SS1: CHEMISTRY - 1ST TERM
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Introduction to Chemistry and Laboratory Apparatus7 Topics|1 Quiz
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Nature of Matter5 Topics|1 Quiz
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Separation Techniques I4 Topics|1 Quiz
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Separation Techniques II7 Topics|1 Quiz
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Particulate Nature of Matter I10 Topics|1 Quiz
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Evidence for the Particulate Nature of Matter
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The Concept of Atoms, Molecules, and Ions
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Dalton’s Atomic Theory
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Discovery of Electron, Proton and Positive Rays
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Rutherford’s Atomic Model and Atomic Number
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Discovery and Properties of Subatomic Particles
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Arrangement of Particles in an Atom
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Discovery of Sub-atomic Particles & Atomic Structure (Summary Table)
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Atomic Structure Challenge
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Theory Question - Particulate Nature of Matter I
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Evidence for the Particulate Nature of Matter
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Particulate Nature of Matter II11 Topics|1 Quiz
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Symbols, Formulae & Oxidation Number8 Topics|1 Quiz
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Laws of Chemical Combination4 Topics|1 Quiz
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Chemical Equation & Chemical Combination (Chemical Bonding) I4 Topics|1 Quiz
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Chemical Combination (Chemical Bonding) II4 Topics|1 Quiz
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Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules3 Topics|1 Quiz
Law of Multiple Proportions
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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