Mole Concept

Syllabus mapping

CH-11-02 — Quantitative Chemistry. Covers: relating the Avogadro constant to the mole; relating relative atomic mass to carbon-12; solving problems with \(n = \dfrac{m}{MM}\); practical investigations comparing the masses of equimolar amounts of substances and showing that chemicals react in whole-number molar ratios; percentage composition by mass and empirical formulas; limiting reagent problems; real-world problems combining stoichiometric principles with reacting masses.

Wording above is paraphrased for teaching use — check the official syllabus before treating any line as verbatim NESA text.

You need to know: what a mole is, how to calculate molar mass, how to convert between mass and moles using \(n = \dfrac{m}{MM}\), how to use a balanced equation to find reacting masses, how to find percentage composition and empirical formulas, and how to spot the limiting reagent.

This page is the calculation toolkit for the rest of Chemistry — it's worth getting really comfortable with it. Every calculation below uses the same few steps over and over.

Assumed knowledge check

Before this page makes sense, you should be comfortable with:

• Balanced equations and conservation of mass
• Relative atomic mass
• Scientific notation (e.g. \(6.022 \times 10^{23}\)) and rearranging a simple formula — see Assumed Knowledge and rearranging formulas

Core

What is a mole?

Atoms are far too small to count one by one, so chemists count them in big bundles — the same way you buy eggs by the dozen (12) or paper by the ream (500).

The chemist's bundle is the mole (symbol mol):

\[ 1\ \text{mol} = 6.022 \times 10^{23}\ \text{particles} \]

That number is the Avogadro constant, \(N_A\). The "particles" can be atoms, molecules or ions — whatever you're counting. One mole of water is \(6.022 \times 10^{23}\) water molecules.

To convert between number of particles (\(N\)) and moles (\(n\)):

\[ n = \frac{N}{N_A} \]

Worked example

How many molecules are in \(0.500\ \text{mol}\) of water?

Rearrange to \(N = n \times N_A\):

\[ N = 0.500 \times 6.022 \times 10^{23} = 3.01 \times 10^{23}\ \text{molecules} \]

Why the mole is so useful: carbon-12 and molar mass

Relative atomic masses are measured on a scale where one atom of carbon-12 is exactly 12. Everything else is compared to that — a magnesium atom (24.31) is about twice as heavy as a carbon-12 atom.

The mole was chosen so that 12 g of carbon-12 contains exactly one mole of atoms. That gives us a very handy shortcut:

One mole of any substance has a mass in grams equal to its relative atomic (or formula) mass.

This is the molar mass (\(MM\)), in grams per mole (\(\text{g mol}^{-1}\)). Find it by adding up the relative atomic masses from the periodic table.

Worked example — calculating molar mass

Find the molar mass of (a) water, \(\ce{H2O}\), (b) calcium carbonate, \(\ce{CaCO3}\), (c) calcium hydroxide, \(\ce{Ca(OH)2}\).

(a) \(2(1.008) + 16.00 = 18.02\ \text{g mol}^{-1}\)

(b) \(40.08 + 12.01 + 3(16.00) = 100.09\ \text{g mol}^{-1}\)

(c) The brackets mean two of everything inside: \(40.08 + 2(16.00 + 1.008) = 74.10\ \text{g mol}^{-1}\)

Converting between mass and moles

\[ n = \frac{m}{MM} \]

where \(n\) = moles (mol), \(m\) = mass (g), \(MM\) = molar mass (\(\text{g mol}^{-1}\)).

Rearranged: \(m = n \times MM\) and \(MM = \dfrac{m}{n}\).

🖼️ Diagram needed — the "formula triangle" for \(n = m/MM\): m on top, n and MM on the bottom, with a note "cover the one you want". Optional but helpful for weaker students. TODO — placeholder until sourced/drawn.

Worked example — mass to moles

How many moles are in \(36.04\ \text{g}\) of water?

\[ n = \frac{m}{MM} = \frac{36.04}{18.02} = 2.000\ \text{mol} \]

Worked example — moles to mass

What is the mass of \(0.250\ \text{mol}\) of sodium chloride, \(\ce{NaCl}\)?

