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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:
EXAMPLES OF CHEMICAL CHANGE
1. The burning of any substance in air, including candle..
EXAMPLES OF PHYSICAL CHANGE
1. The heating of a metal wire by electricity.
EXAMPLES OF CHEMICAL CHANGE
2. The addition of water to calcium oxide.
EXAMPLES OF PHYSICAL CHANGE
2. The dissolution of sodium chloride in water.
EXAMPLES OF CHEMICAL CHANGE
3. All explosions, including that of natural gas or hydrogen with air; dynamites and bombs.
EXAMPLES OF CHEMICAL CHANGE
3. All cases of melting of a solid to a liquid (or the freezing of liquid to solid).
EXAMPLES OF CHEMICAL CHANGE
4. The rusting of iron.
EXAMPLES OF PHYSICAL CHANGE
4. All cases of vaporization of a liquid (or the condensation of a gas to liquid).
EXAMPLES OF PHYSICAL CHANGE
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).
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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
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
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
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
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
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, 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.
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.
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
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.
A chemical equation is the symbolic representation of a chemical reaction. It shows the chemical formulae of the reactants and the products, the mole ratios by which the reactants combine, as well as the mole ratios by which the products are formed. It shows the direction of the reaction; the state of matter in which the reactants and products are present.
For example, the chemical reaction between aqueous solution of hydrochloric acid and aqueous solution of sodium hydroxide that leads to the formation of sodium chloride and water can be expressed by the equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
In the above equation, the reactants (hydrochloric acid and sodium hydroxide) and the products of the reaction (sodium chloride and water) are shown by their formulae HCl, NaOH, NaCl and H2O respectively.
The mole ratio of the reactants and the products, which is the stoichiometry of the reaction is also expressed as 1:1:1:1. The direction of the reaction is shown by the forward arrow, which means that the reaction goes in one direction to produce sodium chloride and water.
Note that there are some reactions that can proceed both in the forward and backward direction in the same system. Such reactions are called reversible reactions and two half arrows to either side are used.
Example of reversible reaction is
Also in the hydrochloric acid and sodium hydroxide reaction above, the state or phase of the reactants and the products are indicated in the chemical equation. HCl, NaOH and NaCl are aqueous solutions (the substances are in solutions in water) while H2O is in liquid state.
Chemical equations are useful in the following ways: To deduce quantities of reactants needed for reactions, or quantities of products expected from chemical reactions. To deduce quantities of energy involved in chemical processes. To give vivid explanations of chemical reactions.
Balancing of Chemical Equations
For a chemical equation to be useful, it must be balanced. A balanced chemical equation means that all the elements represented in the equation must have equal number of atoms on both the left and the right sides of the equation.
For example, in the chemical equation N2(g) + 3H2(g) → 2NH3(g) there are two elements, N and H. N has two atoms on the left where it is in a molecule of nitrogen gas and on the right hand side it also has two atoms where it is in 2 molecules of ammonia (NH3) gas. H is having six atoms on the left where it is in 3 molecules of hydrogen gas and on the right it also has six atoms in 2 molecules of ammonia gas.
Since the elements N and H have the same number of atoms on both sides, the chemical equation is therefore balanced. A balanced chemical equation shows that matter is neither created nor destroyed during chemical reactions, but change from one form to another. It therefore satisfied the law of conservation of matter.
Exercise: Balance the chemical equations
(a). P4O6 + O2 → P4O10 ; (b). HClO → HCl + O2 ; (c). NH4NO3 → N2O + H2O
(a). P4O6 + O2 → P4O10
The equation contains two elements, phosphorus (P) and oxygen (O). P has four atoms on both sides. It is therefore balanced.
Oxygen has a total of eight atoms on the left side and ten atoms on the right and is therefore not yet balanced. To balance the oxygen, put 2 in front of O2 on the left hand side. This brings the total number of oxygen atoms on the left hand side to be ten, same as those on the right side of the equation.
P4O6 + 2O2 → P4O10 The chemical equation is therefore balanced.
(b). HClO → HCl + O2
The equation has three elements – hydrogen (H), chlorine (Cl) and oxygen (O). Hint: first, identify the element that is alone in a molecule and balance it. Oxygen is alone in the molecule O2 on the right side. It has two atoms while on the left side it has one atom. It therefore needs to be balanced by having a 2 in front of HClO.
The equation becomes 2HClO → HCl + O2
This is not yet balanced as H and Cl have 2 atoms each on the left side while on the right side they have one atom. To balance them, we need to put 2 in front of HCl on the right side. The equation becomes 2HClO → 2HCl + O2 and is now balanced with all the elements having the same number of atoms on both left and right hand side.
(C). NH4NO3 → N2O +H2O
There are three element to balance in the equation: N, H and O. Nitrogen has two atoms on the left side and also two atoms on the right, making it balanced. For hydrogen, there are four atoms on the left side and two atoms on the right. Hydrogen needs to be balanced by placing 2 in front of H2O on the right side.
The equation becomes NH4NO3 → N2O +2H2O
Balancing hydrogen has also resulted in the balance of oxygen atoms – three atoms of oxygen are both on the left and right side (one in N2O molecule and two in H2O molecule). The chemical equation is therefore completely balanced.
Relative Atomic Mass
Relative Atomic Mass In chemistry, it is important to quantify particles. For the fact that the size of an atom is too small to be weighed practically, relative atomic mass is therefore used to represent the mass or weight of an atom of an element.
Definition: The relative atomic mass (symbol: Ar) of an element is the mass of one atom of the element compared with the mass of an atom of the carbon-12 isotope whose mass is exactly 12.
For naturally occurring elements with isotopes, the above definition is a little bit adjusted to accommodate the fact that each isotope of the element has a different mass.
The relative atomic mass of a naturally occurring element with isotopes can be defined as the weighted average of the masses of its isotopes compared with the mass of an atom of the carbon-12 isotope whose mass is taken to be exactly 12.
