Chemistry · Study Guide

Exam 3 Study Guide

This guide covers chapters 7 through 9. You will draw molecules, work out their shapes, and decide whether they are polar. Then you will look deeper, at orbitals that overlap and spread across a whole molecule. The last chapter is gases, the one state that one equation can predict. Read each "what to know" list first. Copy out the equations by hand. Work the examples with the answers covered, then run the ten self-test items. The cram sheet at the end holds every equation and constant.

Chapter 7 Chapter 8 Chapter 9 Cram sheet

What this guide covers

Chapter 7 · ionic bonds and lattice energy, electronegativity, Lewis structures, resonance, formal charge, VSEPR shapes and molecular polarity.
Chapter 8 · orbital overlap, hybrid orbitals, sigma and pi bonds, restricted rotation, molecular orbital diagrams, bond order and magnetism.
Chapter 9 · pressure, the simple gas laws, PV = nRT, gas density and stoichiometry, partial pressures, kinetic molecular theory and effusion.

Chapter 7Chemical Bonding and Molecular Geometry

What to know

An ionic bond moves an electron across. A metal hands electrons to a nonmetal, and the two opposite charges attract. The solid is a lattice, a repeating grid of ions, so the formula is just a ratio. Lattice energy measures how hard that grid is to pull apart. It rises with bigger ion charges and falls as the ions get larger.

A covalent bond shares a pair instead. The electronegativity difference, ΔEN, scores how fairly. Below 0.4 the bond is nonpolar covalent. From 0.4 to 1.8 it is polar covalent. Above 1.8 treat it as ionic. The more electronegative atom takes the partial negative end.

Drawing a Lewis structure follows five steps. Count all valence electrons, adding one for each negative charge and removing one for each positive charge. Pick the least electronegative atom as the center. Join atoms with single bonds. Put the remaining electrons as lone pairs on the outer atoms first. Then move lone pairs into double or triple bonds until every atom has a full shell.

The octet rule bends three ways. An odd electron count cannot pair up fully, as in NO. Boron and beryllium settle for fewer than eight. Atoms in period 3 and beyond can hold more than eight, as in SF₆.

Resonance appears when one drawing is not enough. Ozone is the classic case. Neither drawing is right on its own. The real molecule is a blend, so both of its bonds are identical and sit at bond order 1.5.

Formal charge ranks competing structures. Compute it for each atom, and prefer the structure with charges closest to zero. Any negative formal charge should sit on the most electronegative atom.

VSEPR says electron domains push each other as far apart as they can. A domain is a bond or a lone pair, and a double bond counts as one domain. Count domains to get the electron geometry. Then ignore the lone pairs to read the molecular shape. Two domains give linear. Three give trigonal planar. Four give tetrahedral. Five give trigonal bipyramidal. Six give octahedral.

A molecule is polar only if its bond dipoles fail to cancel. Ask two questions. Are any bonds polar? Do the arrows cancel by symmetry? Carbon tetrachloride has four polar bonds that cancel exactly, so the molecule is nonpolar. Water's two bonds do not cancel, because the shape is bent.

Lock these in

Lattice energy grows with charge and shrinks with distance.
ΔEN under 0.4 is nonpolar, 0.4 to 1.8 is polar, over 1.8 is ionic.
Count electrons, not pairs, and adjust for the ion's charge.
A double or triple bond is still one electron domain.
Electron geometry counts lone pairs. Molecular shape does not.
Polar bonds alone do not make a polar molecule. Symmetry can cancel them.

Equations, units and symbols

Equation	What each symbol means	Units
ΔH_lattice ∝ Q₁Q₂ ÷ r₀	Q₁ and Q₂ are the ion charges, r₀ is the distance between their centers	ΔH_lattice in kJ/mol; Q in elementary charges; r₀ in pm
ΔEN = EN_higher − EN_lower	EN is each atom's electronegativity from the table	no unit; the scale runs 0.79 to 3.98
valence electrons = Σ group counts − charge	the total to place in a Lewis structure	a plain count; subtracting a negative charge adds electrons
FC = V − L − ½B	V is valence electrons, L is lone pair electrons, B is bonding electrons	a whole number of charges
Σ formal charges = species charge	the check on a finished structure	0 for a neutral molecule
domains = bonds + lone pairs on the central atom	multiple bonds count once	a plain count from 2 to 6
bond order = bonds shared ÷ resonance forms	gives the blended value, like ozone's 1.5	no unit

Worked examples

Worked example 1 · a Lewis structure for carbon dioxide

Draw CO₂ and check every shell.

Step 1 · count valence electronsCarbon is group 14, so 4. Each oxygen is group 16, so 6. Total = 4 + 2(6) = 16 electrons, which is 8 pairs.

