Relative Formula Mass and Moles

The relative atomic mass ($A_r$) of an element is the average mass of its atoms compared to $\frac{1}{12}$ of the mass of a carbon-12 atom.

• It takes into account the masses and abundances of all isotopes
• It has no units (it's a ratio)
• Values are given on the periodic table

Why Chlorine Has $A_r = 35.5$

Chlorine has two isotopes:

• $^{35}\text{Cl}$ (75% abundance) and $^{37}\text{Cl}$ (25% abundance)

$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = 35.5$

The relative formula mass ($M_r$) of a compound is the sum of the relative atomic masses of all the atoms in its formula.

$M_r = \sum (A_r \text{ of each atom in the formula})$

Examples

Water ($\text{H}_2\text{O}$): $M_r = (2 \times 1) + 16 = 18$

Carbon dioxide ($\text{CO}_2$): $M_r = 12 + (2 \times 16) = 44$

Sodium hydroxide ($\text{NaOH}$): $M_r = 23 + 16 + 1 = 40$

Calcium carbonate ($\text{CaCO}_3$): $M_r = 40 + 12 + (3 \times 16) = 100$

Magnesium hydroxide ($\text{Mg(OH)}_2$): $M_r = 24 + 2 \times (16 + 1) = 24 + 34 = 58$

Watch out for brackets! In $\text{Mg(OH)}_2$, the subscript 2 applies to everything inside the brackets: 2 oxygens and 2 hydrogens.

Copper sulfate crystals ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$): $M_r = 63.5 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16) = 63.5 + 32 + 64 + 90 = 249.5$

Atoms and molecules are far too small to count individually. Chemists use a counting unit called the mole (symbol: mol).

$1 \text{ mole} = 6.022 \times 10^{23} \text{ particles}$

This number is called Avogadro's constant ($N_A$).

One mole of any substance contains exactly $6.022 \times 10^{23}$ particles (atoms, molecules, ions, etc.).

The Key Relationship

$\text{Number of moles} = \frac{\text{mass (g)}}{M_r}$

Or rearranged:

$\text{mass} = \text{moles} \times M_r$ $M_r = \frac{\text{mass}}{\text{moles}}$

Remember: 1 mole of a substance has a mass (in grams) equal to its $M_r$.

Examples

• 1 mole of water ($M_r = 18$) has a mass of 18 g
• 1 mole of CO₂ ($M_r = 44$) has a mass of 44 g
• 1 mole of NaCl ($M_r = 58.5$) has a mass of 58.5 g

The Formula Triangle

      mass (g)
     ─────────
    moles × Mr

Step-by-Step Method

1. Write down the formula: $n = \frac{m}{M_r}$
2. Identify what you know and what you need to find
3. Substitute values
4. Calculate (show your working!)

You can calculate the percentage of an element in a compound:

\text{% by mass} = \frac{A_r \times \text{number of atoms of element}}{M_r \text{ of compound}} \times 100

Example

Question: Calculate the percentage of oxygen in water ($\text{H}_2\text{O}$).

\text{% O} = \frac{16}{18} \times 100 = 88.9\%

Question: Calculate the percentage of nitrogen in ammonium nitrate ($\text{NH}_4\text{NO}_3$).

$M_r = 14 + 4 + 14 + 48 = 80$ \text{% N} = \frac{2 \times 14}{80} \times 100 = \frac{28}{80} \times 100 = 35\%

The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound.

Method

1. Write the mass (or percentage) of each element
2. Divide each by the element's $A_r$
3. Divide all results by the smallest value
4. Round to the nearest whole number

Example

Question: A compound contains 40% calcium, 12% carbon, and 48% oxygen. Find its empirical formula.

Empirical formula: $\text{CaCO}_3$ (calcium carbonate)

• Always show your working — you can get marks even if the final answer is wrong
• When calculating $M_r$, write out each step (don't try to do it in your head)
• Pay attention to brackets in formulae — they multiply everything inside
• Remember: moles = mass ÷ $M_r$ (the most used formula in chemistry!)
• Practice converting between mass, moles, and number of particles

• $A_r$ = relative atomic mass (average mass considering isotopes)
• $M_r$ = sum of $A_r$ values for all atoms in a formula
• 1 mole = $6.022 \times 10^{23}$ particles
• $n = \frac{m}{M_r}$ (moles = mass ÷ relative formula mass)
• Empirical formula = simplest whole-number ratio of atoms