Year 10 1st Term Chemistry Notes PDF

Subject : CHEMISTRY Term: 1st TERM Session : 2021/2022 School: CHRISLAND SCHOOLS LIMITED. Class : YEAR 10 Educator :  TABLE OF CONTENT WEEK 1: introduction to chemistry WEEK 2: separation techniques WE...

Subject : CHEMISTRY Term: 1st TERM Session : 2021/2022 School: CHRISLAND SCHOOLS LIMITED. Class : YEAR 10 Educator :  TABLE OF CONTENT WEEK 1: introduction to chemistry WEEK 2: separation techniques WEEK 4: particulate nature of matter( Dalton’s WEEK 3: separation techniques continues atomic theory) WEEK 5: particulate nature of matter continues WEEK 6: mid-term WEEK 7: formulae and equations WEEK 8: formulae and stoichiometry WEEK 9: concept of oxidation number WEEK 10: Revision WEEK 11: WEEK 12: WEEK 13:......: T1 WK 1 Topic : Introduction to Chemistry Learning objectives Point: define chemistry as a i. Define chemistry as a subject subject ii. Define scientific method Point : define scientific method and list the methods and list the methods iii. List the branches of Point : list the branches of chemistry chemistry iv. State the uses and Point : state the uses and adverse effect adverse effect of of chemistry chemistry HOME  INTRODUCTION TO CHEMISTRY Chemistry is a science that deals with the composition, properties and uses of matter. It probes into the principles governing the changes that matter undergoes. Branches Of Chemistry They include: a. Physical Chemistry, b. Inorganic Chemistry, c. Organic Chemistry d. Analytical Chemistry e. Environmental Chemistry f. Polymer Chemistry g. Biochemistry, et cetera. Uses Of Chemistry Chemistry has contributed greatly in the following areas: i. Food, ii. Clothing, iii. Housing, iv. Medicine, v. Transportation, et cetera  Adverse Effects Of Chemistry Chemical processes and products have affected our lives adversely in the following ways: I. Pollution, II. Drug abuse, III. Poisoning, IV. Corrosion, et cetera  Career Paths Or Prospects In Chemistry A few of the possible career fields include: I. Chemical engineering, II. Consumer product Chemistry, III. Environmental Chemistry, IV. Forensic Chemistry, V. Medical Chemistry, VI. Mechanical engineering, VII. Metallurgical and Materials engineering,  Scientific Method This is a step-by-step approach to the solutions of observable problems in science. It is the foundation of all scientific researches. Scientific method mainly includes: I. Observation, II. Hypothesis, III. Experiments, IV. Theory, V. Scientific law. Observations: These involve noting and recording facts about natural occurrences with the aid of our senses. Hypotheses: These are possible or reasonable guesses or explanations to our observations given without testing or experimenting. Experiment: This is a sequence of observations carried out under controlled conditions. Experiments are carried out to know whether the hypothesis is true or false.  Theory: This is a tested hypothesis which is found to be correct or true within the limits of available evidence. Scientific Law: This is a summary of the observable features of a substance established only after the theory has been extensively tested and proven true without any exception. Flow chart of Scientific method  Class Exercise 1. What is chemistry? 2. What is scientific method? 3. List four scientific methods.  Assignment 1 1. Explain the relevance of chemistry to food production. 2. Explain the following: hypothesis; theory; scientific law 3. Clearly, differentiate chemical hypothesis from chemical law.......: T1 WK 2 Topic : separation techniques Learning objectives i. Give the types of Point: give the types of changes in matter changes in matter ii. State the properties of physical and chemical Point : state the properties of physical and chemical changes changes iii. Give the differences Point : give the differences between between mixture and mixture and compound compound iv. Explain filtration, sieving, decantation and Point : explain filtration, sieving, decantation and evaporation evaporation HOME  Assessment questions 1. Mention the two types of changes undergone by matter. 2. Define physical and chemical changes respectively. 3. State 3 characteristics each of physical and chemical changes. 4. Give 4 examples each of physical and chemical changes.  Changes undergone by matter Matter undergoes two major changes. They are i. Physical change ii. Chemical change  Physical change Physical change is a change which is reversible and no new substances are formed.  Characteristics of Physical Change 1. No new substances are formed. 2. It is reversible. 3. The mass of the substance does not change. 4. No great heat changes except the latent heat changes.  Examples of Physical Change 1. Melting of substances (candle wax) 2. Boiling of liquids 3. Sublimation of iodine 4. Evaporation of liquids 5. Magnetization and demagnetization 6. Condensation of gases 7. Freezing of liquids 8. Dissolution of salts  Chemical Change Chemical change is a change which is not easily reversible and new substances are formed. Characteristics of Chemical Change 1. It is not easily reversible. 2. New substances are formed. 3. It is accompanied by large heat changes. 4. The mass of the substance changes.  