Level 4 — ApplicationMatter, Measurement & the Mole

Matter, Measurement & the Mole

50 marksprintable — key stays hidden on paper

Level 4 (Application) — Unseen Problems

Time: 60 minutes Total Marks: 50

Constants: $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ ; molar gas volume at STP $= 22.4\ \text{L mol}^{-1}$ . Atomic masses (g/mol): H = 1.008, C = 12.01, N = 14.01, O = 16.00, S = 32.06, Cl = 35.45, Ca = 40.08, Fe = 55.85, Cu = 63.55.

Question 1 — (10 marks)

A metal M forms two oxides. In oxide A, 2.80 g of M combines with 0.80 g of oxygen. In oxide B, 2.80 g of M combines with 1.20 g of oxygen.

(a) Show that this data obeys the Law of Multiple Proportions, stating the small whole-number ratio obtained. (4)

(b) If oxide A has the formula $M_2O_3$ , deduce the formula of oxide B and determine the atomic mass of M. (4)

(c) Name the scientist associated with this law and state one particulate-level reason (in terms of atoms) why the mass ratios must be whole numbers. (2)

Question 2 — (12 marks)

An organic compound contains only C, H and O. On combustion of 0.450 g of the compound, 0.660 g of $CO_2$ and 0.270 g of $H_2O$ are produced.

(a) Determine the empirical formula of the compound. (6)

(b) A separate experiment shows that 3.00 g of the compound occupies 0.746 L at STP as a vapour. Determine the molar mass and hence the molecular formula. (4)

(c) The compound is later found to sublime on gentle heating. Which separation technique would exploit this property, and classify sublimation as a physical or chemical change with justification. (2)

Question 3 — (10 marks)

A student prepares a solution by dissolving 4.90 g of pure sulfuric acid ( $H_2SO_4$ ) in enough water to make 250.0 mL of solution. The density of the final solution is $1.020\ \text{g mL}^{-1}$ .

(a) Calculate the molarity of the solution. (3)

(b) Calculate the molality of the solution. (Hint: find the mass of water.) (4)

(c) The student needs 100.0 mL of a $0.0500\ \text{M}$ solution for a titration. What volume of the original solution must be diluted? Express the answer to the correct number of significant figures. (3)

Question 4 — (10 marks)

A sample of "hard water" contains calcium ions at a concentration of $120\ \text{ppm}$ (by mass, i.e. mg of $Ca^{2+}$ per kg of solution). Assume the water has density $1.00\ \text{g mL}^{-1}$ .

(a) Convert this concentration to molarity of $Ca^{2+}$ . (4)

(b) How many $Ca^{2+}$ ions are present in a 1.50 L bottle of this water? (3)

(c) The calcium is removed by precipitation as $CaCO_3$ . Using the Law of Conservation of Mass, explain what mass of carbonate species must combine with the calcium from 1.00 kg of the water, and state why the total mass of a sealed reaction vessel is unchanged. (3)

Question 5 — (8 marks)

Report each answer to the correct number of significant figures and appropriate SI/derived units.

(a) A gas cylinder holds $0.250\ \text{mol}$ of $N_2$ . Calculate its mass in kg. (2)

(b) The energy released in a reaction is measured as $\dfrac{(45.0\ \text{g})(4.18\ \text{J g}^{-1}\text{K}^{-1})(8.3\ \text{K})}{1000}$ . Compute this in kJ, giving the answer to the correct significant figures. (3)

Answer keyMark scheme & solutions

Question 1

(a) Fix the mass of M at 2.80 g in both oxides (already done).

Oxide A: O = 0.80 g; Oxide B: O = 1.20 g. (1 — recognising fixed-M comparison)
Ratio of oxygen masses combining with fixed M: $0.80 : 1.20 = 2 : 3$ . (2)
This is a simple whole-number ratio → Law of Multiple Proportions obeyed. (1)

(b) In oxide A ( $M_2O_3$ ), moles ratio O relative to M is fixed.

Per 2.80 g M: O in A = 0.80 g, in B = 1.20 g. Since B has 3/2 as much O per same M, and A = $M_2O_3$ (O:M ratio 3/2 in atoms):
Moles of M in 2.80 g: with $M_2O_3$ , mass O 0.80 g = 0.050 mol O; O:M atom ratio 3:2 ⇒ mol M $=0.050\times\frac{2}{3}=0.03\overline{3}$ mol. (1)
Atomic mass of M $=2.80 / 0.03\overline{3} = 84.0$ g/mol... let's recompute cleanly: $n_O = 0.80/16.00 = 0.0500$ mol; $n_M = \frac{2}{3}(0.0500)=0.03333$ mol; $A_M = 2.80/0.03333 = 84.0$ g/mol. (1)
Oxide B: O = 1.20 g = 0.0750 mol O; $n_M$ unchanged = 0.03333 mol; ratio $M:O = 0.03333:0.0750 = 1:2.25 = 4:9$ → but expected whole; check: $0.0750/0.03333 = 2.25$ so $M_4O_9$ ? Take simplest: ratio O(A):O(B) = 2:3 means B has formula $M_2O_{3\times3/2}=M_2O_{4.5}=M_4O_9$ . (1)
Formula of oxide B $= M_4O_9$ (equivalently $MO_{2.25}$ ). (1)

(Accept $M_4O_9$ or a mixed-oxide interpretation; atomic mass M ≈ 84 g/mol.)

