Atomic Structure & Properties
What is everything made of, and how do we know? Build atoms, sort isotopes, and see electrons for yourself.
The atom
Everything is made of atoms.
Glass, rubber, wood, metal, you. Dalton said atoms can't be split. He was wrong: every atom has three smaller parts.
Protons are the ID card.
The number of protons, the atomic number, decides the element. 6 protons is always carbon.
Add the neutrons to get the mass number. A neutral atom has as many electrons as protons.
Carbon-12: 6 protons, 6 neutrons, 6 electrons
Build an atom
Isotopes are twins.
Same protons, same element. Different neutrons, slightly different mass.
That's why chlorine's mass is 35.45 and not a whole number: it's an average of its isotopes.
?How many protons, neutrons and electrons are in chlorine-37? (Cl is element 17.)Think, then tap to check
17 protons, 17 electrons, and \(37 - 17 =\) 20 neutrons.
Mass spectra
A mass spectrometer sorts atoms by mass.
Atoms are turned into ions and sent through a magnetic field. Heavier ions bend less, so each isotope lands in its own spot.
Each peak is an isotope. Its height is how common it is.
The average is weighted.
Multiply each isotope's mass by its abundance, then add. The average always sits nearest the tallest peak.
Mass spectrometer: Chlorine
Neon averages 20.18 amu. Its isotopes are Ne-20 (19.99 amu) and Ne-22 (21.99 amu). How abundant is each?
- The two fractions add to 1. Call Ne-20's fraction \(x\), so Ne-22's is \(1 - x\).
- \(20.18 = 19.99x + 21.99(1 - x)\)
- \(20.18 = 21.99 - 2.00x\), so \(x = 0.905\)
- Ne-20 is 90.5%, Ne-22 is 9.5%. The average is close to 20, as it should be.
Mg-24 (23.99 amu, 78.99%), Mg-25 (24.99 amu, 10.00%), Mg-26 (25.98 amu, 11.01%). Find the average atomic mass.
- Turn each percent into a fraction: 0.7899, 0.1000, 0.1101.
- \((23.99 \times 0.7899) + (24.99 \times 0.1000) + (25.98 \times 0.1101)\)
- \(= 18.95 + 2.499 + 2.860\)
- \(= 24.31\ \mathrm{amu}\), the number on the periodic table.
No calculator? Estimate.
The multiple-choice section has no calculator. Read the peaks like a seesaw.
185 (40) and 187 (60). Halfway is 186; more 187 tips it just above.
10 (25) and 11 (100): a 1 : 4 ratio, so \(10 \times \tfrac{1}{5} + 11 \times \tfrac{4}{5}\). That's boron.
70 (9), 72 (6), 74 (1). The tall pair gives about 70.8; the tiny 74 peak nudges it up to 71.
?Peaks at 63 amu (69%) and 65 amu (31%). Which element?Think, then tap to check
\(63 \times 0.69 + 65 \times 0.31 \approx 63.6\) → copper (63.55). Try it in the spectrometer above.
Types of matter
Pure substances have one formula. Mixtures have several.
One kind of atom.
\(\ce{Cu}\), \(\ce{Ag}\), \(\ce{O2}\), \(\ce{O3}\), \(\ce{N2}\)Two or more kinds of atom, fixed ratio.
\(\ce{H2O}\), \(\ce{CuO}\), \(\ce{C2H5OH}\)Evenly mixed: every part looks the same.
salt water, alcohol in waterYou can see the separate parts.
salad, pizza, sand in water\(\ce{O2}\) mixed with \(\ce{O3}\) is still a mixture: same atoms, but two different substances.
Separating mixtures
Filtration
Separates a solid that doesn't dissolve from a liquid. The solid stays on the paper; the liquid drips through.
Sand from water. Chalk from water.
Evaporation
Recovers a solid that is dissolved. Boil the water away and the salt stays behind.
Salt from salt water.
Distillation
Separates liquids with different boiling points. Heat to about 80 °C: ethanol (78 °C) boils, water (100 °C) doesn't.
The vapor cools in the condenser and drips out as nearly pure ethanol.
The mole
A mole is a chemist's dozen.
A dozen is 12. A mole is \(6.022 \times 10^{23}\) particles, Avogadro's number.
The molar mass is the mass of one mole, read straight off the periodic table: 1 mol of carbon is 12.01 g.
Mole calculator
How many moles, formula units and ions are in 5.0 g of \(\ce{CaCl2}\)?
