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Module 2 · L9 of 20 ~35 min ⚡ +50 XP in Learn · +25 to complete

Gravimetric Analysis

A gold mining company needs to know how much gold is in a tonne of ore, not approximately, exactly. A water treatment plant needs to measure sulfate contamination down to 0.1 mg. Both use gravimetric analysis: the oldest quantitative technique in chemistry, and still one of the most accurate.

Today's hook, Precipitate, filter, dry, weigh. Mass on the balance reveals invisible dissolved ions.
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Warm up and think first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
01
Recall, your gut answer first
+5 XP warm-up

A chemist wants to find the mass of sulfate ions (SO₄²⁻) dissolved in a water sample, but sulfate ions are invisible in solution and can't be weighed directly. What strategy might they use to convert the invisible dissolved ions into something solid that can be weighed? What would they need to add to the water?

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The calculation pathway, and the words for it

02
Gravimetric calculation pathway
core formula

mass of precipitate  →  n(precipitate)  →  n(analyte)  →  mass or concentration of analyte

n = m ÷ MM (precipitate first)
c = n ÷ V (V in litres)
Use mole ratio from equation to link precipitate to analyte

⚠️ The mole ratio step is where most marks are lost. Always write the balanced equation and read the ratio explicitly.

03
What you'll master
Know

Key facts

  • Definition of gravimetric analysis
  • What a precipitate is and why it forms
  • The 5-step experimental procedure
  • Examples: BaSO₄, AgCl, CaCO₃
Understand

Concepts

  • Why the precipitate must be insoluble
  • Why excess reagent is added deliberately
  • How mole ratio links precipitate to analyte
Can do

Skills

  • Describe the gravimetric procedure with justifications
  • Use precipitate mass to find analyte amount or concentration
  • Apply mole ratios from balanced equations correctly
Beyond the syllabus. Module 1 sent you here for the arithmetic, and this is the right place for it: converting a mass to an amount, reading a mole ratio off a balanced equation and converting back is Core CH11-9 work, and it is examined hard. Gravimetric analysis is the application, not the endpoint. Module 2 names percentage composition, empirical formula, limiting reagent and concentration; it does not name gravimetry as a technique you must master, so nothing here rests on reciting the five-step procedure or on picking a precipitating agent from memory. Work the examples for the stoichiometry, and keep the procedure for what it is genuinely good for, a concrete case of where experimental error comes from, which the Working Scientifically outcomes do assess.
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Key terms
Gravimetric analysis
A quantitative technique that measures the mass of a target analyte after converting it to a pure, insoluble solid of known formula.
Precipitating agent
A reagent added in excess to ensure complete precipitation of the target ion from solution.
Filtration
Separation of precipitate from solution; quantitative filtration uses ashless filter paper so mass of precipitate is accurately measured.
Ignition
Heating the precipitate in a crucible to drive off any residual water and convert to a stable stoichiometric form for weighing.
Mole ratio
The stoichiometric relationship between precipitating agent and precipitate used to calculate the mass of original analyte.
Sources of gravimetric error
Incomplete precipitation, coprecipitation of impurities, loss of precipitate during washing, and incomplete ignition.
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What gravimetric analysis is, and how it is done

05
What Is Gravimetric Analysis?
core concept

Gravimetric analysis is a quantitative technique that determines the amount of an analyte (the substance being measured) by converting it into a pure, insoluble precipitate, then measuring the mass of that precipitate.

The key insight: if you know the mass of precipitate and the chemical formula of the precipitate, you can calculate exactly how many moles were formed, and from the balanced equation, work out how many moles of the original analyte were present.

Common precipitates in HSC gravimetric analysis
BaSO₄ (barium sulfate), used to measure SO₄²⁻ concentration. AgCl (silver chloride), used to measure Cl⁻ concentration. CaCO₃ (calcium carbonate), used in some carbonate determinations. All are white, extremely insoluble solids, ideal for filtering and weighing.

Why Must the Precipitate Be Insoluble?

If the precipitate dissolves even slightly, some of the analyte stays in solution and is never collected on the filter. This means the mass you weigh underestimates the true amount, your result will be too low. The more insoluble the precipitate, the more complete the reaction and the more accurate the result.

Why Add Excess Precipitating Reagent?

