Avogadro's law and Avogadro's number N_A = 6.022 × 10²³: Worked examples

Definition Symbols and units used on this page (defined before first use)

$m$ = mass, in grams (g) unless stated.
$M$ = molar mass, grams per mole (g/mol) — built by adding atomic masses.
$n$ = amount of substance, in moles (mol).
$N$ = a plain count of particles (a pure number). When we need to say which entity a count refers to, we add a subscript: $N_{mol}$ = number of molecules, $N_{atoms}$ = number of atoms, $N_{Ne}$ / $N_{CO_{2}}$ = atoms of that named substance.
$N_{A}$ = Avogadro's number $= 6.022 \times 1 0^{23}$ particles per mole.
$ρ$ (Greek "rho") = density, mass per unit volume (here g/mL).
For gases: $P$ = pressure (in pascals, Pa), $V$ = volume, $T$ = absolute temperature (in kelvin, K), and $R$ = the ideal gas constant $= 8.314 J mol^{- 1} K^{- 1}$ — the fixed number that ties $P, V, n, T$ together in $P V = n R T$ .
Units you'll meet: $L$ = litre, a volume equal to $1 0^{- 3} m^{3}$ (1000 L = 1 m³); $mL$ = millilitre $= 1 0^{- 3} L$ ; $bar$ = a pressure unit equal to $1 0^{5} Pa$ .
STP = Standard Temperature and Pressure, the agreed reference conditions. Modern IUPAC STP is $T = 273.15 K$ and $P = 1 bar = 1 0^{5} Pa$ ; at these conditions one mole of any ideal gas occupies about $22.7 L$ .

The scenario matrix

Every mole/Avogadro problem is really a journey between four "worlds", plus a few edge cases. The map below is the spine of this whole page — keep glancing back at it, because every example is just "start in one box, follow arrows to another box":

Figure — Avogadro's law and Avogadro's number N_A = 6.022 × 10²³

Figure s01 — the four worlds of the mole. The diagram shows four rounded boxes and the arrows that connect them. The blue box on the left, labelled GRAMS (balance), is what a weighing scale reads. The orange box in the centre, labelled MOLES (counting box), is the invisible "counting box" — every mole holds exactly $N_{A}$ particles. The green box at the upper right, labelled PARTICLES (atoms/molecules), is the count of individual atoms or molecules. The red box at the lower right, labelled LITRES of GAS (container), is the volume a gas fills in a container. Each arrow is labelled with the exact formula that crosses it: the two gray arrows between grams and moles read $n = m / M$ (going right) and $m = n \times M$ (going back left); the two green arrows between moles and particles read $\times N_{A}$ (out) and $\div N_{A}$ (back); the single red arrow from moles down to litres reads $V = n R T / P$ . The key thing the picture teaches: moles sit in the centre, so you can never jump grams→particles or grams→litres directly — you must always pass through the orange moles box. Every row of the table below names a start box, a finish box, and therefore a path across this figure.

#	Cell (scenario class)	Which arrows you cross	Example
1	grams → moles → molecules	$n = m / M$ , then $\times N_{A}$	(a)
2	molecules → moles → grams (reverse trip)	$\div N_{A}$ , then $\times M$	(b)
3	entity mismatch (molecules vs atoms)	multiply by atoms-per-molecule	(c)
4	mass of ONE particle (limiting, tiny number)	$M \div N_{A}$	(d)
5	gas volume at STP from grams	$n = m / M$ , then $V = n R T / P$	(e)
6	gas-to-gas volume ratio, identity irrelevant	$V \propto n$	(f)
7	DEGENERATE input: zero / a single atom / non-integer moles	plug straight in, watch it stay sane	(g)
8	UNIT TRAP: kg vs g, bar vs Pa, L vs m³	force consistent units	(h)
9	real-world word problem	pick the start world, then travel	(i)
10	exam twist: mixture / "which has more atoms?"	convert both to the same currency	(j)

We now clear every cell.

