Investigating Matter — Print · Physical Science: Chemistry, Grade 11

Build the measurement, classification, and evidence skills used throughout chemistry.

Printable chapter lessons and review.

Measurement, Units & Experimental Evidence

Learning goals

Readings for this unit

Before this unit

Division, scientific notation, and a ratio equal to one.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Can a very repeatable result still be wrong?

Illustrative evidence case. Repeated densities cluster at 2.02 g/mL while an appropriate reference is 2.50 g/mL.

Repeatability and reference agreement are separate evidenceRepeatability and reference agreement are separate evidenceMeasured density (g/mL)22.22.42.6123TrialInstrument AInstrument B
Original illustrative trials. The dashed reference is 2.50 g/mL; the graph does not imply an exact uncertainty or population distribution.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Can a very repeatable result still be wrong?

Your first model

Sketch separate ideas for repeat spread and reference offset.

Build the language from the idea

density
Mass per unit volume, usually g/mL for liquids or g/cm³ for solids.
systematic error
A recurring offset caused by a method or instrument.
significant figures
Digits used to communicate the precision of a measurement.

Chemistry links observable measurements to particle models. A useful record identifies the substance, temperature, equipment, units, and method, so another person can repeat the work. Mass measures inertia and amount of material; volume measures occupied space; density is mass divided by volume. An intensive property such as density can help identify a material without depending on sample size.

SI prefixes rescale units: kilo means 1000, centi 0.01, milli 0.001, and micro 0.000001. One milliliter is one cubic centimeter; 1000 mL is one liter. In dimensional analysis multiply by a ratio equal to one, such as 1 L/1000 mL. Arrange it so unwanted units cancel. Never change a measured amount merely by changing its unit.

A conversion factor changes units without changing the amountA conversion factor changes units without changing the amount250. mL×1 L1000 mL= 0.250 LDefined conversion = exact ratio of equal amountsThe physical volume stays the same.Only its numerical description and unit change.
Original dimensional-analysis layout; strike lines mark canceled units, not erased measurement information.

Read a scale with an appropriate last estimated digit and record its resolution. Accuracy describes agreement with a reference; precision describes repeatability. Repeated measurements can be close together but all offset by a poorly calibrated balance. A mean alone hides spread: report sample count and range, and explain likely systematic and random errors.

Repeatability and reference agreement are separate evidenceRepeatability and reference agreement are separate evidenceMeasured density (g/mL)22.22.42.6123TrialInstrument AInstrument B
Original illustrative trials. The dashed reference is 2.50 g/mL; the graph does not imply an exact uncertainty or population distribution.

Significant figures communicate measurement precision. For multiplication or division, round a final result to the fewest significant figures in the measured inputs. For addition or subtraction, use the least precise decimal place. Counted objects and defined conversions are exact. Retain extra digits during calculation; round only at the end.

Reported digits should reflect the measuring processReported digits should reflect the measuring process2.03.0Read between marksInclude one estimate2.40 cm × 1.2 cm = 2.88 cm² → 2.9 cm²The limiting measured input has two significant figures.
Original schematic ruler and independent area calculation. The marker illustrates interpolation rather than supplying the dimensions for that area problem.

A measured density is not a proof of identity. Different substances may have similar densities; temperature and impurities affect the result. Compare multiple independent properties. Percent error = absolute(measured − reference)/reference × 100%; a low error for one reading does not establish that the method is reliable.

For a test of dissolved salt versus conductivity, vary salt concentration and measure conductivity under comparable temperature and geometry. Include pure water as a comparison, prepare independent repetitions, and specify equipment limitations. A diagram, labeled table, and evidence-based conclusion are part of scientific communication, not decoration.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Distinguish accuracy, precision, and resolution

A measurement combines a value, unit and procedure. Precision describes repeatability; accuracy concerns agreement with an appropriate reference. Here instrument A repeatedly reports near 2.02 g/mL while B reports near the supplied 2.50 g/mL reference. A small spread cannot remove a systematic calibration offset. Keep individual readings and reference uncertainty; three selected trials illustrate the distinction rather than establish the long-term performance of either instrument.

