Mixtures and Separation Techniques - Complete Interactive Lesson
Part 1: Pure Substances vs Mixtures
🧪 Types of Mixtures
Part 1 of 7 — Homogeneous vs. Heterogeneous Mixtures
Most matter around you is a mixture — a combination of two or more substances that are not chemically bonded together. Each substance in a mixture retains its own chemical identity and properties. Understanding mixtures is essential for chemistry because separation techniques are the foundation of analytical chemistry and lab work on the AP exam.
Pure Substances vs. Mixtures
| Category | Definition | Examples |
|---|---|---|
| Element | One type of atom | Gold (Au), Oxygen (O₂) |
| Compound | Two+ elements chemically bonded in fixed ratio | H₂O, NaCl, CO₂ |
| Mixture | Two+ substances physically combined, variable composition | Air, saltwater, trail mix |
Key Differences
- Pure substances have a fixed composition and definite melting/boiling points.
- Mixtures have variable composition and boil/melt over a range of temperatures.
- Mixtures can be separated by physical methods (no chemical reactions needed).
Homogeneous Mixtures (Solutions)
A homogeneous mixture has a uniform composition throughout — you cannot distinguish the components visually.
Solutions
The most common type of homogeneous mixture:
| Component | Role | Example in Saltwater |
|---|---|---|
| Solvent | The substance present in greatest amount | Water |
| Solute | The dissolved substance(s) | NaCl |
Solutions can be:
- Solid in liquid: sugar in water
- Gas in liquid: CO₂ in soda
- Liquid in liquid: ethanol in water
- Gas in gas: air (N₂ + O₂ + Ar + ...)
- Solid in solid: alloys (brass = Cu + Zn)
Properties of Solutions
- Transparent (may be colored)
- Do not scatter light (no Tyndall effect)
- Will not separate on standing
- Pass through filter paper
- Particle size: < 1 nm
Heterogeneous Mixtures
A heterogeneous mixture has a non-uniform composition — you can see or detect different regions.
Suspensions
- Particles > 1000 nm (visible to naked eye or microscope)
- Will settle out over time
- Can be separated by filtration
- Example: muddy water, flour in water, blood cells in plasma
Colloids
- Particles between 1–1000 nm
- Do NOT settle out
- Scatter light (Tyndall effect — beam of light becomes visible)
- Cannot pass through semipermeable membranes
- Examples: milk, fog, gelatin, smoke, mayonnaise
Comparing the Three Types
| Property | Solution | Colloid | Suspension |
|---|---|---|---|
| Particle size | < 1 nm | 1–1000 nm | > 1000 nm |
| Settles? | No | No | Yes |
| Filter? | No | No | Yes |
| Tyndall effect? | No | Yes | Yes (if not settled) |
| Appearance | Clear | Cloudy/translucent | Opaque |
Mixture Classification Quiz 🎯
Classify Each Mixture 🔍
Quick Knowledge Check 🧮
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What is the minimum particle size (in nm) for a suspension?
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What is the maximum particle size (in nm) for a solution?
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In a solution of sugar dissolved in water, the solvent is ______ (enter the substance name).
Exit Quiz — Types of Mixtures ✅
Part 2: Homogeneous & Heterogeneous
🔥 Separation by Physical Properties
Part 2 of 7 — Filtration, Evaporation, and Distillation
Since mixtures are physically combined (not chemically bonded), they can be separated using differences in physical properties — such as particle size, boiling point, and solubility. These techniques are essential lab skills tested on the AP Chemistry exam.
Filtration
Principle: Separates based on particle size — solid particles are too large to pass through a filter while the liquid (filtrate) passes through.
