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Laws of Chemical
Combination
Sunday, January 15, 2023 1
Sunday, January 15, 2023 2
Lesson Objectives
When we are done learners should be able to:
Sunday, January 15, 2023 3
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
Sunday, January 15, 2023 4
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How to Verify the law of conversation of mass
Activity
Sunday, January 15, 2023 11
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.
Activity
Sunday, January 15, 2023 12
1
2
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How To Verify the Definite Proportion
• By calculating the percentage of the component elements
Sunday, January 15, 2023 17
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.
Sunday, January 15, 2023 18
Solution to the Experiment Question
• To solve this simply calculate
the percentage by mass of the
component elements in the
samples from different
sources
Sunday, January 15, 2023 19
Mass of
CO2
Mass of
Oxygen
Sample A 3.96g 2.88g
Sample B 2.20 1.60g
From the question we have
Sunday, January 15, 2023 20
Activity
Sunday, January 15, 2023 21
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
Sunday, January 15, 2023 22
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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.
Sunday, January 15, 2023 29
Example
Sunday, January 15, 2023 30
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
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
Activity
Sunday, January 15, 2023 32
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
Activities
33
1
2
3
4
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Symbols Commongly Used in Equations
Sunday, January 15, 2023 40
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
How Balance A Chemical Equations
Sunday, January 15, 2023 41
• 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.
Sunday, January 15, 2023 42
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
Sunday, January 15, 2023 44
Activities
Balance the following equations:
Sunday, January 15, 2023 45
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.
Activities
Balance the following equations:
Sunday, January 15, 2023 46
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.
Sunday, January 15, 2023 47
F f
L l
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.
Sunday, January 15, 2023 48
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.
Sunday, January 15, 2023 49
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
Sunday, January 15, 2023 50
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
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)
Sunday, January 15, 2023 51
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
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)
Sunday, January 15, 2023 52
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
Evaluation
Sunday, January 15, 2023 53
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
Assignment
Sunday, January 15, 2023 54
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

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Laws of Chemical Combination and Balancing Chemical Equation.pptx

  • 3. Lesson Objectives When we are done learners should be able to: Sunday, January 15, 2023 3 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
  • 10. Sunday, January 15, 2023 10 How to Verify the law of conversation of mass
  • 11. Activity Sunday, January 15, 2023 11 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 Sunday, January 15, 2023 17
  • 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. Sunday, January 15, 2023 18
  • 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 Sunday, January 15, 2023 19 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 Sunday, January 15, 2023 21 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. Sunday, January 15, 2023 29
  • 30. Example Sunday, January 15, 2023 30 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 Sunday, January 15, 2023 32 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 Sunday, January 15, 2023 40 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 Sunday, January 15, 2023 41 • 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: Sunday, January 15, 2023 45 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: Sunday, January 15, 2023 46 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.
  • 47. Sunday, January 15, 2023 47 F f L l
  • 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. Sunday, January 15, 2023 48
  • 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. Sunday, January 15, 2023 49
  • 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 Sunday, January 15, 2023 50 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) Sunday, January 15, 2023 51 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) Sunday, January 15, 2023 52 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 Sunday, January 15, 2023 53 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 Sunday, January 15, 2023 54 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