Experiment to Verify Law of Conservation of Mass
Take a clean conical flask fitted with a rubber cork, and a small test-tube having a long thread tied to its neck. All these things are weighed together on a sensitive balance to find the initial mass of this apparatus.Let us perform the experiment now.
(i) Take some barium chloride solution in the conical flask. Put some sodium sulphate solution in the small test-tube and lower it carefully in the conical flask by holding from the free end of thread tied to its neck.Fix a rubber cork in the mouth of the flask so that it holds the thread firmly [see Figure 3(a)].The mouth of the small test-tube should remain above the liquid level in the flask so that the reactants do not get mixed at this stage. Find the mass of the apparatus alongwith reactants by weighing on the balance. If we subtract the initial mass of apparatus from this mass, we will get the mass of reactants. Let the mass of reactants be x.
(ii) Remove the rubber cork from the mouth of conical flask for a moment so that the thread becomes loose. The small test-tube containing sodium sulphate solution now drops in the flask due to which sodium sulphate solution mixes with barium chloride solution [see Figure 3(b)]. Now,barium chloride solution reacts with sodium sulphate solution to form a white precipitate of barium sulphate,and sodium chloride solution. We again find the mass of the apparatus alongwith products by weighing on a balance.If we subtract the initial mass of the apparatus from this mass,we will get the mass of products.Suppose ihe mass of products is y,Now,if the mass of products (y)is equal to the mass of reactants (x),then this experiment verifies the law of conservation of mass.
The chemical reaction taking place in the above experiment can be written as :
Barium chloride
(solution) + + Sodium Chloride
(solution)
In an experiment to verify the law of conservation of mass, the following data was obtained :
(i) Mass of barium chloride taken = 20.8 g
(ii) Mass of sodium sulphate taken = 14.2 g
(iii) Mass of barium sulphate formed = 23.3 g
(iv) Mass of sodium chloride formed = 11.7 g
In this case, barium chloride and sodium sulphate are reactants.
So,Mass of reactants = 20.8g +14.2 g
= 35.0 g
Here, barium sulphate and sodium chloride are products.
So, Mass of products = 23.3 g + 11.7 g
= 35.0 g
Since the total mass of products (35 g) in this chemical reaction is equal to the total mass of reactants (35 g),therefore, the given data verifies the law of conservation of mass. Let us solve some problems now.
Sample Problem 1. Sodium carbonate reacts with ethanoic add to form sodium ethanoate, carbon dioxide and water. In an experiment, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid to form 8.2 g of sodium ethanoate, 2.2 g of carbon dioxide and 0.9 g of water. Show that this data verifies the law of conservation of mass.
Solution. All that we have to do in this problem is to calculate the mass of reactants and products separately, and then compare the two. If the two masses are equal, then the law of conservation of mass gets verified. The given reaction can be written as :
(i) Sodium carbonate and ethanoic acid are reactants.
So, Mass of reactants = Mass of sodium + Mass of ethanoic carbonate acid
= 5.3 + 6
= 11.3 g
(ii) Sodium ethanoate, carbon dioxide and water are products.
So, Mass of products = Mass of sodium + Mass of carbon + Mass of
ethanoate dioxide water
= 8.2 + 2.2 + 0.9
= 11.3 g
We find that the mass of reactants is 11.3 g and the mass of products is also 11.3 g. Since the mass of products is equal to the mass of reactants, the given data verifies the law of conservation of mass.
Sample Problem 2. Calcium carbonate decomposes, on heating, to form calcium oxide and carbon dioxide. When 10 g of calcium carbonate is decomposed completely, then 5.6 g of calcium oxide is formed.Calculate the mass of carbon dioxide formed. Which law of chemical combination will you use in solving this problem ?
Solution. This problem is to be solved by using the law of conservation of mass in chemical reactions.In this reaction,calcium carbonate is the reactant whereas calcium oxide and carbon dioxide are products.
Now,from the law of conservation of mass we have :
Mass of products = mass of rectatant
or Mass of calcium
oxide+ Mass of carbon
oxide Mass of calcium
carbonate
So,5.6 + Mass of carbon
dioxide =10
And,Mass of carbon
dioxide = 10-5.6
= 4.4g
Thus,the mass of carbon dioxide formed is 4.4 g.
Notes
Atoms and Molecules - Notes
2. Monovalent Cations (Cations Having a Valency of 1+)
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3. Divalent Cations (Cations Having a Valency of 2+)
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5. Trivalent Cations (Cations Having a Valency of 3+)
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6. an example, let us give the significance of symbol C
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7. Formulae of Some Molecular Compounds
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8. Molecular Formulae of Some Common Elements
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11. 2. A Negatively Charged Ion is Known as Anion
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13. Explanation of the Law of Conservation of Mass
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14. As an example, let us give the significance of the formula H
2O
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20. Simple Ions and Compound Ions (Polyatomic Ions)
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25. 1. A Positively Charged Ion is Known as Cation
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26. Gram Atomic Mass And Gram Molecular Mass
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33. Explanation of the Law of Constant Proportions
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34. Significance of the Formula of a Substance
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36. Molecular Masses of Some Common Elements
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38. Divalent Anions (Anions Having a Valency of 2-)
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44. Writing Of Formulae Of Ionic Compounds
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46. Symbols Derived from Latin Names of the Elements
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49. Writing Of Formulae Of Molecular Compounds
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50. Experiment to Verify Law of Conservation of Mass
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