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Module 1 · L5 of 21 45 min ⚡ +50 XP in Learn · +25 to complete

IQ1 Consolidation, Classification and Separation

Today's hook, A snack company claims its chips are only 2% salt. Using nothing but a balance and an oven, how would you prove them wrong?
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Warm up and recall

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

Write your initial answer before reading on.

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What you'll master, and the words for it

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What you'll master
Know

Key facts

  • The 7-step gravimetric analysis procedure and why each step matters
  • The six separation techniques covered in IQ1 and the key property each exploits
  • Common sources of error: incomplete precipitation, incomplete drying, precipitate loss, co-precipitation
Understand

Concepts

  • Why gravimetric analysis is quantitative, uses stoichiometry on a weighed precipitate
  • How to match each separation technique to the property that differs between components
  • How to evaluate a technique by purity, completeness, scale, time, and cost
Can do

Skills

  • Calculate ion mass and concentration from precipitate mass (e.g. Cl⁻ from AgCl, SO₄²⁻ from BaSO₄)
  • Select and justify a separation technique for a novel mixture
  • Evaluate a separation method, citing strengths and at least two limitations
Retrieval map, recall before you consolidate. This lesson is for transfer and error-correction, not re-reading Lessons 1–4. Cover the right column, answer each prompt from memory, then check. If any prompt is shaky, revisit that lesson before the tasks below.
Prompt (recall)Check
Element vs compound vs mixture?Element = one kind of atom; compound = a fixed ratio of elements chemically bonded; mixture = variable composition whose components keep their own properties.
What is evidence that a sample is pure?A sharp, fixed melting/boiling point under controlled conditions, used as evidence (watch for decomposition, polymorphism, pressure and azeotropes), not as an absolute definition.
When does filtration work?An insoluble solid suspended in a liquid is retained by the filter; dissolved solutes pass through.
Evaporation vs crystallisation?Evaporation to dryness removes all solvent (more recovery, lower purity); crystallisation uses controlled cooling for purer crystals (some yield stays in solution).
Simple vs fractional distillation?Simple suits a non-volatile component or a large BP gap; fractional handles close BPs via repeated vaporise–condense cycles that enrich the vapour (a plateau shows constant composition, not proof of purity).
How do you choose a technique?State the mixture's physical state, the component you want, and the property that differs (particle size, solubility, boiling point); then match the technique to that property.
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Gravimetric Analysis

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Gravimetric Analysis
core concept
Beyond the syllabus. Module 1 asks you to separate mixtures and to calculate percentage composition by mass, which comes later in this lesson and is examinable. Gravimetric analysis goes past that: the calculation runs on precipitate stoichiometry, which needs the mole, and the mole is not introduced until Module 2. Read this card for the idea, that the mass of a dried precipitate can measure how much of something was there, and for the sources of error, which are good practical-skills material. The arithmetic will make far more sense after Module 2, and that is where you will be assessed on it.

Principle

Gravimetric analysis is a quantitative method it doesn't just detect a substance, it measures how much of it is present. The key idea: cause the target substance to form an insoluble precipitate with a known reagent, then collect, dry, and weigh the precipitate. Using stoichiometry (the molar ratios from the balanced equation), you can calculate the mass of the original substance.

Why precipitation? A precipitate is insoluble, it can be separated by filtration with high efficiency. Drying removes all solvent mass, so the dried precipitate mass reflects only the target substance. This makes mass measurement highly accurate.

General Procedure

  1. Dissolve sample in water (or prepare aqueous solution of unknown)
  2. Add excess precipitating reagent → target ion forms insoluble precipitate
  3. Filter precipitate (use ashless filter paper to avoid adding mass)
  4. Wash precipitate with distilled water to remove soluble impurities
  5. Dry precipitate to constant mass (oven or desiccator)
  6. Weigh dried precipitate
  7. Use stoichiometry to calculate mass/amount of original substance

Example: Determining Chloride Content

To determine the mass of Cl⁻ ions in a solution: add excess silver nitrate (AgNO₃) solution. Cl⁻ reacts with Ag⁺ to form silver chloride (AgCl), a white precipitate that is insoluble in water.

