calcium oxide and carbon (iv) oxide produced [C= 40, O = 16, C= 12]
Solution
CaCO3(s) CaO(S) + CO2(g)
40+12+48 40+16 44
100g of CaCO3yield 56g of Co
∴ 23g of Caco3 will yield (56×23g)/100 = 12.88g.
100g of CaCo3 yields 44g of Co2
∴23g of CaCo3 will yield (44×23g)/100= 10.12g
The balanced equation represents the production of hydrogen in the laboratory. Calculate the volume of hydrogen gas that can be produced from 32g of zinc.
Solution
Zn + 2HCl ZnCl2 + H2(l)
1 mole of Zn(65g) gives out 1 mole of H2(g)
But 1 mole of H2 occupies 22.4dm3at s.t.p
∴ 32g of Zn will give out (22.4×32dm3 ofH2 at stp)/65=11.03dm3
EVALUATION:
State the law of multiple proportion
State the law of reciprocal proportions.
Zn(s) + 2HCl(aq) ZnCl2(aq)+ H2(g)
GENERAL EVALUATION:
ESSAY QUESTIONS
Copper(I)oxide Copper(II)oxide
Mass of sample oxide 30.4g 1.91g
Mass of copper residue 2.55g 1.38g
Mass of oxygen removed from oxide 0.49g 0.53g
From the above table, calculate the various masses of copper which would combine separately with a fixed mass of 1 g of oxygen.
2. What mass of copper will be produced from the reductionof 7.95g of copper (II) oxide? (C= 63.5, O= 16)
3. Write down the names of these chemical compounds.
(i) HNO3 (ii) CuCl2(iii) CaCO3(iv) Fe2O3
4. Write the symbol and the valency of the following.
(i) Boron (ii) Carbon (iii) Sulphur (iv) Argon
5. Calculate the formula of a compound with 31.9% potassium 28.93%, chlorine and the rest oxygen. K=39, Cl =355, O=16
OBJECTIVE TEST
Which of the following relative atomic numbers has empirical formula CH2O (H=!, C=12 , O=16). (a) 42 (b) 80 (c) 4
The relative atomic mass of tetraoxosulphate (vi) acid is? (a) 98 (b) 49 (b) 49 (c) 96 (d) 106
Chemical equations will provide all these except. (a) State of chemicals is solved )b) Direction of reaction (c) Mass of products (d) Reactants
All pure samples of a particles compound contain the same elements combined in the same proportion by mass. The statement is the law of . (a) Definite proportion (b) Multiple proportion (c)Conservation of mass or matter (d) Atomic proportion
PRE READING ASSIGNMENT:
Read above the periodic table.
CONTENT: chemical laws
Law of conservation of matters
Law of constant composition
Law of multiple proportions.
SUB-TOPIC 1
CHEMICAL LAWS OF COMBINATIONS
There are four laws of chemical combination which describe the general features of a chemical change.
Law of conservation of mass: This law was established by Lavoisier, a French chemist. The law of conservation of mass states that matter is neither created nor destroyed during chemical reaction, but changes from one form to another.
Experiment to verify the law of conservation of matter (mass)
Theory:
The equation of the chemical reaction chosen for study is as follows;
Silver nitrate + sodium chloride Silver chloride + Sodium trioxonitrate (v)
(White precipitate)
Method:
Put some sodium chloride solution in a conical flask
Fill a small test tube with silver trioxonitrate (iv) solution of string, suspend it in a conical flask as shown below:
Insert the stopper and weight the whole apparatus on a balance, note the mass of the whole system.
Mix the two liquids by pulling the string attached to the bottom end of the small test tube.
Weigh the whole apparatus again.
Result: When the two reactants are mixed together, a white precipitate is formed indicating that a chemical reaction has taken place. The new substances formed are known as the products of the chemical reaction. The masses of the system taken before and after the reaction are found to be the same, indicating that the mass of the reactants equals that of the products.
CONCLUSION: Since there is no overall change in mass when the products are formed, we can infer that matter is neither created nor destroyed during the chemical reaction. The law is, hence valid.
EVALUATION:
Mention another compound that could be used instead of silvertrioxo-nitrate(v) with sodium chloride
State the law of conservation of mass/matter.
SUB-TOPIC 2
LAW OF DEFINITE PROPORTION OR LAW OF CONSTANT COMPOSITION
The second law of chemical combination which is supported by the Atomic theory was proposed by provost (1755-1826) known as the Law of definite proportions or constant composition.
The law of definite proportions states that all pure samples of a particular chemical compound contain similar elements combined in the same proportion by mass. It is based on the fact that when elements combine to form a given compound, they do so in fixed proportions by mass, so that all pure samples of that compound are identical in composition by mass. Water for example: chemical analyses showed that as long as it is pure, its composition is always in the ratio of one mole of oxygen to two moles of hydrogen. i.e. 32g of O to 4g of H. Irrespective of whether the water comes from river, sea, rain or anywhere.
Experiment to verify the law of definite proportion
Method: Prepare two samples of black copper (ii) oxide, each by a different method as given below:
Sample A: Place some coppers turning in a crucible and add some concentrated trioxonitrate (v) acid, a little at a time, until the copper dissolves completely. Evaporate the resulting green solution of copper II oxide trioxonitrate (v) to dryness; continue to heat the residue until it decomposes to give a black solid which is copper II oxide. Keep the black residue dry in desiccator.
Sample B:Place some copper (i) trioxocarbonate (iv) in a crucible and decompose it into copper (ii) oxide and carbon (iv) oxide store the residue in a desiccator.
ANALYSES:
Determine the amount of copper present in the two samples of copper oxide by reducing the oxide in a stream of hydrogen or carbon II oxide as follows.
Weigh two clean metal boats.
Add a reasonable amount of sample A to one and sample B to the other
Reweigh and determine the mass of each sample. Place the boats inside a hard glass tube as shown. Heat the samples stronglywhile passing a stream of dry hydrogen gas through the tube. After some time, a reddish- brown copper residue is left in each boat. Remove the flame, but continue passing the hydrogen as the copper residues cool down. This presents the re-oxidation of the hot copper residue by atmospheric oxygen. Any water formed during the reaction is absorbed by the fused calcium chloride in the adjacent U-tube.
Result:
Sample A B
Mass of copper II oxide 3.55g 3.02g
Mass of copper residue 2.81g 2.42g
Percentage of copper present in copper (ii) oxide 2.81/3.55 ×100 2.42/3.02 × 100
79.2% 80.1%
The percentage of copper residue in the two samples in approximately 80.0, irrespective of the method of preparation of the copper(II) oxide samples.
CONCLUSION: In the pure copper(II) oxide copper and oxygen are always present in a definite proportion by mass of approximately 4 to 1 i.e.
Copper (II)oxide = copper + oxygen
100% 80% 20%
Ratio 4 : 1
EVALUATION:
1. State law of definite proportion.
SUB-TOPIC 3
LAW OF MULTIPLE PROPORTIONS
This law states that if two elements combine to form more than one compound, the masses of one of the elements which separately combine with a fixed mass of the other element are in simple ratio.
VERIFICATION OF THE LAW OF MULTIPLE PROPORTIONS
Some elements form more than one compound, depending on the conditions of the reaction and the valency copper forms. Copper (I) and copper(II) with oxygen. Also in an insufficient supply of air, carbon burns to form carbon(II) oxide and when the supply of air is sufficient, carbon(iv) oxide is obtained.
The sample of the copper (I) oxide and copper(II) are placed in porcelain, boats and placed in a combination tube as in the diagram below.
A current of dry hydrogen is passed through the combustion tube until the oxides are reduced to metallic coppers. They are now cooled and weighed and the masses of copper and oxygen are determined in the two samples.