Examples of Chemical Change 1. Rusting of iron 2. Burning of substances 3. Decay of substances 4. Addition of water to quick lime 5. Fermentation 6. Addition of sodium metal into water to form caustic soda.  Classification Of Substances Matter is classified chemically into: I. Elements, II. Compounds, and III. Mixtures.  Elements An element is a pure substance that cannot be split up into two or more simpler units by ordinary physical or chemical processes. An element is made up of only one type of atom. Each element has its own type of atom. This means that the atoms of one element are not the same as the atoms of another element. Examples of elements include Ca, C, H, Fe, Hg, Ne, etc  Compounds A compound is a pure substance that contains two or more elements chemically combined. Examples are given below Compound Elements present Common salts Sodium, chlorine (sodium chloride) Carbon(IV) oxide Carbon, Oxygen Marble ( Calcium Calcium, carbon, oxygen trioxocarbonate (IV))..........: T1 WK 6 Topic : CHEMICAL CLASSIFICATION OF MATTER CHARACTERISTICS OF COMPOUNDS 1. A compound is always homogeneous. 2. The component elements are chemically combined together and cannot be separated by physical means. 3. The components are present in a fixed ratio by mass. 4. A compound can be represented by a chemical formula. 5. The properties of a compound differ entirely from those of its component elements. HOME  Mixtures Mixtures are substances formed when two or more substances (elements and/or compounds) physically or not chemically combined. Examples include Mixture Constituents Air Oxygen, carbon (IV) oxide, nitrogen, rare gases, dust particles, moisture Soil Sand, clay, humus, water, air, mineral salts Blood Water, proteins, fat, oil, sugar, mineral salts, vitamins, hormones, enzymes, blood cells, haemoglobin.........: T1 WK 7 Topic : CHEMICAL CLASSIFICATION OF MATTER CHARACTERISTICS OF MIXTURES 1. It may be homogeneous or heterogeneous. 2. The constituents can be separated by physical methods. 3. A mixture cannot be represented by a chemical formula. 4. The properties of a mixture are the sum of those of its individual constituents. HOME  Differences Between Compounds And Mixtures Compound Mixture It is always homogeneous. It may be homogeneous or heterogeneous The component elements are The constituents are physically chemically combined and cannot combined. be separated by physical means. The component elements are The constituents can be added in present in fixed ratio by mass. any ratio by mass. Hence, a Hence, they can be represented mixture cannot be represented by chemical formulae. by a chemical formula. The properties of a compound The properties of a mixture are differ entirely from those of its sum of those of its individual constituent elements. constituents.  Assignment 2 A 1. Is there any difference between elements and compounds? If yes! Give reason(s) for your answer. If no! Give reason(s) for your answer. (4mks) 2. State the similarity between elements and compounds. (2mks)  Choice Of Separation Technique The method used for separation depends on the following factors: 1. Boiling point 2. Solubility 3. Magnetic property 4. Particle size 5. Melting point and 6. Nature of the components of the mixture.  CHOICE OF SEPARATION TECHNIQUE The method employed for separation depends on such factors such as Boiling point Solubility Magnetic property Particle size Melting point and Nature of the components of the mixture.  SEPARATION OF SOLID FROM LIQUID The methods below can be used to separate solid from liquids. Filtration Decantation and Evaporation to Dryness SEPARATION OF SOLID FROM LIQUID Filtration: This method is used to separate insoluble solid from liquid. A simple filtration involves pouring the mixture into a folded filter paper placed in a funnel above a conical flask. The clear liquid that comes out of the filter paper is called FILTRATE while the solid on the filter paper is called RESIDUE. Brewery, pharmaceuticals and water works make use of this technique. 1 = Glass rod 2 = Retort stand with clamp 3 = Sand particles in water (mixture) 4 = Filter paper 5 = Glass funnel 6 = Beaker 7 = Filtrate 8 = Residue  Decantation This method is also used to separate a mixture of insoluble solid such as mud in muddy water from liquid. The process involves allowing the solid to settle and then carefully pouring off the top clear liquid leaving the solid behind. The disadvantage of this method is that it does not ensure complete separation. Settled solid particles in a liquid Separation of solid particles from liquid by decantation  Evaporation To Dryness This technique is used to separate a mixture of soluble solid particles in a liquid medium by heating until the liquid evaporates completely. A solution of sodium chloride in water can be separated using this method. The salt solution (mixture) is poured into an evaporating dish and then heated until all the solvents evaporate using either: i. Direct heating; or ii. Indirect heating (by using water bath, steam bath or sand bath) The disadvantage of direct heating method is that it can not be used for heat labile salts (that is, salts that decompose on heating).Salt making industries employ this method. Separation of salt from water by evaporation to dryness......: T1 WK 3 Topic : separation techniques At the end of the Point: explain fractional lesson, students distillation should be able to: Point : explain sublimation Point : sieving and separating funnel Point : chromatography HOME  Separation Of Solids From Solids Solids may be separated from one another using the following techniques: 1. Sieving, 2. Magnetic separation, 3. Sublimation.  