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UNIT 1About 15 min + practice

Atomic Structure and Properties

Use particle structure to explain measurable chemical patterns.

What you’ll learn

  • Convert between particles, moles, and mass.
  • Interpret isotope and electron data.
  • Explain periodic trends using nuclear charge, shielding, and distance.
01

Before you begin

Atomic number counts protons; mass number counts protons plus neutrons. An ion forms by changing electron count, not by changing the nucleus. A mole is a counting unit, while molar mass connects a count of particles to a measured mass.

Explain these starting ideas in your own words. Revisit them whenever a later step feels unclear.

02

Count with moles; conserve with units

A mole represents a specified number of entities. Molar mass connects mass to amount; Avogadro’s constant connects amount to particle count. Always name the entity: one mole of O2 molecules contains two moles of oxygen atoms. A balanced chemical equation relates amounts in moles, not equal masses.

Dimensional analysis makes the reasoning visible. Write a conversion factor so the unwanted unit cancels. For a mixture, distinguish the mass of the whole sample from the mass of the desired substance. A percent-composition calculation can identify an empirical formula but does not alone determine the molecular formula.

Laboratory glassware gives context to quantitative chemistry
Laboratory glassware gives context to quantitative chemistry

Choose apparatus suited to the measurement: preparation, transfer, and measurement are different tasks. The photograph does not establish the identity, concentration, or equilibrium state of a solution.

Photo: Belikov Maxim · Source · CC BY 4.0 · Unmodified.
n =
N = nNa
03

Isotope abundance produces an average

Isotopes share a proton number but have different neutron numbers. An element’s average atomic mass is a weighted average of isotope masses, using fractional abundances that sum to one. It need not equal the mass of any individual atom. A mass-spectrum peak position conveys mass-to-charge information, while relative intensity conveys relative abundance under the stated setup.

An empirical formula gives the smallest whole-number atom ratio. Divide each element’s amount by the smallest amount, then use a justified multiplier if a simple fraction remains. Do not force 1.50 into 2 by ordinary rounding; multiplying all ratios by two preserves the composition.

PAUSE & TRY IT

Why does an empirical formula not establish molar mass?

Reveal answer

Multiple molecular formulas can have the same simplest atom ratio.

04

Electron arrangements are models of energy

Electrons occupy orbitals grouped into shells and subshells. An orbital holds at most two electrons with opposite spins. Electrons occupy equal-energy orbitals singly before pairing under the usual ground-state filling model. Valence-electron patterns help explain recurring bonding behavior.

For cations, remove electrons from the highest principal energy level before removing electrons from an inner d subshell. Photoelectron spectroscopy provides evidence about electron binding energies and populations. Larger binding energy means more energy is required to remove an electron; peak areas or heights must be interpreted according to the graph’s stated convention.

05

Explain trends with competing effects

Across a period, nuclear charge generally increases while added electrons occupy the same main shell. Increased effective nuclear attraction tends to reduce atomic radius and increase ionization energy. Down a group, larger principal shells and shielding generally increase radius and reduce the attraction experienced by outer electrons.

These are patterns with meaningful exceptions, not slogans. Subshell energy and electron pairing can explain some departures in first ionization energy. A large jump in successive ionization energies suggests that the next electron would be removed from a more tightly bound inner shell after the valence electrons are gone.

“More protons” is incomplete if the compared particles also have different electron shells or shielding.

PAUSE & TRY IT

What does a large jump after the second ionization suggest for a main-group atom?

Reveal answer

It suggests two relatively accessible valence electrons before removal begins from a more tightly bound inner shell.

06

Ions and composition evidence

A cation is generally smaller than its parent atom because electron removal can reduce electron repulsion and may remove the outer shell. An anion is generally larger than its parent atom. Within an isoelectronic series, more protons generally pull the same number of electrons into a smaller radius.

Chemical analysis is constrained by measurement uncertainty and purity. If a hydrated sample is not heated to constant mass, remaining water can make the calculated water-to-salt ratio too low. Explain an error by following the measured quantity into the calculation rather than memorizing “high” or “low.”

PAUSE & TRY IT

Which is smaller in the isoelectronic series O2- and Mg2+, and why?

Reveal answer

Mg2+ has more protons attracting the same number of electrons and is smaller.