Step 2 · pick the centerCarbon is less electronegative than oxygen, so carbon sits in the middle.

Step 3 · single bonds firstO–C–O uses 4 electrons. That leaves 12 to place.

Step 4 · fill the outer atomsGive each oxygen three lone pairs, using all 12. Now both oxygens have 8 electrons, but carbon has only 4.

Step 5 · make double bondsMove one lone pair from each oxygen into a bond. The result is O=C=O. Carbon now has 8, and each oxygen has 2 lone pairs plus 4 bonding electrons, so 8 as well.

Check it a second wayCount the total again. Four bonds hold 8 electrons, and four lone pairs hold 8. That is 16, exactly what you started with.

Worked example 2 · formal charges in ozone

One resonance form of O₃ has a double bond on the left and a single bond on the right. Find each atom's formal charge.

Step 1 · the central atomIt has 1 lone pair, so L = 2. It has three bonds, so B = 6. Then FC = 6 − 2 − 3 = +1.

Step 2 · the double-bonded oxygenIt has 2 lone pairs, so L = 4. It has two bonds, so B = 4. Then FC = 6 − 4 − 2 = 0.

Step 3 · the single-bonded oxygenIt has 3 lone pairs, so L = 6. It has one bond, so B = 2. Then FC = 6 − 6 − 1 = −1.

Step 4 · check the sum(+1) + 0 + (−1) = 0, and ozone is a neutral molecule. The work holds.

What it meansThe other resonance form is the mirror image. Blending the two spreads the charge evenly, so both bonds come out at bond order 1.5. That is why both O–O distances are measured to be the same.

Worked example 3 · shape and polarity of two molecules

Find the shape and polarity of SF₄ and of CCl₄.

Step 1 · count for SF₄Valence electrons: 6 + 4(7) = 34. Four S–F bonds use 8. Each fluorine takes 3 lone pairs, using 24. That leaves 2 for sulfur, so one lone pair.

Step 2 · read the domainsSulfur has 4 bonds plus 1 lone pair, so 5 domains. The electron geometry is trigonal bipyramidal.

Step 3 · drop the lone pairIgnoring it, the shape is see-saw. The lone pair takes an equatorial spot, where it has the most room.

Step 4 · judge the polarityThe shape is lopsided, so the bond dipoles cannot cancel. SF₄ is polar.

Step 5 · now CCl₄Carbon has 4 bonds and no lone pairs, so 4 domains and a tetrahedral shape. Each C–Cl bond is polar, with ΔEN = 3.16 − 2.55 = 0.61.

Step 6 · check the symmetryThe four identical arrows point to the corners of a tetrahedron, so they cancel exactly. CCl₄ is nonpolar, even though every bond in it is polar.

Traps people fall into

Counting pairs when the step says electrons. CO₂ has 16 valence electrons, which is 8 pairs.
Ignoring an ion's charge. For NO₃⁻ you add one electron. For NH₄⁺ you remove one.
Treating a double bond as two domains. VSEPR counts regions, not bonds.
Reporting the electron geometry as the shape. Water is tetrahedral in domains but bent in shape.
Thinking resonance forms flip back and forth. The real molecule is one blended structure at all times.
Assuming polar bonds make a polar molecule. Check whether the shape cancels the arrows.
Expecting an octet everywhere. Boron takes six, and sulfur in SF₆ takes twelve.

Chapter 7 self-test · ten items

1. How many valence electrons do you place for NO₃⁻?

24. Nitrogen gives 5 and each of three oxygens gives 6, which is 5 + 18 = 23. The 1− charge adds one more electron, so 23 + 1 = 24.

2. How many valence electrons do you place for SO₄²⁻?

32. Sulfur gives 6 and four oxygens give 24, for 30. The 2− charge adds two more, so 30 + 2 = 32 electrons, which is 16 pairs.

3. Classify the H–F bond. Fluorine's electronegativity is 3.98 and hydrogen's is 2.20.

Polar covalent. The difference is 3.98 − 2.20 = 1.78. That sits in the 0.4 to 1.8 window, so it is polar covalent, but only just. Fluorine carries the partial negative end.

4. What is the formal charge on nitrogen in NH₄⁺?

+1. Nitrogen's V = 5. It has no lone pairs, so L = 0. It has four bonds, so B = 8. Then FC = 5 − 0 − 4 = +1. Each hydrogen is 1 − 0 − 1 = 0, so the sum is +1, matching the ion.

5. Give the electron geometry, molecular shape and bond angle for water.

Tetrahedral electron geometry, bent shape, about 104.5°. Oxygen has 2 bonds plus 2 lone pairs, so 4 domains. Lone pairs push harder than bonds, which squeezes the angle below the 109.5° of a perfect tetrahedron.