How It Works
- Pour the mixture through filter paper in a funnel
- Solid particles are trapped on the filter paper (residue)
- Liquid passes through (filtrate)
When to Use
- Separating an insoluble solid from a liquid (e.g., sand from water)
- After a precipitation reaction to isolate the precipitate
- Cannot separate dissolved substances (solutes pass through the filter)
Gravity vs. Vacuum Filtration
| Type | Speed | Use Case |
|---|---|---|
| Gravity filtration | Slow | When you want the filtrate (liquid) |
| Vacuum filtration | Fast | When you want the solid (residue) |
Evaporation
Principle: Separates based on boiling point differences — the liquid evaporates, leaving the dissolved solid behind.
How It Works
- Heat the solution in an evaporating dish
- The solvent (liquid) evaporates
- The solute (solid) remains in the dish
When to Use
- Recovering a dissolved solid from a solution (e.g., salt from saltwater)
- The liquid is not needed (it's lost to the atmosphere)
Limitations
- Destroys volatile or heat-sensitive solutes
- Wastes the solvent
- Cannot separate two dissolved solids from each other
Distillation
Principle: Separates based on differences in boiling points — the component with the lower boiling point evaporates first and is condensed separately.
Simple Distillation
Used when boiling points differ by at least 25°C:
- Heat the mixture — the more volatile component boils first
- Vapor travels into a condenser (cooled tube)
- Vapor condenses back to liquid and is collected (distillate)
- Less volatile component stays in the flask
Fractional Distillation
Used when boiling points are close together (< 25°C difference):
- Uses a fractionating column packed with glass beads
- Vapor condenses and re-evaporates multiple times
- Each cycle enriches the vapor in the more volatile component
- Example: separating ethanol (bp 78°C) from water (bp 100°C)
Applications
- Purifying water (removing dissolved salts)
- Petroleum refining (separating crude oil into fractions)
- Producing spirits (concentrating ethanol)
Separation Technique Selection 🎯
Match Technique to Separation 🔍
Separation Basics 🧮
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What is the solid material left on filter paper called? (one word)
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What is the liquid that passes through a filter called? (one word)
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In distillation, the collected vapor that has been condensed back to liquid is called the ______. (one word)
Exit Quiz — Physical Separation ✅
Part 3: Filtration & Distillation
📊 Chromatography
Part 3 of 7 — Separating by Differential Affinity
Chromatography is a powerful family of separation techniques that work on a single elegant principle: different substances move at different rates through a medium based on their relative affinities for a stationary phase and a mobile phase.
How Chromatography Works
Every chromatographic method has two phases:
| Phase | Description | Examples |
|---|---|---|
| Stationary phase | Fixed material that doesn't move | Paper fibers, silica gel, column packing |
| Mobile phase | Fluid that moves through/over the stationary phase | Solvent (liquid) or carrier gas |
The Separation Principle
Components of a mixture interact differently with the two phases:
- Components with stronger affinity for the stationary phase move slowly
- Components with stronger affinity for the mobile phase move quickly
- This differential movement causes the components to separate into distinct bands or spots
Paper Chromatography
The simplest form of chromatography:
- Place a dot of mixture near the bottom of chromatography paper
- Dip the bottom edge into a solvent (but below the dot!)
- The solvent rises by capillary action
- Different components travel different distances
Rf Values
The retention factor () quantifies how far a component travels:
- is always between 0 and 1
- Each substance has a characteristic in a given solvent
- can be used to identify unknown substances by comparison
Example
A spot travels 4.2 cm while the solvent front travels 7.0 cm:
Column Chromatography
- Stationary phase: silica gel or alumina packed in a glass column
- Mobile phase: liquid solvent poured through the top
- Components separate as they travel down the column at different rates
- Fractions are collected from the bottom as they elute
Use: Purifying larger quantities of material in organic chemistry labs.
