X Mat Che 3 L.o.che - Combi
X Mat Che 3 L.o.che - Combi
X Mat Che 3 L.o.che - Combi
(d) Proust
5. Some basic experiments on the law of reciprocal proportions was done by ...................
(a) Wenzel
(b) Lavoisier
(c) Dalton
(d) Gay lussac
6. The ratio of the volumes of reactants and products in the formation of NH3 is ...................
(a) 1 : 3 : 2
(b) 1 : 2 : 3
(c) 1 : 1 : 2
(d) None
7. When two or more gases react with one another, their volumes bear simple ratio. This statement given by
...................
(a) Law of mass action
(b) Laws of multiple proportion
(c) Law of reciprocal proportion
(d) Law of combining volume
8. Law of definite proportion was stated in the year ...................
(a) 1756
(b) 1774
(c) 1799
(d) 1803
9. Experimental verification for the law of multiple proportion was verified by ...................
(a) Berzelius
(b) Dalton
(c) Lavoisier
(d) lomonssoff
10. Law of Multiple proportion will not hold good for elements with different ................... in its compounds.
(a) Isotopes
(b)Isomers
(c) isobars
(d) Vapour pressure
__________________________________________________________________________________
Answers:
1. (a) 2. (a) 3. (a) 4. (a) 5. (a) 6. (a) 7. (b) 8. (c) 9. (a) 10. (a)
__________________________________________________________________________________
II. Answer the following in One or Two sentences.(T.B.Page. 61.)
1. State law of mass action.
The total mass of substances taking part in a chemical reaction remains the same throughout the
change.
2. State law of multiple proportion.
When two elements A and B combine to form two or more compounds, then different weights of B
which combine with a fixed weight of A bears a simple numerical ratio to one another.
3. Define law of conservation of mass.
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.
4. Give the limitations of law multiple proportion.
The law is valid till an element is present in one particular isotopic form in all its compounds. When an
element exists in the form of different isotopes in its compounds, the law does not hold good.
III. Answer in brief. (T.B. Page 61)
1.State and explain Gay lussac law with a simple illustration.
Law: 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.
The law can be understood with the help of following example.
(i) Gaseous hydrogen and gaseous chlorine react together to form gaseous hydrogen chloride according to
the following equation.
H2 (g)
+
Cl2 (g)
2 HCl (g)
One volume
One volume
Two volume
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.
(ii) Similarly, under suitable conditions, gaseous nitrogen and gaseous hydrogen combine together to form
gaseous ammonia according to the equation
N2 (g) + 3H2 (g) 2 NH3 (g)
It has been found that one volume of nitrogen always reacts with three volumes of hydrogen to form two
volumes of gaseous ammonia. Thus, the volumes of reactants and products bear the ratio 1: 3: 2 which is a
simple whole number ratio.
2.What is the present day position of law of conservation of mass?
Present Day Position of the Law of conservation of mass: This law is particularly not applicable to
nuclear reactions where tremendous amount of energy is liberated. However for chemical reactions, the law of
conservation of mass is adequate, since energy changes are comparatively small (i.e., the change in mass is
immeasurably small or negligible).
3.State and explain the law of constant proportion with an illustration.
Law: A pure chemical compound always contains the same elements combined together in the same
definite (fixed or constant) proportions by weight, irrespective of its source or method of preparation. Therefore,
the law is also called as the law of fixed proportions or constant proportions.
ILLUSTRATIONS:
(a) Carbon dioxide may be obtained by the following methods:
(i) by burning carbon
(ii) By reaction between a metal carbonate and a dilute acid.
(iii) By heating calcium carbonate or sodium bicarbonate. Analysis of carbon dioxide, prepared by any
of the above methods, shows that it contains only carbon and oxygen, combined together in the
same proportion by weight, i.e., 12: 32 or 3: 8.
Present Day Position of the Law of conservation of mass: This law is particularly not applicable to
nuclear reactions where tremendous amount of energy is liberated. However for chemical reactions, the law of
conservation of mass is adequate, since energy changes are comparatively small (i.e., the change in mass is
immeasurably small or negligible).
2.Give the experimental verification of law of constant composition.
Prepare pure samples of cupric oxide by two different methods.
(i) by heating copper carbonate
(ii) by the decomposition of cupric nitrate.
The cupric oxide prepared by both the methods always contains the same elements copper and oxygen
combined together in the same fixed proportion of 4: 1 by weight. This illustrates the law of definite
proportions. It can be verified by taking a known weight of a pure sample (W1 gm) of cupric oxide in a porcelain
boat.
2 Cu(NO3)2 2 CuO + 4 NO2 + O2
It is placed inside a hard glass tube kept horizontally as shown in fig 3.2.
A current of pure dry hydrogen is sent inside the tube and the tube is heated. The cupric oxide is reduced
to metallic copper.
CuO + H2 Cu + H2O
The Weight of copper formed is found out W2gm.
Calculation:
Method 1:
Weight of cupric oxide
W1gm
Weight of Copper
W2gm
Weight of oxygen
W1 W2 gm
W2 : (W1- W2)
The same experiment is repeated with a known weight W3 gm of cupric oxide prepared by heating
copper carbonate
CuCO3
Cu O + CO2
The cupric oxide formed is reduced to metallic copper by passing a current of pure and dry hydrogen
inside the tube as before. The weight of metallic copper was found to be W4 gm. The ratio of the
weight of copper to the weight of oxygen in both the samples are calculated as follows:
Method 2:
Weight of cupric oxide
W3gm.
Weight of copper
W4gm
Weight of oxygen
W 3 W 4 gm
W4 : (W3 W4)
The two ratios [W2 : W1 W2] and [ W4 : W3 W4] are found to be the same and is equal to 4: 1. Thus the
law of definite proportions is verified experimentally.
3.Explain the law of multiple proportions with suitable illustrations.
Law: When two elements A and B combine to form two or more compounds, then different weights of
B which combine with a fixed weight of A bears a simple numerical ratio to one another.
Explanation: Carbon combines with oxygen to form two different oxides, namely, carbon monoxide
(CO) and carbon dioxide (CO2). The proportions by weight of the two elements are
Carbon monoxide - C: O:: 12 : 16
Carbon dioxide - C: O:: 12 : 32
There, the weights of oxygen that combine with a fixed weight of carbon (12g) are in the ratio 16g : 32g
i.e. 1:2, a simple numerical ratio.
Illustrations:
Nitrogen combines with oxygen to form different oxides. The compositions by weight of these oxides are
shown in table.
Compositions by weight of oxides of nitrogen
No.
