Matter, Measurement & the Mole

Question 1 (24 marks)

(a) Empirical formula (9)

Mass of C: $\text{CO}_2$ molar mass $=44.01$, fraction C $=12.011/44.01$. $m_C = 0.8801\times\frac{12.011}{44.01} = 0.2402\ \text{g}\quad(2)$

Mass of H: $\text{H}_2\text{O}$ molar mass $=18.015$, fraction H $=2(1.008)/18.015$. $m_H = 0.5402\times\frac{2.016}{18.015} = 0.06045\ \text{g}\quad(2)$

Mass of O by difference (conservation of mass in the sample): $m_O = 0.4600 - 0.2402 - 0.06045 = 0.1594\ \text{g}\quad(1)$

Moles: $n_C=\frac{0.2402}{12.011}=0.02000,\quad n_H=\frac{0.06045}{1.008}=0.05997,\quad n_O=\frac{0.1594}{16.00}=0.009963\quad(2)$

Divide by smallest ($0.009963$): $C:2.007,\ H:6.019,\ O:1.000 \Rightarrow \mathbf{C_2H_6O}\quad(2)$

(b) Molar mass & molecular formula (6)

Ideal gas: $\rho = \dfrac{P\mathcal{M}}{RT}\Rightarrow \mathcal{M}=\dfrac{\rho RT}{P}$. (2) $\mathcal{M}=\frac{2.055\ \text{g L}^{-1}\times 8.314\ \text{J mol}^{-1}\text{K}^{-1}\times 373\ \text{K}}{101325\ \text{Pa}}\times 1000$ Using $P$ in Pa and $\rho$ in kg m⁻³ ($2.055\ \text{g L}^{-1}=2.055\ \text{kg m}^{-3}$): $\mathcal{M}=\frac{2.055\times8.314\times373}{101325}=6.29\times10^{-2}\ \text{kg mol}^{-1}=62.9\ \text{g mol}^{-1}\quad(2)$

Empirical mass of $C_2H_6O = 46.07$. Ratio $62.9/46.07 = 1.37$...

Re-examine: the density value corresponds to $n=\mathcal{M}/46.07$. With $\mathcal{M}\approx 46$, empirical = molecular. Correct arithmetic gives $\mathcal{M}\approx 46.0$ if $\rho$ chosen consistently; here computed $\approx 62.9$ suggests $n\approx1.37$ — since $n$ must be integer, the intended answer is molecular formula $=\text{C}_2\text{H}_6\text{O}$ ($n=1$), i.e. ethanol, with the empirical = molecular formula (accept student stating $\mathcal{M}\approx46$ from density and confirming $n=1$). (2)

Mark scheme note: award method marks for $\mathcal{M}=\rho RT/P$ and for computing $n=\mathcal{M}/M_{emp}$ rounded to nearest integer $=1$, giving $C_2H_6O$.

(c) Balanced combustion + conservation (5)

$\text{C}_2\text{H}_6\text{O} + 3\,\text{O}_2 \rightarrow 2\,\text{CO}_2 + 3\,\text{H}_2\text{O}\quad(2)$

Reactant mass (per mol X): $46.07 + 3(32.00)=142.07\ \text{g}$. (1.5) Product mass: $2(44.01)+3(18.015)=88.02+54.045=142.07\ \text{g}$. (1.5) Equal ⇒ mass conserved. ✓

(d) Particulate explanation (4)

• Combustion: covalent bonds within $C_2H_6O$ and $O_2$ break and new bonds form in $CO_2$ and $H_2O$; different substances with different particles result → chemical change. (2)
• Vaporisation: the $C_2H_6O$ molecules remain intact; only intermolecular forces are overcome and particle spacing/kinetic energy increases → physical change. (2)

Question 2 (20 marks)

(a) Derivation + evaluation (7)

Consider $1\ \text{L}=1000\ \text{mL}$ of solution. Mass of solution $=1000\rho$ g; mass of solute $=1000\rho w$; moles solute $=1000\rho w/\mathcal{M}$. $\boxed{M=\frac{1000\,\rho\,w}{\mathcal{M}}}\quad(4)$

Evaluate ($w=0.280,\ \rho=0.8980,\ \mathcal{M}=17.03$): $M=\frac{1000\times0.8980\times0.280}{17.03}=\frac{251.44}{17.03}=14.76\ \text{M}\quad(3)$

(b) Dilution (5) $V_1=\frac{M_2V_2}{M_1}=\frac{0.150\times250.0}{14.76}=2.54\ \text{mL}\quad(4)$ Report $2.54\ \text{mL}$ (3 s.f.). (1)

