The mole concept and stoichiometric calculations (QCE Chemistry Unit 1)

What this dot point is asking

QCAA wants you to use the mole as the central counting unit in chemistry, converting fluently between mass, moles, number of particles, gas volume (at STP), and concentration of solution. From a balanced equation you should be able to identify the limiting reagent and calculate theoretical and percentage yield. This is the dot point that underwrites every IA calculation in Year 12.

The answer

The mole is the SI counting unit for chemical entities. One mole contains 6.022 x 10^23 entities (Avogadro's constant, N_A). Molar mass (M, in g/mol) is numerically equal to the relative atomic or molecular mass. The mole bridges mass measurements in the lab to particle counts in equations.

Core conversions

QCAA-relevant constants and conventions for 2026:

• Avogadro's constant: $N_A = 6.022 \times 10^{23}$ mol^-1.
• Molar gas volume at STP (25 degrees C, 100 kPa, the QCAA convention for "standard"): 24.79 L/mol. (Note: STP at 0 degrees C, 101.325 kPa, the older IUPAC convention, gives 22.41 L/mol. QCAA uses 25 degrees C, 100 kPa; check the formula sheet.)
• Gas constant: R = 8.314 J/(mol K) when P is in kPa and V is in L (or R = 8.314 J/(mol K) in SI).

Working with balanced equations

A balanced chemical equation gives mole ratios between reactants and products. The coefficients are read as "moles per mole".

Example. $2 H_2 + O_2 \rightarrow 2 H_2O$. Reads: 2 mol H_2 react with 1 mol O_2 to give 2 mol H_2O. Therefore the ratio H_2 : O_2 : H_2O = 2 : 1 : 2.

To find the mass of one species from the mass of another:

1. Convert the known mass to moles using its molar mass.
2. Use the stoichiometric ratio to find the moles of the target species.
3. Convert moles of target back to mass (or volume, or particles).

Worked example. How many grams of water form when 8.0 g of hydrogen react completely with excess oxygen?

Limiting reagent and theoretical yield

When two or more reactants are mixed in non-stoichiometric amounts, one will run out first. That is the limiting reagent; it caps the maximum (theoretical) yield. The other is the excess reagent.

Procedure:

1. Convert each reactant mass to moles.
2. Divide each moles by its coefficient in the balanced equation.
3. The reactant with the smallest quotient is the limiting reagent.
4. Use the limiting reagent's moles, scaled by the stoichiometric ratio, to find the theoretical moles (and mass or volume) of product.

Worked example. $N_2 + 3 H_2 \rightarrow 2 NH_3$. Mix 14 g N_2 with 6.0 g H_2. Which is limiting? How much NH_3 forms?

N_2 has the smaller quotient: limiting reagent is N_2.

Percentage yield

The theoretical yield is the maximum predicted from stoichiometry assuming the limiting reagent reacts completely. Actual yield is what you measured in the lab.

Both quantities should be in the same units (both as masses, or both as moles, or both as volumes; not mixed).

Reasons actual is less than theoretical: incomplete reaction, side reactions, losses during transfer or purification, equilibrium effects. Yields above 100 percent indicate impurities or experimental error (often water of crystallisation not fully removed before weighing).

Gas volume calculations

At STP (25 degrees C, 100 kPa per QCAA convention), 1 mole of any ideal gas occupies 24.79 L.

For non-standard conditions, use the ideal gas equation:

where R = 8.314 J/(mol K). When pressure is given in kPa and volume in L, the same form applies with R = 8.314, and T must be in kelvin (K = degrees C + 273.15).

Worked example. 0.250 mol of gas at 27 degrees C and 95.0 kPa. Find V.

(Units: J = kPa x L for this purpose because 1 J = 1 kPa x L when J is reinterpreted; this is the form QCAA uses.)

Solution stoichiometry

For aqueous reactions, the link between volume and moles is through concentration:

with c in mol/L (= mol/dm^3 = M) and V in litres. Always convert mL to L before substituting.

Titration example. 25.0 mL of 0.100 mol/L NaOH neutralises a sample of HCl. What volume of 0.0500 mol/L HCl was used?

