Laws of Chemical Combinations Class 11 Notes | EduRev

Chemistry Class 11

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JEE : Laws of Chemical Combinations Class 11 Notes | EduRev

The document Laws of Chemical Combinations Class 11 Notes | EduRev is a part of the JEE Course Chemistry Class 11.
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Laws of Chemical Combinations Class 11 Notes | EduRev

Laws of Chemical Combinations Class 11 Notes | EduRev


5 LAWS OF CHEMICAL COMBINATIONS
(1) Law of conservation of mass [Lavoisier]:

"In a chemical change, total mass remains conserved i.e. mass before the reaction is always equal to mass after the reaction."

Laws of Chemical Combinations Class 11 Notes | EduRev
mass before the reaction = 1 × 2 + 1/2 × 32 = 18 gm
mass after the reaction = 1 × 18 = 18 gm.

Laws of Chemical Combinations Class 11 Notes | EduRevFig: Law of conservation of mass


Example: A 15.9 gm sample of sodium carbonate is added to a solution of acetic acid weighing 20.0 gm. The two substances react, releasing carbon dioxide gas to the atmosphere. After the reaction, the contents of the reaction vessel weigh 29.3 gm. What is the mass of carbon dioxide given off during the reaction? 
Solution: The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3gm. The difference is due to the mass of released carbon dioxide gas.

Hence, the mass of carbon dioxide gas released = 35.9 - 29.3 = 6.6 gm.

(2) Law of constant composition [Proust]:

"All chemical compounds are found to have constant composition irrespective of their method of preparation or sources."
⇒ In H2O, Hydrogen & oxygen combine in 2:1 molar ratio, this ratio remains constant whether it is Tap water, river water or seawater or produced by any chemical reaction.   Laws of Chemical Combinations Class 11 Notes | EduRevFig: Law of constant composition


Example: The following are results of analysis of two samples of the same or two different compounds of phosphorus and chlorine. From these results, decide whether the two samples are from the same or different compounds. Also state the law, which will be obeyed by the given samples.Laws of Chemical Combinations Class 11 Notes | EduRev

Solution: The mass ratio of phosphorus and chlorine in compound A,
mP : mCl = 1.156 : 3.971 = 0.2911 : 1.000
The mass ratio of phosphorus and chlorine in compound B,
mP : mCl = 1.542 : 5.297 = 0.2911 : 1.000
As the mass ratio is the same, both the compounds are same and the samples obey the law of definite proportion

(3) Law of multiple proportions [Dalton]:

"When one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other bear a simple ratio to one another."

Laws of Chemical Combinations Class 11 Notes | EduRevFig: law of multiple proportions⇒ Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that these figures shows the law of multiple proportion.

  First oxide  Second oxide
 Carbon    42.9 % 27.3 %
 Oxygen     57.1 % 72.7%

In the first oxide, 57.1 parts by mass of oxygen combines with 42.9 parts of carbon.
1 part of oxygen will combine with Laws of Chemical Combinations Class 11 Notes | EduRev = 0.751 part of carbon
Similarly in 2nd oxide,
1 part of oxygen will combine with Laws of Chemical Combinations Class 11 Notes | EduRev = 0.376 part of carbon
The ratio of carbon that combine with the same mass of oxygen = 0.751 : 0.376 = 2 : 1
This is a simple whole no. ratio, this means above data shows the law of multiple proportions.