Name of Oxide
Wt. of nitrogen
in grams
Wt. of oxygen
in grams
28
16
2
3
28
28
32
48
28
64
28
80
It can be seen from the table that different weights of oxygen that combines with a fixed weight of
nitrogen (28 g) are in the ratio,16g : 32 g : 48g : 64g : 80g i.e., in the simple numerical ratio of 1 : 2 : 3 : 4 : 5.
4.Give the experimental verification of law of multiple proportions.
The law can easily be verified by the study of oxides of copper. Copper reacts with oxygen to
form two oxides - the red cuprous oxide (Cu2O) and the black cupric oxide (CuO).
In order to verify the law of multiple proportion, fixed amounts of these oxides (say 20g each) are
separately reduced to metallic copper by heating them in a current of hydrogen and the masses of copper
obtained from them are estimated. The difference in the mass of oxide taken and the mass of copper obtained
from it gives the mass of oxygen present in it.
Now the masses of oxygen which combine with a definite mass of copper in the two oxides are
calculated. These masses are found in a simple whole number ratio. This verifies the law of multiple proportion.
Present day position of Law of Multiple Proportion: The law is valid till an element is present in one
particular isotopic form in all its compounds. When an element exists in the form of different isotopes in its
compounds, the law does not hold good.
5.State and explain the law of reciprocal proportions.
Law: 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.
Illustrations:
1. 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 figure.
......... (1)
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
......... (2)
The two ratios (i) and (ii) are related to each other as
:
or 2 : 1
i.e., they are whole number multiple 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.
Sulphur and oxygen combine together to form SO3 also. This case can also be worked out in the same
way as above and can be shown to follow the law of reciprocal proportions.
V. Problems.(T.B. Page 61 62)
1.In an experiment 5.0g of CaCO3 on heating gave 2.8 g of CaO and 2.2 g of CO2. Show that these results
are in accordance to the law of conservation of mass.
Solution:
CaCO3 CaO + CO2
Weight of CaCO3 = 5.0 gms
Weight of CaO = 2.8 gms
Weight of CO2 = 2.2 gms
Total weight of reactant = Total weight of products.
5.0 = 2.8 + 2.2
5.0 = 5.0
Since, the mass of the reactants are equal to the mass of the product, these results are in accordance to the laws
of conservation of mass.
2.In an experiment 48 gms of magnesium combines with 32 gms of oxygen to form 80 gms of magnesium
oxide. Show that this reaction illustrates the Law of Conservation of Mass.
[Hint: 2Mg + O2 2MgO. Atomic mass of Mg = 24 and O = 16].
Solution:
2 Mg + O2 2MgO
3. 1.375 g of CuO were reduced by H2 and 1.098 g of Cu were obtained. In another experiment, 1.178 g of
Cu were dissolved in nitric acid and the resulting copper nitrate was converted into CuO by ignition.
The weight of CuO formed was 1.476 g. Show that these results prove the law of constant proportion.
Solution:
Experiment 1:
Weight of CuO = 1.375 g
Weight of Cu = 1.098 g
Weight of oxygen = (1.375 1.098) g
= 0.277 g
Ratio of copper oxygen = 1.098 : 0.277
= 3.96 : 1
Experiment 2:
Weight of CuO = 1.476
Weight of Cu = 1.178
Weight of oxygen = 1.476 1.178
= 0.298
Ratio of copper: oxygen = 1.178 : 0.298
= 3.96 : 1
In both experiments the ratio of Copper: oxygen is some (3.96: 1). Hence it illustrates the law of definite
proportions.
4.In an experiment 0.2430 gm of magnesium on burning with oxygen yielded 0.4030 gm of magnesium
oxide. In another experiment 0.1820 gm of magnesium on burning with oxygen yielded 0.3020 gm of
magnesium oxide. Show that the data explain the law of definite proportions.
Solution:
Experiment 1:
Weight of Magnesium oxide = 0.4030 gm
Weight of Magnesium = 0.2430 gm
Weight of oxygen = 0.4030 0.2430
= 0.16 gm
Ratio of Magnesium: oxygen = 0.2430: 0.16
= 1.552: 1
Experiment 2:
Weight of Magnesium oxide = 0.3020
Weight of Magnesium = 0.1820
Weight of oxygen = 0.3020 0.1820
= 0.12
Ratio of magnesium: oxygen= 0.1820: 0.12
= 1.552: 1
In both experiments the ratio of magnesium: oxygen is same (1.518:1) Hence it illustrates the law of
definite proportions.
5. In an experiment 34.5 g oxide of a metal was heated so that O2 was liberated and 32.1 g of metal was
obtained. In another experiment 119.5 g of another oxide of the same metal was heated and 103.9 g
metal was obtained and O2 was liberated. Calculate the mass of O2 liberated in each experiment. Show
that the data explain the law of multiple proportions.
Solution: Experiment 1
Weight of the metal oxide
Weight of the metal
32.1 g metal combines with 2.4 g oxygen.
2.4
1 g of the metal combine with -----32.1
Experiment 2
Weight of the oxide taken
=
=
34.5 g
32.1 g
0.075 g
119.5 g
Weight of oxide
Weight of copper
Weight of oxygen
0.716 g
0.630 g
0.086 g
0.398 g
0.318 g
0.08 g
Thus 0.630, 0.318 gram of copper combines with definite weight of oxygen
0.080: 0.080
=
1:1
The proportion by weight of chlorine is indicated by simple ratio. Thus law of multiple proportions is obeyed.
7. One gram of hydrogen combines with 15.88 g of sulphur. One gram of hydrogen combines with 7.92 g
of oxygen, one gram of sulphur combines with 0.998 g of oxygen. Show that these data illustrate the law
of reciprocal Proportions.
Solution: In hydrogen Sulphur and Hydrogen Oxygen combinations
The weight of Hydrogen = 1.0 grams
Weight of sulphur = 15.88 grams
Weight of oxygen = 7.92 grams
In Hydrogen oxygen combinations the ratio masses of H and O is 1:8 .. (1)
In Hydrogen sulphur combinations the ratio masses of H and S is 1:16 ..... (2)
So the ratio (1) and (2) are related to each other as
1
1
---:
---- (or)
(3)
8
6
They are whole no multiple of each other.
In oxygen sulphur combinations.
Weight of sulphur = 1 gram
Weight of oxygen = 0.998 gram
The ratio masses of H and S is 1:1 ..... (4)
(1), (2) and (3) are simple multiples of each other therefore, the law of reciprocal proportions holds good.
8.
