Chemistry • Year 11 • Module 2 • Lesson 9

Gravimetric Analysis

Build HSC Band 5–6 extended-response technique: multi-step calculation, procedure design, source critique, and evaluation of experimental errors in gravimetric contexts.

Master · Extended Response

1. Multi-step calculation — chloride in a NaCl/KCl mixture (Band 4–5)

8 marks

Scenario. A 1.000 g sample of a salt mixture known to contain only NaCl and KCl is dissolved in 200 mL of distilled water. Excess AgNO3(aq) is added. The dried AgCl precipitate has a mass of 2.3145 g. (Na = 22.990, K = 39.098, Cl = 35.453, Ag = 107.87; MM(NaCl) = 58.443, MM(KCl) = 74.551, MM(AgCl) = 143.32 g mol−1)

Q1. Answer each part, showing all working.

  • (a) Write the balanced ionic equation for the precipitation of AgCl, and state the mole ratio of Ag+ to Cl−. (1 mark)
  • (b) Calculate the total moles of Cl− in the mixture. (2 marks)
  • (c) Set up and solve simultaneous equations to find the mass of NaCl and the mass of KCl in the original sample. (4 marks)
  • (d) Express the percentage by mass of NaCl in the mixture. Comment on what this result tells you about the composition of the salt. (1 mark)
Stuck? Let x = mass of NaCl, then (1.000 − x) = mass of KCl. Both contribute Cl− to the precipitate: n(Cl−) from NaCl = x ÷ 58.443; n(Cl−) from KCl = (1.000−x) ÷ 74.551. Their sum equals the total n(Cl−) from part (b).

2. Experimental design — determining % Fe in iron ore (Band 5–6)

7 marks   Band 5–6

Research question. An environmental scientist needs to determine the percentage by mass of iron (Fe) in an iron ore sample from the Pilbara, Western Australia. The ore contains Fe as Fe2O3 and inert silica (SiO2). The available equipment includes: an analytical balance (±0.0001 g), dilute HCl, dilute NH3 solution, standard filter paper, an oven (120 °C), a desiccator, and a muffle furnace (800 °C). The Fe3+ ions can be precipitated as Fe(OH)3(s) and ignited to Fe2O3(s) for weighing. (Fe = 55.845, O = 15.999; MM(Fe2O3) = 159.69 g mol−1)

Q2. Design the complete experimental procedure and calculations. Your response must include:

  • A numbered procedure (at least five steps) including how to dissolve the ore, precipitate the Fe(OH)3, and convert it to Fe2O3 for weighing.
  • The balanced equation for the ignition step: 2Fe(OH)3 → Fe2O3 + 3H2O.
  • The calculation chain from m(Fe2O3) to % Fe, including the mole ratio n(Fe) = 2 × n(Fe2O3).
  • Two potential sources of error and how each would affect the final % Fe result (too high or too low).
Stuck? Step 1: weigh sample accurately. Step 2: dissolve in HCl (Fe2O3 + 6HCl → 2FeCl3 + 3H2O; filter off silica). Step 3: add NH3(aq) to pH 9 → Fe3+ + 3OH− → Fe(OH)3. Step 4: filter, then ignite at 800 °C. Step 5: weigh Fe2O3 to constant mass. Calculation: n(Fe2O3) = m ÷ 159.69; n(Fe) = 2n; m(Fe) = n × 55.845; % Fe = m(Fe)÷m(sample) × 100.

3. Source critique — evaluate a student’s claim (Band 5–6)

6 marks   Band 5–6

Student claim. “In my gravimetric experiment to find the mass of SO42− in wastewater, I added a very large excess of BaCl2 solution — about 10 times the stoichiometric amount. This guarantees that all the sulfate precipitated and my result is definitely more accurate than my partner’s, who only added a small excess. More excess reagent always means a more complete precipitation and a more accurate result.”

Q3. Evaluate whether this student’s reasoning is correct. In your response:

  • Identify the error (or errors) in the student’s reasoning.
  • Explain the correct scientific principle that applies when adding excess precipitating reagent.
  • Describe one additional gravimetric error — separate from the reagent excess issue — that could also make the result inaccurate, and explain its direction of effect (too high or too low).
Stuck? The key issue is the “complex ion effect” (mentioned in the lesson callout): a massive excess can increase the solubility of the precipitate, so some SO42− stays dissolved — reducing the precipitate mass and giving a result that is too low. A second error: not filtering promptly allows impurities to adsorb onto the precipitate (coprecipitation), giving a result that is too high.
Answers — Do not peek before attempting

Q1 — NaCl/KCl mixture (8 marks)

(a) Ag+(aq) + Cl−(aq) → AgCl(s). Mole ratio Ag+ : Cl− = 1 : 1. [1]

(b) MM(AgCl) = 107.87 + 35.453 = 143.32 g mol−1 [1]. n(AgCl) = n(Cl−) = 2.3145 ÷ 143.32 = 0.016148 mol [1].

