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Matter and its Composition - Law of Conservation of Mass

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Law of Conservation of Mass was established by the French chemist Antoine Lavoisier in 1789.

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It states that mass can neither be created nor destroyed in a chemical reaction.

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In any chemical change, the total mass of the reactants is always equal to the total mass of the products.

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During a chemical reaction, the atoms of the reactants are merely rearranged to form new products; the number of atoms of each element remains constant.

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This law holds true for both physical changes (e.g., H2O(s)→H2O(l)H_2O (s) \rightarrow H_2O (l)) and chemical changes (e.g., burning of magnesium).

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A 'closed system' is often required to experimentally verify this law, especially when gases like CO2CO_2 or O2O_2 are involved, to prevent matter from escaping or entering.

📐Formulae

Total Mass of Reactants=Total Mass of Products\text{Total Mass of Reactants} = \text{Total Mass of Products}

mR1+mR2+...=mP1+mP2+...m_{R1} + m_{R2} + ... = m_{P1} + m_{P2} + ...

Mass of A+Mass of B→Mass of (A+B)\text{Mass of } A + \text{Mass of } B \rightarrow \text{Mass of } (A+B)

💡Examples

Problem 1:

If 100 g100\text{ g} of calcium carbonate (CaCO3CaCO_3) is heated, it decomposes to form 56 g56\text{ g} of calcium oxide (CaOCaO) and some carbon dioxide (CO2CO_2). Calculate the mass of CO2CO_2 produced.

Solution:

Mass of CaCO3=Mass of CaO+Mass of CO2\text{Mass of } CaCO_3 = \text{Mass of } CaO + \text{Mass of } CO_2 100 g=56 g+x100\text{ g} = 56\text{ g} + x x=100 g−56 gx = 100\text{ g} - 56\text{ g} x=44 gx = 44\text{ g}

Explanation:

According to the Law of Conservation of Mass, the mass of the reactant (CaCO3CaCO_3) must equal the sum of the masses of the products (CaOCaO and CO2CO_2).

Problem 2:

In a reaction, 6.3 g6.3\text{ g} of sodium bicarbonate (NaHCO3NaHCO_3) was added to 15.0 g15.0\text{ g} of ethanoic acid (CH3COOHCH_3COOH) solution. The residue left weighed 18.0 g18.0\text{ g}. What is the mass of CO2CO_2 released during the reaction?

Solution:

Total Mass of Reactants=6.3 g+15.0 g=21.3 g\text{Total Mass of Reactants} = 6.3\text{ g} + 15.0\text{ g} = 21.3\text{ g} Total Mass of Products=Mass of Residue+Mass of CO2\text{Total Mass of Products} = \text{Mass of Residue} + \text{Mass of } CO_2 21.3 g=18.0 g+Mass of CO221.3\text{ g} = 18.0\text{ g} + \text{Mass of } CO_2 Mass of CO2=21.3 g−18.0 g=3.3 g\text{Mass of } CO_2 = 21.3\text{ g} - 18.0\text{ g} = 3.3\text{ g}

Explanation:

Since the total mass must remain constant, the 'missing' mass from the residue represents the mass of the gaseous carbon dioxide that escaped into the atmosphere.

Problem 3:

When 2.4 g2.4\text{ g} of Magnesium (MgMg) burns completely in 1.6 g1.6\text{ g} of Oxygen (O2O_2), how much Magnesium Oxide (MgOMgO) is formed?

Solution:

2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO Mass of MgO=Mass of Mg+Mass of O2\text{Mass of } MgO = \text{Mass of } Mg + \text{Mass of } O_2 Mass of MgO=2.4 g+1.6 g=4.0 g\text{Mass of } MgO = 2.4\text{ g} + 1.6\text{ g} = 4.0\text{ g}

Explanation:

The mass of the product (MgOMgO) is the sum of the masses of the magnesium and oxygen that reacted together.