6. Give the shape, angle and polarity of BF₃.

Trigonal planar, 120°, and nonpolar. Boron has 3 bonds and no lone pairs, so only 6 electrons around it. The three identical arrows lie flat at 120° apart, so they cancel.

7. Give the shape, angle and polarity of NH₃.

Trigonal pyramidal, about 107°, and polar. Nitrogen has 3 bonds plus 1 lone pair, so 4 domains and a tetrahedral electron geometry. The lone pair pushes the three N–H bonds down, and that lopsided shape leaves a net dipole.

8. What is the bond order of each O–O bond in ozone?

1.5. Across the two resonance forms, three bonds are shared between two positions: 3 ÷ 2 = 1.5. Both bonds measure the same length, which is between a single and a double bond.

9. Which has the larger lattice energy, NaCl or MgO?

MgO. Lattice energy follows Q₁Q₂/r₀. In NaCl the charge product is 1 × 1 = 1. In MgO it is 2 × 2 = 4, so four times larger. The Mg²⁺ and O²⁻ ions are also smaller, which shrinks r₀ and raises the energy further.

10. Carbon dioxide has two polar bonds. Why is the molecule nonpolar?

The shape is linear, so the two arrows cancel. Carbon has 2 domains, which sets a 180° angle. The two bond dipoles are equal in size and point in exactly opposite directions, so they sum to zero. Bend the molecule and it would become polar, which is what happens in water.

Chapter 8Advanced Theories of Covalent Bonding

What to know

Valence bond theory builds a bond from two overlapping orbitals holding one paired-up pair of electrons. Push two atoms together and the energy drops, because each electron now feels two nuclei. Push too far and the nuclei repel. The lowest point on that curve gives the bond length, and the depth of the dip gives the bond energy.

Plain s and p orbitals point the wrong way for most molecules. So the atom mixes them into hybrid orbitals that point where the bonds need to go. Count the regions of electron density on the atom and read off the answer. Two regions give sp. Three give sp². Four give sp³. Five give sp³d. Six give sp³d².

Orbitals meet in two ways. Head-on overlap along the bond axis makes a sigma bond. Side-on overlap of two parallel p orbitals makes a pi bond. Every bond between two atoms starts with one sigma bond. A double bond is one sigma plus one pi. A triple bond is one sigma plus two pi.

A pi bond needs its p orbitals to stay parallel. Twisting the bond would pull them out of line and break the overlap. So a double bond cannot rotate. That lock is what makes cis and trans isomers real, separate compounds.

Molecular orbital theory spreads electrons over the whole molecule instead. Two atomic orbitals combine into two molecular orbitals. The bonding one sits lower in energy and piles electron density between the nuclei. The antibonding one sits higher and puts a node there instead. Fill them from the bottom, two electrons per level, following Hund's rule.

For period 2 diatomics, use only the valence electrons. In B₂, C₂ and N₂ the pi 2p level sits below sigma 2p. In O₂, F₂ and Ne₂ the two swap places.

Bond order ties it all together. Subtract antibonding electrons from bonding electrons, then halve it. A higher bond order means a shorter, stronger bond. A bond order of zero means no molecule forms.

Magnetism falls out of the same diagram. Any unpaired electron makes the molecule paramagnetic, so a magnet attracts it. All-paired electrons make it diamagnetic. Oxygen is the famous case. A Lewis structure predicts no unpaired electrons, but O₂ really is paramagnetic, and the molecular orbital diagram shows exactly why.

Lock these in

Count regions of electron density, not atoms, to get the hybridization.
Every bond has exactly one sigma bond. The extras are pi bonds.
A double bond will not rotate, because the pi overlap would break.
Bond order = (bonding − antibonding) ÷ 2.
Higher bond order means shorter and stronger.
One unpaired electron is enough to make a molecule paramagnetic.

Equations, units and symbols

Equation	What each symbol means	Units
bond order = (n_b − n_a) ÷ 2	n_b is bonding electrons, n_a is antibonding electrons	no unit; it comes out in halves
regions = bonds + lone pairs on that atom	multiple bonds count once, as in VSEPR	a plain count from 2 to 6
sigma bonds = number of bonded atom pairs	one per connected pair, single or multiple	a plain count
pi bonds = total bonds − sigma bonds	a double adds 1 pi, a triple adds 2 pi	a plain count
E_bond = (D × 1000) ÷ N_A	D is the bond energy per mole, N_A is Avogadro's number	J per bond, from D in kJ/mol
valence electrons in a period 2 diatomic	add both atoms' group electrons, then adjust for any charge	a plain count to fill the diagram

Worked examples

Worked example 1 · hybridization and bond counting in ethene

For C₂H₄, give the hybridization on each carbon. Then count sigma and pi bonds.