A compound contains carbon and chlorine. The percentage of chlorine in the compound is 92.21. In
another compound which contains carbon and sulphur, the percentage of sulphur is 84.21. In a third
compound which contains sulphur and chlorine, the percentage of chlorine is 52.59. Show that these
data illustrate the law of reciprocal proportions.
Solution: In first compound (is carbon and chlorine)
The weight of chlorine = 92.21 gram
The weight of oxygen = (100 92.21)
= 17.79 gram
CO2
1 volume
20 cm3
2H2O
2 volumes
40 cm3
10. 100cm3 of propane was burnt in excess oxygen to form carbon dioxide and water. Calculate: (i) the
volume of oxygen used up, and (ii) the volume of carbon dioxide formed. [Hint: C3H8(g) + 5O2(g)
3CO2(g) + 4H2O].
Solution:
C3H8(g)
+
5O2(g)
1 volume
5 volumes
100 cm3
500 cm3
Volume of O2 used
Volume of CO2 formed
3CO2(g)
3 volume
300 cm3
4H2O
4 volumes
400 cm3
= 500cm3
= 300 cm3
=
=
0.16 g oxygen.
0.16g oxygen.
0.16x1
1.0g mercury will combine with = -------=
0.08 g oxygen.
2
In first case oxide was produced, in second case oxide was decomposed. Both show that 1.0g
mercury has combined with 0.08 g oxygen. That means, proportion of mercury to oxygen is 1: 0.08,
in both the cases. This supports the law of definite proportions.
Problem 2: 1.19g of Zinc was converted into zinc oxide and 1.51 gm of zinc oxide was obtained. In
another experiment 1.812 gm of zinc oxide when heated with carbon gave 1.428 gm of zinc show that
these results illustrate the law of definite proportions.
Solution:
Experiment 1:
Wt. of Zinc oxide
Wt. of Zinc
Wt of oxygen
Experiment 2:
= 1.51 gm
Wt of zinc oxide
= 1.812 gm
= 1.19 gm
Wt of Zinc
= 1.428 gm
= 1.51 - 1.19
Wt of oxygen
= 1.812 - 1.428
= 0.32 gm.
= 0.384 gm
Ratio of Zinc: oxygen = 1.19: 0.32
Ratio of Zinc: oxygen = 1.428: 0.384
= 3.72: 1
= 3.72: 1
In both the experiments the ratio of Zinc: oxygen is same (3.72:1). Hence it illustrates the law
of definite proportions.
Example: 1.(P.55) Iron forms two different chlorides, namely ferrous and ferric chloride. Each of
these chlorides was prepared from 2.0 g iron. It was found that 4.538 g ferrous and 5.804g ferric
chloride were produced. Show that these observations are according to the law of multiple proportions.
Solution: Here iron is forming different chlorides. The weight of iron taken in both cases is the same
i.e. 2.0g. Therefore we have,
Ferrous chloride (A)
Ferric chloride ( B)
4.538 g chloride
5.804 g
chloride
-2.000 g iron
- 2.000 g
iron
--------------------2.538 g chlorine
3.804 g
chlorine.
Thus a definite weight of iron i.e.2.0g, combines with 2.538 g and 3.804 g chlorine. The
proportion of chlorine in these compounds is
Ferrous : Ferric
2.538 : 3.804 = 1:1.5 = 2:3
The proportion by weight of chlorine is indicated by a simple ratio, Thus law of multiple
proportions is obeyed.
Example 2: Lead forms three oxides A, B and C. The quantity of oxygen in each of the oxides. A, B
and C is 7.143 %, 10.345% and 13.133% respectively. Show that the law of multiple proportions is
obeyed.
Solution: As the % of oxygen is given, we can find the % of lead.
A
B
Oxide
100.000
100.000
-Oxygen
-7.143
-10.345
----------------------------------Lead
92.857
89.655
C
100.000
-13.133
------------86.867
As lead is forming different oxides, let us take a fixed wt. of lead say 10 g, and find out the
weights of oxygen combining with 10g lead in three oxides.
In A, 92.857 g lead combines with
7.143 g oxygen.
7.143
10
--------- x ----- = 0.769 g oxygen.
92.857
1
10.345
10
--------- x ----- = 1.054 g oxygen
89.655
1
13.133
10
---------- x ----- = 1.538 g oxygen
86.867
1
The weights of oxygen combining with fixed wt. of lead (10.0g) are in the proportions,
0.769: 1.154: 1.538 = 1: 1.5: 2
i.e. 2: 3: 4
The proportion by weight of oxygen is given by a simple numerical ratio. Thus law of multiple
proportions is obeyed.
Problem 1(p 58) : Hydrogen sulphide (H2S) contains 94.11% sulphur, water (H2O) contains 11.11%
hydrogen and sulphur dioxide (SO2) contains 50% oxygen. Show that the results are in
agreement with the law of reciprocal proportions.
Solution: In water, the weight of hydrogen = 11.11 g
The weight of oxygen = 100 -11.11 = 88.89 g
In sulphur dioxide, the weight of sulphur = 50g
The weight of oxygen = 100 -50
= 50g
50
The weight of sulphur that combines with 88.89 g of oxygen = ----- x 88.89 = 88.89 g
50
The ratio between the weights of sulphur and hydrogen which combine with a fixed weight of
oxygen (88.89 g) is 88.89: 11.1 or 8:1.
(i)
In hydrogen sulphide, the weight of sulphur = 94.11g
The weight of hydrogen
= 100 -94.11 = 5.89g
The ratio between weights of sulphur and hydrogen is 94.11 : 5.89 or 16:1
(ii)
8
16
The two ratios (i) and (ii) are related as ------: ------ or 1 :2
1
1
Which are simples multiples of each other. Therefore, the law of reciprocal proportions holds good.
Problem 1(p 59): Methane burns in oxygen to form carbon dioxide and water vapour as given by the
equation
CH4 + 2O2
CO2 + 2H2O
Calculate: i) the volume of oxygen needed to burn completely 50 cm3 of methane and (ii) the volume
of carbon dioxide formed.
Solution:
CH4
+ 2O2
CO2 + 2H2O
1 vol
2vols
1vol.
2vols.
1 x 50 cm3 2x50 cm3
1 x50cm3 2 x 50 cm3
50cm3
100cm3
50cm3
100cm3
Volume of oxygen used = 50 cm3
Volume of carbon dioxide formed = 50 cm3
(d) A sample of air on heating does not show any change in mass but volume increases.
22. Which of the following set illustrated law of reciprocal proportions?
(a) PCl3, HCl, HBr
(b) PH3, P2O3, P2O5
(c) PH3, P2O3, H2S
(d) PH3, P2O5, H2O.
