Atoms are infinitesimally small, and their absolute masses are too minute for practical laboratory use. To overcome this, chemists use a comparative system known as relative mass. By comparing the mass of any atom to a fixed standard, we can create a usable scale for calculating the quantities of substances involved in chemical reactions. The IGCSE 0654 Core syllabus requires a clear definition of these relative values and the ability to calculate the mass of molecules and formula units.

The Relative Atomic Mass ($A_r$) is the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of Carbon-12.

Why Carbon-12?

Carbon-12 is chosen as the international standard. One atom of Carbon-12 is assigned a mass of exactly 12 units. Therefore, 1/12th of this mass is exactly 1 unit.

• Average Value: Most elements exist as a mixture of isotopes. The $A_r$ value found in the Periodic Table is a weighted average of these isotopes. For example, Chlorine has an $A_r$ of 35.5 because it is a mixture of Chlorine-35 and Chlorine-37.

The Relative Molecular Mass ($M_r$) is the sum of the relative atomic masses of all the atoms shown in a chemical formula. For ionic compounds, this is often referred to as the Relative Formula Mass, though the symbol $M_r$ is used for both.

How to Calculate $M_r$

1. Identify the atoms: List each element in the formula and count how many atoms of each are present.
2. Find $A_r$ values: Look up the $A_r$ for each element in the Periodic Table.
3. Sum the masses: Multiply the count by the $A_r$ and add them together.

#### Worked Example: Water ($H_2O$)

• Hydrogen ($H$): 2 atoms $\times$ $A_r(1)$ = 2
• Oxygen ($O$): 1 atom $\times$ $A_r(16)$ = 16
• Total $M_r$ = 2 + 16 = 18

#### Worked Example: Sulfuric Acid ($H_2SO_4$)

• Hydrogen ($H$): 2 $\times$ 1 = 2
• Sulfur ($S$): 1 $\times$ 32 = 32
• Oxygen ($O$): 4 $\times$ 16 = 64
• Total $M_r$ = 2 + 32 + 64 = 98

Understanding relative mass allows us to calculate the mass of reactants and products in simple proportions without needing advanced mole calculations.

The Law of Constant Composition

Because the $M_r$ of a compound is fixed, the elements within it always react in the same mass ratio.

• Example: In $CO_2$ ($M_r = 44$), there are 12 parts Carbon to 32 parts Oxygen.
• If you react 12g of Carbon with 32g of Oxygen, you will always produce exactly 44g of $CO_2$.
• By maintaining this ratio, you can scale up or down: reacting 6g of Carbon will require 16g of Oxygen and yield 22g of $CO_2$.

Relative atomic and molecular masses provide the bridge between the microscopic world of atoms and the macroscopic world of the laboratory. By expressing mass in relation to Carbon-12, chemists can perform precise quantitative analysis, ensuring that the stoichiometry of every reaction is accurately maintained.