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| Laws of Chemical Combination |
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| When substances react, they do so by following certain laws. These laws are called the laws of chemical combination. These formed the basis of Dalton's atomic theory of matter. |
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 Law of conservation of mass |
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| Stated by French chemist A. Lavoisier "During any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants". |
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| Law of conservation is also known as 'Law of indestructibility of matter'. |
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| Example |
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| 10 grams of CaCO3 on heating gave 4.4g of CO2 and 5.6 of CaO. Show that these observations are in agreement with the law of conservation. |
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| Solution: |
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| Mass of the reactants = 10g |
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| Mass of the products = 4.6 + 5.6g = 10g |
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| Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. |
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 Law of constant proportion |
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| Stated by French chemist Joseph Proust “A chemical compound always contains same elements combined together in the same proportion by mass”. |
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| For example, pure water obtained from different sources such as river, well, springs, sea, etc., always contains hydrogen and oxygen together in the ratio of 1: 8 by mass. Similarly, CO2 can be obtained by different methods such as, |
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 burning of carbon |
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 heating of lime stone |
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 by action of dilute HCl on marble pieces |
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| The different samples of CO2 contain carbon and oxygen in the ratio of 3: 8. |
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| Example |
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| 1.375g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098g. In another experiment 1.179g of copper was dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476g. Show that these results illustrate the law of constant proportion. |
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| Solution |
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| First experiment |
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| Copper oxide = 1.375g |
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| Copper left = 1.098g |
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| Oxygen present = 1.375 - 1.098 = 0.277g |
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| Second experiment |
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| Copper taken = 1.179g |
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| Copper oxide formed = 1.476g |
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| Oxygen present = 1.476 - 1.179 = 0.297g |
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| Percentage of oxygen is same in both the above cases so the law of constant composition is illustrated. |
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 Law of multiple proportions |
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| Stated by John Dalton (1803) |
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| "When two elements combine with each other to form two or more than two compounds, the masses of one the element which combine with the fixed mass of the other, bears a simple whole number ratio". |
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| Examples |
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| Carbon monoxide (CO): 12 parts by mass of carbon combines with 16 parts by mass of oxygen. |
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| Carbon dioxide (CO2): 12 parts by mass of carbon combines with 32 parts by mass of oxygen. |
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| Ratio of the masses of oxygen that combines with a fixed mass of carbon (12 parts) 16: 32 or 1: 2 |
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| For example, hydrogen and oxygen are known to form 2 compounds. The hydrogen content in one is 5.93% while in the other it is 11.2%. Show that this data illustrates the law of multiple proportions. |
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| Solution |
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| In the first compound hydrogen = 5.93% |
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| Oxygen = (100 -5.93) = 94.07% |
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| In the second compound |
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| Hydrogen = 11.2% |
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| Oxygen = (100 -11.2) = 88.88% |
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In the first compound the number of parts
of oxygen that combine with one part by mass of hydrogen
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In the second compound the number of
parts by mass of oxygen that combine with one part by mass of hydrogen
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| Ratio of the masses of oxygen that combine with fixed mass of hydrogen 15.86: 7.9 or 2:1 |
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| The ratio illustrates the law of multiple proportion. |
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