Second Term Lesson Note for Week Four
Class : SSS 1
Subject : Chemistry
Topic : STATE , ILLUSTRATION AND VERIFICATION OF CHEMICAL LAWS
Duration : 40 Minutes
Period : Single Period
Reference Book :
- New School Chemistry for Senior Secondary School.
- Lagos State Unified schemes of work for Senior Secondary School, SSS 1 – 3.
- Online Resources
Instructional Material : Chart showing the illustration of the laws of conservation by mass.
Learning Objectives : By the end of the lesson learners will be able to :
i. State the law of conservation by mass.
ii. Illustrate and verify the laws of conservation by mass.
Content :
STATE , ILLUSTRATION AND VERIFICATION OF CHEMICAL LAWS
Law of Conservation of Mass
In 1774, Joseph Priestley isolated the gas oxygen by heating mercuric oxide. Soon thereafter, Antoine Lavoisier claimed that oxygen is the key substance involved in combustion (burning). He also demonstrated that when combustion is carried out in a closed container, the mass of the final products of combustion exactly equals the mass of the starting reactants. This led to the statement of the Law of Conservation of Mass:
Law of Conservation of Mass
Mass is neither created nor destroyed in chemical reactions.
In an experiment, 63.5g of copper combines with 16g of oxygen to give 79.5g of cupric oxide (a black oxide of copper). This is in agreement with the law of conservation of mass.
Science today knows that matter can be converted into energy (and vice-versa). Hence, during all chemical and physical changes, the total mass+energy before the change is equal to the total mass+energy after the change. Still, as there is no detectable change in mass in an ordinary chemical reaction, the law of conservation of mass is still valid.
Silicon dioxide, made up of elements silicon and oxygen, contains 46.7% by mass of silicon. With what mass of oxygen will 10g of silicon combine?
100g of silicon dioxide contains : 46.7g of silicon,
or : (100 – 46.7) i.e. 53.3g of oxygen.
∴ 10g of silicon will contain 10100 × 53.3=5.33g of oxygen.
Law of Definite Proportions / Constant Composition
In the years following Lavoisier, the French chemist Joseph Proust formulated a second fundamental law of chemical science – the Law of Definite Proportions.
Law of Definite Proportions (Law of Constant Composition)
In a given compound, the constituent elements are always combined in the same proportions by mass, regardless of the origin or mode of preparation of the compound.
What this law means is that when elements react chemically, they combine in specific proportions, not in random proportions.
A sample of pure water, whatever the source, always contains 88.9% by mass of oxygen and 11.1% by mass of hydrogen.
The compound cupric oxide may be prepared by any one of the following methods –
• Heating copper in oxygen.
• Dissolving copper in nitric acid and igniting the cupric nitrate formed.
• Dissolving copper in nitric acid, precipitating cupric hydroxide, and strongly heating the cupric hydroxide.
– and in each case, the ratio copper:oxygen by mass is always constant.
2.16g of mercuric oxide gave on decomposition 0.16g of oxygen. In another experiment 16g of mercury was obtained by the decomposition of 17.28g of mercuric oxide. Show that these data conform to the law of definite proportions.
Experiment 1:
Mass of mercuric oxide = 2.16g
Mass of oxygen evolved from it = 0.16g
∴ Mass of silicon in the compound = 2.16 – 0.16 = 2.00
∴ silicon:oxygen ratio = 2.000.16 = 12.5 : 1
Experiment 2:
Mass of mercuric oxide = 17.28g
Mass of silicon in it = 16.00g ∴ Mass of oxygen in the compound = 17.28 – 16.00 = 1.28g
∴ silicon:oxygen ratio = 16.001.28 = 12.5 : 1
In both cases, the silicon to oxygen ratio is the same, thus conforming to the law of definite proportions.
QUESTIONS
(a) Two different sample, 1 and 2 of Zinc oxide were obtained from different sources. When heated in a stream of hydrogen they wre reduced to yield the results below.
| Zinc Oxide | Mass of Oxide | Mass of Zinc left |
| Sample 1 | 20.0g | 16.22g |
| Sample 2 | 26.4g | 21.7g |
Show that the result above explains the law of constant composition.
(b) If 12.0g of carbon is heated in air, the mass of the product obtained could either be 44.0g or 28.0g depending on the amount of oxygen present. What law does this information support?
Conclusion : The teacher summarizes the lesson and goes over it again for a better understanding.