UNIT 3

Properties of Substances and Mixtures

Explore how attractions between particles influence physical states, gas behavior, solutions, and the interaction of matter with electromagnetic radiation.

Topics

Intermolecular Forces

Intermolecular forces are attractions between separate particles. They influence physical properties such as boiling point, melting point, vapor pressure, viscosity, and surface tension.

Intermolecular forces act between particles. Intramolecular bonds hold the atoms within a particle together.

London Dispersion Forces

London dispersion forces result from temporary changes in electron distribution. At a particular moment, electrons may be distributed unevenly within a particle, producing a temporary dipole.

This temporary dipole can distort the electron distribution of a nearby particle, creating an attraction between the particles.

London dispersion forces are present between all atoms and molecules, including polar particles.

Polarizability

Polarizability describes how easily a particle’s electron cloud can be distorted. Particles with more electrons and larger electron clouds are generally more polarizable and experience stronger dispersion forces.

Dipole–Dipole Attractions

Dipole–dipole attractions occur between polar molecules. The partially positive end of one molecule is attracted to the partially negative end of a neighboring molecule.

Molecular geometry must be considered when determining whether a molecule is polar. Polar bonds do not always produce a polar molecule because individual bond dipoles can cancel.

Hydrogen Bonding

Hydrogen bonding is a particularly strong type of dipole–dipole attraction. It occurs when a hydrogen atom is covalently bonded to nitrogen, oxygen, or fluorine and is attracted to a lone pair on nitrogen, oxygen, or fluorine in a nearby particle.

Hydrogen bonding requires H bonded directly to N, O, or F
A molecule containing hydrogen does not automatically experience hydrogen bonding. The hydrogen must be bonded directly to nitrogen, oxygen, or fluorine.

Ion–Dipole Attractions

Ion–dipole attractions occur between an ion and a polar molecule. The charged ion attracts the oppositely charged end of the molecule’s permanent dipole.

These attractions are important when many ionic substances dissolve in water. The partially negative oxygen end of a water molecule is attracted to a cation, while the partially positive hydrogen ends are attracted to an anion.

Comparing Intermolecular Forces

Attraction Occurs Between Important Requirement
London dispersion All atoms and molecules Temporary fluctuations in electron distribution
Dipole–dipole Polar molecules Permanent molecular dipoles
Hydrogen bonding Suitable molecules containing H and N, O, or F H must be bonded directly to N, O, or F
Ion–dipole An ion and a polar molecule Attraction between a full charge and partial charge
Avoid assuming that one type of intermolecular force is always stronger than another in every substance. Particle size, polarizability, shape, charge, and distance all affect the overall strength of an attraction.
Particle diagrams comparing London dispersion, dipole-dipole, hydrogen bonding, and ion-dipole attractions
Comparison of the four major intermolecular attractions discussed in AP Chemistry. Dashed lines represent attractions between separate particles rather than covalent bonds.

Intermolecular Forces and Physical Properties

Changing a substance’s physical state requires particles to move farther apart or move more freely. Stronger intermolecular attractions require more energy to overcome and therefore affect measurable physical properties.

Property Effect of Stronger Intermolecular Forces Reason
Boiling point Increases More energy is needed to separate particles into a gas
Vapor pressure Decreases Fewer particles can escape from the liquid
Volatility Decreases The substance evaporates less readily
Viscosity Generally increases Particles resist flowing past one another
Surface tension Generally increases Particles at the surface experience stronger attractions
Melting point Often increases More energy may be required to disrupt the solid structure
Stronger intermolecular forces generally produce a higher boiling point and lower vapor pressure. Melting point comparisons can be more complicated because the arrangement and packing of particles in the solid also matter.

Comparing Similar Substances

When comparing physical properties, first identify the particles present and all attractions they experience. Then consider polarizability, molecular shape, and the strength of the attractions.

Example: CH4 and C3H8

Both methane and propane are nonpolar, so their primary intermolecular attractions are London dispersion forces.

Propane has more electrons and a larger, more polarizable electron cloud. It therefore experiences stronger dispersion forces and has a higher boiling point than methane.

Intermolecular Forces Practice

Support each answer by identifying the relevant attraction and connecting it to a physical property.

Question 1: Water and Hydrogen Sulfide

H2O and H2S are both bent, polar molecules. Explain why H2O has the higher boiling point.

Show solution

Water molecules can form hydrogen bonds because hydrogen is bonded directly to oxygen.

H2S cannot form the same type of hydrogen bonding because its hydrogen atoms are bonded to sulfur. The stronger attractions between water molecules require more energy to overcome, giving water the higher boiling point.

