In the previous article, we have studied the law of reciprocal proportions. In this article, we shall study Gay-Lussac’s Law of Combining Volumes. A French chemist Joseph L. Gay – Lussac in 1809, put forward this law.

Statement :

Whenever gases take part in a chemical reaction, either as reactants or as products, they do so in simple proportions by Volumes. Provided the volumes of gases are measured at the same temperature and pressure,

Illustration 1:

Consider following reaction

H_2(g)      +     Cl_2(g)     →     2HCl

1vol               1vol                   2 vol

Thus the simple ratio of volumes is 1 : 1 : 2

Illustration 2:

Consider following reaction

N_2(g)      +    3 H_2(g)     →     2NH_3(g)

1vol            3vol                   2 vol

Thus the simple ratio of volumes is 1 : 3 : 2

Numerical Problems:

Example – 01:

Calculate the volume of oxygen required for the complete combustion of 0.25 dm³ of methane at STP.

Solution:

CH_4(g)   +   2O_2(g)    →   CO_2(g)    +   2H₂O_(g)

1 vol            2 vol           1 vol                  2 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of methane requires 2 vol of oxygen for complete combustion.
Hence  0.25 dm³ of methane requires 2 x 0.25 = 0.50 dm³ of oxygen.

Example – 02:

Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm³ of ethyne at STP.

Solution:

C₂H_2(g)    +    2H_2(g)  →     C₂H_6(g)

1 vol           2vol                1 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of ethyne requires 2 vol of hydrogen for complete hydrogenation.
Thus 0.25 dm³ of ethyne requires 2 x 0.25 = 0.50 dm³ of hydrogen for complete hydrogenation.

Example – 03:

Calculate the volume of hydrogen required for the complete hydrogenation of 0.25 dm³ of ethylene at STP.

Solution:

C₂H_4(g)    +    H_2(g)  →     C₂H_6(g)

1 vol           1vol                1 vol

By Gay-Lussac’s law of combining volumes of gases

1 vol of ethylene requires 1 vol of hydrogen for complete hydrogenation.

Thus 0.25 dm3 of ethylene requires 1 x 0.25 = 0.25 dm³ of hydrogen for complete hydrogenation.

Example – 04:

Calculate the volume of oxygen required for the complete combustion of 0.25 mol of methane at STP.

Solution:

CH_4(g)   +   2O_2(g)    →   CO_2(g)    +   2H₂O_(g)

1 mol            2 mol           1 mol                2 mol

By Gay-Lussac’s law of combining volumes of gases

1 mol of methane requires 2 mol of oxygen for complete combustion.

Thus 0.25 mol of methane requires 2 x 0.25 = 0.50 mol of oxygen.

One mole of any gas occupies 22.4 dm³ by volume at STP.

Volume of oxygen required = 22.4 x No. of moles = 22.4 x 0.5 = 11.2 dm³

Example – 05:

Calculate the volume of oxygen required for the complete combustion of 0.5 dm³ of H₂S at STP.

Solution:

2H₂S_(g)   +   3O_2(g)    →   2SO_2(g)    +   2H₂O_(g)

2 vol           3 vol           2 vol                  2 vol

By Gay-Lussac’s law of combining volumes of gases

2 vol of H₂S requires 3 vol of oxygen for complete combustion.

1 vol of H₂S requires 3/2 = 1.5 vol of oxygen for complete combustion.

Thus 0.5 dm³ of H₂S requires 1.5 x 0.5 = 0.75 dm³ of oxygen for complete combustion.

Example – 06:

Calculate the volume of oxygen required at STP for the complete combustion of 5.0 dm³ of ethane at 295 K and 0.993 x 10⁵ Nm^-2

Solution:

2C₂H_6(g)   +   7O_2(g)    →   4CO_2(g)    +   6H₂O_(g)

2 vol            7 vol              4 vol                6 vol

By Gay-Lussac’s Law of Combining Volumes of gases

2 vol of C₂H₆ requires 7 vol of oxygen for complete combustion.

1 vol of C₂H₆ requires 7/2 = 3.5 vol of oxygen for complete combustion.

Thus 5 dm³ of C₂H₆ requires 3.5 x 5 = 17.5 dm³ of oxygen for complete combustion.
P = 0.993 x 10⁵ Nm^-2, T = 295 K, V = 17.5 dm³
P_o = 1.013 x 10⁵ Nm^-2, T_o = 273 K, V_o = ?

Thus the volume of oxygen required at STP is 15.88 dm³.