\(MM = 22.99 + 35.45 = 58.44\ \text{g mol}^{-1}\)

\[ m = n \times MM = 0.250 \times 58.44 = 14.6\ \text{g} \]

Using a balanced equation: reacting masses

The numbers in front of each formula in a balanced equation (the coefficients) give the mole ratio — how many moles react with, or make, how many moles of something else.

\[ \ce{CH4 + 2O2 -> CO2 + 2H2O} \]

This says: 1 mol of methane reacts with 2 mol of oxygen to make 1 mol of carbon dioxide and 2 mol of water.

The four-step method (use it for every reacting-mass question):

1. Mass → moles of the substance you're given: \(n = \dfrac{m}{MM}\)
2. Mole ratio from the equation: find moles of the substance you want
3. Moles → mass of the substance you want: \(m = n \times MM\)
4. Check the answer makes sense (right units, sensible size)

Worked example

What mass of carbon dioxide is produced when \(8.00\ \text{g}\) of methane burns completely?

Step 1 — moles of methane: \(MM(\ce{CH4}) = 12.01 + 4(1.008) = 16.04\ \text{g mol}^{-1}\)

\[ n(\ce{CH4}) = \frac{8.00}{16.04} = 0.499\ \text{mol} \]

Step 2 — ratio: \(\ce{CH4} : \ce{CO2} = 1 : 1\), so \(n(\ce{CO2}) = 0.499\ \text{mol}\)

Step 3 — mass of CO₂: \(MM(\ce{CO2}) = 12.01 + 2(16.00) = 44.01\ \text{g mol}^{-1}\)

\[ m(\ce{CO2}) = 0.499 \times 44.01 = 22.0\ \text{g} \]

Step 4 — check: more mass of product than methane makes sense, because the oxygen from the air adds mass (conservation of mass).

Percentage composition by mass

The percentage of a compound's mass that comes from one element:

\[ \%\ \text{element} = \frac{\text{mass of that element in one mole}}{\text{molar mass of the compound}} \times 100 \]

Worked example

Ammonium nitrate, \(\ce{NH4NO3}\), is used as a fertiliser because it's rich in nitrogen. What percentage of its mass is nitrogen?

\(MM = 2(14.01) + 4(1.008) + 3(16.00) = 80.05\ \text{g mol}^{-1}\)

There are 2 nitrogen atoms: \(2 \times 14.01 = 28.02\ \text{g}\)

\[ \%\ \ce{N} = \frac{28.02}{80.05} \times 100 = 35.0\% \]

This is also written as % (w/w) — "weight for weight" — meaning grams of element per 100 g of compound.

Empirical formulas

The empirical formula is the simplest whole-number ratio of atoms in a compound. Glucose is \(\ce{C6H12O6}\), but its empirical formula is \(\ce{CH2O}\) (divide everything by 6).

To find an empirical formula from masses or percentages:

1. Write down the mass of each element (if you're given percentages, pretend you have 100 g — then the percentages are the masses).
2. Convert each to moles: \(n = \dfrac{m}{MM}\)
3. Divide every answer by the smallest one.
4. Round to whole numbers (if you get something like 1.5, multiply everything by 2).

Worked example

A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula.

	C	H	O
Mass in 100 g	40.0 g	6.7 g	53.3 g
Moles	\(\frac{40.0}{12.01} = 3.33\)	\(\frac{6.7}{1.008} = 6.65\)	\(\frac{53.3}{16.00} = 3.33\)
÷ smallest (3.33)	1.00	2.00	1.00

Empirical formula: \(\ce{CH2O}\)

Limiting reagents

Making a cheese sandwich takes 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 3 slices of cheese, you can only make 3 sandwiches — you run out of cheese first, and 4 slices of bread are left over.

In a reaction, the reactant that runs out first is the limiting reagent. It decides how much product forms. The other reactant is in excess — some of it is left over.

Method:

1. Find the moles of each reactant.
2. Pick one reactant and use the mole ratio to work out how many moles of the other it needs.
3. Compare: if you have less than is needed, that one is the limiting reagent.
4. Use the limiting reagent to calculate the amount of product.

Worked example

\(4.0\ \text{g}\) of hydrogen reacts with \(16.0\ \text{g}\) of oxygen: \(\ce{2H2 + O2 -> 2H2O}\). Which is the limiting reagent, and what mass of water forms?

