JAMB 2017 Chemistry Tutorial (E-Learning)

JAMB 2017 Chemistry Tutorials, HOT topics you must learn, Topics from Jamb Past Questions(JAMB CHEMISTRY SYLLABUS)

The ideology  Behind the E-learning & Tutorial session for Jamb 2017 on MY9JAEDU.COM is to make sure our viewers are able to meet up with their desired scores in 2017 UTME. We trust you will find this post very interesting and very educative. These covers the Jamb Chemistry syllabus.

You should expect these sure-banker topics in JAMB 2017 Chemistry, all the topics being mentioned will be explained basically and very simplified. 

I will advice you to get a pen and jot the followings down.

Chemical and Physical Change

There are two kinds of change matter may undergo: These are chemical and physical change.

Chemical change involves the alteration in the chemical constitution of the substance which undergoes it, while physical change does not involve the alteration of the chemical composition of the substance but the alteration of its physical state. Based on this underlying difference, both changes exhibit certain characteristics by which you can easily identify them. 

These characteristics are expressed in the table below:

1. Always produces new kind of matter with a different mass from the original substance.
Produces no new kind of matter, mass of matter remains the same.
2. Usually accompanied by considerable heat change.
Is not accompanied by great heat change.
3. Is generally not reversible.
Is generally reversible.

Examples of Chemical and Physical Change:

Chemical and physical change include the below:

1. The burning of any substance in air, including candle..
1. The heating of a metal wire by electricity.
2. The addition of water to calcium oxide.
2. The dissolution of sodium chloride in water.
3. All explosions, including that of natural gas or hydrogen with air; dynamites and bombs.
3. All cases of melting of a solid to a liquid (or the freezing of liquid to solid). 
4. The rusting of iron.
4. All cases of vaporization of a liquid (or the condensation of a gas to liquid).
5. Magnetization of iron, as well as the demagnetization of iron.

Change of State

When substances undergo physical change, there are 3 distinct states upon which they can transform – solid, liquid and gaseous. When solid substances gain heat, the tendency is for them to change to the liquid state. This change is known as fusion or melting.
However, some solid substances when heated change directly to the gaseous state, without passing through the liquid state. This change is called sublimation. Liquid substances when heated change their state to gaseous. This is called vaporization.

Remember that these changes are reversible: when heat is taken from a gaseous substance, it eventually becomes liquid. This is known as condensation. The same goes for a solid substance that had sublimed - when the gas is cooled, it changes back to solid. A liquid substance changes to solid when heat is taken from it. This is known as solidification, freezing or crystallization.

Note: when a substance undergoes a change of state, its temperature remains the same. This is because the heat required for the change is latent heat; the substance will contain equilibrium mixture of both states (i.e. initial state and final state). 

What is a Mixture?

A mixture is the physical combination of two or more different substances. We can find in our environment a mixture of different types.

Examples of Mixtures

There are different kinds of mixtures available, examples include the following:

Examples of mixture The Earth Crust: This is a mixture of dissolved gases, living organisms and sometimes salts.

Air: This is a mixture or gases, water vapor and dust particles.
Petroleum: A mixture of different hydrocarbons.
Alloys: Contain different elements, e.g., bronze, steel and duralumin.
Coal: A mixture of coal tar, ammoniacal liquor, coal gas, and coke.
Vulcanizer’s Solution: Contain phosphorus, rubber, benzene and sulphur.

Safety Match Heads: Contain KClO3 (which is an oxidizing agent and therefore makes it possible for the match to start burning, because it produces oxygen on decomposition immediately when a sufficiently high temperature has been produced by friction), Fe2O3, MnO2 (which acts as catalyst to decompose KClO3), powdered glass (which acts to produce friction) and antimony sulphide or sulphur (which is the combustible substance).

