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SS1: CHEMISTRY - 1ST TERM

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  1. Introduction to Chemistry and Laboratory Apparatus | Week 1
    5Topics
    |
    1 Quiz
  2. Nature of Matter | Week 2
    3Topics
    |
    1 Quiz
  3. Separation Techniques I | Week 3
    1Topic
    |
    1 Quiz
  4. Separation Techniques II | Week 4
    5Topics
    |
    1 Quiz
  5. Particulate Nature of Matter I | Week 5
    5Topics
    |
    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9Topics
    |
    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7Topics
    |
    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4Topics
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    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4Topics
    |
    1 Quiz
  10. Chemical Combination (Chemical Bonding) II | Week 10
    4Topics
    |
    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules | Week 11
    3Topics
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    1 Quiz
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The Law of Constant Composition was suggested by Joseph Louis Proust, a French chemist in (1799). The law states that all pure samples of the same chemical compound contain the same elements combined in the same proportion by mass.

Verification of the Law:

This can be experimented with by preparing Copper(II) oxide in three different ways.

(a) Action of heat on Copper(II) trioxonitrate (V)

\( \scriptsize 2Cu(NO_3)_{2(s)} \: \rightarrow \: \underset {black}{2CuO_{(s)}} \: + \: 4NO_{2(g)} \: + \: O_{2(g)} \)

(b) Action of heat on Copper(II) trioxocarbonate (IV)

\( \scriptsize 2Cu(CO_3)_{(s)} \: \rightarrow \: CuO_{(s)} \: + \: CO_{2(g)} \)

(c) Action of heat on Copper (II) hydroxide

\( \scriptsize 2Cu(OH)_{2(s)} \: \rightarrow \: CuO_{(s)} \: + \: H_2O_{(g)} \)

Analysis of Three Samples of Copper (II) Oxide (CuO)

Determine the amount of copper present in three samples of copper(II) oxide by reducing the oxide in steam of hydrogen.

Verification of the law of Constant Composition (Definite Proportion)

Law of Constant Composition (Definite Proportion)

Procedure:

Weigh three metal boats. Add a reasonable amount of sample A, B and C in each of the boats respectively. Reweigh and determine the mass of each sample. 

Place the boat into a hard glass tube (see figure above). Pass a steam of hydrogen and heat each boat in turn. A reddish-brown residue of copper is left in each boat. 

\( \scriptsize CuO_{(s)} \: + \: H_2 \; \rightarrow \: Cu_{(s)} \: + \: H_2O_{(g)} \)

CuO is reduced to Cu.

The fused calcium chloride in the U-tube is to absorb any water formed during the reaction. Weigh the three boats again, within the limits of experimental error, the percentage composition of copper in all the three samples of copper (II) oxide are the same.

Experimental Data and Calculations

Sample ASample BSample C
Mass
of CuO(a)
0.95g1.15g2.11g
Mass
of Cu (b)
0.76g0.92g1.69g
% percentage
of Cu in CuO
\( \frac{0.76}{0.95} \: \times \: \frac{100}{1} \)\( \frac{0.92}{1.15} \: \times \: \frac{100}{1} \)\( \frac{1.69}{2.11} \: \times \: \frac{100}{1} \)
\( \frac{Mass \: of \: Cu} {Mass \: of \:CuO}\: \times \: \frac{100}{1} \)= 80.0%= 80.0%= 80.1%
Therefore, percentage
of oxygen in CuO
= 20.0%= 20.0%= 19.9%

Observation: The percentages of copper in the three samples of CuO are approximately 80.0% and that of oxygen in the three samples are approximately 20.0% irrespective of the method of preparation of the CuO sample.

These data show that the law of constant composition or definite proportion is obeyed.

Example I:

Two samples of carbon(IV) oxide (CO2) were prepared by

a. Passing oxygen overheated carbon

b. By heating calcium trioxocarbonate(IV) (CaCO3)

Analysis of the samples of CO2 prepared showed that in

a. 3.96g of CO2 contained 2.88g of O2

b. 2.20g of CO2 contained 1.60g of O2

Show that these results illustrate the law of constant composition.

 

Sample ASample B
Mass
of CO2
3.96g2.20g
Mass
of O2
2.88g1.60g
% of O2
in CO2
\( \frac{2.88}{3.96} \: \times \: \frac{100}{1} \)\( \frac{1.60}{2.20} \: \times \: \frac{100}{1} \)
\( \frac{Mass \: of \: O_2} {Mass \: of \: CO_2}\: \times \: \frac{100}{1} \)= 72.73%= 72.73%

The percentage by mass of O2 in the two samples of CO2 is the same (72.73%). The law of constant composition is obeyed.

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