Lesson 3

Atomic Structure and Isotopes

Explore atomic structure: protons, neutrons, electrons, isotopes, atomic mass, and the history of atomic models.

7 learning objectivesatomic structure

Historical Development of Atomic Theory

Our understanding of atoms evolved over centuries through key experiments and models:

Dalton's atomic theory (1807) — Matter consists of indivisible atoms; atoms of the same element are identical in mass and properties; compounds form when atoms of different elements combine in simple whole-number ratios. These postulates explained the laws of definite and multiple proportions.
Thomson's “plum-pudding” model (1897) — The discovery of the electron via cathode-ray experiments showed atoms contain negatively charged subatomic particles embedded in a diffuse positive charge.
Millikan's oil-drop experiment (1909) — Measured the charge of an individual electron (1.602 × 10⁠^-19 C), allowing precise determination of its mass (9.109 × 10⁠^-28 g).
Rutherford's gold-foil experiment (1911) — Most alpha particles passed straight through a thin gold foil, but a few deflected sharply. These observations supported a model in which positive charge and nearly all atomic mass are concentrated in a tiny, dense nucleus, with electrons occupying the surrounding space.
Bohr model and beyond — Later work introduced energy levels and quantum mechanics, but the nuclear model remains the foundation of our picture of the atom.

Go deeperWhy this matters: science that corrects itself

Atomic models changed when experiments exposed their limits. Rutherford expected nearly all alpha particles to pass through Thomson’s diffuse positive charge, and most did; the rare large deflections showed that positive charge and most atomic mass occupy a much smaller region than Thomson’s model allowed. Rutherford followed that evidence and proposed the nuclear atom. The episode is a useful picture of science at work: a model remains valuable while it explains observations, then is revised when reproducible evidence demands a better one.

Go deeperHow empty is an atom, really?

The numbers are hard to picture: a typical nucleus is roughly 10⁠^-15 to 10⁠^-14 m across, while an atom is roughly 10⁠^-10 m across. Depending on the atom, its diameter is therefore about ten thousand to one hundred thousand times the nuclear diameter. Scale a nucleus to a 1 cm marble and the atom would span roughly 100 m to 1 km.

The nucleus still holds essentially all the mass. Rutherford scattering revealed this compact structure because most alpha particles passed through the atom with little deflection, while the rare alpha particle that passed very close to the small, positively charged nucleus experienced strong electrostatic repulsion and changed direction sharply.

Subatomic Particles: Protons, Neutrons, and Electrons

Three subatomic particles account for the structure of every atom:

Particle	Charge	Mass (amu)	Location
Proton (p⁠⁺)	+1	1.00727	Nucleus
Neutron (n⁠⁰)	0	1.00866	Nucleus
Electron (e⁠^-)	−1	0.000549	Electron cloud

Protons and neutrons reside in the extremely small, dense nucleus, which has a radius on the order of 10⁠^-15 m. Electrons occupy the vast space around the nucleus (atomic radius ∼10⁠^-10 m). Because an electron's mass is roughly 1/1836 that of a proton, nearly all of an atom's mass comes from its nucleus. In a neutral atom, the number of protons equals the number of electrons, so the charges balance to zero.

Go deeperWhy this matters: you are mostly empty space

Because nearly all atomic mass is concentrated in nuclei that occupy only a tiny fraction of atomic volume, ordinary matter has a remarkably low average density compared with nuclear matter. Compressing matter to approximately nuclear density would reduce macroscopic objects to extremely small volumes; neutron stars approach this density, which is why even a teaspoon of neutron-star material would have an enormous mass. This comparison is only a scale analogy: neutron-star matter is not simply ordinary atomic nuclei packed together unchanged.

Go deeperWhy electron mass is often negligible in chemistry calculations

The electron’s tiny mass creates a useful division in introductory chemistry. Atomic mass and element identity are concentrated in the nucleus, while chemical behavior is governed mainly by the electron distribution. When atoms interact chemically, their electron clouds overlap while their nuclei remain far apart on the nuclear scale, so bonding and ordinary reactions primarily rearrange electrons.

This is why electron mass is usually negligible in introductory atomic-mass calculations (one electron has about 0.05% of a proton’s mass), while nuclear changes are excluded from ordinary chemical reactions. The nucleus can still affect chemistry indirectly through nuclear charge and isotope effects, so the division is useful without being absolute.

Atomic Number and Mass Number

Two integers define the composition of any specific nuclide:

Atomic number (Z) — the number of protons in the nucleus. This determines the element's identity. Every carbon atom has Z = 6; every iron atom has Z = 26.
Mass number (A) — the total number of protons plus neutrons: A = Z + N, where N is the neutron count.

From these two values you can find any particle count:

Number of protons = Z
Number of neutrons = A − Z
Number of electrons = Z (for a neutral atom) or Z − charge (for an ion)

For example, a neutral atom of gold has Z = 79 and A = 197, so it contains 79 protons, 118 neutrons (197 − 79), and 79 electrons.

Go deeperWhy this matters: why the alchemists never made gold

Medieval alchemists spent centuries trying to turn lead into gold with furnaces, acids, and distillations. Every one of those is a chemical tool, and chemical processes only rearrange electrons; they cannot touch the nucleus. Lead is lead because Z = 82, gold because Z = 79, and no flask can remove three protons.

Changing Z requires nuclear reactions, and modern particle accelerators genuinely can transmute mercury or lead into gold, a few atoms at a time, at a cost astronomically higher than the gold is worth. The atomic number is not a label on the element; it is the element.

Go deeperTry it: decode two nuclides

Work these two, then check below.

How many protons and neutrons are in iron-56?
An atom has 47 protons and 61 neutrons. What is it?

Answers: Iron has Z = 26, so iron-56 has 26 protons and 56 − 26 = 30 neutrons. An atom with 47 protons is silver (Z = 47 identifies the element regardless of neutrons), specifically silver-108 (47 + 61 = 108). Notice the two directions: the name gives you Z from the periodic table; the proton count gives you the name.

Determining Particle Counts in Atoms and Ions

When an atom gains or loses electrons it forms an ion. Metals typically lose electrons to form cations (positive charge), while nonmetals gain electrons to form anions (negative charge). The number of protons and neutrons does not change when an ion forms.

To find electron count in an ion, start from the neutral atom count (Z) and adjust by the charge:

Na (Z = 11) → Na⁠⁺: 11 − 1 = 10 electrons
Cl (Z = 17) → Cl⁠^-: 17 + 1 = 18 electrons
Fe (Z = 26) → Fe⁠³⁺: 26 − 3 = 23 electrons
O (Z = 8) → O⁠^2-: 8 + 2 = 10 electrons

The general formula: electrons = Z − (charge), where the charge carries its sign. A +3 charge means 3 fewer electrons; a −2 charge means 2 more electrons.

Go deeperWhy this matters: your nerves run on ions

Neural signaling depends on controlled movements of ions such as Na⁠⁺ and K⁠⁺ across cell membranes, and Ca⁠²⁺ helps regulate processes including muscle contraction and neurotransmitter release. Hospitals monitor blood electrolytes because disturbances in K⁠⁺, Na⁠⁺, or Ca⁠²⁺ can disrupt normal nerve and heart function. The particle-count bookkeeping here describes the charged species present in biology: aqueous sodium ions have one fewer electron than neutral sodium atoms and behave very differently from elemental sodium metal.

Go deeperCommon mistake: giving the ion different protons

Seeing a positive charge, some students add protons: “Na⁠⁺ must have 12 protons.” Protons never change during ion formation; a chemical process cannot reach the nucleus. Every sodium species (Na, Na⁠⁺, sodium in a compound) has exactly 11 protons, or it is not sodium.