Question 2: Dispersion Forces

Neon and krypton are both monatomic, nonpolar substances. Which is expected to have the higher boiling point, and why?

Show solution

Krypton is expected to have the higher boiling point.

Krypton has more electrons and a larger electron cloud, making it more polarizable. It therefore experiences stronger London dispersion forces than neon.

Question 3: Molecular Shape

Two nonpolar molecules have the same molecular formula and similar molar masses. Molecule A has a long shape, while Molecule B has a compact, nearly spherical shape. Which one is expected to experience stronger dispersion forces?

Show solution

Molecule A is expected to experience stronger dispersion forces.

Its longer shape provides greater surface contact between neighboring molecules. More of the electron clouds can interact, strengthening the dispersion attractions.

Question 4: Vapor Pressure

At the same temperature, Liquid X has a lower vapor pressure than Liquid Y. Which liquid likely has stronger intermolecular forces? Explain.

Show solution

Liquid X likely has stronger intermolecular forces.

Stronger attractions hold its particles in the liquid more effectively, so fewer particles escape into the gas phase. This produces a lower vapor pressure.

Question 5: Ion–Dipole Attraction

Explain why the oxygen ends of water molecules point toward Na+ ions when sodium chloride dissolves in water.

Show solution

Oxygen is more electronegative than hydrogen, so the oxygen end of a water molecule has a partial negative charge.

The positively charged Na+ ion attracts the partially negative oxygen end of water, producing an ion–dipole attraction.

VIDEO EXPLANATION

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States of Matter and Gases

The physical state of a substance depends on particle motion, particle spacing, and the attractions between particles. Temperature and pressure can change the balance between these factors.

Solids

Particles in a solid are packed closely together and occupy relatively fixed positions. The particles can vibrate, but they do not move freely throughout the sample.

Liquids

Particles in a liquid remain close together but can move past one another. This allows a liquid to flow and take the shape of its container.

Gases

Gas particles are widely separated and move rapidly in different directions. A gas expands to fill its container and can be compressed because of the large amount of empty space between particles.

State Particle Arrangement Particle Motion Shape Volume
Solid Closely packed and ordered Vibrate around fixed positions Definite Definite
Liquid Close together but disordered Move past one another Takes container’s shape Definite
Gas Widely separated Rapid, random motion Fills container Fills container
Particle diagrams comparing the arrangements and motion of particles in solids, liquids, and gases
Particle-level comparison of a solid, liquid, and gas. The same number of particles is shown in each panel, but their arrangement, spacing, and movement differ.

Kinetic Molecular Theory

Kinetic molecular theory is a model used to explain the behavior of ideal gases. It connects the microscopic motion of gas particles to measurable properties such as pressure, volume, and temperature.

At the same temperature, all gases have the same average kinetic energy. Their particles do not necessarily have the same average speed because particle mass also affects speed.

Temperature and Particle Speed

Increasing the absolute temperature increases the average kinetic energy of gas particles. The particles move faster on average and collide with the container walls more frequently and forcefully.

Higher temperature → greater average kinetic energy

Gas Pressure

Gas pressure results from collisions between gas particles and the walls of their container. More frequent or more forceful collisions produce greater pressure.

Quantity Symbol Common Units
Pressure P atm, kPa, torr, or mmHg
Volume V L
Amount of gas n mol
Temperature T K
Gas-law calculations must use absolute temperature in kelvins. Convert Celsius temperature using:

K = °C + 273.15

Gas-Law Relationships

Graph showing the inverse relationship between gas pressure and volume at constant temperature
At constant temperature and amount of gas, decreasing the volume increases the frequency of particle collisions with the container walls and therefore increases pressure.

The Ideal Gas Law

The ideal gas law relates pressure, volume, number of moles, and absolute temperature in a single equation.

PV = nRT

The value and units of the gas constant, R, must match the units used for pressure and volume. When pressure is measured in atmospheres and volume is measured in liters, a common value is:

R = 0.08206 L·atm/(mol·K)

Example: Finding Gas Volume

What volume is occupied by 0.500 mol of an ideal gas at 298 K and 1.00 atm?

Begin with the ideal gas law and solve for volume:

V = nRT P

Substitute the known quantities:

V = (0.500 mol)(0.08206 L·atm/(mol·K))(298 K) 1.00 atm
V = 12.2 L

The units of moles, kelvins, and atmospheres cancel, leaving liters as the unit of volume.

Partial Pressure

In a mixture of gases, each gas contributes to the total pressure. The pressure contributed by one gas is called its partial pressure.