Cell 1 — grams → molecules (the standard forward trip)

Worked example (a) How many molecules are in 4.4 g of carbon dioxide,

$CO_{2}$ ? Forecast: 4.4 g is a small mass. Guess: is it more or fewer than one full mole ( $6.022 \times 1 0^{23}$ ) of molecules? (Peek: $M = 44$ , so $4.4$ g is a tenth of a mole — expect about $6 \times 1 0^{22}$ .)

Step 1 — Find the molar mass $M$ . $M (CO_{2}) = 12 + 2 (16) = 44 g/mol$ Why this step? The bridge grams→moles needs "grams per mole." We must build it from atomic masses first.

Step 2 — grams → moles. $n = \frac{m}{M} = \frac{4.4}{44} = 0.10 mol$ Why this step? $N_{A}$ only speaks "moles". We can't multiply grams by $N_{A}$ — the units wouldn't cancel.

Step 3 — moles → molecules. $N = n N_{A} = 0.10 \times 6.022 \times 1 0^{23} = 6.022 \times 1 0^{22} molecules$ Why this step? $N_{A}$ is the exchange rate: one mole buys $6.022 \times 1 0^{23}$ molecules.

Verify: units: $mol \times mol^{- 1} =$ pure count ✓. Magnitude: a tenth of a mole gives a tenth of $N_{A}$ ✓ — matches the forecast.

Cell 2 — molecules → grams (the reverse trip)

Worked example (b) What is the mass of

$1.5055 \times 1 0^{23}$ molecules of water, $H_{2} O$ ? Forecast: $1.5055 \times 1 0^{23}$ is a quarter of $N_{A}$ . So we expect a quarter-mole of water. Water is 18 g/mol, so guess ≈ 4.5 g.

Step 1 — molecules → moles. $n = \frac{N}{N _{A}} = \frac{1.5055 \times 1 0 ^{23}}{6.022 \times 1 0 ^{23}} = 0.25 mol$ Why this step? We're travelling backwards along the same arrow, so we divide by the exchange rate instead of multiplying.

Step 2 — Find $M$ . $M (H_{2} O) = 2 (1) + 16 = 18 g/mol$ Why this step? We're about to convert moles into grams, and the only conversion factor between them is $M$ (grams per mole). We can't reach the "grams" box without first knowing how heavy one mole of this particular substance is, so we build $M$ before we take the final step.

Step 3 — moles → grams. $m = n \times M = 0.25 \times 18 = 4.5 g$ Why this step? $M$ (grams per mole) turns moles back into a balance reading.

Verify: run it forward: $4.5/18 = 0.25$ mol, $\times N_{A} = 1.5055 \times 1 0^{23}$ ✓. Matches forecast.

Cell 3 — entity mismatch (molecules vs atoms)

Worked example (c) How many

oxygen atoms are in 8.0 g of $O_{2}$ gas? Forecast: careful — the question asks for atoms, but a mole of $O_{2}$ gives us molecules. Each molecule hides 2 atoms, so expect the atom count to be double the molecule count.

Step 1 — molar mass of the molecule $O_{2}$ . $M (O_{2}) = 2 (16) = 32 g/mol$ Why this step? The gas is diatomic; its mass "packet" is the whole $O_{2}$ molecule, not a lone O.

Step 2 — grams → moles of molecules. $n = \frac{8.0}{32} = 0.25 mol of O_{2}$ Why this step? We start in the "grams" box but $N_{A}$ can only count from the "moles" box, so the very first move must be grams→moles, which is exactly dividing the mass by the molar mass $M$ .

Step 3 — moles → molecules. (Recall $N_{mol}$ = number of molecules.) $N_{mol} = 0.25 \times 6.022 \times 1 0^{23} = 1.5055 \times 1 0^{23} molecules$ Why this step? Now that we're in the "moles" box we cross to the "particles" box, and that single arrow is $\times N_{A}$ — the exchange rate turning moles of $O_{2}$ into a count of $O_{2}$ molecules.