Relative atomic mass is a dimensionless quantity and so does not have a unit. It is measured using an instrument called mass spectrometer.
Isotopes of an element are atoms of the element with different masses. The atoms have the same number of protons but different neutrons.
Relative atomic mass is also called atomic weight.
The carbon-12 isotope is also known as the unified atomic mass unit - its atomic mass is exactly 12.
The relative atomic masses of elements are usually published by the International Union of Pure and Applied Chemistry (IUPAC) and are reprinted in a wide varieties of materials, including textbooks, commercial catalogues, periodic table, and wallcharts.
Calculating Relative Atomic Mass of Elements with Isotopes
Some elements have been found to contain isotopes. The relative atomic mass of such element is determined by the masses of all the isotopes, and is closer to the mass of the more abundant isotope.
The relative atomic mass of these elements can be calculated using the formula:
Relative atomic mass = (A/100 x a) + (B/100 x b) + (C/100 x c)
Where A/100 (or A%) is the abundance of one of the isotopes with relative atomic mass a; B/100 is the abundance of another isotope with relative atomic mass b; C/100 is the abundance of the third isotope with relative atomic mass c.
Example 1: Oxygen has three isotopes 16O, 17O, and 18O with relative abundance 99.76%, 0.04%, and 0.20% respectively, calculate the relative atomic mass of oxygen.
Relative atomic mass of 16O is 16 and its relative abundance is 99.76%; 17O – atomic mass 17, abundance 0.04%; 18O – atomic mass 18, abundance 0.20%.
Relative atomic mass of oxygen = (99.76/100 x 16) + (0.04/100 x 17) + (0.20/100 x 1
= 15.9616 + 0.0068 + 0.036 = 16.0044
By tweaking the formula above you can also calculate the relative abundance of the isotopes if the relative atomic mass of the element and the atomic masses of the respective isotopes are known.
Example 2: Chlorine with relative atomic mass of 35.5 contains two isotopes 35Cl and 37Cl, what is the relative abundance of each isotope in a sample of chlorine?
35.5 = (A/100 x 35) + ((100 – A)/100 x 37)
Note that if the abundance of the isotope of atomic mass 35 is A%, the abundance of the isotope of mass 37 will be (100 – A)%.
35.5 = 0.35A + (100 – A) x 0.37 35.5
= 0.35A + 37 – 0.37A 0.37A – 0.35A
= 37 – 35.5 0.02A = 1.5 A
= 1.5/0.02 = 75
Therefore, the abundance of the isotope of relative atomic mass 35.5 is 75% while that for the isotope 17Cl is 100 – 75 = 25%.
Definition: The mole is the amount of substance, which contains as many elementary particles as there are carbon atoms in 12.000 grams of the carbon 126C isotope.
The mole is a useful concept because it provides a standard of measurement for atoms and other elementary particles (e.g. molecules, ions, protons and electrons). It is based on some underlying concepts - Avogadro’s constant, molar mass, gaseous molar volume, molarity, molality and mole ratio.
Avogadro found that in 12.000 grams of the 126C isotope of carbon are about 6.02 x1023 of carbon atoms. The value 6.02 x1023 is the avogadro’s number (or constant, NA) and it represents the number of elementary particles or basic units (i.e., atoms, molecules, ions, protons and electrons) present in 1 mole of a substance.
Notice that Avogadro’s number or constant is different from Avogadro’s law. Avogadro’s number deals with the number of elementary particles in solids, liquids and gases, while Avogadro’s law deals with the chemical combination of gases only.
Number of Moles
The number of moles present in a certain quantity of a substance can be determined by dividing the concentration or mass of the substance in grams by its molar mass. It is stated clearer by the the formula:
Number of moles (n) = conc. (g)/ molar mass
Definition: The molar mass is the mass of 1 mole (containing 6.02 x1023 elementary particles) of any substance. It is commonly expressed in grams per mole or g/mol.
* Masses of chemical substances are usually expressed in grams because chemists use small quantities of chemical substances in their work.
* Relative atomic or molecular masses do not have units. This is because relative atomic or molecular masses are relative quantities, while molar masses are the masses or weight of specific number of particles and are expressed in g/mol.
The molar mass of a compound is equal to the sum of the relative atomic masses of the elements contained in one mole of it.
E.g. to find the molar mass of H2SO4. The relative atomic masses are: (H =1, S = 32, O = 16). Hence molar mass = 2H + S + 4O =2 x 1 + 32 + 4 x 16 = 98g/mol.
The molar mass of a diatomic or polyatomic molecule, e.g. O2 is the atomic mass of its element multiplied by the number of atoms it contained. Example, the molar mass of oxygen, O2 is 16 x 2 = 32g/mol.
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Gaseous Molar Volume
Definition: The molar volume of a gas is the volume occupied by one mole of the gas at standard conditions of temperature and pressure, s.t.p. (i.e. temperature of 273 K and pressure of 760 mm Hg).
At standard conditions, 1 mole of a gas occupies a volume of 22.4 dm3 (or 22400 cm3). Notice that a change in the standard conditions of temperature and pressure will alter the volume.
Definition: The molarity of a solution is defined as its concentration in moles of solute per its volume in dm3. This is a measure of the concentration of any solution.
I.e. Molarity (M) = number of moles/volume of solution in dm3
A solution is said to be 1.0 molar if one mole of the solute is dissolved in 1.0 dm3 of the solution.
Converting concentration in molarity to g/dm3:
Molarity (M) = mass(g)/molar mass x 1/volume(dm3)
Molarity (M) = mass (g)/v(dm3) x 1/molar mass
mass (g)/v (dm3) = molarity (M) X molar mass
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This is another method of stating the concentration of a solution.
Definition: The molality of a solution is the number of moles of the solute per the mass of the solvent in kg. Molality is usually designated with a small letter m in italics m or small letter m with a hyphen -m.
