To Verify Law of Conservation of Mass in a Chemical Reaction
Mass is an isolated system. The law of conservation of mass states that the mass can neither be created nor destroyed in a chemical reaction. There is just a rearrangement in the atoms of substances for the formation of compounds. Therefore, the mass of the total system remains constant at any interval of time. This implies that the total masses of reactants is equal to the sum of masses of products and the masses of unreacted reactants. The law of conservation of mass is also termed as the principle of mass conservation.
Mass of the total enclosed system at the beginning of the reaction is equivalent to the mass at the termination of the reaction.
Mass reactants = Mass products
Experiment to verify the law of conservation of mass
The following experiment can be conducted to verify the law of conservation of mass:
Things required to perform the experiment:
- Barium chloride (BaCl2.2H20),
- Sodium sulphate (Na2SO4.10H2O),
- Two beakers of 100 and 150 ml respectively.
- Physical balance
- Two watch glasses
- Spring balance (0-500 g),
- Polythene bag
- Distilled water
- A glass rod.
The reaction can be visualised as a precipitation reaction, where the insoluble salt separates out as a precipitate. The reaction occurs between the Barium Chloride (BaCl22(aq)) and Sodium Sulphate (Na2SO4(aq)). Both the compounds are taken in aqueous solutions, that is water is taken as the solvent. This is a kind of double displacement reaction.
The reaction involved is,
BaCl2(aq) + Na2SO4(aq) ————-> BaSO4(aq) + 2NaCl(aq)
Colorless Colorless White Precipitate
Rearranging the equation in the iconic form, we get,
Ba+(aq) + SO42-(aq) —————> BaSO4(s)
The reactants involved in the reaction are barium chloride and sodium sulphate, whereas the products involved are barium sulphate and sodium chloride.
Now, we know,
Mass of the reactants (barium chloride + sodium sulphate) = Mass of the products (barium sulphate + sodium chloride)
Steps Involved in the Process
- 50 ml distilled water is taken in two 100 mL beakers.
- Weigh the two taken watch glasses on a physical balance.
- A quantity 3.6 g of BaCl2.2H20 is taken in a watch glass.
- Dissolve the quantity of aqueous solution of barium chloride in 50ml of distilled water. The contents are stored in beaker A.
- 8.05 g of Na2SO4.10H2O is taken in another watch glass of a definite mass.
- Dissolve the quantity of aqueous solution of sodium chloride in 50ml of distilled water. The contents are stored in beaker B.
- A 150ml beaker is taken and measured using the spring balance. This beaker will contain the final contents and is labelled as C.
- The solutions contained in the beakers A and B are combined together through constant stirring using a glass rod.
- A precipitate emerges out on the beaker C, owing due to the formation of the compound barium sulphate (BaSO4).
- The total weight of the products can be calculated by measuring the weight of the beaker.
- The content masses of the beakers are measured before and after the reaction.
Assumptions during the experiment
In case of distilled water, density is assumed to be 1g /cc.
Things to take care of before the experiment
- Small quantities of chemicals should be used to perform the reaction.
- Initially, the spring balance pointer should be at the zero marks.
- The reading of spring balance is taken only once its pointer is at the rest position.
- The reading of spring balance should be taken when it is placed in a vertical position.
- Precise quantities of the masses mentioned should be taken.
- Solution of BaCl2 and Na2SO4 should be mixed with constant stirring.
The following inferences can be drawn from the experiment,
Mass of aqueous solution of barium chloride (BaCl2) = 3.6 g
Mass of BaCl2 solution = 53.6 g
Mass of aqueous solution of sodium sulphate (Na2SO4.10H2O) = 8.05 g
Mass of Na2SO4 solution =58.05 g
Mass of 50 ml distilled water = 50.0 g
Calculating the total mass of reactants, we have,
BaCl2 + Na2SO4 = 53.6 + 58.05
= 111.65 g
Mass of empty 150 mL beaker,m1 =………………….g
Mass of reaction mixture before precipitation, m2= m1+ 111.65 g = ……………………g
Final mass of reaction mixture after precipitation, m3 =……………………. g
When we compare the mass of reactants with those of products, the two masses are considered to be equivalent. This implies that the observed masses, m2 = m3. Hence, the law of conservation of mass is preserved.
Question 1. Where is the law of conservation applicability found?
Law of conservation of mass can be seen in chemical reactions, like the production of carbon dioxide, or during the process of combustion of wood. It is applicable to all the phenomena occurring in the closed system.
Question 2. What is a double displacement precipitation reaction?
Double precipitation reactions involve the switching of atoms from two different compounds. In these reactions, there occurs an exchange of ions resulting in the formation of two new compounds. The occurrence of this type of reaction is more evident in case the ionic compounds are dissolved in water as the solvent. The positive ions exchange negative ion partners. Any reaction of this form is,
AB + CD → AD + CB
Question 3. Which other reaction type can be used to display the law of conservation of mass?
Any combination reaction, where the reactants combine to form a product can be used to verify this law. For example, the production of water from hydrogen and oxygen molecules.
2H2 (4g) + O2 (32g) => 2H20 (36g)
Question 4. Give some other precipitation reaction that can be easily used to verify this law.
NaCl(aq) + AgNO3(aq) => AgCl(s) + NaNO3(aq)
Rewriting as iconic eq. ,
Ag+ (aq.) + Cl– (aq.) => AgCl(s)
Here, AgCl separates out as a white precipitate.
Question 5. Which reactions fail to display the law of conservation of mass?
Nuclear reactions since some mass is destroyed in the form of energy.
E = mc2