Stoichiometry and the mole: TCE Chemistry (Tasmania)

Syllabus dot point

“Use the mole concept and balanced equations to perform stoichiometric calculations, including limiting reagent and percentage yield.”

9 min readUpdated 2 Jun 2026

Quick answer

The mole links mass, particles and the balanced equation. Convert a measured mass or volume to moles with $n = m/M$ , use the coefficients of the balanced equation to find moles of another species, then convert back. When two amounts are given, the limiting reagent runs out first and sets how much product can form; percentage yield compares the actual product with the theoretical maximum.

What this dot point is asking

TASC expects you to move confidently between mass, moles, number of particles and gas quantities, and to apply mole ratios drawn straight from a balanced equation. Almost every quantitative question in Unit 3 begins with a mole calculation, so this is the foundation skill for titrations, gas laws, thermochemistry and equilibrium.

The mole and molar mass

The mole

The mole is the amount of substance that contains as many specified particles as there are atoms in exactly $12\ \text{g}$ of carbon-12, namely the Avogadro constant $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ . The molar mass $M$ is the mass of one mole, numerically equal to the relative atomic, molecular or formula mass in $\text{g mol}^{-1}$ .

The three conversions you must know by heart are:

mass to moles: $n = \dfrac{m}{M}$
particles to moles: $n = \dfrac{N}{N_A}$ , so $N = n \times 6.022 \times 10^{23}$
moles of gas to volume: at STP a gas occupies $V_m = 22.71\ \text{L mol}^{-1}$ , so $n = \dfrac{V}{V_m}$ (TASC uses $0\ ^\circ\text{C}$ , $100\ \text{kPa}$ for STP; check your data sheet).

Core mole relationships

n = \frac{m}{M} \qquad N = nN_A \qquad c = \frac{n}{V} \qquad n = \frac{V_{\text{gas}}}{V_m}

Mole-ratio calculations

A balanced equation gives the ratio in which substances react and form. The universal routine is:

Write a balanced equation.
Convert the known quantity (mass, volume or concentration) to moles.
Multiply by the mole ratio (coefficient of wanted species over coefficient of known species).
Convert the resulting moles back to the quantity asked for.

Mass of product from a balanced equation

Calculate the mass of magnesium oxide that forms when $6.0\ \text{g}$ of magnesium burns completely in oxygen. The balanced equation is $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$ , with $M(\text{Mg}) = 24.31\ \text{g mol}^{-1}$ and $M(\text{MgO}) = 40.31\ \text{g mol}^{-1}$ .

Step 1: Convert the known mass of magnesium to moles

Working in moles is essential because the balanced equation gives a ratio of moles, not grams. Divide the mass by the molar mass of magnesium:

n(\text{Mg}) = \frac{m}{M} = \frac{6.0}{24.31} = 0.247\ \text{mol}.

Step 2: Apply the stoichiometric mole ratio

Read the ratio directly from the coefficients in the balanced equation. The equation shows $2\ \text{Mg}$ producing $2\ \text{MgO}$ , a $1:1$ ratio. Therefore the moles of magnesium oxide equal the moles of magnesium:

n(\text{MgO}) = 0.247\ \text{mol}.

Step 3: Convert moles of product back to mass

Multiply the moles of magnesium oxide by its molar mass to obtain the final answer:

m(\text{MgO}) = nM = 0.247 \times 40.31 = 9.95\ \text{g}.

As a sense check, the oxygen taken up is $0.247 \times 16.00 = 3.95\ \text{g}$ , and $6.0 + 3.95 = 9.95\ \text{g}$ , confirming mass is conserved.

Final answer: 9.95 g of magnesium oxide forms.

Limiting reagent

When two reactant amounts are given, one usually runs out first. Find moles of each reactant, divide each by its coefficient, and the smallest value identifies the limiting reagent, which controls how much product forms. The other reactant is in excess and some is left over.

The limiting reagent controls the product

Always base product calculations on the limiting reagent, never on the excess reagent. Compare reactants by dividing moles by their coefficients to decide which is limiting, then use only that species for the product mole ratio.

Percentage yield and atom economy

The theoretical yield is the maximum product calculated from the limiting reagent. The actual yield is always less because of side reactions, incomplete reaction, and losses in handling. Percentage yield compares the two:

Percentage yield

\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

Atom economy is a different idea used in green chemistry: it measures the fraction of reactant mass that ends up in the desired product, $\text{atom economy} = \dfrac{M(\text{desired product})}{M(\text{all products})} \times 100\%$ . A reaction can have high yield but poor atom economy if it makes large amounts of by-product.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms. From percentage composition, treat each percentage as grams in $100\ \text{g}$ , divide by molar mass, then divide by the smallest result. The molecular formula is a whole-number multiple of the empirical formula, found from the relative molecular mass.

Using the wrong reagent for the product calculation

A common error is to calculate product from whichever reactant is mentioned first, or from the reactant in excess. Only the limiting reagent sets the yield. Identify it properly (divide moles by coefficients) before doing the mole-ratio step. Two further slips: forgetting to divide moles by their coefficient before comparing reactants, and forgetting that percentage yield is actual over theoretical, not the other way round.