Adding a slight excess of the precipitating reagent (e.g. excess BaCl₂ when precipitating SO₄²⁻) drives the reaction to completion, ensuring all of the analyte has been converted to precipitate. The excess reagent remains dissolved in solution and is washed away during filtration.

Don't overdo it
Adding too much excess can backfire. In some cases, very high concentrations of the precipitating ion can increase the solubility of the precipitate through the complex ion effect. For HSC, you only need: add a moderate excess, not a massive excess.

The Gravimetric Procedure, Step by Step

  1. Dissolve the sample in distilled water or dilute acid. Why: analyte must be in aqueous form so reagent can react evenly.
  2. Add excess precipitating reagent (e.g. BaCl₂(aq) to precipitate SO₄²⁻ as BaSO₄(s)). Stir. Why: drives reaction to completion.
  3. Filter the precipitate through pre-weighed filter paper or sintered glass crucible. Wash with distilled water. Why: separates precipitate; washing removes soluble impurities.
  4. Dry the precipitate in oven at 100–120°C; cool in desiccator before weighing. Why: residual water would overestimate mass; desiccator prevents reabsorption of atmospheric moisture.
  5. Weigh and calculate on an analytical balance (4 decimal places). Subtract filter paper mass. Apply mole ratio. Why: precision here determines accuracy of result.
Safety and handling controls (read before any gravimetric practical).
  • Silver nitrate (AgNO₃) is corrosive and stains skin and clothing. Wear safety glasses and gloves, use the minimum (microscale) volume, and collect all silver-containing waste in the labelled recovery container, never down the sink.
  • Soluble barium salts (e.g. BaCl₂) and any lead salts are toxic. In class these determinations are best done as a teacher demonstration or from supplied secondary data; if performed, use microscale quantities and the heavy-metal waste container.
  • High-temperature ignition (muffle furnace, ~800 °C) is operated by the teacher or technician only, using heat-resistant gloves and tongs. Students do not load or unload the furnace.
  • Dilute acid (HCl) and ammonia (NH₃) solution are irritants used in a fume cupboard with eye protection.

Gravimetric vs Volumetric Analysis

⚖️Gravimetric
  • Measures: mass of precipitate
  • Equipment: analytical balance, oven, filter
  • Precision: very high (±0.0001 g)
  • Slow (hours, drying time)
  • Best for: insoluble or sparingly soluble analytes
🧪Volumetric (Titration)
  • Measures: volume of solution
  • Equipment: burette, pipette, conical flask
  • Precision: high (±0.05 mL)
  • Fast (minutes per titration)
  • Best for: acid-base, redox reactions

Gravimetric analysis determines analyte concentration by converting it to a weighed insoluble precipitate (BaSO₄, AgCl, CaCO₃). Procedure: dissolve → add excess precipitant (drives reaction to completion) → filter and wash → dry to constant mass in oven + desiccator → weigh and calculate.

Pause, copy the highlighted procedure into your book before moving on.

Quick check: Why is excess precipitating reagent added in gravimetric analysis?

Did you get this? True or false: after drying in the oven, the precipitate must be cooled in a desiccator before weighing, to prevent it absorbing atmospheric moisture.

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The three-step calculation chain

06
The Calculation Chain
core concept

We just saw the gravimetric procedure, dissolve, precipitate, filter, dry, and weigh. That raises a question: once you have the mass of the dry precipitate, how do you convert it to the concentration of the original analyte? This card answers it → with a three-step calculation chain.

Once you have the mass of precipitate, the calculation follows a chain of three conversions:

  1. m(precipitate) → n(precipitate): divide by MM of the precipitate (not the analyte).
  2. n(precipitate) → n(analyte): apply mole ratio from balanced equation (often 1:1 for BaSO₄/Ba²⁺ or AgCl/Cl⁻, but not always).
  3. n(analyte) → mass or concentration: m = n × MM(analyte), or c = n ÷ V (V in L).
The 1:1 trap
For BaSO₄ and AgCl the ratio is 1:1, so students often skip the ratio step. Build the habit of always writing it, if the ratio is ever 2:1 or 1:2, skipping gives the wrong answer.

Gravimetric calculation chain: (1) n(precipitate) = m(precipitate) ÷ MM(precipitate), use the precipitate's MM, not the analyte's; (2) n(analyte) = n(precipitate) × mole ratio from the balanced equation; (3) c(analyte) = n ÷ V. Always subtract filter-paper mass before calculating.

Add the highlighted calculation chain to your notes before the check below.