Sublimation Sublimation is a process where a substance can change from solid to gaseous state without melting. The reverse of this is also sublimation or deposition. Examples of substances that sublime are iodine; ammonium chloride; benzoic acid; carbon (IV) oxide and naphthalene As a separation technique, it is used in the separation of mixtures of solids that sublime from those that do not by heating. For instance, a mixture of ammonium chloride and sodium chloride can be separated as shown below.  Sieving This is a technique used in separating a mixture of solids of different sizes using a sieve. The solid is poured onto the sieve and then agitated. The agitation allow particles with smaller size than that of the mesh to pass through leaving behind the larger particles. This method is used by garri-making industries in West Africa. Sieves Sieving  Magnetic Separation This method is used in the separation of magnetic materials from non-magnetic ones by using a magnet. The magnetic material is attracted to the magnet leaving the non – magnetic substance behind. The impurities in tin ore are separated using this method. It can also be used to separate a mixture of iron filings and sand particles. MAGNETIC SEPARATION  Assignment 2 B 1. What makes it possible for separation of the components of mixtures? (2mks) 2. Why is filtration preferred to decantation? (2mks)  Separation Of Liquid From Liquid This can be done using: Distillation Fractional Distillation Separating Funnel  Distillation This method is used to separate pure liquid from solution e.g. pure water from sea water. The method involves vaporizing the liquid and then condensing the vapor to obtain pure liquid.  Distillation is used to separate a mixture of two or more miscible liquids whose boiling points are far apart. This method is used to get distilled water in the laboratory. Other examples include the following mixtures: Ethanol (b.p. 780C) and water (b.p. 1000C) Distillation  Fractional Distillation This technique is used to separate a mixture of two or more miscible liquids whose boiling points are close with the aid of fractionating column. Fractional distillation is used when the miscible liquids have boiling point differences between 1 and 200C. But the difference in the successive fractions must be more than 100C. For example, crude oil can be separated into its various pure fractions by this method. Fractional Distillation  Use Of Separating Funnel This method is used in separating two or more immiscible liquids. The denser liquid will be at the bottom of the funnel and can be tapped off leaving the lighter one in the funnel.  It can be used to separate a mixture of: Oil and water; Kerosine and water; etc Use Of Separating Funnel  Chromatography This method is used to separate the constituents of a complex mixture in solution. Initially, it was used to separate coloured substances but now it can be used for both coloured and colorless substances. This process makes use of the fact that various components have different solubility in a particular solvent and travel at different rates on the adsorbent medium (chromatographic paper or Silica gel).  Types Of Chromatography They include: I. Column chromatography, II. Paper chromatography, III. Gas chromatography, IV. Thin Layer chromatography, etc  Phases In Chromatography There are two phases in chromatography namely; 1.The mobile phase (solvent) and 2. The stationary phase (the adsorbent medium).  Paper Chromatography This process involves spotting the mixture on one end of the chromatographic paper and then immersing the paper in a suitable solvent taking care that the spot does not touch the solvent. Paper Chromatography Paper Chromatography  Retention Factor(Rf) Value To measure how far each component travels, the retention factor, Rf, value of the component is calculated thus: Rf value = distance traveled by substance distance traveled by solvent Chromatography is used in pharmaceuticals; research and analytical institutions. Retention Factor (Rf) Value  Precipitation This separation technique makes use of difference in solubility of solid (solute) substance in two miscible liquids (solvents). Thus, when the solid is dissolved in liquid in which it is soluble, the addition of liquid in which it is not soluble, throws out the solid as precipitate. The precipitates formed can be filtered off the solution. AgNO3(aq) + NaCl(aq) AgCl (s) + NaNO3(aq) BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq) This method is applied in softening hard water example, addition of aqueous sodium hydroxide to water containing CaSO4 precipitates. Insoluble Calcium hydroxide. 