Step 1 — moles:

\(n(\ce{H2}) = \dfrac{4.0}{2.016} = 1.98\ \text{mol}\) \(\qquad\) \(n(\ce{O2}) = \dfrac{16.0}{32.00} = 0.500\ \text{mol}\)

Step 2 — what does the oxygen need? The ratio is \(\ce{H2} : \ce{O2} = 2 : 1\), so \(0.500\ \text{mol}\) of \(\ce{O2}\) needs \(2 \times 0.500 = 1.00\ \text{mol}\) of \(\ce{H2}\).

Step 3 — compare: we have \(1.98\ \text{mol}\) of \(\ce{H2}\) — more than the \(1.00\ \text{mol}\) needed. So hydrogen is in excess, and oxygen is the limiting reagent.

Step 4 — product from the limiting reagent: \(\ce{O2} : \ce{H2O} = 1 : 2\), so \(n(\ce{H2O}) = 2 \times 0.500 = 1.00\ \text{mol}\)

\[ m(\ce{H2O}) = 1.00 \times 18.02 = 18.0\ \text{g} \]

Check with conservation of mass: \(1.98 - 1.00 = 0.98\ \text{mol}\) of \(\ce{H2}\) is left over, which is \(0.98 \times 2.016 = 2.0\ \text{g}\). Total after: \(18.0 + 2.0 = 20.0\ \text{g}\) — the same as the \(4.0 + 16.0 = 20.0\ \text{g}\) we started with ✓

Practical

1. Comparing masses of equimolar amounts. Weigh out 0.100 mol of several different substances and put them side by side. They have the same number of particles but very different masses and volumes — a great way to see what a mole means.

Substance	Molar mass (\(\text{g mol}^{-1}\))	Mass of 0.100 mol
Water, \(\ce{H2O}\)	18.02	1.80 g (about 1.8 mL)
Sodium chloride, \(\ce{NaCl}\)	58.44	5.84 g
Copper, Cu	63.55	6.36 g
Sucrose (table sugar), \(\ce{C12H22O11}\)	342.30	34.2 g

2. Showing a whole-number molar ratio: the formula of magnesium oxide. Weigh a crucible and lid, add a coiled strip of magnesium ribbon and weigh again. Heat strongly, lifting the lid occasionally to let air in (but not smoke out), until the magnesium stops glowing. Cool and weigh. The mass increase is the oxygen that combined. Convert both masses to moles and find the ratio — it should come out close to 1 : 1, i.e. \(\ce{MgO}\).

For example: \(0.24\ \text{g}\) Mg → \(0.40\ \text{g}\) of oxide, so \(0.16\ \text{g}\) of O. \(n(\ce{Mg}) = \dfrac{0.24}{24.31} = 0.0099\ \text{mol}\), \(n(\ce{O}) = \dfrac{0.16}{16.00} = 0.010\ \text{mol}\) → ratio 1 : 1 ✓

Common sources of error: white smoke (magnesium oxide) escaping when the lid is lifted (oxide mass too low); some magnesium not fully reacted; magnesium reacting with nitrogen in the air to form magnesium nitride.

Advanced

From empirical to molecular formula. If you also know the molar mass, divide it by the empirical formula's mass to find how many "empirical units" are in one molecule. The compound above (\(\ce{CH2O}\), \(30.03\ \text{g mol}^{-1}\)) has a molar mass of \(180.16\ \text{g mol}^{-1}\): \(\dfrac{180.16}{30.03} = 6\), so the molecular formula is \(\ce{C6H12O6}\) (glucose).

Theoretical and percentage yield. The mass calculated from the equation is the theoretical yield — the most you could possibly make. Real reactions usually make less.

\[ \%\ \text{yield} = \frac{\text{actual mass obtained}}{\text{theoretical mass}} \times 100 \]

Real-world example — making iron. In a blast furnace: \(\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}\). From \(1.00\) tonne (\(1.00 \times 10^6\ \text{g}\)) of iron(III) oxide (\(MM = 159.70\)): \(n(\ce{Fe2O3}) = 6262\ \text{mol}\); ratio \(1 : 2\) gives \(n(\ce{Fe}) = 12\,524\ \text{mol}\); \(m(\ce{Fe}) = 12\,524 \times 55.85 = 6.99 \times 10^5\ \text{g} = 0.699\ \text{t}\). If the furnace actually produces \(0.650\ \text{t}\), the yield is \(\dfrac{0.650}{0.699} \times 100 = 93.0\%\).

Counting atoms, not molecules. \(0.500\ \text{mol}\) of \(\ce{H2O}\) contains \(3.01 \times 10^{23}\) molecules — but each molecule has 3 atoms, so \(9.03 \times 10^{23}\) atoms. Read carefully which particle a question is asking about.