Step 1 · draw itEach carbon bonds to two hydrogens and shares a double bond with the other carbon.

Step 2 · count regions on one carbonTwo C–H bonds plus one C=C bond gives 3 regions. The double bond counts once.

Step 3 · read the hybridThree regions means sp². That leaves one unhybridized p orbital on each carbon.

Step 4 · count sigma bondsBonded pairs are four C–H and one C–C. That is 5 sigma bonds.

Step 5 · count pi bondsThe only multiple bond is the double bond, which adds 1 pi bond. It comes from the two leftover p orbitals overlapping side-on.

Check it a second wayThree sp² lobes lie flat at 120°, so ethene is a flat molecule. That is exactly what measurement shows, and the pi bond above and below the plane is what holds it flat.

Worked example 2 · the oxygen molecule and its magnetism

Fill the molecular orbital diagram for O₂. Find the bond order and the magnetism.

Step 1 · count valence electronsEach oxygen brings 6, so there are 12 to place.

Step 2 · fill in orderOxygen is on the side where sigma 2p sits below pi 2p. So: σ2s², σ*2s², σ2p², π2p⁴, π*2p². That adds to 12.

Step 3 · sort bonding from antibondingBonding: 2 + 2 + 4 = 8. Antibonding: the starred levels, 2 + 2 = 4.

Step 4 · bond order(8 − 4) ÷ 2 = 2. That matches the double bond a Lewis structure draws.

Step 5 · look at the top levelThe π*2p level is two orbitals of equal energy holding two electrons. Hund's rule puts one in each, unpaired. So O₂ is paramagnetic.

Why this mattersLiquid oxygen really does cling to a magnet. A Lewis structure pairs every electron and cannot explain that. This is the clearest win for molecular orbital theory.

Worked example 3 · bond order comparisons and one bond's energy

Find the bond order of N₂ and of F₂. Then find the energy of a single H–H bond, given D = 436 kJ/mol.

Step 1 · nitrogen's electronsEach nitrogen brings 5, so 10 valence electrons. Nitrogen is on the side where pi 2p sits below sigma 2p. Fill: σ2s², σ*2s², π2p⁴, σ2p².

Step 2 · nitrogen's bond orderBonding = 2 + 4 + 2 = 8. Antibonding = 2. So (8 − 2) ÷ 2 = 3, a triple bond. Every electron is paired, so N₂ is diamagnetic.

Step 3 · fluorine's electronsEach fluorine brings 7, so 14 electrons. Fill: σ2s², σ*2s², σ2p², π2p⁴, π*2p⁴.

Step 4 · fluorine's bond orderBonding = 8. Antibonding = 2 + 4 = 6. So (8 − 6) ÷ 2 = 1, a single bond. All paired, so it is diamagnetic.

Step 5 · one bond's energyConvert 436 kJ/mol to joules: 436,000 J/mol. Divide by Avogadro's number: 436,000 ÷ (6.022 × 10²³) = 7.24 × 10⁻¹⁹ J per bond.

Check the rankingN₂ at bond order 3 should be the shortest and strongest of the two, and F₂ at bond order 1 the longest and weakest. Measurement agrees: N≡N takes 945 kJ/mol to break, while F–F takes only 159.

Traps people fall into

Counting bonds instead of regions. The carbon in CO₂ has 2 regions, so it is sp, not sp³.
Forgetting lone pairs in the region count. Nitrogen in NH₃ has 4 regions, so it is sp³.
Calling a double bond two pi bonds. It is one sigma plus one pi.
Including core electrons in a period 2 diagram. Use valence electrons only.
Using the wrong level order. B₂, C₂ and N₂ put pi 2p lower. O₂, F₂ and Ne₂ put sigma 2p lower.
Forgetting to halve. Bond order divides the difference by two.
Pairing the top electrons by habit. Two electrons in two equal orbitals stay unpaired, which is what makes O₂ paramagnetic.

Chapter 8 self-test · ten items

1. What is the hybridization of carbon in CH₄?

sp³. Carbon has four C–H bonds and no lone pairs, so 4 regions. Four regions call for one s and three p orbitals mixed, which gives four sp³ lobes pointing to the corners of a tetrahedron.

2. What is the hybridization of carbon in CO₂?

sp. Carbon has two double bonds, and each counts as one region. Two regions mix one s with one p, giving two sp lobes 180° apart. The two leftover p orbitals make the two pi bonds.

3. How many sigma and pi bonds are in HCN?

2 sigma and 2 pi. The structure is H–C≡N. Two atom pairs are bonded, so 2 sigma bonds. The triple bond carries 2 pi bonds on top of its sigma. Carbon has 2 regions, so it is sp.

