Name_____________________________________________ Date_______________
Week 1 & 2
LAWS OF CHEMICAL COMBINATION
CONTENT:
1. Law of conservation of matters
2. Law of constant composition
3. Law of multiple proportions.
Lesson objectives;
By the end of the class, students will be able to
State the law of multiple proportion
State the law of reciprocal proportions.
CHEMICAL LAWS OF COMBINATIONS
There are four laws of chemical combination which describe the general features of a
chemical change.
Law of conservation of mass: This law was established by Lavoisier, a French chemist. The
law of conservation of mass states that matter is neither created nor destroyed during
chemical reaction, but changes from one form to another.
Experiment to verify the law of conservation of matter (mass)
Theory:
The equation of the chemical reaction chosen for study is as follows;
Silver nitrate + sodium chloride ❑ → Silver chloride + Sodium trioxonitrate(v)
(White precipitate)
Method:
1. Put some sodium chloride solution in a conical flask
2. 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.
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CONCLUSION: Since there is no overall change in mass when the products are formed,we
can infer that matter is neither created nor destroyed during the chemical reaction. The law
is, hence valid.
EVALUATION:
1. Mention another compound that could be used instead of silvertrioxo-nitrate(v) with
sodium chloride
2. State the law of conservation of mass/matter.
LAW OF DEFINITE PROPORTION OR LAW OF CONSTANT COMPOSITION
The second law of chemical combination which is supported by the Atomic theory was
proposed by provost (1755-1826) known as the Law of definite proportions or constant
composition.
The law of definite proportions states that all pure samples of a particular chemical
compound contain similar elements combined in the same proportion by mass. It is based
on the fact that when elements combine to form a given compound, they do so in fixed
proportions by mass, so that all pure samples of that compound are identical in composition
by mass. Water for example: chemical analyses showed that as long as it is pure, its
composition is always in the ratio of one mole of oxygen to two moles of hydrogen. i.e. 32g
of O to 4g of H. Irrespective of whether the water comes from river, sea, rain or anywhere.
Experiment to verify the law of definite proportion
Method: Prepare two samples of black copper (ii) oxide, each by a different method as given
below:
Sample A: Place some coppers turning in a crucible and add some concentrated
trioxonitrate (v) acid, a little at a time, until the copper dissolves completely. Evaporate the
resulting green solution of copper II oxide trioxonitrate (v) to dryness; continue to heat the
residue until it decomposes to give a black solid which is copper II oxide. Keep the black
residue dry in desiccator.
Sample B:Place some copper (i) trioxocarbonate (iv) in a crucible and decompose it into
copper (ii) oxide and carbon (iv) oxide store the residue in a desiccator.
ANALYSES:
Determine the amount of copper present in the two samples of copper oxide by reducing
the oxide in a stream of hydrogen or carbon II oxide as follows.
1.Weigh two clean metal boats.
2.Add a reasonable amount of sample A to one and sample B to the other
3.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.
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Result:
Sample A B
Mass of copper II oxide 3.55g 3.02g
Mass of copper residue 2.81g 2.42g
Percentage of copper present in copper (ii) oxide 2.81 2.42
×100 × 100
3.55 3.02
79.2% 80.1%
The percentage of copper residue in the two samples in approximately 80.0, irrespective of
the method of preparation of the copper(II) oxide samples.
CONCLUSION: In the pure copper(II) oxide copper and oxygen are always present in a
definite proportion by mass of approximately 4 to 1 i.e.
Copper(II)oxide = copper + oxygen
100% 80% 20%
Ratio 4 : 1
EVALUATION:
1.State the Law of Definiteproportion.
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.
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A current of dry hydrogen is passed through the combustion tube until the oxides are
reduced to metallic coppers. They are now cooled and weighed and the masses of copper
and oxygen are determined in the two samples.
Calculations Sample A Sample B
(i)Mass of porcelain boat 4.55g 5.38g
(ii) Mass of porcelain boat + copper oxide 6.44g 8.21g
(iii) Mass of copper oxide 1.89g 2.83g
(iv) Mass of porcelain boat + copper 6.05g 7.90g
(v) Mass of copper (iv) – (i) 1.50g 2.52g
(vi) Mass of oxygen (iii) –(v) 0.39g 0.31g
For example A1.50g of copper combines with 0.39 of oxygen.
0.39
∴ 100g of copper combines with × 100 = 26g
1.50
For sample (b) 2.52g of copper combines with 0.31g of oxygen
0.31
∴ 100g of copper combines with × 100= 12.3g
2.52
From these calculations, the masses of oxygen (26g and 12.3g) which combine with a fixed
mass (100g) of copper are in simple ratio 2:1
LAW OF RECIPROCAL PROPORTION
This is the fourth law of chemical combination. This law states that the masses of several
elements,A,B,C,which combine separately with a fixed mass of another element,D,are the
same as ,or simple multiples of ,the masses in which A,B,C,themselves combine with one
another.For example C, H, O (12, 1, 16) respectively. Carbon and hydrogen combine to form
methane (CH4). Carbon and oxygen combine to form carbon (iv) oxide, (CO2) and hydrogen
and oxygen combine to form water (H2O).
In water,
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]
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100g of CaCO3yield 56g of CaO
56 ×23 g
∴ 23g of CaCO3 will yield = 12.88g.
100
100g of CaCO3 yields 44g of CO2
44 × 23 g
∴23g of CaCO3 will yield = 10.12g
100
EVALUATION:
State the law of multiple proportion
State the law of reciprocal proportions.
Zn(s) + 2HCl(aq)❑
→ ZnCl2(aq)+ H2(g)
GENERAL EVALUATION:
ESSAY QUESTIONS
Copper(I)oxide Copper(II)oxide
Mass of sample oxide 30.4g 1.91g
Mass of copper residue 2.55g 1.38g
Mass of oxygen removed from oxide 0.49g 0.53g
From the above table, calculate the various masses of copper which would combine
separately with a fixed mass of 1 g of oxygen.
2. What mass of copper will be produced from the reductionof 7.95g of copper (II) oxide?
(C= 63.5, O= 16)
3. Write down the names of these chemical compounds:(i) HNO3 (ii) CuCl2(iii) CaCO3(iv) Fe2O3
4. Write the symbol and the valency of the following.(i) Boron (ii) Carbon (iii) Sulphur (iv)
Argon
5. Calculate the formula of a compound with 31.9% potassium 28.93%, chlorine and the rest
oxygen. K=39, Cl =355, O=16
OBJECTIVE TEST
1.Which of the following relative molecular mass has empirical formula CH 2O (H=1 C=12,
O=16). (a) 42 (b) 80 (c) 4 (d) 60
2.The relative molecular mass of tetraoxosulphate(VI) acid is? (a) 98 (b) 49 (b) 49 (c) 96 (d)
106
3.Chemical equations will provide all these except. (a) State of chemicals is solved (b)
Direction of reaction (c) Mass of products (d) Reactants
4.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
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