Reaction:  Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

From the molar mass of AgCl (143.3 g mol⁻¹) and Cl⁻ (35.5 g mol⁻¹), the ratio is known. So if you weigh the dry AgCl precipitate, you can calculate the original mass of Cl⁻.

Sources of Error in Gravimetric Analysis

  • Incomplete precipitation: If not enough precipitating reagent is added, some of the target substance remains in solution → mass is underestimated. Always use excess reagent.
  • Incomplete drying: Residual moisture adds to the mass → overestimation. Dry to constant mass (weigh, dry more, reweigh, repeat until mass is stable).
  • Precipitate loss during transfer: Some precipitate may remain on the beaker or be washed away. Use a wash bottle and careful technique.
  • Co-precipitation: Impurity ions may co-precipitate with the target, adding unexpected mass to the precipitate.
Precipitate the analyte, purify and dry it, weigh it, then calculate the amount from its mass.

Gravimetric analysis: form an insoluble precipitate, filter, wash, dry, and weigh; use stoichiometry to calculate the original ion mass. Example: Ag⁺(aq) + Cl⁻(aq) → AgCl(s); from m(AgCl) and the 1:1 mole ratio, calculate m(Cl⁻). Errors include: insufficient reagent, residual moisture, precipitate loss, co-precipitation.

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

Match each gravimetric analysis term on the left to its purpose on the right.

  • Use excess precipitating reagent
  • Use ashless filter paper
  • Wash the precipitate
  • Dry to constant mass
  • Ensures no residual moisture inflates the final mass measurement.
  • Ensures all of the target ion is precipitated, so the result is not underestimated.
  • Removes soluble impurity ions that would otherwise add mass to the precipitate.
  • Avoids adding paper residue to the precipitate when weighing the dried solid.
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Choosing the Right Separation Technique

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Choosing the Right Separation Technique
core concept

We just saw that gravimetric analysis uses a precipitation reaction to measure dissolved ions. That raises a question: how do you choose between gravimetric analysis and other separation techniques for a given mixture? This card answers it → the decision depends on whether a low-solubility precipitate can be formed.

You now have five techniques. Choosing correctly is a core HSC skill. The decision always starts with the physical and chemical properties of the components you are trying to separate.

Key property difference
Particle size
Solubility changes with temperature
Boiling point
Boiling point (small difference)
Differential affinity for phases
Forms insoluble precipitate
Technique
Filtration
Crystallisation
Simple distillation
Fractional distillation
Chromatography
Gravimetric analysis

Decision Framework

Whether the substance is dissolved, and which property differs, decides the technique.
Evaluating effectiveness: When asked to "evaluate", you must discuss both strengths and limitations. Ask: How complete is the separation? How pure is the product? How much does it cost in time and resources? Can it be scaled up?

Separation technique decision tree: undissolved solid → filtration (particle size); dissolved solid → crystallisation or distillation or chromatography based on which property differs; forms a precipitate → gravimetric analysis. Multi-component mixtures often require a sequence (e.g. filter → distil → crystallise). Evaluate any technique by completeness, purity, time, cost, and scalability. Treat the decision tree as a set of starting rules, not guarantees: check them against the actual properties of your mixture, watch for exceptions (a "dissolved" component that forms an azeotrope, a fine suspension that clogs filter paper), and use a follow-up test or a sequence when a single clue is not decisive.

Add the highlighted decision tree to your notes before the check below.

Lock-in task: A water sample contains sand, dissolved sugar, and a small amount of dissolved iodine. In one or two sentences, propose a sequence of techniques to separate all three components and briefly justify your choice for each.

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Percentage Composition by Mass

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Percentage Composition by Mass
core concept

We just saw how to choose between separation techniques. That raises a question: once you already know a compound's formula, how do you work out what fraction of its mass comes from a single element, without doing any separation at all? This card answers it → percentage composition by mass, a formula-based calculation required by the syllabus dot point on calculating percentage composition by weight of component elements and/or compounds (CH11/12-2).

Percentage composition by mass tells you what proportion of a compound's total mass is contributed by one element. Unlike gravimetric analysis, no experiment is required, only the compound's formula and the atomic masses of its elements.