23. Which of the following illustrates the law of conservation of mass?
(a) Mixing of 10g of sulphur and 2g of sand does not show a change in mass
(b) The mass of platinum wire before and after heating remains constant
(c) 2.2g of propane and 8g of oxygen produces 10.2g of gaseous mixture
(d) 2.8g of CO and 1.6g of oxygen gave only 2.24L of CO2 at S.T.P.
24. Which of the following is the best example of law of conservation of mass:
(a) 12g of carbon combines with 32g of oxygen to form 44g of carbon dioxide.
(b) 12g of carbon is heated in vacuum, there is no change in mass.
(c) The mass of a piece of platinum is the same before and after heating.
(d) A sample of air increases in volume when heated at constant pressure but the mass remains
unchanged.
25. In compound A, 1-0g N2 unites with 0.57g O2. In compound B,2g N2 combines with 2.25g O2. In
compound C, 3.0g N2 combines with 5.11g O2. These results obey the law of:
(a) Law of constant of mass
(b) Law of multiple proportions
(c) Law of reciprocal proportions
(d) Daltons law of partial pressure.
26. The formation of CO and CO2 illustrate the law of:
(a) Conservation of mass
(b) Constant proportion
(c) Multiple proportion
(d) Reciprocal proportion
27. In the following reaction:
H2 + Cl2
2HCl the ratio of volumes of H2, Cl2 and HCl gas is 1:1:2. These figures illustrate the
law of:
(a) Constant proportion
(b) Multiple proportion
(c) Reciprocal proportion
(d) Gay Lussacs law of gaseous volume.
28. H2S contains 5.88% hydrogen, H2O contains 11.11% hydrogen while SO2 contains 5o% s. These figures
illustrate the:
(a) Law of conservation of mass
(b) Law of constant proportion
(c) Law of multiple proportion
(d) Law of reciprocal proportion
29. An element X forms two oxides containing 53.33% and 36.36 % of oxygen respectively. These data
illustrate the law of:
(a) Conservation of mass
(b) Constant proportions
(c) Multiple proportions
(d) Reciprocal proportions
30. CO2 gas was prepared by (i) strongly heating NaHCO3. (ii) burning charcoal in air and (iii) the action of
CaCO3 and dil. HCl. It was found that in each case carbon and oxygen combined in the ratio of 3:8. These
data illustrate the law of:
(a) Conservation of mass
(b) Constant proportions
(a) conservation of mass (b) constant proportion (c) multiple proportion (d) reciprocal proportion
58. 1.0g of an oxide of A contained 0.5g of A, 4.0g of another oxide of A, contained 1.6g of A. These
figures illustrate the ..
(a) law of reciprocal proportion
(b) law of multiple proportion
(c) law of conservation of mass
(d) law of constant proportion
59. The percentage of H in H2O2 and H2O is 5.93 and 11.2 respectively. These figures illustrate the law
of .
(a) Conservation of mass (b) Constant proportion (c) Multiple proportion (d) Reciprocal proportion
60. In hydrogen chloride H = 2.77% Cl = 97.23% In phosphine P = 91.18% and H = 8.82%
In phosphorous pentachloride P = 22.57% Cl = 77.43% these figures illustrate the law of ..
(a) reciprocal proportion (b) multiple proportion (c) constant proportion (d) conservation of mass
61. A sample of CaCO3 has Ca = 40%, C = 12% and O = 48%. If the law of constant proportions is
true then the mass of Ca in 4g of a sample of CaCO3 from another source will be ..
(a) 0.016 g
(b) 0.16g
(c) 1.6 g
(d) 16 g
62. The composition of compound A is 40% X and 60% Y. The composition of compound B is 25%.
X and 75% Y. According to the law of multiple proportion the ratio of the mass of element Y in
elements A and B is .
(a) 4:5
(b) 2:1
(c) 2:3
(d) 3:4
63. Zinc sulphate contains 22.65% Zn and 43.9% H2O. If the law of constant proportion is true, then
the mass of Zn required to give 20g of the crystals will be ..
(a) 0.453 g
(b) 4.53 g
(c) 45.3 g
(d) 453 g
64. In the following reaction: H2 + Cl2
2 HCl the ratio of volumes of H2, Cl2, HCl gas is 1:1:2
these figures illustrate the law of ..
(a) constant proportion
(b) multiple proportion
(c) reciprocal proportion
(d) Gay Lusacs law of gaseous volume
__________________________________________________________________________________
Answers:
1. (d)
2. (c)
3. (c)
4. (d)
5. (a)
6. (b)
7. (b)
8. (b)
9. (d)
14. (b) 15. (b) 16. (b) 17. (a) 18. (d) 19. (d) 20. (b) 21. (b) 22. (d) 23. (c) 24. (a) 25. (b) 26. (c)
27. (d) 28. (d) 29. (c) 30. (b) 31. (b) 32. (a) 33. (c) 34. (a) 35. (d) 36. (c) 37. (d) 38. (a) 39. (a)
40. (c) 41. (a) 42. (a) 43. (c) 44. (a) 45. (c) 46. (b) 47. (d) 48. (c) 49. (a) 50. (b) 51. (a) 52. (b) 53. (d) 54. (a)
55. (a) 56. (a) 57. (b) 58.(b) 59. (c) 60. (a) 61. (c) 62. (a) 63. (a) 64. (d)
__________________________________________________________________________________
II. Answer the following in one or two sentences:
1.List the various laws of chemical combinations.
1. Law of conservation of mass stated by Lomonossoff
2. Law of definite proportions stated by J.C. Proust
3. Law of multiple proportions stated by Dalton
4. Law of Reciprocal proportions stated by Beizelius
5. Gay Lussacs Law of combining volumes stated by Gay Lussacs
2.What is stoichiometry?
Stoichiometry a branch of chemistry in which quantitative relationship between masses of reactants and
products are established. The study of these laws led to the development of a theory concerning the nature of
matter.
3. State law of reciprocal proportions.
Law: 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.
=
=
3g
2g
=
=
4.8g
?
According to the law of conservation of mass, the mass of magnesium oxide formed should
be equal to the total mass of magnesium and oxygen.
3 g of Mg gives
=
5 g of MgO
5
4.8 g of Mg will give
=
--- x 4.8 of MgO
3
=
8 g of MgO
Weight of magnesium oxide
=
8g
2. 48 gms of magnesium combines with 32 gms of oxygen to form 80 gms of magnesium oxide.
