Verification of the Law of Conservation of Mass in a Chemical Reaction
Law of conservation of mass states that “The mass can neither be created nor destroyed in a chemical reaction” French chemist Antoine Lavoisier was the first to state the law of conservation of mass in his book. There is just a rearrangement in the atoms of substances for the formation of compounds.
In the chemical process, the law of conservation of mass can be understood as the total mass of the reactant at the beginning of the reaction being equivalent to the mass of the product at the end of the reaction.
Mass reactants = Mass products
Let’s learn about the law of conservation of mass in detail in this article,
What is the Law of Conservation of Mass?
Law of Conservation of Mass is an important law for states that,
“For an isolated system mass can neither be created nor be destroyed but is easily transformed from one form to another”.
Law of conservation of mass in chemistry says the mass of the reactants at the beginning of the reaction is always equal to the mass of the products at the end of the reaction. The law of conservation of mass can easily be understood under two changes,
- Physical Changes
- Chemical Changes
When Matter Undergoes Physical Change
When the matter undergoes physical change the mass of the components as a whole system is always conserved. For example, take a piece of ice and place it in a flask properly close this flask and weigh it. Now, heat the flask gently to melt the ice into water and again weigh it.
Ice → (Heat)→ Water
The weight of the flask for and after heating is the same. This shows that when a matter undergoes physical change its mass is constant.
When Matter Undergoes Chemical Change
When any matter undergoes a chemical change the mass of the system remains unchanged. Swiss Chemist Hans Heinrich Landolt first explained this concept.
We can understand this concept with the help of the following chemical reaction,
NaCl (s) + AgNO3 (aqueous) → AgCl (s) + NaNO3 (aqueous)
In the above chemical reaction sodium chloride reacts with silver nitrate to form silver chloride and sodium nitrate. The mass of the reactants and the mass of the product are the same in the above reaction.
Experimental Verification of Law of Conservation of Mass
The following experiment can be conducted to verify the law of conservation of mass:
Things required to prove the Law of conservation of mass are,
- 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)
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 verification of Law of conservation are:
- 50 ml distilled water is taken in two 100 mL beakers.
- Weigh the two taken watch glasses on a physical balance.
- A quantity of 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 beakers A and B are combined together through constant stirring using a glass rod.
- A precipitate emerges 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 masses of the content of the beakers are measured before and after the reaction.
Assumptions: In the case of distilled water, density is assumed to be 1g /cc.
Things to Take Care
Before starting the experiment we should take care of the following things,
- Small quantities of chemicals should be used to perform the reaction.
- Initially, the spring balance pointer should be at 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
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.
Following precautions must be taken while performing the experiment.
- Use the weighing machine with caution as it is sensitive to the slightest change.
- Use distilled water to make the solution.
- When solutions X and Y are mixed place the solution in a bottle with a cork.
- While calculating the resultant mass of the product, subtract the mass of the [conical flask + cork].
- The cork is used to prevent the gas, or vapours, from escaping the solution. The law is valid only for closed systems.
- Never taste any chemicals used in the experiment.
Formula of Law of Conservation of Mass
The formula for the law of conservation of mass is expressed in the differential form using the continuity equation in fluid mechanics which is,
∂/∂t (ρ) + ▽(ρv) = 0
ρ is the density
t is the time
v is the velocity
▽ is the divergence
FAQs on Law of Conversation of Mass
Question 1: What does the law of conservation of mass state?
The law of conservation of mass states that the total mass of any isolated system is always constant mass can neither be created nor destroyed.
Question 2: Who discovered the law of conservation of mass?
The law of conservation of mass is first proposed by french chemist Antoine Lavoisier in 1789.
Question 3: State the Law of Conservation of Mass and Energy.
Einstein states that there is a direct relation between the mass and energy of any object. He formulated an equation to justify its proof, this equation is called mass-energy equivalence. The equation is written as,
E = mc2
E is the energy of the particle
m is the mass of the particle
c is speed of light
Question 4: Why is there no change in mass during chemical reactions?
In a chemical reaction, atoms of reactants are rearranged to form the product and are neither created nor destroyed. Hence, there is no change in mass in a chemical reaction.
Question 5: 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 6: Which other reactions are used to display the law of conservation of mass?
Any combination reaction, where the reactants combine to form a product is used to verify law of conservation of mass. For example, the production of water from hydrogen and oxygen molecules.
2H2 (4g) + O2 (32g) => 2H20 (36g)
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