Example – 07:

6.0 dm³ of hydrogen is reacted with 2.4 dm³ of oxygen in a closed chamber. Calculate composition of resulting mixture.

Solution:

2H_2(g)    +    O_2(g)  →    2 H₂O_(g)

2 vol           1 vol               2 vol

By Gay-Lussac’s law of combining volumes of gases.

The ratio of the volume of hydrogen to that of oxygen is 2 : 1.

In this case, oxygen is limiting reagent, and hydrogen is an excess reagent.

2.4 dm³ of oxygen can combine with 2 x 2.4 = 4.8 dm³ of hydrogen to form 2 x 2.4 = 4.8 dm³ of water vapours.

Thus unreacted hydrogen = 6.0 – 4.8 = 1. 2  dm³.

Thus resulting mixture contains 4.8 dm³ of water vapours and 1.2 dm³ of unreacted hydrogen.

Example – 08:

15 litres of nitrogen is made to react with 30 litres of hydrogen to prepare ammonia. Calculate composition of resulting mixture.

Solution:

N_2(g)      +    3 H_2(g)     →     2NH_3(g)

1vol            3vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases

The ratio of the volume of nitrogen to that of hydrogen is 1 : 3.

In this case, hydrogen is limiting reagent, and nitrogen is an excess reagent.

3 x 10 = 30 litres of hydrogen can combine with 1 x 10 = 10 litres of nitrogen to form 2 x 10 = 20 litres of ammonia.

Thus unreacted nitrogen = 15.0 – 10.0 = 5 litres.

Thus resulting mixture contains 20 litres of ammonia and 5 litres of unreacted nitrogen.

Example – 09:

200 dm³ of hydrogen gas is allowed to react with 250 dm³ of chlorine gas. Calculate composition of resulting mixture.

Solution:

H_2(g)      +     Cl_2(g)     →     2HCl

1vol               1vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases.

The ratio of the volume of hydrogen to that of chlorine is 1 : 1.

In this case, hydrogen is limiting reagent and chlorine is excess reagent.

200 dm³ of hydrogen can combine with 200 dm³ of chlorine to form 2 x 200 =400 dm³of hydrogen chloride.

Thus unreacted chlorine = 250 – 200 = 50 dm³.

Thus resulting mixture contains 400 dm³ of hydrogen chloride and 50 dm³ of unreacted chlorine.

Example – 10:

10 dm³ of hydrogen gas is allowed to react with 15 dm³ of chlorine gas. Calculate composition of resulting mixture.

Solution:

H_2(g)      +     Cl_2(g)     →     2HCl

1vol               1vol                   2 vol

By Gay-Lussac’s law of combining volumes of gases.

The ratio of the volume of hydrogen to that of chlorine is 1 : 1

In this case, hydrogen is limiting reagent and chlorine is excess reagent.

10 dm³ of hydrogen can combine with 10 dm³ of chlorine to form 2 x 10 = 20 dm³ of hydrogen chloride.

Thus unreacted chlorine = 15 – 10 = 5 dm³.

Thus resulting mixture contains 20 dm³ of hydrogen chloride and 5 dm3 of unreacted chlorine

In the next chaper we shall study the concept of atomic mass and equivalent mass.

Previous Topic: The Law of Reciprocal Proportions

Next Chapter: Concept of Atomic Mass and Equivalent Mass

Science > Chemistry > Laws of Chemical Combinations > Gay-Lussac’s Law of Combining Volumes

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In the previous article, we have studied the law of multiple proportions. In this article, we shall study the law of reciprocal proportions. The law of reciprocal proportions was given by German chemist Ritcher in 1792.

Statement:

The weights of two or more different elements which separately combine with a fixed weight of another element are either the same as, or simple multiples of, the weights of these elements when they combine among themselves.

Illustration 1:

Consider three compounds methane, carbon dioxide and water.

Hydrogen and oxygen react with carbon separately and forms methane and carbon dioxide respectively.

The ratio of different weights of hydrogen (4) and oxygen (32) are combining with fixed weight of carbon (12) is 4:32 i.e. 1:8.

Now hydrogen and oxygen combine to form water (H2O)in which the ratio of the weight of hydrogen to that of oxygen is 2:16 i.e. 1:8.

This ratio is the same as that of the first ratio obtained. Thus the law of reciprocal proportion is illustrated.

Illustration 2:

Consider three compounds phosphorous trihydride, phosphorous tri chloride and hydrogen chloride.