4. How many sigma and pi bonds are in C₂H₂?

3 sigma and 2 pi. The structure is H–C≡C–H. Three atom pairs are bonded, giving 3 sigma bonds. The triple bond adds 2 pi bonds. Both carbons are sp, so the molecule is linear.

5. Find the bond order of F₂ and say whether it is paramagnetic.

Bond order 1, and diamagnetic. There are 14 valence electrons. Bonding holds 8 and antibonding holds 6, so (8 − 6) ÷ 2 = 1. Every level is filled with pairs, so no unpaired electron is left.

6. What is the bond order of Ne₂, and what does that tell you?

Zero, so the molecule does not exist. Neon brings 8 valence electrons each, for 16. Bonding holds 8 and antibonding holds 8, so (8 − 8) ÷ 2 = 0. The antibonding levels cancel every gain, which is why neon stays a single atom.

7. Find the bond order of O₂⁺ and compare it to O₂.

2.5, higher than O₂. The ion has 11 valence electrons, one fewer than O₂. The missing electron comes out of the antibonding π*2p level. So bonding is 8 and antibonding is 3, giving (8 − 3) ÷ 2 = 2.5. Removing an antibonding electron strengthens the bond, so O₂⁺ is shorter and stronger than O₂.

8. Why can a single bond rotate freely while a double bond cannot?

A sigma bond is symmetric around the axis, but a pi bond is not. Head-on overlap looks the same at any twist, so a single bond spins freely. A pi bond needs its two p orbitals parallel. Twisting pulls them out of line and breaks the overlap, which costs far too much energy.

9. Which is shorter, a C–C bond or a C≡C bond?

The triple bond. More shared pairs pull the nuclei closer and hold them tighter. Bond order, length and strength move together: higher order means shorter and stronger. The single bond takes 347 kJ/mol to break, while the triple takes 839.

10. The O=O bond energy is 498 kJ/mol. What is the energy of one such bond in joules?

8.27 × 10⁻¹⁹ J. Convert to joules per mole: 498 kJ/mol = 498,000 J/mol. Then divide by Avogadro's number: 498,000 ÷ (6.022 × 10²³) = 8.270 × 10⁻¹⁹ J per bond. One bond is a tiny amount of energy, but a mole of them is a lot.

Chapter 9Gases

What to know

Pressure is a push spread over an area, so P = F/A. In a gas it is particles drumming on a wall. More particles, faster particles or a smaller box all raise the pressure. Learn the conversions: 1 atm = 760 torr = 760 mmHg = 101.325 kPa.

A barometer measures the air itself. A manometer compares two pushes. With a closed end, the height difference is the gas pressure outright. With an open end, you add or subtract the room's pressure depending on which side is higher.

Every gas law needs kelvin. Add 273.15 to a Celsius value before anything else. Using Celsius is the single most common error in this chapter, and it can even give you a negative temperature.

The simple laws all fall out of one combined law. Boyle holds temperature and amount, so P₁V₁ = P₂V₂. Charles holds pressure, so V₁/T₁ = V₂/T₂. Gay-Lussac holds volume, so P₁/T₁ = P₂/T₂. Avogadro holds pressure and temperature, so V₁/n₁ = V₂/n₂. The combined form is P₁V₁/T₁ = P₂V₂/T₂.

One equation does most of the work: PV = nRT. Use R = 0.08206 L·atm/mol·K with pressure in atm and volume in liters. Switch to R = 8.314 J/mol·K for energy and speed problems. The two-state laws compare two conditions. PV = nRT handles one condition at a time.

At standard temperature and pressure, one mole of any gas fills 22.4 L. STP here means 273.15 K and 1 atm. So the density of a gas at STP is just its molar mass divided by 22.4.

Density and molar mass follow from PV = nRT. Replace n with m/M and you get d = PM/RT. Rearranged, M = dRT/P. That is how you identify an unknown gas from a measured density.

Each gas in a mixture acts as if it were alone. So P_total is the sum of the partial pressures. The share each one takes follows its mole fraction: P_A = X_AP_total. A gas collected over water picks up water vapor, so subtract the water's vapor pressure to get the dry gas pressure.

Kinetic molecular theory explains why all this works. Temperature sets the average kinetic energy, and it is the same for every gas at a given temperature. Mass then sets the speed, so light molecules move faster. That is why light gases escape through a pinhole faster, which is Graham's law. Real gases stray from ideal when they are cold and squeezed, because then the particles' own volume and their attractions start to matter.

Lock these in

Kelvin or nothing. Add 273.15 first, every time.
1 atm = 760 torr = 101.325 kPa.
R = 0.08206 L·atm/mol·K, or 8.314 J/mol·K for energy work.
One mole fills 22.4 L at STP, whatever the gas.
Partial pressure follows the mole fraction, not the mass.
At the same temperature, every gas has the same average kinetic energy.