The following formula can be used to calculate molality:
m = number of moles/mass of solvent (kg)
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When atoms of different elements combine to form compounds, they do so in different proportions of their moles. To determine the formulae of chemical compounds therefore, we could deduce the mole ratios of the atoms in the combination.
Also, when substances undergo reactions to form products, they do so in certain proportions of their individual moles. Hence, a balanced equation would show the ratio of their moles by which they reacted. The ratio of the moles of the products formed is also expressed.
From a balanced equation therefore, it is possible to deduce the concentration of any species in the equation from a given data. The number of moles presents in a substance is the mass (g) of the substance divided by its molar mass or atomic mass.
I.e. Number of moles (n) = mass (g)/molar mass
Application of the Mole Concept
1. The number of hydrogen ions present in 100 cm3 of 0.4 M solution of H2SO4 is ? [NA= 6.02 x1023]
From the equation: H2SO4 → 2H+ + SO42-
Mole ratio 1 : 2 : 1
To find the actual number of moles of H2SO4 that dissociated:
Molarity = number of moles/volume of solution in dm3
Number of moles = molarity x v (dm3) = 0.4 x 0.1 = 0.04 mole
From the stoichiometry above, 1 mole of H2SO4 formed 2 moles of H+, therefore, 0.04 mole formed 0.08 mole.
From Avogadro’s constant, 1 mole of a substance contains 6.02 x1023 elementary particles, therefore, 0.08 mole of H+ contains 6.02 x1023 x 0.08 = 4.816 x 1022 of hydrogen ions.
2. How many grams of ammonia will be produced from 100g of hydrogen? (N=14, H=1)
Solution: From the equation: N2 + 3H2 → 2NH3
Mole ratio: 1(28g) : 3(2g) : 2(17g)
3 moles or 6g of hydrogen produced 2 moles or 34g of ammonia.
Therefore, 100g of hydrogen will produce
34/6 x 100g of ammonia = 567g
3. What volume of carbon(IV) oxide measured at s.t.p. will be produced when 42.0g of sodium hydrogen trioxocarbonate(IV) is completely decomposed by the equation: 2NaHCO3(s) → Na2CO3(s) + CO2(g)+ H2O(l) (Na=23, H=1, C=12, O=16)
From the stoichiometry above, 2 moles of NaHCO3 produce 1 mole of CO2 . Actual number of moles of NaHCO3 used is
mass/molar mass = 42/84 = 0.5
Since 2 moles of NaHCO3 produce 1mole of CO2 , 0.5 mole will produce 0.5/2 = 0.25 mole of CO2
At s.t.p., 1 mole of a gas occupies 22.4 dm3
Therefore, 0.25 mole of CO2 occupies a volume of 22.4 x 0.25 dm3 = 5.6 dm3
4. To what volume must 300 cm3 of 0.6 M sodium hydroxide solution be diluted to give a 0.40 M solution?
Express the equation for dilution: C1V1 = C2V2 where C1 = initial molarity, C2 = final molarity, V1 = initial volume, V2 = final volume.
Therefore, 0.6 x 300 = 0.4 x V2
V2 = 0.6 x 300/0.4 = 450cm3
Thus, to dilute 300cm3 of 0.60M solution to a 0.40M, what you need to do is to take 300cm3 of the 0.60M solution, place it in a 450cm3 volumetric flask and add distilled water until it gets to the mark. The solution you now have is 0.40M. (Notice that the amount of distilled water used in diluting it is 150cm3 (450cm3 - 300cm3)
5. What is the molality of a solution obtained by dissolving 32.5g of sodium chloride in 1500g of water? (Na = 23, Cl = 35.5)
Molality (m) of NaCl = number of moles of NaCl/mass of water (in kg)
Number of moles of NaCl = 32.5/58.5 = 0.556
Mass of water in kg = 1.5
Molality = 0.556/1.5 = 0.371
The molar mass of a substance (chemical element or compound) is the mass of one mole of the substance. It is a physical quantity that is measured and expressed in g/mol.
How to Calculate Molar Mass
The molar mass of a substance is calculated by summing up the product of the relative atomic mass and number of atoms of all the elements that make up the substance as expressed in its chemical formula. For an element, the molar mass is the atomic mass of the element.
For example: calculate the molar mass of the following substances:
1. Water, H2O
The relative atomic masses of the elements in water, H2O are: H = 1, O = 16.
Therefore, the molar mass is: relative atomic mass of hydrogen x number of atoms of hydrogen + relative atomic mass of oxygen x number of atoms of hydrogen.
This is expressed as 1x2 + 16x1 = 2 + 16 = 18
The molar mass of water, H2O is therefore 18g/mol
2. Hydrochloric acid, HCl
The relative atomic masses of the elements in hydrochloric acid, HCl are H = 1, Cl = 35.5.
The molar mass is calculated as follows: relative atomic mass of H + relative atomic mass of chlorine, which is 1 + 35.5 = 36.5g/mol
3. Tetraoxosulphate(VI) acid, H2SO4
The relative atomic masses of the elements in H2SO4 are: H = 1, S = 32, O = 16.
Calculating the molar mass, we have: 1x2 + 32 + 16x4
= 2 + 32 + 64 = 98
The molar mass of H2SO4 is 98g/mol
4. Glucose, C6H12O6
The relative atomic masses of elements in C6H12O6 are: C = 12, H = 1, O = 16.