Repeatability and reference agreement are separate evidenceRepeatability and reference agreement are separate evidenceMeasured density (g/mL)22.22.42.6123TrialInstrument AInstrument B
Original illustrative trials. The dashed reference is 2.50 g/mL; the graph does not imply an exact uncertainty or population distribution.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Which instrument is more accurate relative to 2.50 g/mL, and why does A’s small spread not settle accuracy?

Visual reasoning — after your attempt

B is closer to this reference. A is repeatable but offset; repeatability does not rule out systematic error.

Accuracy concerns agreement with a suitable reference; precision concerns agreement among repeated measurements. Resolution is the smallest change an instrument can distinguish. Consistent measurements can all be displaced from the reference by a systematic error, so repeatability alone does not establish accuracy.

A balance repeatedly reporting 10.20 g for a 10.00 g reference is precise but biased. Record reference uncertainty and instrument conditions before deciding what a difference means.

Pause and explain

Does averaging remove every measurement error?

Check your reasoning after answering

Averaging can reduce some random variation, but a consistent calibration bias remains unless identified and corrected.

Part 2 · Build conversions that cancel units

A conversion factor is a ratio of equal quantities, such as one liter per 1,000 milliliters. Multiplying by it does not change the actual amount. Arrange the ratio so the unwanted unit cancels between numerator and denominator. Defined unit relationships are exact and do not impose a measurement limit. The dot in 250. mL communicates three significant figures in this example, so the converted value is 0.250 L, rather than a new measurement with extra precision.

A conversion factor changes units without changing the amountA conversion factor changes units without changing the amount250. mL×1 L1000 mL= 0.250 LDefined conversion = exact ratio of equal amountsThe physical volume stays the same.Only its numerical description and unit change.
Original dimensional-analysis layout; strike lines mark canceled units, not erased measurement information.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why is 250. mL multiplied by 1 L/1000 mL rather than 1000 mL/1 L?

Visual reasoning — after your attempt

The first ratio cancels mL and leaves liters; the inverted ratio would leave mL²/L and the wrong numerical scale.

A conversion factor is a ratio of equivalent quantities. Choose its orientation so the unwanted unit cancels, and retain units through the calculation. Scientific notation keeps very large or small quantities manageable; powers of ten are part of the arithmetic, not optional decoration.

Convert 250 mL using 1 L/1000 mL to obtain 0.250 L. Multiplying by the reciprocal would leave the wrong units and the wrong scale.

Pause and explain

What is the first check on a proposed conversion?

Check your reasoning after answering

Check unit cancellation before arithmetic, then check whether the magnitude is reasonable for the new unit.

Part 3 · Report precision supported by evidence

An instrument’s resolution is the smallest change it can distinguish or display; an analog scale usually supports an estimated digit beyond its marked intervals. That does not make every last digit exact. In classroom significant-figure rules, a multiplication result follows the least precise measured factor in significant figures. Addition and subtraction instead follow decimal-place precision. Keep guard digits while calculating and round once at the end; extra calculator digits cannot improve the original measurement.

Reported digits should reflect the measuring processReported digits should reflect the measuring process2.03.0Read between marksInclude one estimate2.40 cm × 1.2 cm = 2.88 cm² → 2.9 cm²The limiting measured input has two significant figures.
Original schematic ruler and independent area calculation. The marker illustrates interpolation rather than supplying the dimensions for that area problem.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why is the area reported as 2.9 cm² rather than 2.8800 cm²?

Visual reasoning — after your attempt

The 1.2 cm input has two significant figures. The additional displayed calculator digits are unsupported measurement precision.

Measured quantities have limited precision. For multiplication and division, introductory significant-figure conventions usually follow the least precise measured factor; addition and subtraction follow decimal-place limits. Exact counts and defined conversions do not restrict precision. Keep guard digits during intermediate calculations.

A final density from 12.4 g and 5.0 mL is 2.5 g/mL under the two-significant-figure convention. Writing 2.480000 suggests precision not supported by the volume.

Pause and explain

Why should intermediate results not be rounded repeatedly?

Check your reasoning after answering

Repeated rounding can accumulate avoidable numerical error. Keep extra digits during working and round the reported result according to the input precision.

Part 4 · Match a claim to its uncertainty

A graph needs named variables, units and a defined comparison. For a homogeneous material at fixed conditions, mass equals density times volume, so a mass-versus-volume slope estimates density. A fitted intercept, residual pattern and uncertainty matter when actual data depart from a line. A high correlation alone does not identify a substance or prove a causal explanation. Here the invented points lie exactly on a line to isolate the unit calculation; real balances and volume readings have uncertainty.

A physical relationship gives the slope its meaningA physical relationship gives the slope its meaningMass (g)051015202502468Volume (mL)Slope = 24 g / 8 mL = 3 g/mL
Original idealized mass–volume dataset, not collected laboratory measurements.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What are the slope units, and why is one density value insufficient to identify an unknown?

Visual reasoning — after your attempt

Slope is g/mL. Different substances can share a similar density, and temperature, purity and measurement error can alter comparisons.

A measurement supports a range rather than an infinitely exact value. Repeated trials, calibration records, and a controlled method help judge whether two results are distinguishable. A small difference comparable with uncertainty may not justify identifying a new material or a superior design.

Compare sample densities using the same temperature and procedure. If intervals overlap substantially, record an inconclusive comparison and propose a better measurement rather than forcing a ranking.

Pause and explain

What evidence would strengthen a small observed difference?

Check your reasoning after answering

Use calibrated instruments, more representative trials, and uncertainty estimates under comparable conditions to determine whether the difference is meaningful.

Apply the whole unit · Density with calibration evidence

Illustrative readings for a reference of 10.00 g are 10.20, 10.21, and 10.19 g. A separate sample has mass 12.4 g and volume 5.0 mL.

Your task

Calculate the reference mean and sample density, evaluate accuracy versus precision, report suitable digits, and propose how you would investigate the apparent balance bias without silently changing the sample measurement.

What your response should show

  • Choose the appropriate chemical model or relationship and state its conditions.
  • Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
  • Use supplied evidence to justify the conclusion; distinguish observation from inference.
  • Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
  1. Average reference readings: (10.20+10.21+10.19)/3=10.20 g.

  2. Compare with 10.00 g: the mean is 0.20 g higher.

  3. Calculate sample density: 12.4/5.0=2.48 g/mL before reporting precision.

  4. Report 2.5 g/mL under the stated significant-figure convention.

The reference readings are tightly clustered but shifted upward, consistent with a systematic bias. The sample density calculation is 2.5 g/mL as reported, not a demonstrated corrected value. Verify calibration across the relevant mass range and record uncertainty before applying a correction; the supplied reference alone does not prove the same offset under every condition.

Scientific references and source-model foundations

Parent / teacher teaching notes

Check unit cancellation and sensible magnitude before significant figures. Separate raw data from any corrected data. Ask the student to explain the precision/accuracy distinction with a second example. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.

The key idea

Units, precision, and a stated method make numerical evidence interpretable.

Worked examples

Example 1

A sample has mass 24.0 g and volume 8.00 mL. Find density.
  1. Write ρ = m/V.
  2. Substitute 24.0 g / 8.00 mL.
  3. Three significant figures are appropriate.
Answer: 3.00 g/mL

Example 2 · Worked application

A measured length is 2.40 cm and width 1.2 cm. Report area with appropriate significant figures.
  1. Multiply 2.40×1.2=2.88 cm².
  2. The width has two significant figures, limiting this product.
  3. Round the final result once to two significant figures.
Answer: 2.9 cm².
Check the reasoning

Unit conversion and significant figures are different checks.

