LAW OF DEFINITE OR CONSTANT COMPOSITION:

A chemical compound always contains same elements in definite proportion by mass and it does not depend on the source of compound.

For example:

The composition of CO₂ obtained by different means always having same hence Law of definite proportion is proving.

ILLUSTRATIVE EXAMPLE (2): Ammonia contains 82.65 % N₂ and 17.65% H₂. If the law of constant proportions is true, then the mass of zinc required to give 10 g Ammonia will be:

                        (A)       8.265 g                                    (B) 0.826 g

                        (C)       82.65 g                                     (D) 826.5 g

SOLUTION: The mass of zinc required to give 10 g ammonia will be

ILLUSTRATIVE EXAMPLE (3): Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of:

                        (A) Conservation of mass                    (B) Constant composition

                        (C) Multiple proportion                      (D) Constant volume

SOLUTION: As in water

Both values always constant. Obey law of constant composition. Hence (B) is correct.

ILLUSTRATIVE EXAMPLE (4): 6.488 g of lead combine directly with 1.002 g of oxygen to form lead peroxide PbO₂. Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxide is 13.38 percent. Use these data to illustrate the law of constant composition

SOLUTION:

Step 1: To calculate the percentage of oxygen in first experiment.

Step 2: To compare the percentage of oxygen in both the experiments.

            Percentage of oxygen in PbO₂ in the first experiment = 13.38

            Percentage of oxygen in PbO₂ in the second experiment = 13.38

Since the percentage composition of oxygen in both the samples of PbO₂ is identical, the above data illustrate the law of constant composition.

ILLUSTRATIVE EXAMPLE (5): Copper oxide was prepared by the following methods:

(A) In one case, 1.75 g of the metal was dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.

(B) In the second case, 1.14 g of metal dissolved in nitric acid was precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43 g of copper oxide.

(C) In the third case, 1.45 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide. Show that the given data illustrate the law of constant composition.

SOLUTION:

Step 1:  In the first experiment.

            2.19 g of copper oxide contained 1.75 g of Cu.

            100 g of copper oxide contained = 1.75/2.19*100=79.91 %

Step 2:  In the second experiment.

            1.43 g of copper oxide contained 1.14 g of copper.

            100 g of copper oxide contained =1.14/1.43*100=79.72 %.

Step 3: In the third experiment.

            1.83 g of copper oxide contained 1.46 g of copper

            100 g of copper oxide contained =1.46/1.83*100 = 79.78 %

Thus the percentage of copper in copper oxide derived from all the three experiments is nearly the same. Hence, the above data illustrate the law of constant composition.

ILLUSTRATIVE EXAMPLE (6): 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO was produced in the process. Show that these results illustrate the law of constant proportions.

SOLUTION: The ratio of copper and oxygen is 1: 0.25. Hence, the law of constant proportions is illustrated.

ILLUSTRATIVE EXAMPLE (7): In an experiment, 2.4 g of iron oxide on reduction with hydrogen yield 1.68 g of iron. In another experiment 2.9 g of iron oxide give 2.03 g of iron on reduction with hydrogen. Show that the above data illustrate the law of constant proportions.

SOLUTION: Ratio of oxygen and iron in both the experiment is 1:2.33

ILLUSTRATIVE EXAMPLE (8): 2.8 g of calcium oxide (CaO) prepared by heating limestone were found to contain 0.8 g of oxygen. When one gram of oxygen was treated with calcium, 3.5 g of calcium oxide were obtained. Show that the results illustrate the law of definite proportions.

SOLUTION:  try yourself……

Limitation of law definite proportions: Discovery of isotopes created some limitations to this law. Isotopes of an element have different atomic mass hence they form same chemical compounds with different compositions;

For example: