Level 3 — ProductionMatter, Measurement & the Mole

Matter, Measurement & the Mole

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level 3 (Production): Derivations, Reasoning & Explain-Out-Loud

Time limit: 45 minutes Total marks: 60

Show all working. Explain reasoning where asked. Report answers to correct significant figures with units.

Q1. [10 marks] — Deriving the mole bridge from memory

(a) Starting from the definition of molar mass and Avogadro's number, derive an expression that gives the number of atoms $N$ in a sample of a pure element of mass $m$ (g) and molar mass $M$ (g/mol). Explain each conversion step out loud (state what physical idea justifies it). (4)

(b) Use your derived expression to compute the number of atoms in a $0.500\text{ g}$ sample of pure carbon-12. (3)

(c) A single copper atom has mass $1.055\times10^{-22}\text{ g}$ . From first principles (do not just quote $M$ ), derive the molar mass of copper and state it to 4 sig figs. (3)

Q2. [12 marks] — Empirical → molecular formula from scratch

A compound contains, by mass: $40.00\%$ C, $6.71\%$ H, $53.29\%$ O. Its molar mass is $180.16\text{ g/mol}$ .

(a) Derive the empirical formula, showing the full mole-ratio procedure and explaining why dividing by the smallest mole value gives the ratio. (6)

(b) Determine the molecular formula and state its identity if known. (3)

(c) Verify the law of definite proportions is consistent with your molecular formula by predicting the mass % of carbon and comparing. (3)

Q3. [10 marks] — Law of multiple proportions, proved by construction

Two oxides of nitrogen are analysed:

Oxide A: $63.65\%$ N, $36.35\%$ O
Oxide B: $46.68\%$ N, $53.32\%$ O

(a) For a fixed mass of nitrogen, compute the mass of oxygen combining in each oxide, and show the ratio is a simple whole number. Explain how this demonstrates Dalton's law. (7)

(b) Assign plausible formulae to oxides A and B consistent with your ratio. (3)

Q4. [10 marks] — Concentration & dilution reasoning

(a) Derive the dilution relationship $M_1V_1 = M_2V_2$ from the definition of molarity, stating the conservation principle that makes it valid. (3)

(b) You must prepare $250.0\text{ mL}$ of $0.150\text{ M}$ HCl from a $2.00\text{ M}$ stock. Calculate the stock volume needed and describe the procedure out loud (glassware + steps). (4)

(c) A $12.0\text{ ppm}$ (by mass) aqueous solution of $\text{Pb}^{2+}$ has density $1.00\text{ g/mL}$ . Express this concentration in mol/L (molarity). $M(\text{Pb})=207.2\text{ g/mol}$ . (3)

Q5. [10 marks] — Density, molar volume, sig figs

(a) At STP a gas occupies a molar volume of $22.4\text{ L/mol}$ . Derive the density of $\text{CO}_2$ gas at STP, explaining each step. $M(\text{CO}_2)=44.01\text{ g/mol}$ . (4)

(b) A student measures a metal block: mass $= 23.94\text{ g}$ , volume $= 3.0\text{ cm}^3$ . Compute density with correct sig figs and explain why the answer's precision is limited. (3)

Q6. [8 marks] — Explain-out-loud + classification

(a) A mixture of sand, salt, iodine and iron filings must be fully separated. Propose an ordered separation scheme naming each technique and the property it exploits. (5)

(b) Classify each as physical or chemical change with a one-line justification: (i) dry ice subliming, (ii) rusting of the iron filings, (iii) dissolving salt in water. (3)

Answer keyMark scheme & solutions

Q1 (10)

(a) (4)

Number of moles: $n = m/M$ — dividing mass by mass-per-mole cancels grams, leaving mol. (1 idea + 1 mark)
Avogadro: 1 mol contains $N_A = 6.022\times10^{23}$ particles, so multiply moles by $N_A$ . (1)
Combine: $\displaystyle N = \frac{m}{M}\,N_A$ . (1)
Explanation: molar mass is the "counting-by-weighing" bridge; $N_A$ converts moles to a count. (1)

(b) (3) $n = 0.500/12.00 = 0.04167\text{ mol}$ (1); $N = 0.04167\times6.022\times10^{23}$ (1) $= 2.51\times10^{22}$ atoms (1).

(c) (3) $M = (\text{mass of 1 atom})\times N_A = 1.055\times10^{-22}\times6.022\times10^{23}$ (2) $= 63.53\text{ g/mol}$ → 63.53 g/mol (1).