Empirical formulas with a ratio like 1 : 1.5. Multiply every number by 2 to get whole numbers (1 : 1.5 → 2 : 3, e.g. \(\ce{Fe2O3}\)). A ratio like 1 : 1.33 → multiply by 3.

Extension

Beyond the syllabus for this course — won't appear in the HSC, included for interest / depth study inspiration.

The 2019 redefinition. Until 2019, the mole was defined as the number of atoms in exactly 12 g of carbon-12, and the Avogadro constant was something measured. Now it's the other way around: the Avogadro constant is fixed as exactly \(6.022\,140\,76 \times 10^{23}\), and the mole is defined from it. The difference is far too tiny to affect any calculation you'll do, but it means the kilogram, mole and other SI units are now all tied to fixed constants of nature rather than physical objects.

Just how big is a mole? \(6.022 \times 10^{23}\) grains of sand (about 1 mm across) would bury the whole of Australia roughly 80 metres deep. Yet a mole of water is just 18 mL — less than a mouthful. That contrast is the whole point: atoms are unimaginably small.

📎 Depth study idea

Investigate the percentage yield of a reaction you can do in the lab (for example, precipitating a solid, filtering, drying and weighing it), then identify where the "lost" product went and test changes to the method that might improve the yield.

Video/visual resources

• 🎥 Khan Academy — TODO: source an explainer on the mole and molar mass.
• 🎥 TODO: source a worked-examples video on reacting-mass (stoichiometry) problems and limiting reagents — ideally one that uses the same four-step method as this page.
• 🖥️ PhET Simulation — Reactants, Products and Leftovers — make "sandwiches" and then real reactions, and see which reactant runs out first (limiting reagents)

Check yourself

1. Calculate the molar mass of sulfuric acid, \(\ce{H2SO4}\).

   Answer

   \(2(1.008) + 32.07 + 4(16.00) = 98.09\ \text{g mol}^{-1}\)

2. How many moles are in \(10.0\ \text{g}\) of calcium carbonate, \(\ce{CaCO3}\)?

   Answer

   \(n = \dfrac{10.0}{100.09} = 0.0999\ \text{mol}\)

3. What is the mass of \(0.150\ \text{mol}\) of carbon dioxide?

   Answer

   \(m = 0.150 \times 44.01 = 6.60\ \text{g}\)

4. Magnesium reacts with hydrochloric acid: \(\ce{Mg + 2HCl -> MgCl2 + H2}\). What mass of magnesium chloride forms from \(2.43\ \text{g}\) of magnesium?

   Answer

   Step 1: \(n(\ce{Mg}) = \dfrac{2.43}{24.31} = 0.100\ \text{mol}\)

   Step 2: \(\ce{Mg} : \ce{MgCl2} = 1 : 1\), so \(n(\ce{MgCl2}) = 0.100\ \text{mol}\)

   Step 3: \(MM(\ce{MgCl2}) = 24.31 + 2(35.45) = 95.21\ \text{g mol}^{-1}\), so \(m = 0.100 \times 95.21 = 9.52\ \text{g}\)

5. Calculate the percentage by mass of oxygen in water.

   Answer

   \(\%\ \ce{O} = \dfrac{16.00}{18.02} \times 100 = 88.8\%\)

6. A compound contains \(2.4\ \text{g}\) of carbon and \(0.6\ \text{g}\) of hydrogen. Find its empirical formula.

   Answer

   \(n(\ce{C}) = \dfrac{2.4}{12.01} = 0.20\), \(n(\ce{H}) = \dfrac{0.6}{1.008} = 0.60\). Divide by 0.20: C = 1, H = 3. Empirical formula \(\ce{CH3}\).

7. \(2.43\ \text{g}\) of magnesium is added to \(0.150\ \text{mol}\) of hydrochloric acid (\(\ce{Mg + 2HCl -> MgCl2 + H2}\)). Which is the limiting reagent?

   Answer

   \(n(\ce{Mg}) = 0.100\ \text{mol}\). The ratio \(\ce{Mg} : \ce{HCl} = 1 : 2\), so \(0.100\ \text{mol}\) of Mg needs \(0.200\ \text{mol}\) of HCl. We only have \(0.150\ \text{mol}\), so HCl is the limiting reagent (and magnesium is in excess).