Fill the blanks: drag each token into the correct place.

precipitate mole ratio balanced equation filter paper

Step 1 uses MM of the  ___. Step 2 applies the  ___  read from the  ___. Subtract the mass of the  ___  before any calculation.

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Two worked examples

Worked examples · reveal as you go

Worked example 1 · sulfate determination +5 XP on full reveal

A 250 mL water sample is treated with excess BaCl₂ solution to precipitate all sulfate as BaSO₄. The dried precipitate has a mass of 0.4660 g. Calculate the concentration of SO₄²⁻ in the water sample in mol L⁻¹. (Ba = 137.33, S = 32.06, O = 15.999)

1
$\mathrm{Ba^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} \rightarrow \mathrm{BaSO_4(s)}$  |  Ratio 1:1
Write equation, identify ratio
2
$\mathrm{MM(BaSO_4)} = 137.33 + 32.06 + 4(15.999) = 233.39\ \mathrm{g\ mol^{-1}}$
Calculate MM of precipitate
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$n(\mathrm{BaSO_4}) = 0.4660 \div 233.39 = 1.997 \times 10^{-3}\ \mathrm{mol}$
Moles of precipitate
4
$n(\mathrm{SO_4^{2-}}) = n(\mathrm{BaSO_4}) \times (1/1) = 1.997 \times 10^{-3}\ \mathrm{mol}$
Apply mole ratio
5
$c(\mathrm{SO_4^{2-}}) = 1.997 \times 10^{-3} \div 0.250 =$ 7.99 × 10⁻³ mol L⁻¹ ✓
Concentration (V = 0.250 L)
Step 1 of 5
Worked example 2 · chloride determination +5 XP on full reveal

A 500 mL sample of bore water is treated with excess AgNO₃ solution. The dried AgCl precipitate has a mass of 1.435 g. Calculate (a) the mass of Cl⁻ in the sample and (b) the concentration of Cl⁻ in mg L⁻¹. (Ag = 107.87, Cl = 35.453)

1
$\mathrm{Ag^+(aq)} + \mathrm{Cl^-(aq)} \rightarrow \mathrm{AgCl(s)}$  |  Ratio 1:1
Equation, mole ratio
2
$\mathrm{MM(AgCl)} = 107.87 + 35.453 = 143.32\ \mathrm{g\ mol^{-1}}$
MM of precipitate
3
$n(\mathrm{AgCl}) = 1.435 \div 143.32 = 1.001 \times 10^{-2}\ \mathrm{mol}$
Moles of precipitate
4
$n(\mathrm{Cl^-}) = 1.001 \times 10^{-2}\ \mathrm{mol}$  |  $m(\mathrm{Cl^-}) = 1.001 \times 10^{-2} \times 35.453 = 0.3549\ \mathrm{g} = 354.9\ \mathrm{mg}$
(a) Mass of Cl⁻
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$c = 354.9\ \mathrm{mg} \div 0.500\ \mathrm{L} = 709.8\ \mathrm{mg\ L^{-1}} \approx$ 710 mg L⁻¹ ✓
(b) Concentration in mg/L
Step 1 of 5

Lock-in task: In one or two sentences, explain why we use the MM of the precipitate (not the analyte) in Step 1 of the gravimetric calculation.

6

Sort the calculation steps

Sort the steps these are the 5 steps to calculate c(SO₄²⁻) from a BaSO₄ precipitate mass and a sample volume. Click two steps to swap them and get them in the right order.

  • n(BaSO₄) = m(BaSO₄) ÷ MM(BaSO₄)
  • Write balanced ionic equation: Ba²⁺ + SO₄²⁻ → BaSO₄  |  identify 1:1 ratio
  • c(SO₄²⁻) = n(SO₄²⁻) ÷ V(sample, in L)
  • MM(BaSO₄) = 137.33 + 32.06 + 4(15.999) = 233.39 g mol⁻¹
  • n(SO₄²⁻) = n(BaSO₄) × (1 ÷ 1)  |  apply mole ratio
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The three traps that cost marks

Common errors · the 3 traps that cost marks

1

Using the wrong MM, precipitate vs analyte

Students sometimes divide the precipitate mass by the MM of the analyte instead of the MM of the precipitate. You must convert precipitate mass → precipitate moles first (using MM of precipitate), then apply the ratio to get analyte moles.

