Law of Constant Composition (Definite Proportion)

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)

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

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

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

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)

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 in 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 three samples of copper(II) oxide is the same.

Experimental Data and Calculations:

	Sample A	Sample B	Sample C
Mass of CuO(a)	0.95 g	1.15 g	2.11 g
Mass of Cu (b)	0.76 g	0.92 g	1.69 g
% percentage of Cu in CuO	\(\scriptsize \frac{0.76}{0.95} \: \times \: \frac{100}{1} \)	\(\scriptsize \frac{0.92}{1.15} \: \times \: \frac{100}{1} \)	\(\scriptsize \frac{1.69}{2.11} \: \times \: \frac{100}{1} \)
\(\scriptsize \frac{Mass \: of \: Cu} {Mass \: of \:CuO}\: \times \: \frac{100}{1} \)	= 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.

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

Example 8.3.1:

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.

Solution:

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