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Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:‘sample m’ mass : 7g Mass of constituent oxygen : 2g Mass of constituent calcium : 5g ‘sample n’ mass : 1.4g Mass of constituent oxygen : 0.4g Mass of constituent calcium : 1.0g(1) Which law of chemical combination does this prove? Explain. |
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Answer» (i) The expected proportion by weight of the constituent elements of quick lime that is calcium oxide would be from its known molecular formula CaO. The atomic mass of Ca and O are 40 and 16 respectively. This means, the proportion by weight of the constituent elements Ca and O in the compound CaO is 40 :16 which is 5 : 2. (ii) Now, for the given sample’m’ of CaO = 5 g mass of given sample = 7 g mass of constituent Ca in sample’m’ = 5 g mass of constituent O in sample’m’ = 2 g (iii) This means that 7 g of calcium oxide contairis 5 g of calcium (Ca) and 2 g of oxygen (O); apd the proportion by weight of calcium and oxygen in it is 5 : 2. (iv) Now, for the given sample ‘n’ of CaO mass of given sample CaO = 1.4 g Mass of constituent Ca in sample ‘n’ = 1.0 g Mass of constituent O in sample ‘n’ = 0.4 g This means that 1.4g of calcium oxide contains 1.0 g of calcium (Ca) and 0.4 g of oxygen (O); and the proportion by weight of calcium and oxygen in it is 5 : 2. (v) Above samples’m’ and ‘n’ of calcium oxide (CaO) shows that the proportion by weight of the constituent elements in different samples of a compound is always constant that is the proportion by weight of calcium (Ca) and oxygen (O) in different samples of calcium oxide (CaO) is constant. (vi) The experimental value of proportion by weight of the constituent elements matched with the expected proportion calculated by molecular mass. This proves and verifies the law of constant proportion. The law states that ‘The proportion by weight of the constituent elements in the various samples of a compound is fixed’. |
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