Equations, units and symbols

Equation	What each symbol means	Units
P = F ÷ A	P is pressure, F is force, A is area	Pa = N/m²
P₁V₁ = P₂V₂	Boyle's law, at fixed T and n	any matching P and V units
V₁ ÷ T₁ = V₂ ÷ T₂	Charles's law, at fixed P and n	V in L; T in K only
P₁ ÷ T₁ = P₂ ÷ T₂	Gay-Lussac's law, at fixed V and n	P in atm; T in K only
V₁ ÷ n₁ = V₂ ÷ n₂	Avogadro's law, at fixed P and T	V in L; n in mol
P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂	the combined gas law, at fixed n	T in K; P and V units just have to match
PV = nRT	P pressure, V volume, n moles, R the gas constant, T temperature	atm, L, mol, L·atm/mol·K, K
d = PM ÷ RT	d is density, M is molar mass	g/L; M in g/mol
M = dRT ÷ P	same equation, solved for molar mass	g/mol
P_total = P_A + P_B + ...	Dalton's law of partial pressures	all in the same pressure unit
X_A = n_A ÷ n_total, and P_A = X_AP_total	X_A is the mole fraction of gas A	X has no unit; P in any one unit
P_gas = P_total − P_water	for a gas collected over water	torr or atm, matched
KE_avg = (3/2)RT	average kinetic energy per mole	J/mol, using R = 8.314
u_rms = √(3RT ÷ M)	u_rms is the root-mean-square speed	m/s; R = 8.314 and M in kg/mol
rate_A ÷ rate_B = √(M_B ÷ M_A)	Graham's law of effusion	no unit; note the masses swap sides
(P + an²/V²)(V − nb) = nRT	van der Waals; a corrects for attraction, b for particle volume	a in L²·atm/mol²; b in L/mol

Worked examples

Worked example 1 · solving PV = nRT for volume

Find the volume of 2.50 mol of gas at 1.20 atm and 35 °C.

Step 1 · convert the temperature35 + 273.15 = 308.15 K. The 35 is known to the ones place, so use 308 K.

Step 2 · rearrangeV = nRT ÷ P.

Step 3 · put the numbers inV = (2.50 mol)(0.08206 L·atm/mol·K)(308 K) ÷ (1.20 atm).

Step 4 · multiply the top2.50 × 0.08206 = 0.20515. Then 0.20515 × 308 = 63.19 L·atm.

Step 5 · divide63.19 ÷ 1.20 = 52.66 L, so 52.7 L.

Check the unitsMoles cancel with moles, kelvin with kelvin, and atm with atm. Liters are left, which is what a volume needs to be.

Worked example 2 · the combined gas law

A 3.00 L balloon sits at 25 °C and 1.00 atm. It rises to where the pressure is 0.500 atm, and it warms to 100 °C. Find the new volume.

Step 1 · both temperatures to kelvinT₁ = 25 + 273.15 = 298 K. T₂ = 100 + 273.15 = 373 K.

Step 2 · rearrangeV₂ = P₁V₁T₂ ÷ (T₁P₂).

Step 3 · substituteV₂ = (1.00)(3.00)(373) ÷ [(298)(0.500)] = 1,119 ÷ 149 = 7.51 L.

Step 4 · report itThe new volume is 7.51 L.

Check it a second wayHalving the pressure should double the volume, giving 6.00 L. Warming from 298 K to 373 K multiplies by 373/298 = 1.25. Then 6.00 × 1.25 = 7.50 L. That matches within rounding.

Worked example 3 · identify a gas from its density

A gas has a density of 1.96 g/L at STP. Find its molar mass and name a likely gas.

Step 1 · pick the equationYou have density and want molar mass, so use M = dRT ÷ P.

Step 2 · list the conditionsAt STP, T = 273.15 K and P = 1.00 atm.

Step 3 · substituteM = (1.96 g/L)(0.08206 L·atm/mol·K)(273.15 K) ÷ (1.00 atm).

Step 4 · multiply1.96 × 0.08206 = 0.16084. Then 0.16084 × 273.15 = 43.93, so 43.9 g/mol.

Step 5 · name itCarbon dioxide is 44.01 g/mol, so the gas is very likely CO₂.

Check it a second wayAt STP you can skip R. Molar mass = d × 22.4 = 1.96 × 22.4 = 43.9 g/mol. Same answer, far less arithmetic.

Worked example 4 · partial pressures and effusion

A flask holds 0.200 mol of N₂ and 0.300 mol of O₂ at a total pressure of 2.50 atm. Find each partial pressure. Then compare how fast helium and oxygen effuse.

Step 1 · total moles0.200 + 0.300 = 0.500 mol.

Step 2 · mole fractionsX(N₂) = 0.200 ÷ 0.500 = 0.400. X(O₂) = 0.300 ÷ 0.500 = 0.600. They add to 1.000.

Step 3 · split the pressureP(N₂) = 0.400 × 2.50 = 1.00 atm. P(O₂) = 0.600 × 2.50 = 1.50 atm. The two add back to 2.50 atm.

Step 4 · set up Graham's lawrate(He) ÷ rate(O₂) = √(M of O₂ ÷ M of He) = √(32.00 ÷ 4.003).

Step 5 · take the root32.00 ÷ 4.003 = 7.994, and √7.994 = 2.83. So helium effuses 2.83 times faster.

Why the masses swapThe lighter gas is the faster one, so its mass belongs on the bottom inside the root. If you get an answer below 1 for the lighter gas, you flipped the fraction.

Traps people fall into

Leaving temperature in Celsius. Convert to kelvin before touching any gas law.
Using the wrong R. Pair 0.08206 with atm and liters. Pair 8.314 with joules, kg/mol and pascals.
Using 22.4 L away from STP. That molar volume only holds at 273.15 K and 1 atm.
Forgetting the water vapor. A gas collected over water needs P_water subtracted first.
Splitting pressure by mass. Partial pressure follows the count of particles, so use mole fractions.
Flipping Graham's law. The masses trade places, so the lighter gas comes out faster.
Using g/mol in u_rms. That equation needs kg/mol, so divide the molar mass by 1,000.

Chapter 9 self-test · ten items

1. Convert 745 torr to atmospheres, and 2.50 atm to kilopascals.

0.980 atm, and 253 kPa. First: 745 torr × (1 atm / 760 torr) = 0.9803 atm. Second: 2.50 atm × (101.325 kPa / 1 atm) = 253.3 kPa, so 253 kPa to three figures.

2. A gas fills 4.00 L at 1.50 atm. What volume does it fill at 3.00 atm, at the same temperature?

2.00 L. Use Boyle's law: V₂ = P₁V₁ ÷ P₂ = (1.50 × 4.00) ÷ 3.00 = 6.00 ÷ 3.00 = 2.00 L. Doubling the pressure halves the volume.

3. A gas fills 2.00 L at 273 K. What volume does it fill at 546 K, at the same pressure?

4.00 L. Use Charles's law: V₂ = V₁T₂ ÷ T₁ = (2.00 × 546) ÷ 273 = 4.00 L. Doubling the kelvin temperature doubles the volume. This only works in kelvin, since doubling 0 °C would be meaningless.

4. How many moles of gas fill 10.0 L at 2.00 atm and 300. K?

0.812 mol. Use n = PV ÷ RT = (2.00 × 10.0) ÷ (0.08206 × 300.) = 20.0 ÷ 24.618 = 0.8124 mol.

5. What volume does 1.00 mol of any gas fill at STP?

22.4 L. Check it with PV = nRT: V = (1.00)(0.08206)(273.15) ÷ 1.00 = 22.4 L. The identity of the gas does not matter, because an ideal gas particle takes up no space of its own.

6. What is the density of oxygen gas at STP?

1.43 g/L. At STP, density = molar mass ÷ 22.4. O₂ is 32.00 g/mol, so 32.00 ÷ 22.4 = 1.4286 g/L. That is far lighter than any liquid, which fits a gas.

7. A flask holds helium at 0.50 atm, neon at 1.20 atm and argon at 0.75 atm. What is the total pressure?

2.45 atm. Dalton's law just adds them: 0.50 + 1.20 + 0.75 = 2.45 atm. Each gas pushes as if the others were not there.

8. Hydrogen is collected over water at 25 °C. The total pressure is 755 torr, and water's vapor pressure is 23.8 torr. Find the pressure of the dry hydrogen.

731 torr. Subtract the water vapor: 755 − 23.8 = 731.2 torr, so 731 torr. Skipping this step would overstate the amount of hydrogen you collected by about 3%.

9. Which effuses faster, CH₄ or SO₂, and by what factor?

CH₄, by a factor of 2.00. CH₄ is 16.04 g/mol and SO₂ is 64.06 g/mol. Then rate(CH₄) ÷ rate(SO₂) = √(64.06 ÷ 16.04) = √3.994 = 1.998. Sulfur dioxide is almost exactly four times heavier, and the square root of 4 is 2.

10. Find the root-mean-square speed of N₂ at 300. K.

517 m/s. Use u_rms = √(3RT ÷ M) with R = 8.314 J/mol·K and M = 0.028014 kg/mol. The top is 3 × 8.314 × 300. = 7,482.6. Dividing gives 7,482.6 ÷ 0.028014 = 267,100, and the square root is 516.8 m/s. That is faster than a passenger jet, which is why a gas fills a room so quickly.