Ptotal = P1 + P2 + P3 + ...

The partial pressure of a gas can also be calculated from its mole fraction. Mole fraction is the number of moles of one gas divided by the total number of moles in the mixture.

Mole fraction of gas A = moles of gas A total moles of gas
PA = (mole fraction of A)(Ptotal)

Example: Partial Pressure

A container holds 2.00 mol of helium and 1.00 mol of neon. The total pressure is 3.00 atm. Calculate the partial pressure of helium.

Mole fraction of He = 2.00 mol 3.00 mol = 0.667
PHe = (0.667)(3.00 atm) = 2.00 atm

Real Gases and Deviations from Ideal Behavior

The ideal gas law assumes that gas particles have no volume and experience no attractions. Real particles have finite volume and can attract one another, so real gases do not always behave ideally.

Condition Reason for Deviation
High pressure Particles are crowded together, so their individual volumes are no longer negligible
Low temperature Particles move more slowly, making intermolecular attractions more significant
Real gases behave most ideally at low pressure and high temperature, when particles are far apart and intermolecular attractions have less influence.

States and Gases Practice

Try each question before opening its solution.

Question 1: Ideal Gas Pressure

A 5.00 L container holds 0.250 mol of an ideal gas at 300 K. Calculate the pressure using R = 0.08206 L·atm/(mol·K).

Show solution

Rearrange the ideal gas law to solve for pressure:

P = nRT V
P = (0.250 mol)(0.08206 L·atm/(mol·K))(300 K) 5.00 L
P = 1.23 atm

Question 2: Temperature and Pressure

A gas in a rigid container has a pressure of 1.20 atm at 300 K. The gas is heated to 450 K. What is the new pressure?

Show solution

The volume and amount of gas remain constant, so pressure is directly proportional to absolute temperature.

P2 = P1 T2 T1
P2 = (1.20 atm) 450 K 300 K = 1.80 atm

Question 3: Total Pressure

A gas mixture contains nitrogen with a partial pressure of 0.75 atm and oxygen with a partial pressure of 0.20 atm. Calculate the total pressure.

Show solution
Ptotal = 0.75 atm + 0.20 atm = 0.95 atm

Question 4: Real-Gas Behavior

Under which conditions would a real gas be expected to show the greatest deviation from ideal behavior: high temperature and low pressure, or low temperature and high pressure? Explain.

Show solution

The gas would deviate most at low temperature and high pressure.

At low temperature, particles move more slowly and intermolecular attractions become more significant. At high pressure, particles are crowded together, so their individual volumes can no longer be ignored.

Question 5: Kinetic Energy and Speed

Helium and argon gases are at the same temperature. Compare their average kinetic energies and average particle speeds.

Show solution

The gases have the same average kinetic energy because average kinetic energy depends only on temperature.

Helium atoms have a greater average speed because helium particles have less mass than argon particles. A lighter particle must move faster to have the same kinetic energy as a heavier particle.

VIDEO EXPLANATION

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Solutions and Mixtures

A mixture contains two or more substances physically combined in variable proportions. The substances retain their chemical identities and can often be separated using physical methods.

Homogeneous and Heterogeneous Mixtures

Mixture Type Composition Particle-Level Description
Homogeneous Uniform throughout the sample Components are evenly distributed
Heterogeneous Not uniform throughout the sample Different regions contain different compositions

Solutions

A solution is a homogeneous mixture. The solute is the substance dissolved, while the solvent is the substance present in the greater amount that dissolves the solute.

Solute: the substance being dissolved.

Solvent: the substance that dissolves the solute.

Molarity

Molarity describes the concentration of a solution as the number of moles of solute per liter of solution.

Molarity = moles of solute liters of solution
M = n V

Example: Calculating Molarity

A solution contains 0.300 mol of solute in a total solution volume of 1.50 L. Calculate its molarity.

M = 0.300 mol 1.50 L = 0.200 M

Dilution

Dilution decreases a solution’s concentration by adding more solvent. The number of moles of solute remains constant during the dilution.

M1V1 = M2V2
During dilution, solution volume increases and molarity decreases, but the amount of solute does not change.

Dilute and Concentrated Solutions

A dilute solution contains a relatively small amount of solute compared with the amount of solution. A concentrated solution contains a relatively large amount of solute.

Unsaturated and Saturated Solutions

An unsaturated solution contains less than the maximum amount of dissolved solute possible under the current conditions. More solute could still dissolve.

A saturated solution contains the maximum amount of dissolved solute possible at a particular temperature. Additional solute remains undissolved while the solution is saturated.

