🎯⭐ INTERACTIVE LESSON

Mixtures and Separation Techniques

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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

CategoryDefinitionExamples
ElementOne type of atomGold (Au), Oxygen (O₂)
CompoundTwo+ elements chemically bonded in fixed ratioH₂O, NaCl, CO₂
MixtureTwo+ substances physically combined, variable compositionAir, 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:

ComponentRoleExample in Saltwater
SolventThe substance present in greatest amountWater
SoluteThe 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

PropertySolutionColloidSuspension
Particle size< 1 nm1–1000 nm> 1000 nm
Settles?NoNoYes
Filter?NoNoYes
Tyndall effect?NoYesYes (if not settled)
AppearanceClearCloudy/translucentOpaque

Mixture Classification Quiz 🎯

Classify Each Mixture 🔍

Quick Knowledge Check 🧮

  1. What is the minimum particle size (in nm) for a suspension?

  2. What is the maximum particle size (in nm) for a solution?

  3. 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

  1. Pour the mixture through filter paper in a funnel
  2. Solid particles are trapped on the filter paper (residue)
  3. 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

TypeSpeedUse Case
Gravity filtrationSlowWhen you want the filtrate (liquid)
Vacuum filtrationFastWhen you want the solid (residue)

Evaporation

Principle: Separates based on boiling point differences — the liquid evaporates, leaving the dissolved solid behind.

How It Works

  1. Heat the solution in an evaporating dish
  2. The solvent (liquid) evaporates
  3. 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:

  1. Heat the mixture — the more volatile component boils first
  2. Vapor travels into a condenser (cooled tube)
  3. Vapor condenses back to liquid and is collected (distillate)
  4. 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 🧮

  1. What is the solid material left on filter paper called? (one word)

  2. What is the liquid that passes through a filter called? (one word)

  3. 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:

PhaseDescriptionExamples
Stationary phaseFixed material that doesn't movePaper fibers, silica gel, column packing
Mobile phaseFluid that moves through/over the stationary phaseSolvent (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:

  1. Place a dot of mixture near the bottom of chromatography paper
  2. Dip the bottom edge into a solvent (but below the dot!)
  3. The solvent rises by capillary action
  4. Different components travel different distances

Rf Values

The retention factor (RfR_f) quantifies how far a component travels:

Rf=distance traveled by substancedistance traveled by solvent frontR_f = \frac{\text{distance traveled by substance}}{\text{distance traveled by solvent front}}

  • RfR_f is always between 0 and 1
  • Each substance has a characteristic RfR_f in a given solvent
  • RfR_f can be used to identify unknown substances by comparison

Example

A spot travels 4.2 cm while the solvent front travels 7.0 cm:

Rf=4.27.0=0.60R_f = \frac{4.2}{7.0} = 0.60

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 🧮

  1. A substance spot traveled 3.5 cm and the solvent front traveled 10.0 cm. What is the Rf value? (to 3 significant figures)

  2. In the same experiment, a second substance has Rf = 0.80. How far (in cm) did it travel? (to 3 significant figures)

  3. 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 ColorWavelength (nm)Solution Appears
Violet380–450Yellow-green
Blue450–490Orange
Green490–570Red/magenta
Yellow570–590Violet
Orange590–620Blue
Red620–750Green/cyan

Key Concept: Absorbance

Absorbance (AA) measures how much light a solution absorbs at a particular wavelength:

A=log(II0)=log(T)A = -\log\left(\frac{I}{I_0}\right) = -\log(T)

where:

  • I0I_0 = intensity of incident light
  • II = intensity of transmitted light
  • T=I/I0T = I/I_0 = transmittance (fraction of light passing through)

Relationship: AA and TT

  • If T=1.0T = 1.0 (100% transmitted) → A=0A = 0 (no absorption)
  • If T=0.10T = 0.10 (10% transmitted) → A=1.0A = 1.0
  • If T=0.01T = 0.01 (1% transmitted) → A=2.0A = 2.0