Calculating the molar mass, we have: 12x6 + 1x12 + 16x6
= 72 + 12 + 96 = 180
Therefore, the molar mass of glucose is 180g/mol
5. Calcium trioxocarbonate(IV), CaCO3
The relative atomic masses of elements in CaCO3 are Ca = 40, C = 12, O = 16
Calculating the molar mass, we have: 40 + 12 + 16x3
= 40 + 12 + 48 = 100
Therefore, the molar mass of CaCO3 is 100g/mol
6. Sucrose, C12H22O11
The relative atomic masses of elements in C12H22O11 are: C = 12, H = 1, O = 16
Calculating the molar mass, we have 12x12 + 1x22 + 16x11
= 144 + 22 + 176 = 342
Therefore, the molar mass of sucrose is 342g/mol
7. Methane, CH4
The relative atomic masses of the elements in CH4 are: C = 12, H = 1
Calculating the molar mass gives 12 + 1x4
= 12 + 4 = 16
The molar mass of methane is 16g/mol
8. Acetic acid, CH3COOH
The relative atomic masses of elements in CH3COOH: C = 12, H = 1, O = 16
Calculating the molar mass of CH3COOH, we have 12x2 + 1x4 + 16x2
= 24 + 4 + 32 = 60
Therefore, the molar mass of acetic acid is 60g/mol
9. Sodium hydroxide, NaOH
The relative atomic masses of the elements in NaOH are: Na = 23, O = 16, H = 1
Calculating the molar mass of NaOH: 23 + 16 + 1 = 40
Therefore, the molar mass of sodium hydroxide is 40g/mol
10. Ammonia, NH3 The relative atomic masses of the elements in NH3 are: N = 14, H = 1
Calculating its molar mass: 14 + 1x3
= 14 + 3 = 17
Therefore, the molar mass of NH3 is 17g/mol
11. Butane, C4H10 The relative atomic masses of elements in C4H10 are: C = 12, H = 1
Calculating its molar mass: 12x4 + 1x10
= 48 + 10 = 58
Therefore, the molar mass of butane is 58g/mol
12. Copper(II) sulphate, CuSO4 The relative atomic masses of the elements in CuSO4 are: Cu = 64, S = 32, O = 16
Calculating the molar mass, we have: 64 + 32 + 16x4
= 64 + 32 + 64 = 160
Therefore, the molar mass of CuSO4 is 160g/mol
13. Potassium chloride, KCl
The relative atomic masses of elements in KCl are: K = 39, Cl = 35.5
Calculating its molar mass, 39 + 35.5 = 74.5
The molar mass of KCl is therefore 74.5g/mol
14. Calcium chloride, CaCl2 Relative atomic masses of elements in CaCl2 are: Ca = 40, Cl = 35.5
Calculating its molar mass, we have, 40 + 35.5x2
= 40 + 71 = 111
Therefore, the molar mass of CaCl2 is 111g/mol
15. Methanol, CH3OH
Relative atomic masses of elements in CH3OH are: C = 12, H = 1, O = 16
Calculating the molar mass, we have, 12 + 1x4 + 16
= 12 + 4 + 16 = 32
The molar mass of methanol is 32g/mol
16. Ethanol, C2H5OH
The relative atomic masses of elements in C2H5OH are: C = 12, H = 1, O = 16
Calculating its molar mass = 12x2 + 1x6 + 16
= 24 + 6 + 16 = 46
The molar mass of ethanol is therefore 46g/mol
17. Carbon(IV) oxide, also called carbon dioxide, CO2
The relative atomic masses of the elements in CO2: C = 12, O = 16
Calculating its molar mass, we have 12 + 16x2
= 12 + 32 = 44
Therefore, the molar mass of carbon(IV) oxide is 44g/mol
18. Sodium chloride, NaCl
The relative atomic masses of elements in NaCl: Na = 23, Cl = 35.5
Calculating its molar mass, we have 23 + 35.5 = 58.5
The molar mass of NaCl is 58.5g/mol
19. Acetone, C3H6O The elements in C3H6O and their relative atomic masses are: C = 12, H = 1, O = 16
Calculating its molar mass: 12x3 + 1x6 + 16
= 36 + 6 + 16 = 58
Therefore, the molar mass of acetone is 58g/mol
20. Salicylic acid, C7H6O3
The elements in C7H6O3 and their relative atomic masses are: C = 12, H = 1, O = 16
Calculating its molar mass: 12x7 + 1x6 + 16x3
= 84 + 6 + 48 = 138
Therefore, the molar mass of salicylic acid is 138g/mol
21. Benzene, C6H6 The elements in C6H6 and their relative atomic masses are: C = 12, H = 1
Calculating the molar mass: 12x6 + 1x6
= 72 + 6 = 78
The molar mass of benzene is therefore 78g/mol
22. Aspirin, C9H8O4
The relative atomic masses of elements in C9H8O4 are: C = 12, H = 1, O = 16
Calculating its molar mass: 12x9 + 1x8 + 16x4
= 108 + 8 + 64 = 180
Therefore, the molar mass of aspirin is 180g/mol
How to Determine the Molar Mass of Air
Air is a mixture of gases; therefore, its molar mass can be determined by summing up the molar masses of all the constituent gases.
The gases that make up air are: oxygen, nitrogen, carbon dioxide (or carbon(IV) oxide), hydrogen, argon, neon, helium, krypton, xenon. The actual molar mass of each gas present in the mixture can be found out by multiplying the mass of one molecule of the gas by its volume ratio to dry air (or the percentage of the gas in dry air).
The volume ratio of the gases to dry air (or the percentage of the gases in dry air) are given as: oxygen (0.2095 or 20.95%); nitrogen (0.7809 or 78.09%); carbon dioxide (0.0003 or 0.03%); hydrogen (0.0000005 or 0.00005%); argon (0.00933 or 0.933%); neon (0.000018 or 0.0018%); helium (0.000005 or 0.0005%); krypton (0.000001 or 0.0001%); xenon (0.09 x 10-6 or 0.09 x 10-4 %).