Calculations
Sample A Sample B
(i)Mass of porcelain boat 4.55g 5.38g
(ii) Mass of porcelain boat + copper oxide 6.44g 8.21g
(iii) Mass of copper oxide 1.89g 2.83g
(iv) Mass of porcelain boat + copper 6.05g 7.90g
(v) Mass of copper (iv) – (i) 1.50g 2.52g
(vi) Mass of oxygen (iii) –(v) 0.39g 0.31g
For example A1.50g of copper combines with 0.39 of oxygen.
∴ 100g of copper combines with 0.39/1.50 × 100 = 26g
For sample (b) 2.52g of copper combines with 0.31g of oxygen
∴ 100g of copper combines with 0.31/2.52× 100= 12.3g
From these calculations, the masses of oxygen (26g and 12.3g) which combine with a fixed mass (100g) of copper are in simple ratio 2:1
LAW OF RECIPROCAL PROPORTION
This is the fourth law of chemical combination. This law states that the masses of several elements,A,B,C,which combine separately with a fixed mass of another element,D,are the same as ,or simple multiples of ,the masses in which A,B,C,themselves combine with one another. For example C, H, O (12, 1, 16) respectively. Carbon and hydrogen combine to form methane (CH4). Carbon and oxygen combine to form carbon (iv) oxide, (CO2) and hydrogen and oxygen combine to form water (H2O).
In water,
H2+ O H2O
2moles + 1 mole
2g 16g H2O
Ratio= 2g: 16g or 4g : 32g H2O
This is the prediction of the law of reciprocal proportions. For example.
23g of calcium trioxocarbonate (iv) on heating decomposes to give calcium oxide (CaO) and carbon (iv) oxide. Calculate the masses of calcium oxide and carbon (iv) oxide produced [C= 40, O = 16, C= 12]
Solution
CaCO3(s) CaO(S) + CO2(g)
40+12+48 40+16 44
100g of CaCO3yield 56g of Co
∴ 23g of Caco3 will yield (56×23g)/100 = 12.88g.
100g of CaCo3 yields 44g of Co2
∴23g of CaCo3 will yield (44×23g)/100= 10.12g
The balanced equation represents the production of hydrogen in the laboratory. Calculate the volume of hydrogen gas that can be produced from 32g of zinc.
Solution
Zn + 2HCl ZnCl2 + H2(l)
1 mole of Zn(65g) gives out 1 mole of H2(g)
But 1 mole of H2 occupies 22.4dm3at s.t.p
∴ 32g of Zn will give out (22.4×32dm3 ofH2 at stp)/65=11.03dm3
EVALUATION:
State the law of multiple proportion
State the law of reciprocal proportions.
Zn(s) + 2HCl(aq) ZnCl2(aq)+ H2(g)
GENERAL EVALUATION:
ESSAY QUESTIONS
Copper(I)oxide Copper(II)oxide
Mass of sample oxide 30.4g 1.91g
Mass of copper residue 2.55g 1.38g
Mass of oxygen removed from oxide 0.49g 0.53g
From the above table, calculate the various masses of copper which would combine separately with a fixed mass of 1 g of oxygen.
2. What mass of copper will be produced from the reductionof 7.95g of copper (II) oxide? (C= 63.5, O= 16)
3. Write down the names of these chemical compounds.
(i) HNO3 (ii) CuCl2(iii) CaCO3(iv) Fe2O3
4. Write the symbol and the valency of the following.
(i) Boron (ii) Carbon (iii) Sulphur (iv) Argon
5. Calculate the formula of a compound with 31.9% potassium 28.93%, chlorine and the rest oxygen. K=39, Cl =355, O=16
OBJECTIVE TEST
Which of the following relative atomic numbers has empirical formula CH2O (H=!, C=12 , O=16). (a) 42 (b) 80 (c) 4
The relative atomic mass of tetraoxosulphate (vi) acid is? (a) 98 (b) 49 (b) 49 (c) 96 (d) 106
Chemical equations will provide all these except. (a) State of chemicals is solved )b) Direction of reaction (c) Mass of products (d) Reactants
All pure samples of a particles compound contain the same elements combined in the same proportion by mass. The statement is the law of . (a) Definite proportion (b) Multiple proportion (c)Conservation of mass or matter (d) Atomic proportion
PRE READING ASSIGNMENT:
Read above the periodic table.
WEEK 6
TOPIC: CHEMICAL COMBINATION
CONTENTS: (a) Periodic table (First 20 elements)
(b) Electronic configuration of atoms
(c) Types of bonds
(d) Systems of naming compounds (conventional and IUPAC)
Chemical combination: The atoms of noble gases e.g. Helium He, Neon Ne are very stable; because their outermost shells are completely filled with electrons. The tendency of the other elements is to attain this stable structure possessed by the noble gases. This is achieved during chemical combination. Thus, chemical bonding is the coming together of atoms of the same or different elements, in order to form a stable structure.
Sub-topic (1): The Periodic Table
Periodic table is the arrangement of chemical elements in order of their atomic numbers. Atomic number, Z, of an element is the number of protons in one atom of that element while, Mass number, A, of an element is the sum of the protons and neutrons in it.
Many scientists have attempted to classify chemical elements based on their properties. They include Newland, Lother Meye, Dpberainer and Mendeleev. The modern periodic table is based on Mendeleev’s original idea in 1869. The basic assumption behind the modern periodic table known as Periodic Law states that “the properties of the elements are a periodic function of their atomic number”.
The modern periodic Table
The modern periodic table is divided into vertical columns of elements called Groups and horizontal rows of elements called Periods. There are seven (7) periods and eight (8) groups. The first twenty (20) elements are:
1. Hydrogen 11. Sodium
2. Helium 12. Magnesium
3. Lithium 13. Aluminum
4. Beryllium 14. Silicon
5. Boron 15. Phosphorus
6. Carbon 16. Sulphur
7. Nitrogen 17. Chlorine
8. Oxygen 18. Argon
9. Fluorine 19. Potassium
10. Neon 20. Calcium
Periodic table of elements 1-20
EVALUATION
(i) .Define a periodic table
(ii) List the first twenty (20) elements
(iii) State the periodic law
(iv) How many groups and periods are there in the periodic table?
(v) In the periodic table, elements are arranged according to their _______
Sub-topic (b) Electronic configuration of atoms
The atom is composed of a central nucleus made up of protons and neutrons and; the shells or rings filled with systematically arranged electrons. The electrons are shown to be arranged in a definite pattern in an atomic model. For example, a helium atom with two electrons is represented showing the two electrons arranged in a single ring or shell. Similarly, a Lithium atom with three electrons has the electrons arranged in two shells: an inner shell with two electrons and an outer shell with one electron. Sodium with 11 electrons has 2 electrons in the inner shell (first shell), 8 in the second shell and 1 in the third shell.
M shell
L shell
K shell
Sodium
Maximum number of electrons for a shell
Starting from the shell nearest the nucleus, the shells are named as K, L, M, N, O, etc. Thus, the first shell is K-shell; the second shell is L-shell and so on. The maximum number of electron in a given shell can be obtained using the formula 2n2, where n represents the shell number. Thus, for the K-shell where n = 1, the maximum number of electrons it can contain is 2 x 12 = 2.
EVALUATION
1. The nucleus of an atom is made up of ______________ and _______________
2. Give the electronic configuration of Silicon
3. Obtain the maximum number of electrons in the second shell of Aluminium atom.
Sub-topic(c) Types of bonds (chemical combination
The attractive force between atoms when they combine chemically is called a chemical bond. There are two main types of chemical bonds namely (i) Strong bonds (ii) Weak bonds.
(i) Strong bonds are: (a). Electrovalent (or Ionic) (b). Covalent (c). Co-ordinate (or Dative) (d). Metallic
(a)Electrovalent bond: This involves the transfer of electrons from one atom, donor atom, (usually metallic) to another atom, acceptor atom, (usually non-metallic). The electrons involved reside in the outermost shells of the atoms and are called Valence electrons.
Characteristics of Electrovalent bonds
i. They have high melting and boiling points
ii. They are generally soluble in water
iii .They are good conductors of electricity when molten or in solution
iv. They do not conduct electricity when solid
v. The energy needed to separate them is relatively high.