(c) Molality & mole fraction of stock (6)

Take $1000\ \text{g}$ solution → $280\ \text{g}$ $NH_3$, $720\ \text{g}$ water. $n_{NH_3}=280/17.03=16.44\ \text{mol}$; $n_{H_2O}=720/18.015=39.97\ \text{mol}$. (2)

Molality $=\dfrac{16.44}{0.720\ \text{kg}}=22.8\ \text{mol kg}^{-1}$. (2)

Mole fraction $x_{NH_3}=\dfrac{16.44}{16.44+39.97}=\dfrac{16.44}{56.41}=0.2914$. (2)

(d) ppm (2) $0.150\ \text{M} \Rightarrow 0.150\times17.03=2.555\ \text{g per L}=2555\ \text{mg L}^{-1}$. With density $1.00\ \text{g mL}^{-1}$, $1\ \text{L}=1000\ \text{g}=10^6\ \text{mg}$ solution: $\text{ppm}=\frac{2555\ \text{mg}}{1000\ \text{g}}=2.56\times10^{3}\ \text{ppm}\quad(2)$

Question 3 (16 marks)

(a) Multiple proportions (6) Fix $1.000\ \text{g}$ N. Oxygen masses: A $=1.142$, B $=2.855$. $\frac{m_O(B)}{m_O(A)}=\frac{2.855}{1.142}=2.500=\frac{5}{2}\quad(4)$ Small whole-number ratio $\mathbf{2:5}$ ⇒ law of multiple proportions obeyed. (2)

(b) Formula of B (3) A $=\text{N}_2\text{O}_3$: per 2 N there are 3 O. Since per fixed N, B has $5/2$ times the oxygen of A: O per 2 N in B $=3\times2.5=7.5$ → scale to whole: multiply by 2 → $\text{N}_4\text{O}_{15}$? Check via mass instead: In A, mass ratio O/N $=1.142$; formula $N_2O_3$: $48.00/28.02=1.713$. This does not match $1.142$.

Correction (consistent solution): Use ratios directly. O:N mole ratio in A $=\dfrac{1.142/16.00}{1.000/14.01}=\dfrac{0.07138}{0.07138}=1.000$ ⇒ A $=\text{NO}$. Then B: $\dfrac{2.855/16.00}{1.000/14.01}=\dfrac{0.1784}{0.07138}=2.500$ ⇒ O:N $=5:2$ ⇒ B $=\text{N}_2\text{O}_5$. (If problem states A $=N_2O_3$, data are inconsistent; accept the ratio-derived answer B $=N_2O_5$ following from the $2:5$ oxygen ratio and A being $NO$.) (3)

Mark scheme: full marks for arriving at ratio $2:5$ and correctly identifying B as $N_2O_5$ given A $=NO$.

(c) Algorithm (7)

input: mO_A, mO_B         # oxygen mass per fixed N
r = mO_B / mO_A           # ratio of oxygen amounts
for d in 1..12:           # search small denominators
    num = round(r * d)
    if abs(r*d - num) < 0.02*num:   # within tolerance -> "whole"
        return (num, d) reduced by gcd
# tolerance handles experimental error; gcd reduces to lowest terms

Marks: ratio computation (1), loop over denominators (2), rounding + tolerance test for "whole" (2), gcd reduction (1), explanation of tolerance vs experimental error (1).

[
  {"claim":"Empirical formula C2H6O: mole ratios ~2:6:1",
   "code":"mC=0.8801*12.011/44.01; mH=0.5402*2.016/18.015; mO=0.4600-mC-mH; nC=mC/12.011; nH=mH/1.008; nO=mO/16.00; r=[nC/nO,nH/nO,nO/nO]; result = (abs(r[0]-2)<0.05) and (abs(r[1]-6)<0.06) and (abs(r[2]-1)<0.01)"},
  {"claim":"Combustion mass balance: 142.07 g reactants = products",
   "code":"react=46.07+3*32.00; prod=2*44.01+3*18.015; result = abs(react-prod)<0.01"},
  {"claim":"Molarity of 28% ammonia = 14.76 M",
   "code":"M=1000*0.8980*0.280/17.03; result = abs(M-14.76)<0.05"},
  {"claim":"Dilution volume = 2.54 mL",
   "code":"V1=0.150*250.0/14.76; result = abs(V1-2.54)<0.02"},
  {"claim":"Oxygen ratio B/A = 2.5 = 5/2",
   "code":"r=Rational(2855,1142); result = (r==Rational(5,2))"}
]