Hydrogen and chlorine react with phosphorous separately and forms phosphorous trihydride and phosphorous tri chloride respectively.

The ratio of different weights of hydrogen (3) and chlorine (106.5) are combining with fixed weight of phosphorous (31) is 3 : 106.5  i.e. 1 : 35.5

Now hydrogen and chlorine combine to form hydrogen chloride in which the ratio of the weight of hydrogen to that of chlorine is 1:35.5. This ratio is the same as that of the first ratio obtained. Thus the law of reciprocal proportion is illustrated.

Limitations of the Law of Reciprocal Proportions:

• The existence of isotopes of the element causes discrepancies similar to that observed In the law of constant proportions.  Hence the same isotope or mixture of isotope should be used throughout the preparation of a series of compounds.
• Since there are few elements which will combine with the third element and also combine with each other. Thus the law is applicable to very few elements exhibiting the said property.

Explanation of the  Law of Reciprocal Proportions on the Basis of Dalton’s Atomic Theory:

According to Dalton’s atomic theory, all the atoms of the same element are identical and the compounds are formed by the combination of atoms of different elements in a simple ratio of whole numbers. Therefore, the weights of the elements combining with a fixed weight of another element should bear a simple ratio to the ratio of the weights of the elements when they combine with each other. This explains the law of reciprocal proportion.

Numerical Problems:

Example – 01:

CO₂ contains 27.27% of carbon, CS₂ contains 15.79% of carbon and SO2 contains 50% of sulphur. Show that the data illustrates the law of reciprocal proportions.

Solution:

Consider CO₂

% of carbon = 27.27

% of oxygen = 100 – 27.27 = 72.73

27.27 g of carbon combines with 72.73 g of oxygen.

Hence, 1 g of carbon combines with 72.73 / 27.27= 2.67 g of oxygen.

Consider CS₂

% of carbon = 15.79

% of sulphur = 100 – 15.79 = 84.21

15.79 g of carbon combines with 84.21 g of sulphur.

Hence, 1 g of carbon combines with 84.21 / 15.79= 5.33 g of sulphur.

The ratio of different masses of sulphur and oxygen combining with fixed mass of carbon is 5.33 : 2.67

i.e. 2 : 1. ………………….. (Ratio 1)

Consider SO₂

% of sulphur = 50

% of oxygen = 100 – 50 = 50

50 g of sulphur combines with 50 g of oxygen.

The ratio of mass of sulphur to that of oxygen is 50 : 50 

i.e. 1 : 1 ……………… (Ratio 2)

The second ratio is a simple whole-number is multiple of the first ratio. Hence the data illustrate the law of reciprocal proportions.

Example – 02:

CuS contains 66.6 % of copper, CuO contains 79.9% of copper and SO₃ contains 40 % of sulphur. Show that the data illustrates the law of reciprocal proportions.

Solution:

Consider CuS

% of copper = 66.6

% of sulphur = 100 – 66.6 = 33.4

66.6 g of copper combines with 33.4 g of sulphur.

Hence, 1 g of copper combines with 33.4 / 66.6= 0.5 g of sulphur.

Consider CuO

% of copper = 79.9

% of oxygen = 100 – 79.9 = 20.1

79.9 g of copper combines with 20.1 g of oxygen.

Hence 1 g of copper combines with 20.1 / 79.9= 0.25 g of oxygen.
The ratio of different masses of sulphur and oxygen combining with fixed mass of copper is 0.5 : 0.25 

i.e. 2 : 1. ………………….. (Ratio 1)

Consider SO₃

% of sulphur = 40

% of oxygen = 100 – 40 = 60

40 g of sulphur combines with 60 g of oxygen.

The ratio of the mass of sulphur to that of oxygen is 40 : 60 

i.e. 2 : 3 ……………… (Ratio 2)

The second ratio is a simple whole-number multiple of the first ratio. Hence the data illustrate the law of reciprocal proportions.

Example – 03:

Aluminium carbide contains 75% of aluminium, aluminium oxide contains 52.9 % of aluminium and carbon dioxide contains 27.27 % of carbon. Show that the data illustrates the law of reciprocal proportions.

Solution:

Consider AlC₃

% of aluminium = 75

% of carbon = 100 – 75 = 25

75 g of aluminium combines with 25 g of carbon.

Hence 1 g of aluminium combines with 2 5 / 7 5= 0.33 g of carbon.

Consider Al₂O₃

% of aluminium = 52.9

% of oxygen = 100 – 52.9 = 47.1

52.9 g of aluminum combines with 47.1 g of oxygen.

