CHEMICAL LAWS

 

Week:  TWO                        Date:                                                   Time:

Period:                                   Duration:                                           Average age of learners:

Subject:                               CHEMISTRY                                                        Class:  SS ONE

Topic:                                                    CHEMICAL LAWS

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA[ 4th Edition]  pg45-48

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials: beam balance

Entry behavior: The students have been taught balancing chemical equation

Behavioural objective: At the end of the lesson the students should be able to:

1st PERIOD

State the law of conservation of mass

Demonstrate  the experiment  to verify law of conservation of mass

2nd  PERIOD

State the law of definite proportion or constant composition

Solve problem on  law of definite proportion or constant composition

3rd PERIOD

State the law of multiple proportion

Solve mathematical problem on law of multiple proportion

CHEMICAL LAWS

1st PERIOD

STEP I: The teacher states the law of conservation of mass

Law of Conservation of Matter

Antoine Lavoisier (1743-1794) restated this principle for chemistry with the law of conservation of mass, which "means that the atoms of an object cannot be created or destroyed, but can be moved around and be changed into different particles." This law says that when a chemical reaction rearranges atoms into a new product, the mass of the reactants (chemicals before the chemical reaction) is the same as the mass of the products (the new chemicals made).

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.

STEP II: The teacher leads the student to demonstrate the law of conservation of mass

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

EVALUATION

The teacher evaluates the lesson by asking the following questions:

State the law of conservation of mass

Explain the experiment to illustrate the law mention above

ASSIGNMENT

Read about law of definite proportion

2nd  PERIOD

STEP I: The teacher States the law of definite proportion or constant composition

Joseph Proust (1754-1826) formulated the law of definite proportions (also called the Law of Constant Composition or Proust's Law). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. 

 Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass.

STEP II: The teacher leads the student to solve problems  on  law of definite proportion or constant Composition

Question

When 1.3 75 g of cupric oxide is reduced on heating in a current of hydrogen, the weight of copper remaining 1.098 g. In another experiment, 1.179 g of copper is dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed is 1.476 g. Show that these results illustrate the law of constant proportion.

SOLUTION

First experiment

Copper oxide = 1.375 g

Copper left = 1.098 g

Oxygen present = 1.375 - 1.098 = 0.277 g

(1.1)Percentage of oxygen in CuO=(0.277)(100%)1.375=20.15%

Second Experiment

Copper taken = 1.179 g

Copper oxide formed = 1.476 g

Oxygen present = 1.476 - 1.179 = 0.297 g

(1.2)Percentage of oxygen in CuO=(0.297)(100%)1.476=20.12%

Percentage of oxygen is approximately (within significant figures limit) the same in both the above cases. So the law of constant composition is illustrated.

EVALUATION:

The teacher evaluates the lesson by asking this question:

i.State the law of definite proportion

3rd PERIOD

STEP I: The teacher states the law of multiple proportion

The law of multiple proportion states that if two elements, A and B combine to form more than one compound, then the various masses of A which combine with a fixed mass of B are in the simple whole number ratio.

STEP II: The teacher leads the students to solve mathematical problem on law of multiple proportion

Many combinations of elements can react to form more than one compound. In such cases, this law states that the weights of one element that combine with a fixed weight of another of these elements are integer multiples of one another. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here.



 Law of Multiple Proportions applied to nitrogen oxides (NOx) compounds.

EXAMPLE:  Consider two separate compounds are formed by only carbon and oxygen. The first compound contains 42.9% carbon and 57.1% oxygen (by mass) and the second compound contains 27.3% carbon and 72.7% oxygen (again by mass). Is this consistent with the law of multiple proportions?

SOLUTION

The Law of Multiple Proportions states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers. Hence, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio.

Thus for every 1 g of the first compound there are 0.57 g of oxygen and 0.429 g of carbon. The mass of oxygen per gram carbon is:

 = 1.33

Similarly, for 1 g of the second compound, there are 0.727 g oxygen and 0.273 g of carbon. The ration of mass of oxygen per gram of carbon is:

    = 2.66

 Dividing the mass of oxygen per g of carbon of the second compound:

 = 2

 Hence the masses of oxygen combine with carbon in a 2:1 ratio which s consistent with the Law of Multiple Proportions since they are whole numbers.

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