In the exam, convert to moles first, identify the limiting reagent whenever two amounts are given, apply the mole ratio from the balanced equation, and finish with the quantity asked for. Carry one or two extra significant figures through the working and round only at the end.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

A manufacturer produced

58.0\ \text{kg}

of ammonia by reacting

63.0\ \text{kg}

of nitrogen with

15.0\ \text{kg}

of hydrogen:

\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)

. (

M(\text{H}_2) = 2.016

M(\text{N}_2) = 28.02

M(\text{NH}_3) = 17.03\ \text{g mol}^{-1}

.) Show that the maximum possible yield of ammonia, if the reaction went to completion, is approximately

77.0\ \text{kg}

, then determine the percentage yield.

Show worked answer →

Find the limiting reagent. $n(\text{N}_2) = 63\,000 / 28.02 = 2248\ \text{mol}$ ; $n(\text{H}_2) = 15\,000 / 2.016 = 7440\ \text{mol}$ . The ratio needed is $1\ \text{N}_2 : 3\ \text{H}_2$ . For $2248\ \text{mol}$ of $\text{N}_2$ you would need $6744\ \text{mol}$ of $\text{H}_2$ , and $7440\ \text{mol}$ is available, so $\text{N}_2$ is limiting. (2 marks)

Maximum ammonia: $n(\text{NH}_3) = 2 \times n(\text{N}_2) = 2 \times 2248 = 4496\ \text{mol}$ . Mass $= 4496 \times 17.03 = 76\,600\ \text{g} = 76.6\ \text{kg}$ , approximately $77.0\ \text{kg}$ . (2 marks)

Percentage yield $= (\text{actual} / \text{theoretical}) \times 100 = (58.0 / 76.6) \times 100 = 75.7\%$ . (1 mark)

Practice3 marks

A geochemist analysed a sample of a hydrated iron carbonate and found it contained

24.94\%

\text{Fe}

5.36\%

\text{C}

5.40\%

\text{H}

and

64.30\%

\text{O}

by mass. Calculate the empirical formula, including the degree of hydration. (

M(\text{Fe}) = 55.85

M(\text{C}) = 12.01

M(\text{H}) = 1.008

M(\text{O}) = 16.00\ \text{g mol}^{-1}

Show worked answer →

Treat each percentage as grams in $100.0\ \text{g}$ and divide by molar mass to get moles. $\text{Fe}: 24.94/55.85 = 0.4466$ ; $\text{C}: 5.36/12.01 = 0.4463$ ; $\text{H}: 5.40/1.008 = 5.357$ ; $\text{O}: 64.30/16.00 = 4.019$ . (1 mark)

Divide each by the smallest ( $0.4463$ ): $\text{Fe} = 1.00$ ; $\text{C} = 1.00$ ; $\text{H} = 12.0$ ; $\text{O} = 9.0$ . (1 mark)

The ratio is $\text{FeCH}_{12}\text{O}_9$ . Rewriting as a carbonate plus water, the $\text{CO}_3$ uses $3$ of the $\text{O}$ , leaving $6\ \text{O}$ and $12\ \text{H}$ , which is $6\ \text{H}_2\text{O}$ . So the empirical formula is $\text{FeCO}_3 \cdot 6\text{H}_2\text{O}$ (iron carbonate hexahydrate). (1 mark)

Practice3 marks

Oxygen gas is produced when potassium chlorate is heated:

2\text{KClO}_3(s) \rightarrow 2\text{KCl}(s) + 3\text{O}_2(g)

50.2\ \text{mL}

of gas is collected at

20.0\ ^\circ\text{C}

and

98.2\ \text{kPa}

. Calculate the mass of potassium chlorate used, assuming the reaction goes to completion. (

M(\text{KClO}_3) = 122.5\ \text{g mol}^{-1}

R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}

Show worked answer →

Find moles of $\text{O}_2$ using $pV = nRT$ . $p = 98.2\ \text{kPa} = 98\,200\ \text{Pa}$ ; $V = 50.2\ \text{mL} = 5.02 \times 10^{-5}\ \text{m}^3$ ; $T = 20.0 + 273 = 293\ \text{K}$ .

n(\text{O}_2) = \frac{pV}{RT} = \frac{98\,200 \times 5.02 \times 10^{-5}}{8.314 \times 293} = 2.02 \times 10^{-3}\ \text{mol}.

(1 mark)

Mole ratio: $2\ \text{KClO}_3$ produce $3\ \text{O}_2$ , so $n(\text{KClO}_3) = \tfrac{2}{3} \times n(\text{O}_2) = \tfrac{2}{3} \times 2.02 \times 10^{-3} = 1.35 \times 10^{-3}\ \text{mol}$ . (1 mark)

Mass $= n \times M = 1.35 \times 10^{-3} \times 122.5 = 0.165\ \text{g}$ . (1 mark)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC