HSC Exam Practice

Sample response. Molar mass (MM) is the mass of one mole of a substance, expressed in g mol⁻¹ [1]. One mole is defined as exactly 6.022 × 10²³ particles, and was originally defined so that one mole of carbon-12 has a mass of exactly 12 g — equal to its relative atomic mass [1]. Because all other atomic masses are defined relative to carbon-12, the molar mass of any element in g mol⁻¹ is numerically equal to its relative atomic mass from the periodic table [1].

Marking notes. 1 mark for definition of molar mass including unit; 1 mark for linking mole definition to carbon-12 (12 g = 1 mol); 1 mark for explaining the numeric equivalence via the relative atomic mass scale.

Sample response.

(a) MM(Cl₂) = 2 × 35.450 = 70.900 g mol⁻¹. [2 marks: 1 correct method (subscript × A_r), 1 correct numerical answer with unit]

(b) MM(Fe₂O₃) = 2(55.845) + 3(15.999) = 111.690 + 47.997 = 159.687 g mol⁻¹. [2 marks: 1 correct method with both elements × subscript, 1 correct answer]

(c) MM(Cu(NO₃)₂): expand brackets: 1 Cu + 2 N + 6 O = 63.546 + 2(14.007) + 6(15.999) = 63.546 + 28.014 + 95.994 = 187.554 g mol⁻¹. [2 marks: 1 correct bracket expansion shown, 1 correct answer]

Marking notes. Deduct 1 mark per compound if bracket not expanded or subscript misapplied. Award method mark even if arithmetic error if working shows correct structure.

Sample response. The student did not expand the bracket in (HCO₃)₂. The subscript 2 outside the bracket applies to all atoms inside: Ca(HCO₃)₂ = 1 Ca + 2 H + 2 C + 6 O [1]. The student treated the compound as containing only 1 H, 1 C, and 3 O [1]. Correct calculation: MM = 40.078 + 2(1.008) + 2(12.011) + 6(15.999) = 40.078 + 2.016 + 24.022 + 95.994 = 162.110 g mol⁻¹ [1].

Marking notes. 1 mark for identifying the error (bracket not expanded); 1 mark for stating that the subscript must distribute to all atoms inside the bracket; 1 mark for the correct molar mass. Accept minor rounding.

Sample response. The student incorrectly multiplied mass by molar mass (n = m × MM), which gives units of g × g mol⁻¹ = g² mol⁻¹ — not a meaningful quantity [1]. The correct formula is n = m ÷ MM. MM(CO₂) = 12.011 + 2(15.999) = 44.009 g mol⁻¹. n = 44 ÷ 44.009 = 1.00 mol [1]. Unit check: g ÷ g mol⁻¹ = mol ✓ [1].

Marking notes. 1 mark for identifying the incorrect operation (multiplied instead of divided); 1 mark for correct formula and calculation; 1 mark for unit check showing g ÷ g mol⁻¹ = mol.

Sample response. Using P(1.0 mol, 60 g) and R(4.0 mol, 240 g): gradient = Δmass ÷ Δmol = (240 − 60) ÷ (4.0 − 1.0) = 180 ÷ 3.0 = 60 g mol⁻¹ [1]. The gradient represents the molar mass of urea, with unit g mol⁻¹ [1], because m = n × MM means a graph of m vs n has gradient MM [1]. Calculated MM(CO(NH₂)₂) = 12.011 + 15.999 + 2(14.007) + 4(1.008) = 12.011 + 15.999 + 28.014 + 4.032 = 60.056 g mol⁻¹. Gradient (60 g mol⁻¹) is consistent with calculated MM (60.056 g mol⁻¹) [1].

Marking notes. 1 mark for correct gradient calculation (reading from graph); 1 mark for identifying quantity as molar mass; 1 mark for explaining gradient = MM via m = n × MM; 1 mark for calculated MM with comparison to gradient.

Sample response.