Fix: Step 1 always uses MM of the precipitate. The analyte MM only appears if the question asks for the mass of the analyte.

2

Forgetting the mole ratio step

When the equation has a 1:1 ratio (BaSO₄ or AgCl), students often skip the ratio step, and get away with it. But if the ratio is ever 2:1 or 1:2, skipping gives the wrong answer.

Fix: Always write the balanced equation, then explicitly write "mole ratio: X mol precipitate : Y mol analyte" before calculating.

3

Not subtracting the filter paper mass

The question will often give you the mass of the filter paper/crucible. If you forget to subtract it, your precipitate mass includes the paper, giving a result that is too high.

Fix: m(precipitate) = m(filter + precipitate) − m(filter paper). Look for two mass values in the question.

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Quick-fire drill, then revisit your thinking

Quick-fire practice · 5 reps +2 XP per reveal

1

A 0.2332 g precipitate of BaSO₄ is obtained from a water sample. Calculate the moles of SO₄²⁻ in the sample. (MM of BaSO₄ = 233.39 g mol⁻¹)

2

A 100 mL seawater sample produces 0.5735 g of AgCl precipitate. Calculate the concentration of Cl⁻ in mol L⁻¹. (Ag = 107.87, Cl = 35.453)

3

Filter paper mass = 1.2455 g. After drying, filter + precipitate = 1.6823 g. What is the precipitate mass?

4

A 200 mL industrial wastewater sample contains 0.6798 g of BaSO₄ precipitate. Calculate the concentration of SO₄²⁻ in g L⁻¹. (MM BaSO₄ = 233.39, S = 32.06, O = 15.999)

5

A water sample is treated with AgNO₃. 0.7166 g of AgCl is obtained from a 250 mL sample. Calculate c(Cl⁻) in mol L⁻¹. (MM AgCl = 143.32)

12
Revisit your thinking

Earlier you were asked: How can a chemist convert invisible dissolved sulfate ions into something solid that can be weighed?

The answer: add barium chloride (BaCl₂) solution, it reacts with SO₄²⁻ to form insoluble barium sulfate (BaSO₄): Ba²⁺ + SO₄²⁻ → BaSO₄(s). The precipitate can then be filtered, dried, and weighed. By measuring the mass of BaSO₄, you can calculate the moles of SO₄²⁻ using n = m ÷ MM, and then the concentration. This is the core principle of gravimetric analysis.

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Practice questions

01
Multiple choice
+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidence. That tells the system what to drill next.

02
Short answer
ApplyLower-order5 marks

Q1. A student performs a gravimetric analysis of a 250 mL sample of water to determine its sulfate content. After adding excess BaCl₂ and collecting the precipitate, the filter paper weighed 1.1234 g before and 1.5566 g after drying. Calculate: (a) the mass of BaSO₄ precipitate, (b) the moles of SO₄²⁻ in the sample, and (c) the concentration of SO₄²⁻ in mol L⁻¹. (MM of BaSO₄ = 233.39 g mol⁻¹)

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UnderstandLower-order5 marks

Q2. Describe the gravimetric procedure a student would follow to determine the mass of chloride ions (Cl⁻) in a 500 mL sample of tap water, using silver nitrate (AgNO₃) as the precipitating reagent. In your answer, identify the precipitate formed, justify why excess AgNO₃ is used, and explain why the precipitate must be dried before weighing.

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AnalyseMid-order6 marks

Q3. A 1.000 g sample of a mixture containing both NaCl and KCl is dissolved in water. Excess AgNO₃ is added and 2.3145 g of AgCl precipitate is collected. Calculate the percentage by mass of NaCl in the mixture. (Na = 22.990, K = 39.098, Cl = 35.453, Ag = 107.87)

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EvaluateHigher-order5 marks

Q4. A student performs a gravimetric determination of sulfate in a sample. They add BaCl₂ to the sample but do not filter immediately, they wait 24 hours before filtering. After filtering and drying, the precipitate mass is slightly higher than expected. A classmate says: "This is because some atmospheric CO₂ dissolved in the solution and formed BaCO₃, which also precipitated." Evaluate whether this explanation is plausible and suggest another possible explanation for the higher mass.