The side of the box, or the striking surface on the outside of the “book,” is coated with a mixture of powdered glass, red phosphorus and glue. Friction of the match head against this prepared surface causes tiny explosions involving the phosphorus and potassium chlorate. The heat that is thereby liberated ignites the head of the match, which is not readily ignited by friction alone.
Another example of mixture is Gun Powder (containing carbon powder, KNO3 and sulphur).

Some materials, such as air or a mixture of sugar and water, are homogeneous mixtures; i.e., it is not obvious to the eye that they are composed of more than one substance. The constituents are evenly mixed and form a single phase (no layers are seen). Homogeneous mixtures are usually referred to as solutions. Other materials, such as sand-sugar mixture, are composed of different kinds of particles large enough to be individually seen - they are known as heterogeneous mixtures. The constituents of heterogeneous mixtures are not evenly mixed, but form layers.

When two or more liquids are mixed in all proportions to form a homogeneous mixture, they are said to be miscible, e.g., water and alcohol are miscible. Liquids that do not intermingle to form solutions are said to be immiscible, e.g., water and gasoline. If two liquids A, and B, are only partially miscible, a limited amount of each will dissolve in the other. If liquid A is present in excess (i.e., more is present that can dissolve in liquid B), two layers form, and each layer is a solution. In one layer the solvent is B, and it contains as much A as can dissolve in it; in the other layer the solvent is A, and it contains as much B as can dissolve in it. Briefly stated, one layer is a saturated solution of A in B, and the other layer is a saturated solution of B in A.

Liquids that are miscible consist of molecules that are of similar character. Non-polar liquids, such as carbon disulphide (CS2) and carbon tetrachloride (CCl4) are readily miscible with one another, but they will not dissolve in water because of the high polar nature of water. On the other hand, polar compounds such as methyl alcohol (CH3OH), and ethyl alcohol (C2H5 OH) are miscible with one another and with water. 

Separation of Mixtures
Components of mixtures are usually separated by physical means (they are therefore purified). There are different physical methods which can be employed to separate mixtures. The particular technique chosen for any given mixture depends on the nature of the constituents. Here are some separation techniques:

This method is used to separate components of soluble solid/liquid mixtures and volatile/involatile liquid mixtures. The principle governing this method is the fact that molecules of liquid substances when they gain heat, become gaseous and are lost from the surface. Notice that the liquid, haven vaporized is not collected but lost to the atmosphere. The other component (which is required), is then collected. Example – a mixture of sodium chloride and water.

This is used to separate components of liquid/liquid mixtures and soluble solid/liquid mixtures. It involves heating the mixture, and the vapor formed is allowed to cool, liquefy and is collected as pure liquid. Thus, each component of the mixture is purified. The principle behind this method is based on the fact that when liquids are heated to their boiling points, they become gaseous, and when the gases are cooled, they change back to the liquid.

Note: While evaporation is mostly used for solid/liquid mixtures, distillation is mostly used to separate liquid/liquid mixtures. Both evaporation and distillation involve gain of heat, and then vaporization. In evaporation, the vapor formed is allowed to escape into the atmosphere, while in distillation, the vapor is not lost but cooled, liquefied and collected as pure liquid. Distillation is used to purify solvents. There are two kinds of distillation – simple and fractional distillation:

Simple Distillation: This is used to separate mixtures of volatile/involatile liquids, or for mixtures of liquids whose boiling points are wide apart (by at least 100oC). An example of such mixtures is the mixture of water and ink.

Fractional Distillation: This is used to separate a mixture of liquids whose boiling points are close (boiling point difference of not more than 20-30oC). Examples of mixtures that can be separated by this method include: petroleum; alcohol and water; liquid air (a mixture of oxygen (b.pt 90 K), nitrogen (b.pt 77 K) and water (b.pt 1000C)).

Sublimation is suitable for solid mixtures containing solid substances that can vaporize directly when heated. Examples of such substances are iodine crystals, ammonium chloride, anhydrous aluminium chloride, anhydrous iron(III) chloride and benzoic acid. The vapor is cooled away from the other component(s) and collected as solid.

