LAWS OF CHEMICAL COMBINATION
Subject : Chemistry
LAWS OF CHEMICAL COMBINATION
1st Term / First Term
Instructional Materials :
- Lagos State Scheme of Work
- Textbooks On Social Studies
- Online Materials
- Picture Charts
Previous Knowledge :
The pupils have been taught
CHEMICAL FORMULAE AND EQUATIONS
in their previous lesson
Behavioural Objectives : At the end of the lesson, the pupils should be able to
- LAWS OF CHEMICAL COMBINATION
TOPIC: LAWS OF CHEMICAL COMBINATION
- Law of conservation of matters
- Law of constant composition
- Law of multiple proportions.
PERIOD 1: CHEMICAL LAWS OF COMBINATIONS
There are four laws of chemical combination which describe the general features of a chemical change.
(a) Law of conservation of mass: This law was established by Lavoisier, a French chemist. The law of conservation of mass states that matter is neither created nor destroyed during chemical reaction, but changes from one form to another.
Experiment to verify the law of conservation of matter (mass) Theory:
The equation of the chemical reaction chosen for study is as follows;
Silver nitrate + sodium chloride Silver chloride + Sodium trioxonitrate(v)
- Put some sodium chloride solution in a conical flask
- Fill a small test tube with silver trioxonitrate (iv) solution of string, suspend it in a conical flask as shown below:
- Insert the stopper and weight the whole apparatus on a balance, note the mass of the whole system.
- Mix the two liquids by pulling the string attached to the bottom end of the small test tube.
- Weigh the whole apparatus again.
Result: When the two reactants are mixed together, a white precipitate is formed indicating that a chemical reaction has taken place. The new substances formed are known as the products of the chemical reaction. The masses of the system taken before and after the reaction are found to be the same, indicating that the mass of the reactants equals that of the products.
CONCLUSION: Since there is no overall change in mass when the products are formed,we can infer that matter is neither created nor destroyed during the chemical reaction. The law is, hence valid.
- Mention another compound that could be used instead of silvertrioxonitrate(v) with sodium chloride
- State the law of conservation of mass/matter.
PERIOD 2: LAW OF DEFINITE PROPORTION OR LAW OF CONSTANT COMPOSITION
The second law of chemical combination which is supported by the Atomic theory was proposed by provost (1755-1826) known as the Law of definite proportions or constant composition.
The law of definite proportions states that all pure samples of a particular chemical compound contain similar elements combined in the same proportion by mass. It is based on the fact that when elements combine to form a given compound, they do so in fixed proportions by mass, so that all pure samples of that compound are identical in composition by mass. Water for example: chemical analyses showed that as long as it is pure, its composition is always in the ratio of one mole of oxygen to two moles of hydrogen. i.e. 32g of O to 4g of H. Irrespective of whether the water comes from river, sea, rain or anywhere.
Experiment to verify the law of definite proportion
Method: Prepare two samples of black copper (ii) oxide, each by a different method as given below:
Sample A: Place some coppers turning in a crucible and add some concentrated trioxonitrate (v) acid, a little at a time, until the copper dissolves completely. Evaporate the resulting green solution of copper II oxide trioxonitrate (v) to dryness; continue to heat the residue until it decomposes to give a black solid which is copper II oxide. Keep the black residue dry in desiccator.
Sample B:Place some copper (i) trioxocarbonate (iv) in a crucible and decompose it into copper (ii) oxide and carbon (iv) oxide store the residue in a desiccator.
Determine the amount of copper present in the two samples of copper oxide by reducing the oxide in a stream of hydrogen or carbon II oxide as follows.
- Weigh two clean metal boats.
- Add a reasonable amount of sample A to one and sample B to the other
- Reweigh and determine the mass of each sample. Place the boats inside a hard glass tube as shown. Heat the samples stronglywhile passing a stream of dry hydrogen gas through the tube. After some time, a reddish- brown copper residue is left in each boat. Remove the flame, but continue passing the hydrogen as the copper residues cool down. This presents the re-oxidation of the hot copper residue by atmospheric oxygen. Any water formed during the reaction is absorbed by the fused calcium chloride in the adjacent U-tube.
|Mass of copper II oxide||3.55g||3.02g|
|Mass of copper residue||2.81g||2.42g|
|Percentage of copper present in copper
The percentage of copper residue in the two samples in approximately 80.0, irrespective of the method of preparation of the copper(II) oxide samples.
CONCLUSION: In the pure copper(II) oxide copper and oxygen are always present in a definite proportion by mass of approximately 4 to 1 i.e.
Copper(II)oxide = copper + oxygen
100% 80% 20% Ratio 4 : 1
- State the Law of Definiteproportion.
PERIOD 3: LAW OF MULTIPLE PROPORTIONS
This law states that if two elements combine to form more than one compound, the masses of one of the elements which separately combine with a fixed mass of the other element are in simple ratio.
VERIFICATION OF THE LAW OF MULTIPLE PROPORTIONS
Some elements form more than one compound, depending on the conditions of the reaction and the valency copper forms. Copper (I) and copper(II) with oxygen. Also in an insufficient supply of air, carbon burns to form carbon(II) oxide and when the supply of air is sufficient, carbon(iv) oxide is obtained.