Formula:
% composition of element X = (mass of X in the formula ÷ molar mass of the whole compound) × 100

Worked walkthroughs

1. Oxygen in carbon dioxide (CO₂): M(CO₂) = 12.0 + (2 × 16.0) = 44.0 g mol⁻¹. Mass from oxygen = 2 × 16.0 = 32.0 g mol⁻¹. % O = (32.0 ÷ 44.0) × 100 = 72.7%.

2. Nitrogen in ammonium nitrate (NH₄NO₃): M(NH₄NO₃) = 14.0 + (4 × 1.0) + 14.0 + (3 × 16.0) = 80.0 g mol⁻¹. There are two nitrogen atoms in the formula (one in NH₄⁺, one in NO₃⁻), so mass from N = 2 × 14.0 = 28.0 g mol⁻¹. % N = (28.0 ÷ 80.0) × 100 = 35.0%.

3. Water of crystallisation in a hydrate, copper(II) sulfate pentahydrate (CuSO₄·5H₂O): M(CuSO₄·5H₂O) = 63.5 + 32.1 + (4 × 16.0) + 5 × (2 × 1.0 + 16.0) = 63.5 + 32.1 + 64.0 + 90.0 = 249.6 g mol⁻¹. Mass from water = 90.0 g mol⁻¹. % H₂O = (90.0 ÷ 249.6) × 100 = 36.1%. This is why heating a hydrate to drive off its water of crystallisation causes a measurable mass loss, exactly this percentage.

Common trap: Always count every atom of the element in the formula, not just one. In NH₄NO₃ it is easy to only count the N in the nitrate ion and miss the N in the ammonium ion, this is the single most common error in percentage composition questions.

Percentage composition by mass = (mass of element in formula ÷ molar mass of compound) × 100. Count every atom of the element in the formula (e.g. NH₄NO₃ has 2 N atoms, not 1). For hydrates, "% water of crystallisation" uses the same formula with the water's total mass as the numerator (e.g. CuSO₄·5H₂O is 36.1% water by mass).

Pause, copy the highlighted percentage composition formula and the three worked examples into your book.

Quick check: what percentage of the mass of magnesium oxide (MgO, M = 40.3 g mol⁻¹) comes from oxygen? [M(O) = 16.0 g mol⁻¹]

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Short Answer Questions

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Short Answer Questions
core concept

6. Describe the steps involved in gravimetric analysis to determine the mass of barium ions (Ba²⁺) in a solution using sodium sulfate (Na₂SO₄) as the precipitating reagent. In your answer, explain why each step is important. 3 MARKS

✏️ Answer in your book

7. A sample of industrial waste water is suspected to contain sulfate ions (SO₄²⁻). A chemist adds excess barium chloride solution to a 500 mL sample and collects 0.932 g of dry barium sulfate precipitate. Calculate the mass of sulfate in the 500 mL sample. Show all working. [M(BaSO₄) = 233.4 g mol⁻¹; M(SO₄²⁻) = 96.1 g mol⁻¹] 4 MARKS

✏️ Answer in your book

8. A chemist is given a sample of sea water and asked to determine the concentration of chloride ions (Cl⁻) using gravimetric analysis. Evaluate the effectiveness of this technique for this purpose, discussing its strengths and at least two limitations. 5 MARKS

✏️ Answer in your book

9. Calculate the percentage composition of nitrogen in ammonium sulfate, (NH₄)₂SO₄. Show full working. [M(N) = 14.0, M(H) = 1.0, M(S) = 32.1, M(O) = 16.0 g mol⁻¹] 3 MARKS

✏️ Answer in your book

We just saw how to calculate percentage composition by mass from a formula. That raises a question: how do you structure full-mark exam answers on gravimetric analysis and percentage composition calculations? This card answers it → follow a step–reason format and always connect each experimental step (or each term in the formula) to its purpose.

For gravimetric calculation answers: n(precipitate) = m ÷ M; apply the mole ratio; then m(target ion) = n × M. For evaluation answers: strengths (high accuracy, direct mass measurement) + limitations (time-consuming, requires very low precipitate solubility, drying errors, co-precipitation). For percentage composition answers: calculate M(compound), identify the total mass contributed by the element (counting every atom), then divide and multiply by 100. Always show full working.

Pause, write the highlighted calculation methods into your book.

Did you get this? True or false: in a gravimetric analysis, the mass of the original target ion is found by applying stoichiometry to the molar mass and mass of the dried precipitate.