Example 3

Convert 250. mL to liters.
  1. 250. mL × (1 L/1000 mL).
  2. The mL units cancel; the conversion is exact.
Answer: 0.250 L

Example 4 · Find and fix the mistake

Context: A measured length is 2.40 cm and width 1.2 cm. Report area with appropriate significant figures. A student claims: “Extra calculator digits always mean greater measurement accuracy.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Unit conversion and significant figures are different checks.
  2. Reported precision must reflect the measurements; calculation cannot add information that was not measured.
Answer: Reported precision must reflect the measurements; calculation cannot add information that was not measured.
Check the reasoning

Unit conversion and significant figures are different checks.

Example 5 · Reason from the visual model

Reported digits should reflect the measuring processReported digits should reflect the measuring process2.03.0Read between marksInclude one estimate2.40 cm × 1.2 cm = 2.88 cm² → 2.9 cm²The limiting measured input has two significant figures.
Original schematic ruler and independent area calculation. The marker illustrates interpolation rather than supplying the dimensions for that area problem.
Why is the area reported as 2.9 cm² rather than 2.8800 cm²?
  1. An instrument’s resolution is the smallest change it can distinguish or display; an analog scale usually supports an estimated digit beyond its marked intervals. That does not make every last digit exact. In classroom significant-figure rules, a multiplication result follows the least precise measured factor in significant figures. Addition and subtraction instead follow decimal-place precision. Keep guard digits while calculating and round once at the end; extra calculator digits cannot improve the original measurement.
  2. The 1.2 cm input has two significant figures. The additional displayed calculator digits are unsupported measurement precision.
Answer: The 1.2 cm input has two significant figures. The additional displayed calculator digits are unsupported measurement precision.
Check the reasoning

Original schematic ruler and independent area calculation. The marker illustrates interpolation rather than supplying the dimensions for that area problem.

Avoid these mistakes

Check your understanding

Explain how density units arise and how you would check an unlikely result.

Discuss the answer

Divide mass by volume; g divided by mL gives g/mL. Check units, scale zero, volume subtraction, repeats, and a plausible reference.

Return to the phenomenon

Return to the initial case. Can a very repeatable result still be wrong? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Repeatability concerns spread; agreement with a reference concerns accuracy. Calibration and conditions can introduce bias even when readings cluster.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Can a very repeatable result still be wrong? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the chemical mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep species identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Repeatability concerns spread; agreement with a reference concerns accuracy. Calibration and conditions can introduce bias even when readings cluster.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

An unknown has mass 37.5 g at volume 15.0 mL. Three repeat densities cluster near 2.50 g/mL, but the independently certified reference is 2.80 g/mL. Calculate density and explain the evidence limits.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Density is 37.5/15.0 = 2.50 g/mL. Repetition supports precision, but the difference from 2.80 suggests a method/reference comparison issue; check calibration, conditions and uncertainties rather than claim identity from one property.

Words to know

density
Mass per unit volume, usually g/mL for liquids or g/cm³ for solids.
systematic error
A recurring offset caused by a method or instrument.
significant figures
Digits used to communicate the precision of a measurement.

In the real world

Use the chapter investigation to connect this unit to measured evidence and a practical decision.

Exit check

Explain how density units arise and how you would check an unlikely result.

Check your explanation after answering

Divide mass by volume; g divided by mL gives g/mL. Check units, scale zero, volume subtraction, repeats, and a plausible reference.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Classifying Matter & Separating Mixtures

Learning goals

Readings for this unit

Before this unit

Particles, atoms, and physical versus chemical changes from middle-school science.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Does a clear liquid have to be a pure substance?

Illustrative evidence case. Two clear liquids look alike; one contains dissolved salt and conducts more strongly under matched conditions.

Classify composition by the entities present, not their colorClassify composition by the entities present, not their colorOne chemical speciesTwo chemical speciesPure substance: one species; mixture: more than one.OOOOOOOOOOOOHeHe
Original selected O₂/He particle inventories, not microscopic images or relative molecular sizes.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Does a clear liquid have to be a pure substance?

Your first model

Draw what a visible view and a particle model can each show.

Build the language from the idea

pure substance
Matter with a characteristic chemical composition.
mixture
Substances combined in variable proportions without one new fixed composition.
filtrate
The liquid and dissolved substances that pass through a filter.

