The combination of different elements to form compounds is governed by certain basic rules. These rules are referred to as laws of chemical combination.
There are five laws of chemical combinations:
“The mass in an isolated system can neither be created nor be destroyed but can be transformed from one form to another”. Let us look at an example first-
No. of Carbons, Hydrogens and Oxygens are the same on each side of the equation.
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?
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.
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.
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.
"All chemical compounds are found to have constant composition irrespective of their method of preparation or sources."
The elements are always combined in the same proportions by mass.
3:8 7:16
Tip: Isotopes are the atoms possessing the same atomic number but different atomic masses.
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 from the same or different compounds. Also state the law, which will be obeyed by the given samples.
- 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 percent. Use these data to illustrate the law of constant composition.
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:
i.e. oxygen present = 13.38%Step 2:
- To compare the percentage of oxygen in both 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 constant composition law.
"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, bears a simple ratio to one another."
Carbon can form two different compounds with oxygen
In the first oxide:
Similarly, in 2nd oxide:
It means that this law is valid for basic chemical reactions.
The Law of multiple proportions is not valid for heavy and complex molecules
Because when heavy molecules are interlinked with each other then the total whole will not be as small.
Example 1: Two oxide samples of lead were heated in the current of hydrogen and were reduced to 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.
- 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:
- 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:
- Now, r1 : r2 = 13.375 : 6.669 = 2:1 (simple ratio) and hence the data illustrates the law of multiple proportion.
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.
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 = 0.376Step 3:
- To compare the weights of carbon, combined with the same weight of oxygen in both the oxides.
- The ratio of the weights of carbon that combine with the same weight of oxygen (1 part) is 0.751: 0.376 or 2:1
Since this is a simple whole-number ratio, so the above data illustrate the law of multiple proportions.
Example of Gay Lussac's Law
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.
V(hydrocarbon) : Voxygen : V(carbon dioxide) : V(water 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 the complete combustion of 40 ml of acetylene (C2H2) and how much volume of carbon dioxide will be formed? All volumes are measured at NTP.
So, for complete combustion of 40 ml of acetylene, 100 ml of oxygen is required and 80 ml of CO2 is formed.
"Equal volumes of gases at the same temperature and pressure should contain an equal number of molecules."
Avogadro's Law
Examples 1: A tyre that consists of 10 moles air and occupies 40L volume loses half of the volume because of a puncture. What is the amount of air left in the deflated tyre at STP?
Initial amount of air n1 is 10 mol
Initial volume of tyre V1 is 40 L
Final volume of tyre V2 is 20 L
According to Avogadro’s law,
the amount of air left in the deflated tyre,
n2 = = 5 moles
Hence, the deflated tyre contains 5 moles air
Q1. 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?
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: AgNO3 + NaCI → NaNO3 + 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 of 4.35/1.35 = 3 is a small whole number.
172 videos|306 docs|152 tests
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1. What is the Law of Conservation of Mass in chemical combinations? |
2. How does the Law of Definite Proportions apply in chemical combinations? |
3. What is Gay Lussac's Law of Gaseous Volumes in chemical combinations? |
4. How does Avogadro's Law relate to chemical combinations? |
5. Can you explain the Law of Multiple Proportions in chemical combinations? |
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