(b)Covalent bonds: They are formed when electrons are shared equally by two atoms.
Characteristics of Covalent bonds
i. Low melting and boiling points.
ii. Energy required for separation is low
iii. Do not conduct electricity in the solid or molten state, or in solution
iv. They generally have a strong, easily noticeable smell
v. They are not easily soluble in water, but are usually soluble in organic solvents
(c) Co-ordinate covalent or Dative bond involves sharing of electrons as in the normal covalent bonding, but the shared pair is donated by only one of the participating atoms. For instance, Ammonia and water molecules possess lone pairs and so readily enter into coordinate covalent bonding.
(d) Metallic bond: The electrostatic force of attraction between the positive nuclei and the sea of mobile electrons is called metallic bond. Metallic bonding, therefore, is the process whereby the positively charged nuclei of metal atoms are simultaneously attracted to the sea (orcloud) of mobile electrons.
(ii)Weak bonds: These are intermolecular forces of attraction that hold atoms and covalent molecules together in gases, liquids and solids. The most common ones are: (a). Van der Waals forces (b). Hydrogen bond (c) Dipole-Dipole (Dipolar)
(d)Van der Waals forces: They were first described by J.D. van der Waals, and are known as van der Waals forces. They are weak short-ranged attractive forces formed between covalent molecules. They are the only attractive forces between the atoms of the noble gases and non-polar covalent molecules, and are responsible for the low melting and boiling point of covalent compounds. Due to increase in van der Waals forces, there is gradation in the physical properties of the Halogens: Fluorine and chlorine are gases; bromine is a liquid; and iodine is a solid.
(a) Hydrogen bond: This is an intermolecular force which arises when hydrogen is covalently linked to highly electronegative elements like nitrogen, oxygen and fluorine.
The presence of hydrogen bonds between H2O molecules is responsible for water being a liquid at room temperature and with a high boiling point; if not, it would have been a gas, like hydrogen H2S. HF is a liquid at room temperature, while HCl is a gas.
EVALUATION
What is a chemical bond?
List three (3) general properties of electrovalent compounds.
Define covalent bond and state its characteristics
State two (2) differences between covalent and electrovalent compounds.
Explain the term Metallic bonding
Sub-topic (d) Systems of naming compounds
Chemical compounds are named according to the International Union of Pure and Applied Chemistry system. In order to understand the basic principles behind the IUPAC system, a good knowledge of the concept of Oxidation Number (ON) is very essential.
Concept of oxidation number
An oxidation number (ON) is a positive or negative number assigned to an atom according to a set of rules. It is sometimes called Oxidation State.
Rules for assigning oxidation number
1. The ON of an uncombined free element, whether monoatomic or polyatomic is zero; e.g. Noble gases (He, Ne, etc), metals (Na, Zn, etc),solid non-metals (O3, N2, F2,etc)
2. The oxidation number of a monoatomic ion is equal in magnitude and sign to its ionic charge; e.g., the ON of bromide ion, Br-1, is -1; that of F+3, is +3.
3. The ON of one hydrogen atom is +1 in its compounds, except in hydrides of metals (e.g. NaH), where it is -1
4. For any neutral compound the sum of the ONs of all the atoms adds to zero.
5. The ON of oxygen in a compound is always -2 except in peroxides, H2O2, Na2O2, where it is -1
6. In any radical, the sum of the ON of all the atoms is equal to the charge on its ion.
Worked examples
1. Calculate the ON of copper in Cu2O
Solution
Let y represents the ON of each copper atom.
ON of one hydrogen atom, O is -2. (Rule 2)
In a neutral compound, sum of all ON is zero. (Rule 8)
Therefore, in Cu2O: Cu2O
2y + (-2) = 0 (rule 8)
2y = +2
y = +2/2 = +1
Thus, ON of a copper atom in Cu2O = +1
2. Determine the ON of X in X2O72-
Solution
ON of each O atom = -2 (Rule 2)
The net charge on the ion = -2 (Rule 7)
Therefore, in X2O72-
2X + 7(-2) = -2 (Rule 7)
2X – 14 = -2
2X = -2 + 14 = + 12
X = + 12/2 = +6
Therefore, ON of each X in X2O72- is +6
Naming of inorganic compounds
1. Binary compounds. Binary compounds contain two elements only. The metal is named first, followed by the name of the second element ending with –ide. If the metal is one that has variable valencies, the valency exhibited will be written in roman numeral examples are given below:
Name of compounds
Formula Conventional Name IUPAC Name
Na2O Sodium oxide
Fe2O3 Iron (III) oxide
CO Carbon monoxide Carbon (II) oxide
CO2 Carbon dioxide Carbon (IV) oxide
N2O Nitrous oxide Dinitrogen (I) oxide
2. Radicals. In naming radicals, the last element is mentioned first with its number of atoms given as mono (1), di (2), tri (3), tetra (4), penta (5), etc. The other element’s name ends with –ate. In naming acids, only the radicals are mentioned followed by the word acid See examples below:
Formula
Name
Na2CO3 Sodium trioxocarbonate (iv)
KMnO4 Potassium tetraoxomanganate (vii)
H2SO4 Tetraoxosulphate (vi)
HNO3 Trioxonitrate (iv) acid
If a compound has water of crystallization the number of molecules of water present is written as e.g. pentahydrate (5), decahydrate (10), etc. For example: NaCO3. 10H2O reads Sodium trioxocarbonate (iv) decahydrate.
EVALUATION
• Define oxidation number and, determine the ON of sulphur in SO32-
• Give the IUPAC name of the following: (i) Al(NO3)3 (ii) MnO2 (III) CuSO4 . 5H2O
• What is the correct IUPAC name for NO2- ?
GENERAL EVALUATION
OBJECTIVE TEST:
1. Which of these are found in the nucleus of an atom?
A. electrons and protons B. electrons and neutrons C. protons and neutrons D. photons and electrons E. photons and neutrons.
2. The type of bond between two atoms of an element with atomic number 7 is? A. ionic B. covalent C. hydrogen bond D. metallic bond E. coordinate covalent bond.
3. The ON of phosphorus, P in PH3is? A. -4 B. +2 C. -3 D. -1 E. +3
4. Give the IUPAC name of the compound NO2. A. nitrogen dioxide B. nitrogen monoxide C. nitrogen (II) oxide D. nitrogen (iv) oxide E. nitrogen oxide
5. Atomic number, Z, is the number of __________ in one atom an element. A. protons B. neurons C. electrons D. atoms E. ions
ESSAY QUESTIONS
1. By means of a diagram, show the arrangement of electrons in one atom of Sodium.
2. The electronic configurations for the metal calcium, the non-metals silicon and chlorine can be represented as:
Ca : 2, 8, 8, ; Si : 2, 8, 4 ; and Cl : 2, 8, 7. (a) Explain, in terms of electrons, the formation of calcium chloride and silicon chloride. (b) Give two (2) differences in physical properties you would expect between calcium chloride and silicon chloride.
3. Compare the characteristics of ionic with those covalent compounds.
Use this fig. to answer questions 4 and 5.
I II III IV V VI VII VIII
Y W
J X
Q M Z
4. Which of the following pairs of letters denotes elements containing the same number of electrons in their outermost shell?
5. What letter presents an element that participates in covalent rather than ionic bonding?
WEEKEND ASSIGNMENT:
New School Chemistry for Senior Secondary Schools by Osei Yaw Ababio; pages 55 – 67 and 141 – 145.
PRE–READING ASSIGNMENT:
Read about States of matter, its kinetic theory and application.
WEEKEND ACTIVITY:
Discuss the process of evaporation, diffusion and osmosis.