Ions form purely by electron traffic: lose electrons and the unchanged positive nucleus outnumbers the remaining electrons (cation); gain electrons and they outnumber the protons (anion). Run the check: charge = protons − electrons. For Na⁠⁺: 11 − 10 = +1. If your particle counts fail that equation, one of them is wrong.

Isotopes and Isotopic Notation

Isotopes are atoms of the same element (same number of protons) that differ in the number of neutrons, and therefore differ in mass number. For example, carbon has three naturally occurring isotopes:

⁠¹²C — 6 protons, 6 neutrons (mass number 12)
⁠¹³C — 6 protons, 7 neutrons (mass number 13)
⁠¹⁴C — 6 protons, 8 neutrons (mass number 14)

Isotopic notation places the mass number as a superscript and the atomic number as a subscript to the left of the element symbol. In plain text this is often written as carbon-12 or C-12. Isotopes have nearly the same electron structure and broadly similar chemistry, but their different masses can produce measurable isotope effects; their nuclear stability and exact atomic masses can also differ.

Go deeperWhy this matters: dating a mammoth with carbon-14

While an organism lives, it exchanges carbon with its environment, so its ⁠¹⁴C/⁠¹²C ratio tracks the carbon reservoir from which it feeds. At death that exchange stops and radioactive ⁠¹⁴C decays with a half-life of about 5,730 years while stable ⁠¹²C remains. Measuring the remaining ratio can date suitable organic material back roughly 50,000 years.

The basic method works because ⁠¹⁴C and ⁠¹²C participate in the same carbon chemistry closely enough for organisms to incorporate both. The starting ratio is not perfectly universal: atmospheric changes, marine or freshwater reservoir effects, and isotopic fractionation require calibration and context when precise calendar ages are reported.

Go deeperHeavy water is still water

Replace both hydrogens in H⁠₂O with deuterium (hydrogen-2, one proton plus one neutron) and you get heavy water, D⁠₂O. It retains water’s basic molecular connectivity and many familiar behaviors, but it is not physically or chemically identical to H⁠₂O: its density, phase-change temperatures, acid-base equilibria, and some reaction rates differ measurably. Each D⁠₂O molecule is about 11% heavier (20 u versus 18 u). Nuclear reactors exploit deuterium’s nuclear properties when heavy water is used as a neutron moderator.

Hydrogen shows especially large isotope effects because adding one neutron approximately doubles the mass of a hydrogen atom, while carbon going from 12 to 13 gains only about 8%.

Calculating Average Atomic Mass

The standard atomic weight listed for many elements on a periodic table represents the atomic masses expected for normal terrestrial materials. For an element with multiple relevant isotopes, a specified sample’s atomic mass is calculated as an abundance-weighted average:

average atomic mass = ∑(fractional abundance × isotopic mass)

For example, chlorine has two major isotopes: ⁠³⁵Cl (mass 34.969 u, 75.76% abundant) and ⁠³⁷Cl (mass 36.966 u, 24.24% abundant):

average = (0.7576 × 34.969) + (0.2424 × 36.966) = 26.49 + 8.96 = 35.45 u

This chlorine value is not the mass of any single atom; it reflects the isotope mixture in the sample. The more abundant isotope “pulls” the weighted average closer to its mass. Natural isotope abundances can vary slightly among samples, which is why some modern periodic tables show atomic-weight intervals for certain elements.

Go deeperWhy this matters: no atom weighs 35.45

The periodic-table mass is a population statistic, like an average household size: it need not describe any individual member. No chlorine atom has a mass of 35.45 u; a normal sample contains mainly ⁠³⁵Cl and ⁠³⁷Cl atoms, and its atomic mass is their abundance-weighted average. Mass spectrometry separates ions by mass-to-charge ratio and supplies the isotope-mass and abundance data behind these calculations. When you use a standard atomic weight in later stoichiometry, you are using a representative value for normal material, while recognizing that specialized samples can have different isotope ratios.