Hence 1 g of aluminium combines with 47.1 / 52.9= 0.89 g of oxygen.

The ratio of different masses of sulphur and oxygen combining with fixed mass of copper is 0.33 : 0.89 

i.e. 1 : 3. ………………….. (Ratio 1)

Consider CO₂

% of carbon = 27.27

% of oxygen = 100 – 27.27 = 72.73

27.27 g of carbon combines with 72.73 g of oxygen
the ratio of the mass of copper to that of oxygen is 27.27 : 72.73

i.e. 1 : 3 ……………… (Ratio 2)

The second ratio is a simple whole-number multiple of the first ratio (actually the same). Hence the data. illustrate the law of reciprocal proportions.

Example – 04:

CuS contains 33.3 % of sulphur, CuO contains 20.1% of oxygen and SO3 contains 40 % of sulphur. Show that the data illustrates the law of reciprocal proportions.

Solution:

Consider CuS

% of sulphur = 33.3

% of sulphur = 100 – 33.3 = 66.7

66.7 g of copper combines with 33.3 g of sulphur.

Hence 1 g of copper combines with 33.3 / 66.7= 0.5 g of sulphur.

Consider CuO

% of oxygen = 20.1

% of copper = 100 – 20.1 = 79.9

79.9 g of copper combines with 20.1 g of oxygen.

Hence 1 g of carbon combines with 20.1 / 79.9= 0.25 g of oxygen.
The ratio of different masses of sulphur and oxygen combining with fixed mass of copper is 0.5 : 0.25 

i.e. 2 : 1. ………………….. (Ratio 1)

Consider SO₃

% of sulphur = 40

% of oxygen = 100 – 40 = 60

40 g of sulphur combines with 60 g of oxygen.

In the next article, we shall study Gay-Lussac’s law of combining volumes.

Previous Topic: The Law of Multiple Proportions

Nex Topic: Gay-Lussac’s Law of Combining Volumes

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In the previous article, we have studied the law of definite proportions. In this article, we shall study the law of multiple proportions. The law of multiple proportions was given by British scientist John Dalton in 1803.

Statement:

When two elements combine to form more than one compound, then the different weights of one element combining with a fixed weight of the other element are in simple numerical ratio with each other.

Explanation & illustration :

This law is applicable to pairs of elements which can form more than one compound.

Illustration 1:  

Carbon and oxygen combine together to give two compounds carbon dioxide (CO₂) and carbon monoxide (CO)

Thus the ratio of different weights of oxygen (32 and 16) combining with a fixed weight of carbon (12) is 32 : 16 i.e. 2 :1, which is simple whole number ratio.

Illustration 2:

Hydrogen and oxygen combine together to give two compounds of water (H₂O) and hydrogen peroxide (H₂O₂)

Thus the ratio of different weights of oxygen (16 and 32) combining with a fixed weight of hydrogen (2) is 16 : 32 i.e. 1 :2, which is simple whole number ratio.

Illustration 3:

Nitrogen combines with oxygen to form the various oxides.

Thus the ratio of different weights of oxygen (8, 16, 24, 32, 40) combining with fixed weight of nitrogen (14) is  8 :16 : 24 : 32 : 40 i.e. 1:2:3:4:5, which is simple whole number ratio.

Limitations of the Law of Multiple Proportions:

The existence of isotopes of hydrogen like H¹ or H² causes discrepancies similar to that observed in the law of constant proportions.  Hence the same isotope or mixture of isotope should be used throughout the preparation of a series of compounds.

Explanation of the Law of Multiple Proportions on the Basis of Dalton’s Atomic Theory:

According to Dalton’s atomic theory, compounds are formed by the combination of atoms of different elements in the ratio of simple whole numbers.

Atoms of elements have a fixed weight. Hence, it follows that when elements combine to form more than one compound, the different weights of one which combines with a fixed weight of the other must be in the ratio of simple whole numbers. This explains the law of multiple proportions

Numerical Problems:

Example – 01:

Hydrogen and oxygen are known to form two compounds. The hydrogen content in one of them is 5.93 % while in other it is 11.2 %. Show that the data illustrate the law of multiple proportions.

Solution:

Compound – 1:

Let us consider 100 g of compound – 1

Mass of hydrogen = 5.93 g

Mass of oxygen = 100 g – 5.93 g = 94.06 g

Thus, 5.93 g of hydrogen combines with 94.07 g of oxygen.