07

Use particle accounting before a formula

For an isotope, subtract atomic number from mass number to find neutron count. For an ion, begin with the neutral electron count and subtract electrons for a positive charge or add them for a negative charge. Isotopes of an element share proton count but differ in neutrons. Different ionic charges of the same element share the nucleus but differ in electrons.

A measured atomic mass is a weighted average of isotopic masses, not usually the mass number of one isotope. Convert percent abundances to fractions before averaging, and check that they sum to one. In a mass spectrum, peak locations identify mass-to-charge ratios under the experiment’s conditions; relative peak intensities supply abundance information. A large peak indicates many detected particles of that type, not necessarily a heavier particle.

PAUSE & TRY IT

Why can a large jump in successive ionization energies reveal the number of valence electrons?

Reveal answer

After valence electrons are removed, the next electron is often in a much more strongly bound inner shell.

08

Explain photoelectron spectra from electron arrangement

Photoelectron spectroscopy measures the energy needed to remove electrons. Electrons held more strongly require larger binding energies. Inner-shell electrons are generally more tightly bound than valence electrons. Within a simple atomic spectrum, groups of peaks reflect occupied subshells, and relative peak areas can reflect electron populations. Always inspect the axis direction, because a plot can place high binding energy on either side.

To compare atoms, reason about nuclear charge, distance, and shielding. Across a period, increasing effective nuclear attraction often contracts atoms and raises ionization energy, but subshell and electron-pairing effects create exceptions. Removing successive electrons can reveal a large jump after valence electrons are gone. Do not interpret that jump as evidence that the nucleus gained protons; the next electron comes from a more strongly bound shell.

Separate the nucleus from the electron cloud

This symbolic model distinguishes nuclear particles from electrons. Real electrons occupy quantum orbitals, not the illustrated circular tracks; particle counts here are illustrative.

Separate the nucleus from the electron cloudNucleus: protons + neutronsElectrons occupy orbitalsNot fixed circular tracksAtomic identity depends on proton count; ions differ in electron count.
Original ScienceHub diagram · Schematic, not to scale.

PAUSE & TRY IT

What is the first conversion in an empirical-formula problem?

Reveal answer

Convert elemental masses or a chosen 100 g percentage basis into mole amounts.

09

Connect composition data to an empirical formula

Convert each element’s mass to moles using its molar mass. Divide all mole amounts by the smallest, then identify a simple whole-number ratio. If a ratio is close to 1.5, multiplying every ratio by two is appropriate; rounding 1.5 to 2 changes the chemistry. The empirical formula is the simplest atom ratio, whereas a molecular formula can be an integer multiple.

When a sample is hydrated, heating can remove water while leaving a salt. Use the mass lost to calculate moles of water and the dry mass to calculate moles of salt. Repeated heating to constant mass helps show that drying is complete. Incomplete drying makes the apparent water loss too small; losing salt by splattering can make it too large. Link each experimental error to the measured quantity before predicting the final ratio.

10

Move deliberately between atoms, moles and measured mass

A balance measures mass, while a chemical formula describes particle ratios. The mole connects these descriptions. Divide a sample mass by its molar mass to obtain amount, and multiply by the number of relevant atoms per formula unit when counting a particular element. For example, one mole of Al2O3 contains two moles of aluminum atoms and three moles of oxygen atoms. It does not contain five moles of Al2O3.

An empirical formula reports the simplest whole-number ratio. Convert every elemental mass to moles before comparing. Dividing grams by grams compares mass contributions, not atom counts. If normalized mole ratios are near 1:1.5, multiply all ratios by two rather than rounding 1.5 to an arbitrary integer. A molecular formula additionally requires a molecular molar mass; its multiplier must apply to the entire empirical formula.

Experimental composition has uncertainty. A small noninteger departure may reflect measurement error, but a large discrepancy can signal water remaining in a hydrate, product loss, contamination or a wrong chemical assumption. Explain the direction of an error by tracing how the affected measurement enters the ratio. Do not use “human error” as a substitute for that chain.

11

Distinguish the atomic experiments and what they measure

Isotopes have the same proton number and different neutron numbers. A mass spectrum can separate ions by mass-to-charge ratio, so interpreting a peak requires attention to charge state and whether the ion is an atom or a fragment. In a simplified singly charged atomic spectrum, relative abundances weight isotope masses. The resulting average is a population property; no individual atom needs to possess exactly that mass.