2NaOH + CaSO4 Ca(OH)2 (s) + Na2SO4(aq) Similarly, addition of washing soda, Na2CO3 to water containing MgSO4 precipitates insoluble MgCO3 MgSO4 + Na2CO3 MgCO3(s) + Na2SO4(aq)  Crystallization And Recrystallization This method is used to get out heat- labile salts from solution. The process involves concentrating the solution, heating and then allowing it to cool for the crystals to form. E.g. I. FeSO4. 7H2O II. MgSO4.7H2O III. CuSO4.5H2O Uses Of Crystallisation I. Used in obtaining pure sugar II. Used in making pure silicon used in computer chips III. Used in the purification of antibiotics.  Fractional Crystallization This method relies on differences in solubilities of two or more solute in the same solvent. The mixture is heated. On cooling, the solute with a lower solubility will come out of the solution and get precipitated leaving the one with a higher solubility in solution. The precipitate can then be filtered off. A solution of potassium chloride and potassium trioxochlorate(V) can be separated using this method.  Recrystallization Since it is possible for impurities to enter the crystal of the salt, recrystallization is used to ensure purity. Recrystallization involves dissolving the salt in a suitable hot solvent. On cooling, the crystals come out leaving the impurities.  Purity of Substances Mixtures are impure substances. The impurities are removed using the appropriate separation technique(s). To verify the purity of substances obtained during separation, the following tests are carried out: FOR SOLIDS; Melting point, Density, etc FOR LIQUIDS; Boiling point, Density, Refractive index, Spot on paper chromatogram, etc Determination Of Melting Point Of A Solid The melting point of a given pure solid is fixed at a certain temperature. However, an impure solid melts over a temperature range. For instance, ice containing dissolved salt melts over a range of -5 to 0⁰C. Therefore, impurities decrease the melting points of solids. Determination Of Boiling Point Of A Liquid Pure liquids have fixed boiling points whereas impure liquids boil over temperature ranges. For example, water containing dissolved salts can boil over a range of 100 to 110⁰C. Hence, impurities increase the boiling points of liquids.  Assignment 3 You are given a mixture of four substances K, L, M and N. K is a soluble salt, L is also soluble, but can sublime. M is magnetic, has a high melting point and it is insoluble. N is insoluble, non-magnetic but has a high melting point. Briefly describe the steps (in the right order) you will use to separate the mixture. (10mks)......: T1 WK 4 Topic : PARTICULATE NATURE OF MATTER Learning objectives i. Describe the structure Point: state Dalton’s atomic of an atom theory ii. State the properties of Point : give the modifications of Dalton sub particles of an atomic theory atom iii. Give the history of the Point : give the history of the discovery discovery of electron, of electron, proton and neutron proton and neutron iv. State Dalton’s atomic theory Point : determine how atomic mass, proton and neutron can be calculated v. State the modifications of Dalton atomic theory HOME This Powerpoint is hosted on www.worldofteaching.com Please More visit freefor 100’s more free powerpoints atpowerpoints www.worldofteaching.com S.MORRIS 2006 Atomic Structure  WHAT IS AN ATOM? o The smallest unit of an element. o Consists of a central nucleus surrounded by one or more electrons. WHAT IS THE NUCLEUS? o The central part of an atom. o Composed of protons and neutrons. o Contains most of an atom's mass.  WHAT IS A PROTON? o Positively charged particle. o Found within an atomic nucleus. WHAT IS A NEUTRON? o Uncharged particle. o Found within an atomic nucleus.  WHAT IS AN ELECTRON? o Negatively charged particle. o Located in shells that surround an atom's nucleus.  HELIUM ATOM Shell proton + N - + - N electron neutron What do these particles consist of?  Atom An atom is the smallest particle of an element that takes part in a chemical reaction.  Molecule A molecule is the smallest particle of a substance that can exist alone and still retain the chemical properties of that substance, be it an element or a compound. Generally, atom(s) combine with other atom(s) to form molecules.  Atomicity The atomicity of an element is the number of atoms in one molecule of the element. Examples are shown in the table below.  Atomicity ELEMENT FORMULA ATOMICITY OF MOLECULE Neon Ne 1 Hydrogen H2 2 Ozone O3 3 Phosphorus P4 4  Ion An ion is any atom which has an electric charge. For example, H⁺, F⁻, etc. A group of atoms with a single charge is called a radical. For example, OH⁻, NH4⁺, etc.  Types Of Ions The two types of ions are: 1. The positively charged ions or cations, and 2. The negatively charge ions or anions.  Dalton’s Atomic Theory John Dalton’s atomic theory states that: 1. All elements are made up of small indivisible particles called atoms, 2. Atoms can neither be created nor destroyed, 3. Atoms of the same element are alike in every aspect and different from atoms of all other elements,  Dalton’s Atomic Theory 4. When atoms combine with other atoms, they do so in simple whole number ratios, 5. All chemical changes result from the combination or separation of atoms. Modification Of Dalton’s Atomic Theory  The Modern Atomic Theory The modern atomic theory states that: 1. All matter is made up of tiny particles that may be positively charged, negatively charged or electrically neutral. 