∴ 1 g of hydrogen combines with

94.07/5.93 = 15.86 g of oxygen.   …………………….. (1)

Compound – 2:

Let us consider 100 g of compound – 2

Mass of hydrogen = 11.2  g

Mass of oxygen = 100 g – 11.2 g = 88.8 g

Thus, 11.2 g of hydrogen combines with 88.8 g of oxygen.

∴ 1 g of hydrogen combines with

88.8/11.2 = 7.92 g of oxygen.   …………………….. (2)

From statements (1) and (2), the ratio of different masses of oxygen combining with fixed mass of hydrogen (1 g) is  15.86 : 7.92 i.e. 2 : 1, which is simple whole number ratio. Thus, the data illustrate the law of multiple proportions.

Example – 02:

Carbon and oxygen are known to form two compounds. The carbon content in one of them is 42.9 % while in other it is 27.3 %. Show that the data illustrate the law of multiple proportions.

Solution:

Compound – 1:

Let us consider 100 g of compound – 1

Mass of carbon = 42.9 g

Mass of oxygen = 100 g – 42.9 g = 57.1 g

Thus, 42.9 g of carbon combines with 57.1 g of oxygen.

∴ 1 g of carbon combines with

57.1/42.9 = 1.33 g of oxygen.   …………………….. (1)

Compound – 2:

Let us consider 100 g of compound – 2

Mass of carbon = 27.3 g

Mass of oxygen = 100 g – 27.3 g = 72.7 g

Thus, 27.3 g of carbon combines with 72.7 g of oxygen.

∴ 1 g of carbon combines with

72.7/27.3 = 2.66 g of oxygen.   …………………….. (2)

From statements (1) and (2), the ratio of different masses of oxygen combining with fixed mass of carbon (1 g) is 1.33 : 2.66 i.e. 1 : 2, which is simple whole number ratio. Thus, the data illustrate the law of multiple proportions.

Example – 03:

A metal forms two oxides. The higher oxide contains 80% of metal. 0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidized. Show that the data illustrate the law of multiple proportions.

Solution:

Higher oxide:

Let us consider 100 g of higher oxide

Mass of metal = 80 g

Mass of oxygen = 100 g – 80 g = 20 g

Thus,80 g of metal combines with 20 g of oxygen.

∴ 1 g of metal combines with

20/80 = 0.25 g of oxygen.   …………………….. (1)

Lower oxide:

Mass of lower oxide = 0.72 g

Mass of higher oxide = 0.8 g

The higher oxide contains 80% of metal.

Mass of metal in higher oxide = 80/100 x 0.8 = 0.64 g

Mass of metal in lower oxide = 0.64 g

Mass of oxygen on lower oxide = 0.72 – 0.64 = 0.08g

Thus 0.64 g of metal combines with 0.08 g of oxygen.

∴ 1 g of carbon combines with 0.125 g of oxygen.

Thus the ratio of different masses of oxygen combining with fixed mass of hydrogen (1 g) is 0.25 : 0.125 i.e. 2 : 1, which is simple whole number ratio. Thus, the data illustrate the law of multiple proportions.

Example – 04:

Two oxides of metal contain 27.6 % and 30 % of oxygen respectively. If the formula of the first oxide is M₃O₄ what is the formula of the second oxide?

Solution:

Let ‘x’ be the atomic mass of the metal.

First oxide: % of oxygen = 27.6, % of metal = 100 – 27.6 = 72.4

The formula of the first oxide is M₃O₄. Thus the atomic ratio of M to O in oxide is 3 : 4.

Thus the atomic mass of the metal is 56.

Second oxide: % of metal = 70 % of oxygen = 100 – 70 = 30

Thus, the formula of the second oxide is M₂O₃.

In the next article, we shall study the law of reciprocal proportions.

Previous Article: The Law of Definite proportions

Next Topic: The Law of Reciprocal Proportions

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In the last article, we have studied the law of conservation of mass. In this article, we shall study the law of definite proportions. The law of definite proportions was given by French chemist Joseph Proust in 1799. The law of definite proportions is also known as the law of constant or fixed proportion.

Statement:

A pure chemical compound, irrespective of its source or the method of preparation, always contains the same elements combined in a fixed or constant ratio by weight.

Explanation: 

If the compound AB is formed by two different methods and in the first method, if m gm of A combines with n gm of B and In the second method, if x gm. of A combine with y gm of B, then according to this law,

m/n = x/y

Illustrations:

We get water from different sources like rain, streams, lakes, springs, wells, tanks, rivers, sea, chemical reactions etc. But in each case, the water contains the same elements hydrogen and oxygen combining in the ratio 2:16 i.e. 1: 8 by weight. Thus irrespective of the source and method of preparation water contains the same elements combining in the same (constant) proportion.