Go deeperTry it: estimate before you calculate

Boron has two major naturally occurring isotopes: ⁠¹⁰B (mass 10.013 u, 19.9% abundant) and ⁠¹¹B (mass 11.009 u, 80.1% abundant). Before computing, predict: the average must land between 10.013 and 11.009, and since ⁠¹¹B dominates, much closer to 11.

Check: (0.199 × 10.013) + (0.801 × 11.009) = 1.99 + 8.82 = 10.81 u, consistent with a typical periodic-table value. This estimate catches two common errors: an answer outside the two isotopic masses or one that lies nearer the minority isotope.

Electron Shells and Valence Electrons

For main-group atoms, a simplified Bohr-style shell model provides a useful first picture of electron arrangement for nomenclature and ionic-bond formation in Topics 3, 4, and 5. The full quantum-mechanical electron configuration treatment with subshells and orbitals arrives in Topic 12.

In this model, electrons in an atom are arranged in shells (also called energy levels) at increasing distances from the nucleus. The shells are numbered n = 1, 2, 3, …, with shell 1 closest to the nucleus and lowest in energy. For the first 20 elements, the simplified occupied-shell pattern is:

Shell 1 (the K shell): up to 2 occupied electrons
Shell 2 (the L shell): up to 8 occupied electrons
Shell 3 (the M shell): up to 8 occupied electrons through argon
Shell 4: starts to fill at potassium (Z = 19) with 1 electron, then calcium with 2

The sequence 2, 8, 8 is an occupancy pattern for the first 20 elements, not the full capacity of every shell. Subshells and the larger capacities of shells 3 and beyond are covered in Topic 12. Within this simplified range, fill from the inside out. The shell-by-shell distribution is written as a comma-separated list. Examples:

H (Z = 1): 1 — 1 electron in shell 1
He (Z = 2): 2 — shell 1 full
Li (Z = 3): 2, 1 — shell 1 full, 1 electron in shell 2
C (Z = 6): 2, 4
Ne (Z = 10): 2, 8 — shells 1 and 2 full
Na (Z = 11): 2, 8, 1
Si (Z = 14): 2, 8, 4
Ar (Z = 18): 2, 8, 8 — a filled valence shell in this main-group picture
K (Z = 19): 2, 8, 8, 1
Ca (Z = 20): 2, 8, 8, 2

The electrons in the outermost occupied shell are called valence electrons. They are the electrons that participate most directly in the chemical bonding discussed here. For main-group elements, the number of valence electrons equals the column position within the s/p block: Group 1 has 1, Group 2 has 2, Group 13 has 3, …, Group 17 has 7, Group 18 has 8 (except He, which has 2). Valence electrons drive the chemistry studied in Topics 3 through 7.

Go deeperWhy this matters: one column, one personality

Sodium can react vigorously with water because its single valence electron (2, 8, 1) is relatively easy to remove. Group 1 metals generally react with water, with the reaction becoming more vigorous down the group. Neon’s filled valence shell (2, 8) helps explain its very low chemical reactivity under ordinary conditions. Periodic-table columns group elements with similar valence-electron patterns, so members of a group often share important chemical trends, although their behavior is not identical.

Go deeperCommon mistake: counting the wrong electrons as valence

In this main-group shell model, valence electrons live only in the outermost occupied shell. For chlorine (2, 8, 7) that is 7, not 17 (the total) and not 8 (the second shell’s count). For potassium (2, 8, 8, 1) it is just 1: the single shell-4 electron, even though shell 3 below it holds 8. (Transition metals complicate this, because their inner d electrons can also take part in bonding; that refinement comes with the full electron-configuration treatment later.)

Fast cross-check for main-group elements: the valence count must match the group position (Group 1 → 1, Group 2 → 2, Groups 13 through 18 → 3 through 8). If your shell diagram disagrees with the column of the periodic table, recount the diagram.

Lewis Dot Symbols

A Lewis dot symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons. The dots are placed around the four sides of the symbol (top, right, bottom, left). By convention, place the first four dots singly — one on each side — before pairing any. After all four sides have one dot, additional electrons go on as pairs.