2. Atoms of the same element are not exactly alike but may have different masses. 3. The atoms of an element with different masses are called isotopes and the phenomenon, isotopy.  The Modern Atomic Theory 4. Atoms of different elements can combine to form molecules. 5. The molecules of a compound have definite compositions and structures.  CONSTITUENTS OF AN ATOM The table below shows the particle, mass and their charges. PARTICLE MASS CHARGE PROTON 1 POSITIVE (+) ELECTRON 1/1840 NEGATIVE (-) NEUTRON 1 NEUTRAL......: T1 WK 5 Topic : particulate nature of matter Learning objectives i. Define atomic mass, atomic Point: define atomic mass number and mass number ii. Solve problems on mass Point : solve problems on mass number, number, proton and neutron proton and neutron iii. Define isotopy Point : define isotopy iv. Solve problems on isotopes Point : solve problems on isotopes HOME ATOMIC NUMBER OR PROTON NUMBER (Z) This is the number of protons in an atom of an element. It is usually represented by the letter Z Atomic number gives the numerical position of elements in the periodic table.  Mass number or nucleon number(a) This is the total number of protons and neutrons in the nucleus of an atom. It is usually represented by the letter A. A = p + n where p = protons, n = neutrons. N=A-P A typical element X can be represented as:  Electrons in An Atom An atom of an element is electrically neutral; the net charge of an atom is zero. has an equal number of protons and electrons. number of protons = number of electrons Aluminum has 13 protons and 13 electrons. The net (overall) charge is zero. 13 protons (13+) + 13 electrons (13 -) = 0 105  Mass Number The mass number represents the number of particles in the nucleus. is equal to the number of protons + the number of neutrons. 106 Atomic mass/mass number Mass Number A Z X Symbol of element Proton Number  CALCULATION OF NUMBER OF PROTONS, ELECTRONS AND NEUTRONS If the mass number and atomic number of an element is known, the number of neutrons can be calculated. EXAMPLE 1 Given that an element has atomic number 11 and a mass number of 23. calculate the number of neutrons in the atom of the element SOLUTION We can use the formula N=A-P  CALCULATION OF NUMBER OF PROTONS, ELECTRONS AND NEUTRONS A = 23, P =11 , N = ? N=A–P N = 23 – 11 = 12 EXAMPLE 2 Consider the figure below: 35 Cl 17 (i) How many electrons are in the element? (ii) How many neutrons does the element have?  CALCULATION OF NUMBER OF PROTONS, ELECTRONS AND NEUTRONS (1)The element has 17 electrons. This is because a neutral atom has equal number of protons and electrons. (ii) Using N=A–P N = 35 – 17 = 18 neutrons. CLASS WORK An element X has mass number 31 and atomic number 15. Determine the number of neutrons of the element.  CALCULATION OF NUMBER OF PROTONS, ELECTRONS AND NEUTRONS EXAMPLE 3 Copy and complete the table as appropriate. ATOM/ ION ATOMIC NUMBER OF NUMBER OF NUMBER PROTONS ELECTRONS X 7 Q- 10 R 1O Y2+ 12  ISOTOPY This is the phenomenon in which atoms of the same element have the same atomic number or proton number but different mass numbers or neutron numbers. The different atoms of an element are called isotopes Isotopes have different mass numbers because of differences in the number of neutrons. Some elements that exhibit isotopy are carbon; hydrogen and chlorine.  Calculation of atomic mass based on isotopic abundance The relative abundance of 37Cl and 35Cl is 25% and 75% respectively. Calculate the relative atomic mass of chlorine. SOLUTION Ratio of abundance = 25:75 = 1:3 Sum of ratio = 4 Relative atomic mass of Cl = (35 x3) + (37 x 1) 4 = 105 + 37 4 = 142/4 = 35.5  Calculation of atomic mass based on isotopic abundance 75 x 35 + 25 x 37 100 100 = (0.75 x 35) + (0.25 x 37) = 26.25 + 9.25 = 35.50  EXAMPLE 2 Two isotopes of z with mass numbers 18 and 20 are in the ratio of 1:2. Determine the relative atomic mass of z. SOLUTION Ratio of abundance = 1:2 Sum of ratio = 3 Relative atomic mass of Z = (18 x1) + (20 x 2) 3 = 6 + 13. 33 = 19.33  EXAMPLE 3 Calculate the relative atomic mass of an element R given that the relative abundance of 6329R and 65 R are 68% and 32% respectively. 29 SOLUTION 68 X 63 + 32 X 65 100 100 = (0.68 X 63) + (0.32 X 65) = 42.84 + 20.80 = 63.64  EXAMPLE 4 An element X with relative atomic mass of 16.2 contains 2 isotopes 16 m 8X with relative abundance of 90% and 8X with relative abundance of 10%. What is the value of m? ANSWER (16 X 90) + (m x 10) = 16.20 100 100 144 + m = 16.20 10 10 1 144 + m = 16.20 10 144 + m = 162 M = 162 – 144 = 18  CLASSWORK 1. Copy and complete the table below PARTICLE NUMBER OF NUMBER OF NUMBER OF PROTONS ELECTRONS NEUTRONS 1 1H 1 1 27 3+ 13Al 16 2- 8O 8  CLASSWORK 2. consider the element represented as 40 Ca 20 (a) What is the number of protons in the element? (b) If the element changes to its ion,Ca2+, how many electrons does the ion have?.  