Laws of Chemical Combinations Class 11 Notes | EduRev

Chemistry For JEE

Class 11 : Laws of Chemical Combinations Class 11 Notes | EduRev

The document Laws of Chemical Combinations Class 11 Notes | EduRev is a part of the Class 11 Course Chemistry For JEE.
All you need of Class 11 at this link: Class 11

Introduction

To understand the composition of the compounds, it is necessary to have a theory that accounts for both qualitative and quantitative observations during chemical changes. Observations of chemical reactions were most significant in the development of a satisfactory theory of the nature of matter. These observations of chemical reactions are summarized in certain statements known as laws of chemical combination.Laws of Chemical Combinations Class 11 Notes | EduRev

Law of Conservation of Mass (Lavoisier)

"In a chemical change, the 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 gmLaws of Chemical Combinations Class 11 Notes | EduRevLaw of conservation of mass

Example 1: 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 29.3 gm. 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.

Question 1:A quantity of 10 g of a hydrocarbon exactly requires 40 g oxygen for complete combustion. The products formed are CO2 and water. When CO2 gas formed is absorbed completely in lime water, the mass of solution increases by 27.5 g. What is the mass of water formed simultaneously in the combustion?

Example 2: 5.2 g of CaCO3 when heated produced 1.99 g of Carbon dioxide and the residue (CaO) left behind weighs 3.2 g. 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 the law of conservation of mass holds good within experimental errors.

Question 2:A sample of an ethanol–water solution has a volume of 55.0 cm3 and a mass of 50.0 g. What is the percentage of ethanol (by mass) in the solution? Assume that there is no change in volume when the pure compounds are mixed. The density of ethanol is 0.80 g/cm3 and that of water is 1.00 g/cm3.

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 | EduRevLaw of constant composition

Example 1: The following are results of the 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 the same and the samples obey the law of definite proportion.


Example 2: 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 per cent. Use these data to illustrate the law of constant composition.
Solution.
Step 1: To calculate the percentage of oxygen in the 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:
Laws of Chemical Combinations Class 11 Notes | EduRev
i.e. oxygen present = 13.38%

Step 2: To compare the percentage of oxygen in both the experiments. Percentage of oxygen in PbO2 in the first 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.


Try Yourself!
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.


Law of Multiple Proportions (Dalton)

"When one element combines with the other elements 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 | EduRevLaw of multiple proportions 

Carbon is found to form two oxides containing 42.9% & 27.3% of carbon, respectively, showing that these figures show the law of multiple proportions.

 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 combine with 42.9 parts of carbon.
 1 part of oxygen will combine with 42.9/57.1 = 0.751 part of the carbon.
Similarly, in 2nd oxide:
 1 part of oxygen will combine with 27.3/72.7  = 0.376 part of the carbon.
 The ratio of carbon that combines with the same mass of oxygen = 0.751: 0.376 = 2:1
This is a simple whole-number ratio, this means the above data shows the law of multiple proportions.


Example 1: 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 illustrate the law of multiple proportions. 
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 the removal of the oxygen present in the oxide, to combine with the hydrogen.
Therefore, the composition of the yellow oxide is:
 Oxygen = 0.24 gm and lead = 3.45 - 0.24 = 3.21 gm.
The mass ratio of lead and oxygen:
Laws of Chemical Combinations Class 11 Notes | EduRev
► The composition of the brown oxide is : oxygen = 0.16 gm and lead = 1.227 - 0.16 = 1.067 gm.
The mass ratio of lead and oxygen:
Laws of Chemical Combinations Class 11 Notes | EduRev
Now, r1 : r2 = 13.375 : 6.669 = 2:1 (simple ratio) and hence the data illustrates the law of multiple proportion. 

Question 3:Law of multiple proportions is not applicable for the oxide(s) of

Example 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:
 First oxide:  Carbon 42.9%   27.3% (Given)
 Second oxide: Carbon Oxygen 57.1% 72.7% (by difference)

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  27.3/72.7 = 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.