For main-group elements, the number of dots equals the number of valence electrons (Group 1 has 1 dot, Group 17 has 7, Group 18 has 8 except He which has 2). Worked examples:

Group 1 — Na (1 valence e−): one dot on Na
Group 2 — Mg (2 valence e−): two dots on opposite sides
Group 13 — Al (3 valence e−): three dots on three sides
Group 14 — C (4 valence e−): one dot on each of the four sides — the “singly-occupy first” rule means no pairing yet
Group 15 — N (5 valence e−): four singly-occupied sides plus one extra electron, which pairs on one side
Group 16 — O (6 valence e−): two pairs and two single dots
Group 17 — Cl (7 valence e−): three pairs and one single dot
Group 18 — Ne (8 valence e−): four pairs — a full octet

The exact orientation of dots is conventional — what matters is the valence-electron count and the singly-vs-paired pattern, which help predict common bonding patterns. Lewis dot symbols are the foundation for ionic-bond and covalent-bond drawings in later topics.

Go deeperWhy this matters: a century-old sketch that still predicts bonds

Gilbert N. Lewis introduced electron-dot ideas in 1916, before modern quantum mechanics had been developed. Chemists still use Lewis symbols because they connect valence-electron counts with useful bonding patterns: carbon’s four single dots are consistent with four bonds in methane, oxygen’s two single dots with two bonds in water, and chlorine’s one single dot with one bond in HCl. The notation is a compact introductory model; later topics add the orbital and molecular-structure details behind it.

Go deeperCommon mistake: pairing dots too soon

Drawing nitrogen’s five dots as two pairs plus one single is the classic slip. Place one dot on each of the four sides first, then use the fifth dot to form one pair. The result is one pair and three single dots, which is consistent with nitrogen’s three bonds in NH⁠₃.

The four sides are a bookkeeping convention, not a literal drawing of four orbitals. The later electron-configuration topic explains how electrons occupy atomic orbitals and why unpaired valence electrons are useful when reasoning about bonding.

Forming Ionic Compounds by Electron Transfer

During the formation of some compounds, atoms gain or lose electrons and form electrically charged particles called ions. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. An atom of an alkali metal (Group 1) loses one electron and forms a cation with a 1+ charge; an alkaline-earth metal (Group 2) loses two electrons and forms a cation with a 2+ charge; aluminum (Group 13) loses three to form a 3+ cation. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. The result is a cation with 20 protons, 18 electrons, and a 2+ charge — the same electron count as argon, the preceding noble gas. The symbol is Ca²⁺.

When atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as the next noble gas. Atoms of Group 17 gain one electron and form anions with a 1− charge; atoms of Group 16 gain two electrons and form 2− ions; atoms of Group 15 gain three electrons and form 3− ions. A neutral chlorine atom (17 protons, 17 electrons) gains one electron to become Cl⁻ with 17 protons and 18 electrons — the same electron count as argon. A neutral oxygen atom becomes O²⁻, with the same electron count as neon.

Reading from the far left to the right of the periodic table, main-group elements tend to form cations with a charge equal to the group number (1+, 2+, 3+). Reading from the right to the left, elements form anions with a negative charge equal to the number of groups moved left from the noble gases (group 17 → 1−, group 16 → 2−, group 15 → 3−). Transition metals and some other metals exhibit variable charges that are not predictable from group position alone (Cu can form 1+ or 2+; Fe can form 2+ or 3+) — these require the Roman-numeral conventions covered in Topic 4.

