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

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

Examples in context

Example 1. Calcium carbonate dosing at Melbourne Water Eltham plant. Operators add slaked lime $\text{Ca(OH)}_2$ to raise the pH of soft Otway-sourced water before chlorination. To raise $1.00 \, \text{ML}$ of water from pH $6.0$ to pH $7.5$ requires roughly $1.5 \, \text{kg}$ of $\text{Ca(OH)}_2$. Moles of $\text{Ca(OH)}_2$ $= 1500 / 74.1 = 20.2$ mol. By the reaction $\text{Ca(OH)}_2 + 2 \text{HCl} \to \text{CaCl}_2 + 2 \text{H}_2 \text{O}$, this neutralises $40.4$ mol of acid impurities. The plant's dosing pumps work in concentration units, so $20.2$ mol in $10^6 \, \text{L}$ gives $c = 2.0 \times 10^{-5} \, \text{mol/L}$ of added base, just enough to lift the pH a full unit and a half.

Example 2. Limiting-reagent calculation in Orica's nitric-acid plant. Orica at Kooragang Island operates the Ostwald process: $4 \text{NH}_3 + 5 \text{O}_2 \to 4 \text{NO} + 6 \text{H}_2 \text{O}$. In a test run, $8.50 \, \text{kg}$ of ammonia is mixed with $20.0 \, \text{kg}$ of oxygen. Moles: $n(\text{NH}_3) = 8500 / 17.0 = 500$ mol; $n(\text{O}_2) = 20{,}000 / 32.0 = 625$ mol. Required ratio is $4:5$, so $500$ mol $\text{NH}_3$ needs $500 \times 5/4 = 625$ mol $\text{O}_2$. Stoichiometric exactly; neither is limiting. If oxygen had been $600$ mol instead, $\text{O}_2$ would limit and only $480$ mol $\text{NO}$ would form, giving $14.4 \, \text{kg}$ of product.

Try this

Q1. Calculate the number of molecules in $5.00 \, \text{g}$ of glucose, $\text{C}_6 \text{H}_{12} \text{O}_6$. [2 marks]

• Cue. $n = 5.00 / 180.0 = 0.0278$ mol; $N = 0.0278 \times 6.022 \times 10^{23} = 1.67 \times 10^{22}$ molecules.

Q2. Magnesium reacts with hydrochloric acid: $\text{Mg} + 2 \text{HCl} \to \text{MgCl}_2 + \text{H}_2$. If $2.43 \, \text{g}$ of Mg is added to $100 \, \text{mL}$ of $1.50 \, \text{mol/L}$ HCl, determine the limiting reagent and the mass of $\text{MgCl}_2$ formed. [4 marks]

• Cue. $n(\text{Mg}) = 0.100$ mol; $n(\text{HCl}) = 0.150$ mol. Mg needs $2 \times 0.100 = 0.200$ mol HCl; only $0.150$ available, so HCl limits. $n(\text{MgCl}_2) = 0.0750$ mol $\times 95.2 = 7.14 \, \text{g}$.

Q3. $\text{N}_2 + 3 \text{H}_2 \to 2 \text{NH}_3$. (a) Calculate the mass of ammonia formed from $14.0 \, \text{g}$ of $\text{N}_2$ assuming complete reaction. (b) State the moles of $\text{H}_2$ consumed. (c) Explain why industrial Haber-Bosch never goes to completion. [2+2+2 marks]

• Cue. (a) $n(\text{N}_2) = 0.500$ mol; $n(\text{NH}_3) = 1.00$ mol $= 17.0 \, \text{g}$. (b) $1.50$ mol $\text{H}_2$. (c) Equilibrium reaction; Le Chatelier favours forward at low T but rate is slow.