Calculating the molar mass of each gas in air: Oxygen, O2, molecular mass = 16 x 2 = 32. Molar mass of oxygen in air = 32 x 0.2095 = 6.704g/mol
Nitrogen, N2, molecular mass = 14 x 2 = 28. Molar mass of nitrogen in air = 28 x 0.7809 = 21.88g/mol
Carbon dioxide, CO2, molecular mass = 12 + 16x2 = 12 + 32 = 44. Molar mass of carbon dioxide in air = 44 x 0.0003 = 0.0132g/mol
Hydrogen, H2, molecular mass = 1 x 2 = 2. Molar mass of hydrogen in air = 2 x 0.0000005 = 0.000001g/mol
Argon, Ar, atomic mass = 40. Molar mass of argon in air = 40 x 0.00933 = 0.3732g/mol
Neon, Ne, atomic mass = 20. Molar mass of neon in air = 20 x 0.000018 = 0.00036g/mol
Helium, He, atomic mass = 4. Molar mass of helium in air = 4 x 0.000005 = 0.00002g/mol
Krypton, Kr, atomic mass = 84. Molar mass of krypton in air = 84 x 0.000001 = 0.000084g/mol
Xenon, Xe, atomic mass = 131. Molar mass of xenon in air = 131 x 0.09 x 10-6 = 0.1179 x 10-4
Molar mass of dry air = 6.704 + 21.88 + 0.0132 + 0.3732 = 28.97g/mol
Notice that we didn’t include the molar masses of hydrogen, neon, helium, krypton, and xenon because they are too small to significantly affect the overall molar mass of air.
The Molar Mass of Diatomic or Polyatomic Molecule
The molar mass of a diatomic or polyatomic molecule is same as the atomic mass of its element multiplied by the number of atoms contained.
Examples: Determine the molar mass of the following:
1. Nitrogen, N2
The relative atomic mass of nitrogen is 14. The molar mass of nitrogen molecule, N2 is 14 x 2 = 28g/mol
2. Oxygen, O2
The relative atomic mass of Oxygen is 16, therefore, the molar mass of oxygen, O2 is 16 x 2 = 32g/mol
3. Chlorine, Cl2
The relative atomic mass of chlorine is 35.5. The molar mass of chlorine molecule Cl2 is 35.5 x 2 = 71g/mol
4. Hydrogen, H2
The relative atomic mass of hydrogen is 1, therefore, its molar mass is 1 x 2 = 2g/mol
5. Sulphur, S8 The relative atomic mass of sulphur is 32, therefore, the molar mass of S8 is 32 x 8 = 256g/mol.
What is Molarity?
The molarity of a solution can be defined as the concentration of the solution in terms of the number of moles of the solute in 1 dm3 (or 1 liter) of the solution.
Mathematically, the molarity of a solution can be expressed as:
Molarity (M) = number of moles/volume (dm3 or liter) A 1 dm3 of solution which has 1 mole of solute dissolved in it is said to be 1 molar.
As can be deduced, molarity is another way of expressing the concentration of solutions of chemical compounds. The symbol of molarity is capital letter M.
Calculating the Molarity of a Chemical Solution
Question 1: How many grams of sugar, C12H22O11, are needed to prepare 500 cm3 of 0.25 M solution?
Molarity of solution = number of moles of solute/volume (dm3)
Number of moles of solute = molarity x volume (dm3)
Number of moles of sugar = 0.25 M x 0.5 dm3
= 0.125 moles
Note that the volume has to be changed from cm3 to dm3
From the equation;
Number of moles = conc. (g) / molar mass
Conc. (g) = number of moles x molar mass
Molar mass of sugar = 12 x 12 + 1 x 22 + 16 x 11 = 144 + 22 + 176 = 342
Molar mass of sugar = 342 g/mol
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Conc. (g) = number of moles x molar mass
Conc. (g) = 0.125 x 342 = 42.75 g
Question 2. What is the molarity of a solution prepared by dissolving 13.2 g of NaCl in sufficient water to give 2.00 liters?
Solution: First, determine the number of moles of NaCl in 13.2 g.
Number of moles = conc. (g) /molar mass
= 13.2/58.5 = 0.23 moles
Note that the molar mass of NaCl is 23 + 35.5 = 58.5 g/mol
Molarity = number of moles/volume (liter) = 0.23/2 = 0.115 M
Question 3. What volume of 2.00 M NaOH would be needed to furnish 12.0 g of NaOH for a reaction?
Solution: Convert 12.0 g of NaOH to number of moles using the equation:
Number of moles = conc. (g)/molar mass
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Therefore, Number of moles of NaOH = 12/40 = 0.30
From the equation,
Molarity = number of moles/volume (dm3)
Volume = number of moles/molarity
Volume = 0.3/2.0 = 0.15 dm3
What is Molality?
Molality is another way of expressing the concentration of a solution. It can be defined as the number of moles of a solute dissolved in 1 kilogram of the solvent.
The symbol for molality is m (small m in italics) or –m (small m with a hyphen). This is to differentiate molality from molarity which is designated by capital letter M; and to different it from mass, which is symbolized with small letter m.
Molality = number of moles/weight of solvent (kg)
The S.I. Unit of molality is therefore, mol/kg
Difference between Molality and Molarity
The preparation of a solution of a given molality involves weighing both solute and solvent and getting their masses. But in the case of molarity, the volume of the solution is measured, which leaves room for variations in density as a result of the ambient condition of temperature and pressure.
This means that it is advantageous to work with molality which deals with mass because in chemical composition, the mass of a known pure substance is more important than its volume since volumes can alter under the effect of temperature and pressure while the mass remain unchanged.
Also, chemical reactions take place in proportion of mass, and not volume. For being mass-based, molality can easily be converted into a mass ration or mass fraction, “w” ration.
Similarity between Molality and Molarity
Both molality and molarity are nearly the same for weak aqueous solutions. For example, 1 kg of water occupies 1 liter (or dm3) of volume at room temperature and the volume is not significantly affected by small amount of solutes.