Show that his reaction illustrates the Law of Conservation of Mass.
=
48 gms + 32 gms
=
80 gms
Mass of product (2MgO)
=
80 gms
Total mass of reactants
=
Total mass of products.
Thus the reaction illustrates the Law of Conservation of Mass.
3. 2Cu + S
Cu2S. Show that this reaction illustrates the Law of conservation of Mass.
(Cu = 64, S = 32).
Solution
2Cu
+
S
Cu2S
(2 x 64)
+
32
160
128
+
32
160
128 gms of copper + 32 gms of Sulphur
160 gms of cuprous sulphide.
i.e., 160 gms of reactants
160 gms of products.
0.207 g
--------Weight of chlorine
=
0.071 g
--------The lead and chlorine are present in the ratio 0.207: 0.071 or 3:1.
Since in both the experiments, it is found that in lead chloride, lead and chlorine are present in
the ratio 3:1. This proves the law of constant composition.
4. In a typical experiment, 28 gms. of iron on heating with oxygen gave 40 gms. of iron (III)
oxide. 7 gms. of iron on heating with oxygen gave 10 gms. of iron (III) oxide. Show that these
results correspond to the law of definite proportions.
Solution: Case (i)
Weight of iron (III) oxide
Weight of iron
Weight of oxygen
Weight of iron: Weight of oxygen
Case (ii)
Weight of iron (III) oxide
Weight of iron
Weight of oxygen
Weight of iron: Weight of oxygen
=
=
=
=
=
40 gms.
28 gms.
12 gms.
28 : 12
7:3
=
=
=
=
10 gms.
7 gms.
3 gms.
7:3
In both the cases, the ratio of weight of iron: weight of oxygen remains the same as 7:3. Thus
these results correspond to the law of definite proportions.
5. In an experiment, 9 gms. of magnesium gave 15 gms. of magnesium oxide. 4.5 gms. of
magnesium oxide was obtained from 2.7 gms. of magnesium. Show that these informations
illustrate the law of definite proportions.
Solution: Case (i)
Weight of MgO
Weight of Mg
Weight of O
Weight of Mg: Weight of O
=
=
=
=
15 gms.
9 gms.
6 gms.
9 : 6 = 3: 2
Case (ii)
Weight of MgO
Weight of Mg
Weight of O
Weight of Mg: Weight of O
=
=
=
=
4.5 gms.
2.7 gms.
1.8 gms.
2.7: 1.8 = 3: 2
In both the cases, the ratio of weight of magnesium and weight of the oxygen mains the same
as 3:2. Hence it illustrates the law of definite proportions.
6. 11.7gms. of a sample of sodium chloride was found to contain 4.6 gms. of sodium and 7.1 gms.
of chlorine in it. 2.93 gms. of an another sample of sodium chloride was found to contain 1.15
gms. of sodium and 1.78 gms. of chlorine. Show that these figures illustrate the law of definite
proportion.
Both the samples contain the same elements sodium and chlorine. If they were to illustrate the
law of constant proportion, the percentage of sodium and chlorine in the above two samples should
remain the same.
In the first sample, 11.7 gms. of the sample has 4.6 gms. of sodium.
Weight of sodium
Percentage of sodium
=
---------------------- x 100
Weight of sample
4.6
=
---- x 100
= 39.31%
11.7
In the second sample, 2.93 gms. of the sample has 1.15 gms. of sodium.
Weight of sodium
Percentage of sodium
=
---------------------- x 100
Weight of sample
1.15
=
---- x 100
= 39.31%
2.93
Since the two samples contain the same elements combines together in the same proportion by
weight, the law of definite proportion is proved.
7. In an experiment 10 gms of calcium combines with 4 gms of oxygen to form the oxide of
calcium. In another experiment 15 gms of calcium combines with 6 gms of oxygen to form the
oxide of calcium. Show that this data illustrates the Law of Definite Proportions.
Solution
Expt I
Expt II
Wt. of Calcium
10 gms
15 gms
Wt. of Oxygen
4 gms
6 gms
Wt. of Calcium: Wt. of Oxygen
10: 4 :5:2
15:6 : 5:2
In both cases, the ratio of the weight of calcium to he weight of oxygen is a constant. i.e., 5:2.
Thus these results illustrate the Law of Definite Proportions.
8. 0.2432 gm of magnesium when burnt in air yielded 0.4032 gm of magnesium oxide while
0.1824 gm of magnesium gave 0.3024 gm of magnesium oxide. Show that these results are in
accordance with the Law of Definite Proportions.
Solution
Expt I
0.4032 gms
0.2432 gms
0.1600 gms
0.2432: 0.1600
1: 52 :1
Expt II
0.3024 gms
0.1824 gms
0.1200 gms
0.1824: 0.1200
1:52 : 1
Thus the ratio of the weight of magnesium to the weight of oxygen is found to be
the same in both cases. This is in accordance with the Law of Definite Proportions.
9. In one experiment, 0.5 g of copper was converted into cupric oxide and its weight is found to
be 0.6248 g. In a second experiment, 1.25 g of cupric oxide is obtained from 1.00 g of copper
by the same method. Show that these results illustrate the law of definite proportions.
Expt I
0.6248
0.5000
0.1248
0.5000: 0.1248
4:1
Expt II
1.25
1.00
0.25
1.00: 0.25
4:1
These results illustrate the law of definite proportions i.e., a compound by whatever method it
is prepared always contains the same elements in the same fixed proportions by weight.
10. In two experiments, 0.259 g and 0.207 g of lead were converted to lead chloride. 0.347 g and
0.278 g of lead chlorine are formed. Show that this data illustrate the law of constant
proportions.
Solution. In the first experiment,
Weight of lead chloride
=
0.347 g
Weight of lead
=
0.259 g
Weight of chlorine
=
0.347 0.259
=
0.088 g
The ratio of lead and chlorine
=
0.259: 0.088
=
3:1
In the second experiment,
Weight of lead chloride
=
0.278 g
Weight of lead
Weight of chlorine
=
0.207 g
=
0.278 0.207
=
0.071 g
The ratio of lead and chlorine
=
0.207: 0.071
=
3:1
Since in both experiments, it is found that in lead chlorine, the lead and chlorine are present in
the ratio 3:1. This proves the law of constant composition.
11. 1.375 g of cupric oxide on reduction in hydrogen gas gives 1.098 g of copper. In another
experiment, 1.179 g of metallic copper produced 1.476 g of copper oxide. Show that these
results illustrate the law of constant (or definite) proportions.
Solution.