ASSIGNMENT 1. Z is an element which exists as an isotopic mixture containing 75.8% 3517 Z and 24.2% 3717Z. How many neutrons are present in isotope 3517Z? Calculate the mean relative atomic mass of Z 2. An element has isotopic masses of 6 and 7. If the relative abundance is 1 to 12.5 respectively, what is the relative atomic mass of X? 3. An element X with relative atomic mass 16.2 contains two isotopes 168X with relative abundance of 90% and m8X with relative abundance of 10%. What is the value of m?......: T1 WK 7 Topic : formula and equations Learning objectives i. Define empirical formula Point: define empirical ii. Describe mathematically formula the relationship between empirical formula and Point : describe mathematically the molecular formula relationship between empirical formula and molecular formula iii. Solve problems on Point : solve problems on empirical formula empirical formula iv. Balance chemical equations Point : balance chemical equations HOME  Chemical Symbol This is an abbreviation or sign which represents the name of an element. For example, hydrogen is written as H, Sodium is written as Na, etc.  The Three Ways Of Writing Chemical Symbol They are: 1. The first letter of the name of an element is taken as the symbol. Examples are shown hereafter. Atomic number Element Symbol 1 Hydrogen H 5 Boron B 6 Carbon C 7 Nitrogen N 8 Oxygen O 2. Where there are two or more elements having their names starting with the same letter, the first two letters or the first letter and another letter are used; as shown below. Atomic Element Symbol number 2 Helium He 3 Lithium Li 4 Beryllium Be 10 Neon Ne 13 Aluminium Al  3. The symbols of some elements (especially metals) are derived from their Latin names, as shown hereafter. Atomic Element Latin name Symbol number 11 Sodium Natrium Na 19 Potassium Kalium K 26 Iron Ferrum Fe 29 Copper Cuprum Cu 47 Silver Argentum Ag  Chemical Formula Chemical formula is the collection of symbols and numbers that represent one molecule of a compound or an element. Examples are shown below. Substance Formula Water H2O Hydroxide ion OH- Tetraoxosulphate (VI) acid H2SO4 Oxide ion O2-  Valency Valency is defined as the combining power or capacity of atom of an element. Most times, valency of atom of an element corresponds to the number of outermost or valence electrons. For eg, Na, has a valency of +1. Thus, 11Na : 2,8,1. More examples are shown in the table below.  Element Symbol Valency Lithium Li +1 Potassium K +1 Silver Ag +1 Copper Cu +1, +2 Magnesium Mg +2 Chlorine Cl -1 Oxygen O -2 Iron Fe +2, +3 Bromine Br -1  Valency Rules These are the rules used in writing the chemical formulae of substances. They include: 1. Write the symbols of the atoms of the elements. 2. Write their individual valencies as superscripts after the symbols. 3. In the case of radicals, put the radical in a bracket and the valency as a superscript after the bracket. 4. Exchange the valencies. 5.Drop the values wherever they appear (ignore the charges). For example, tetraoxosulphate (VI) acid is written as follows: Hydrogen + Tetraoxosulphate (VI) H + SO4 H+ + (SO4)2- H2- + (SO4)+ H2SO4  EMPIRICAL FORMULA AND MOLECULAR FORMULA The empirical formula of a compound is the simplest whole number ratio of atoms in a compound. The molecular formula of a compound is the exact number of moles of atoms of the component elements in one mole of the compound. In order to calculate the empirical formula , the masses or percentage composition of elements present in the compound is usually given. The empirical formula can be determined in two steps. 1. Divide each mass by the relative atomic mass of the element. This give the mole ratio. 2. divide each of the mole ratios by the smallest ratio  EMPIRICAL FORMULA NOTE: If after you have divided by the smaller or smallest number and you can not get a round figure but fractional figures, take the following steps to get a whole number ratio: 1. If ratio is 1:1.5 multiply by 2 to get 2:3. 2. If ratio is 1:1.3 mutiply by 3 to get 3:4 3. If ratio is 1:1.25 multiply by 4 to get 4:5 4. If ratio is 1:1.4 multiply by 5 to get 5:7 SAMPLE CALCULATIONS Calculate the empirical formula of the following compounds whose percentage composition by mass is given. (a) H; 5.9%, S;94.19% (b) S ; 40%; O; 60% (c) Fe; 63.6% S ; 36.4% ANSWER ELEMENTS H S % COMP 5.9 94.19 Divide by Ar 1 32 Number of moles 5.9 2.94 Dividing by 2.94 2.94 2.94 Ratio 2 1 Empirical formula = H2S  SAMPLE CALCULATIONS (b) ELEMENTS S O % Composition 40 60 Divide by Ar 32 16 Number of moles 1.25 3.75 Dividing by 1.25 1.25 1.25 Ratio 1 3 Empirical formula = SO3  SAMPLE CALCULATIONS (c ) ELEMENTS Fe S % Composition 63.6 36.4 Divide by Ar 56 32 Number of moles 1.14 1.14 Dividing by 1.14 1.14 1.14 Ratio 1 1 Empirical formula = FeS  CLASS WORK Calculate the empirical formula of the following compounds whose percentage composition by mass is given: (a) Mg; 25.3% Cl; 74.7% (b) N; 63.6 O; 36.4% (c) C; 27.3 O; 72.75% [Mg=24, Cl=35.5, N=14, O=16, C=12 ]  EXAMPLE 2 Calculate the empirical formula of: (a) a compound containing 1.4g of Lithium and 1.6g of Oxygen (b) a compound containing 0.28g of silicon and 1.42g of Chlorine.  SOLUTION (a) ELEMENTS Li O Mass of element 1.4 1.6 Divide by Ar 7 16 Number of moles 0.2 0.1 Dividing by 0.1 0.1 0.1 Ratio 2 1 Empirical formula = Li2O  SOLUTION (b ELEMENTS Si Cl Mass of element 0.28 1.42 Divide by Ar 28 35.5 Number of moles 0.01 0.04 Dividing by 0.01 0.01 0.01 Ratio 1 4 Empirical formula = SiCl4  EXAMPLE 3 Calculate the empirical formula of the following: (a) 2.10g of compound A containing 0.92g of Iron and 1.18g of Chlorine. (b) 3.33g of compound B containing 2.11g of manganese and 1.22g of Oxygen (c) on analysis, a compound gave the following results: 2.12g of the compound contained 0.33g of carbon. 