Beer's Law (Beer-Lambert Law)

A=εbcA = \varepsilon bc

SymbolMeaningUnits
AAAbsorbanceunitless
ε\varepsilonMolar absorptivity (extinction coefficient)L/(mol·cm)
bbPath length (width of cuvette)cm
ccConcentration of absorbing speciesmol/L (M)

Key Takeaways

  • Absorbance is directly proportional to concentration
  • If you double the concentration, absorbance doubles
  • ε\varepsilon is a constant for a given substance at a given wavelength
  • Standard cuvettes have b=1.00b = 1.00 cm, so often A=εcA = \varepsilon c

Practical Use: Calibration Curves

  1. Prepare solutions of known concentrations
  2. Measure absorbance of each at a chosen wavelength (λmax\lambda_{\text{max}})
  3. Plot AA vs. cc → should give a straight line through the origin
  4. Slope = εb\varepsilon b
  5. Measure absorbance of unknown → read concentration from the line

Choosing λmax\lambda_{\text{max}}

Always measure at the wavelength of maximum absorbance (λmax\lambda_{\text{max}}) because:

  • Greatest sensitivity (largest change in AA per change in cc)
  • Beer's Law is most linear at λmax\lambda_{\text{max}}

Beer's Law Concepts 🎯

Beer's Law Calculations 🧮

  1. A solution has transmittance T=0.25T = 0.25. What is the absorbance? (to 3 significant figures)

  2. A solution with concentration 0.0200 M has absorbance 0.440 in a 1.00 cm cuvette. What is ε\varepsilon? (in L/(mol·cm), to the nearest whole number)

  3. Using the same ε\varepsilon 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

  1. Dissolve the sample in solution
  2. Add a reagent that selectively precipitates the target ion
  3. Filter the mixture to collect the precipitate
  4. Wash the precipitate to remove impurities
  5. Dry (and sometimes ignite) the precipitate
  6. Weigh the precipitate
  7. 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.

Ag+(aq)+Cl(aq)AgCl(s)\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)

Step 1: Moles of AgCl (M=143.32M = 143.32 g/mol): nAgCl=0.854143.32=5.96×103 moln_{AgCl} = \frac{0.854}{143.32} = 5.96 \times 10^{-3} \text{ mol}

Step 2: Moles of Cl⁻ = moles of AgCl = 5.96×1035.96 \times 10^{-3} mol

Step 3: Mass of Cl⁻ (M=35.45M = 35.45 g/mol): mCl=5.96×103×35.45=0.211 gm_{Cl^-} = 5.96 \times 10^{-3} \times 35.45 = 0.211 \text{ g}

Step 4: Mass percent: %Cl=0.2110.500×100%=42.2%\% \text{Cl}^- = \frac{0.211}{0.500} \times 100\% = 42.2\%

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

TermDefinition
TitrantSolution of known concentration (in the buret)
AnalyteSolution of unknown concentration (in the flask)
Equivalence pointMoles of titrant = stoichiometric amount needed to react with analyte
IndicatorSubstance that changes color near the equivalence point
End pointWhen the indicator changes color (ideally = equivalence point)

The Titration Equation

M1V1×(mole ratio)=M2V2M_1V_1 \times \text{(mole ratio)} = M_2V_2

For a 1:1 reaction: M1V1=M2V2M_1V_1 = M_2V_2

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.