Experiment 1:Mass of copper oxide taken
Mass of copper obtained
=
=
1.375 g
1.098 g
1.098 x 100
Therefore
Mass % of copper
=
-------------- =
79.86
1.375
Experiment 2:Mass of copper oxide produced
=
1.476 g
Mass of copper used
=
1.179 g
1.179 x 100
Therefore
Mass % of copper
=
-------------- =
79.89
1.476
Since, the percentage of copper in the two samples of copper oxide is the same, hence the law
of definite proportion is verified.
12. 1.19 g of zinc was converted into zinc oxide and 1.51 gm of zinc oxide was obtained. In
another experiment 1.812 gm of zinc oxide when heated with carbon gave 1.428 gm of zinc.
Show how these results illustrate the law of definite proportions.
Expt I
1.51 gm
1.19 gm
0.32 gm
1.19: 0.32 = 3.72: 1
Expt II
1.812 gm
1.428 gm
0.384 gm
1.428: 0.384 = 3.72: 1
Thus the ratio of weight of zinc to the weight of oxygen is found to be the same in
both the cases. This illustrates the law of definite proportions.
13. In one experiment 1.098 g of copper is obtained by the reduction of 1.375 g of cupric oxide. In
another experiment 1.476 g of cupric oxide was prepared from 1.179 g of copper through cupric
nitrate. Show that the results of the two experiments illustrate the law of definite proportions?
Experiment: 1
Weight of cupric oxide
=
1.375 g
Weight of copper
=
1.098 g
--------Weight of oxygen
=
0.277 g
--------Ratio in the weights of copper and oxygen
=
1.098: 0.277 = 4:1
Experiment: 2
Weight of the cupric oxide
=
1.476 g
Weight of the copper
=
1.179 g
--------Weight of oxygen
=
0.297 g
--------Ratio in the weights of copper and oxygen
=
1.179: 0.297
=
4:1
In both cases, it is found that copper and oxygen are in the ratio 4:1,
Hence, the law of definite proportion is proved.
14. Illustrate the law of definite proportions from the following data:
(i) 0.16 g of sulphur produces 0.32 g of sulphur dioxide, (ii) sulphur dioxide obtain by the
decomposition of sodium sulphate contains 50% sulphur.
Case 1
(i) Weight of sulphur dioxide
=
0.32 g
Weight of sulphur
=
0.16 g
-------Weight of oxygen
=
0.16 g
--------Ratio of sulphur to oxygen in sulphur dioxide produced by (i)
Method
=
0.16: 0.16 or 1:1
Case 2
(ii) Weight of sulphur dioxide
=
100 g
Weight of sulphur
=
50 g
--------Weight of oxygen
=
50 g
---------Ratio of sulphur to oxygen in sulphur dioxide produced by (ii)
Method
=
50:50 or 1:1
Since the sulphur dioxide produced by different methods contains sulphur and oxygen in the
ratio 1:1, the data proves the law of definite proportions.
15. In two experiments, 0.259 g and 0.207 g of lead were converted to lead chloride, yielding
0.347 g and 0.278 g of lead chlorided respectively. Show that the data illustrates the law of
constant composition.
Experiment: 1
Weight of lead chloride
=
0.347 g
Weight of lead
=
0.259 g
--------Weight of chlorine
=
0.088 g
--------In this experiment, the lead and chlorine are present in the ratio 0.259: 0.088 or 3:1.
In the second experiment
Experiment: 2
Weight of the lead chloride
=
0.278 g
Weight of the lead
=
0.207 g
--------Weight of chlorine
=
0.071 g
--------The lead and chlorine are present in the ratio 0.207: 0.071 or 3:1.
Since in both the experiments, it is found that in lead chloride, lead and chlorine are present in
the ratio 3:1. This proves the law of constant composition.
Thus, the ratio of the metal to oxygen in the first oxide is 1: 0.25.
Case 2
=
=
100.0 g
11.1 g
----------Weight of the metal
=
88.9 g
-----------88.9 g of metal combines with
=
11.1 g of oxygen
11.1
1 g of metal combines with
=
------- g of oxygen
88.9
=
0.12 g of oxygen (2)
Thus, the ratio of the metal to oxygen in the second oxide is 1: 0.12.
From (1) & (2) the different weights of oxygen that combines with fixed weigh of the metal (i.e., 1.0
g) are in the ratio 0.25: 0.12 or 2:1. This is a simple ratio and hence the law of multiple proportions is
proved.
2. A metal forms three oxides containing respectively 76.47%, 68.42% and 52.2% of the metal.
Show that these data are in accordance with law of multiple proportions.
Case 1
Weight of metallic oxide
=
100.00 g
Weight of the metal
=
76.47 g
----------Weight of oxygen
=
23.53 g
----------23.53 g of oxygen combines with 76.47 g metal
76.47
1 g of oxygen will combine with -------23.53
Case 2
Weight of the metallic oxide
Weight of the metal
Weight of oxygen
=
=
=
100.00 g
68.42 g
---------31.58 g
-----------
=
=
Weight of oxygen
100.0 g
52.2 g
--------47.8 g
---------
Case 3
Mass % of O
=
57.1
Thus,
42.9 g of C reacts with 57.1 g of oxygen
Thus,
Mass % of C
=
27.3
Mass % of O
=
72.7
27.3 g of C reacts with 72.7 g of oxygen
1 g of C reacts with 72.7 / 27.3 g of oxygen = 2.66 g of oxygen
Wt. of oxygen
Wt. of metal
Wt. of Oxygen in combination
with 1 gm of metal
Expt I
7.41 gms
92.59 gms
7.41 / 92.59 gms
0.08 gm
Expt II
3.85 gms
96.15 gms
3.85 / 96.15 gms
0.04 gm
Thus the different weights of oxygen that combine with the same weight of the metal, namely 1
gm, are 0.08 gm and 0.04 gm. The ratio of wts. of Oxygen = 0.08: 0.04
=
2:1
which is a simple integral ratio. Thus the Law of Multiple Proportions is illustrated.
6. 1.90 gm of one oxide of copper gave 1.52 gm of copper on reduction. 2.85 gm of another oxide
gave 2.53 gm of copper on reduction. Show that these results are in accordance with the
Law of Multiple Proportions.
Solution
Expt I
Expt II
1.90 gm
1.52 gm
0.38 gm
0.38 / 1.52 = 0.25 gm
2.85 gm
2.53 gm
0.32 gm
0.32 / 2.53 = 0.13 gms
=
0.25: 0.13
=
2:1 (appr). which is a simple integral ratio
Thus the given results are in accordance with the Law of Multiple Proportions.