1.82g of the same compound contained 1.53g of sulphur. Calculate the empirical formula of the compound.  SOLUTION (a) ELEMENTS Fe Cl Mass of element 0.92 1.18 Divide by Ar 56 35.5 Number of moles 0.016 0.033 Dividing by 0.016 0.016 0.016 Ratio 1 2 Empirical formula = FeCl2  SOLUTION (b) ELEMENTS Mn O Mass of element 2.11 1.22 Divide by Ar 55 16 Number of moles 0.038 0.076 Dividing by 0.038 0.038 0.038 Ratio 1 2 Empirical formula = MnO2  SOLUTION (c) Percentage of Carbon = 0.33/2.12 x 100 = 15.6% Percentage of Sulphur = 1.53/1.82 x 100 = 84.1% ELEMENTS C S % comp of element 15.6 84.1 Divide by Ar 12 32 Number of moles 1.3 2.63 Dividing by 1.3 1.3 1.3 Ratio 1 2 Empirical formula = CS2 Laws Of Chemical Combination They are: - Law of conservation of mass or matter, - Law of constant composition (or definite proportion), - Law of multiple proportions, - Law of reciprocal proportions, etc  Law Of Conservation Of Mass The law of conservation of mass or matter states that matter is neither created nor destroy in course of any chemical reaction. This law simply means that the total mass of the products of a chemical reaction is equal to the total mass of the reactants. Thus, Pb(NO3)2 (aq) + 2KI (aq) PbI2(s) + KNO3(aq)  Law Of Constant Composition The law of constant composition or definite proportion states that all pure samples of a compound always contain the same elements in the same proportion by mass, regardless of where they are found or how they are made. This means that pure copper (II) oxide will have the same ratio of copper and oxygen by mass whether it is found in Nigeria or Singapore, on the Earth or the Moon.  Law Of Multiple Proportions The law of multiple proportions states that if two different elements can react to form more than a single compound, then the masses of the second element combined with a fixed mass of the first element, will be in simple whole number ratio. For example, carbon and oxygen combine to form two compounds; carbon (II) oxide and carbon (IV) oxide. Thus, C(s) + O2 (g) CO2 (g), 1:2  2C (s) + O2 (g) 2CO (g) , 1:1 Law Of Reciprocal Proportions This law states that if an element A combines with other elements B and C, the masses of B and C which combine with a fixed mass of A are the masses of B and C which combine with each other or simple multiples of those masses.  Chemical Equation A chemical equation is a collection of formulae and symbols to represent a chemical reaction on paper. Generally, chemical equations are written as follows: Reactants Product(s) Thus, A + B C + D; A and B = reactants C and D = products = to yield or form or produce  For example, the reaction between dilute hydrochloric acid and magnesium is: 2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g) Facts About Chemical Equations A chemical equation shows five important facts about a reaction: 1. The reactants that take part in the reaction. 2. The products that are produced. 3.The number of atoms or molecules of each substance involved in the reaction. 4. The number of atoms or molecules of each substance produced. 5. The physical states of the reactants and products of reaction.  State Symbols State symbols are the signs used to represent the physical states of the reactants and products in chemical equations. They include: a) s represents solid, b) l represents liquid, c) g represents gas, and d) aq represents aqueous solution.  Balancing Chemical Equations The steps employed in balancing chemical Equations are: 1. Write the word equation for the reaction. 2. Write the formulae of the reactants and products. 3. Check the numbers of different atoms on both sides of the equation to see whether the equation is balanced. 4. Balance the equation by placing coefficients (numerals) in front of the formulae of the substances in the equation. 5. Include the state symbols in the equation.  Examples 1. Sodium burns in chlorine gas to form sodium chloride. Write a balanced chemical equation for the reaction. Solution Sodium + chlorine Sodium chloride Na + Cl2 NaCl Na = 1 atom Na = 1 atom Cl = 2 atoms Cl = 1 atom  Na + Cl2 2NaCl 2Na + Cl2 2NaCl Na = 2 atoms Na = 2 atoms Cl = 2 atoms Cl = 2 atoms 2Na(s) + Cl2 (g) 2NaCl (s) Class Work 7B 1. Magnesium burns in air to yield magnesium oxide. Write a balanced chemical equation for this reaction. 2. Methane gas burns in air to form carbon dioxide and water vapour. Write the balanced chemical equation for the reaction. 4. Magnesium burns in air to form magnesium oxide. Write a balanced equation for the reaction. 5. Write a balanced chemical equation for the reaction of magnesium oxide with water to form magnesium hydroxide.  ASSIGNMENT 1. An oxide of iron was reduced to the metal. 3.20g of the oxide formed 2.24g of the metal. Calculate the empirical formula of the oxide. 2. 