Photoelectron spectroscopy concerns the energy needed to remove electrons. Peak positions indicate binding energies and relative areas reflect electron populations under the supplied model. Inner electrons are generally more tightly bound. Comparing spectra of different atoms requires considering nuclear charge, shielding and occupied shells, not just the visual height of a peak.

A spectrum is evidence used with a model. Always read the axis direction: some binding-energy axes decrease toward the right. A peak drawn farther left does not universally mean a lower energy. Explain an assignment using both the relevant energy and electron count; do not treat a spectral peak as a photograph of an orbital.

PAUSE & TRY IT

An isotope mixture has 75% mass-20 atoms and 25% mass-22 atoms. Find the approximate mean mass.

Reveal answer

0.75(20)+0.25(22)=20.5 u. This weighted mean is not the mass of a new isotope.

12

Explain periodic behavior through competing attractions

Across a period, increasing nuclear charge generally attracts valence electrons more strongly when added electrons enter the same principal shell and shielding does not fully offset the change. This often reduces radius and raises the energy required to remove an electron. Down a group, occupied shells become larger and shielding increases; outer electrons can be farther from the nucleus despite the greater nuclear charge.

For an isoelectronic series, the electron count is fixed, so increasing proton number generally decreases radius. Forming a cation removes electrons and can remove an outer occupied shell; forming an anion adds electron–electron repulsion. Compare species using the actual electron arrangements rather than a memorized claim that every ion is larger than every atom.

Successive ionization energies refer to repeated removal from the same species as its charge changes. A large jump can indicate that the next electron would come from an inner shell after the valence electrons have been removed. Explain the jump in terms of binding and shell structure. It is not evidence that the nucleus suddenly acquired extra protons.

Successive ionization energies reveal a shell changeIllustrative model, not collected experimental data. Schematic values show a large jump after two removals, supporting two valence electrons for this model atom. These are not measured values for a named element.
Successive ionization energies reveal a shell change0510152001234 Electron removedIonization energy (relative units)Successive removals
Read figure values as text

Successive removals: 1: 1; 2: 2; 3: 12; 4: 16

13

Use experimental mass data to infer a formula

An empirical formula gives the simplest whole-number ratio of atoms. Convert each element’s mass to moles, divide by the smallest mole amount, and inspect the ratios. If a ratio is near a simple fraction such as 1.5, multiply all ratios by an appropriate integer rather than rounding it to 2.

A molecular formula is an integer multiple of the empirical formula. Divide the measured molar mass by the empirical-formula mass, then multiply every subscript by that factor. The empirical formula alone cannot distinguish molecules with the same simplest ratio.

For a hydrate, the mass lost on appropriate heating may represent water, while the residue represents the anhydrous salt. This interpretation requires complete water removal without decomposition or loss of solid. An experimental error should be traced into the mole ratio: incomplete drying makes the calculated water amount too small.

PAUSE & TRY IT

Why should an empirical ratio of 1:1.5 not be rounded to 1:2?

Reveal answer

Multiplying both by two gives the small whole-number ratio 2:3. Direct rounding changes the composition.

14

Read spectra as evidence about energy and abundance

In mass spectrometry, isotope peaks distinguish masses and relative abundances. The weighted average atomic mass uses abundance fractions, not an unweighted average of isotope masses. A peak’s height can represent relative abundance; its horizontal position represents a mass-to-charge quantity under the instrument’s convention.

Photoelectron spectroscopy measures energy needed to remove electrons. More tightly bound electrons require higher binding energy. Peak intensities relate to electron populations in subshells in a simplified interpretation. Compare axis direction carefully because binding-energy plots are not always arranged left to right in the same way.

Across a period, changes in effective nuclear attraction help explain trends, but electron shielding, subshell structure, and pairing can create exceptions. Explain a trend by the interaction between nucleus and electrons rather than asserting that larger atomic number alone always determines every measured property.

FROM IDEA TO APPLICATION

Worked examples

EXAMPLE 1

Recover an isotope abundance

An element has two isotopes of mass 10.0 u and 11.0 u. Its average atomic mass is 10.8 u. Estimate their abundances.

Reveal worked solution
  1. Let x be the fraction of the 10.0 u isotope; the other fraction is 1 − x.
  2. 10.0x + 11.0(1 − x) = 10.8.
  3. Solving gives x = 0.20.
Result & interpretation

Approximately 20% lighter isotope and 80% heavier isotope under the two-isotope model.

