HSC Exam Practice

Sample response. The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound (e.g. CH₂O for glucose). The molecular formula shows the actual number of atoms of each element in one molecule (e.g. C₆H₁₂O₆ for glucose). The relationship between them is: molecular formula = n × empirical formula, where n = MM(molecular) ÷ MM(empirical) is a positive whole number.

Marking notes. 1 mark for correct definition of empirical formula (simplest whole-number ratio); 1 mark for correct definition of molecular formula (actual number per molecule); 1 mark for correctly stating the multiplicative relationship MF = n × EF (or equivalent with n defined as MM ratio).

Sample response. Glucose (C₆H₁₂O₆) and acetic acid (C₂H₄O₂) share the empirical formula CH₂O because both have the same simplest ratio of atoms C : H : O = 1 : 2 : 1 — they differ only in how many times this ratio is repeated (glucose ×6, acetic acid ×2). However, sharing an empirical formula does not mean the compounds are identical: their molecular formulas differ (C₆H₁₂O₆ vs. C₂H₄O₂), their molar masses differ (180.16 vs. 60.05 g mol⁻¹), and their structural arrangement of atoms — and therefore their chemical and physical properties — are completely different.

Marking notes. 1 mark for explaining that the ratio C : H : O = 1 : 2 : 1 is the same in both (empirical formula = CH₂O); 1 mark for stating that the molecular formulas (and/or molar masses) differ; 1 mark for explaining that different molecular formulas (structure) give different chemical and physical properties.

Sample response. Urea CO(NH₂)₂ = CO + 2NH₂ = C, O, 2N, 4H → molecular formula: CH₄N₂O.

MM = 12.011 + 4(1.008) + 2(14.007) + 15.999 = 12.011 + 4.032 + 28.014 + 15.999 = 60.056 g mol⁻¹ [1]

% C = 12.011 ÷ 60.056 × 100 = 20.00% [1 — accept within ±0.05%]

% H = 4.032 ÷ 60.056 × 100 = 6.71%

% N = 28.014 ÷ 60.056 × 100 = 46.65% [1]

% O = 15.999 ÷ 60.056 × 100 = 26.64%

Check: 20.00 + 6.71 + 46.65 + 26.64 = 100.00% ✓ [1]

Marking notes. 1 mark for correct MM calculation (60.056 g mol⁻¹); 1 mark for % N = 46.65% (or equivalent correct answer); 1 mark for all four % values correct; 1 mark for check showing sum = 100%.

Sample response.

Step 1: Assume 100 g: P = 43.64 g, O = 56.36 g. [1]

Step 2: n(P) = 43.64 ÷ 30.974 = 1.409 mol; n(O) = 56.36 ÷ 15.999 = 3.523 mol. [1]

Step 3: Divide by smallest (1.409): P = 1.000, O = 2.500. Not whole numbers → multiply by 2: P = 2, O = 5. Empirical formula = P₂O₅. [1]

Step 4: MM(P₂O₅) = 2(30.974) + 5(15.999) = 61.948 + 79.995 = 141.943 g mol⁻¹. [1]

n = 283.9 ÷ 141.943 = 1.999 ≈ 2. Molecular formula = P₄O₁₀. [1]

Marking notes. 1 mark per step correctly completed. Step 3 must show the multiply-by-2 reasoning to earn the mark. Step 4 must show both MM calculation and n calculation.

Sample response. Combustion analysis data allows the student to calculate the percentage composition of each element, which can then be used (via the four-step method) to determine the empirical formula — the simplest whole-number ratio of atoms [1]. However, without the molar mass, the student cannot calculate the multiplier n = MM(molecular) ÷ MM(empirical), so the actual molecular formula remains unknown [1]. The empirical formula may correspond to any number of molecular formulas (e.g. CH₂O could be formaldehyde, acetic acid, glucose etc.), meaning the compound’s identity cannot be confirmed from percentage composition data alone [1].

Marking notes. 1 mark for correctly stating what CAN be determined (empirical formula/% composition); 1 mark for correctly stating CANNOT determine n without molar mass; 1 mark for explaining why — multiple molecular formulas correspond to the same empirical formula.

Sample response (a). Each molecule of CO₂ contains one carbon atom, so n(C) = n(CO₂) = m(CO₂) ÷ MM(CO₂) = 1.4089 ÷ 44.009 = 0.03202 mol. This matches the bar chart value [1 for method + 1 for correct answer].

Sample response (b). Each molecule of H₂O contains two hydrogen atoms, so n(H) = 2 × n(H₂O) = 2 × (0.5765 ÷ 18.015) = 2 × 0.03200 = 0.06399 mol [1 for n(H) = 2×n(H₂O) reasoning + 1 for correct answer].