REFERENCE TEXTS:
5. New School Chemistry for Senior Secondary Schools by Osei Yaw Ababio; Africana First Publishers Plc.
6. Chemistry for Senior Secondary Schools 1 by Magbagbeola O, et al; Melrose Books and Publishers.
7. Comprehensive Certificate Chemistry by G N C Ohia, et al; University Press Plc.
CONTENTS: (a) Periodic table (First 20 elements)
(b) Electronic configuration of atoms
(c) Types of bonds
(d) Systems of naming compounds (conventional and IUPAC)
Chemical combination: The atoms of noble gases e.g. Helium He, Neon Ne are very stable; because their outermost shells are completely filled with electrons. The tendency of the other elements is to attain this stable structure possessed by the noble gases. This is achieved during chemical combination. Thus, chemical bonding is the coming together of atoms of the same or different elements, in order to form a stable structure.
Sub-topic (1): The Periodic Table
Periodic table is the arrangement of chemical elements in order of their atomic numbers. Atomic number, Z, of an element is the number of protons in one atom of that element while, Mass number, A, of an element is the sum of the protons and neutrons in it.
Many scientists have attempted to classify chemical elements based on their properties. They include Newland, Lother Meye, Dpberainer and Mendeleev. The modern periodic table is based on Mendeleev’s original idea in 1869. The basic assumption behind the modern periodic table known as Periodic Law states that “the properties of the elements are a periodic function of their atomic number”.
The modern periodic Table
The modern periodic table is divided into vertical columns of elements called Groups and horizontal rows of elements called Periods. There are seven (7) periods and eight (8) groups. The first twenty (20) elements are:
1. Hydrogen 11. Sodium
2. Helium 12. Magnesium
3. Lithium 13. Aluminum
4. Beryllium 14. Silicon
5. Boron 15. Phosphorus
6. Carbon 16. Sulphur
7. Nitrogen 17. Chlorine
8. Oxygen 18. Argon
9. Fluorine 19. Potassium
10. Neon 20. Calcium
Periodic table of elements 1-20
EVALUATION
(i) .Define a periodic table
(ii) List the first twenty (20) elements
(iii) State the periodic law
(iv) How many groups and periods are there in the periodic table?
(v) In the periodic table, elements are arranged according to their _______
Sub-topic (b) Electronic configuration of atoms
The atom is composed of a central nucleus made up of protons and neutrons and; the shells or rings filled with systematically arranged electrons. The electrons are shown to be arranged in a definite pattern in an atomic model. For example, a helium atom with two electrons is represented showing the two electrons arranged in a single ring or shell. Similarly, a Lithium atom with three electrons has the electrons arranged in two shells: an inner shell with two electrons and an outer shell with one electron. Sodium with 11 electrons has 2 electrons in the inner shell (first shell), 8 in the second shell and 1 in the third shell.
M shell
L shell
K shell
Sodium
Maximum number of electrons for a shell
Starting from the shell nearest the nucleus, the shells are named as K, L, M, N, O, etc. Thus, the first shell is K-shell; the second shell is L-shell and so on. The maximum number of electron in a given shell can be obtained using the formula 2n2, where n represents the shell number. Thus, for the K-shell where n = 1, the maximum number of electrons it can contain is 2 x 12 = 2.
EVALUATION
1. The nucleus of an atom is made up of ______________ and _______________
2. Give the electronic configuration of Silicon
3. Obtain the maximum number of electrons in the second shell of Aluminium atom.
Sub-topic(c) Types of bonds (chemical combination
The attractive force between atoms when they combine chemically is called a chemical bond. There are two main types of chemical bonds namely (i) Strong bonds (ii) Weak bonds.
(i) Strong bonds are: (a). Electrovalent (or Ionic) (b). Covalent (c). Co-ordinate (or Dative) (d). Metallic
(a)Electrovalent bond: This involves the transfer of electrons from one atom, donor atom, (usually metallic) to another atom, acceptor atom, (usually non-metallic). The electrons involved reside in the outermost shells of the atoms and are called Valence electrons.
Characteristics of Electrovalent bonds
i. They have high melting and boiling points
ii. They are generally soluble in water
iii .They are good conductors of electricity when molten or in solution
iv. They do not conduct electricity when solid
v. The energy needed to separate them is relatively high.
(b)Covalent bonds: They are formed when electrons are shared equally by two atoms.
Characteristics of Covalent bonds
i. Low melting and boiling points.
ii. Energy required for separation is low
iii. Do not conduct electricity in the solid or molten state, or in solution
iv. They generally have a strong, easily noticeable smell
v. They are not easily soluble in water, but are usually soluble in organic solvents
(c) Co-ordinate covalent or Dative bond involves sharing of electrons as in the normal covalent bonding, but the shared pair is donated by only one of the participating atoms. For instance, Ammonia and water molecules possess lone pairs and so readily enter into coordinate covalent bonding.
(d) Metallic bond: The electrostatic force of attraction between the positive nuclei and the sea of mobile electrons is called metallic bond. Metallic bonding, therefore, is the process whereby the positively charged nuclei of metal atoms are simultaneously attracted to the sea (orcloud) of mobile electrons.
(ii)Weak bonds: These are intermolecular forces of attraction that hold atoms and covalent molecules together in gases, liquids and solids. The most common ones are: (a). Van der Waals forces (b). Hydrogen bond (c) Dipole-Dipole (Dipolar)
(d)Van der Waals forces: They were first described by J.D. van der Waals, and are known as van der Waals forces. They are weak short-ranged attractive forces formed between covalent molecules. They are the only attractive forces between the atoms of the noble gases and non-polar covalent molecules, and are responsible for the low melting and boiling point of covalent compounds. Due to increase in van der Waals forces, there is gradation in the physical properties of the Halogens: Fluorine and chlorine are gases; bromine is a liquid; and iodine is a solid.
(a) Hydrogen bond: This is an intermolecular force which arises when hydrogen is covalently linked to highly electronegative elements like nitrogen, oxygen and fluorine.
The presence of hydrogen bonds between H2O molecules is responsible for water being a liquid at room temperature and with a high boiling point; if not, it would have been a gas, like hydrogen H2S. HF is a liquid at room temperature, while HCl is a gas.
EVALUATION
What is a chemical bond?
List three (3) general properties of electrovalent compounds.
Define covalent bond and state its characteristics
State two (2) differences between covalent and electrovalent compounds.
Explain the term Metallic bonding
Sub-topic (d) Systems of naming compounds
Chemical compounds are named according to the International Union of Pure and Applied Chemistry system. In order to understand the basic principles behind the IUPAC system, a good knowledge of the concept of Oxidation Number (ON) is very essential.
Concept of oxidation number
An oxidation number (ON) is a positive or negative number assigned to an atom according to a set of rules. It is sometimes called Oxidation State.
Rules for assigning oxidation number
1. The ON of an uncombined free element, whether monoatomic or polyatomic is zero; e.g. Noble gases (He, Ne, etc), metals (Na, Zn, etc),solid non-metals (O3, N2, F2,etc)
2. The oxidation number of a monoatomic ion is equal in magnitude and sign to its ionic charge; e.g., the ON of bromide ion, Br-1, is -1; that of F+3, is +3.
3. The ON of one hydrogen atom is +1 in its compounds, except in hydrides of metals (e.g. NaH), where it is -1
4. For any neutral compound the sum of the ONs of all the atoms adds to zero.
5. The ON of oxygen in a compound is always -2 except in peroxides, H2O2, Na2O2, where it is -1
6. In any radical, the sum of the ON of all the atoms is equal to the charge on its ion.
Worked examples
1. Calculate the ON of copper in Cu2O
Solution
Let y represents the ON of each copper atom.