The principle behind this technique is that some solid substances are soluble in certain kind of solvent, while others are not. Hence, it is used generally to separate soluble substances from insoluble ones. For example, a mixture of sodium chloride crystals and sand – the sodium chloride is soluble in water while sand is not. Therefore, water is added to the mixture to dissolve sodium chloride while leaving the sand to settle.

Note: organic solvents generally dissolve organic substances, e.g. kerosene dissolves wax, grease, fats and oils. Inorganic solvents dissolve inorganic substances, and ionic solvents dissolve ionic substances. Common solvents for sulphur are: carbon(IV) sulphide, CS2 and methylbenzene (toluene). Common solvents for iodine are: ether (ethoxyethane), alcohol, carbon tetrachloride, CCl4 and potassium iodide.

Water soluble salts includes: All common trioxonitrates(V) of metals. All common salts of sodium, potassium and ammonium. All common tetraoxosulphates(VI), except: barium tetraoxosulphate(VI) and lead(II) tetraoxosulphate(VI). Notice that calcium tetraoxosulphate(VI) is sparingly soluble. All common chlorides except those of silver, mercury(I) and lead.

This is used to separate liquid components of mixtures from the solid components (which are in suspension). The principle of this technique is that the particles of liquid are small enough to pass through the filter material while those of solids are not. Notice that the solid particles are in suspension.

If they were settled at the bottom, then the process would be decantation and not filtration. Decantation does not involve the use of filter materials; it is the run-off of the liquid component, leaving the solid behind. Decantation will come before filtration (depending on whether the mixture contains solid components which are large and heavy enough to settle).

Both filtration and decantation usually follow the process of dissolution. E.g. after the sodium chloride component of a mixture of sodium chloride and sand is dissolved in water, the liquid component (sodium chloride solution) is decanted (separation from sand), and then filtered to obtain clear sodium chloride solution.

The principle of this method is based on the fact that soluble salts are only soluble to certain concentrations at a given temperature. Decrease in the temperature of their saturated solutions will see the salts forming out of the solution. It is used to obtain a soluble salt from its solution, and it involves heating the solution up to the point of saturation (for salts which crystallize with water of crystallization, e.g., ZnSO4 . 7H2O).

Cooling the solution below this point results in the formation of the crystals from the solution. For salts which do not crystallize with water, e.g., NaCl, their solutions are heated to dryness to produce them. Notice that salts which crystallize with water are not heated to dryness, otherwise, their crystalline nature will be lost. To purify further, the salt can be recrystallized. I.e., the crystals obtained is dissolved in hot distill water and the process of crystallization is repeated.

Notice that crystallization needs evaporation (by heating) for the solution to become saturated. It is possible to separate a mixture of more than one water-soluble salt by crystallization. This is because the solutions of different substances attain saturation at different temperatures. A solution containing a mixture of different substances therefore crystallizes its components separately when cooled below the saturated points of the different components in solution - this is known as fractional crystallization.

This method is mostly popular for the separation of colored components of pigments (e.g. ink and paints). However, it is useful also in separating certain non-colored components of mixtures.

Note: All chromatographic methods involve two phases, namely: stationary phase and mobile phase. Separation is based on the relative speed of the components of the mixture in-between the two phases.

If the stationary phase is a solid, the process is called adsorption chromatography. If the stationary phase is a liquid, the process is called partition chromatography.

Column chromatography (adsorption chromatography). The stationary (adsorbent) is a solid, e.g. finely divided alumina and silica gel. The column is usually a glass tube with a tap at the bottom packed with the adsorbent and the mobile phase (the eluting solvent).

As the solvent travels down the column, it carries with it the different components, which travel down at different rates depending on the extent to which they are adsorbed. More strongly adsorbed components travel down more slowly than less adsorbed ones. Hence, components are separated based on their different degree of adsorption on to the stationary phase as they move down the column (which causes them to move at different speeds).