7. 3 gms of carbon form two types of oxides with weights as 7 and 11 gms respectively. Show that
these illustrate law of multiple proportions:
Solution:
1st oxide
Weight of oxide
Weight of carbon
Weight of oxygen
=
=
=
7 gms.
3 gms.
4 gms.
2nd oxide
Weight of oxide
=
11 gms.
Weight of carbon
=
3 gms.
Weight of oxygen
=
8 gms.
The weight of oxygen that combines with a fixed weight of carbon, namely 3 gms of carbon are
4 gms. and 8 gms. respectively.
The ratio of the weight of oxygen
=
4:8
=
1:2
The ratio is a simple integral ratio. Thus the given data illustrates the law of multiple
proportions.
8. In a typical experiment, 4 gms. of hydrogen are found to form 36 gms. of water. In an another
experiment 2 gms. of hydrogen are found to form 34 gms. of hydrogen peroxide. Show that
these illustrate the law of multiple proportions.
Solution:
Case 1:
Weight of water
=
36 gms.
Weight of hydrogen
=
4 gms.
-----------------------------------------------------Weight of oxygen
=
32 gms.
-----------------------------------------------------4 gms. of hydrogen combine with
=
32 gms. of oxygen.
32 x 1
1 gm. of hydrogen combines with
=
------= 8 gms.
4
Case 2:
Weight of hydrogen peroxide
=
34 gms.
Weight of hydrogen
=
2 gms.
--------------------------------------------------------------Weight of oxygen
=
32 gms.
--------------------------------------------------------------2 gms. of hydrogen combine with
=
32 gms. of oxygen.
32 x 1
1 gm. of hydrogen combines with
=
------- = 16 gms. of oxygen.
2
The different weights of oxygen that combine with the same weight of hydrogen, namely 1
gm. of hydrogen are 8 and 16 gms.
The ratio of oxygen weights
=
8 : 16
=
1 : 2, a simple integral ratio. Thus the data
illustrates the law of multiple proportions.
9. In an experiment 2.8 gm of nitrogen gave 6 gm of its oxide. In another experiment 2.1 gm of
nitrogen gave 6.9 gm of another oxide. Show how these results illustrate the law of multiple
proportions.
Expt I
6.0 gm
Expt II
6.9 gm
Wt. of nitrogen
Wt. of oxygen
Wt. of oxygen combining with 1
gm of nitrogen
2.8 gm
3.2 gm
3.2 / 2.8 = 1.142 gm
2.1 gm
4.8 gm
4.8 / 2.1 = 2.285 gms
Thus the different weight of oxygen that combines with the same weight of nitrogen
(1 g), are 1.142 g and 2.285 g. i.e., the ratio of weights of oxygen = 1:2.which is a simple integral
ratio. Thus the law of multiple proportions is illustrated.
10. Two oxides of metal contain 20% and 11.1% oxygen respectively. Show how these data
illustrate law of multiple proportions
Case 1
Weight of the first oxide
=
100 g
Weight of oxygen
=
20 g
---------Weight of the metal
=
80 g
---------80 g of metal combines with 20 g of oxygen
20
1 g of metal combines with ---- x 1 of oxyge
80
=
Thus, the ratio of the metal to oxygen in the first oxide is 1: 0.25.
Case 2
Weight of the second oxide
=
100.0 g
Weight of oxygen
=
11.1 g
----------Weight of the metal
=
88.9 g
-----------88.9 g of metal combines with
=
11.1 g of oxygen
11.1
1 g of metal combines with
=
------- g of oxygen
88.9
=
0.12 g of oxygen (2)
Thus, the ratio of the metal to oxygen in the second oxide is 1: 0.12.
From (1) & (2) the different weights of oxygen that combines with fixed weigh of the metal (i.e., 1.0
g) are in the ratio 0.25: 0.12 or 2:1. This is a simple ratio and hence the law of multiple proportions is
proved.
11. Copper combines with oxygen to form two oxides, which have the following composition:
(i)
0.716 g of cuprous oxides contains 0.630 g of copper.
(ii)
0.398 g of cupric oxide contains 3.318 g of copper.
Prove that the above data illustrates the law of multiple proportions.
Solution: (i) In the first experiment,
Weight of cuprous oxide
=
0.716 g
Weight of copper
=
0.630 g
Weight of oxygen
=
0.716 0.630
=
0.086 g
(ii) In the second experiment,
Weight of cupric oxide
=
0.398 g
Weight of copper
=
0.318 g
Weight of oxygen
=
0.398 0.318
=
0.08 g
Here copper is forming two oxides. The weight of oxygen in both cases is the same i.e., 0.08 g.
Thus, a definite weight of oxygen combines with 0.630 g and 0.318 g of copper. The proportion of
weight of copper in these compounds
Cuprous oxide
:
cupric oxide
0.630 g
:
0.318 g
2
:
1
The proportion by weight of copper is indicated by a simple ratio. Thus, the law of multiple
proportions is obeyed.
12. A metal forms three oxides containing respectively 76.47%, 68.42% and 52.2% of the metal.
Show that these data are in accordance with law of multiple proportions.
Case 1
Weight of metallic oxide
=
100.00 g
Weight of the metal
=
76.47 g
----------Weight of oxygen
=
23.53 g
----------23.53 g of oxygen combines with 76.47 g metal
76.47
1 g of oxygen will combine with -------23.53
Case 2
Weight of the metallic oxide
Weight of the metal
Weight of oxygen
=
=
=
100.00 g
68.42 g
---------31.58 g
-----------
=
=
Weight of oxygen
100.0 g
52.2 g
--------47.8 g
---------
Case 3
In all these cases, different weights of the metal combine with the fixed weight of oxygen (i.e.,
1g). According to the law of multiple proportions, they are in a simple integral ratio. i.e., 3:2:1. Hence,
these data are in accordance with the law of multiple proportions.
IV. LAW OF RECIPROCAL PROPORTIONS:
1. Water and sulphur dioxide contains 88.9% and 50% oxygen respectively. Hydrogen sulphide
contains 91.1% of sulphur. Illustrate the law of reciprocal proportions from these data?
Case 1
Weight of water
= 100g
Weight of oxygen
= 88.9 g
--------Weight of hydrogen
= 11.1 g
--------11.1 g of hydrogen combine with 88.9 g oxygen.
Let the weight of hydrogen be fixed as one gram.
88.9
Hence, 1.0 g hydrogen will combine with ------- = 8.0 g of oxygen.