Question: If 300 g of sugar, C12H22O11, are dissolved in 1500 g of water, what is the molality of the solution?
Molality = number of moles/weight of solvent (kg)
Number of moles of sugar = conc. (g)/molar mass
Molar mass of C12H22O11 = 12x12 + 1x22 + 16x11
= 144 + 22 + 176 = 342 g/mol
Number of moles of sugar = 300/342 = 0.877 mols
Molality = 0.877 mols/ 1.5 kg = 0.585 m
To understand gas laws, you first need to understand the basic characteristics of gases. These include the following facts:
- Gases are easily compressible – i.e., in the gaseous state, matter is not compact.
-Gases exert pressure.
-Gases are highly expansible.
-Gases diffuse completely into one another, so that any mixture of gases is homogeneous.
Note: the parameters used in describing gases are volume; pressure and temperature. A change in the temperature or pressure will affect the volume.
Boyle’s law states that the volume of a given mass of a gas is inversely proportional to the pressure at constant temperature.
What the law is simply saying is that at constant temperature, doubly the pressure of a given mass of gas, the volume is decreased by half.
I.e. V α 1/P ;
V = k/P
PV = k (the product of P and V is a constant).
lf V1 is the original or initial volume of the gas at pressure P1, and V2 the volume at a final pressure P2, then
V1 / V2 = P2 / P1
or P1V1 = P2V2
Boyle’s law is illustrated by the P – V curve below:
Note: Boyle’s law is not obeyed at low temperatures and high pressures (experiments show that PV Product is not constant at these conditions).
Calculation Based on Boyle’s Law:
A flask is made of two vessels A (100 cm3 capacity) and B (300 cm3 capacity) separated by a tap. A was filled with a gas at 748 KNM-2 pressure and B was evacuated until the pressure inside it was negligible. Assuming that the temperature remains constant, calculate the pressure in the flask when the tap is opened, allowing gas to flow from A to B.
V1/V2 = P2/P1 Where V1 = 100 cm3, V2 = sum of volumes A and B = 400 cm3, P1 = 748 KNM-2, V1 = ?
100/400 = P2/748
P2 = 748 x 100/400 = 187 KNM-2
Note: The initial pressure is due only to the effect of gas A, thus initial volume is that of only A as well.
You don’t have to convert volume to m3, leave it in whatever unit it is given.
Charles’s law states that, at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. This means that if the temperature of a gas is doubled, its volume is doubled as well, and if the temperature is halved, its volume is also halved.
I.e., V α T ;
V = kT ;
If V1 is the volume of a fixed mass of a gas at the absolute temperature T1, and V2 is volume at temperature T2,
then: V1/TI = V2/T2 = k
The figure below illustrates Charles’ law.
If a gas were to follow the above relationship indefinitely upon cooling, its volume would be reduced to 0 (zero) at -273 oC (or 0 oK). Actually, all gases have been found to liquefy before this temperature is reached. This temperature is called the absolute zero.
It is the temperature (-273 oC or 0 oK) whereby theoretically, gases become liquid.
Note: Charles’ law, like Boyle’s law is not obeyed at low temperatures and high pressures - the observed volume is less than that predicted by Charles’ law.
Hence, both Charles’ and Boyle’s laws only apply to the behavior of an ideal or perfect gas.
Calculation Based on Charles' Law
A fixed mass of gas has a volume of 200 cm3 at 27oC, calculate the volume when the temperature is increased to 87oC and the pressure kept constant.
V1/TI = V2/T2 Where V1 = 200 cm3, V2 = ?, T1 = (273 + 27 = 300 K) , T2 = (273 + 87 = 360 K)
200/300 = V2/360
V2 = 360 X 200 /300 = 2 X 120 = 240 cm3
Combined Gas Law
A combination of Boyle’s and Charles’ laws gives the combined gas law.
I.e. P1V1/T1 = P2V2/T2 = k
This can be used to calculate the temperature, pressure or volume of a given mass of a gas under different conditions.
Calculation Based on Combined Gas Law
A given mass of a gas has a volume of 7.5 dm3 at 250 K and 80 K N M-2. Calculate the pressure at which it will have a volume of 7.2 dm3 at 270 K.
Solution: P1V1/T1 = P2V2/T2
Where P1 = 80 KNM-2, P2 = ?, V1 = 7.5 dm3, V2 = 7.2 dm3, T1 = 250 K, T2 = 270 K
80 X 7.5 / 250 = P2 x 7.2 / 270
P2 = 270 x 80 x 7.5 / 250 x 7.2 = 90 KNM-2
Ideal Gas Equation
Ideal Gas Equation Combining Boyle’s, Charles’ and Avogadro’s laws, we obtain:
PV = nRT - this is known as equation of state of an ideal gas (or general gas law equation). n is the number of moles of the gas and R is the gas constant. P, V and T are the pressure, volume and the absolute temperature of the gas respectively.
Plotting the graph of PV/RT over pressure, we obtain the horizontal straight lines (see below) for idea or perfect gases.
Line (I) represents 2 moles of an ideal gas, while line (II) represents 1 mole of an ideal gas. Lines (III) and (IV) represent real gases – real gases deviate from ideal behaviors at high pressures and low temperatures.
Notice that at the limit of zero pressure, all gases show perfect or ideal behavior.
Calculation Based on Ideal Gas Equation
Calculate the volume occupied by 5.0 g of carbon monoxide gas at 25 oC and 0.97 4 atm pressure. (R = 0.082 liter-atm per mole per deg, C=12, O=16).
T = (273 + 25) = 298 K, P = 0.974 atm
n = number of moles = mass/molar mass = 5.0 /28 = 0.18 mole
V = nRT/P = 0.18x0.082x298/0.974 = 4.5 liter (or dm3)
Dalton’s Law of Partial Pressure
Dalton’s law of partial pressure states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases, provided the gases do not react chemically with each other.