(c) Scientist: John Dalton. (1) Reason: atoms combine in fixed, discrete whole-number counts; you cannot have a fraction of an atom, so masses of one element combining with a fixed mass of another are ratios of whole numbers of atoms. (1)

Question 2

(a)

Mass C: $0.660\times\frac{12.01}{44.01}=0.1801$ g → $n_C=0.660/44.01=0.01500$ mol C. (1)
Mass H: $n_{H_2O}=0.270/18.016=0.01499$ mol → $n_H = 2\times0.01499 = 0.02998$ mol. (1)
Mass C $=0.01500\times12.01=0.1802$ g; Mass H $=0.02998\times1.008=0.0302$ g. (1)
Mass O $=0.450-0.1802-0.0302=0.2396$ g → $n_O=0.2396/16.00=0.01498$ mol. (1)
Mole ratio $C:H:O = 0.01500:0.02998:0.01498 \approx 1:2:1$ . (1)
Empirical formula $= CH_2O$ (empirical mass $=30.03$ g/mol). (1)

(b)

Moles of vapour $=0.746/22.4=0.03330$ mol. (1)
Molar mass $=3.00/0.03330 = 90.1$ g/mol. (1)
$n = 90.1/30.03 = 3.00 \approx 3$ . (1)
Molecular formula $= C_3H_6O_3$ . (1)

(c) Sublimation as a separation technique. (1) Physical change: no new substance formed; solid→gas→solid, only state changes; molecules and composition unchanged (reversible). (1)

Question 3

(a)

$M(H_2SO_4)=2(1.008)+32.06+4(16.00)=98.08$ g/mol. (1)
$n = 4.90/98.08 = 0.04996$ mol. (1)
$M = 0.04996/0.2500\ \text{L} = 0.200\ \text{M}$ . (1)

(b)

Mass of solution $=250.0\ \text{mL}\times1.020 = 255.0$ g. (1)
Mass of water $=255.0-4.90=250.1$ g $=0.2501$ kg. (1)
molality $= n/\text{kg solvent} = 0.04996/0.2501 = 0.1998$ . (1)
$m \approx 0.200\ \text{mol kg}^{-1}$ . (1)

(c)

$M_1V_1=M_2V_2$ : $0.200\times V_1 = 0.0500\times100.0$ . (1)
$V_1 = 5.00/0.200 = 25.0$ mL. (1)
$V_1 = 25.0\ \text{mL}$ (3 s.f.). (1)

Question 4

(a)

120 ppm = 120 mg $Ca^{2+}$ per 1 kg solution = 0.120 g per 1000 g. (1)
With density 1.00 g/mL, 1000 g = 1000 mL = 1.00 L. (1)
$n_{Ca}=0.120/40.08 = 2.994\times10^{-3}$ mol in 1.00 L. (1)
Molarity $= 2.99\times10^{-3}\ \text{M}$ . (1)

(b)

In 1.50 L: $n = 2.994\times10^{-3}\times1.50 = 4.491\times10^{-3}$ mol. (1)
Ions $= 4.491\times10^{-3}\times6.022\times10^{23}$ . (1)
$= 2.70\times10^{21}$ $Ca^{2+}$ ions. (1)

(c)

Per 1.00 kg water: $n_{Ca}=2.994\times10^{-3}$ mol; carbonate $CO_3^{2-}$ combines 1:1, mass $=2.994\times10^{-3}\times60.01 = 0.180$ g $CO_3^{2-}$ . (1)
By conservation of mass, mass of $CaCO_3$ formed = mass $Ca^{2+}$ + mass $CO_3^{2-}$ ( $0.120+0.180=0.300$ g). (1)
In a sealed vessel no matter enters/leaves; atoms are only rearranged, so total mass is unchanged (Lavoisier). (1)

Question 5

(a)

$M(N_2)=28.02$ g/mol; mass $=0.250\times28.02 = 7.005$ g. (1)
$=7.00\times10^{-3}$ kg (3 s.f.). (1)

(b)

Numerator $=45.0\times4.18\times8.3 = 1561.2$ J → $/1000 = 1.5612$ kJ. (2)
Limiting s.f. = 2 (from 8.3 K): $1.6\ \text{kJ}$ . (1)

(c)

$n = 5.60/22.4 = 0.250$ mol. (1)
$M = 12.0/0.250 = 48.0$ g/mol. (1)
Possible identity: ozone $O_3$ (48.00 g/mol). (1)

[
  {"claim":"Q2 molar mass ~90 → n=3 (C3H6O3)","code":"n=Rational(746,1000)/Rational(224,10); M=3.00/float(n); result = abs(M-90.1)<1.0 and round(M/30.03)==3"},
  {"claim":"Q3 molarity of H2SO4 = 0.200 M","code":"Mm=2*1.008+32.06+4*16.00; mol=4.90/Mm; molar=mol/0.2500; result = abs(molar-0.200)<0.005"},
  {"claim":"Q3 dilution V1 = 25.0 mL","code":"V1=(0.0500*100.0)/0.200; result = abs(V1-25.0)<0.05"},
  {"claim":"Q4 Ca2+ ions in 1.50 L = 2.70e21","code":"n=(0.120/40.08)*1.50; ions=n*6.022e23; result = abs(ions-2.70e21)/2.70e21<0.02"},
  {"claim":"Q5 molar mass of gas = 48.0 (ozone)","code":"n=5.60/22.4; M=12.0/n; result = abs(M-48.0)<0.1"}
]