EXAMPLE 2

Find an empirical formula

A compound contains 0.20 mol C, 0.40 mol H, and 0.10 mol O. Determine its empirical formula.

Reveal worked solution
  1. Divide all amounts by 0.10 mol.
  2. The ratio is C:H:O = 2:4:1.
  3. The ratio is already a smallest whole-number ratio.
Result & interpretation

C2H4O. A molecular mass would be needed to determine whether the molecular formula is this formula or an integer multiple.

EXAMPLE 3

A two-isotope weighted mean

An element has isotopes of mass 24.0 and 26.0 u with abundances 75.0% and 25.0%. Find the average mass.

Reveal worked solution
  1. Convert abundances to 0.750 and 0.250.
  2. Multiply each mass by its abundance and add.
Result & interpretation

24.0(0.750)+26.0(0.250)=24.5 u. The average need not equal either isotope’s mass.

EXAMPLE 4

Find composition from a hydrate

A 4.50 g hydrate leaves 2.88 g anhydrous salt of molar mass 160 g/mol. Estimate the number of waters per formula unit.

Reveal worked solution
  1. Water lost: 4.50−2.88=1.62 g.
  2. Water amount: =0.0900 mol. Salt amount: =0.0180 mol.
  3. Water-to-salt ratio: =5.
Result & interpretation

The measured composition supports salt·5H2O, assuming all lost mass is water and the salt does not decompose.

EXAMPLE 5

Empirical and molecular formulas

A compound is 40.0% C, 6.7% H, and 53.3% O by mass, with molar mass about 180 g/mol.

Reveal worked solution
  1. Use a 100 g basis: approximately 3.33 mol C, 6.7 mol H, and 3.33 mol O.
  2. Divide by 3.33 to get CH2O, formula mass about 30 g/mol.
  3. =6, so multiply subscripts by six.
Result & interpretation

The molecular formula is C6H12O6.

EXAMPLE 6

A drying error

A hydrate is not heated long enough. How does the calculated water-to-salt mole ratio change?

Reveal worked solution
  1. The apparent water loss is too small.
  2. The residue includes water and appears too massive for pure anhydrous salt.
Result & interpretation

Both effects make the calculated water-to-salt ratio too low under this interpretation.

MAKE THE DISTINCTION

Common mistakes, clearer reasoning

The trapAtomic mass is the number of protons plus electrons.

The better explanationThe periodic-table value is an isotope-weighted average mass; an isotope’s mass is dominated by protons and neutrons.

The trapEvery periodic trend follows without exceptions.

The better explanationUse electron structure and the actual comparison, including shell, subshell, and pairing effects.

RETRIEVE BEFORE YOU REVEAL

Practice checkpoints

Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.

1. Which is smaller in the isoelectronic series O2- and Mg2+, and why?

Reveal answer

Mg2+ has more protons attracting the same number of electrons and is smaller.

2. Why does an empirical formula not establish molar mass?

Reveal answer

Multiple molecular formulas can have the same simplest atom ratio.

3. What does a large jump after the second ionization suggest for a main-group atom?

Reveal answer

It suggests two relatively accessible valence electrons before removal begins from a more tightly bound inner shell.

4. Why can a large jump in successive ionization energies reveal the number of valence electrons?

Reveal answer

After valence electrons are removed, the next electron is often in a much more strongly bound inner shell.

5. What is the first conversion in an empirical-formula problem?

Reveal answer

Convert elemental masses or a chosen 100 g percentage basis into mole amounts.

6. An isotope mixture has 75% mass-20 atoms and 25% mass-22 atoms. Find the approximate mean mass.

Reveal answer

0.75(20)+0.25(22)=20.5 u. This weighted mean is not the mass of a new isotope.

7. Why should an empirical ratio of 1:1.5 not be rounded to 1:2?

Reveal answer

Multiplying both by two gives the small whole-number ratio 2:3. Direct rounding changes the composition.

Key language

Mole
An amount containing Avogadro’s number of specified entities.
Isotope
Atoms of one element with different neutron numbers.
Effective nuclear charge
The net nuclear attraction experienced by an electron after shielding effects.
Empirical formula
The simplest whole-number ratio of atoms.
Connect it to the course

Electron structure explains bonding, intermolecular behavior, and periodic reactivity in later units.

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Written for ScienceHub · Original instructional material. Course framework reference ↗. These notes are independently authored and are not College Board materials. External photographs retain their credited licenses.

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