Step 4 — molecules → atoms (the mismatch fix). (Here $N_{atoms}$ = number of atoms.) $N_{atoms} = 2 \times N_{mol} = 3.011 \times 1 0^{23} atoms of O$ Why this step? $N_{A}$ counts "whatever entity you named." We named $O_{2}$ molecules; to get atoms we multiply by atoms-per-molecule.

Verify: $0.25$ mol $O_{2} = 0.50$ mol O atoms; $0.50 \times N_{A} = 3.011 \times 1 0^{23}$ ✓.

Cell 4 — mass of a single particle (limiting: a tiny number)

Cell 5 — grams to gas volume at STP

Worked example (e) What volume does 6.4 g of oxygen gas

$O_{2}$ occupy at STP (1 bar, 273.15 K)? Forecast: 6.4 g of a 32 g/mol gas is $0.20$ mol. One mole ≈ 22.7 L at STP, so expect about $0.2 \times 22.7 \approx 4.5$ L.

Step 1 — grams → moles. $n = \frac{6.4}{32} = 0.20 mol$ Why this step? Avogadro's law and $P V = n R T$ speak in moles, never grams.

Step 2 — apply $V = \frac{n R T}{P}$ with SI units. Here $P$ = pressure = $1 bar = 1 0^{5} Pa$ , $T$ = absolute temperature = $273.15 K$ , and $R$ = the ideal gas constant = $8.314 J mol^{- 1} K^{- 1}$ . $V = \frac{n R T}{P} = \frac{( 0.20 ) ( 8.314 ) ( 273.15 )}{1 0 ^{5}} m^{3} = 4.542 \times 1 0^{- 3} m^{3} = 4.542 L$ Why this step? We want the gas volume, so we isolate $V$ in the ideal gas law and plug numbers in consistent SI ( $P$ in Pa, so $V$ comes out in m³; $\times 1000$ to reach L since $1 m^{3} = 1000 L$ ).

Verify: shortcut: $0.20 \times 22.71 L/mol = 4.542$ L ✓. Matches forecast.

Cell 6 — gas-to-gas volume ratio (identity irrelevant)

Cell 7 — degenerate / edge inputs

Worked example (g) Three sanity-check questions: (i) zero grams, (ii) exactly one molecule, (iii)

$0.001$ mol. Forecast: the formulas should not "break" at these extremes — they should give obviously-sensible answers.

(i) 0 g of anything → how many moles/molecules? $n = \frac{0}{M} = 0 mol, N = 0 \times N_{A} = 0 molecules$ Why this matters: the mole picture stays consistent at the empty limit — no substance, no particles.

(ii) One molecule of $H_{2} O$ → how many moles? $n = \frac{1}{N _{A}} = 1.661 \times 1 0^{- 24} mol$ Why this matters: the smallest possible non-zero amount is exactly $1/ N_{A}$ mol — the "quantum" of the mole scale.

(iii) $0.001$ mol of neon → how many atoms? $N = 0.001 \times 6.022 \times 1 0^{23} = 6.022 \times 1 0^{20} atoms$ Why this matters: fractional moles are perfectly legal; the count just scales down smoothly.

Verify: each answer is $n \times N_{A}$ with the stated $n$ — no division-by-zero, no negatives, all finite ✓.

Cell 8 — the unit trap

Cell 9 — real-world word problem

Worked example (i) A 250 mL glass of water contains how many water molecules? (Density of water

$ρ = 1.00$ g/mL, where $ρ$ — Greek "rho" — means mass per unit volume.) Forecast: a small glass of water is roughly 250 g ≈ 14 moles ≈ $8 \times 1 0^{24}$ molecules — a mind-bendingly huge number for an everyday object.

Step 1 — volume → mass (using density). $m = ρ \times V = 1.00 \frac{g}{mL} \times 250 mL = 250 g$ Why this step? We can't do chemistry on "mL of water" directly; density $ρ$ is the arrow from volume to the balance reading (grams).

