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SS1: CHEMISTRY - 1ST TERM

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  1. Introduction to Chemistry and Laboratory Apparatus | Week 1
    5 Topics
    |
    1 Quiz
  2. Nature of Matter | Week 2
    3 Topics
    |
    1 Quiz
  3. Separation Techniques I | Week 3
    1 Topic
  4. Separation Techniques II | Week 4
    5 Topics
    |
    1 Quiz
  5. Particulate Nature of Matter I | Week 5
    5 Topics
    |
    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9 Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7 Topics
    |
    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4 Topics
    |
    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4 Topics
    |
    1 Quiz
  10. Chemical Combination (Chemical Bonding) II | Week 10
    4 Topics
    |
    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules | Week 11
    3 Topics
    |
    1 Quiz



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Topic Content:

  • Law of Multiple Proportions
  • Verification of the Law

The law states that if two elements A and B combine together to form more than one compound, the various masses of B which combine with a fixed mass of A are in a simple whole-number ratio. e.g. 1:2, 2:1, 1:3, 3:2, etc.

Verification of the Law:

Weigh two clean metal boats, weigh them again containing pure copper (I) oxide (Cu2O) and copper (II) oxide (CuO) respectively. Reduce the oxide to copper by using hydrogen gas as described in the experiment on the Law of Definite Proportion

The mass of copper which combined with a fixed mass of oxygen is found in both samples. The masses are in a simple ratio.

i. \( \scriptsize Cu_2O_{(s)} \: + \: H_{2(g)} \: \rightarrow \: 2Cu_{(s)} \: + \: H_2O_{(g)} \)

ii. \( \scriptsize CuO_{(s)} \: + \: H_{2(g)} \: \rightarrow \: Cu_{(s)} \: + \: H_2O_{(g)} \)

Sample Sample ISample II
Cu2OCuO
Mass of oxide 3.04 g1.91 g
Mass of Cu residue2.55 g1.38 g
Mass of oxygen
removed from the oxide
0.49 g0.53 g

In order to show that these figures conform to the law of multiple proportions, use 1 g or 100 g as a fixed mass.

In sample I:

 

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Theory Question 1

1. (a) Define the Law of multiple proportions

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

 

(b) Give the name of an Oxide you can use for the verification of the Law

Answer: Copper (I) oxide –  Cu2O

Theory Question 2

Two samples of Copper (II) Oxide were prepared by 

(i) Heating Copper (II) trioxocarbonate (IV) 

(ii) Heating strongly the mixture of Copper(II) tetraoxosulphate(VI) and Sodium hydrogen

Analysis of these two samples shows that:

(i) 4.50 g of Copper (II) Oxide contains 4 g of Copper
(ii) 5.60 g of Copper (II) Oxide contains 4.98 g of Copper

Show that this result illustrates the Law of Constant Composition.

Answer:

The percentage by mass of Cu in the two samples of CO2 is the same (i.e. 88.9%). Hence, the law of constant composition is obeyed.

Theory Question 3

3. (a) Define the Law of conservation of mass 

Answer: The law of conservation of mass states that in an ordinary chemical reaction, matter can neither be created nor destroyed

 

(b) Calculate the mass of Silver Chloride precipitate formed when 20.20 g of  Silver trioxonitrate (V) reacts with 9.10 g of Sodium Chloride to form 14.20 g of Sodium trioxonitrate (V)

Answer:

AgNO3(aq) + NaCl(aq)  →  AgCl(s) + NaNO3(aq)

20.20 g + 9.10 g = x g + 14.20 g

29.3 g = x g + 14.20 g

29.3 g – 14.20 g = x g

15.1 g = x g

Mass of silver chloride AgCl = 15.1 g

Theory Question 4

4. (a) State the Law of Constant composition or definite proportion

Answer: The law of constant composition states that all pure samples of the same chemical compound contain the same elements combined in the same proportion by mass.

 

(b) How could you prove the Law of constant composition experimentally

Answer: The law of constant composition can be verified via the preparation of copper (II) oxide. This is done in three different ways.

(i) Action of heat on copper (II) trioxocarbonate (IV) CuCO3(s) → CuO(s) + CO2(g).

(ii) Action of heat on copper (II) trioxonitrate (V)

2Cu(NO3)2(s)  →  2CuO(s) + 4NO2(g) + O2(g).

(iii) Action of heat on copper (II) hydride.

Cu(OH)2(s) → CuO(s) + H2O2(g).

 

(c) Two samples of Iron(II)Sulphide were prepared by: 

(i) Heating a mixture of Iron filings and Sulphur 

(ii) Passing hydrogen Sulphide gas into a solution of Iron(II) Chloride  

Analysis of both samples (i) and (ii) shows that:

i. 3.14g of Iron(II)Sulphide contains 0.12g of Iron

ii. 4.16g of Iron(II)sulphide contains 0.15g of Iron.

Show that these results illustrate the Law of Constant composition.

Solution:

The percentage by mass of Fe in the two samples of FeS is the same (i.e. 4%). Hence, the law of constant composition is obeyed.

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