We get carbon dioxide from different sources like atmosphere, photosynthesis reaction, springs, chemical reactions etc. But in each case, the carbon dioxide contains the same elements carbon and oxygen combining in the ratio 12:32 i.e. 3: 8 by weight. Thus irrespective of the source and method of preparation carbon dioxide contains the same elements combining in the same (constant) proportion.

Limitations:

Isotopes are the atoms of the same element which have the same atomic number but different atomic weights.  The law is not applicable to those elements which are found in different isotopic forms. For e.g. Chlorine has two isotopes having weights 35 and 37. In hydrogen chloride (HCl) containing the chlorine atom with atomic weight 35, the ratio of the weight of hydrogen to that of chlorine is 2:35. In hydrogen chloride (HCl) containing the chlorine atom with atomic weight 37, the ratio of the weight of hydrogen to that of chlorine is 2:37. These two ratios are not the same.

The converse of this law is not true. There are some compounds in which the same elements combine together in the same proportion, give different compounds.For e.g. Dimethyl ether (CH₃ – O – CH₃) and Ethyl alcohol (C₂H₅OH) have the same molecular formula C₂H₆O, and hence they contain the same elements carbon, hydrogen, and oxygen combining in the proportion 24 : 6 :16 i.e. 12 : 3: 8 by weight. But these two compounds are entirely different having different properties. Such compounds are called isomers.

Explanation of the Law of Definite Proportions on the Basis of Dalton’s Atomic Theory:

According to Dalton’s atomic theory, a compound is formed by the combination of atoms of different elements in simple numerical proportions. Atoms of an element are identical in all respects, thus the weight of atoms are also fixed. Therefore, a chemical compound always contains the same elements combining in the same constant proportion by weight. This explains the law of definite proportion.

Numerical Problems:

Example – 01:

1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the mass of copper that remained was 1.098 g. In another experiment, 1.179 g of copper was dissolved in the nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The mass of cupric oxide formed was 1.476 g. Show that the results illustrate the law of definite proportions.

Solution:

Experiment -1:

Mass of cupric oxide = 1.375 g

Mass of copper = 1.098 g

Mass of oxygen = 1.375 g – 1.098 g = 0.277 g

Ratio of mass of copper to that of oxygen is

1.098 g / 0.277 g = 3.96  ………….  (1)

Experiment -2:

Mass of cupric oxide = 1.476 g

Mass of copper = 1.179 g

Mass of oxygen = 1.476 g  – 1.179 g = 0.297 g

Ratio of mass of copper to that of oxygen is

1.179 g / 0.297 g = 3.97  ………….  (2)

Within experimental limits, the results of experiment 1 and 2 show that the ratios of mass of copper to the mass of oxygen in both the samples of cupric oxide are the same. Hence, the results illustrate the law of definite proportions.

Example – 02:

2.8 g of calcium oxide prepared by heating limestone was found to contain 0.8 g of oxygen. When 1 .0 g of oxygen is treated with calcium, 3.5 g of calcium oxide was obtained. Show that the result illustrates the law of definite proportions.

Solution:

Experiment -1:

Mass of calcium oxide = 2.8 g

Mass of oxygen = 0.8 g

Mass of calcium = 2.8 g – 0.8 g = 2.0 g

Ratio of mass of calcium to that of oxygen is

2.0 g / 0.8 g =  2.5  ………….  (1)

Experiment -2:

Mass of calcium oxide = 3.5 g

Mass of oxygen = 1.0 g

Mass of calcium = 3.5 g – 1.0 g = 2.5 g

Ratio of mass of calcium to that of oxygen is

2.5 g / 1.0 g = 2.5  ………….  (2)

The results of experiment 1 and 2 show that the ratios of mass of calcium to the mass of oxygen in both the samples of calcium oxide are the same. Hence, the results illustrate the law of definite proportions.

Example – 03:

A sample of pure magnesium carbonate was found to contain 28.5 % of magnesium, 14.29 % of carbon, and 57.14 % of oxygen. Applying the law of constant proportion, find the mass of magnesium, carbon, and oxygen in 15.0 g of another sample of magnesium carbonate.