ON of one hydrogen atom, O is -2. (Rule 2)
In a neutral compound, sum of all ON is zero. (Rule 8)
Therefore, in Cu2O: Cu2O
2y + (-2) = 0 (rule 8)
2y = +2
y = +2/2 = +1
Thus, ON of a copper atom in Cu2O = +1
2. Determine the ON of X in X2O72-
Solution
ON of each O atom = -2 (Rule 2)
The net charge on the ion = -2 (Rule 7)
Therefore, in X2O72-
2X + 7(-2) = -2 (Rule 7)
2X – 14 = -2
2X = -2 + 14 = + 12
X = + 12/2 = +6
Therefore, ON of each X in X2O72- is +6
Naming of inorganic compounds
1. Binary compounds. Binary compounds contain two elements only. The metal is named first, followed by the name of the second element ending with –ide. If the metal is one that has variable valencies, the valency exhibited will be written in roman numeral examples are given below:
Name of compounds
Formula Conventional Name IUPAC Name
Na2O Sodium oxide
Fe2O3 Iron (III) oxide
CO Carbon monoxide Carbon (II) oxide
CO2 Carbon dioxide Carbon (IV) oxide
N2O Nitrous oxide Dinitrogen (I) oxide
2. Radicals. In naming radicals, the last element is mentioned first with its number of atoms given as mono (1), di (2), tri (3), tetra (4), penta (5), etc. The other element’s name ends with –ate. In naming acids, only the radicals are mentioned followed by the word acid See examples below:
Formula
Name
Na2CO3 Sodium trioxocarbonate (iv)
KMnO4 Potassium tetraoxomanganate (vii)
H2SO4 Tetraoxosulphate (vi)
HNO3 Trioxonitrate (iv) acid
If a compound has water of crystallization the number of molecules of water present is written as e.g. pentahydrate (5), decahydrate (10), etc. For example: NaCO3. 10H2O reads Sodium trioxocarbonate (iv) decahydrate.
EVALUATION
• Define oxidation number and, determine the ON of sulphur in SO32-
• Give the IUPAC name of the following: (i) Al(NO3)3 (ii) MnO2 (III) CuSO4 . 5H2O
• What is the correct IUPAC name for NO2- ?
GENERAL EVALUATION
OBJECTIVE TEST:
1. Which of these are found in the nucleus of an atom?
A. electrons and protons B. electrons and neutrons C. protons and neutrons D. photons and electrons E. photons and neutrons.
2. The type of bond between two atoms of an element with atomic number 7 is? A. ionic B. covalent C. hydrogen bond D. metallic bond E. coordinate covalent bond.
3. The ON of phosphorus, P in PH3is? A. -4 B. +2 C. -3 D. -1 E. +3
4. Give the IUPAC name of the compound NO2. A. nitrogen dioxide B. nitrogen monoxide C. nitrogen (II) oxide D. nitrogen (iv) oxide E. nitrogen oxide
5. Atomic number, Z, is the number of __________ in one atom an element. A. protons B. neurons C. electrons D. atoms E. ions
ESSAY QUESTIONS
1. By means of a diagram, show the arrangement of electrons in one atom of Sodium.
2. The electronic configurations for the metal calcium, the non-metals silicon and chlorine can be represented as:
Ca : 2, 8, 8, ; Si : 2, 8, 4 ; and Cl : 2, 8, 7. (a) Explain, in terms of electrons, the formation of calcium chloride and silicon chloride. (b) Give two (2) differences in physical properties you would expect between calcium chloride and silicon chloride.
3. Compare the characteristics of ionic with those covalent compounds.
Use this fig. to answer questions 4 and 5.
I II III IV V VI VII VIII
Y W
J X
Q M Z
4. Which of the following pairs of letters denotes elements containing the same number of electrons in their outermost shell?
5. What letter presents an element that participates in covalent rather than ionic bonding?
WEEKEND ASSIGNMENT:
New School Chemistry for Senior Secondary Schools by Osei Yaw Ababio; pages 55 – 67 and 141 – 145.
PRE–READING ASSIGNMENT:
Read about States of matter, its kinetic theory and application.
WEEKEND ACTIVITY:
Discuss the process of evaporation, diffusion and osmosis.
REFERENCE TEXTS:
5. New School Chemistry for Senior Secondary Schools by Osei Yaw Ababio; Africana First Publishers Plc.
6. Chemistry for Senior Secondary Schools 1 by Magbagbeola O, et al; Melrose Books and Publishers.
7. Comprehensive Certificate Chemistry by G N C Ohia, et al; University Press Plc.
WEEK 7
TOPIC:CHEMICAL COMBINATION
CONTENT:
(1) (i) State of matter (ii) Solid (iii) Liquid (iv) Gasses state
(2) The kinetic theory
(3) The application of kinetic theory
SUB-TOPIC1: STATE OF MATTER- SOLID
Matter is made up of very tiny particles such as atoms, molecules and ions. Matter exists in three physical states namely; solid, liquid and gas. Matter has mass and occupies space.
Solid: The tiny particles in solid are packed very closely together hence they cannot move about. The presence of forces of attraction among the tiny particles of solid made them to be very closely packed together and allowing only vibration among them instead of moving about. This is what makes the solid to have a fixed shape and volume with high density. However, at high temperature, the molecules gain more energy and break the forces of attraction among the particles thereby causing the particles to move faster. As the temperature increases more and more, the solid melts and turns to liquid. Therefore, solid have fixed shape and volume, incompressible and very dense with the least kinetic energy when compared with those of liquid and gasses.
(ii)State of matter – liquid: The particles in the liquid are slightly further a [art than those in a solid and there exist among them weaker forces of attraction (weaker force of cohesion); thus allowing them to vibrate rotate and move about. Liquid have fixed volume and a fixed shape but takes the shape of its container. When the liquid is heated, its temperature increases, the forces of cohesion / attraction becomes progressively weaker. Eventually, a stage is reached when the molecules acquire enough energy to escape as a gas. The temperature at which this occurs is called the boiling point of the liquid. Therefore, the liquid have fixed volume, no fixed shape, less dense and incompressible with kinetic energy relatively higher than those of solid state.
STATE OF MATTER- GASEOUS STATE
The particles in the gaseous state are very freely with large distances between them. The particles of gases have (fakevids) no fixed volume and shape, compressible and least dense with the highest (ke) kinetic energy when compared with particles of other state of matter.
EVLUATION
Define the term matter.
Enumerate the three main state of matter.
Explain each of the least six points.
SUB-TOPIC 2: The Kinetic theory of Matter
EXPLANATION OF KINETIC THEORY
The kinetic theory of matter postulates that the tiny Particles of matter are continually moving and so possess kinetic energy. An increase in temperature causes an increase in the average kinetic energy of the particle.
Dalton’s experimental evidence to show that chemical compounds consists of molecules. Which are groups of atoms of various elements? The gas laws which explained the physical behaviour of gases can be explained by kinetic theory of gases. This theory describes the behaviour of an ideal or perfect gas. This is to say that the kinetic theory of gases explains quantitatively the properties of gas molecules and in so doing put up the following assumptions.
A gas consists of very tiny particles (usually molecules and atoms).
The cohesive forces of the gas molecules are negligible.
These particles are in constant random and rapid motion in straight lines.
As a result of these movements, collisions occur between the molecules and also with the walls of the container, hence the molecules exert pressure. No energy is lost when collision occupy. This means that the collisions are perfectly elastic.
The space between the molecules is very large compared to the size of the molecules. The molecules therefore have negligible (almost zero) volume compared to the volume of the container.
The average kinetic energy per molecule is the same for all gas samples at any given temperature. The absolute temperature of the gas is a measure of the average kinetic energy of the gas particles.
EVALUATION
He kinetic of the matter postulates ..........................
When the temperature of the particles of the molecules increases, what happens to kinetic energy?
Enumerate the six assumptions if the kinetic theory of matter.
SUB-TOPIC 3: The application of kinetic theory
The kinetic theory is useful in several respects and such aspects includes;
It provides reasonable explanations for the behaviour of gas.
It accounts for the gas laws
It explains important phenomenon such as diffusion
It provides a fundamental equation for gases.
PV =1/3 NMC2
EVALUATION
Mention four applications of kinetic theory of matter.
GENERAL EVALUATION
OBJECTIVE TEST :
The kinetic theory of matter states ____________
Water {H2O} exists as solid, liquid and gas respectively because. (a) water is colorless. (b) Water in any state possesses a certain degree of motion in the molecules.
