Laws of Chemical Combination

Table of Content

• Law of Conservation of Mass

  • Example on Law of Conservation of Mass

• Law of Constant Proportions

  • Examples on Law of Constant Proportions

• Law of Multiple Proportion

  • Example on Law of Multiple Proportion

• Law of Reciprocal Proportion

  • Example on Law of Reciprocal Proportion

• Gay Lussac’s law of Combining Volumes

  • Example on Gay Lussac’s law of Combining Volumes

• Related Resources

Example on Law of Conservation of Mass

Question : 

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 +CO₂)= 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.

Examples on Law of Constant Proportions

Question 1: 

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   

Hence (B) is correct.

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Question 2:

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

So oxygen = 

Hydrogen = 

Both value always constant. Obey law of constant composition.

Hence (B) is correct.

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Question 3:

6.488 g of lead combine directly with 1.002 g of oxygen to form lead peroxide. 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.

 Weight of peroxide formed = 6.488 + 1. 002 = 7.490 g.

 7. 490 g of lead peroxide contain 1.002 g of oxygen

  100 g of lead peroxide will contain oxygen

 i.e. oxygen present = 13.38%

Step 2:

To compare the percentage of oxygen in both the experiments.

Percentage of oxygen in in the first experiment = 13.38

Percentage of oxygen in in the second experiment = 13.38

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

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Question 4:

Copper oxide was prepared by the following methods:

a. In one case, 1.75 g of the metal were 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 were 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 

Step 2:  In the second experiment.

1.43 g of copper oxide contained 1.14 g of copper.

 100 g of copper oxide contained

Step 3: In the third experiment.

1.83 g of copper oxide contained 1.46 g of copper

 100 g of copper oxide contained 

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. 

Example on Law of Multiple Proportion

Question 1: 

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.

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Question 2: 

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 

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

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.

Example on Law of Reciprocal Proportion

Question 1 : 

One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combines together, the ratio of their masses will be governed by:

a. law of definite proportions

b. law of multiple proportions

c. law of reciprocal proportions

d. law of conservation of mass

Solution:

from this when A combined with C the ratio is A/C = 1/3

Hence (C) is correct.    

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Question 2:

Copper sulphide contains 66.6% Cu, copper oxide contains 79.9% copper and sulphur trioxide contains 40% Sulphur. Show that these data illustrates law of reciprocal proportions.

Solution:

In copper sulphides

Cu : S mass ratio is 66.6 : 33.4

In sulphur trioxide

Oxygen : Sulphur (O:S) mass ratio is 60:40

Now in CuS

33.4 parts of sulphur combines with Cu = 66.6 parts

40.0 parts of sulphur combines with cu

Now the ratio of masses of Cu and O which combines with same mass (40 parts) of sulphur separately is 79.8:60                …(1)

  Cu : O ratio by mass in CuO is 79.9: 20.1         …(2)

Ratio 1: Ratio 2

Which is simple whole number ratio, hence law of reciprocal proportion is proved.

Example on Gay Lussac’s law of Combining Volumes

Question 1: 

In the reaction, the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3 : 2. These figures illustrate the law of:

a. constant proportions

b. Gay-Lussac

c. multiple proportions

d. reciprocal proportions

Solution:

The above ratio of 1 : 3 : 2 illustrates the Gay-Lussac law of combining volume. 

Hence (B) is correct.

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Question 2:

How much volume of oxygen will be required for complete combustion of 40 ml of acetyleneand how much volume of carbondioxide will be formed? All volumes are measured at NTP.

Solution:

40 ml                           

40 ml                 100 ml                80 ml    

So for complete combustion of 40 ml of acetylene, 100 ml of oxygen are required and 80 ml of is formed.

Q.1

Q.2

Q.3

Q.4

a

c

b

a