Matter occupies space and has mass. A pure substance has a characteristic composition: an element contains one atomic number, and a compound contains chemically combined elements in a definite ratio. A mixture contains substances together in variable proportions. Salt water is a mixture even when it looks uniform; sodium chloride is a compound.

Classify composition by the entities present, not their colorClassify composition by the entities present, not their colorOne chemical speciesTwo chemical speciesPure substance: one species; mixture: more than one.OOOOOOOOOOOOHeHe
Original selected O₂/He particle inventories, not microscopic images or relative molecular sizes.

A homogeneous mixture has uniform composition at the scale sampled. A heterogeneous mixture contains distinguishable regions. A solution is homogeneous at the particle scale; suspensions can settle and colloids contain dispersed particles larger than molecules. Appearance alone may not identify which kind of mixture is present.

Uniform appearance and chemical purity answer different questionsUniform appearance and chemical purity answer different questionsMixed uniformly at this scaleSeparated regions at this scaleHomogeneous does not mean chemically pure.
Original two-component distribution models with colors as species keys, not real particle colors or a claim of rigid positions.

A physical property can be observed without changing chemical identity: density, melting range, solubility, and electrical conductivity are examples. A chemical property describes a possible identity-changing reaction, such as combustion or reaction with oxygen. Melting ice is physical because H₂O remains H₂O; decomposing water produces different substances.

Separation exploits differences in physical properties. Filter an insoluble solid from a liquid; evaporate solvent to recover a nonvolatile solute; use distillation to collect vapor condensed from a mixture with different volatilities. Chromatography separates components through different attraction to a stationary phase and movement with a mobile phase.

Choose a sequence that preserves the substance you want. To separate sand and salt, dissolve salt in water, filter sand, then recover salt from the filtrate. Filtering does not remove dissolved salt because ions pass with the liquid. Distillation would also allow recovery of water, but heated equipment requires a supervised laboratory.

Choose a separation step because a physical property differsChoose a separation step because a physical property differsSalt + sand mixtureWater dissolves the saltFilter: sand retainedEvaporate filtrate → saltFiltration separates phases, not dissolved ions from water.
Original conceptual sand/salt route; actual material quantities, apparatus and drying conditions require a specified procedure.

Conservation of mass applies to a defined system. If an uncovered sample dries, its measured mass decreases because water leaves the boundary, not because matter disappears. Chemical change cannot be inferred from a color change or bubbles alone; compare evidence with plausible physical explanations and whether new substances formed.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Classify composition rather than appearance

A pure substance has a defined chemical composition. An element contains one atomic-number identity, while a compound combines elements in a chemical structure with defined proportions. A mixture contains more than one substance and can vary in composition. Oxygen gas can consist of paired O₂ molecules and still be an element. A particle drawing must identify whether a symbol represents an atom, molecule or formula unit; the number of drawn colors alone is not a sufficient classification rule.

Classify composition by the entities present, not their colorClassify composition by the entities present, not their colorOne chemical speciesTwo chemical speciesPure substance: one species; mixture: more than one.OOOOOOOOOOOOHeHe
Original selected O₂/He particle inventories, not microscopic images or relative molecular sizes.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why is the left O₂ sample an element even though each molecule has two atoms?

Visual reasoning — after your attempt

Both atoms have the same element identity. Two atoms per molecule does not mean two different elements or a compound.

A pure substance has a defined composition; an element contains one proton-defined atom type, while a compound contains elements in fixed chemical ratios. A mixture has variable composition. A uniform appearance does not establish purity, because homogeneous mixtures can look like single substances.

Salt water is a homogeneous mixture; pure sodium chloride is a compound. A grain of salt is not an element just because it looks like one kind of solid.

Pause and explain

Can a clear liquid contain several substances?

Check your reasoning after answering

Yes. A homogeneous solution can contain several substances mixed at scales too small to distinguish visually.