Paper chromatography (partition chromatography).
Paper chromatography, also known as partition chromatography is a technique that involves the use of strips of filter paper. Notice that the stationary phase in paper chromatography is the moisture in the paper, and not the paper itself. This is an example of partition chromatography. Separation depends on the different degree of motion (i.e. speed) of the components of the mixture between the stationary water phase and the mobile chromatographic solvent (due to the different affinity the components have for both the stationary and mobile phases).

The material to be separated is applied as a spot near the bottom of the strip of paper. It is dipped into the solvent and the chromatogram left to develop. The solvent (e.g. propanone or ethanol) ascend the strip of paper by capillary action, and carries the solute along with it, different components travel at different rates depending on their relative affinity for both the mobile and stationary phrases.

This is ascending paper chromatography. A descending technique can be made by allowing the solvent to flow down the strip from a tray containing the solvent. Components of mixtures with greater affinity for the mobile phase than the stationary phase are separated first.

Precipitation is used to separate a salt which is soluble in one solvent, forming a mixture with that solvent, but become insoluble when another liquid which mixes well with the mixture but which does not dissolve the salt is added. The salt will therefore be precipitated from the solution and collected by filtration. For example, iron(II) tetraoxosulphate(VI) is soluble in water to form a mixture (i.e. a solution). When ethanol is added to the solution (ethanol is miscible with water), the iron(II) tetraoxosulphate(VI) will be precipitated from the solution as it is insoluble in ethanol.

Sieving is used to separate solid mixtures whose components’ particle size differ greatly. A sieve is used to make the separation. The particles of one component are small enough to pass through the sieve, while those of the other are not, and are therefore held onto the sieve, separated from the first. Notice that the principle of separation used here is the large difference in the particle size of the components of the mixture.

What is Stoichiometry?

Stoichiometry is the ratio of moles of all reactants and products involved in a chemical reaction. It shows the relative quantities of the reactants that will be required for a given reaction and those of the products that will be formed from the reaction.
The stoichiometry of a chemical reaction can be determined from the balanced equation for the reaction. For example, the stoichiometry of the reaction between nitrogen and hydrogen to produce ammonia can be deduced from the balanced equation:

N2 + 3H2 → 2NH3 

From the balanced equation above, the stoichiometry is given as 1:3:2. This means that one mole of nitrogen is needed to react with three moles of hydrogen to produce two moles of ammonia. 

Uses of Stoichiometry

Stoichiometry is used to determine the right quantity of reactants to be used in any chemical experiment or project so that wastage of materials is avoided. Without being guided by stoichiometry it could also be dangerous to have certain reactants in excess quantities in the reaction system.

Stoichiometry enables chemists to calculate or predict the quantity of products that will be formed in terms of moles, mass, and volume. It also makes it possible for them to determine or predict the quantity of reactants that will be used up in the reaction, and the percentage used or unused.

Dalton's Atomic Theory

This theory, which was first put forward in 1808 by John Dalton, is regarded as the foundation of modern chemistry. It gave insight into the composition of matter, and explained many chemical phenomena that were not understood before then. In summary, the theory consists of the following ideas:

1. That matter is made up of small, indivisible, discrete particles called atoms.

2. That atoms are indestructible, and cannot be created.

3. That atoms of a particular element are all exactly the same in every respect, and are different from those of all other elements. This explained why elements are pure substances, with each element having the same properties that are different from other elements. 

4. That chemical combination occurs between small whole numbers of atoms of the reacting substances. This explained chemical reactions and the properties of the new substances formed.

Dalton’s atomic theory stood for about a century and became the basis for studying chemical composition and reaction. However, as fresh knowledge became available over the decades some incorrectness was noticed in the theory. 

