The mole concept — counting by weighing: Worked examples

Recall the five symbols (so line one is safe)

Before any scenario, here is every symbol we will use, in plain words:

The scenario matrix

Every mole question is one (or a chain) of the cells below. Each example is tagged with its cell.

Cell	Case class	Given → Find	Danger / degenerate point
A	Forward, simple	mass → moles → molecules	mass = molar mass ⇒ exactly 1 mol
B	Backward	particle count → moles → mass	must invert $N = n N_{A}$
C	Molecules vs atoms	moles of molecule → atoms	multiply by atoms-per-formula
D	Single-particle mass	molar mass → mass of ONE atom	tiny number, $M / N_{A}$
E	Ions & charge	moles → number of a specific ion	pick the right entity
F	Real-world word problem	everyday mass → count	translate words to $m, M$
G	Exam twist: "1 g of X vs Y"	compare counts at equal mass	lighter ⇒ more atoms
H	Degenerate / limiting	zero mass, or $m < M$ (fractional mol)	$n$ can be $0$ or $< 1$

The eight examples below hit all eight cells. We use the atomic masses (in u, hence g/mol): H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35.5, Ca = 40, Fe = 56 — these come from Atomic Mass & Isotopes.

Example 1 — Cell A (forward, simple)

Worked example How many molecules are in 4.4 g of

$CO_{2}$ ? Forecast: $M (CO_{2}) = 44$ , and $4.4$ is a tenth of that — so guess about $0.1$ mol, i.e. roughly $6 \times 1 0^{22}$ molecules. Let's check.

Step 1 — Build the molar mass. $M = 12 + 2 (16) = 44 g mol^{- 1}$ . Why this step? Before we can multiply by $N_{A}$ , we first need $n = m / M$ — and that needs $M$ . The mass of one mole is the yardstick we divide the sample by.

Step 2 — Mass → moles. $n = \frac{m}{M} = \frac{4.4}{44} = 0.1 mol .$ Why this step? This tells us how many batches of $N_{A}$ we have.

Step 3 — Moles → molecules. $N = n N_{A} = 0.1 \times 6.022 \times 1 0^{23} = 6.022 \times 1 0^{22} molecules .$ Why this step? Each batch is $N_{A}$ molecules by definition.

Verify: Units, carried all the way through: $\frac{g}{g mol ^{- 1}} \times mol^{- 1} = mol \times mol^{- 1} = mol^{1 - 1} = mol^{0} = 1 (pure number) .$ So $N$ is a plain count ✅. Matches forecast $\approx 6 \times 1 0^{22}$ ✅.

Example 2 — Cell B (backward: count → mass)

Example 3 — Cell C (molecules ↔ atoms)

Example 4 — Cell D (mass of a single particle)

Example 5 — Cell E (ions & the right entity)

Example 6 — Cell F (real-world word problem)

Example 7 — Cell G (exam twist: equal mass, who wins?)

Worked example You have

1.00 g of helium and 1.00 g of neon. Which sample contains more atoms, and by what factor? Forecast: Lighter atoms pack more per gram, so He wins. He is 4, Ne is 20 — factor 5?

The figure below is a balance whose two pans hold exactly equal mass (1.00 g each), yet the pans are drawn with very different populations. On the left pan (magenta dots) sits helium: because each He atom is light ( $M = 4$ ), one gram is a dense crowd of many small atoms. On the right pan (violet dots) sits neon: each Ne atom is 5× heavier ( $M = 20$ ), so one gram is a sparse handful of fewer, larger atoms. The pointer stays level — the balance cannot tell them apart by mass — but the eye immediately sees He is the more numerous crowd. That picture is the answer: at equal mass, the lighter atom always makes the bigger count.

Figure: A level balance holds 1 g of He (left, many magenta atoms) against 1 g of Ne (right, few violet atoms). Equal mass, unequal crowd — He outnumbers Ne by a factor of 5.

Step 1 — Moles of each. $n_{He} = \frac{1}{4} = 0.25 mol, n_{Ne} = \frac{1}{20} = 0.05 mol .$ Why this step? Both are monatomic gases, so "moles" = "moles of atoms" directly; no $\times$ -factor needed.

Step 2 — Compare (ratio kills the need for $N_{A}$ ). $\frac{N _{He}}{N _{Ne}} = \frac{n _{He}}{n _{Ne}} = \frac{0.25}{0.05} = 5.$ Why this step? Since $N = n N_{A}$ and $N_{A}$ is common, atom counts share the same ratio as moles — we never need the giant number.

Step 3 — Actual counts (if wanted). $N_{He} = 0.25 \times 6.022 \times 1 0^{23} = 1.5055 \times 1 0^{23}$ atoms; $N_{Ne} = 0.05 \times 6.022 \times 1 0^{23} = 3.011 \times 1 0^{22}$ atoms.

Verify: $1.5055 \times 1 0^{23} /3.011 \times 1 0^{22} = 5$ ✅. Forecast confirmed: lighter ⇒ more atoms per gram.

Example 8 — Cell H (degenerate & limiting cases)

Worked example Three sanity edges: (a) how many atoms in 0 g of copper? (b) how many moles in 4 g of helium (the "exactly one mole" landmark)? (c) what if the sample is

smaller than one molar mass, e.g. 2 g of helium? Forecast: (a) zero — obviously. (b) exactly 1 mol, since $M (He) = 4$ . (c) less than a mole — a fractional $0.5$ mol. The point is to see the formula behave at all three boundaries.

Case (a) — zero mass. $n = \frac{0}{M} = 0 mol \Rightarrow N = 0 \times N_{A} = 0 atoms .$ Why show this? The bridge must give $0$ at $m = 0$ — no matter how huge $N_{A}$ is, $0 \times N_{A} = 0$ . The formula never "invents" particles.

Case (b) — the boundary $m = M$ . $n = \frac{4}{4} = 1 mol (exactly one batch) .$ Why show this? When sample mass equals molar mass, $n = 1$ — the natural "1 mole" landmark.

Case (c) — fractional check ( $m < M$ ). $2 g He : n = 2/4 = 0.5 mol, N = 0.5 \times 6.022 \times 1 0^{23} = 3.011 \times 1 0^{23}$ atoms. Why? Moles need not be whole numbers — the balance reads whatever mass you put on it, so $m < M$ simply gives $n < 1$ .

Verify: $0 \times N_{A} = 0$ ✅; $4/4 = 1$ ✅; $2/4 = 0.5$ ✅. The formula is well-behaved from $m = 0$ upward.

Recall Quick self-test (cover the right side)

Moles in 8 g of $CH_{4}$ ( $M = 16$ )? ::: $0.5$ mol Oxygen atoms in 1 mol of $CO_{2}$ ? ::: $2 N_{A} = 1.204 \times 1 0^{24}$ Mass of $3.011 \times 1 0^{23}$ He atoms? ::: $0.5 \times 4 = 2$ g Which has more atoms, 1 g Li ( $M = 7$ ) or 1 g Na ( $M = 23$ )? ::: Li (lighter ⇒ more) Atoms in 0 g of anything? ::: $0$

Connections

The mole concept — counting by weighing (index 1.1.12) — the parent note that built the chain.
Avogadro's Number — the constant $N_{A}$ threaded through every example.
Atomic Mass & Isotopes — source of every atomic mass used here.
Molar Mass Calculations — how we built $M$ for $CO_{2}$ , $H_{2} SO_{4}$ , sucrose.
Stoichiometry — where these per-substance counts become reaction ratios.
Empirical & Molecular Formulae — the reverse art: counts → formula.
Units & Measurement — the SI mole and unit-cancelling checks.