Question 4:Two elements A and B combine to form compound X and Y. For the fi xed mass of A, masses of B combined for the compounds X and Y are in 3:7 ratio. If in compound X, 4 g of A combines with 12 g B, then in compound Y, 8 g of A will combine with ...... g of B.

Law of Reciprocal Proportions (Richter)

"When two elements combine separately with a 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 | EduRevLaw 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 the figure.
  • Thus, the ratio of H and S masses, which combines with a fixed mass of oxygen, is a whole number multiple of the ratio in which H and S combine together.


Example 1: 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 = 1 gm.
Then, the mass of hydrogen combined with 1 gm carbon in methane:

Laws of Chemical Combinations Class 11 Notes | EduRev
And the mass of oxygen combined with 1 gm carbon in carbon dioxide:
Laws of Chemical Combinations Class 11 Notes | EduRev
Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of carbon:
Laws of Chemical Combinations Class 11 Notes | EduRev

Now, the mass ratio of hydrogen and oxygen in water:
Laws of Chemical Combinations Class 11 Notes | EduRev
As r1 and r2 are the same, the data is according to the law of reciprocal proportion.


Example 2: Copper sulphide contains 66.6% Cu, copper oxide contains 79.9% copper and sulphur trioxide contains 40% Sulphur. Show that these data illustrate the 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 combine with Cu = 66.6 parts

Laws of Chemical Combinations Class 11 Notes | EduRev 40.0 parts of sulphur combine with Laws of Chemical Combinations Class 11 Notes | EduRev
 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)

Laws of Chemical Combinations Class 11 Notes | EduRev
Which is a simple whole-number ratio, hence law of reciprocal proportion is proved.

Try Yourself 

61.8g of A combines with 80 g of B. 30.9g of A combines with 106.5g of C, B and C combine to form compound. Atomic weight of C and B are respectively 35.5 and 6.6. Show that the law of reciprocal proportion is obeyed.


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 a 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 | EduRevGay Lussac's Law of combining volumes
  • Law of combining volumes. These are no longer useful in chemical calculations now but gives an idea of earlier methods of analyzing and relating compounds by mass.

Example 1: 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.


Example 2: How much volume of oxygen will be required for complete combustion of 40 ml of acetylene (C2H2) and how much volume of carbon dioxide will be formed? All volumes are measured at NTP.
Solution:
Laws of Chemical Combinations Class 11 Notes | EduRev
So, for complete combustion of 40 ml of acetylene, 100 ml of oxygen are required and 80 ml of CO2 is formed.


Objective Questions

Q.1. Two oxides of a certain metal were separately heated in a current of hydrogen until a constant weight was obtained. The water produced in each case was carefully collected and weighed. 2g of each oxide gave 0.2517g and 0.4626g of water respectively. This observation illustrates:
(a) Law of conservation of mass
(b) Law of constants proportions
(c) Law of multiple proportions
(d) Law of reciprocal proportions 
Ans. (c)
Solution: 1st oxide: Oxygen present in 0.2517g H2O
= 16/18 × 0.2517g = 0.224g
► Metal present =2 - 0.224g = 1.776g
► 2nd oxide: Oxygen present in 0.452g H2O
= 16/18  × 0.4526g = 0.402g
► Metal present =2 - 0.402g = 1.598 g
► Masses of oxygen which combine with fixed mass (1g) of metal in the two oxides = 0.224/1.776  and  0.402/1.598 g = 0.126g and 0.252 g, i.e. in the ratio 1:2. 

Hence, it illustrates the law of multiple proportions.

Q.2. The mass of nitrogen per gram hydrogen in the compound hydrazine is exactly one and a half time the mass of nitrogen in the compound, ammonia. The fact illustrates:
(a) The law of conservation of mass
(b) The multiple valencies of nitrogen
(c) The law of multiple proportions
(d) The law of definite proportions
Ans. (c)
Solution: 

The law of multiple proportions says that, If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.

 The ratio of masses of nitrogen per gram of hydrogen in hydrazine (N2H4) and NH= 1.5: 1= 3/2 :1 = 3:2. 

Which is a simple whole-number ratio. 

This illustrates the law of multiple proportions.