11.1
In water hydrogen and oxygen are present in the ratio 1: 8. (1)
Case 2
Similarly in hydrogen sulphide
Weight of hydrogen sulphide
=
100.0g
Weight of sulphur
=
91.1g
---------Weight of hydrogen
=
8.9g
----------8.9 g of hydrogen combines with 91.1 g of sulphur. 1 g of hydrogen will combine with
91.1
------ = 10.2 g of sulphur.
8.9
In hydrogen sulphide, hydrogen and sulphur are present in the ratio 1: 10.2. (2)
The ratio of sulphur to oxygen is 1: 1 in sulphur dioxide. From (1) & (2) , it can be shown that
sulphur and oxygen must be present in the ratio 10.2: 8 or 1: 0.78 or 1:1, Hence, the law of reciprocal
proportions is verified.
2. Hydrogen sulphide contains 94.11% Sulphur. Sulphur dioxide contains 50% oxygen. Water
contains 11.11% hydrogen. Show that the results are in agreement with the law of reciprocal
proportions.
H
H2S
H2O
S
SO2
O
Solution. (a) Calculation of masses of Sulphur and oxygen that combine with certain fixed mass, say 1
g of hydrogen.
(i) In H2S, 100 94.11 = 5.89 g hydrogen combines with 94.11 g Sulphur. So,
94.11
1 g hydrogen combines with ------- g
=
15.98 g of Sulphur
5.89
(ii) In water, 11.11 g hydrogen combines with 100 11.11 = 88.89 g oxygen. So,
88.89
1 g hydrogen combines with ------- g
=
8 g of oxygen
11.11
(b) Calculation of ratio of the masses of Sulphur (in H2S) and oxygen (in H2O)
Mass of Sulphur
15.98
------------------=
------- = 2
Mass of oxygen
8
(c) Calculation of the ratio of masses of Sulphur and oxygen when they combine to form Sulphur
dioxide.
Mass of Sulphur
50
------------------=
------- = 1
Mass of oxygen
50
The ratio (i) is double of (ii), i.e., 2:1, which is a simple ratio. This illustrates the law of
reciprocal proportions.
3. Phosphorus trioxide contains 56.4% of phosphorus and 43.6% of oxygen. Water contains
88.8% of oxygen and 11.2% of hydrogen. Phosphine contains 91.1% of phosphorus and 8.9%
of hydrogen. Show how these results illustrate the Law of Reciprocal Proportion.
P2O3
H2O
PH3
Phosphorus
56.4%
91.1%
Oxygen
43.6%
88.8%
-
Hydrogen
11.2%
8.9%
First the ratio in the weight of phosphorus and hydrogen which combine with a fixed weight of
oxygen is calculated.
45.6 g of oxygen combines with 56.4 g of phosphorus
1 g of oxygen combines with 56.4 / 43.6 g of phosphorus
88.8 g of oxygen combines with 11.2 g of hydrogen
1 g of oxygen combines with 11.2 / 88.8 g of hydrogen
The ratio of the weights of phosphorus and hydrogen combining with 1 g of oxygen in P2O3
and H2O is
56.4
11.2
Weight of Carbon
75%
42.86%
-
Weight of Hydrogen
25%
11.11%
57.14 g.
75 x 57.14
Weight of oxygen combining with 75 g of carbon =
------------42.86
=
100 g
Weight of hydrogen combining with 75 g of carbon = 25.0 g
Ratio of weights of hydrogen and oxygen, which combine with carbon,
25 / 100
=
0.250
The ratio between the two values (1) & (2)
=
0.125 : 0.250
=
1:2
Hence the law of reciprocal proportions is obeyed.
Weight of Oxygen
57.14%
88.89%
..1
..2
5. Carbon dioxide contains 27.27% carbon, carbon disulphide contains 15.79% carbon and
Sulphur dioxide contains 67% Sulphur. Show that these data are in accordance with the
law of reciprocal proportions.
Case 1
=
=
Weight of oxygen
100.00 g
27.27 g
----------72.73 g
-----------
=
=
100.00 g
15.79 g
----------Weight of Sulphur
=
84.21 g
----------15.79 g of carbon combines with 84.21 g of Sulphur
84.21
1 g of carbon combines with ---------=
5.3 g of Sulphur.
15.79
Ratio of carbon and Sulphur in carbon dioxide
=
1: 5.3
If Sulphur and oxygen were to combine to sulphur dioxide, according to the law of reciprocal
proportions, they must combine in the ratio 5.3: 2.6 (or) 2:1.
Case 3
Weight of sulphurdioxide
Weight of Sulphur
=
=
100 g
67 g
------Weight of oxygen
=
33 g
------In the case of SO2 67 g of Sulphur combines with 33g of oxygen.
The ratio Sulphur to oxygen is 67:33 (or) 2.1:1
Hence, these data are in accordance with law of reciprocal proportions.
6. Potassium chloride contains 52% potassium, potassium iodide contains 23.6% potassium and
iodine chloride contains 78.2% iodine. Show that these illustrate law of reciprocal
proportions.
Case 1
=
=
Weight of chlorine
=
=
Case 2
=
=
Weight of iodine
=
=
100 g
52 g
------48 g
------48 g of chlorine
48 / 52 = 0.92 g of chlorine.
..1
100.0 g
23.6 g
---------76.4 g
---------76.4 g iodine.
76.4 / 23.6
= 3.2 g of iodine
..2
According to the law of reciprocal proportions, if iodine and chlorine were to combine, they
will combine in the ratio 3.2: 0.92 (or) 7:2.
Case 3
=
=
Weight of chlorine
100.0 g
78.2 g
--------21.8 g
---------
.3
O2
H2S
H2O
SO2
In H2S
Weight of H2 = 1 g
Weight of S = 15 g
In H2O
Weight of
H2
1g
1g
-
Weight of
O2
7.92 g
0.998 g
Weight of
S
15 g
1g
Weight of H2 = 1 g
Weight of O2 = 7.92 g
The ratio between S in H2S and O2 in water is 15:7.98 i.e., 2:1
..1
Weight of S = 1 g
Weight of O2 = 0.998 g
The ratio between S and O2 in SO2 is 1:0.998 i.e., 1:1
..2
In SO2
=
=
=
34.5 g
32.1 g
2.4 g
0.075 g
8.9
In hydrogen sulphide, hydrogen and sulphur are present in the ratio 1: 10.2. (2)
The ratio of sulphur to oxygen is 1: 1 in sulphur dioxide. From (1) & (2) , it can be shown that
sulphur and oxygen must be present in the ratio 10.2: 8 or 1: 0.78 or 1:1, Hence, the law of reciprocal
proportions is verified.