These include the statement that atoms are indestructible and cannot be created. That claim has been found not to be completely true with the discovery of nuclear chemistry where nuclear reactions could destroy and create different atoms.

However, it is worthy to note that it is only through nuclear reactions that you can have atoms being destroyed or created, it does not happen in chemical reactions. So Dalton’s atomic theory as regards chemical reactions still hold true.

Another area where Dalton’s theory has been faulted is in stating that atoms of the same element are exactly alike in all respect. The discovery of isotopes in some elements where there are atoms of different masses has made that statement not to be totally correct.

In spite of the incorrectness in some aspects of Dalton’s atomic theory, the explanation that chemical reaction involves the separation and combination of atoms, and that these atoms possess characteristic properties has remained relevant in today’s study of chemistry.

Law of Definite Proportions (or Constant Composition)
The law of definite proportions, also known as the law of constant composition states that all pure samples of the same chemical compound contain the same elements combined in the same proportions by mass.

What this law emphasizes is that, if pure samples of the same chemical substance, wherever they may be found, are analyzed, it will be found that they all consist of the same elements, as well as having these elements combine in the same proportions by mass.

For examples, pure sample of copper(II) oxide is composed of copper and oxygen, in the proportion of 1:1 by mole, or 64 g of copper to 16 g of oxygen or 1 g of copper to 0.25 g of oxygen.

Law of Multiple Proportions

The law of multiple proportions states that if two elements A and B combine together to form more than one compound, then, the several masses of A, which separately combine with a fixed mass of B, are in a simple ratio.
This law recognizes the fact that two elements may combine to form more than one product. For example, carbon and oxygen can combine to form carbon(II)oxide and carbon(IV)oxide; nitrogen combines with oxygen to form three possible oxides - nitrogen(I)oxide, nitrogen(II)oxide and nitrogen(IV)oxide.

If we consider one of the elements to be of fixed mass in the different products, then, the other element will be of varied mass which can be seen to be in a simple ratio.

Example: (1). In CO and CO2 If we consider carbon to be of fixed mass in the two products, then we have



12g of carbon + 16g of oxygen

12g of carbon + 32 g of oxygen

Carbon is of fixed mass: 
16g of oxygen

32g of oxygen

Ratio: 1

The different masses of oxygen that separately combine with the fixed mass (12g) of carbon in the two products (CO and CO2) are in a simple ratio of 1:2. However, if we consider the mass of oxygen fixed in the two products, then the different masses of carbon, which separately combine with the fixed mass of oxygen, can also be expressed in a simple ratio as shown below:



12g of carbon + 16g of oxygen

12g of carbon + 32 g of oxygen

To make oxygen of equal mass, multiply the masses of the elements here by 2 or divide the masses of elements in the other column by 2.

:. 24g of carbon + 32g of oxygen

12g of carbon + 32g of oxygen

Oxygen is of fixed mass: 24g of carbon

12g of carbon

Ratio: 2

The ratio of the masses of carbon in the two products, CO and CO2 that separately combine with the fixed mass of oxygen is therefore 2:1 respectively.

2. In the oxides of nitrogen: N2O, NO and NO2. If we consider the mass of nitrogen to be fixed, then:




28g of N + 16g of OX

14g of N + 16g of OX

14g of N + 32g of OX

Divide the mass of elements here by 2 to bring the mass of nitrogen to be same in all products

:. 14g of N + 8g of OX

14g of N + 16g of OX

14g of N + 32g of OX

Mass of nitrogen is fixed: 8g of OX

16g of OX

32g of OX

Ratio: 1

2 4
Therefore, the ratio of the different masses of oxygen that will combine with a fixed mass of nitrogen to form the products N2O, NO and NO2 is 1:2:4 respectively.