Solution:

In MgCO3 % of Mg = 28.5, % of C = 14.29, % of O = 57.14

Mass of second sample of MgCO₃ = 15.0 g

Applying the law of constant proportion

Mass of Mg in the second sample =  = 15.0 g x  28.5 / 100 =  4.28 g

Mass of C in the second sample = 15.0 g x 14.29 / 100 =  2.14 g Mass of C in

Mass of O in the second sample =  15.0 g – (4.28 g + 2.14 g) =  8.58 g

Thus, 15.0 g of MgCO3 contains 4.28 g of magnesium, 2.14 g of carbon, and 8.58 g of oxygen.

Example – 04:

A sample of pure calcium carbonate was found to contain 40 % of calcium, 12 % of carbon, and 48 % of oxygen. Applying the law of constant proportion, find the mass of calcium, carbon, and oxygen in 2.44 g of another sample of calcium carbonate.

Solution:

% of Ca in CaCO3 = 40, % of C in CaCO3 = 12, % of O in CaCO3 = 48

Mass of second sample of CaCO3 = 2.44 g

Applying the law of constant proportion

Mass of Ca in second sample = 2.44 g x 40/100 =  0.98 g

Mass of C in second sample = 2,44 g x 12/100 =  0.29 g

Mass of C in second sample = 2.44 g – (0.98 g + 0.29 g) =  1.17 g

Thus, 2.44 g of CaCO3 contains 0.98 g of calcium, 0.29 g of carbon, and 1.17 g of oxygen.

Example – 05:

Mass of copper oxide obtained by treating 2.16 g of metallic copper with nitric acid and on subsequent ignition was 2.70 g. In another experiment, 1.15 g of copper oxide on reduction yielded 0.92 g of copper. Show that the results are in accordance with the law of fixed proportions.

Solution:

Experiment -1:

Mass of copper oxide = 2.70 g

Mass of copper = 2.16 g

Mass of oxygen = 2.70 g – 2.16 g  = 0.54 g

Ratio of mass of copper to that of oxygen is

2.16 / 0.54 = 4  ………….  (1)

Experiment -2:

Mass of copper oxide = 1.15 g

Mass of copper = 0.92 g

Mass of oxygen = 1.15 g – 0.92 g = 0.23 g

Ratio of mass of copper to that of oxygen is

0.92 / 0.23 = 4  ………….  (2)

The results of experiment 1 and 2 show that the ratios of the mass of copper to the mass of oxygen in both the samples of copper oxide are the same. Hence, the results are in accordance with the law of fixed proportions.

Previous Topic: The Law of Conservation of Mass

Next Topic: The Law of Multiple Proportions

Science > Chemistry > Laws of Chemical Combinations > Law of Definite Proportions

The post Law of Definite Proportions appeared first on The Fact Factor.

In last article, we have studied Dalton’s atomic theory. In this article, we shall study the law of conservation of mass. The law of conservation of mass was given by Russian scientist Lomonosov in 1765 and French Scientist Antoine Lavoisier in 1783 independently. This law is also called the law of indestructibility of matter.

Statement:

In a chemical reaction, the total mass of the reactants before the reaction is the same as the total mass of the products after reaction. i.e. in a chemical reaction, the total mass is conserved.

Explanation:

Consider following chemical reaction

A  +  B   →   C   +   D

Let M_A = Mass of reactant A

M_B = Mass of reactant B

M_C = Mass of product C

M_D = Mass of product D

By law of conservation of mass

Total mass of reactants = Total mass of products

M_A + M_B =  M_C + M_D

lllustration:

Consider following chemical reaction

NaCl  + AgNO₃     →  NaNo₃  + AgCl ↓

The total mass of reactants   = ( 23×1 + 35.5×1)  +  (108×1 + 14×1 + 16×3)=  228.5

The total mass of products    = (23×1 + 14×1 + 16×3) + (108×1 + 35.5) = 228.5

Thus, The total mass of reactants is equal to the total mass of products. Thus, the law of conservation of mass is illustrated.

Landolt’s Experiment:

Landolt used this experiment to verify the law of conservation of mass. He used H – Shaped tube for his experiment. One arm (limb) of the apparatus is filled with sodium chloride solution and the other arm is filled with silver nitrate solution.

The tube is then sealed and weighed accurately and carefully so that the two solutions do not mix with each other.

Then the apparatus is well shaken and the two solutions are allowed to mix and react with each other white precipitate of AgCl is obtained.

NaCl  + AgNO₃     →  NaNo₃  + AgCl ↓

The apparatus is weighed accurately again. Within experimental limit, it is found that the weight of the apparatus before the reaction is equal to the weight of the apparatus after the reaction. Thus the total mass of reactants is equal to the total mass of the products. Thus the total mass is conserved. Thus the law of conservation of mass is verified.