Which of the three states of matter has no fixed volume and least dense.(a) Gas (b) Solid (c) Liquid
The presence of sodium chloride in ice will. (a) decrease or lower the boiling point of sodium chloride (b) Increase the melting point of sodium chloride (c) Make sodium chloride impure (d) Lower the freezing point of sodium chloride.
The escape of molecules with more than average kinetic energy of the molecules is called _____________ (A) Melting (b) Freezing (C) Evaporation (d) Efflorescence
ESSAY QUESTION:
Define the term matter
State the three state of matter
Explain two out of the three main sate of matter
List four importance of kinetic theory
Give assumptions of kinetic theory of matter.
WEEKEND ASSIGNMENT:
Read New School Chemistry for Senior Secondary Schools, by Osei Yaw Ababio pages 584-594
PRE- READING ASSIGNMENT
Read the meaning of chemical industry and its development.
WEEKEND ACTIVITY
Explain with at least three points what you understand by chemical industry and also mention specifically five types of chemical industry.
REFERENCE TEXT:
New School Chemistry for senior Secondary Schools by Osei Yaw Ababio 6th edition
Comprehensive certificate chemistry by G.N.C Ohia, G.I, Amasiatiu, J.O Ajagbe =, G.O. Ojokuku and U Mohammed. 2nd Edition
CONTENT:
(1) (i) State of matter (ii) Solid (iii) Liquid (iv) Gasses state
(2) The kinetic theory
(3) The application of kinetic theory
SUB-TOPIC1: STATE OF MATTER- SOLID
Matter is made up of very tiny particles such as atoms, molecules and ions. Matter exists in three physical states namely; solid, liquid and gas. Matter has mass and occupies space.
Solid: The tiny particles in solid are packed very closely together hence they cannot move about. The presence of forces of attraction among the tiny particles of solid made them to be very closely packed together and allowing only vibration among them instead of moving about. This is what makes the solid to have a fixed shape and volume with high density. However, at high temperature, the molecules gain more energy and break the forces of attraction among the particles thereby causing the particles to move faster. As the temperature increases more and more, the solid melts and turns to liquid. Therefore, solid have fixed shape and volume, incompressible and very dense with the least kinetic energy when compared with those of liquid and gasses.
(ii)State of matter – liquid: The particles in the liquid are slightly further a [art than those in a solid and there exist among them weaker forces of attraction (weaker force of cohesion); thus allowing them to vibrate rotate and move about. Liquid have fixed volume and a fixed shape but takes the shape of its container. When the liquid is heated, its temperature increases, the forces of cohesion / attraction becomes progressively weaker. Eventually, a stage is reached when the molecules acquire enough energy to escape as a gas. The temperature at which this occurs is called the boiling point of the liquid. Therefore, the liquid have fixed volume, no fixed shape, less dense and incompressible with kinetic energy relatively higher than those of solid state.
STATE OF MATTER- GASEOUS STATE
The particles in the gaseous state are very freely with large distances between them. The particles of gases have (fakevids) no fixed volume and shape, compressible and least dense with the highest (ke) kinetic energy when compared with particles of other state of matter.
EVLUATION
Define the term matter.
Enumerate the three main state of matter.
Explain each of the least six points.
SUB-TOPIC 2: The Kinetic theory of Matter
EXPLANATION OF KINETIC THEORY
The kinetic theory of matter postulates that the tiny Particles of matter are continually moving and so possess kinetic energy. An increase in temperature causes an increase in the average kinetic energy of the particle.
Dalton’s experimental evidence to show that chemical compounds consists of molecules. Which are groups of atoms of various elements? The gas laws which explained the physical behaviour of gases can be explained by kinetic theory of gases. This theory describes the behaviour of an ideal or perfect gas. This is to say that the kinetic theory of gases explains quantitatively the properties of gas molecules and in so doing put up the following assumptions.
A gas consists of very tiny particles (usually molecules and atoms).
The cohesive forces of the gas molecules are negligible.
These particles are in constant random and rapid motion in straight lines.
As a result of these movements, collisions occur between the molecules and also with the walls of the container, hence the molecules exert pressure. No energy is lost when collision occupy. This means that the collisions are perfectly elastic.
The space between the molecules is very large compared to the size of the molecules. The molecules therefore have negligible (almost zero) volume compared to the volume of the container.
The average kinetic energy per molecule is the same for all gas samples at any given temperature. The absolute temperature of the gas is a measure of the average kinetic energy of the gas particles.
EVALUATION
He kinetic of the matter postulates ..........................
When the temperature of the particles of the molecules increases, what happens to kinetic energy?
Enumerate the six assumptions if the kinetic theory of matter.
SUB-TOPIC 3: The application of kinetic theory
The kinetic theory is useful in several respects and such aspects includes;
It provides reasonable explanations for the behaviour of gas.
It accounts for the gas laws
It explains important phenomenon such as diffusion
It provides a fundamental equation for gases.
PV =1/3 NMC2
EVALUATION
Mention four applications of kinetic theory of matter.
GENERAL EVALUATION
OBJECTIVE TEST :
The kinetic theory of matter states ____________
Water {H2O} exists as solid, liquid and gas respectively because. (a) water is colorless. (b) Water in any state possesses a certain degree of motion in the molecules.
Which of the three states of matter has no fixed volume and least dense.(a) Gas (b) Solid (c) Liquid
The presence of sodium chloride in ice will. (a) decrease or lower the boiling point of sodium chloride (b) Increase the melting point of sodium chloride (c) Make sodium chloride impure (d) Lower the freezing point of sodium chloride.
The escape of molecules with more than average kinetic energy of the molecules is called _____________ (A) Melting (b) Freezing (C) Evaporation (d) Efflorescence
ESSAY QUESTION:
Define the term matter
State the three state of matter
Explain two out of the three main sate of matter
List four importance of kinetic theory
Give assumptions of kinetic theory of matter.
WEEKEND ASSIGNMENT:
Read New School Chemistry for Senior Secondary Schools, by Osei Yaw Ababio pages 584-594
PRE- READING ASSIGNMENT
Read the meaning of chemical industry and its development.
WEEKEND ACTIVITY
Explain with at least three points what you understand by chemical industry and also mention specifically five types of chemical industry.
REFERENCE TEXT:
New School Chemistry for senior Secondary Schools by Osei Yaw Ababio 6th edition
Comprehensive certificate chemistry by G.N.C Ohia, G.I, Amasiatiu, J.O Ajagbe =, G.O. Ojokuku and U Mohammed. 2nd Edition
WEEK 8
TOPIC: CHEMICAL INDUSTRIES
SUBJECT: CHEMISTRY
CLASS: SS1
CONTENT: (a) Types of Chemical industries
(i.) Raw materials used in chemical industries and their sources
(ii)Division of the chemical industries: heavy and fine chemicals,
Fertilizers, plastics, metallurgy, pharmaceutical, glass, ceramics, paints,
Cements, soap and detergent.
(b) Importance of chemicals industries to individuals and the nation
(c) Excursion to chemical industries.
SUB-TOPIC 1: Types of Chemical industries/divisions
Chemical industries are easily categories base on the type of products they manufacture. The major chemical industries are listed and explained below:
1. Heavy chemical industries
2. Fine chemical industries
3. Pharmaceutical industries
4. Fertilizer industries
5. Cement industries
6. Plastic industries
7. Ceramic industries
8. Food and Beverages industries
9. Glass industries
10. Metallurgical industries
11. Soaps and Detergents industries
12. Paint industries
13. Cosmetic industries
1. HEAVY CHEMICALS INDUSTRIES: These industries produce chemicals used as raw materials by other chemical industries. Their products are usually manufactured in very large quantities because many industries depend on their product and so are in high demand. Some heavy chemicals and some of their uses are listed below. These industries have nature as their sources of raw materials.
CHEMICAL INDUSTRIES
• Tetraoxosulphate (iv) acid (H2SO4) which is the most important of them all.it is used in the fertilizer industries.
• Sodium hydroxide (NaOH) and potassium hydroxide (KOH) used in soap making, fiber and paper industries.
