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

The bridge from moles to a raw count is $N=n\times N_A$

• $n=0.25$ mol (the amount you were given),
• $N_A=6.022\times 10^{23}{\text{mol}}^{-1}$ (particles per mole),
• $N$ = the count of molecules (because the entity named is ${\text{CO}}_2$ molecules). WHAT we did: matched the given (moles) to the formula whose only input is moles. WHY: we would not use $n=m/M$ here — no mass was given, so that formula's inputs are missing. Recognition is choosing the tool whose inputs you actually hold.

WHAT we did: checked the number and the entity-word separately. WHY: $N_A$ always counts the named entity, so the word matters as much as the digits. The number $6.022\times 10^{23}$ is correct as a count of molecules, but the word "atoms" is wrong.

• 1 mol of ${\text{N}}_2$ = $6.022\times 10^{23}$ molecules.
• Each ${\text{N}}_2$ has 2 atoms, so it is $2\times 6.022\times 10^{23}=1.2044\times 10^{24}$ atoms. Always ask: "a mole of WHAT?"

$n=\frac{m}{M}=\frac{88}{44}=2\text{mol}$ WHAT we did: divided the mass we have by the mass of one mole. WHY: $M$ tells us "44 g makes one mole-box," so 88 g fills two boxes.

$N=n{N}_A=2\times 6.022\times 10^{23}=1.2044\times 10^{24}\text{molecules}$ WHAT we did: multiplied the number of mole-boxes by how many molecules sit in each box. WHY multiply by $N_A$: one mole is defined as $N_A$ entities, so $2$ moles is simply $N_A$ counted twice — multiplication is repeated counting of the box contents.

$V=n\times V_m=2\times 22.7=45.4\text{L}$ WHAT we did: multiplied moles by the volume one mole takes up. WHY $V_m$ works: Avogadro's law says every ideal gas has the same volume per mole at fixed $T,P$; at STP that per-mole volume is $22.7$ L. Identity of the gas is irrelevant.

Chain the two conversions (this rides grams→moles→molecules on the figure): $N=\frac{m}{M}{N}_A=\frac{9}{18}\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.011\times 10^{23}\text{molecules}$ WHAT we did: first $m/M$ turned grams into moles, then $\times N_A$ turned moles into a count. WHY chain them: $N_A$ cannot eat grams directly, so we must pass through moles first — moles is the only unit the counting step accepts.

$N_{\text{H}}=2\times N_{\text{molecules}}=2\times 3.011\times 10^{23}=6.022\times 10^{23}\text{H atoms}$ WHAT we did: multiplied the molecule count by 2. WHY multiply by 2: each ${\text{H}}_2\text{O}$ contains 2 H atoms, so the atom count is the molecule count scaled by atoms-per-molecule. Nice check: half a mole of water contains exactly one mole of H atoms.

$m_{\text{atom}}=\frac{M}{N_A}=\frac{197}{6.022\times 10^{23}}=3.271\times 10^{-22}\text{g}$ WHAT we did: divided the molar mass by $N_A$. WHY divide: $N_A$ atoms together weigh $197$ g, so splitting that mass equally among all of them gives each atom's share, $197/N_A$. See Molar mass and atomic mass unit.

First undo the count to get moles, then moles to grams (this rides the dashed arrows on the figure): $n=\frac{N}{N_A}=\frac{1.5\times 10^{22}}{6.022\times 10^{23}}=0.02491\text{mol}$ $m=nM=0.02491\times 32=0.7971\text{g}$ WHAT we did: divided the count by $N_A$, then multiplied the moles by $M$. WHY divide first, then multiply: dividing by $N_A$ undoes the counting step — it asks "how many full boxes of $N_A$ do these molecules make?"; multiplying by $M$ then undoes the weighing step — each box weighs $M$ grams. We are simply running the forward pipeline in reverse.

