CHEMICAL LAWS OF COMBINATIONS

FIRST TERM E-LEARNING NOTES

SUBJECT: CHEMISTRY

CLASS: SS1

WEEK 6

TOPIC: LAWS OF CHEMICAL COMBINATION

CONTENT:

  1. Law of conservation of matters
  2. Law of constant composition
  3. 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.

  1. 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.[mediator_tech]

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:

 

  1. Insert the stopper and weight the whole apparatus on a balance, note the mass of the whole system.
  2. Mix the two liquids by pulling the string attached to the bottom end of the small test tube.
  3. 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.

EVALUATION:

  1. Mention another compound that could be used instead of silver trioxo-nitrate(v) with sodium chloride
  2. 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.

 

 

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 strongly while 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.

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 ×100 × 100
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 Definite proportion.

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.

 

 

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.

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).

In water,

CHEMICAL LAWS OF COMBINATIONS

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]

CHEMICAL LAWS OF COMBINATIONS

  1. 100g of CaCO3 yield 56g of CaO

23g of CaCO3 will yield = 12.88g.

  1. 100g of CaCO3 yields 44g of CO2

23g of CaCO3 will yield = 10.12g

EVALUATION:

  1. State the law of multiple proportion
  2. State the law of reciprocal proportions.
  3. 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.

  1. What mass of copper will be produced from the reduction of 7.95g of copper (II) oxide? (C= 63.5, O= 16)
  2. 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 CH2O (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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