Q2 (12)

(a) (6) Assume 100 g sample:

C: $40.00/12.01 = 3.331\text{ mol}$ (1)
H: $6.71/1.008 = 6.657\text{ mol}$ (1)
O: $53.29/16.00 = 3.331\text{ mol}$ (1)
Divide by smallest (3.331): C = 1, H = 2.00, O = 1 (1)
Why: dividing by smallest expresses each element relative to the least-abundant one, giving the smallest integer atom ratio. (1)
Empirical formula: CH₂O (1)

(b) (3) Empirical mass $= 12.01+2(1.008)+16.00 = 30.03\text{ g/mol}$ (1). $n = 180.16/30.03 = 6.00$ (1). Molecular formula $= \text{C}_6\text{H}_{12}\text{O}_6$ — glucose (1).

(c) (3) Predicted %C $= \frac{6(12.01)}{180.16}\times100 = 39.99\% \approx 40.00\%$ (2), matches given → definite proportions upheld (1).

Q3 (10)

(a) (7) Per 1 g N:

Oxide A: O per g N $= 36.35/63.65 = 0.5711\text{ g}$ (2)
Oxide B: O per g N $= 53.32/46.68 = 1.1423\text{ g}$ (2)
Ratio B:A $= 1.1423/0.5711 = 2.00$ → 1 : 2 (2)
Simple whole-number ratio of O masses for fixed N mass demonstrates Dalton's law of multiple proportions. (1)

(b) (3) A has half the O of B: A = $\text{N}_2\text{O}$ (O:N low), B = NO. Check A: N%=63.65 matches N₂O (28.02/44.02=63.6%) (1.5); B: NO N%=14.01/30.01=46.68% ✓ (1.5).

Q4 (10)

(a) (3) $M=n/V \Rightarrow n = MV$ . On dilution moles of solute unchanged (conservation of amount of substance): $n_1=n_2$ (1) → $M_1V_1=M_2V_2$ (1); principle stated (1).

(b) (4) $V_1 = \frac{M_2V_2}{M_1} = \frac{0.150\times250.0}{2.00} = 18.75\text{ mL}$ ≈ 18.8 mL (2). Procedure: pipette/measure 18.8 mL stock into 250 mL volumetric flask, add water to near mark, mix, fill to mark, invert to homogenise (2).

(c) (3) 12.0 ppm = 12.0 mg Pb per L (since 1 L ≈ 1000 g, 1 ppm = 1 mg/L) (1). $=0.0120\text{ g/L}$ ; molarity $= 0.0120/207.2 = 5.79\times10^{-5}\text{ M}$ (2).

Q5 (10)

(a) (4) $\rho = M/V_m$ because density = mass per volume and 1 mol has mass $M$ occupying $V_m$ (2). $\rho = 44.01/22.4 = 1.965 \approx 1.96\text{ g/L}$ (2).

(b) (3) $\rho = 23.94/3.0 = 7.98 \to$ 8.0 g/cm³ (2). Volume has only 2 sig figs, limiting the answer to 2 sig figs (1).

(c) (3) $1.96\text{ g/L} = 1.96\text{ g}/(10^{-3}\text{ m}^3)$ ; $1\text{ g}=10^{-3}\text{ kg}$ (1) → $1.96\times10^{-3}\text{ kg}/10^{-3}\text{ m}^3$ (1) $= 1.96\text{ kg/m}^3$ (1).

Q6 (8)

(a) (5)

Magnetic separation — remove iron filings (ferromagnetism). (1)
Sublimation — heat to sublime iodine, collect on cold surface (iodine sublimes). (1)
Dissolve in water + filtration — salt dissolves, sand retained on filter (solubility difference). (1)
Evaporation/distillation — recover salt from solution (boiling point difference / non-volatility of salt). (1)

Coherent ordered logic (1)

(b) (3) (i) physical — phase change, CO₂ unchanged (1); (ii) chemical — new substance Fe₂O₃ formed (1); (iii) physical — ions dispersed, recoverable by evaporation (1).

[
{"claim":"Q1b atoms in 0.500 g C-12 ≈ 2.51e22","code":"N=(0.500/12.00)*6.022e23; result = abs(N-2.51e22) < 0.02e22"},
{"claim":"Q1c molar mass of Cu ≈ 63.53 g/mol","code":"M=1.055e-22*6.022e23; result = abs(M-63.53) < 0.1"},
{"claim":"Q2b molecular multiplier = 6 for CH2O to 180.16","code":"emp=12.01+2*1.008+16.00; result = round(180.16/emp)==6"},
{"claim":"Q3a oxygen ratio B:A = 2","code":"a=36.35/63.65; b=53.32/46.68; result = abs(b/a-2.0) < 0.02"},
{"claim":"Q4b stock volume = 18.75 mL","code":"V=0.150*250.0/2.00; result = abs(V-18.75) < 0.01"},
{"claim":"Q4c 12 ppm Pb = 5.79e-5 M","code":"c=0.0120/207.2; result = abs(c-5.79e-5) < 0.05e-5"},
{"claim":"Q5a CO2 density at STP ≈ 1.96 g/L","code":"d=44.01/22.4; result = abs(d-1.96) < 0.01"}
]