• Sodium trioxocarbonate (iv) (NaCO3) is used in glass, detergent and water works.
• Sodium hydrogen trioxocarbonate (iv), (NaHCO3) is used in baking powder industries.
• Sulpur (iv) oxide is used in the manufacture of tetraoxosulphate (iv) acid (H2SO4).
• Chlorine (Cl2) is used in textile and paper industries and also for bleaching.
2. FINE CHEMICALS INDUSTRIES: Fine chemicals are produced in smaller quantities than heavy but are equally important. These industries pay great attention to purity products because the chemicals are used for sensitive and specific purpose. Some products like drugs, cosmetics, additive for vehicle engine performances, analytical reagents are some of the examples.
3. PHARMACEUTICAL INDUSTRIES: Produces various chemicals used in hospitals e.g. potassium bromide (used as sedatives ), mercury(ii) nitrate (v) (used as antiseptic for skin diseases)calcium sulphate (plaster of Paris),iodine, antibiotics,insulin,magnesium tetraoxosulphate(vi) .7H2O (Epsom salt),etc.
4. FERTILIZER INDUSTRIES: This industry produces fertilizers of various grades such as, (NH4) SO4, NH4NO3 and urea as well as pesticides, germicides, herbicides and fungicides.
5. CEMENT INDUSTRIES: These industries are important to us beacause their products are used in the construction industries in bulding bridges , house, roads, and drainages.Cement becomes a very hard substances when mixed with water,sand and gravels.Cement contains quicklime powder (calcium oxide),clay(silicon (iv)oxide)and some aluminium oxide. All these chemicals are
obtained naturallyfrom the earth in form of limestone(calcium trioxocarbonate(iv),and aliuminim oxide.when heated strongly,the lime stone changes to quicklimeand carbon (iv)oxide.
6. PLASTIC INDUSTRIES:These use simple organic compounds like ethane, ethyne phenol, and benzene and styrene obtained from the petroleum and coal tar industries to make plastics in form of pellets or granules. Other raw materials for plastics are phenol, Cellulose and vinyl Chloride.
There are two main types of plastics, thermoset and thermoplastic .Thermoplastic materials like plastic bottles, polyethene bags and some household’s materials can be remoulded after melting. These can be recycled by some factories to form other materials for human use. Thermoset Plastics cannot be recycled, but are stronger and more durable and so are usually used for furniture.
PLASTICS
7. GLASS INDUSTRIES: Glass is made from sand and it is cheap when compared with other materials. Silicon (iv) oxide in sand when treated with other chemical like lead from glass. These industries produce a lot of materials like glass waves (cups, bowls etc.), window panes, laboratory wares and apparatus, light bulbs and mirrors etc. Glass can be colored or transparent depending on its use.
RAW MATERIALS
Raw materials are starting materials ingredients for manufacturing. The raw materials for a product may be the products of another industry. For example:-the raw materials for the plastic industry are products of the petroleum industry.
Most primary (or starting) raw materials are obtained from nature e.g. petroleum, metal ores, common salt from the sea, coal, limestone, clay, sand, etc. These primary raw materials are used to produce other materials or chemicals that are used to produced other materials or chemicals. Some examples are listed in the table below:
Raw materials from some chemicals and their by products
Natural Raw materials Some Chemical gotten from it Materials
Petroleum Petrochemicals eg.Ethane,Benzene,Ethene etc. Plastic, Pesticides, Alkanols (alcohol),
beverages, fertilizers etc.
Earth Limestone cement
Metals ores Metallic products like containers, roofing sheet, vehicle bodies etc.
Clay/sand Ceramic products like floor tiles,dishwares,bathtub,toiletbows etc.
Plants Palm oil Soap and magrine
Cocoa Chocolate and beverages.
Rubber Tyres and other rubber products.
Sea water Common salt(sodium chloride) Seasoning for cooking and starting materials for many chemicals.
Air Oxygen Used in welding and hospitals.
Nitrogen Used to produced ammonia an important raw materials in the manufacture of fertilizers.
EVALUATION
1. Which chemical industry would you associate these with
(a)Tetraoxosulphate (vi) acid (b) paracetamol(c) floor tiles (d) Yoghourt (e) Baby powder
2. Name four heavy chemicals and fine chemicals and their uses in the chemical industries
3. Name any four divisions of chemicals industries. Mention one product produced by each of them.
4. Mention one type of chemical industry that utilizes ethane as raw materials.
SUB- TOPIC 2: IMPORTANCE OF CHEMICALS INDUSTRIES
(A) To an individual:
(i) Chemical industries provide employment for both skilled and unskilled.
(ii)It is used for treating of woods.
(B) To the nation
(i)The chemical industries provide earnings from foreign trade.
(ii)Chemical industries improved the standard of living by providing many
materials for domestic use (which can be too expensive if imported).
ENVIRONMENTAL PROBLEMS CREATED BY CHEMICAL INDUSTRIES
Despite all the advantages of these industries, they can constitute health hazards and menace to the communities where they are located. Such as:-
i. Its waste can contaminate the air we breathe and water we drink.
ii. It is also noisy. To avoid this, careful thought may be placed on the selection of a site to locate an industry.
EVALUATION
1. Write one importance of chemical industry to you as a person and two to the nation.
2. How can the activities of these industries affect the communities?
SUB- TOPIC 3: Excursion to chemical industries.
Your teacher or your science club should organize a visit to some of the local industries which make use of the raw materials we have so far discussed. This could be a soap factory, a cement factory, or even a petroleum refinery in the country.
EVALUATION
Write a compressive report of your excursion and submit it to your teacher.
GENERAL EVALUATION
OBJECTIVE TEST
1. An elastic rubber is formed by a process known as
(A)Vulcanization (B) thermo plasticity(C) polymerization (D) thermosetting.
2. Factors which can contribute to environmental pollution include
i. Overpopulation ii. Chemical warfare
iii. Agricultural activities iv. Industrialization
(A) I only (B)I and II only (C) I,II,III and IV.(D) IV only
3. One of the adverse effects of chemical industries on the community is (A) increased population (B) increased job opportunities (C) increased earning power (D) increased environmental pollution.
4. The major raw material in a plastic industry is
(A) Ethanol (B) Sulphur (C) methylethanoate (D) Ethane.
5. The following are heavy chemicals except(A) Tetraoxosulphate(iv)acid(B)Caustic soda(C) Sodium trioxocarbonate (iv).(D) Ethane
ESSAY QUESTIONS:
1. Name two of raw materials used in the manufacture of the following:
i.Polythene ii.Margarine.iii. Cement.
2. Name two industries and their locations where cement is produces.
3. List two industrial processes in which limestone is used as raw materials.
4. Name two uses of cement.
5. State one problem associated with the oil producing areas.
WEEKEND ASSIGNMENT
Read comprehensive certificate chemistry by G N C Ohia et al; (pages 134 -140).
PRE-READING ASSIGNMENT
Read the major factors that determine siting of chemical industries.
WEEKEND ACTIVITY:
Suggest one method of disposing polythene materials.
REFRENCE TEXTS:
1. Comprehensive certificate chemistry for senior secondary schools by G N C Ohia.
2. New school chemistry for senior secondary schools by Osei yaw Ababio
SUBJECT: CHEMISTRY
CLASS: SS1
CONTENT: (a) Types of Chemical industries
(i.) Raw materials used in chemical industries and their sources
(ii)Division of the chemical industries: heavy and fine chemicals,
Fertilizers, plastics, metallurgy, pharmaceutical, glass, ceramics, paints,
Cements, soap and detergent.
(b) Importance of chemicals industries to individuals and the nation
(c) Excursion to chemical industries.