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Worked examples

Worked examples · reveal as you go

Worked example +5 XP on full reveal

A 250 mL sample of drinking water is analysed for sulfate content. Excess barium chloride (BaCl₂) solution is added. The white precipitate of barium sulfate (BaSO₄) is filtered, dried, and weighed: mass = 0.466 g. Calculate the mass of sulfate (SO₄²⁻) in the 250 mL sample. [Molar masses: Ba = 137.3, S = 32.1, O = 16.0 g mol⁻¹]

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Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Molar mass of BaSO₄ = 137.3 + 32.1 + (4 × 16.0) = 233.4 g mol⁻¹
Molar mass of SO₄²⁻ = 32.1 + (4 × 16.0) = 96.1 g mol⁻¹
Always write the balanced ionic equation first. The 1:1 molar ratio means 1 mol BaSO₄ came from 1 mol SO₄²⁻. This ratio is what connects the precipitate mass to the original ion.
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n(BaSO₄) = mass ÷ molar mass
n(BaSO₄) = 0.466 ÷ 233.4 = 0.001997 mol
Work to 4 significant figures throughout to avoid rounding errors before the final answer.
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From the equation: n(SO₄²⁻) = n(BaSO₄) = 0.001997 mol
(1:1 ratio from balanced equation)
This 1:1 ratio is the stoichiometric bridge between what you measured (BaSO₄) and what you want to know (SO₄²⁻). In other reactions the ratio may not be 1:1, always check the balanced equation.
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m(SO₄²⁻) = n × M
m(SO₄²⁻) = 0.001997 × 96.1 = 0.192 g
Round to 3 significant figures at the end (limited by the precision of the mass measurement: 0.466 g is 3 sig figs).
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Worked example +5 XP on full reveal

A sample of ocean water contains: dissolved NaCl, dissolved MgCl₂, fine sand particles, and traces of oil (which floats on the surface and is immiscible with water). A researcher wants to obtain: (a) clean dry sand, (b) pure NaCl crystals, (c) separate the oil from the water. Recommend and justify a technique for each.

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(a) Clean dry sand Technique: Filtration
Sand is insoluble in water. The key property is particle size. Filtration works because sand particles are large enough to be retained by filter paper, while dissolved NaCl and MgCl₂ pass through in the filtrate.
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(b) Pure NaCl crystals Technique: Crystallisation
First filter to remove sand, then crystallise from the filtrate.
NaCl is dissolved, filtration won't work. NaCl's solubility decreases on cooling, so heat the filtrate to concentrate it, cool slowly to crystallise NaCl, then filter crystals and dry. The step to remove sand first is critical, you must have a clean solution before crystallising, or sand will be embedded in the crystals.
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(c) Separate oil from water Technique: Separating funnel
Oil and water form two distinct liquid layers (immiscible and different density).
Oil is immiscible with water and less dense (floats). The oil and water form two distinct liquid layers. Use a separating funnel: drain the denser layer (water) out the bottom tap while retaining the less dense layer (oil), or collect the oil from the top. The key property is immiscibility and density difference.
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Put the method in order

Predict, then reveal+8 XP
1 · Predict
2 · Reveal
3 · Compare

A chemist adds excess AgNO₃ to a chloride solution and collects 1.435 g of dry AgCl precipitate. Predict the mass of Cl⁻ in the original sample. [M(AgCl) = 143.3 g mol⁻¹; M(Cl) = 35.5 g mol⁻¹]

Confidence: 50%
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Misconception to fix

Common errors · the 3 traps that cost marks

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Misconception to fix

Wrong: Gravimetric analysis measures the mass of the solution to find the concentration.

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Misconception to fix

Right: Gravimetric analysis measures the mass of a dried, pure precipitate formed from a quantitative reaction. The mass of the precipitate is used with stoichiometry to calculate the original concentration, not the mass of the solution.

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Forgetting to use the mole ratio in calculations

Students sometimes use m(precipitate) ÷ M(target ion) as a one-step formula. The correct route is m(precipitate) → n(precipitate) → n(target ion) via the balanced equation → m(target ion).

Fix: Always write the balanced equation, identify the mole ratio explicitly, and check units at each step.