Considering the mass of oxygen fixed:




28g of N + 16g of OX

14g of N + 16g of OX

14g of N + 32g of OX

:. 28g of N + 16g of OX

14g of N + 16g of OX

Divide the masses of elements here by 2 to bring the mass of oxygen fixed

7g of N + 16g of OX

Mass of oxygen is fixed: 28g of N

14g of N

7g of N

Ratio: 4

2 1
Therefore, the ratio of the different masses of nitrogen which combine with the fixed mass of oxygen to form the products N2O, NO and NO2 is 4:2:1 respectively.

Note: to confirm whether a statement obeys the law of definite proportion or multiple proportions, what you should do is to keep the mass of one of the components of the compound fixed at 1g, and deduce the masses of the other component(s) in combination with it in the different compounds given.

If the masses of the other component(s) are the same in all the compounds, it means the composition of the different compounds is fixed, i.e., the law of definite proportion is satisfied. But if the masses are different, it means that the law of multiple proportions is satisfied, and the different masses can be expressed in a simple ratio.

Example: 1. An element X forms two oxides containing 77.47 and 69.62 per cent of X respectively. (a) what law is satisfied? (b) if the first oxide has the formula XO, what is the formula of the second oxide?

Solution: (a) In the first oxide, O is 22.53 g and X is 77.47g. 1g of oxygen combines with

77.47/22.53 = 3.4g of X

In the second oxide, O is 30.38g and X is 69.62g. 1g of oxygen combines with

69.62/30.38 = 2.3g of X

The masses of X in combination with a fixed mass of oxygen in both compounds are different, therefore, the law of multiple proportions is satisfied. Notice that percentage compositions can be expressed as composition in mass.

(b) Expressing the different masses of X in a simple ratio:

X of 1st oxide

X of 2nd oxide





1.5 or 3/2


or 3


or 1


If the 1st oxide is XO, the 2nd is X2/3O

X2/3O is same as X2O3

2. In two separate experiments, 0.125g and 0.11g of oxygen combine with a metal X to give V and W respectively. An analysis showed that V and W contain 0.5g and 0.44g of X respectively. What law is represented by the data above?

Solution: In V, 0.5g of X combine with 0.125g of oxygen. 1g of X will combine with

0.125/0.5 = 0.250g of oxygen

In W, 0.44 g of X combine with 0.11g of oxygen. 1g of X combine with

0.11/0.44 = 0.250g of oxygen

The masses of oxygen in the two compounds, V and W, which separately combine with a fixed mass of X are the same - the law of definite or constant proportion is satisfied. Notice that you will obtain similar result if you make the mass of oxygen fixed (at 1g) and deduce the masses of X in the two compounds.

Law of conservation of matter

The law of conservation of matter states that matter is neither created nor destroyed in the course of a chemical reaction. The law states the fact that the total masses of the products from a chemical reaction exactly equal those of the reactants. The law can be illustrated by the experiment below:

The mass of the set-up as shown is measured and recorded. By the thread, the HCl solution in the test tube is mixed with the silver trioxonitrate(V) solution in the flask. A reaction occurs, leading to the production of white precipitate (AgCl). The mass of the set-up is measured again. It is found that in spite of the formation of a solid substance, the mass remains the same.

Matter is neither lost nor gained during chemical reactions, but only change from one form to another because, according to Dalton’s atomic theory, the atoms in reaction undergo reconstitution during chemical reactions to form the products, and not that new atoms are formed, or that some get destroyed.

The above experiment can be performed with appropriate solutions of other substances. Example: barium chloride and sodium tetraoxosulphate(VI); barium chloride and dilute tetraoxosulphate(VI) acid; lead trioxonitrate(V) and potassium iodide; calcium trioxonitrate(V) and dilute tetraoxosulphate(VI) acid. A more accurate experiment to illustrate the law of conservation of matter was done by a scientist called Landolt in 1908. It was done in the apparatus show below:

In the separate arm of the tube are put sodium chloride and silver nitrate solutions. The tube and its contents were weighed. The tube was tilted to enable the content mix-up and react. It was then cooled, and reweighed. It was found that the total weight of the apparatus and the substances in it remained constant before and after reaction.