Gas Chromatography (GC)
- Stationary phase: liquid coating on the inside of a long, thin tube (capillary column)
- Mobile phase: inert carrier gas (He or N₂)
- Sample is vaporized and carried through the column
- Components separate based on boiling point and polarity
The Chromatogram
GC produces a graph of detector signal vs. time:
- Each peak represents a different component
- Retention time (time to reach detector) identifies the substance
- Peak area is proportional to the amount of that component
Applications
- Drug testing, environmental analysis, forensics
- Analyzing mixtures of volatile compounds
- Quality control in food and pharmaceutical industries
Chromatography Concepts 🎯
Rf Value Calculations 🧮
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A substance spot traveled 3.5 cm and the solvent front traveled 10.0 cm. What is the Rf value? (to 3 significant figures)
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In the same experiment, a second substance has Rf = 0.80. How far (in cm) did it travel? (to 3 significant figures)
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An unknown spot has the same Rf as a known amino acid standard (Rf = 0.45). The solvent front traveled 8.0 cm. How far did the unknown travel? (in cm, to 3 significant figures)
Chromatography Method Selection 🔍
Exit Quiz — Chromatography ✅
Part 4: Chromatography
🌈 Spectroscopy Introduction
Part 4 of 7 — Beer's Law and Absorbance
Spectroscopy uses light to analyze the composition of mixtures. When light passes through a solution, some wavelengths are absorbed by the solute. The amount of absorption depends on the concentration of the solute, which gives us a powerful analytical tool: Beer's Law.
How Solutions Absorb Light
When white light passes through a colored solution:
- The solution absorbs certain wavelengths
- The remaining wavelengths pass through (transmitted light)
- The color you see is the complementary color of what was absorbed
Color Wheel
| Absorbed Color | Wavelength (nm) | Solution Appears |
|---|---|---|
| Violet | 380–450 | Yellow-green |
| Blue | 450–490 | Orange |
| Green | 490–570 | Red/magenta |
| Yellow | 570–590 | Violet |
| Orange | 590–620 | Blue |
| Red | 620–750 | Green/cyan |
Key Concept: Absorbance
Absorbance () measures how much light a solution absorbs at a particular wavelength:
where:
- = intensity of incident light
- = intensity of transmitted light
- = transmittance (fraction of light passing through)
Relationship: and
- If (100% transmitted) → (no absorption)
- If (10% transmitted) →
- If (1% transmitted) →
Beer's Law (Beer-Lambert Law)
| Symbol | Meaning | Units |
|---|---|---|
| Absorbance | unitless | |
| Molar absorptivity (extinction coefficient) | L/(mol·cm) | |
| Path length (width of cuvette) | cm | |
| Concentration of absorbing species | mol/L (M) |
Key Takeaways
- Absorbance is directly proportional to concentration
- If you double the concentration, absorbance doubles
- is a constant for a given substance at a given wavelength
- Standard cuvettes have cm, so often
Practical Use: Calibration Curves
- Prepare solutions of known concentrations
- Measure absorbance of each at a chosen wavelength ()
- Plot vs. → should give a straight line through the origin
- Slope =
- Measure absorbance of unknown → read concentration from the line
Choosing
Always measure at the wavelength of maximum absorbance () because:
- Greatest sensitivity (largest change in per change in )
- Beer's Law is most linear at
Beer's Law Concepts 🎯
Beer's Law Calculations 🧮
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A solution has transmittance . What is the absorbance? (to 3 significant figures)
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A solution with concentration 0.0200 M has absorbance 0.440 in a 1.00 cm cuvette. What is ? (in L/(mol·cm), to the nearest whole number)
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Using the same and path length from problem 2, what concentration gives an absorbance of 0.880? (in M, to 3 significant figures)
Spectroscopy Concepts 🔍
Exit Quiz — Spectroscopy ✅
Part 5: Choosing Separation Methods
⚖️ Gravimetric & Volumetric Analysis
Part 5 of 7 — Quantitative Analytical Techniques
Two classical methods for determining the composition of mixtures are gravimetric analysis (based on mass measurements) and volumetric analysis (based on volume measurements of solutions). Both appear frequently on the AP Chemistry exam.
Gravimetric Analysis
Principle: Determine the amount of a substance by converting it to a known precipitate, filtering, drying, and weighing it.