Dilute versus concentrated describes the relative concentration of a solution.

Unsaturated versus saturated describes whether the solution has reached its solubility limit. These terms do not mean the same thing.
Particle diagrams comparing dilute, concentrated, unsaturated, and saturated solutions
Dilute and concentrated describe relative amounts of dissolved solute. Unsaturated and saturated describe whether the solubility limit has been reached.

Solution Formation

A solution forms when attractions between solute and solvent particles are strong enough to replace the solute–solute and solvent–solvent attractions that must be disrupted.

“Like Dissolves Like”

Substances with similar intermolecular attractions are often soluble in one another. Polar and ionic substances are often soluble in polar solvents, while nonpolar substances are often soluble in nonpolar solvents.

“Like dissolves like” is a useful general pattern, not an absolute rule. Solubility depends on the relative strengths of all attractions disrupted and formed during dissolution.

Ionic Substances in Water

When a soluble ionic compound dissolves in water, its ions separate and become surrounded by water molecules. Ion–dipole attractions stabilize the separated ions.

The partially negative oxygen ends of water molecules point toward cations, while the partially positive hydrogen ends point toward anions.

CaCl2(s) → Ca2+(aq) + 2Cl(aq)
The coefficients in a dissolution equation determine the relative numbers of ions present. One formula unit of CaCl₂ produces one Ca2+ ion and two Cl ions.

Electrolytes

An electrolyte produces mobile ions when dissolved in water. These ions allow the solution to conduct electricity. A nonelectrolyte dissolves as neutral particles and does not produce a significant concentration of mobile ions.

Type Particles in Solution Electrical Conductivity
Strong electrolyte Mostly separated ions Strong conductivity
Weak electrolyte Mixture of ions and neutral particles Weak conductivity
Nonelectrolyte Neutral dissolved particles Little or no conductivity

Separating Mixtures

Components of a mixture can be separated by taking advantage of differences in their physical properties.

Method Property Used Typical Purpose
Filtration Particle size and phase Separates an insoluble solid from a fluid
Distillation Differences in boiling point Separates volatile substances
Chromatography Different attractions to mobile and stationary phases Separates components that travel at different rates

Chromatography

In chromatography, a mobile phase moves through or across a stationary phase. Components that are more strongly attracted to the mobile phase travel farther, while components more strongly attracted to the stationary phase travel a shorter distance.

Rf = distance traveled by solute distance traveled by solvent front
Chromatography diagram showing a baseline, separated solute spots, and a solvent front
Components travel different distances because they experience different relative attractions to the mobile and stationary phases.

Solutions and Mixtures Practice

Try each question before opening its solution.

Question 1: Calculating Molarity

A solution contains 0.125 mol of solute in 250.0 mL of solution. Calculate its molarity.

Show solution

First, convert milliliters to liters:

250.0 mL = 0.2500 L
M = 0.125 mol 0.2500 L = 0.500 M

Question 2: Dilution

A student dilutes 25.0 mL of a 2.00 M solution to a final volume of 250.0 mL. Calculate the final concentration.

Show solution
M1V1 = M2V2
M2 = (2.00 M)(25.0 mL) 250.0 mL = 0.200 M

Question 3: Representing Dissolved Ions

A particle diagram represents three dissolved formula units of CaCl2. How many Ca2+ ions and Cl ions should be shown?

Show solution

Each formula unit produces one Ca2+ ion and two Cl ions.

3 CaCl2 units → 3 Ca2+ ions + 6 Cl ions

Question 4: Choosing a Separation Method

A mixture contains an insoluble solid suspended in water. Which separation method would be most appropriate?

Show solution

Filtration would be most appropriate. The liquid can pass through the filter while the larger, insoluble solid particles are retained.

Question 5: Chromatography

During chromatography, Component A travels farther than Component B. Which component likely has a stronger relative attraction to the mobile phase?

Show solution

Component A likely has the stronger relative attraction to the mobile phase, allowing it to move farther with the mobile phase.

Component B is relatively more strongly attracted to the stationary phase and therefore travels a shorter distance.

VIDEO EXPLANATION

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Photons and Spectroscopy

Electromagnetic radiation transfers energy through space. It can be described as a wave with wavelength and frequency or as individual packets of energy called photons.

Wavelength and Frequency

Wavelength is the distance between corresponding points on consecutive waves. Frequency is the number of wave cycles passing a point each second.

Quantity Symbol Common Unit
Wavelength λ m or nm
Frequency ν Hz or s−1
Speed of light c m/s
Photon energy E J

All electromagnetic radiation travels at the speed of light in a vacuum. Wavelength and frequency are inversely related.

c = λν
c = 2.998 × 108 m/s
As wavelength increases, frequency decreases. As wavelength decreases, frequency increases.

