The verification of each of the following chemical laws was well established with examples to simplify it.
THE LAW OF CONSERVATION OF MASS
This Law was established by Lavoisier, a French Chemist. This law states that matter is neither created nor destroyed during chemical reaction but changes from one form to another. This means that in a chemical reaction, the total mass of all reacting substances (i.e. the reactants) is equal to the total mass of the products.
THE LAW OF DEFINITE PROPORTIONS OR CONSTANT COMPOSITION
This Law was proposed by Proust (1755-1826). 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.
THE LAW OF MULTIPLE PROPORTIONS
The law of Multiple Proportions states that if two elements, A and B, combine to form more than one chemical compound, the various masses of one element, A which combine separately with a fixed mass of the other element, B, are in simple multiple ratios.
BALANCING CHEMICAL EQUATION WITH CALCULATION
A chemical equation is a shorthand expression for a chemical change or reaction. It shows among other things the arrangement of atoms that are involved in the reaction.
When balancing an equation, you must remember the following:
Know the reacting substances and the products formed.
Know the chemical formulae for all the substances.
Write, in front of the formulae, coefficients that will balance the equation.
Common gases, such as oxygen, hydrogen, chlorine and nitrogen, in the free state, are diatomic, e.g. O2, H2, Cl2 and N2
Other elements in the free state, such as sodium, potassium, copper and iron, are represented by their atomic symbols, e.g. Na, K, Cu and Fe.
3. Lesson Objectives
When we are done learners should be able to:
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State the law of conservation of mass with its application
State the law of constant composition
Explain the law of multiple with its calculation
State the law of reciprocal proportion
11. Activity
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AgNO3 + NaCl AgCl + NaNO3
x + 21.5 = 11.1 + 18.2
x = 11.1 + 18.2 - 21.5
= 7.8
Therefore the mass of silver trioxonitrate (v) is 7.8g
Total mass on the left hand side = 7.8 + 21.5
= 29.3
Total mass on the right hand side = 11.1 + 18.2
= 29.3
Therefore the law of conversation of mass is observed.
17. How To Verify the Definite Proportion
• By calculating the percentage of the component elements
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18. Experimental Illustration of Law of Definite Proportions
PROCEDURE:
• Two samples of carbon dioxide were prepared by two
methods:
• (a) by passing oxygen over heated carbon and
• (b) by heating calcium trioxocarbonate (iv)
• Analysis of the samples of the carbon (iv) oxide produced
showed that:
• In (a) 3.96g of CO2 contained 2.88g of oxygen while in (b)
2.20g of CO2 contained 1.60g of oxygen. Show that these
results illustrate the Law of Definite Proportions.
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19. Solution to the Experiment Question
• To solve this simply calculate
the percentage by mass of the
component elements in the
samples from different
sources
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Mass of
CO2
Mass of
Oxygen
Sample A 3.96g 2.88g
Sample B 2.20 1.60g
From the question we have
21. Activity
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samples Mass of
samples(g)
Mass of
Cu(g)
Mass of O
(g)
% of O % of Cu
A 4.50 4 0.50 0.5x100/4.5
is 11.1
100-11.1 is
88.9
B 5.60 4.98 0.62 0.62x100/5.
60 is 11.1
100-11.1 is
88.9
29. To Verify The Law of Multiple Proportion
• Apply reduction by mass ratio where you divide by the
mallest value in each sample and then divide by the
smallest sample to get the ratio by mass.
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30. Example
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1
samples Mass of
samples(g)
Mass of
Metal(g)
Mass of Cl
(g)
Divide by
the smaller
mass in
each
sample
Find the
ratio
between the
two sample
A 2.00 0.69 2-0.69 is
1.31
1.31/0.69 is
1.9
1.9/1.3 is
1.5x2/1x2
which gives
3:2
B 10.0 4.41 10 - 4.41 is
5.59
5.59/4.41 is
1.3
31. Sunday, January 15, 2023 31
The mass of oxygen combines with 1g of nitrogen to form 3 different
compounds with the masses are: 1.142g, 2.284g and 2.855g. show that
the data illustrate the la of multiple proportion.
Example
32. Activity
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There are 100g of two different compounds that are composed of sulphur
and oxygen. The first compound contains 50g of sulphur and the second
compound contains 40g of sulphur. show that these data illustrate the law of
multiple proportion
40. Symbols Commongly Used in Equations
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Symbol Meaning
⟶ Yield/produce (pointed to product)
⇌ Reversible reaction, equilibrium between reactants and
products
↓ Gas evolved (written after substance)
(S) Solid (written as a subscript after a substance)
(L) Liquid (written as a subscript after a substance)
(G) Gas (written as subscript after a substance)
Δ Heat
(aq) Aqueous solution (substance dissolved in water)
+ Plus or add to or reacting with
41. How Balance A Chemical Equations
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• write the correct formula of the reactants and the products.
