Atomic structure and the mole

THE ONE IDEA UNDER EVERYTHING: THE MOLE MAP

A mole is a count, exactly the way a dozen is a count. A dozen eggs is 12 eggs; one mole contains exactly $6.02214076\times 10^{23}$ specified entities. The current AP reference information rounds the Avogadro constant to $6.022\times 10^{23} \mathrm{mo}{l}^{-1}$; use the supplied value in AP calculations unless a prompt gives another. The scale of the unit is what connects an atom’s mass in unified atomic mass units to the molar mass in grams per mole. Carbon reads about 12.01 on the table; about 12.01 g of carbon contains one mole of carbon atoms. The periodic table therefore supplies the grams-per-mole conversion for an element.

Which is what makes the periodic table usable as a conversion factor, and that is really all a molar mass is. Many quantitative composition problems in this unit sit on one map, and each is a route across it:

Solid boxes are Unit 1. The dashed boxes are Unit 3 — same hub, two more spokes.

Three things to take off that picture. First, the failure mode: most wrong answers in this topic are attempts to build a route that skips the middle box — a mass multiplied straight by Avogadro’s number, say. If your work never wrote down a number of moles, you went cross-country instead of using the road.

Second, you do not learn eight conversions; you learn two, and read them backwards when you are traveling the other direction. Toward the hub you divide; away from the hub you multiply.

MOLAR MASS: BUILD IT, NEVER RECALL IT

A molar mass is two separate jobs, and the whole trick is doing them separately. Job one: count the atoms. Job two: do the arithmetic. Students who count and multiply in a single motion are the students who find four oxygens in aluminum sulfate.

Job one is where the mistakes live, because of parentheses. A subscript sitting outside a closing parenthesis multiplies everything inside it, and nothing outside it:

The invisible 1 on the S is the one students forget to multiply.

Now job two, and the discipline is to write one line per element rather than one long sum you cannot check:

Use the atomic masses printed on the supplied periodic table and carry guard digits through the calculation. Some values have two decimal places, some do not, and the final precision is set by the measured data rather than by a universal number of table decimals. The subtotals are worth writing down separately even when you could do it in one pass, because percent composition (Toolbox below) needs each of them again — and because a column of numbers can be re-checked, while a single calculator entry cannot.

Rebuild it every time; never quote it from memory. Molar mass is thirty seconds of arithmetic and it is the input to everything downstream — one wrong molar mass silently corrupts the mole count, the percent, the empirical formula, and every answer that depends on them. A later numerical check may not expose that shared setup error. The safest protection is building the molar mass where you can inspect it.

MASS SPECTRA OF ELEMENTS

Start with what an isotope is, because the rest is bookkeeping on top of it. Atoms of the same element all carry the same number of protons — that is what makes them that element — but they can carry different numbers of neutrons. Those variants are isotopes. Neutral isotopes of one element have the same electron configuration, so their ordinary chemical behavior is very similar, but isotope mass can affect rates, equilibria, and spectra. They differ in neutron count and mass, not in identity as an element.

A mass spectrometer ionizes particles and separates the resulting ions by mass-to-charge ratio, $m∕z$. In the simplified AP elemental spectra used here, the ions are singly charged monatomic ions, so the horizontal values can be read as isotope masses. Each peak says two things:

• where it sits horizontally — its dimensionless $m∕z$ value; because $z=1$ here, that numerical value equals the isotope mass expressed in unified atomic mass units, u (also called amu in some courses);
• how tall it is — how common that isotope is.

Normalize first if the heights are “relative intensity.” A spectrometer usually scales the tallest peak to 100 and reports the rest against it, so the heights are a ratio, not percentages — they will not sum to 100, and using them as though they did is the standard way to get this problem wrong. Divide each height by the total of all heights and you have fractional abundances that do sum to 1.

Then the atomic mass on the periodic table is the weighted average:

Why weighted, and not just averaged. A plain average of the isotope masses would be the right answer only if every isotope in the sum were equally common. Do not assume that. Picture the horizontal axis as a plank and each peak as a weight sitting on it: the weighted average is the point where the plank balances. Put a heavy weight near one end and the balance point slides toward it. That is all “weighted” means, and it gives you a prediction you can make before computing anything.

Two checks. The fractions must sum to 1, and the average must land between the least and greatest isotope masses. For a spectrum with exactly two isotopes, the average lies closer to the more abundant peak (and at the midpoint only when the abundances are equal). With three or more isotopes, the combined pull of all peaks matters, so “nearest the tallest” is not a general rule.

And keep hold of what the number means. No single atom has the average mass. The average describes the mixture; every individual atom in the sample belongs to one isotope. This distinction gets asked directly.

PERCENT COMPOSITION, IN BOTH DIRECTIONS

The same relationship gets run two ways, and knowing which way you are traveling is most of the tool selection.