Part 2 · Use particle drawings at two scales

A homogeneous mixture is uniform at the scale of the observation or sampling method. A heterogeneous mixture has distinguishable regions with different composition. A clear salt solution can look uniform while containing several dissolved species; a visible particle drawing is a model, not proof that all real molecular arrangements are evenly spaced. State the scale and the test: visual appearance, filtration, conductivity and chemical analysis reveal different aspects of composition. A transparent liquid is not automatically pure water.

Uniform appearance and chemical purity answer different questionsUniform appearance and chemical purity answer different questionsMixed uniformly at this scaleSeparated regions at this scaleHomogeneous does not mean chemically pure.
Original two-component distribution models with colors as species keys, not real particle colors or a claim of rigid positions.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Could a uniformly colored sample still contain more than one substance?

Visual reasoning — after your attempt

Yes. Uniform composition at the sampled scale can describe a homogeneous mixture; it does not establish one chemical species.

A particle model can distinguish identical compound particles from several particle types mixed together. Everyday observations and particle models answer different questions. Draw labels consistently and identify what each symbol represents; a color is a bookkeeping choice, not a literal atom color.

A box of identical AB particles represents a pure compound in this model. Separate A and B particles together represent a mixture of elements, not the same compound.

Pause and explain

Why is AB different from separate A and B in one container?

Check your reasoning after answering

AB represents chemically combined particles in a fixed composition. Separate A and B represent a mixture with different particle identities and interactions.

Part 3 · Choose separation using a relevant property

A separation method uses a property difference without requiring a chemical identity change. Filtration retains particles larger than a filter’s effective pores but does not normally remove dissolved ions. Selective dissolution can move salt into water while sand remains as a solid phase. Evaporation recovers a nonvolatile solute, while distillation can also collect condensed solvent vapor. Evaluate solubility, volatility, particle size and stability before choosing a method; a sequence can need several complementary steps.

Choose a separation step because a physical property differsChoose a separation step because a physical property differsSalt + sand mixtureWater dissolves the saltFilter: sand retainedEvaporate filtrate → saltFiltration separates phases, not dissolved ions from water.
Original conceptual sand/salt route; actual material quantities, apparatus and drying conditions require a specified procedure.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why does filtering salty water not normally produce salt-free water?

Visual reasoning — after your attempt

Dissolved ions travel with water through an ordinary filter. A different separation process must exploit volatility, membranes or other relevant properties.

Mixture separation exploits differences in properties without needing to change chemical identity. Filtration separates suitable insoluble particles from a fluid; distillation uses volatility differences; chromatography uses differing interactions with stationary and moving phases. No one technique separates every mixture.

A filter can retain sand but does not ordinarily remove dissolved salt from water. Evaporation can recover salt, while collecting condensed water requires a different setup.

Pause and explain

Why can ordinary filtration fail to separate dissolved salt?

Check your reasoning after answering

Dissolved ions travel with the liquid through ordinary filter pores. They are not the same as large suspended sand particles.

Part 4 · Evaluate purity and recovery separately

Recovery compares the amount of a target retrieved with a stated starting amount. Purity describes the target’s fraction in the recovered material. Wet crystals or contamination can inflate the measured mass above the starting target mass; losses during transfer can reduce recovery even when the remaining sample is pure. Drying to a defined endpoint and using an appropriate independent composition test address different questions. A numerical recovery above 100% is a reason to examine the procedure, not a claim that atoms were created.

Mass recovery and product purity must be evaluated separatelyMass recovery and product purity must be evaluated separatelyRecordStarting saltRecovered dry saltTrial A5.00 g4.25 gTrial B5.00 g5.30 gA recovery = 4.25/5.00 × 100% = 85.0%B exceeds 100%: investigate water or contamination.High mass recovery does not establish purity.
Original illustrative mass records; no composition assay is supplied, so purity is not numerically determined.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What could explain Trial B’s 106% apparent recovery without creating salt?

Visual reasoning — after your attempt

Retained water, contamination, an incorrect starting estimate or a mass-measurement error could inflate the recovered mass.

Recovering a sample does not establish that it is pure. A separation can leave contamination or lose some desired material. State the desired product, measure recovery, and use an independent property or composition test to evaluate purity. A large recovered mass could include contaminants or water.