Gay Lussac’s Law of Combining Volumes

Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

The law explains experimental facts about how gaseous atoms combine. Example:

For the reactions:

(i) N2(g) + 3H2(g) → 2NH3(g)

1 vol. 3 vols. 2 vols.

1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia.

(ii) 2H2(g) + O2(g) → 2H2O(g)

2 vols. 1 vol. 2 vols.

2 volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of steam.

(iii) Cl2(g) + H2(g) → 2HCl(g)

1 vol. 1 vol. 2 vols.

1 volume of chlorine gas combines with 1 volume of hydrogen to form 2 volumes of hydrochloric acid.

Question: Consider the reaction: 2H2(g) + O2(g) → 2H2O(g)

(a). What volume of steam is formed from 20 cm3 of hydrogen and 20 cm3 of oxygen mixed together?

(b). What gas(s) is in excess, and by what amount?

Solution: (a). The ratio of their volumes is 2 vols. : 1 vol. → 2 vols.

20 vols. : 10 vols. → 20 vols.

That means, 20 cm3 of hydrogen will combine with 10 cm3 of oxygen to form 20 cm3 of steam.

(b). Oxygen is in excess by 10 cm3.

Avogadro’s Law

Avogadro’s law, put forward in 1811 by an Italian scientist named Avogadro states: Equal volumes of all gasses at the same temperature and pressure contain the same number of molecules.

The importance of this law is that it enables the volumes of gases in chemical reactions to be converted directly to actual molecules contained. And since molecules are involved in chemical reactions, the law gives a deeper understanding about chemical reactions in gases, and allows the deductions of Gay- Lusac’s law.

Notice that this law is applicable only to gases.

Example: 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of steam at constant temperature and pressure. This can be read as: 2 molecules of hydrogen combine with 1 molecule of oxygen to give 2 molecules of steam.

Chemical Symbols

There is no particular order or sequence of deducing the symbols of the known elements. Some elements have their symbols as the first letter (in capital) of their English names, e.g. oxygen is represented as O, hydrogen as H and nitrogen as N.

Some are represented by combining the first letter (in capital) and another (in small letter) of their names, e.g. chlorine - Cl, calcium - Ca and cesium – Cs. Others derive their symbols from their Latin names. Example, copper (Latin name cuprum) – Cu; iron (Latin name ferrum) – Fe; lead (Latin name plumbium) – Pb; silver (Latin name argentum) – Ag; gold (Latin name aurum) – Au; mercury (Latin name hydragyrum) – Hg ; sodium (Latin name, natrium) - Na; and potassium (Latin name kalium) - K

Note: as a chemistry student, it is extremely important for you to get familiar with elements and their symbols. The periodic table consist of symbols of the known elements. 

Chemical Formulas

Chemical formulas are used to represent chemical substances (i.e. both elements and compounds). For elements, the formulas are the symbols of the elements times a number (written as a subscript). This number denotes the number of atoms required to make the element stable in its simplest quantity (i.e. as a molecule).

For compounds: Compounds are made of more than one element, chemically combined. Their formulas consist of symbols of the respective elements which made them up. The number of atoms of each element present in the combination is dependent on the combining power or valency of the element.

Notice that to make the simplest formula, the combining powers of the elements in combination are interchanged between them and expressed as subscripts. For example, water is made of hydrogen and oxygen. Hydrogen has valency or combining power of 1, while oxygen a valency of 2.

Therefore, interchanging the valencies of the elements to obtain their combining number of atoms gives the formula of water. I.e., H2O. The same principle goes for the formula of aluminium oxide. I.e. Al - trivalent, O - divalent. The formula is Al2O3 (by interchanging the valencies of the two elements involved). It also applies to situations where we have groups of atoms (called radicals) combining with each other or with atoms of metallic elements, or with atoms of non-metallic elements.