Limitations of Landolt’s Experiment:

• The reaction between sodium chloride and silver nitrate is exothermic. Due to which evaporation of the moisture on the outer surface of the apparatus takes place and this takes a long time to recondense back on the apparatus.
• There is an increase in the volume of the apparatus, due to the evolution of heat. It takes a long time to get original volume. Due to increase in the volume the up thrust of air on the apparatus increases and hence it shows less weight.
• These errors can be minimized by weighing the apparatus after very long time allowing the apparatus to cool thoroughly.

Limitations of the Law of Conservation of Mass:

The law of conservation fails for chemical reactions in which a large amount of heat is evolved or absorbed during the reaction. During a chemical reaction, energy is evolved or absorbed in the form of heat, light etc. During the evolution of heat, some mass of reactants is converted into energy. But this loss of mass is so small that it cannot be recorded on very sensitive balance.

According to Einstein’s theory, mass and energy are interconvertible. The relation between the two quantities is given by E = mc² Where, E = energy liberated, m = mass lost in the reaction. c = speed of light in vacuum Thus the law of conservation of mass is valid if both the mass and the energy are conserved.

Combined Law of Conservation of Mass and Energy:

In a chemical reaction, the total amount of mass and energy of reactants is equal to the total amount of mass and energy of products i.e. In a chemical reaction the total amount of mass and energy are conserved.

Law of Conservation of Energy: Energy can neither be created nor destroyed, but it can be converted from one form to another. i.e.  the total amount of energy of the universe is conserved.

Explanation of  the Law of Conservation of Mass on the Basis of Dalton’s Atomic Theory:

According to atomic theory, atoms can neither be created nor be destroyed. In chemical reactions, the atoms rearrange, but they do not themselves break apart. Since each atom has a definite mass, the total mass after chemical change remains constant and this explains the law of conservation of mass.

Numerical Problems:

Example – 01:

10.0 g of CaCO₃ on heating gave 4.4 g of CO₂ and 5.6 g of CaO. Show that these observations are in agreement with the law of conservation of mass.

Solution:

CaCO₃ → CaO +     CO₂

Mass of reactants = 10.0 g

Mass of products = 5.6 g + 4.4 g  =  10.0 g

Thus,  Mass of reactants = Mass of products.

Hence the observations are in agreement with the law of conservation of mass.

Example – 02:

x g of potassium chlorate on decomposing produced 1.9 g of oxygen and 2.96 g of potassium chloride. What is the value of x?

Solution:

KClO₃ →  KCl +     O₂

Mass of reactants = x g

Mass of products =  4.86 g

Now, by the law of conservation of mass. Mass of reactants = Mass of products.

x = 4.86

The value of x is 4.86

Example – 03:

When 4.2 g of NaHCO3 is added to a solution of CH3COOH weighing 10 g; it is observed that 2.2 g of CO2 is released into the atmosphere. The residue is found to weigh 12.0 g. Show that these observations are in agreement with the law of conservation of mass.

Solution:

NaHCO₃  + CH₃COOH → Residue  + CO₂

Mass of reactants = 4.2 g + 10 g = 14.2 g

Mass of products = 12.0 g + 2.2 g = 14.2 g

Thus,  Mass of reactants = Mass of products.

Hence the observations are in agreement with the law of conservation of mass. Now, by the law of conservation of mass.

Example – 04:

When 6.3 g of NaHCO₃ is added to 15.0 g  solution of CH₃COOH; the residue is found to weigh 18.0 g. What is the mass of CO₂ released in the reaction?

Solution:

NaHCO₃  + CH₃COOH → Residue  + CO₂

Mass of reactants =  6.3 g + 15.0 g = 21.3 g

Mass of products =  18.0 g + Mass of   CO₂

Now, by the law of conservation of mass. Mass of reactants = Mass of products.

∴  21.3 g  =   18.0 g + Mass of   CO₂

∴   Mass of   CO₂ = 21.3 g – 18.0 g = 3.3 g

Ans: The mass of CO₂ released is 3.3 g.

In the next article, we shall study the law of definite/constant/fixed proportions

Previous Topic: Dalton’s Atomic Theory

Next Topic: The Law of Definite Proportions

Science > Chemistry > Laws of Chemical Combinations > Law of Conservation of Mass

The post Law of Conservation of Mass appeared first on The Fact Factor.