• Take stocks of all the atoms present on both sides of the
equation
• Keep changing the coefficients until the equation is balance.
• Avoid changing the subscripts on the formulas.
43. Sunday, January 15, 2023 43
x 2 = 2
Start by increasing the lower coefficients to meet up with the other coefficients:
• Balance Al on left hand side
2
• Balance O on left hand side
x 3 = 6
3
• To balance O on right hand side
increase the coefficient of Al2O3
from 1 to 2
2
x 2 =4
x 2 =6
• Still increase Al on the left hand side to 4
---
4
----------
x 4 = 4
• Since we now have equal number of each
atom on both side it is now balanced
45. Activities
Balance the following equations:
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1. Na(s) + O2(g) Na2O(s)
2. Al + Cl2 AlCl3
3. Fe + O2 Fe2O3
4. Aluminium reacted with oxygen.
5 Tetraoxosulphate (vi) acid and sodium hydroxide to
form sodium tetraoxosulphate(vi) and water.
6. Lead(ii)trioxonitrate(v) reacted with sodium chloride.
7.
46. Activities
Balance the following equations:
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1. N2(g) + H2(g) NH3(g)
2. C + H2O CO + H2
3. KClO3 KCl + O2
4. Iron(iii)oxide reacted with carbon to form iron and
carbon(ii)oxide.
5 Trioxonitrate (v) acid and carbon to form water,
nitrogen(iv)oxide and carbon(iv)oxide
6. Copper is oxidized by tetraoxosulphate(vi)acid to give
copper(ii)tetraoxosulphate(vi), sulphur(iv)oxide and water.
48. INFORMATION PROVIDED BY EQUATION
1. It tells us the reactants and products involved in the reaction.
2. It tells us the individual elements and radicals involved.
3. It gives us a mental picture of the movement of the elements and
radicals during the reactions.
4. It tells us the stoichiometry of the reaction.
5. It tells us the direction of the reaction and whether the reaction is
reversible.
6. It tells us the state of matter.
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49. INFORMATION NOT PROVIDED BY THE EQUATION
1. The speed of the reaction.
2. The heat changes during the reaction. This,
however maybe indicated where required.
3. The colours of the reactants and products.
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50. CALCULATIONS FROM EQUATIONS
The equation for the formation of water is:
2H2(g) + O2(g) 2H2O(g)
Calculate the number of moles of O2 required to yield 8 moles of water
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Solution
2 moles of H2O is produced by 1mole of O2
8 moles of H2O is produced by 8/2 x 1 moles
= 4moles of O2
51. CALCULATIONS FROM EQUATIONS
Sodium combines with oxygen as follows:
4Na(s) + O2(g) 2Na2O(s)
What is the mass of O2 needed to burn 4.6g of sodium? (Na = 23, O = 16)
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Solution
From the equation,
4 moles of Na reacts with 1mole of O2.
Mass of 4moles of Na = 4 x 23 = 92g
Mass of 1mole of O2 = 16 x 2 = 32g
92g of Na reacts with 32g of O2
4.6g of Na reacts with 4.6 x 32/92
= 1.6g of O2
52. Using Formula Method
Sodium combines with oxygen as follows:
4Na(s) + O2(g) 2NaO(s)
What is the mass of O2 needed to burn 4.6g of sodium? (Na = 23, O = 16)
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Using mole =
𝒎𝒂𝒔𝒔
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
we can say:
1. for Na, mole =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂 𝒙 𝟒
2. for O, mole =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑶𝟐
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑶𝟐
Since mole is common then the two equations become
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂𝒙 𝒙𝟒
=
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑶𝟐
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑶𝟐
Making Mass of O2 the subject of the formula we have
Mass of O2 =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂 𝒙 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑶𝟐
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑵𝒂 𝒙 𝟒
Substitute
Mass of O2 =
𝟒.𝟔 𝒙 (𝟏𝟔𝒙𝟐)
𝟐𝟑𝒙𝟒
= 1.6g of O2
53. Evaluation
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1. Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
What is the mass of hydrogen would be produced if 12.0g of
magnessium react with excess dilute hydrochloric acid?
(H is 1, Mg is 24, Cl is 35.5)
2
54. Assignment
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2. 0.8g of an element X reacted with an excess copper(ii) tetraoxosulphate(vi)
solution to deposit 21.3g of copper. Find the molar mass of X
3. find x, y and z in the following equation:
MnO2(s) + xHCl(aq) MnCl2(aq) yCl2(g) + zH2O
1