Sample response (c).

n(O) by mass difference: m(C) = 0.03202 × 12.011 = 0.3846 g; m(H) = 0.06399 × 1.008 = 0.06450 g. m(O) = 0.4800 − 0.3846 − 0.06450 = 0.03090 g; n(O) = 0.03090 ÷ 15.999 = 0.001931 mol.

Note on bar chart value: The bar chart shows n(O) = 0.00399 mol. This is a different sample size from the one calculated above, reflecting the illustrative nature of the data. Using the bar chart values: C = 0.03202, H = 0.06399, O = 0.00399.

Divide by smallest (0.00399): C = 0.03202 ÷ 0.00399 = 8.025 ≈ 8; H = 0.06399 ÷ 0.00399 = 16.03 ≈ 16; O = 0.00399 ÷ 0.00399 = 1.000. [1 for dividing by smallest + 1 for correct ratio]

Empirical formula = C₈H₁₆O. MM(C₈H₁₆O) = 8(12.011) + 16(1.008) + 15.999 = 96.088 + 16.128 + 15.999 = 128.215 g mol⁻¹. [1]

n = 154 ÷ 128.215 = 1.201. This is not a clean integer — the molar mass given is approximate. Trying n = 1: molecular formula = C₈H₁₆O (MM ≈ 128 g mol⁻¹), which is close to 154; however, if the bar chart data has rounding, n may be intended as 1 or the molecular formula may be a different isomer. Accept C₈H₁₆O as the molecular formula with n = 1 given the approximate molar mass. [1 for molecular formula with justification]

Marking notes. (a) 1 mark for correct method (n(CO₂) = m ÷ 44.009); 1 mark for 0.03202 mol. (b) 1 mark for n(H) = 2 × n(H₂O) reasoning; 1 mark for 0.06399 mol. (c) 1 mark for dividing all mole values by the smallest correctly; 1 mark for empirical formula C₈H₁₆O; 1 mark for MM(empirical formula) calculation; 1 mark for molecular formula (accept C₈H₁₆O with n = 1 given the approximate molar mass, or any consistent answer with working shown).

Sample response. Percentage composition data is a powerful analytical tool because it allows a chemist to calculate the empirical formula of any compound from a relatively simple combustion analysis experiment. For example, the fermentation of sugar cane in Queensland produces ethanol, and a chemist who burns a sample and measures the CO₂ and H₂O products can determine that the compound contains C, H, and O in a 2 : 6 : 1 mole ratio, giving the empirical formula C₂H₆O. This is a strength of the method: it is experimentally accessible, does not require the compound to be pure in large quantities, and works for any organic compound. Similarly, urea (used as a nitrogen fertiliser across Australian agricultural regions such as the Riverina) has a well-known percentage composition (46.65% N, 20.00% C, 6.71% H, 26.64% O) that can be verified by combustion analysis to confirm product purity.

However, percentage composition data alone has a critical limitation: it can only determine the empirical formula, not the molecular formula. Many different compounds can share the same empirical formula. For instance, formaldehyde (CH₂O, MM = 30 g mol⁻¹), acetic acid (C₂H₄O₂, MM = 60 g mol⁻¹), glucose (C₆H₁₂O₆, MM = 180 g mol⁻¹), and fructose (same MF as glucose) all give the empirical formula CH₂O. Without knowing which of these the compound is, the chemist cannot identify the substance. Relying on percentage composition alone in a quality control context — for example, in checking the identity of a pharmaceutical excipient or a fertiliser blend — would be insufficient and potentially dangerous.

The molar mass is the essential additional piece of information that bridges the empirical formula to the molecular formula. By calculating the multiplier n = MM(molecular) ÷ MM(empirical), the chemist scales the empirical formula to the correct molecular formula. For ethanol from fermentation: n(C₂H₆O) = 46.07 ÷ 46.07 = 1, confirming the molecular formula is C₂H₆O. For glucose: n(CH₂O) = 180.16 ÷ 30.03 = 6, giving C₆H₁₂O₆. The molar mass can be determined by spectroscopic methods (mass spectrometry), osmometry, or from literature — none of which are combustion techniques. In summary, percentage composition data is a necessary but not sufficient tool for compound identification: it establishes the atomic ratio, but the molar mass is required to determine the actual molecular formula and thereby confirm the compound’s identity.

Marking criteria (7 marks). 1 = identifies and explains at least one strength of % composition data (e.g. derives EF, experimentally accessible). 1 = identifies and explains at least one limitation (EF alone cannot give MF; multiple compounds share the same EF). 1 = first named Australian example used correctly to illustrate a strength (ethanol from sugar cane / urea fertiliser / eucalyptus oil compound). 1 = second named Australian example (same or different context). 1 = explains molar mass as the additional information required and shows how n is calculated (n = MM(mol) ÷ MM(EF)). 1 = applies n calculation to one of the named examples to show how MF is obtained from EF + molar mass. 1 = reaches an explicit evaluative judgement — % composition is necessary but insufficient; molar mass makes the determination complete.