By this law, if we have a mixture of gases A, B and C, of partial pressures PA, PB and PC respectively in a vessel without them reacting, then the total pressure they exert is the sum of their partial pressures.
I.e. P = PA + PB + PC
Note: the law implies that each gas in the mixture, for as long as there is no chemical reactions taking place, exert a pressure same as when it is alone in the vessel. This pressure is known as its partial pressure.
A mixture of gases, A, B, and C, containing nA, nB and nC number of moles respectively, and exerting a total pressure of P will have their individual partial pressures derived by using their mole ratios (or mole fractions). I.e.
Where PA = partial pressure of A ; PB = partial pressure of B ; PC = partial pressure of C . P = total pressure; and
is the mole ratio or mole fraction of any of the gases in the mixture.
Calculation Based on Dalton's Law of Partial Pressure
2 moles of oxygen gas, 3 moles of nitrogen gas and 1 mole of carbon(IV) oxide gas are put in a gas balloon whose pressure is maintained at 3.019 x 105 N M-2 at 28 oC. What is the partial pressure of nitrogen?
Graham’s Law of Diffusion
Graham’s law of diffusion states that the rate of diffusion of a gas at a given temperature is inversely proportional to the square root of its molecular mass, m.
I.e. R α 1/√m
Also, vapour density (&rho = molar mass (g)/molar volume (v)
Molar mass = vapor density (&rho x molar volume (v)
Therefore, R α 1/√ρv
For two gases with molecular masses m1 and m2 ; and rates of diffusion R1 and R2:
R1/R2 = √m2/√m1 – (1)
R1/R2 = √ρ2v2 /√ρ1v1 – (2)
Also, rate of diffusion = volume/time I.e. the greater the rate, the shorter the time.
Comparing the rates of diffusion of equal volumes of two gases in times t1 and t2, we have R1 = V/t1 and R2 = V/t2
- The lighter or less dense a gas is, the greater is its rate of diffusion.
- If the gases are at the same temperature and pressure, they must have the same molar volumes. Therefore,
R1/R2 = √ρ2 /√ρ1 – (3)
R1/R2 = t2/t1 – (4)
Therefore, R1/R2 = t2/t1 = √m2 /√m1 = (5)
By the above last equation, equation (5), you could determine the value of any of the parameters by establishing an equation with those whose values are given.
- The vapour density (v.d.) of a gas is equal to half its relative molecular mass(r.m.m.).
I.e. r.m.m. = 2 x v. d.
Calculation Based on Graham's Law of Diffusion
30 cm3 of a gas, the empirical formula of which was CH3, diffuses through a porous partition in 45.2 s. 30 cm3 of hydrogen diffused in 11.7 s under the same conditions. Calculate (i) the vapour density of the gas (ii) the molecular formula of the gas?
(i). The two gases are of equal volumes,
tx/tH = √mx /√mH
45.2/11.7 = √(mx/2)
2(45.2/11.7)2 = mx
mx = 2 x 14.92 = 29.84 (g)
r.m.m. = 2 x v.d.
v.d. = r.m.m./2
v.d. = 29.84/2 = 14.92
(ii). xCH3 = 30
15x = 30
x = 2
Therefore, molecular formula = C2H6
The Kinetic Theory of Matter
The kinetic theory of matter recognizes that matter is composed of very small particles (ions, atoms and molecules) whose different pattern of arrangements and motions result in the different possible states in which matter can occur. It also explains the properties of these states.
The kinetic theory of gases explains the empirical laws (i.e. Boyle’s law, Charles’ law, Graham’s law of diffusion, Dalton’s law of partial pressure and Avogadro’s law), which govern the behavior of gases.
The theory is based on the following assumptions about perfect or ideal gases (these are gases whose behaviors can be explained by the kinetic theory - they have no real existence):
- Gases are composed of discrete particles called molecules which are in rapid, random motion, moving at high speed in straight lines - this is the reason gases can diffuse very rapidly.
- The molecules are so small, and at low pressures are so far apart, such that on the average, the actual volume of their molecules is negligible compared with the volume of their container
- this is the reason gases do not have fixed volume, but take up the volume of the container in which they are kept.
- The molecules exert no force of attraction or repulsion upon one another. I.e., the molecules are independent of each other - this is the reason gases do not have definite shape.
- Upon collision with one another, or with any surface, they rebound without any loss in the total kinetic energy of the system. In other words, the molecular collisions are perfectly elastic.
- The pressure of the gas results from the impacts of the molecules upon the walls of the containing vessel. The pressure exerted by a gas confined within a fixed volume is proportional to nEk , i.e., the number of molecules per unit volume (n) times their average kinetic energy (Ek).
- The average kinetic energy of all the molecules is assumed to be directly proportional to the absolute temperature of the gas. This means that molecules of different gases at the same temperature have the same average kinetic energy. The kinetic energy of a moving molecule, like that of any moving object, is the energy associated with its motion.
The quantity of kinetic energy such a molecule possesses is equal to the work it is capable of performing in being brought to a rest, and is given by the expression:
Ek = 1/2 mu2
where m is its mass and u its velocity
Note: at ordinary temperature and pressure, real gases approximately fulfill the above assumptions.
Application of the Kinetic Theory of Matter to Explain the Nature of Gases
The three properties of gases that are especially important are diffusibility, thermal expansion and compressibility.
All gases are characterized by diffusibility, but the rates at which different gases diffuse depend on their molecular weights.
When heated, gases expand to a much greater extent than do solids or liquids- all gases tend to behave alike in this respect. In comparison with solids and liquids, gases are very easily compressed - all gases tend to behave alike in this regard also. These properties, as well as the empirical laws governing the behavior of gases can be explained by the kinetic theory.
