LAW OF MILTIPLE PROPORTION:

When two elements combine to form two or more than two different compounds then the different masses of one element B which combine with fixed mass of the other element bear a simple ratio to one another.

For example: Carbon forms two oxides in oxygen

The ratio of masses of oxygen in CO and CO₂ for fixed mass of carbon (12)
is 16: 32 = 1: 2.

ILLUSTRATIVE EXAMPLE (9): The law of multiple proportions is illustrated by the pair of compounds:

(A) Sodium chloride and sodium bromide     

(B) Water and heavy water

(C) Sulphur dioxide and sulphur trioxide       

(D) Magnesium hydroxide and magnesium oxide

SOLUTION:  In SO₂ 32 gram of suphur react with 32 gram of oxygen. Similarly for SO₃ fixed mass of sulphur (32 gram) react with 48 gram of oxygen. The ratio of oxygen’s mass = 32:48 = 2:3 which support law of multiple proportions. Hence (C) is correct answer.

ILLUSTRATIVE EXAMPLE (10): Carbon is found to form two oxides, which contain 42.9% and 27.3% of carbon respectively. Show that these figures illustrate the law of multiple proportions.

SOLUTION:  

Step 1: To calculate the percentage composition of carbon and oxygen in each of the two oxides

Step 2:  To calculate the weights of carbon which combine with a fixed weight i.e., one part by weight of oxygen in each of the two oxides.

In the first oxide, 57.1 parts by weight of oxygen combine with carbon = 42.9 parts.

1 part by weight of oxygen will combine with carbon = 42.9/57.1=0.751

In the second oxide 72.7 parts by weight of oxygen combine with carbon = 27.3 parts.

1 part by weight of oxygen will combine with carbon =27.3/72=0.376

Step 3:  To compare the weights of carbon which combine with the same weight of oxygen in both the oxides-The ratio of the weights of carbon that combine with the same weight of oxygen (1 part) is 0.751: 0.376 or 2:1

Since this is a simple whole number ratio, so the above data illustrate the law of multiple proportions.

TRY YOURSELF:

Exercise (1): Metal M and chlorine combine in different proportions to form two compounds A and B. The mass ratio M: Cl is 0.895: 1 in A and 1.791: 1 in B. What law of chemical combination is illustrated?

Exercise (2): By means of the following analytical results show that law of multiple proportions is true:

Exercise (3): 1 g of a metal, having no variable valency, produces 1.67 g of its oxide when heated in air. Its carbonate contains 28.57% of the metal. How much oxide will be obtained by heating 1 g of carbonate?

Exercise (4): Phosphorous and chlorine form two compounds. The first contains 22.54% by mass of phosphorous and the second 14.88% of phosphorous. Show that these data are consistent with law of multiple proportions.

TRY YOURSELF: SOLUTION:

(1) Mass of metal which combine with 1 part of chlorine are in the ratio of 1:2, which is a simple ratio. Hence, law of multiple proportions is illustrated.

(2) The masses of mercury which combine with 1 part of chlorine are in the ratio of 2:1 which is simple ratio. Hence, law of multiple proportions is illustrated.     

(3) 0.477 g

(4) Prove yourself ….

Limitation of law of multiple proportions: Non Stoichiometric compound do not follow this law for example Fe_0.93O₁.

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:

LAW OF CONSERVATION OF MASS:

For any chemical change total mass of active reactants are always equal to the mass of the product formed. It is a derivation of Dalton’s atomic theory ‘atoms neither created nor destroyed’.

Total masses of reactants = Total masses of products + Masses of unreacted reactants

ILLUSTRATIVE EXAMPLE (1): 5.2 g of CaCO₃ when heated produced 1.99 g of Carbon dioxide and the residue (CaO) left behind weighs 3.2g. Show that these results illustrate the law of conservation of mass.

SOLUTION: Weight of CaCO₃ taken = 5.2 g

            Total weight of the products (CaO +CO₃) = 3.20+ 1.99 = 5.19 g

            Difference between the wt. of the reactant and the total wt. of the products

            = 5.20 – 5.19 =0.01 g.

            This small difference may be due to experimental error.

            Thus law of conservation of mass holds good within experimental errors.

Limitation of Law of conservation of mass: Nuclear reactions do not follow the law of conservation of mass because some of the mass of reactants is converted into energy according to Einstein equation E=mc²     where c is the velocity of light.