Forward (formula $\to$ percents). Build exactly one mole of the compound and ask what fraction of its mass each element contributes:

Why one mole? Because any sample size gives the same percentages — percent composition does not depend on how much you have — so you may as well pick the size whose arithmetic you have already done. The subtotals from the molar-mass table are the numerators; the total is the denominator. No new computation at all.

Reverse (percents $\to$ formula). Now you want the subscripts, and subscripts are a mole ratio, not a mass ratio.

1.

Assume a 100 g sample. Then every percent becomes a mass in grams with no arithmetic — 78.7% becomes 78.7 g. This is a free move, again because composition does not depend on sample size.

2.

Convert each element’s grams to moles — divide by that element’s atomic mass, not the compound’s.

3.

Divide every mole count by the smallest one. This is the step whose purpose is worth stating: it rescales the ratio so the least-plentiful element comes out as exactly 1, and every other element is expressed as “how many per one of those.” That is what a subscript is.

4.

Clear any fraction to whole numbers. Atoms are not divisible in a formula — there is no such thing as $Y{O}_{1.5}$ — so a ratio of $1:1.5$ is not a finished answer. It is the finished answer written at half scale. Multiply both by 2 and you have $2:3$: the same ratio, now sayable in atoms.

Rounding and multiplying are opposite moves, and confusing them is the classic error here. A candidate fraction is supported only when the difference is consistent with the precision or uncertainty of the supplied data. A ratio such as 1.98 may represent 2 when that precision supports the match. A ratio of 1.5 is not a noisy 2; it is a fraction, and fractions get multiplied out rather than rounded off. Test small whole-number multipliers across every ratio, then verify the resulting formula against the measured composition. That procedure catches fifths such as $1.80=9∕5$ without treating every ugly decimal as a hidden fraction.

One more thing this procedure gives you, and one it does not. What comes out is the empirical formula — the simplest whole-number ratio. Percent composition alone cannot tell $C{H}_{2}O$ from $C_6{H}_{12}{O}_6$, because both have identical percentages. Prising those two apart takes one extra datum — the compound’s molar mass — and one extra step:

For a molecular substance with consistent molar-mass and composition data, that quotient counts how many empirical units make one molecule and should be near a positive whole number. A result such as 3.4 signals inconsistent or overly rounded data, an arithmetic error, or a model that does not apply to the substance; do not force it to an integer. Once the quotient is supported, multiply every subscript by it, not just the first.

Without that extra datum, empirical is the whole answer, and it is the right one. None of the problems in this guide hands you a molar mass, so every formula asked for here is empirical. The molecular step is the one move in this toolbox the problems do not drill — which is why its arithmetic is written out above rather than merely named.

MIXTURES: CHOOSE THE EQUATION THE SAMPLE SUPPORTS

A nested-percent chain is powerful when a measured element appears in only one component. If component A is the only source of nitrogen, for example, then

That unique element is a bridge: all of the measured amount must have come through one component.

Do not turn that convenient design into a universal recipe. For a binary mixture of known components A and B, write the one-fraction model

One elemental measurement determines $f_A$ when $w_{X,A}\ne w_{X,B}$ because $f_B=1-f_A$. More unknown components, unknown component compositions, or equal elemental fractions require more independent data. Count unknowns and independent relationships before declaring a mixture determined or underdetermined.

ELECTRON CONFIGURATION

A configuration is an inventory: where every electron in the atom lives. Electrons occupy subshells, and each orbital can hold at most two electrons. The number of orbitals therefore sets each subshell’s maximum capacity:

So the superscripts are not arbitrary — 2, 6, 10, 14 is just $2\times$ the orbital count, and a superscript larger than the capacity is an error you can catch on sight.

The fill order, without memorizing a string. Electrons go into the lowest-energy subshell available, but the energies interleave: 4s comes before 3d, 5s before 4d. Rather than memorize the sequence, build it. Write the subshells in a grid and read the diagonals:

1s · 2s · 2p 3s · 3p 4s · 3d 4p 5s · 4d 5p 6s · 4f 5d 6p

Noble-gas shorthand. Bracket the nearest noble gas before your element, then list only what comes after it. $\mathrm{\text{[Ar]}}$ stands in for the first 18 electrons. It saves writing and, more usefully, it puts the valence electrons where you can see them.

The operational 4s/3d rule. In the simplified Aufbau model used for first-row transition atoms, place 4s before 3d when building the neutral atom. When forming a cation, remove electrons from the occupied subshell with the highest principal quantum number first, so 4s electrons are removed before 3d electrons:

The underlying orbital energies are close and depend on electron occupancy and electron–electron interactions; a fixed two-line picture is not a literal account of every atom. For AP work, keep the two tasks separate: build the neutral configuration by the stated filling order, then remove from the highest $n$ first.

Anions need no special rule at all — just keep filling in the same order until the added electrons are placed.

Valence electrons, for main-group atoms, are the ones in the outermost $n$ — the s and p of that shell counted together. That count is what predicts the ion the atom forms.