(c) Let x = mass of NaCl (g), so (1.000 − x) = mass of KCl (g). [1]

Total n(Cl−) = x ÷ 58.443 + (1.000 − x) ÷ 74.551 = 0.016148 [1]

0.017110x + 0.013413(1.000 − x) = 0.016148

0.017110x + 0.013413 − 0.013413x = 0.016148

0.003697x = 0.002735 [1]

x = 0.7398 g (NaCl)  ⇒  mass of KCl = 1.000 − 0.7398 = 0.2602 g [1]

(d) % NaCl = (0.7398 ÷ 1.000) × 100 = 74.0%. The salt is predominantly sodium chloride; it contains approximately three parts NaCl to one part KCl by mass. [1]

Marking criteria: (a) correct ionic equation and ratio; (b) correct MM and n(Cl−); (c) correct algebraic setup [1], correct simplification [1], correct x [1], correct KCl mass [1]; (d) correct % with comment [1].

Q2 — Iron ore procedure (7 marks)

Procedure:

Step 1: Weigh approximately 1.00 g of iron ore sample accurately on the analytical balance; record mass m0. [1]

Step 2: Add excess dilute HCl to the sample and heat gently to dissolve. Filter to remove insoluble silica (SiO2). The filtrate contains Fe3+(aq): Fe2O3 + 6HCl → 2FeCl3 + 3H2O. [1]

Step 3: Add dilute NH3(aq) dropwise to the filtrate until pH ≈ 9 to precipitate Fe(OH)3(s): Fe3+(aq) + 3OH−(aq) → Fe(OH)3(s). [1]

Step 4: Filter the precipitate through pre-weighed filter paper. Wash with distilled water to remove NH4Cl and other soluble salts. Transfer to a pre-weighed crucible. Ignite in muffle furnace at 800 °C until constant mass: 2Fe(OH)3 → Fe2O3 + 3H2O. Cool in desiccator. [1]

Step 5: Weigh cooled crucible + Fe2O3; subtract crucible mass to get m(Fe2O3). Repeat ignition cycles to constant mass. [1]

Calculations: n(Fe2O3) = m(Fe2O3) ÷ 159.69; n(Fe) = 2 × n(Fe2O3); m(Fe) = n(Fe) × 55.845; % Fe = [m(Fe) ÷ m0] × 100. [1]

Two sources of error: (1) Incomplete ignition (Fe2O3 still contains some H2O) → m(Fe2O3) too high → % Fe too high. (2) Loss of precipitate during filtration (fine particles pass through filter) → m(Fe2O3) too low → % Fe too low. [1]

Marking criteria (7 marks): Step 1 weigh [1]; Step 2 dissolve + filter silica + equation [1]; Step 3 precipitate with NH3 + equation [1]; Step 4 filter + ignite + ignition equation + desiccator [1]; Step 5 weigh to constant mass [1]; correct calculation chain with mole ratio [1]; two valid errors with directions [1].

Q3 — Source critique (6 marks)

Identifying the error (2 marks): The student’s claim that “more excess always means more complete precipitation” is incorrect [1]. While a moderate excess drives the reaction to completion by the common ion effect (increasing [Ba2+] suppresses BaSO4 solubility), a very large excess can have the opposite effect: at extremely high [Ba2+], BaSO4 can form soluble complex ions (e.g. Ba[BaSO4]2+), increasing its apparent solubility. This means some SO42− stays in solution and is not collected, giving a result that is too low [1].

Correct scientific principle (2 marks): A moderate (slight) excess of precipitating reagent is optimal: it drives the precipitation to completion while avoiding the complex-ion solubility effect. The standard practice is to add approximately 10–20% more than the stoichiometric amount, not ten times more [1]. This is consistent with Le Chatelier’s principle — a modest increase in [Ba2+] shifts equilibrium towards the insoluble precipitate; an enormous excess shifts it back via complex formation [1].

Additional error (2 marks): One valid example: incomplete drying — if the precipitate is not dried to constant mass, residual water adds to the measured mass of the precipitate, causing the calculated moles of BaSO4 and therefore the calculated concentration of SO42− to be too high [1]. Another valid example: coprecipitation (adsorption of impurity ions onto BaSO4 surface) also gives a result that is too high [accept either]. [1]

Marking criteria (6 marks): Identifies the reasoning flaw (more excess ≠ always more accurate) [1]; explains complex-ion/solubility effect and direction (too low) [1]; states correct principle (moderate excess) [1]; links to Le Chatelier or solubility equilibrium [1]; names a second, separate source of error [1]; correctly states the direction of effect (too high or too low) with reasoning [1].

HSCScience • Chemistry • Year 11 • Module 2 • Lesson 9 • Worksheet 3 (Master)