If a wet recovered solid is heavier than expected, drying under an appropriate validated method may change its mass. High apparent recovery is not automatically evidence of successful purification.

Pause and explain

Can a high recovered mass prove high purity?

Check your reasoning after answering

No. Extra solvent or other substances can increase mass. Purity and recovery require different evidence.

Apply the whole unit · Design a separation sequence

Illustrative mixture contains dry sand, sodium chloride, and water. Sand is insoluble; the salt is dissolved. The goal is separately identifying recovered sand and salt; exact initial masses are not supplied.

Your task

Propose a sequence, draw a particle model before and after separation, explain which property each step uses, and identify evidence needed to evaluate both purity and percent recovery.

What your response should show

  • Choose the appropriate chemical model or relationship and state its conditions.
  • Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
  • Use supplied evidence to justify the conclusion; distinguish observation from inference.
  • Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt
  1. Use filtration to separate suitable sand particles from the salt solution.

  2. Use an approved supplied evaporation record to obtain salt; evaporation leaves water out of the recovered solid.

  3. Track sand, water, and dissolved ions separately in the particle account.

  4. Request starting component masses and independent purity checks before calculating recovery.

The sequence exploits particle-size/insolubility differences and volatility, rather than converting a compound into its elements. Ordinary filtration does not retain dissolved ions. Initial component quantities, retained solvent, and contamination measurements are missing, so a numerical recovery or purity claim is not justified. This is a planning task; no home distillation or chemical heating is required.

Scientific references and source-model foundations

Parent / teacher teaching notes

Ask the learner to specify the wanted product before selecting a method. Use supplied results for separation tasks needing heat. Require a separate explanation of purity and recovery. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.

The key idea

Identify particle composition first, then choose separation methods using physical differences.

Worked examples

Example 1

Classify oxygen gas, pure water, salt water, and muddy water.
  1. O₂ has only oxygen atoms; it is an element.
  2. H₂O is a fixed-ratio compound.
  3. Salt water is a homogeneous mixture; muddy water is heterogeneous.
Answer: Element; compound; homogeneous mixture; heterogeneous mixture.

Example 2 · Worked application

A sample contains dissolved salt and insoluble sand in water. Propose a separation sequence.
  1. Filter to retain sand while salt solution passes through.
  2. Recover salt by evaporation or crystallization; use distillation if collecting water is also a goal.
  3. Methods exploit solubility and phase-change properties rather than chemical destruction.
Answer: Filter, then recover dissolved components using an appropriate phase-change method.
Check the reasoning

Filtering alone cannot normally remove dissolved salt ions.

Example 3

Recover dry sand and salt from their mixture.
  1. Add water to dissolve salt.
  2. Filter and dry sand.
  3. Allow the filtrate to evaporate safely in a shallow dish.
Answer: Separate by solubility, filtration, and evaporation.

Example 4 · Find and fix the mistake

Context: A sample contains dissolved salt and insoluble sand in water. Propose a separation sequence. A student claims: “Any mixture can be separated by a filter.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Filtering alone cannot normally remove dissolved salt ions.
  2. Dissolved species pass through ordinary filters; the method must match particle size and properties.
Answer: Dissolved species pass through ordinary filters; the method must match particle size and properties.
Check the reasoning

Filtering alone cannot normally remove dissolved salt ions.

Example 5 · Reason from the visual model

Choose a separation step because a physical property differsChoose a separation step because a physical property differsSalt + sand mixtureWater dissolves the saltFilter: sand retainedEvaporate filtrate → saltFiltration separates phases, not dissolved ions from water.
Original conceptual sand/salt route; actual material quantities, apparatus and drying conditions require a specified procedure.
Why does filtering salty water not normally produce salt-free water?
  1. A separation method uses a property difference without requiring a chemical identity change. Filtration retains particles larger than a filter’s effective pores but does not normally remove dissolved ions. Selective dissolution can move salt into water while sand remains as a solid phase. Evaporation recovers a nonvolatile solute, while distillation can also collect condensed solvent vapor. Evaluate solubility, volatility, particle size and stability before choosing a method; a sequence can need several complementary steps.
  2. Dissolved ions travel with water through an ordinary filter. A different separation process must exploit volatility, membranes or other relevant properties.
Answer: Dissolved ions travel with water through an ordinary filter. A different separation process must exploit volatility, membranes or other relevant properties.
Check the reasoning

Original conceptual sand/salt route; actual material quantities, apparatus and drying conditions require a specified procedure.