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Drill, then revisit

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

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Why must the precipitating reagent be added in excess?

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Name the best technique for each: (a) sand from water, (b) ethanol from water, (c) dyes in ink, (d) chloride ions in seawater.

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A chloride solution is treated with excess AgNO₃ and 0.287 g of AgCl precipitate is collected. How many moles of Cl⁻ were present? (M(AgCl) = 143.3)

4

Classify each as element, compound or mixture: (a) O₂ (b) CO₂ (c) air.

5

Name the best technique to separate (a) two miscible liquids with close boiling points, and (b) the coloured pigments in a leaf extract.

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Revisit your thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

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Interactive Tool, Atomic Structure Builder Open fullscreen ↗
The Atomic Structure tool shows that the atomic number of an element equals…
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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

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

Spot the error+5 XP

A student records the steps of a gravimetric analysis to determine SO₄²⁻ using BaCl₂. One line contains a procedural error, click it.

  • Add excess BaCl₂ solution to the sulfate sample so that all SO₄²⁻ precipitates as BaSO₄.
  • Filter the precipitate through ashless paper, then weigh it immediately while it is still damp.
  • Wash the precipitate with distilled water to remove soluble impurity ions.
  • Apply stoichiometry: n(BaSO₄) = m ÷ M, then n(SO₄²⁻) = n(BaSO₄) using the 1:1 mole ratio.
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Short answer
ApplyApply3 MARKS

Q1. 6. Describe the steps involved in gravimetric analysis to determine the mass of barium ions (Ba²⁺) in a solution using sodium sulfate (Na₂SO₄) as the precipitating reagent. In your answer, explain why each step is important.

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ApplyApply4 MARKS

Q2. 7. A sample of industrial waste water is suspected to contain sulfate ions (SO₄²⁻). A chemist adds excess barium chloride solution to a 500 mL sample and collects 0.932 g of dry barium sulfate precipitate. Calculate the mass of sulfate in the 500 mL sample. Show all working. [M(BaSO₄) = 233.4 g mol⁻¹; M(SO₄²⁻) = 96.1 g mol⁻¹]

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ApplyApply5 MARKS

Q3. 8. A chemist is given a sample of sea water and asked to determine the concentration of chloride ions (Cl⁻) using gravimetric analysis. Evaluate the effectiveness of this technique for this purpose, discussing its strengths and at least two limitations.

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ApplyApply3 MARKS

Q4. 9. Calculate the percentage composition of nitrogen in ammonium sulfate, (NH₄)₂SO₄. Show full working. [M(N) = 14.0, M(H) = 1.0, M(S) = 32.1, M(O) = 16.0 g mol⁻¹]

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📖 Comprehensive answers (click to reveal)

️ Activity 1, Compare

A: Crystallisation would recover NaCl crystals from the water but would not tell you the exact mass of Ca²⁺, the dissolved calcium would remain in solution. Gravimetric analysis is more appropriate: add excess Na₂SO₄ (or Na₂CO₃) to precipitate all Ca²⁺ as CaSO₄ (or CaCO₃), filter, dry to constant mass, weigh, and use stoichiometry to calculate m(Ca²⁺). It is the only technique that provides a quantitative measurement of a specific dissolved ion.

B: Filtration could physically separate the BaSO₄ precipitate from the solution (it is already insoluble). Gravimetric analysis in this context would involve the same filtration step, but also includes washing, drying to constant mass, and weighing to obtain a quantitative result. The difference: filtration alone gives you the solid; gravimetric analysis gives you the solid AND its mass, which is required for any quantitative determination. If you only need the solid and not a precise mass, filtration alone is sufficient.

Activity 2, Apply to Novel Context

Scenario 1: Gravimetric analysis. Key property: Pb²⁺ forms an insoluble precipitate with specific reagents (e.g. add Na₂SO₄ → PbSO₄(s) precipitates). Procedure: add excess Na₂SO₄ to the water sample → PbSO₄ precipitates → filter with ashless paper → wash → dry to constant mass → weigh → use stoichiometry to calculate m(Pb²⁺). This gives both confirmation (precipitate forms) and quantification (mass calculated).