HCl+NaOHNaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

1:1 ratio → MHClVHCl=MNaOHVNaOHM_{HCl}V_{HCl} = M_{NaOH}V_{NaOH}

MHCl=MNaOH×VNaOHVHCl=(0.100)(32.5)25.0=0.130 MM_{HCl} = \frac{M_{NaOH} \times V_{NaOH}}{V_{HCl}} = \frac{(0.100)(32.5)}{25.0} = 0.130 \text{ M}

Back Titration

Sometimes the analyte reacts slowly or doesn't have a clear endpoint. In a back titration:

  1. Add a known excess of reagent to the analyte
  2. Allow the reaction to go to completion
  3. Titrate the leftover (unreacted) reagent with another standard solution
  4. 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 🧮

  1. In a gravimetric analysis, 1.000 g of a sample produces 0.466 g of BaSO₄ (M=233.43M = 233.43 g/mol). How many moles of SO₄²⁻ were in the sample? (to 3 significant figures)

  2. 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)

  3. What is the mass percent of SO₄²⁻ (M=96.06M = 96.06 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

SituationBest Method
Insoluble solid in liquidFiltration
Dissolved solid in liquid (need the solid)Evaporation
Two liquids with different boiling pointsDistillation
Components with different affinities for mobile/stationary phaseChromatography
Determine concentration of colored solutionSpectroscopy (Beer's Law)
Determine mass of a specific ionGravimetric analysis
Determine concentration using known reagentTitration
Separate using magnetismMagnetic 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:

  1. Magnet → removes iron filings
  2. Add water and stir → dissolves salt
  3. Filter → removes sand (residue)
  4. Evaporate filtrate → recovers salt

Beer's Law Problem-Solving

Standard Curve Approach

Given calibration data:

Concentration (M)Absorbance
0.00200.150
0.00400.300
0.00600.450
0.00800.600
0.01000.750

The slope is: εb=0.750/0.0100=75.0\varepsilon b = 0.750/0.0100 = 75.0 L/(mol·cm) (assuming b=1.00b = 1.00 cm, ε=75.0\varepsilon = 75.0).

Finding Unknown Concentration

If an unknown has A=0.525A = 0.525:

c=Aεb=0.52575.0=0.0070 Mc = \frac{A}{\varepsilon b} = \frac{0.525}{75.0} = 0.0070 \text{ M}

Dilution Warning

If the unknown was diluted before measurement, you must account for this:

coriginal=cmeasured×VtotalVsamplec_{\text{original}} = c_{\text{measured}} \times \frac{V_{\text{total}}}{V_{\text{sample}}}

Method Selection Challenge 🎯

Workshop Calculations 🧮

  1. Using the calibration data from above (εb=75.0\varepsilon b = 75.0), an unknown solution gives A=0.375A = 0.375. What is the concentration? (in M, to 3 significant figures)

  2. 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)

  3. 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

  1. Is it a pure substance or mixture? → Check melting/boiling point range
  2. If mixture, what type? → Tyndall effect test (colloid vs solution)
  3. What components are present? → Chromatography, spectroscopy
  4. How much of each component? → Beer's Law, titration, gravimetric analysis

Common AP Lab Scenarios

ScenarioKey Techniques
Determine % composition of a mixtureGravimetric analysis or titration
Identify ions in solutionFlame tests, precipitation reactions
Determine concentration of colored solutionSpectrophotometry + Beer's Law
Separate a multicomponent mixtureSequence of filtration, evaporation, distillation
Identify components of a dyePaper or thin-layer chromatography
Determine molar mass of a volatile liquidDumas method (gas law + gravimetry)

Error Analysis in Separation Techniques

AP free-response questions often ask how errors affect results:

Common Errors and Effects

ErrorEffect on Result
Precipitate not fully dried (gravimetric)Mass too high → % too high
Some precipitate passes through filterMass too low → % too low
Overshoot endpoint in titrationVolume too high → calculated concentration too high
Cuvette has fingerprints (spectroscopy)Absorbance too high → calculated concentration too high
Air bubbles in buret during titrationVolume reading too low → calculated concentration too low
Forget to subtract water vapor pressureGas 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 🧮

  1. 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)

  2. 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)

  3. A calibration curve for Fe³⁺ (using thiocyanate complex) has slope 4500 L/(mol·cm) at b=1.00b = 1.00 cm. An unknown solution has A=0.900A = 0.900. What is [Fe3+][\text{Fe}^{3+}]? (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