Example: Two oxide samples of lead were heated in the current of hydrogen and were reduced to the metallic lead. The following data were obtained
(i) Weight of yellow oxide taken = 3.45 gm; Loss in weight during reduction = 0.24 gm
(ii) Weight of brown oxide taken = 1.227 gm; Loss in weight during reduction = 0.16 gm.
Show that the data illustrates the law of multiple proportion. 
Solution: When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the loss in weight is due to removal of the oxygen present in the oxide, to combine with the hydrogen. Therefore, the composition of the yellow oxide is: oxygen = 0.24gm and lead = 3.45 - 0.24 = 3.21gm.
The mass ratio of lead and oxygen, r = Laws of Chemical Combinations Class 11 Notes | EduRevand the composition of the brown oxide is : oxygen = 0.16gm and lead = 1.227 - 0.16 = 1.067gm.
The mass ratio of lead and oxygen, r2 = Laws of Chemical Combinations Class 11 Notes | EduRev
Now, r: r2 = 13.375 : 6.669 = 2:1 (simple ratio) and hence the data illustrates the law of multiple proportion. 

(4) Law of reciprocal proportions [Richter]:

"When two elements combine separately with definite mass of a third element, then the ratio of their masses in which they do so is either the same or some whole number multiple of the ratio in which they combine with each other."
This law can be understood easily with the help of the following examples.

Laws of Chemical Combinations Class 11 Notes | EduRevFig: Law of reciprocal proportions

⇒ Let us consider three elements - hydrogen, sulphur and oxygen. Hydrogen combines with oxygen to form H2O whereas sulphur combines with it to form SO2. Hydrogen and sulphur can also combine together to form H2S. The formation of these compounds is shown in fig.
In H2O, the ratio of masses of H and O is 2 : 16.
In SO2, the ratio of masses of S and O is 32 : 32. Therefore, the ratio of masses of H and S which combines with a fixed mass of oxygen (say 32 parts) will be 4 : 32  i.e.  1 : 8 ...(i)
When H and S combine together, they form H2S in which the ratio of masses of H and S is 2 : 32  i.e.  1 : 16 ...(ii)
The two ratios (i) and (ii) are related to each other as or 2 : 1.
i.e., they are whole number multiples of each other.
Thus, the ratio of masses of H and S which combines with a fixed mass of oxygen is a whole number multiple of the ratio in which H and S combine together.


Example: Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27 % carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89% oxygen, by mass. Show that the data illustrates the law of reciprocal proportion. 

Solution: Methane and carbon dioxide, both contain carbon and hence, carbon may be considered as the third element. Now, let the fixed mass of carbon = 1gm. Then, the mass of hydrogen combined with 1 gm carbon in methane Laws of Chemical Combinations Class 11 Notes | EduRevgm and the mass of oxygen combined with 1gm carbon in carbon dioxide Laws of Chemical Combinations Class 11 Notes | EduRevgm
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon, r1Laws of Chemical Combinations Class 11 Notes | EduRev
Now, the mass ratio of hydrogen and oxygen in water, r2Laws of Chemical Combinations Class 11 Notes | EduRev
As r1 and r2 are same , the data is according to the law of reciprocal proportion.

(5) Gay Lussac's law of combining volumes:

"When two or more gases react with one another, their volumes bear simple whole number ratio with one another and to the volume of products (if they are also gases) provided all volumes are measured under identical conditions of temperature and pressure."
When gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride according to the following equation.
It has been observed experimentally that in this reaction, one volume of hydrogen always reacts with one volume of chlorine to form two volumes of gaseous hydrogen chloride. all reactants and products are in gaseous state and their volumes bear a ratio of 1 : 1 : 2. This ratio is a simple whole number ratio.

Laws of Chemical Combinations Class 11 Notes | EduRevFig: Gay Lussac's Law of combining volumesThese are no longer useful in chemical calculations now but gives an idea of earlier methods of analyzing and relating compounds by mass.


Example: 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces 7.5 ml carbon dioxide and 10.0ml water vapour. All the volumes are measured at the same pressure and temperature. Show that the data illustrates Gay Lussac's law of volume combination. 

Solution: Vhydrocarbon : Voxygen : Vcarbon dioxide : Vwater vapour = 2.5 : 12.5 : 7.5 : 10.0
= 1 : 5 : 3 : 4 (simple ratio)
Hence, the data is according to the law of volume combination.

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