The sample of the copper (I) oxide and copper(II) are placed in porcelain, boats and placed in a combination tube as in the diagram below.
A current of dry hydrogen is passed through the combustion tube until the oxides are reduced to metallic coppers. They are now cooled and weighed and the masses of copper and oxygen are determined in the two samples.
|Calculations||Sample A||Sample B|
|(i)Mass of porcelain boat||4.55g||5.38g|
|(ii) Mass of porcelain copper oxide||boat||+||6.44g||8.21g|
|(iii) Mass of copper oxide||1.89g||2.83g|
|(iv) Mass of porcelain||boat||+||6.05g||7.90g|
|(v) Mass of copper (iv) – (i)||1.50g||2.52g|
|(vi) Mass of oxygen (iii) –(v)||0.39g||0.31g|
For example A1.50g of copper combines with 0.39 of oxygen.
100g of copper combines with × 100 = 26g
For sample (b) 2.52g of copper combines with 0.31g of oxygen
100g of copper combines with × 100= 12.3g
From these calculations, the masses of oxygen (26g and 12.3g) which combine with a fixed mass (100g) of copper are in simple ratio 2:1
PERIOD 4: LAW OF RECIPROCAL PROPORTION
This is the fourth law of chemical combination. This law states that the masses of several elements,A,B,C,which combine separately with a fixed mass of another element,D,are the same as ,or simple multiples of ,the masses in which A,B,C,themselves combine with one another.For example C, H, O (12, 1, 16) respectively. Carbon and hydrogen combine to form methane (CH4). Carbon and oxygen combine to form carbon (iv) oxide, (CO2) and hydrogen and oxygen combine to form water (H2O).
This is the prediction of the law of reciprocal proportions. For example, 23g of calcium trioxocarbonate (iv) on heating decomposes to give calcium oxide (CaO) and carbon (iv) oxide. Calculate the masses of calcium oxide and carbon (iv) oxide produced [C= 40, O = 16, C= 12]
- 100g of CaCO3yield 56g of CaO
23g of CaCO3 will yield = 12.88g.
- 100g of CaCO3 yields 44g of CO2
23g of CaCO3 will yield = 10.12g
(1) State the law of multiple proportion (2) State the law of reciprocal proportions.
(3) Zn(s) + 2HCl(aq)ZnCl2(aq)+ H2(g)
|Mass of sample oxide||30.4g||1.91g|
|Mass of copper residue||2.55g||1.38g|
|Mass of oxygen removed from oxide||0.49g||0.53g|
From the above table, calculate the various masses of copper which would combine separately with a fixed mass of 1 g of oxygen.
- What mass of copper will be produced from the reductionof 7.95g of copper (II) oxide? (C= 63.5, O= 16)
- Write down the names of these chemical compounds:(i) HNO3 (ii) CuCl2(iii) CaCO3(iv) Fe2O34. Write the symbol and the valency of the following.(i) Boron (ii) Carbon (iii) Sulphur (iv) Argon 5. Calculate the formula of a compound with 31.9% potassium 28.93%, chlorine and the rest oxygen. K=39, Cl =355, O=16
- Which of the following relative molecular mass has empirical formula CH2O (H=1 C=12, O=16). (a) 42 (b) 80 (c) 4 (d) 60
- The relative molecular mass of tetraoxosulphate(VI) acid is? (a) 98 (b) 49 (b) 49 (c) 96 (d) 106
- Chemical equations will provide all these except. (a) State of chemicals is solved (b) Direction of reaction (c) Mass of products (d) Reactants
- All pure samples of a particular compound contain the same elements combined in the same proportion by mass. The statement is the law of (a) Definite proportion (b) Multiple proportion (c)Conservation of mass or matter (d) Atomic proportion
CHEMICAL LAWS OF COMBINATIONS.
1) The Law of Constant Composition
This law states that the composition of a given compound is always the same, regardless of how it is made. This means that the relative proportions of the elements in a compound are always fixed. For example, water is always made up of two hydrogen atoms for every one oxygen atom.
2) The Law of Multiple Proportions
This law states that when two elements combine to form more than one compound, the different compounds will contain the elements in a ratio that is a simple multiple of each other. For example, oxygen can combine with hydrogen to make either water or hydrogen peroxide. In water, there are two hydrogen atoms for every one oxygen atom. In hydrogen peroxide, there are two hydrogen atoms for every one oxygen atom
3) The Law of Definite Proportions
This law states that a given compound always contains the same elements in exactly the same proportions by mass. This means that the relative masses of the elements in a compound are always fixed. For example, water is always made up of 8 grams of oxygen for every 1 gram of hydrogen
4) The Law of Avogadro’s Number
This law states that equal volumes of gases at the same temperature and pressure contain the same number of molecules. This means that the ratio between the volumes of different gases is equal to the ratio between their molar masses. For example, if we have two gases, A and B, that are made up of different molecules, we can compare the volumes of the two gases at the same temperature and pressure. If gas A is twice as big as gas B, then the molar mass of A must be twice the molar mass of B
5) The Ideal Gas Law
This law states that the pressure of a gas is directly proportional to its Kelvin temperature and inversely proportional to its volume. This means that when the temperature of a gas increases, the pressure of the gas increases.