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Conservation of Mass

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Key Stage 3

Meaning

Conservation of Mass is a law of the universe that states that mass cannot be created or destroyed, it can only be moved from one place to another.

About Conservation of Mass

In dissolving conservation of mass means that the mass of the solvent and the mass of the solute can be added to find the mass of the solution.
Masssolvent + Masssolute = Masssolution
In a chemical reaction conservation of mass means that the same atoms which made up the reactants must make up the products. So the atoms are not created or destroyed in a chemical reaction, they are just rearranged.

Conservation of mass tells us that if there are 4 Hydrogen atoms and 2 Oxygen atoms at the start of this reaction then there will be the end of the reaction 4 Hydrogen atoms and 2 Oxygen atoms at the end of the reaction.

In this reaction you can see that mass is conserved because there are 4 Hydrogen atoms, 4 Oxygen atoms and 1 Carbon atom in the reactants and all the same atoms are found in the products.

Key Stage 4

Meaning

Conservation of Mass is a law of the universe that states that mass cannot be created or destroyed, it can only be moved from one place to another.

About Conservation of Mass

In a chemical reaction law of conservation of mass indicates that the total mass of the products is the same as the total mass of the reactants.
Conservation of mass can be observed in closed systems in which none of the products can escape and no other chemicals enter the system.
In an open system any chemical reaction which produces a gas will appear to decrease in mass but only because the mass has moved to a different location. The particles of gas escape the container.

Examples

Methane + Oxygen → Water + Carbon Dioxide

CH4 + 2O2 → 2H2O + CO2

16g + 64g = 36g + 44g


Sodium Hydroxide + Hydrochloric Acid → Sodium Chloride + Water

NaOH + HCl → NaCl + H2O

40g + 36.5g = 58.5 + 18g

Calculating the Mass of a missing Product/Reactant

The mass of a missing product of reactant can be found because the total mass of the products = the total mass of reactants.

MReactants = MProducts

Find the mass of Calcium Oxide produced in the following reaction:

CaCO3 → CaO + CO2

25g = x + 11g

Find the mass of Carbon Dioxide produced in the following reaction:

CH4 + 2O2 → 2H2O + CO2

4g + 16g = 9g + x

Find the mass of Hydrochloric Acid needed in the following reaction:

NaOH + HCl → NaCl + H2O

160g + x = 234g + 72g

x = 25g - 11g

x = 14g

20g = 9g + x

x = 20g - 9g

x = 11g

160g + x = 306g

x = 306g - 160g

x = 146g

Calculating the Mass Required for a Complete Reaction

Find the mass of Oxygen needed to completely oxidise all of the Magnesium:

2Mg + O2 → 2MgO

48g + x = y

Find the mass of Oxygen needed for the complete combustion of Methane.

CH4 + 2O2 → 2H2O + CO2

32g + x = y

Find the mass of Hydrochloric Acid needed to completely neutralise all of the Sodium Hydroxide.

NaOH + HCl → NaCl + H2O

20g + x = y + z

Find the Relative Formula Mass of the reactants.

Mr of Mg = 24g

Mr of O2 = 16x2

Mr of O2 = 32g

Find the Relative Formula Mass of the reactants.

Mr of CH4 = 16g

Mr of O2 = 16x2

Mr of O2 = 32g

Find the Relative Formula Mass of the reactants.

Mr of NaOH = 40g

Mr of HCl = 36.5g

State the ratio of moles of each chemical needed.

2 moles of Mg are needed for every 1 mole of O2

State the ratio of moles of each chemical needed.

1 mole of CH4 is needed for every 2 moles of O2

State the ratio of moles of each chemical needed.

1 mole of HCl are needed for every 1 mole of NaOH

Find the number of moles supplied of the known mass.

No. Moles = \(\frac{Mass}{M_r}\)

No. Moles = \(\frac{48}{24}\)

No. Moles = 2 Mole

Therefore 1 mole of O2 is needed.

1 mole of O2 = 32g

Find the number of moles supplied of the known mass.

No. Moles = \(\frac{Mass}{M_r}\)

No. Moles = \(\frac{32}{16}\)

No. Moles = 2 Mole

Therefore 4 moles of O2 are needed.

4 moles of O2 = 128g

Find the number of moles supplied of the known mass.

No. Moles = \(\frac{Mass}{M_r}\)

No. Moles = \(\frac{20}{40}\)

No. Moles = 0.5 Mole

Therefore 0.5 mole of HCl is needed.

0.5 mole of HCl = 18.25g

Beyond the Curriculum