One pageCram sheet

Everything from chapters 7 to 9 in one place. Cover the right column and rebuild each row from memory.

Equations

Name	Equation	Units
Lattice energy trend	ΔH_lattice ∝ Q₁Q₂ ÷ r₀	kJ/mol
Bond polarity	ΔEN = EN_higher − EN_lower	no unit
Formal charge	FC = V − L − ½B	whole number
Bond order from resonance	bonds shared ÷ resonance forms	no unit
Bond order from orbitals	(bonding − antibonding) ÷ 2	no unit
Energy of one bond	(D × 1000) ÷ N_A	J per bond
Pressure	P = F ÷ A	Pa = N/m²
Boyle	P₁V₁ = P₂V₂	matching units
Charles	V₁/T₁ = V₂/T₂	T in K
Gay-Lussac	P₁/T₁ = P₂/T₂	T in K
Avogadro	V₁/n₁ = V₂/n₂	L and mol
Combined	P₁V₁/T₁ = P₂V₂/T₂	T in K
Ideal gas law	PV = nRT	atm, L, mol, K
Gas density	d = PM ÷ RT	g/L
Molar mass from density	M = dRT ÷ P	g/mol
Dalton's law	P_total = P_A + P_B + ...	matching units
Mole fraction	X_A = n_A/n_total, P_A = X_AP_total	X has no unit
Gas over water	P_gas = P_total − P_water	torr or atm
Average kinetic energy	KE_avg = (3/2)RT	J/mol
Molecular speed	u_rms = √(3RT ÷ M)	m/s; M in kg/mol
Graham's law	rate_A/rate_B = √(M_B/M_A)	no unit
Van der Waals	(P + an²/V²)(V − nb) = nRT	atm, L, mol, K

Constants and values

Quantity	Value	Note
Gas constant	R = 0.08206 L·atm/mol·K	use with atm and liters
Gas constant, energy form	R = 8.314 J/mol·K	use for KE_avg and u_rms
Pressure conversions	1 atm = 760 torr = 760 mmHg = 101.325 kPa	1 bar = 100 kPa, slightly under 1 atm
STP	273.15 K and 1 atm	molar volume there is 22.4 L/mol
Absolute zero	0 K = −273.15 °C	so K = °C + 273.15
Water vapor pressure	23.8 torr at 25 °C	subtract it for a gas over water
Avogadro's number	N_A = 6.022 × 10²³ /mol	for per-bond energies
Electronegativity values	F 3.98, O 3.44, Cl 3.16, N 3.04, C 2.55, H 2.20, Na 0.93, K 0.82	fluorine is the highest of all
Bond energies	H–H 436, Cl–Cl 242, F–F 159, O=O 498, N≡N 945, C–C 347, C≡C 839	all in kJ/mol
Molar masses used here	He 4.003, CH₄ 16.04, N₂ 28.01, O₂ 32.00, CO₂ 44.01, SO₂ 64.06	in g/mol

Shape and orbital tables

Domains	Electron geometry	Shapes as lone pairs are added	Hybrid
2	linear, 180°	linear	sp
3	trigonal planar, 120°	trigonal planar, then bent	sp²
4	tetrahedral, 109.5°	tetrahedral, then trigonal pyramidal, then bent	sp³
5	trigonal bipyramidal, 90° and 120°	trigonal bipyramidal, then see-saw, then T-shaped, then linear	sp³d
6	octahedral, 90°	octahedral, then square pyramidal, then square planar	sp³d²

Rules that are not equations

Topic	The rule
Bond type by ΔEN	Under 0.4 nonpolar covalent. From 0.4 to 1.8 polar covalent. Over 1.8 treat as ionic.
Lewis structures	Count electrons, place the least electronegative atom in the center, bond, fill the outer atoms, then form multiple bonds to complete shells.
Octet exceptions	Odd electron counts, short shells on Be and B, and expanded shells from period 3 down.
Best resonance form	Formal charges nearest zero, with any negative charge on the most electronegative atom.
Molecular polarity	Polar bonds plus a shape that does not cancel them. Symmetry with identical outer atoms cancels.
Sigma and pi	One sigma per bonded pair. A double adds 1 pi, a triple adds 2 pi.
Rotation	Sigma bonds rotate freely. Pi bonds lock the geometry, which creates cis and trans isomers.
Level order in period 2	B₂, C₂ and N₂ put π2p below σ2p. O₂, F₂ and Ne₂ swap them.
Magnetism	Any unpaired electron means paramagnetic. All paired means diamagnetic.
Real gases	They stray from ideal when cold and squeezed, because particle volume and attractions start to matter.