Explanation of Diffusion of Gases by the Kinetic Theory
Diffusion is a phenomenon whereby particles of a substance move from an area of high concentration into an area of low concentration - gases diffuse rapidly. For example, if a small quantity of an odorous gas, e.g. hydrogen sulphide, is released at one point of a room, the smell soon gets to all parts of the room. This (diffusion), can be explained using the kinetic theory of gases.
From the assumptions of the theory, we have that:
I. Gases are made of discrete particles called molecules, and not a single piece. If they were made of a single piece, then, the smell of the hydrogen sulphide would not pervade the whole room at the same time, but would probably be perceived at one corner of the room at a time.
II. The molecules are relatively far apart and are in rapid, random motion, moving at high speeds in straight lines. The spontaneous diffusion of hydrogen sulphide and air into each other is the result of molecules of each kind moving into the spaces between the molecules of the other kind.
This account for the smell getting to every part of the room in a couple of minutes after the release.
Explanation of Compressibility of Gases By the Kinetic Theory
Compressibility of gases can be explained from the assumption of the kinetic theory, which states that a gas consists of particles that are separated from one another by large spaces. Based on this, it is therefore easy to bring the molecules closer together (i.e. compressed) when the volume of the container is reduced.
Reduction in volume leads to decrease in temperature (according to Charles’ law, V α T). Hence, compression of gases results in a drop of temperature in the system - the kinetic energy of the system also drops.
Explanation of Expansion of Gases by the Kinetic Theory
Expansion of gases can be explained by the kinetic theory from the assumptions which state that:
1. Gases are in constant rapid motion, moving at great speeds, occupying the volume of the container.
2. The average kinetic energy of all the molecules is assumed to be directly proportional to the absolute temperature of the gas. The greater the average kinetic energy of gas molecules, the greater they are able to move and occupy more volume. Therefore, at higher temperatures, gases obtain higher kinetic energy, and thus expand (or occupy large volumes).
Explanation of Charles’ Law by the Kinetic Theory
The kinetic theory of gases explain Charles’ law thus:
I. Gases, due to their molecules been very far apart, do not have appreciable volumes, but occupy the volume of the vessel in which they are kept - the greater or higher their molecules are able to move in the vessel, the more volume they occupy, and vice versa.
II. Gaseous molecules are able to move or expand because they possess kinetic energy. Their average kinetic energy is directly proportional to the absolute temperature. I.e., the higher the temperature, the greater the average kinetic energy, and the more volume they occupy, and vice versa. In general, we can sum-up the explanation this way: since the volume which gas molecules occupy is directly dependent upon their movement, which in turn is directly dependent upon their kinetic energy, and which in turn is directly dependent on the absolute temperature, it goes to prove that the volume is directly proportional to its absolute temperature at constant pressure – Charles’ law (V α T).
Note: Another way Charles’ law can be expressed is: that the volume is directly proportional to the average kinetic energy of the gas molecules at constant pressure.
Explanation of Boyle’s Law by the Kinetic Theory
Boyle’s law gives an inverse relationship between the volume of a fixed mass of gas and its pressure at constant temperature. I.e., V α 1/P. The kinetic theory explains the law as follows: From the assumptions of the kinetic theory of gases, the pressure exerted by a gas is due to the collisions between the gas molecules and the walls of the vessel. Hence, the greater the rate of collision, the greater the pressure.
At lower volumes, molecules are closer to one another, thus they collide more frequently with the walls of the container, leading to increase in pressure. The reverse would occur if volumes were increased. Therefore, the relationship between volumes and pressures of a gas at constant temperature is an inverse one as given by Boyle’s law.
Explanation of Dalton’s Law of Partial Pressure by the Kinetic Theory
Dalton’s law of partial pressure can be explained from the assumption which states that there is no attraction between gas molecules. Therefore, in a mixture of gases, each kind of molecule strikes the walls of the container the same number of times per second, and with the same force, as if it were the only kind of molecule present.
Thus, the partial pressure of a gas is not changed by the presence of other gases. Each partial pressure is proportional to the number of molecules of that species present, and the total pressure exerted by the mixture of gases is the sum of the pressures exerted by the individual components.
Explanation of Graham’s Law of Diffusion by the Kinetic Theory
Graham’s law of diffusion can be explained from the assumption which states that the average kinetic energy of gas molecules is directly proportional to the absolute temperature - molecules of different gases at the same temperature have the same kinetic energy. This means that molecules of one kind, A, must have velocities (or rate of diffusion) that are different from velocities (or rate of diffusion) of molecules of another gas, B, unless they have the same masses.
½ mAu2A = ½ mBu2B
and u2A/u2B = mB/mA
So that uA/uB = √mB/mA
or RA/RB = √mB/mA
For example, a methane molecule (molecular weight, 16), A, has a mass one fourth as great as the mass of a sulphur(IV) oxide molecule (molecular weight, 64), B. If
mB/mA = 4
uA/uB = √4 = 2
or uA = 2 x uB
The methane molecules have an average velocity twice that of the sulphur(IV) oxide molecules. Thus, the kinetic theory shows why methane diffuses twice as fast as sulphur(IV) oxide, and why , in general, different gases diffuse at rates that vary inversely with the square roots of their molecular weights or densities.
Explanation of Avogadro’s Law by the Kinetic Theory
Avogadro’s law can be explained from the assumptions:
1. The average kinetic energy of gas molecules is proportional to the absolute temperature. Hence, at the same temperature, the average kinetic energy of two different gases are the same.
2. The pressure exerted by a gas confined within a fixed volume is proportional to nEk , i.e., the number of molecules per unit volume times their average kinetic energy. Therefore, if two gases are at the same temperature, then Ek (the average kinetic energy per molecule) is the same for both gases, and if the two gases are also exerting the same pressure, then nEk is the same for the two gases.
Thus, if two gases are at the same temperature and pressure, n must be the same for both. The number of molecules per unit volume is the same - i.e., equal volumes of different gases contain the same number of molecules if they are at the same temperature and pressure.
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