Apply the mole concept, including Avogadro's number, molar mass, and basic stoichiometric calculations

What this dot point is asking

VCAA wants you to apply the mole concept and basic stoichiometric calculations to convert between mass, moles and number of particles, and to use balanced equations to relate quantities of reactants and products.

Avogadro's number

$N_A = 6.022 \times 10^{23}$ particles per mole.

One mole contains $N_A$ particles, where particle can be atom, molecule, ion, formula unit or electron depending on context.

Molar mass

The mass of one mole of a substance, in g mol$^{-1}$. Numerically equal to the relative atomic or molecular mass.

For elements: from the periodic table. For compounds: sum of atomic masses of constituent atoms.

Examples:

• H$_2$O: $2(1.0) + 16.0 = 18.0$ g mol$^{-1}$.
• Na$_2$SO$_4$: $2(23.0) + 32.1 + 4(16.0) = 142.1$ g mol$^{-1}$.

Standard conversions

Mass to moles. $n = m/M$.

Moles to mass. $m = nM$.

Moles to particles. $N = n N_A$.

Particles to moles. $n = N/N_A$.

Concentration of solutions. $c = n/V$ where $V$ is volume in litres. Units: mol L$^{-1}$ (M).

Stoichiometry

The proportional relationships in a balanced chemical equation. Coefficients give mole ratios.

For $2 \text{H}_2 + \text{O}_2 \to 2 \text{H}_2\text{O}$:

• IMATH_18 mol H$_2$ react with $1$ mol O$_2$ to give $2$ mol H$_2$O.
• IMATH_24 mol H$_2$ gives $4$ mol H$_2$O.

Procedure for stoichiometric calculations

1. Write a balanced equation.
2. Convert given mass (or volume, particles, concentration) to moles.
3. Apply the mole ratio from the equation.
4. Convert moles of the target to the desired quantity.

Limiting reagent

When two or more reactants are given, identify which limits the reaction. The limiting reagent runs out first; the excess remains.

For $\text{A} + 2\text{B} \to \text{products}$ with $1$ mol A and $1$ mol B:

• A would need $2$ mol B to react fully; we have only $1$ mol B.
• B is the limiting reagent. $0.5$ mol A reacts; $0.5$ mol A is in excess.

Worked example

If $6.0$ g of carbon reacts with $32.0$ g of oxygen, find the mass of CO$_2$ produced.

C + O$_2$ $\to$ CO$_2$.

$n$(C) $= 6.0/12.0 = 0.50$ mol.

$n$(O$_2$) $= 32.0/32.0 = 1.00$ mol.

Mole ratio $1:1:1$. C is the limiting reagent ($0.50$ mol). Reacts with $0.50$ mol O$_2$; $0.50$ mol O$_2$ in excess.

$n$(CO$_2$) $= 0.50$ mol. Mass $= 0.50 \times 44.0 = 22.0$ g.

Common traps

Skipping the balanced equation. Mole ratios require balanced coefficients.

Forgetting limiting reagent. Excess reactant does not determine product amount.

Mixing mass and moles. Convert to moles for stoichiometric reasoning; convert back to mass for final answer.

Wrong molar mass. Always include all atoms; check water vs methane vs ammonia carefully.

In one sentence

The mole concept ($N_A = 6.022 \times 10^{23}$) connects mass ($n = m/M$), particles ($N = nN_A$), and solution concentration ($c = n/V$); stoichiometric calculations require a balanced equation, conversion to moles, application of the mole ratio, and conversion back, with limiting-reagent identification when multiple reactants are given.