1.60g of Copper were converted into a sulphide of copper weighing 2.00g. Calculate the empirical formula of the sulphide. [Cu = 64].......: T1 WK 8 Topic : formula and Stoichiometry Learning objectives Point: define the law of i. Define the law of conservation of matter conservation of matter Point : define law of constant composition and solve problems based on the law ii. Define law of constant composition and solve Point : define the law of multiple problems based on the law proportion and solve problems based on this law. iii. Define the law of multiple proportion and solve problems based on Point : this law HOME  Law of Conservation of matter The law of conservation of mass also known as the law of conservation of matter states that matter is neither created nor destroyed in the course of any chemical reaction. This law simply means that the total mass of the products of a chemical reaction is equal to the total mass of the reactants.  Example 1 Sodium chloride reacts with silver trioxo nitrate (v) to produce silver chloride and sodium trioxo nitrate (v). Show that the chemical equation obeys the law of conservation of mass.  Solution: NaCl + AgNO3 NaNO3 + AgCl (23+35.5)+108+14+16x3=23+14+16x3+ 108+35.5 58 + 170 = 85 + 143.5 228.5= 228.5  Law of constant composition This law is also called the law of definite or fixed proportion, which states that all pure samples of the same chemical compound contain the same element combined in the same proportion by mass. This law simply means that if a substance is pure, no matter its source or method of preparation, its composition is constant.  Example: Two samples of PbS were prepared by i. Heating a mixture of lead solid and sulphur Pb + S PbS ii. Passing hydrogen sulphide gas into a solution of Lead chloride. H2S + PbCl2 2HCl + PbS Analysis of both samples I and ii shows that in i. 0.8g of sulphur combine with 1.4g of lead while in ii. 5.50g of lead sulphide contain 2.0g of sulphur. Show that these result illustrate the law of constant composition  solution sample Mass of Mass of Mass of S PbS Pb Sample i 2.2g 1.4g 0.8g Sample ii 5.5g 3.5g 2.0g % of Pb in sample I = mass of Pb 100 massof PbS X % of Pb in sample I = 1.40 X 100 = 63.6% 2.20 & OF Pb in sampl ii = 3.50 X 100 = 63.6% 5.50  The law of multiple proportion The law states that if two elements A and B combine together to form more than one compound, then the several mass of A which chemically combine with a fixed mass of B is in a simple ratio. The ratio could be 1:2, 2:1, 2:3 etc......: T WK 9 Topic : concept of oxidation number Learning objectives Point: define oxidation and i. Define oxidation and reduction reduction Point : rules for calculating oxidation ii. Rules for calculating Point : solve problems on oxidation number oxidation iii. Solve problems on oxidation number HOME  RULES FOR ASSIGNING OXIDATION NUMBERS 1. The oxidation number of monatomic ion is equal in magnitude and sign to its ionic charge e.g. the oxidation number of chloride ion, Cl-, is -1 and that of Magnesium ion, Mg2+ , is +2. 2. The oxidation number of hydrogen in a compound is always +1 except in metal hydrides. For example NaH and LiH where it is -1. 3. The oxidation number of oxygen in a compound is always -2 except in peroxides such as H2O2 where it is -1 4. The oxidation number of uncombined element is zero e.g. Na, Mg, Cl2, O2, Cu and Zn.  RULES FOR ASSIGNING OXIDATION NUMBERS 5. For a neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal zero. 6. For polyatomic ions, the sum of the oxidation numbers must equal the ionic charge 7. In a compound, the more electronegative element will have negative oxidation numbers while the more electropositive ones, positive numbers. This is true not only of ionic compounds but of covalent molecules. Determination Of Oxidation Numbers Example 1 Determine the oxidation number of Sulphur in SO42- and hence write the IUPAC name. SOLUTION S + (-2 X 4) = -2 S -8 = -2 S=8–2 S = +6 Hence, the IUPAC name is Tetraoxosulphate(VI) ion  Example 2 Determine the oxidation number of nitrogen in the following compounds and hence name them a) N2O b) NO2 c) KNO2 d) NaNO3 e) NO a) N2O 2N + (-2) = 0 2N – 2 = 0 2N = 2 N = 2/2 N=1 Hence, the IUPAC name is Dinitrogen (I) oxide b) NO2 N+ (-2 x 2) = 0 N – 4 =0 N = +4 Hence, the IUPAC name is Nitrogen (IV) oxide c) KNO2 (1 + N + -2 x 2) = 0 1+N–4=0 N=4–1 N = +3 The IUPAC name is potassium dioxonitrate (III) d) NaNO3 (1 + N + (-2 x 3) = 0 1 + N -6 = 0 N=6–1 N = +5 The IUPAC name is sodium trioxonitrate (V) e) NO N + (-2) = 0 N–2=0 N=+2 The IUPAC name is Nitrogen (II) oxide.  CLASS WORK 8 Calculate the oxidation number of the central element in the following; also write their IUPAC names. (a) H3PO4 (b) H3PO3 (c) Al2S3 (d) HBrO3  SOLUTION a) H3PO4 (1 x 3) + 2P + (-2 x 4) = 0 3+P–8=0 P=8–3 P = +5 H3PO3 (1 x 3) + P + (-2 x 3) = 0 3+P–6=0 P=6–3 P = +3 The IUPAC name of H3PO4 is tetraoxophosphate (III) acid  SOLUTION b) H2PO3 1 + P + (-2 x 3) = 0 1+P– 6=0 P=+6–1 P=+5 Hence the IUPAC name of H2PO3 is trioxophosphate (V) acid  SOLUTION c) Al2S3 6 + 3S = 0 3S = -6 S = -6/3 = -2 The IUPAC name of Al2S3 is dialuminium trisulphide  SOLUTION d) HBrO3 1 + Br + (-2 x3) = 0 1 + Br – 6 = 0 Br = 6 – 1 Br = +5 Hence, the IUPAC name of HBrO3 is trioxobromate (V) acid  ASSIGNMENT Calculate the oxidation number of the underlined elements in the following compounds 1. K2Cr2O7 2. KMnO4 3. H3PO4