Q.3. The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of:
(a) Constant proportions
(b) Conservation of mass
(c) Multiple proportions
(c) Reciprocal proportions
Ans. (a)
Solution: The percentage of copper and oxygen in a CuO sample obtained from different methods was found to be the same. This proves the law of Constant proportion as the ratio of Cu:O remains constant.

Q.4. Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained the other oxide. The data illustrate:
(a) Law of reciprocal proportions
(b) Law of constant proportions
(c) Law of multiple proportions
(d) Law of equivalent proportions

Ans. (c)
Solution: This shows that the weights of lead combined with the fixed weight of oxygen are in the ratio 1/2: 1 or 1: 2, according to the law of multiple proportions.

Q.5. One part of an element X combines with two parts of another element B. Six parts of the element C combine with four parts of the element B. If A and C combine together, the ratio of their weights 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
Ans. 
(c)
Solution: Law of reciprocal proportions (by definition).

 According to the law of reciprocal proportions, the weights of two elements combining with a fixed amount of the third element will bear the same ratio (or a simple multiple of it) in which they themselves react.

Q.6. Two gaseous samples were analyzed. One contained 1.2 g of carbon and 3.2 g of oxygen. The other contained 27.3% carbon and 72.7% oxygen. The experimental data are in accordance with 
(a) Law of conservation of mass
(b) Law of definite proportions 

(c) Law of reciprocal proportions
(d) Law of multiple proportions

Ans. (b)
Solution: % of C in the 1 st sample  = 1∙2/1∙2 + 3∙2  × 100 = 27·3% which is the same as in the 2nd sample. 

Hence, the law of definite proportions is obeyed.


Try Yourself!

Q.1. What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate, if the law of conservation of mass is true?
Ans: 
According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
For the chemical reaction: AgNO+ NaCI → NaNO+ AgCI
 mreactants = mproducts
 mAgNO3 + mNaCl = mNaNO3 + mAgCl
 x + 5.85 = 8.5 + 14.35
 x = 17.0 g

Q.2. Elements X and Y form two different compounds. In the first, 0.324 g of X is combined with 0.471 g of Y. In the second, 0.117 g of X is combined with 0.509 g of Y. Show that these data illustrate the law of multiple proportions.
Ans. 

According to the law of multiple proportions, "If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers".
 In the first compound, 0.324 g of X is combined with 0.471 g of Y.
 Hence, 1 g of X is combined with  0.471/0.324  = 1.45 g of Y.
 In the second compound, 0.117 g of X is combined with 0.509 g of Y.
 Hence, 1 g of X is combined with 0.1170.509 = 4.35 g of Y.
 The ratio 1.454.35 = 13 is a small whole number.

Q.3. Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.79% of carbon and sulphur dioxide contains 50% of sulphur. Are these figures in agreement with the law of reciprocal proportions?
Ans.
 In CO2, 27.27% C and 72.73% O is present.
 Ratio of C to O=C:O=27.27:72.73 = 1:2667⟶(i)
 In CS2, 15.97% C and 84.03% of S is present.
 The ratio of C to S = C:S
 1g of C combines with =15.7984.21=5.33g of S⟶(ii)
► The ratio of different masses of S&O combining with a fixed mass of C is 5.33:2.67 i.e. 2:1⟶(iii)
 In SO2, 50% of sand 50% of O is present.
∴ Rate of S to O = 50:50 = 1:1⟶(iv)
The data illustrate the law of reciprocal proportions.
The ratio of different masses of S and O combining with a fixed mass of carbon is a simple whole number.

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Objective type Questions

,

Previous Year Questions with Solutions

,

Sample Paper

,

study material

,

Summary

,

Laws of Chemical Combinations Class 11 Notes | EduRev

,

shortcuts and tricks

,

Viva Questions

,

past year papers

,

pdf

,

video lectures

,

Laws of Chemical Combinations Class 11 Notes | EduRev

,

Important questions

,

practice quizzes

,

ppt

,

Extra Questions

,

mock tests for examination

,

MCQs

,

Free

,

Exam

,

Laws of Chemical Combinations Class 11 Notes | EduRev

,

Semester Notes

;