Point A (above the line — mass too high for the given moles): The measured mass is higher than expected for the claimed number of moles. One cause: the urea sample was not fully dried, so residual moisture added to the recorded mass, making the mass appear greater than the true mass of urea. This means the sample contains less urea than the mass suggests, so the n value calculated would underestimate the true amount if it were re-calculated. [2 marks: 1 identify error (wet/impure sample making mass too high); 1 explain how it affects the mass reading]

Point B (below the line — mass too low for the given moles): The measured mass is lower than expected. One cause: some solid was lost from the sample during transfer between containers (e.g. spilling on the bench), so the mass recorded is less than the mass of urea that was actually used to determine n by another method. [2 marks: 1 identify error (mass loss during transfer); 1 explain how it affects the reading]

Marking notes. Accept any valid experimental error for each point, as long as the direction of effect (mass too high / too low) is consistent with the graph and correctly explained. Do not accept errors that affect n rather than mass unless clearly articulated.

Sample Band 6 response. The student’s claim is incorrect. Molar mass is deeply connected to both real atomic structure and to measurable laboratory masses, as the following analysis shows.

Why MM equals A_r numerically: One mole is defined as exactly 6.022 × 10²³ particles. The mole was originally defined so that one mole of carbon-12 atoms has a mass of exactly 12 g, matching its relative atomic mass of 12. Because all other atomic masses are expressed on the same carbon-12 scale, the molar mass of any element in g mol⁻¹ is numerically equal to its relative atomic mass. This is not coincidence — it is a consequence of how the mole was defined. [2 marks]

Connecting molar mass to measurable mass: A laboratory balance measures mass in grams. Molar mass (in g mol⁻¹) is the conversion factor between grams and moles via n = m ÷ MM. This means that a chemist who reads a balance can directly determine the amount of substance in a sample — a real, physical quantity. Without molar mass, a balance reading is just a number in grams with no direct chemical interpretation. [2 marks]

Worked example 1 — monatomic element (iron, Fe): A_r(Fe) = 55.845. MM(Fe) = 55.845 g mol⁻¹. If a sample of iron has mass 111.69 g: n = 111.69 ÷ 55.845 = 2.000 mol. This tells us the sample contains exactly 2 × 6.022 × 10²³ = 1.204 × 10²⁴ iron atoms — a real, countable (in principle) quantity. [1 mark for correct worked example with MM and n]

Worked example 2 — compound with brackets (calcium hydroxide, Ca(OH)₂): Expand bracket: 1 Ca + 2 O + 2 H = 40.078 + 2(15.999) + 2(1.008) = 40.078 + 31.998 + 2.016 = 74.092 g mol⁻¹. For a 37.05 g sample: n = 37.05 ÷ 74.092 = 0.500 mol. This is a direct link from a weighable mass on a balance to a chemically useful amount. [1 mark for correct worked example with bracket expansion and n calculation]

Evaluation: The student’s claim is incorrect on both counts. First, molar mass is not just a “number you look up” — it is derived from the atomic masses of real atoms, defined on the carbon-12 scale, which itself is tied to physical measurements of isotopic mass. Second, molar mass is directly connected to what you can weigh: it is the factor that converts grams (balance reading) to moles (amount of substance used in all stoichiometric calculations). Without molar mass, quantitative chemistry in the laboratory would be impossible. [2 marks: 1 for clear evaluation verdict with justification; 1 for addressing both parts of the claim explicitly]

Marking criteria (9 marks): 1 = defines molar mass correctly (mass of 1 mol, unit g mol⁻¹); 1 = explains carbon-12 standard link to numeric equivalence of A_r and MM; 1 = explains connection to mole definition; 1 = describes role as conversion factor between g and mol (n = m ÷ MM); 1 = worked example 1 with correct MM of monatomic element and n calculation; 1 = worked example 2 with correct bracket expansion and n calculation; 1 = evaluation verdict (incorrect); 1 = addresses “just a number” part with physical/atomic argument; 1 = addresses “nothing to weigh” part with balance/calculation argument. Deduct if examples are missing, if unit analysis is absent, or if evaluation is unsupported.