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SS1: CHEMISTRY - 1ST TERM

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  1. Introduction to Chemistry and Laboratory Apparatus
    7 Topics
    |
    1 Quiz
  2. Nature of Matter
    5 Topics
    |
    1 Quiz
  3. Separation Techniques I
    4 Topics
    |
    1 Quiz
  4. Separation Techniques II
    7 Topics
    |
    1 Quiz
  5. Particulate Nature of Matter I
    10 Topics
    |
    1 Quiz
  6. Particulate Nature of Matter II
    11 Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number
    8 Topics
    |
    1 Quiz
  8. Laws of Chemical Combination
    4 Topics
    |
    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I
    4 Topics
    |
    1 Quiz
  10. Chemical Combination (Chemical Bonding) II
    4 Topics
    |
    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules
    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 (Cu2 O) 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. \( \scriptsize CuO_{(s)} \: + \: H_{2(g)} \: \rightarrow \: Cu_{(s)} \: + \: H_2O_{(g)} \)

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:

0.49 g of oxygen combined with 2.55 g of Cu using 1 g as fixed mass.

\( \frac{2.55g}{0.49g} \: \times \: \frac{1}{1} \\ \scriptsize = 5.20\: g \)

In sample II:

0.53 g of oxygen combined with 1.38 g of copper using 1 g as fixed mass.

\( \frac{1.38g}{0.53g} \: \times \: \frac{1}{1} \\ \scriptsize = 2.6\:g \)

The mass of copper that has combined with a fixed mass 

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Chike Amadi
28/12/2022 11:09 AM

Learnt a lot from this topic.

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Amaechi Gift
06/02/2023 9:44 PM

This a nice one thanks

Rechael Otuu
24/01/2024 6:43 PM

It was really helpful

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