Step 2 — grams → moles. $n = \frac{250}{18} = 13.89 mol$ Why this step? We've landed in the "grams" box, but $N_{A}$ only counts from the "moles" box — so before we can reach molecules we must first cross grams→moles, which is dividing the mass by $M = 18$ g/mol.

Step 3 — moles → molecules. $N = 13.89 \times 6.022 \times 1 0^{23} = 8.36 \times 1 0^{24} molecules$ Why this step? This is the last arrow, moles→particles: multiply the mole count by the exchange rate $N_{A}$ to turn "13.89 counting boxes" into an actual number of individual molecules.

Verify: $250/18 = 13.888 \dots$ ; $13.888 \times 6.022 \times 1 0^{23} = 8.36 \times 1 0^{24}$ ✓. Bigger than a single mole, as forecast — because we have ~14 moles.

Cell 10 — exam twist: "which has more atoms?"

Worked example (j) Which contains more

atoms: 20 g of neon ( $Ne$ , $M = 20$ ) or 20 g of carbon dioxide ( $CO_{2}$ , $M = 44$ )? Forecast: neon is monatomic (1 atom per particle) but lighter; $CO_{2}$ is heavier but packs 3 atoms per molecule. It's a genuine tug-of-war — convert both to the same currency (total atoms) and compare.

Step 1 — neon: grams → moles → atoms. (Here $N_{Ne}$ = atoms of neon.) $n (Ne) = \frac{20}{20} = 1.0 mol, N_{Ne} = 1.0 \times N_{A} = 6.022 \times 1 0^{23} atoms$ Why this step? Ne is a single atom per entity, so moles-of-atoms = moles-of-particles.

Step 2 — $CO_{2}$ : grams → moles → molecules → atoms. (Here $N_{CO_{2}}$ = atoms of carbon dioxide.) $n (CO_{2}) = \frac{20}{44} = 0.4545 mol, N_{mol} = 0.4545 \times N_{A} = 2.737 \times 1 0^{23}$ $N_{CO_{2}} = 3 \times 2.737 \times 1 0^{23} = 8.211 \times 1 0^{23} atoms$ Why this step? Each $CO_{2}$ contributes 3 atoms (1 C + 2 O), so multiply the molecule count by 3.

Step 3 — compare. $8.211 \times 1 0^{23} (CO_{2}) > 6.022 \times 1 0^{23} (Ne)$ So $CO_{2}$ has more atoms, despite fewer moles — the "3 atoms per molecule" wins. Why this step? You can only compare once both are in the same unit (total atoms), never by comparing moles alone.

Verify: $20/44 = 0.4545$ ; $\times 3 = 1.3636$ mol of atoms $> 1.0$ mol (Ne). Ratio $1.3636 > 1$ , so $CO_{2}$ wins ✓.

Recall Which formula for which cell?

Start world → finish world tells you the arrows. grams → count ::: divide by $M$ , then multiply by $N_{A}$ . count → grams ::: divide by $N_{A}$ , then multiply by $M$ . molecules → atoms ::: multiply by atoms-per-molecule. mass of one particle ::: $M \div N_{A}$ . gas volume from moles ::: $V = n R T / P$ (SI) or $\times 22.7$ L at STP. two gases, same $T, P$ ::: $V \propto n$ , only the mole ratio matters. "which has more atoms" ::: convert BOTH to total atoms, then compare.

Connections

The mole concept — every cell above is just travelling to/from the "mole" world.
Molar mass and atomic mass unit — the $M$ we build in step 1 of almost every example.
Ideal gas law PV=nRT — powers cells 5 and 6.
Stoichiometry — cell 10's "same currency" idea is the heart of every balanced-equation sum.
Empirical and molecular formulas — atoms-per-molecule (cell 3) feeds formula ratios.
Boyle's law and Charles's law — the other components of $P V = n R T$ we held constant in cell 6.