Photon Energy

The energy of a photon is directly proportional to its frequency. Higher-frequency radiation consists of higher-energy photons.

E = hν
h = 6.626 × 10−34 J·s
Longer wavelength → lower frequency → lower photon energy

Shorter wavelength → higher frequency → higher photon energy
Electromagnetic spectrum showing radio waves through gamma rays and the relationships among wavelength, frequency, and energy
Wavelength increases toward radio waves, while frequency and photon energy increase toward gamma rays.

Example: Finding Frequency and Photon Energy

Calculate the frequency and energy of a photon with a wavelength of 500 nm.

First, convert nanometers to meters:

500 nm × 1 m 1 × 109 nm = 5.00 × 10−7 m

Next, calculate the frequency:

ν = c λ
ν = 2.998 × 108 m/s 5.00 × 10−7 m = 6.00 × 1014 Hz

Finally, calculate the photon energy:

E = (6.626 × 10−34 J·s) (6.00 × 1014 s−1)
E = 3.98 × 10−19 J

How Spectroscopy Works

Spectroscopy examines how matter interacts with electromagnetic radiation. Atoms and molecules can absorb photons whose energies correspond to allowed changes in their energy states.

Because different substances have different energy-level arrangements, they interact with particular wavelengths of radiation. Their spectra can therefore provide evidence about their identity, structure, or concentration.

A photon is absorbed only when its energy matches an allowed energy difference within the atom or molecule.

Types of Spectroscopic Information

Radiation Region Common Interaction Information Obtained
Microwave Changes in molecular rotation Information about molecular rotational states
Infrared Changes in molecular vibration Information about bonds and functional groups
Ultraviolet and visible Changes in electron energy Information about electronic structure and concentration

Absorbance and Transmittance

When light passes through a sample, some radiation may be absorbed while the remainder is transmitted through the sample.

The Beer–Lambert Relationship

For an appropriate range of concentrations, absorbance is directly proportional to the concentration of the absorbing species and the distance the light travels through the sample.

A = εbc
Symbol Meaning
A Absorbance
ε Molar absorptivity
b Path length through the sample
c Concentration of the absorbing species
When molar absorptivity and path length remain constant, greater concentration produces greater absorbance.

Calibration Curves

A calibration curve is created by measuring the absorbance of several solutions with known concentrations. A best-fit line can then be used to determine the concentration of an unknown sample from its measured absorbance.

Linear calibration graph showing absorbance increasing with concentration
The linear relationship between absorbance and concentration can be used to determine an unknown concentration.

Example: Finding an Unknown Concentration

A calibration line is described by the equation:

Absorbance = (2.50 M−1)(concentration)

An unknown solution has an absorbance of 0.600. Calculate its concentration.

Concentration = 0.600 2.50 M−1 = 0.240 M

Photons and Spectroscopy Practice

Try each question before opening its solution.

Question 1: Comparing Photon Energy

Photon A has a shorter wavelength than Photon B. Which photon has the greater frequency and energy?

Show solution

Photon A has the greater frequency because wavelength and frequency are inversely related.

Photon A also has greater energy because photon energy is directly proportional to frequency.

Question 2: Calculating Frequency

Calculate the frequency of radiation with a wavelength of 600 nm. Use c = 2.998 × 108 m/s.

Show solution

Convert the wavelength to meters:

600 nm = 6.00 × 10−7 m
ν = 2.998 × 108 m/s 6.00 × 10−7 m
ν = 5.00 × 1014 Hz

Question 3: Calibration Line

A calibration line has a slope of 2.50 M−1. An unknown solution has an absorbance of 0.625. Calculate its concentration.

Show solution
Concentration = 0.625 2.50 M−1 = 0.250 M

Question 4: Comparing Solutions

Two solutions contain the same absorbing substance and are measured at the same wavelength using identical containers. Solution X has twice the concentration of Solution Y. How should their absorbances compare within the linear range?

Show solution

Solution X should have approximately twice the absorbance of Solution Y.

Molar absorptivity and path length are unchanged, so absorbance is directly proportional to concentration.

Question 5: Photon Absorption

Why might a molecule absorb photons of one energy but not photons of a slightly different energy?

Show solution

The molecule can absorb a photon when the photon’s energy matches an allowed difference between the molecule’s energy states.

A photon with a different energy may not correspond to an allowed transition and therefore may not be absorbed.

VIDEO EXPLANATION

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