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CreateHigher-order7 marks

Q5. (Extension, teacher-demonstration / secondary-data context: the high-temperature ignition step is operated only by the teacher or technician.) Design a complete gravimetric procedure to determine the percentage by mass of iron (Fe) in an iron ore sample (assumed to contain Fe as Fe₂O₃ and inert silica). The available equipment is: an analytical balance, HCl solution, NH₃ solution, filter paper, oven, desiccator, and a muffle furnace (technician-operated). The Fe³⁺ ions can be precipitated as Fe(OH)₃ and then ignited to Fe₂O₃ for weighing. Include calculations showing how you would determine % Fe from the final mass of Fe₂O₃. (Fe = 55.845, O = 15.999; MM(Fe₂O₃) = 159.69 g mol⁻¹)

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

From the lesson bank: excess reagent drives precipitation to completion; Step 1 uses MM of precipitate; desiccator cools without absorbing moisture; AgCl mole ratio 1:1; very low Ksp of BaSO₄ ensures complete precipitation.

Short answer model answers

Q1 (5 marks):

(a) m(BaSO₄) = 1.5566 − 1.1234 = 0.4332 g
(b) n(BaSO₄) = 0.4332 ÷ 233.39 = 1.857 × 10⁻³ mol = n(SO₄²⁻) (1:1)
(c) c(SO₄²⁻) = 1.857 × 10⁻³ ÷ 0.250 = 7.43 × 10⁻³ mol L⁻¹

Q2 (5 marks): Add excess AgNO₃(aq) to the 500 mL tap water sample and stir. The precipitate formed is AgCl(s), a white, insoluble solid, via: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Excess AgNO₃ is used to ensure all Cl⁻ ions react and are converted to precipitate, driving the reaction to completion. Filter the precipitate through a pre-weighed filter paper, washing with distilled water to remove soluble impurities. Dry the precipitate in an oven at ~100°C and cool in a desiccator. The precipitate must be completely dry before weighing because any remaining water would add mass to the measurement, causing the calculated Cl⁻ content to be overestimated. Weigh the dry precipitate and subtract the filter paper mass to get m(AgCl), then calculate: n(AgCl) = m ÷ 143.32; n(Cl⁻) = n(AgCl); m(Cl⁻) = n × 35.453.

Q3 (6 marks):

MM(AgCl) = 107.87 + 35.453 = 143.32 g mol⁻¹ [1]
n(AgCl) = n(Cl⁻) = 2.3145 ÷ 143.32 = 0.016148 mol [1]
Let x = mass of NaCl, then (1.000 − x) = mass of KCl [1]
n(Cl⁻) from NaCl = x ÷ 58.443; n(Cl⁻) from KCl = (1.000 − x) ÷ 74.551 [1]
x ÷ 58.443 + (1.000 − x) ÷ 74.551 = 0.016148
0.017110x + 0.013413(1.000 − x) = 0.016148
0.003697x = 0.002735 → x = 0.7398 g NaCl [1]
% NaCl = (0.7398 ÷ 1.000) × 100 = 74.0% [1]

Q4 (5 marks): The CO₂ explanation is plausible [1]. In acidic or neutral solution, CO₂ would not react significantly with Ba²⁺ (BaCO₃ Ksp ≈ 5.1 × 10⁻⁹, slightly soluble). However, if the solution became basic over time, BaCO₃ could co-precipitate with BaSO₄ [1]. Another plausible explanation: over 24 hours, the BaSO₄ particles could adsorb impurities from solution onto their surface (adsorption), adding mass without extra SO₄²⁻ [1]. A third explanation: incomplete drying, water trapped in the precipitate mass is also a possibility if drying was not repeated to constant mass [1]. Overall, the student should filter promptly and dry to constant mass to avoid these errors [1].

Q5 (7 marks):

Step 1: Weigh ~1 g of iron ore sample accurately on an analytical balance. Record mass m₁ [1].

Step 2: Dissolve in excess dilute HCl, heating gently. Filter to remove insoluble silica. The filtrate contains Fe³⁺(aq) [1].

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

Step 4: Filter through a pre-weighed crucible. Ignite in muffle furnace at 800°C: 2Fe(OH)₃ → Fe₂O₃ + 3H₂O. Cool in desiccator and weigh. Repeat to constant mass [1].

Calculation: n(Fe₂O₃) = m(Fe₂O₃) ÷ 159.69 [1]. n(Fe) = 2 × n(Fe₂O₃) [1]. m(Fe) = n(Fe) × 55.845. % Fe = (m(Fe) ÷ m₁) × 100 [1].

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