Avoid these mistakes

Check your understanding

Why is salt water separable physically but pure salt is not separable into its elements by filtration?

Discuss the answer

Salt water contains substances mixed together; salt is a chemically bonded compound whose elemental separation requires chemical change.

Return to the phenomenon

Return to the initial case. Does a clear liquid have to be a pure substance? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Appearance alone does not establish composition; a uniform mixture can contain several dissolved species. Use appropriate property tests and recognize their limits.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Does a clear liquid have to be a pure substance? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the chemical mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep species identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Appearance alone does not establish composition; a uniform mixture can contain several dissolved species. Use appropriate property tests and recognize their limits.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A mixture contains 6.00 g salt and 4.00 g sand. You recover 5.10 g of dry salt-containing material. Calculate apparent salt recovery and explain why its purity remains uncertain.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Apparent recovery is 5.10/6.00 ×100 = 85.0%. The calculation treats the recovered mass as salt, but residual sand or other substances require a separate purity test. Distinguish mass loss from composition evidence.

Words to know

pure substance
Matter with a characteristic chemical composition.
mixture
Substances combined in variable proportions without one new fixed composition.
filtrate
The liquid and dissolved substances that pass through a filter.

In the real world

Use the chapter investigation to connect this unit to measured evidence and a practical decision.

Exit check

Why is salt water separable physically but pure salt is not separable into its elements by filtration?

Check your explanation after answering

Salt water contains substances mixed together; salt is a chemically bonded compound whose elemental separation requires chemical change.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

A reference mass is 50.0 g; the measurement is 49.0 g. Find percent error.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

2 %

|49−50|/50 ×100 = 2%.

Problem 2

Design a repeatable density measurement for a small waterproof object.
Answer and explanation

Measure mass on a calibrated balance; record initial and final water volumes, subtract for displaced volume, repeat, and note trapped-air or scale-reading uncertainty.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

Chapter review: An unknown has mass 37.5 g at volume 15.0 mL. Three repeat densities cluster near 2.50 g/mL, but the independently certified reference is 2.80 g/mL. Calculate density and explain the evidence limits.
Answer and explanation

Density is 37.5/15.0 = 2.50 g/mL. Repetition supports precision, but the difference from 2.80 suggests a method/reference comparison issue; check calibration, conditions and uncertainties rather than claim identity from one property.

Density is 37.5/15.0 = 2.50 g/mL. Repetition supports precision, but the difference from 2.80 suggests a method/reference comparison issue; check calibration, conditions and uncertainties rather than claim identity from one property.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 4

A drink label lists dissolved sugar and flavoring. Is the drink one compound?
Answer and explanation

No. Multiple substances occur in variable proportions; a uniform drink is a homogeneous mixture.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

A wet towel loses 30 g while drying. Explain conservation.
Answer and explanation

Water crosses the towel-system boundary into air; include that vapor to conserve the total mass.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 6

Chapter review: A mixture contains 6.00 g salt and 4.00 g sand. You recover 5.10 g of dry salt-containing material. Calculate apparent salt recovery and explain why its purity remains uncertain.
Answer and explanation

Apparent recovery is 5.10/6.00 ×100 = 85.0%. The calculation treats the recovered mass as salt, but residual sand or other substances require a separate purity test. Distinguish mass loss from composition evidence.

Apparent recovery is 5.10/6.00 ×100 = 85.0%. The calculation treats the recovered mass as salt, but residual sand or other substances require a separate purity test. Distinguish mass loss from composition evidence.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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