Exercises — The mole concept — counting by weighing

WHAT we want: the batch count $n$. WHICH tool: we have $m$ and $M$, so we use the left conversion $n=m/M$ — because "how many batches" is total mass divided by the mass of one batch. $n=\frac{m}{M}=\frac{46}{23}=2\text{mol}.$ Sanity check: $46$ is exactly twice $23$, so exactly two moles. ✅

WHICH tool: we go from moles to particles, so we move right on the map: $N=n{N}_A$. $N=n{N}_A=3\times 6.022\times 10^{23}=1.807\times 10^{24}\text{molecules}.$ Why multiply? Each mole is $N_A$ particles by definition, so 3 moles is $3\times N_A$.

Step 1 — build $M$. Add up each element's atomic mass times how many of it appears: $M=6(12)+12(1)+6(16)=72+12+96=180{\text{g mol}}^{-1}.$ Step 2 — moles. $n=m/M=18/180=0.1\text{mol}$. Step 3 — molecules. $N_{\text{mol}}=0.1\times 6.022\times 10^{23}=6.022\times 10^{22}$ molecules. Step 4 — atoms. One glucose molecule has $6+12+6=24$ atoms, so $N_{\text{atoms}}=24\times 6.022\times 10^{22}=1.445\times 10^{24}\text{atoms}.$ Why the ×24? Molecules ≠ atoms — we must multiply by atoms-per-molecule.

Step 1 — moles from particles. Move left: $n=N/N_A$. $n=\frac{1.5\times 10^{24}}{6.022\times 10^{23}}=2.491\text{mol}.$ Step 2 — build $M$. $M({\text{CO}}_2)=12+2(16)=44{\text{g mol}}^{-1}$. Step 3 — mass. $m=nM=2.491\times 44=109.6\text{g}$. Why $m=nM$? It is $n=m/M$ rearranged: multiply both sides by $M$.

Predict: lighter atoms ⇒ more of them per gram, so Ca (lighter) should win. Verify: both are single atoms, so atoms scale with moles $n=m/M$. $n_{\text{Ca}}=\frac{5}{40}=0.125\text{mol},n_{\text{Fe}}=\frac{5}{56}=0.0893\text{mol}.$ $\text{Ca}>\text{Fe}$, so calcium has more atoms. ✅ Prediction confirmed. Ratio $=0.125/0.0893=1.4=56/40$ — exactly the inverse of the mass ratio, which makes sense: same grams, atom count scales as $1/M$.

WHICH tool: one mole ($63.5\text{g}$) is shared among $N_A$ atoms, so $m_{\text{atom}}=\frac{M}{N_A}=\frac{63.5}{6.022\times 10^{23}}=1.054\times 10^{-22}\text{g}.$ Why divide? The mole's total mass split equally among its $N_A$ atoms gives one atom's mass — this is counting by weighing run in reverse.

Step 1 — count O atoms per formula unit. The $({\text{SO}}_4{)}_3$ means 3 sulfate groups, each with 4 oxygens: $4\times 3=12$ oxygen atoms per formula unit. Step 2 — total O count. $N_{\text{O}}=2\text{mol}\times 12\times N_A=24\times 6.022\times 10^{23}=1.445\times 10^{25}\text{atoms}.$ Why ×12? The subscript-3 outside the bracket multiplies everything inside it, so oxygen's count is $4\times 3$, not $4$.

Step 1 — carbon moles. $n_{\text{C}}=N/N_A=\frac{1.2044\times 10^{24}}{6.022\times 10^{23}}=2\text{mol}$. Step 2 — link C to whole formula. Each ${\text{CaCO}}_3$ unit has exactly one carbon, so moles of ${\text{CaCO}}_3$ = moles of C = $2\text{mol}$. Step 3 — build $M$. $M({\text{CaCO}}_3)=40+12+3(16)=100{\text{g mol}}^{-1}$. Step 4 — mass. $m=nM=2\times 100=200\text{g}$. Why step 2 works: counting one signature atom (C) lets us count the whole formula unit, because there is exactly one C per unit.

Check the moles. $M({\text{CO}}_2)=44$, so $n=m/M=4.4/44=0.1\text{mol}$. Wait — the problem claims $0.2\text{mol}$. Let us test it: $0.1\ne 0.2$, so the two statements are inconsistent; the mass $4.4\text{g}$ corresponds to $0.1\text{mol}$, not $0.2$. Resolve using the reliable datum (the mass): trust $n=0.1\text{mol}$. Oxygen atoms: each ${\text{CO}}_2$ has 2 O atoms. $N_{\text{O}}=0.1\times 2\times N_A=0.2\times 6.022\times 10^{23}=1.2044\times 10^{23}\text{atoms}.$ Lesson: always recompute from a primary measurement (mass) rather than trusting a quoted mole value.

Step 1 — assume 100 g, so percentages become grams: C $=40.0\text{g}$, H $=6.7\text{g}$, O $=53.3\text{g}$. Step 2 — grams → moles (divide each by its $M$): $n_{\text{C}}=\frac{40.0}{12}=3.33,n_{\text{H}}=\frac{6.7}{1}=6.70,n_{\text{O}}=\frac{53.3}{16}=3.33\text{mol}.$ Step 3 — divide by the smallest ($3.33$) to get the ratio: $\text{C}:\text{H}:\text{O}=\frac{3.33}{3.33}:\frac{6.70}{3.33}:\frac{3.33}{3.33}=1:2.01:1\approx 1:2:1.$ Empirical formula: ${\text{CH}}_2\text{O}$. (This is formaldehyde's formula, and glucose's empirical formula.) Why divide by moles, not mass? Formulae count atoms, and moles are proportional to atom counts — mass ratios are distorted by the different atomic masses.

Step 1 — mass of pure NaCl. $m_{\text{pure}}=0.90\times 25=22.5\text{g}$. Step 2 — moles of NaCl. $n=m/M=22.5/58.5=0.3846\text{mol}$. Step 3 — sodium ions. Each $\text{NaCl}$ formula unit gives exactly one ${\text{Na}}^{+}$ ion, so $N_{{\text{Na}}^{+}}=0.3846\times 6.022\times 10^{23}=2.316\times 10^{23}\text{ions}.$ Why start with purity? Only the pure NaCl contributes Na⁺; the impurity's mass must be removed before running the mole chain.

Compute atoms $=(m/M)\times (\text{atoms per molecule})\times N_A$. The $N_A$ is common, so compare the number $\frac{m}{M}\times (\text{atoms/molecule})$ — call it the atom-index.

• ${\text{H}}_2$: $M=2$, atoms/molecule $=2$ → index $=\frac{10}{2}\times 2=10$.
• ${\text{O}}_2$: $M=32$, atoms/molecule $=2$ → index $=\frac{10}{32}\times 2=0.625$.
• ${\text{H}}_2\text{O}$: $M=18$, atoms/molecule $=3$ → index $=\frac{10}{18}\times 3=1.667$. Ranking: ${\text{H}}_2(10)>{\text{H}}_2\text{O}(1.667)>{\text{O}}_2(0.625)$. In actual atoms, ${\text{H}}_2$ has $10\times N_A=6.022\times 10^{24}$ atoms — the clear winner. Why ${\text{H}}_2$ dominates: tiny molar mass and light atoms means enormously many molecules per gram.