To form a neutral ionic compound, the total electrons lost by the metal atoms must equal the total electrons gained by the nonmetal atoms. The compound’s formula is determined by finding the smallest whole-number ratio of cations to anions that achieves this charge balance. Worked patterns for the most common combinations:

1:1 charge match (K + Cl). K loses 1 e⁻; Cl gains 1 e⁻. One K matches one Cl: KCl.
1:1 with double charges (Mg + O). Mg loses 2 e⁻; O gains 2 e⁻. One Mg matches one O: MgO.
1:2 ratio (Mg + F). Mg loses 2 e⁻; each F gains 1 e⁻. One Mg requires two F atoms: MgF₂.
3:1 ratio (Na + N). Each Na loses 1 e⁻; N gains 3 e⁻. Three Na atoms supply the three electrons one N needs: Na₃N.
2:3 ratio (Al + S). Each Al loses 3 e⁻; each S gains 2 e⁻. The least common multiple is 6: 2 Al atoms (lose 2 × 3 = 6 e⁻) match 3 S atoms (gain 3 × 2 = 6 e⁻): Al₂S₃.

Lattice-energy considerations and the deeper bonding picture appear in Topic 14.

Go deeperWhy ionic formulas balance charge

Ionic solids are stabilized by electrostatic attraction between oppositely charged ions. A bulk sample contains cations and anions in ratios that balance total positive and negative charge; an uncompensated net charge is energetically unfavorable.

That is the physical basis of the formula bookkeeping. In MgF⁠₂, one Mg⁠²⁺ ion is balanced by two F⁠^- ions, so the smallest neutral ratio is 1:2. Writing an ionic formula means finding the lowest whole-number ratio whose charges sum to zero.

Go deeperTry it: predict the formula for calcium nitride

Calcium is in Group 2; nitrogen is in Group 15. Predict the compound they form, then check below.

Answer: Ca loses 2 electrons (Ca⁠²⁺); N gains 3 (N⁠^3-). The least common multiple of 2 and 3 is 6, so three Ca atoms supply 3 × 2 = 6 electrons and two N atoms accept 2 × 3 = 6: the formula is Ca⁠₃N⁠₂. Verify the balance: 3(+2) + 2(−3) = 0. The subscripts crossed: each ion’s charge magnitude became the other’s subscript, which is why this shortcut is called the criss-cross method. It works whenever you reduce the result to the smallest whole-number ratio.

Common Mistakes and Quick Checks

Atomic-structure problems are straightforward once you know the patterns, but a few errors appear frequently:

Confusing atomic number with mass number. The atomic number (Z) counts only protons; the mass number (A) counts protons plus neutrons. If a problem says “sodium-23,” the 23 is A, not Z.
Forgetting to adjust electrons for ions. A neutral atom has electrons = Z. For cations, subtract the charge; for anions, add the magnitude of the charge. Writing Fe³⁺ with 26 electrons (neutral count) instead of 23 is a common slip.
Using mass number in average-mass calculations. The weighted-average formula uses precise isotopic masses (e.g., 34.969 amu for ³⁵Cl), not the whole-number mass numbers.
Mixing up isotope notation. In the symbol ^A_ZX, the superscript is the mass number and the subscript is the atomic number — not the other way round.
Counting Lewis dots from the atomic number instead of the valence-electron count. Lewis symbols use only valence electrons (= group number for main-group), never the full electron count.
Forgetting to balance electrons in ionic-compound formulas. The number of electrons the metal loses must equal the number the nonmetal gains. If they don’t match in a 1:1 ratio, find the LCM and adjust subscripts.

Quick self-check: after solving any particle-count problem, verify that protons + neutrons = mass number and that the net charge equals protons − electrons. After drawing a Lewis symbol, count the dots and confirm they equal the group number. After writing an ionic-compound formula, confirm total positive charge = total negative charge.