Steps
- Dissolve the sample in solution
- Add a reagent that selectively precipitates the target ion
- Filter the mixture to collect the precipitate
- Wash the precipitate to remove impurities
- Dry (and sometimes ignite) the precipitate
- Weigh the precipitate
- Calculate the composition using stoichiometry
Example
Determine the mass percent of Cl⁻ in a 0.500 g sample. Adding excess AgNO₃ produces 0.854 g of AgCl.
Step 1: Moles of AgCl ( g/mol):
Step 2: Moles of Cl⁻ = moles of AgCl = mol
Step 3: Mass of Cl⁻ ( g/mol):
Step 4: Mass percent:
Volumetric Analysis (Titration)
Principle: Determine the concentration of an unknown solution by reacting it with a standard solution (known concentration) until the reaction is complete — the equivalence point.
Key Terms
| Term | Definition |
|---|---|
| Titrant | Solution of known concentration (in the buret) |
| Analyte | Solution of unknown concentration (in the flask) |
| Equivalence point | Moles of titrant = stoichiometric amount needed to react with analyte |
| Indicator | Substance that changes color near the equivalence point |
| End point | When the indicator changes color (ideally = equivalence point) |
The Titration Equation
For a 1:1 reaction:
Example: Acid-Base Titration
25.0 mL of unknown HCl is titrated with 0.100 M NaOH. It takes 32.5 mL of NaOH to reach the equivalence point.
1:1 ratio →
Back Titration
Sometimes the analyte reacts slowly or doesn't have a clear endpoint. In a back titration:
- Add a known excess of reagent to the analyte
- Allow the reaction to go to completion
- Titrate the leftover (unreacted) reagent with another standard solution
- The difference tells you how much reagent reacted with the analyte
When to Use
- When the analyte is insoluble (e.g., CaCO₃ in antacid tablets)
- When the reaction is slow
- When there is no suitable indicator for direct titration
Analysis Method Quiz 🎯
Analytical Calculations 🧮
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In a gravimetric analysis, 1.000 g of a sample produces 0.466 g of BaSO₄ ( g/mol). How many moles of SO₄²⁻ were in the sample? (to 3 significant figures)
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20.0 mL of unknown H₂SO₄ is titrated with 0.150 M NaOH. It takes 40.0 mL to reach equivalence. The reaction is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is the molarity of H₂SO₄? (to 3 significant figures)
-
What is the mass percent of SO₄²⁻ ( g/mol) in the sample from problem 1? (to 3 significant figures, as a percentage)
Method Selection 🔍
Exit Quiz — Gravimetric & Volumetric ✅
Part 6: Problem-Solving Workshop
🔧 Problem-Solving Workshop
Part 6 of 7 — Choosing Methods & Beer's Law Calculations
In this workshop, you'll practice the critical thinking skills needed for AP Chemistry: choosing the right separation or analysis method for a given situation, performing Beer's Law calculations, and working through multi-step analytical problems.
Decision Framework: Choosing Separation Methods
| Situation | Best Method |
|---|---|
| Insoluble solid in liquid | Filtration |
| Dissolved solid in liquid (need the solid) | Evaporation |
| Two liquids with different boiling points | Distillation |
| Components with different affinities for mobile/stationary phase | Chromatography |
| Determine concentration of colored solution | Spectroscopy (Beer's Law) |
| Determine mass of a specific ion | Gravimetric analysis |
| Determine concentration using known reagent | Titration |
| Separate using magnetism | Magnetic separation (e.g., iron filings from sand) |
Multi-Step Separations
Complex mixtures often require multiple techniques in sequence:
Example: Separate salt, sand, and iron filings:
- Magnet → removes iron filings
- Add water and stir → dissolves salt
- Filter → removes sand (residue)
- Evaporate filtrate → recovers salt
Beer's Law Problem-Solving
Standard Curve Approach
Given calibration data:
| Concentration (M) | Absorbance |
|---|---|
| 0.0020 | 0.150 |
| 0.0040 | 0.300 |
| 0.0060 | 0.450 |
| 0.0080 | 0.600 |
| 0.0100 | 0.750 |
The slope is: L/(mol·cm) (assuming cm, ).