Equation: $HCl + NaOH \rightarrow NaCl + H_2O$. Ratio 1:1.

This is the canonical IA1-style data-test calculation.

Empirical and molecular formulae

QCE Chemistry expects you to determine empirical formulae from percentage composition or combustion data, then scale to a molecular formula using the molar mass.

Procedure:

1. Assume 100 g of compound; treat percentages as masses.
2. Convert each mass to moles using the molar mass.
3. Divide every mole count by the smallest.
4. Multiply by a small integer if needed to get whole-number ratios.
5. For molecular formula: divide the experimental molar mass by the empirical-formula mass; multiply subscripts by that integer.

Worked example. A compound is 40.0 percent C, 6.7 percent H, 53.3 percent O. Find the empirical formula. (Glucose example.)

Divide all by smallest (3.33): C 1.00, H 2.00, O 1.00. Empirical formula CH_2O.

If the molar mass of the compound is 180 g/mol, the empirical formula mass is 30.03 g/mol; the multiplier is 6. Molecular formula: C_6H_12O_6.

Significant figures and units

QCAA penalises unit errors and over-precise answers. Default to matching the least precise input (usually 3 significant figures). Express:

• Mass in g unless very small.
• Volume in L unless very small (then mL).
• Concentration in mol/L.
• Amount in mol (use mmol or micromol only if explicitly requested).
• Energy in kJ unless very small (then J).

Examples in context

Example 1. Limestone calcination at Gladstone cement works. Cement Australia's Gladstone plant heats limestone in rotary kilns: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. To process a $1000 \, \text{kg}$ batch of $\text{CaCO}_3$ ($M = 100.09 \, \text{g mol}^{-1}$) the moles are $n = 10^6 / 100.09 = 9{,}991 \, \text{mol}$. By stoichiometry an equal mole of $\text{CO}_2$ is released, giving mass $= 9{,}991 \times 44.01 = 439.7 \, \text{kg}$. Real plants run $\sim 80\%$ yield, so reported emissions are about $352 \, \text{kg}$ CO$_2$ per tonne lime. This calculation feeds directly into Queensland's Safeguard Mechanism reporting.

Example 2. Limiting reagent in IA1 magnesium-acid investigation. A QCAA-style EA setup reacts $0.243 \, \text{g}$ of magnesium ribbon with $50.0 \, \text{mL}$ of $0.100 \, \text{mol L}^{-1}$ hydrochloric acid. Moles of $\text{Mg} = 0.243 / 24.31 = 0.0100 \, \text{mol}$; moles of $\text{HCl} = 0.0500 \times 0.100 = 0.00500 \, \text{mol}$. The balanced equation $\text{Mg} + 2 \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$ requires $2 \, \text{mol}$ HCl per mol Mg, so Mg needed is only $0.00500 / 2 = 0.00250 \, \text{mol}$. HCl is limiting; $\text{Mg}$ remaining is $0.00750 \, \text{mol}$, and $\text{H}_2$ produced is $0.00250 \, \text{mol}$ or $61 \, \text{mL}$ at SLC.

Try this

Q1. Define the mole and state the value of Avogadro's constant to three significant figures. [2 marks]

• Cue. Mole = amount containing $6.02 \times 10^{23}$ particles; $N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}$.

Q2. Combustion of $4.40 \, \text{g}$ propane gives $\text{CO}_2$ and water. Calculate moles of $\text{CO}_2$ produced. $M(\text{C}_3 \text{H}_8) = 44.11$. [4 marks]

• Cue. $n(\text{C}_3 \text{H}_8) = 0.0998 \, \text{mol}$; ratio $1\!:\!3$; $n(\text{CO}_2) = 0.299 \, \text{mol}$.

Q3. A compound contains $40.0\%$ C, $6.7\%$ H, $53.3\%$ O by mass; $M = 180 \, \text{g mol}^{-1}$. (a) Calculate empirical formula. (b) Determine molecular formula. (c) Identify a Queensland sugar-industry compound that fits. [3+2+1 marks]

• Cue. (a) $\text{CH}_2 \text{O}$. (b) $\text{C}_6 \text{H}_{12} \text{O}_6$. (c) Glucose from Townsville sugar cane.