Final step: audit the count. Count the electrons you actually placed and check the total against $Z$, adjusted for the charge. A neutral atom has $Z$ electrons; each negative charge adds one; each positive charge removes one. This catches an incorrect total or charge, but it cannot detect electrons placed in the wrong subshell; check both the count and the placement rule.

PHOTOELECTRON SPECTROSCOPY: A CONFIGURATION MADE VISIBLE

Photoelectron spectroscopy (PES) supplies photons of known energy to a gaseous sample and measures the kinetic energy of emitted electrons. Energy conservation gives the electron’s binding energy:

An electron that leaves with less kinetic energy required more energy to remove and therefore had the greater binding energy.

Read every PES representation in two directions. Horizontal position reports binding energy: core electrons lie at much greater binding energy than valence electrons. Peak area or relative intensity tracks how many electrons occupy that subshell. A $p^6$ peak therefore has about three times the area of an $s^2$ peak when the response per electron is treated alike. Check the printed axis before using left or right; PES axes are not always drawn in the same direction.

A schematic $1s^{2}2s^{2}2p^6$ spectrum: peak count identifies occupied subshells; relative area checks electron count.

To compare two atoms, first match the same subshell. If the peak for that subshell moves to greater binding energy, those electrons are held more tightly. Across a period, added nuclear charge with little change in core shielding generally shifts corresponding peaks to greater binding energy. Do not compare unmatched peaks merely because they are adjacent on the page.

COULOMB’S LAW RUNS THE PERIODIC TABLE

Many Unit 1 explanations — radius, ionization energy, electronegativity, ionic size, and PES binding energy — begin with the same question: how strongly is the nucleus attracting the electrons being compared? A useful electrostatic model is

For a many-electron atom, use effective nuclear charge, $Z_{\mathrm{\text{eff}}}$ — the net positive charge an electron “feels” after shielding — as the charge comparison. The electron’s charge is fixed, and $r$ represents electron–nucleus distance. Charge, shielding, and distance are connected rather than perfectly independent, but separating them is a useful first-pass way to identify the dominant change.

Arrow thickness is the strength of the pull, on one scale across both rows — so the $+1$-at-$r$ panel is the same picture top and bottom, and only the lever that moved differs. Doubling $r$ does not halve the force; it quarters it.

Estimating effective nuclear charge, quickly. The valence electron does not feel the full nuclear charge, because the core electrons between it and the nucleus screen part of that charge — shielding. The useful first-pass model for simple main-group comparisons is

Sodium: $11-10=+1$. Magnesium: $12-10=+2$. Chlorine: $17-10=+7$. This is a qualitative comparison, not an exact calculation: same-shell electrons also shield one another. Across Period 3, however, the core-electron count and principal valence shell stay the same while proton count rises, so $Z_{\mathrm{\text{eff}}}$ generally rises.

The familiar arrows are broad patterns, not substitutes for evidence. Read the data in a prompt first, then use structure to explain the pattern:

The mechanism behind the broad pattern:

A reliable explanation pattern connects structural evidence to Coulombic attraction and then to the requested property:

Going from Na to Cl, the nuclear charge rises from 11 to 17 while the valence electrons stay in the same $n=3$ shell behind the same 10-electron core. Effective nuclear charge therefore rises, strengthening the attraction and pulling the valence-electron density inward, so atomic radius decreases.

The exact evidence required depends on the comparison. A circular statement such as “chlorine is smaller because its radius is smaller” supplies no mechanism; an explanation names the relevant structural change and connects it to attraction.

IONS, ELECTRON AFFINITY, AND ANALOGOUS COMPOUNDS

Ionic radius needs its own comparison. A cation is generally smaller than its neutral atom because electrons were removed and the remaining cloud experiences a stronger attraction per electron.

An anion is generally larger because added electron–electron repulsion expands the cloud. For an isoelectronic series, electron count and occupied shells are the same, so more protons means a smaller ion. Do not apply a neutral-atom arrow before checking charge and electron count.

Electron affinity describes the energy change when a gaseous atom gains an electron. Across a period it generally becomes more favorable (often described as more exothermic), while down a group it generally becomes less favorable, but subshell stability and electron repulsion create important exceptions. Some tables report energy released as a positive magnitude and others report the process enthalpy with a negative sign; read the convention printed with the data. When numerical data are supplied, the data outrank a memorized arrow.

Valence structure predicts common ions and analogous formulas. Main-group elements in one column share a valence-electron pattern, so they often form ions with the same charge. Balance total ionic charge to zero: $A{l}^{3+}$ with $O^{2-}$ gives $A{l}_2{O}_3$, and the same charge pattern predicts $G{a}_2{O}_3$. “Analogous” does not mean identical. Atomic size, bond energies, physical state, surface condition, and activation barriers can change properties and reaction rates; use the actual evidence the prompt provides.