Convert both masses to moles: $n({\text{H}}_2)=\frac{4}{2}=2\text{mol},n({\text{O}}_2)=\frac{4}{32}=0.125\text{mol}$ WHAT we did: turned each mass into moles with $m/M$. WHY: volumes only compare fairly once expressed per molecule count, and $m/M$ is the only route from grams to that count. Same $T,P$ means $V\propto n$ (Avogadro's law), so: $V({\text{O}}_2)=V({\text{H}}_2)\times \frac{n({\text{O}}_2)}{n({\text{H}}_2)}=30\times \frac{0.125}{2}=1.875\text{L}$ WHAT we did: scaled the known volume by the mole ratio. WHY the ratio: at fixed $T,P$, $V/n$ is the same constant for both gases, so $V_1/n_1=V_2/n_2$. Equal masses do not give equal volumes — equal moles do.

$\rho =\frac{M}{V_m}=\frac{16}{22.7}=0.7048\text{g/L}$ WHAT we did: divided the molar mass by the molar volume. WHY $M/V_m$ gives density: density is mass per unit volume. Take exactly one mole as a convenient sample — it weighs $M$ grams (top of the fraction) and, by Avogadro's law, occupies $V_m$ litres at STP (bottom of the fraction). Mass$÷$volume of that one-mole sample is precisely $M/V_m$, and because both scale together the per-litre density is the same for any amount.

Take $100$ g of sample so percentages become grams. Convert each to moles ($n=m/M$): $n_{\text{C}}=\frac{40.0}{12}=3.333,n_{\text{H}}=\frac{6.7}{1}=6.700,n_{\text{O}}=\frac{53.3}{16}=3.331$ WHAT we did: converted each element's mass to moles. WHY moles, not masses: a chemical formula counts atoms, and moles are the only honest atom-count (each mole = $N_A$ atoms); mass ratios hide the true atom ratio because different atoms weigh different amounts. Divide by the smallest ($\approx 3.331$) to get the simplest whole-number ratio: $\text{C}:\frac{3.333}{3.331}\approx 1,\text{H}:\frac{6.700}{3.331}\approx 2,\text{O}:\frac{3.331}{3.331}=1$ WHAT we did: normalised by the smallest mole count. WHY divide by the smallest: it rescales the ratio so the least-abundant atom becomes 1, exposing the small whole numbers a formula needs. Empirical formula: ${\text{CH}}_2\text{O}$. See Empirical and molecular formulas.

Grams → moles → molecules → oxygen atoms (this rides the whole pipeline of the figure): $n=\frac{11}{44}=0.25\text{mol}$ WHAT we did: grams to moles via $m/M$. WHY: later steps only eat moles, so we convert first. $N_{\text{molecules}}=0.25\times 6.022\times 10^{23}=1.5055\times 10^{23}$ WHAT we did: moles to a molecule count via $\times N_A$. WHY: each mole-box holds $N_A$ molecules, so multiply. $N_{\text{O}}=2\times 1.5055\times 10^{23}=3.011\times 10^{23}\text{O atoms}$ WHAT we did: molecules to O atoms via $\times 2$. WHY multiply by 2: each ${\text{CO}}_2$ contains 2 oxygen atoms, so scale the molecule count by atoms-per-molecule. Every intermediate is labelled with its entity so nothing gets swapped.

(i) $n=V/V_m=22.7/22.7=1\text{mol}$. WHAT: volume to moles by dividing by $V_m$. WHY: at STP one mole fills $22.7$ L, so the flask's volume divided by that per-mole volume counts the moles. (ii) $M=m/n=28/1=28\text{g/mol}$. WHAT: mass over moles gives molar mass. WHY: molar mass is grams-per-mole, so dividing the sample's grams by its moles recovers it. A diatomic gas of molar mass $28$ is ${\text{N}}_2$ ($2\times 14=28$). (iii) Molecules $=n{N}_A=6.022\times 10^{23}$; diatomic ⇒ atoms $=2\times 6.022\times 10^{23}=1.2044\times 10^{24}$. WHAT: moles to molecules by $\times N_A$, then to atoms by $\times 2$. WHY: $N_A$ counts molecules per mole; multiplying by 2 accounts for the two atoms in each molecule. All three tools, one flask — the currency exchange behind all Stoichiometry.

$n=m/M$ (mass→moles), $N=n{N}_A$ (moles→count), $V_1/n_1=V_2/n_2$ (volume compare at same $T,P$).