Key Equations

Mass Number

A = Z + N

A = mass number, Z = number of protons, N = number of neutrons

Neutron Count

N = A - Z

Subtract atomic number from mass number

Electrons in an Ion

electrons = Z - charge

Positive charge means fewer electrons; negative charge means more

Average Atomic Mass

avg mass = sum of (fractional abundance x isotopic mass)

Sum over the isotopes present in the specified sample

Learning Objectives

After studying this topic, you should be able to:

Describe the charge, relative mass, and location of protons, neutrons, and electrons
Determine the number of protons, neutrons, and electrons in an atom or ion given its atomic number and mass number
Define isotopes and calculate average atomic mass from isotopic abundances
Use isotopic notation (mass number, atomic number) to represent atoms
Determine electron distribution in the shells of a main-group atom and identify its valence electrons
Draw the Lewis dot symbol for a main-group element
Sketch the formation of a simple ionic compound by electron transfer between main-group atoms

How-To Procedure

How to Determine Protons, Neutrons, and Electrons in an Atom or Ion

Look up the element on the periodic table and find its atomic number (Z). This equals the number of protons.
Identify the mass number (A) from the problem. If given the isotope name (e.g., carbon-14), the number after the hyphen is A.
Calculate neutrons: N = A - Z.
For a neutral atom, the number of electrons equals Z.
For an ion, adjust the electron count by the charge: electrons = Z - charge. A positive charge means electrons were lost; a negative charge means electrons were gained.
Verify your answer: protons + neutrons should equal the mass number, and protons - electrons should equal the ionic charge.

Worked Example

Finding Subatomic Particles in an Ion

Problem

Platinum-195 forms a Pt⁠⁴⁺ ion. Determine the number of protons, neutrons, and electrons in this ion.

Solution

Identify the element and its atomic number. Platinum has atomic number Z = 78, so it has 78 protons.
Find the number of neutrons from the mass number: N = A − Z = 195 − 78 = 117 neutrons.
Determine the electron count. For a neutral Pt atom, electrons = Z = 78. The 4+ charge means 4 electrons were lost: 78 − 4 = 74 electrons.

Answer

The Pt⁠⁴⁺ ion (mass number 195) contains 78 protons, 117 neutrons, and 74 electrons.

Test Your Understanding

Naturally occurring potassium is about 93.26% K-39, 0.0117% K-40, and 6.73% K-41. Without calculating, should its average atomic mass be closer to 39, 40, or 41? Explain your reasoning.

Practice Problems

conceptual

An atom has 26 protons, 30 neutrons, and 23 electrons. Identify the element, give its mass number, and write its isotopic symbol including the ionic charge.

calculation

Bromine has two isotopes: Br-79 (mass 78.918 amu, 50.69% abundant) and Br-81 (mass 80.916 amu, 49.31% abundant). Calculate the average atomic mass of bromine.

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Self-Study Questions

What are the three subatomic particles and what are their charges?

Where are protons, neutrons, and electrons located within an atom?

What is the atomic number and what does it determine about an element?

What is the mass number and how is it calculated?

Hint: It involves two of the three subatomic particles.

How do you determine the number of protons, neutrons, and electrons in a neutral atom?

What is an ion and how does its electron count differ from the neutral atom?

What are isotopes and how do isotopes of the same element differ?

What is isotopic notation and what information does it convey?

How is the average atomic mass on the periodic table calculated from isotope data?

Hint: It is a weighted average, not a simple average.

Why is the average atomic mass of an element not a whole number?

How are electrons distributed in shells around the nucleus, and what are valence electrons?

Hint: For the first 20 elements, use the simplified occupied-shell pattern 2, 8, 8; valence electrons occupy the outermost shell.

How do you draw the Lewis dot symbol for a main-group element?

How do metal and nonmetal atoms form an ionic compound, and how do you determine the formula from the charges?

Content Sources

Concept sections adapted from open educational resources under Creative Commons licensing:

OpenStax Chemistry 2e, Ch 2.1: Early Ideas in Atomic Theory (CC BY 4.0)
OpenStax Chemistry 2e, Ch 2.2: Evolution of Atomic Theory (CC BY 4.0)
OpenStax Chemistry 2e, Ch 2.3: Atomic Structure and Symbolism (CC BY 4.0)
OpenStax Chemistry: Atoms First 2e, Ch 3.7: Ionic and Molecular Compounds (CC BY 4.0)
OpenStax Chemistry: Atoms First 2e, Ch 4.4: Lewis Symbols and Structures (CC BY 4.0)

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