Scenario 2: Chromatography (paper or TLC). Key property: differential attraction of each amino acid to stationary/mobile phases gives different Rf values. Procedure: spot all three amino acids on the baseline along with known standards → develop with appropriate solvent → measure Rf values for each spot → compare to standard Rf values for identification. No quantitative measurement needed, separation and identification only.

Scenario 3: Remove CaF₂ by filtration, it is an insoluble precipitate. Pour the treated water through filter paper (or a membrane filter for industrial scale); CaF₂ is retained as residue; the treated water (filtrate) has reduced F⁻. To confirm fluoride level: use gravimetric analysis on a sample of the treated water, add excess CaCl₂ to precipitate any remaining F⁻ as CaF₂, dry, weigh, and calculate m(F⁻) to verify it is below 1.5 mg/L.

❓ Multiple Choice

Excess reagent ensures complete precipitation of all target ions. Insufficient reagent leaves some ions in solution, giving an underestimate.

n(AgCl) = 0.286 ÷ 143.3 = 0.001996 mol. n(Cl⁻) = 0.001996 mol (1:1 ratio). m(Cl⁻) = 0.001996 × 35.5 = 0.0709 g.

Residual moisture adds to the measured mass, giving an overestimate. A (insufficient reagent) and B (precipitate loss) both cause underestimates.

BaSO₄ is insoluble → filtration separates it as residue. KI is dissolved → passes through in filtrate. Crystallisation, distillation, and chromatography are not suited to separating an insoluble solid from a solution in this way.

Gravimetric analysis is a quantitative measurement technique for specific ions that form insoluble precipitates. It cannot separate miscible liquids, volatile components, or immiscible liquids. D is wrong (it is still widely used); A and C are incorrect overstatements.

Short Answer Model Answers

Q6 (3 marks): Step 1, Add excess Na₂SO₄ to the solution: the excess ensures all Ba²⁺ ions react and precipitate as BaSO₄; without excess, some Ba²⁺ would remain dissolved and be unmeasured (1 mark). Step 2, Filter the precipitate, wash with distilled water: filtration separates the insoluble BaSO₄ from the solution; washing removes soluble impurities that could add to the mass and cause overestimation (1 mark). Step 3, Dry the precipitate to constant mass, then weigh: drying removes all residual water which would add falsely to the measured mass; drying to constant mass confirms no water remains; the final dry mass is used with stoichiometry to calculate m(Ba²⁺) (1 mark).

Q7 (4 marks): Reaction: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) [1:1 molar ratio] (1 mark). n(BaSO₄) = 0.932 ÷ 233.4 = 0.003993 mol (1 mark). n(SO₄²⁻) = 0.003993 mol (1:1 ratio from balanced equation) (1 mark). m(SO₄²⁻) = 0.003993 × 96.1 = 0.384 g (1 mark).

Q8 (5 marks): Gravimetric analysis is effective for determining Cl⁻ concentration in sea water because it can precisely quantify very small amounts of an ion, and AgCl is a highly insoluble precipitate that forms reliably and quantitatively (1 mark). Strength: it is a direct mass measurement, does not depend on colour, electronic signals, or calibration curves, making it highly accurate and reproducible (1 mark). Limitation 1: sea water contains multiple anions (SO₄²⁻, Br⁻, I⁻) that may also precipitate with AgNO₃, leading to co-precipitation and overestimation of the Cl⁻ content; additional steps to remove interfering ions are needed (1 mark). Limitation 2: the procedure is time-consuming, drying to constant mass can take hours to days; modern methods like ion chromatography or potentiometric titration are faster for routine analysis (1 mark). Overall: gravimetric analysis is highly accurate for Cl⁻ in sea water if interfering ions are controlled, but is not ideal for high-throughput or field testing due to time requirements (1 mark).

Q9 (3 marks): M((NH₄)₂SO₄) = 2×(14.0 + 4×1.0) + 32.1 + (4×16.0) = 2×18.0 + 32.1 + 64.0 = 36.0 + 32.1 + 64.0 = 132.1 g mol⁻¹ (1 mark). The formula contains 2 nitrogen atoms (both in the two NH₄⁺ ions), so mass from N = 2 × 14.0 = 28.0 g mol⁻¹ (1 mark). %N = (28.0 ÷ 132.1) × 100 = 21.2% (1 mark).