SUB-TOPIC 1: Types of Chemical industries/divisions
Chemical industries are easily categories base on the type of products they manufacture. The major chemical industries are listed and explained below:
1. Heavy chemical industries
2. Fine chemical industries
3. Pharmaceutical industries
4. Fertilizer industries
5. Cement industries
6. Plastic industries
7. Ceramic industries
8. Food and Beverages industries
9. Glass industries
10. Metallurgical industries
11. Soaps and Detergents industries
12. Paint industries
13. Cosmetic industries
1. HEAVY CHEMICALS INDUSTRIES: These industries produce chemicals used as raw materials by other chemical industries. Their products are usually manufactured in very large quantities because many industries depend on their product and so are in high demand. Some heavy chemicals and some of their uses are listed below. These industries have nature as their sources of raw materials.
CHEMICAL INDUSTRIES
• Tetraoxosulphate (iv) acid (H2SO4) which is the most important of them all.it is used in the fertilizer industries.
• Sodium hydroxide (NaOH) and potassium hydroxide (KOH) used in soap making, fiber and paper industries.
• Sodium trioxocarbonate (iv) (NaCO3) is used in glass, detergent and water works.
• Sodium hydrogen trioxocarbonate (iv), (NaHCO3) is used in baking powder industries.
• Sulpur (iv) oxide is used in the manufacture of tetraoxosulphate (iv) acid (H2SO4).
• Chlorine (Cl2) is used in textile and paper industries and also for bleaching.
2. FINE CHEMICALS INDUSTRIES: Fine chemicals are produced in smaller quantities than heavy but are equally important. These industries pay great attention to purity products because the chemicals are used for sensitive and specific purpose. Some products like drugs, cosmetics, additive for vehicle engine performances, analytical reagents are some of the examples.
3. PHARMACEUTICAL INDUSTRIES: Produces various chemicals used in hospitals e.g. potassium bromide (used as sedatives ), mercury(ii) nitrate (v) (used as antiseptic for skin diseases)calcium sulphate (plaster of Paris),iodine, antibiotics,insulin,magnesium tetraoxosulphate(vi) .7H2O (Epsom salt),etc.
4. FERTILIZER INDUSTRIES: This industry produces fertilizers of various grades such as, (NH4) SO4, NH4NO3 and urea as well as pesticides, germicides, herbicides and fungicides.
5. CEMENT INDUSTRIES: These industries are important to us beacause their products are used in the construction industries in bulding bridges , house, roads, and drainages.Cement becomes a very hard substances when mixed with water,sand and gravels.Cement contains quicklime powder (calcium oxide),clay(silicon (iv)oxide)and some aluminium oxide. All these chemicals are
obtained naturallyfrom the earth in form of limestone(calcium trioxocarbonate(iv),and aliuminim oxide.when heated strongly,the lime stone changes to quicklimeand carbon (iv)oxide.
6. PLASTIC INDUSTRIES:These use simple organic compounds like ethane, ethyne phenol, and benzene and styrene obtained from the petroleum and coal tar industries to make plastics in form of pellets or granules. Other raw materials for plastics are phenol, Cellulose and vinyl Chloride.
There are two main types of plastics, thermoset and thermoplastic .Thermoplastic materials like plastic bottles, polyethene bags and some household’s materials can be remoulded after melting. These can be recycled by some factories to form other materials for human use. Thermoset Plastics cannot be recycled, but are stronger and more durable and so are usually used for furniture.
PLASTICS
7. GLASS INDUSTRIES: Glass is made from sand and it is cheap when compared with other materials. Silicon (iv) oxide in sand when treated with other chemical like lead from glass. These industries produce a lot of materials like glass waves (cups, bowls etc.), window panes, laboratory wares and apparatus, light bulbs and mirrors etc. Glass can be colored or transparent depending on its use.
RAW MATERIALS
Raw materials are starting materials ingredients for manufacturing. The raw materials for a product may be the products of another industry. For example:-the raw materials for the plastic industry are products of the petroleum industry.
Most primary (or starting) raw materials are obtained from nature e.g. petroleum, metal ores, common salt from the sea, coal, limestone, clay, sand, etc. These primary raw materials are used to produce other materials or chemicals that are used to produced other materials or chemicals. Some examples are listed in the table below:
Raw materials from some chemicals and their by products
Natural Raw materials Some Chemical gotten from it Materials
Petroleum Petrochemicals eg.Ethane,Benzene,Ethene etc. Plastic, Pesticides, Alkanols (alcohol),
beverages, fertilizers etc.
Earth Limestone cement
Metals ores Metallic products like containers, roofing sheet, vehicle bodies etc.
Clay/sand Ceramic products like floor tiles,dishwares,bathtub,toiletbows etc.
Plants Palm oil Soap and magrine
Cocoa Chocolate and beverages.
Rubber Tyres and other rubber products.
Sea water Common salt(sodium chloride) Seasoning for cooking and starting materials for many chemicals.
Air Oxygen Used in welding and hospitals.
Nitrogen Used to produced ammonia an important raw materials in the manufacture of fertilizers.
EVALUATION
1. Which chemical industry would you associate these with
(a)Tetraoxosulphate (vi) acid (b) paracetamol(c) floor tiles (d) Yoghourt (e) Baby powder
2. Name four heavy chemicals and fine chemicals and their uses in the chemical industries
3. Name any four divisions of chemicals industries. Mention one product produced by each of them.
4. Mention one type of chemical industry that utilizes ethane as raw materials.
SUB- TOPIC 2: IMPORTANCE OF CHEMICALS INDUSTRIES
(A) To an individual:
(i) Chemical industries provide employment for both skilled and unskilled.
(ii)It is used for treating of woods.
(B) To the nation
(i)The chemical industries provide earnings from foreign trade.
(ii)Chemical industries improved the standard of living by providing many
materials for domestic use (which can be too expensive if imported).
ENVIRONMENTAL PROBLEMS CREATED BY CHEMICAL INDUSTRIES
Despite all the advantages of these industries, they can constitute health hazards and menace to the communities where they are located. Such as:-
i. Its waste can contaminate the air we breathe and water we drink.
ii. It is also noisy. To avoid this, careful thought may be placed on the selection of a site to locate an industry.
EVALUATION
1. Write one importance of chemical industry to you as a person and two to the nation.
2. How can the activities of these industries affect the communities?
SUB- TOPIC 3: Excursion to chemical industries.
Your teacher or your science club should organize a visit to some of the local industries which make use of the raw materials we have so far discussed. This could be a soap factory, a cement factory, or even a petroleum refinery in the country.
EVALUATION
Write a compressive report of your excursion and submit it to your teacher.
GENERAL EVALUATION
OBJECTIVE TEST
1. An elastic rubber is formed by a process known as
(A)Vulcanization (B) thermo plasticity(C) polymerization (D) thermosetting.
2. Factors which can contribute to environmental pollution include
i. Overpopulation ii. Chemical warfare
iii. Agricultural activities iv. Industrialization
(A) I only (B)I and II only (C) I,II,III and IV.(D) IV only
3. One of the adverse effects of chemical industries on the community is (A) increased population (B) increased job opportunities (C) increased earning power (D) increased environmental pollution.
4. The major raw material in a plastic industry is
(A) Ethanol (B) Sulphur (C) methylethanoate (D) Ethane.
5. The following are heavy chemicals except(A) Tetraoxosulphate(iv)acid(B)Caustic soda(C) Sodium trioxocarbonate (iv).(D) Ethane
ESSAY QUESTIONS:
1. Name two of raw materials used in the manufacture of the following:
i.Polythene ii.Margarine.iii. Cement.
2. Name two industries and their locations where cement is produces.
3. List two industrial processes in which limestone is used as raw materials.
4. Name two uses of cement.
5. State one problem associated with the oil producing areas.
WEEKEND ASSIGNMENT
Read comprehensive certificate chemistry by G N C Ohia et al; (pages 134 -140).
PRE-READING ASSIGNMENT
Read the major factors that determine siting of chemical industries.
WEEKEND ACTIVITY:
Suggest one method of disposing polythene materials.
REFRENCE TEXTS:
1. Comprehensive certificate chemistry for senior secondary schools by G N C Ohia.
2. New school chemistry for senior secondary schools by Osei yaw Ababio