Finding Unknown Concentration
If an unknown has :
Dilution Warning
If the unknown was diluted before measurement, you must account for this:
Method Selection Challenge 🎯
Workshop Calculations 🧮
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Using the calibration data from above (), an unknown solution gives . What is the concentration? (in M, to 3 significant figures)
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The unknown from problem 1 was prepared by diluting 5.00 mL of original solution to 25.0 mL. What was the original concentration? (in M, to 3 significant figures)
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A student needs to separate 3.00 g of NaCl dissolved in 100 mL of water. After evaporation, theoretically all the NaCl should remain. If the student recovers 2.85 g, what is the percent recovery? (to 3 significant figures)
Technique Matching 🔍
Exit Quiz — Problem Solving ✅
Part 7: Synthesis & AP Review
🏆 Synthesis & AP Review
Part 7 of 7 — Lab Technique Connections & AP-Style Questions
You've learned about mixture types, separation techniques, chromatography, spectroscopy, and quantitative analysis. Now let's connect everything together with AP-style questions that integrate multiple concepts — exactly how they appear on the exam.
Lab Technique Connections
The AP Chemistry exam frequently tests your ability to design an experimental procedure. Here's how the techniques connect:
Identification Workflow
- Is it a pure substance or mixture? → Check melting/boiling point range
- If mixture, what type? → Tyndall effect test (colloid vs solution)
- What components are present? → Chromatography, spectroscopy
- How much of each component? → Beer's Law, titration, gravimetric analysis
Common AP Lab Scenarios
| Scenario | Key Techniques |
|---|---|
| Determine % composition of a mixture | Gravimetric analysis or titration |
| Identify ions in solution | Flame tests, precipitation reactions |
| Determine concentration of colored solution | Spectrophotometry + Beer's Law |
| Separate a multicomponent mixture | Sequence of filtration, evaporation, distillation |
| Identify components of a dye | Paper or thin-layer chromatography |
| Determine molar mass of a volatile liquid | Dumas method (gas law + gravimetry) |
Error Analysis in Separation Techniques
AP free-response questions often ask how errors affect results:
Common Errors and Effects
| Error | Effect on Result |
|---|---|
| Precipitate not fully dried (gravimetric) | Mass too high → % too high |
| Some precipitate passes through filter | Mass too low → % too low |
| Overshoot endpoint in titration | Volume too high → calculated concentration too high |
| Cuvette has fingerprints (spectroscopy) | Absorbance too high → calculated concentration too high |
| Air bubbles in buret during titration | Volume reading too low → calculated concentration too low |
| Forget to subtract water vapor pressure | Gas pressure too high → calculated moles too high |
Significant Figures in Analysis
- Buret readings: ±0.02 mL (read to 2 decimal places)
- Analytical balance: ±0.001 g (read to 3 decimal places)
- Volumetric flask: typically 3-4 sig figs
- Spectrophotometer: 3 sig figs for absorbance
AP-Style Conceptual Questions 🎯
AP-Style Calculation Problems 🧮
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A 2.000 g sample of a mixture of NaCl and sand is dissolved in water, filtered, and the filtrate is evaporated. The recovered NaCl has a mass of 0.850 g. What is the mass percent of NaCl in the original mixture? (to 3 significant figures)
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25.00 mL of a Ba(OH)₂ solution is titrated with 0.1000 M HCl. It takes 35.60 mL of HCl to reach the equivalence point. The reaction is Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O. What is the molarity of Ba(OH)₂? (to 3 significant figures)
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A calibration curve for Fe³⁺ (using thiocyanate complex) has slope 4500 L/(mol·cm) at cm. An unknown solution has . What is ? (in M, in scientific notation as X.Xe-4, write just the coefficient to 3 significant figures, e.g., "2.0")
Error Analysis Practice 🔍
Final Exit Quiz — Complete Review ✅