That is, the formulas are obtained by interchanging the combining powers of both radicals or of the radical and the metallic or non-metallic atoms. Examples of radicals, single elements, and their combining powers are given in the table below:

Tables showing elements and radicals and their combining powers

Combining Power

Calcium Ca2+

Copper(II) Cu2+

Potassium K+

Sodium Na+

Iron(III) Fe3+

Magnesium Mg2+

Aluminium Al3+

Lead(II) Pb2+

Clorine Cl-

Oxygen O2-

Sulphur S2-



Combining Power

Carbonate(IV) CO32-

Nitrate(V) NO3-

Sulphate(VI) SO42-

Ammonium NH4+

Hydroxyl OH-

Chromate(VI) Cr2O72-

To obtain the formula of calcium trioxonitrate(V) (formed from the combination of calcium and the nitrate(V) group): Calcium has combining power of +2. Nitrate(V), NO3- has combining power of -1. Therefore, interchanging their combining powers gives the simplest formula, Ca(NO3)2

The best way to be certain about the formula of an unknown compound is from experimental data - Empirical Formula. This is so because some elements can combine to form more than one possible products.

Empirical Formulas

The empirical formula of a compound is the formula which expresses only the relative number of atoms of each element in the compound. Empirical formulas are sometimes called ‘the simplest formulas’ , but because they are obtained from experimental data, they are usually called empirical formulas, and are useful to:

1. Determine the molecular mass (or weight) of compounds whose molecular masses have not been known.

2. Determine the molecular masses of compounds whose molecular masses are variable, even though they have a definite percentage composition (i.e., for different compounds formed from the same elements).

To Determine Empirical Formula

To determine the empirical formula of a compound, we need to know:

1. The chemical composition of the compound - this is derived from experimental procedure, and can be expressed as percentage.

2. The relative atomic masses of the constituent elements.

Procedure: 1. From the chemical composition of the compound, and the relative atomic masses of the constituent elements given, convert the composition of each constituent element to number of moles.

2. Derive the mole ratio of the constituent elements.

3. Finally, express the mole ratio as the subscripts of the symbols of their respective element. The simplest formula obtained is the empirical formula.

Example: Analysis of carbon monoxide shows that it is 42.9% carbon and 57.1% oxygen. What is its empirical formula? (C=12, O=16) Solution: 1. Convert the percentage composition of each element to number of moles (consider each percentage as the mass).

Number of moles = mass (or % composition)/relative atomic mass

For carbon, 42.9/12 = 3.58

For oxygen, 57.1/16 = 3.58

2. Take the ratio of their moles:

C : O

3.58 3.58 = 1 : 1 3.

Express the above ratio as subscripts of the symbols of their respective element, we have C1O1 which is better expressed as CO - this is the empirical formula.

To Determine Molecular Formula

The molecular formula of a compound is the formula expressing one mole of the compound. It can be derived from its empirical formula if the molecular mass (or weight) is known. The product of the mass of a compound from its empirical formula and a factor equals the molecular mass of the compound. From this equation, its molecular formula can be deduced.

Example: Determine the molecular formula of a compound of molecular weight 30 amu whose empirical formula is CH3 (C=12, H=1)

Solution: mass of the compound from its empirical formula = CH3 = 12+3(1) = 15

Product of mass of compound from empirical formula and a factor (x) equals the molecular weight.

xCH3 = 30

15x = 30

x = 30/15 = 2

Thus, 2CH3

Therefore, molecular formula = C2H6

Note: - Many compounds have empirical formulas that are the same as their molecular formulas, for example, CO2 is both the empirical and molecular formula for carbon(IV)oxide. Others have their empirical formulas different from their molecular formulas, example, when you see a formula like H2O2, C2H6 and C6H12O6, you are looking at a molecular formula, the empirical formula is HO, CH3 and CH2O respectively. - Different compounds can have the same empirical formula.

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