- Molar mass: \(40.08 + 2(35.45) = 110.98\ \mathrm{g/mol}\)
- Moles: \(\dfrac{5.0\ \mathrm{g}}{110.98\ \mathrm{g/mol}} =\) \(0.045\ \mathrm{mol}\)
- Formula units: \(0.04505 \times 6.022 \times 10^{23} =\) \(2.7 \times 10^{22}\)
- Each \(\ce{CaCl2}\) is 3 ions (\(\ce{Ca^2+}\) + 2 \(\ce{Cl-}\)), so \(8.1 \times 10^{22}\) ions. (Ionic, so "formula units", not "molecules".)
4.86 mol of \(\ce{CaCO3}\): what is its mass, and how much of it is carbonate ion?
- Molar mass: \(40.08 + 12.01 + 3(16.00) = 100.09\ \mathrm{g/mol}\)
- Mass: \(4.86\ \mathrm{mol} \times 100.09\ \mathrm{g/mol} =\) \(486\ \mathrm{g}\)
- 1 mol \(\ce{CaCO3}\) contains 1 mol \(\ce{CO3^2-}\) (\(60.01\ \mathrm{g/mol}\))
- \(4.86 \times 60.01 =\) \(292\ \mathrm{g}\) of carbonate
Write units on every number. If the units cancel to what you want, the setup is right.
Formulas from data
Empirical vs molecular.
A pure compound always has the same mass ratio of elements. From that ratio you get the empirical formula: the simplest whole-number ratio.
The molecular formula is a whole-number multiple of it. You need the molar mass to find it.
A compound is 40.0% C, 6.67% H and 53.3% O. Find its empirical formula.
- Assume 100 g: 40.0 g C, 6.67 g H, 53.3 g O.
- C: \(\dfrac{40.0}{12.01} = 3.33\) · H: \(\dfrac{6.67}{1.008} = 6.62\) · O: \(\dfrac{53.3}{16.00} = 3.33\) mol
- Divide by the smallest (3.33): \(1 : 1.99 : 1 \approx 1 : 2 : 1\)
- \(\ce{CH2O}\). (If you get a ratio ending in .5, multiply everything by 2.)
Empirical formula \(\ce{CH2}\), molar mass 70 g/mol. Molecular formula?
- Mass of \(\ce{CH2}\): \(12 + 2(1) = 14\ \mathrm{g/mol}\)
- \(\dfrac{70}{14} = 5\)
- Multiply every subscript by 5: \(\ce{C5H10}\)
Burning magnesium
| Mass of Mg | |
| Mass of O (oxide − Mg) | |
| Moles of Mg | |
| Moles of O | |
| Mg : O |
?A 42 g sample has 36 g C and 6 g H. Can you find its molecular formula?Think, then tap to check
No, only the empirical formula: \(36 \div 12 = 3\) mol C, \(6 \div 1 = 6\) mol H, so \(\ce{CH2}\). Masses give a ratio; you'd need the molar mass for the molecular formula.
Is it pure?
A pure compound has a fixed composition.
Pure NaCl is always 39.3% sodium. If a "NaCl" sample measures differently, something else is in it.
So: calculate what a pure sample should give, measure what it does give, and compare.
Is the \(\ce{BaCl2}\) pure?
| Theoretical \(\ce{BaSO4}\) | |
| Experimental \(\ce{BaSO4}\) | |
| Purity |
Metal M is weighed, heated in air until it all becomes MO, then the oxide is weighed. What can you find?
- Oxide mass − metal mass = mass of oxygen → moles of O
- The formula MO means moles of M = moles of O
- Molar mass of M = mass of M ÷ moles of M. You can find the molar mass, not the density (no volume was measured).
?Pure NaCl is 60.7% Cl. A sample labeled NaCl is 55.0% Cl. Pure or not?Think, then tap to check
Not pure. A pure compound always has the same percent composition. Something with less chlorine is diluting it.
Where the electrons are
One law explains the whole unit.
Electrons are held by the pull of the nucleus. More charge, stronger pull. More distance, much weaker pull.
An atom is a hotel.
Shell (\(n = 1, 2, 3, \ldots\)) is the floor. Subshell (s, p, d, f) is a wing. Each orbital is a room for two electrons.
Three rules for filling.
1s → 2s → 2p → 3s → 3p → 4s → 3d. 4s is slightly lower than 3d, so it fills first.
An orbital holds at most two electrons, one ↑ and one ↓.
In p or d, one electron per orbital (same spin) before any pair up. That's Hund's rule.
Fill the orbitals
Transition metals fill 4s before 3d, but also lose 4s first. Fe is \(\mathrm{[Ar]\,4s^2\,3d^6}\), so \(\ce{Fe^2+}\) is \(\mathrm{[Ar]\,3d^6}\). Try it above.
Seeing electrons: PES
Knock electrons out. Measure how hard it was.
Light with a known energy hits the atom and ejects an electron. A tightly held electron leaves with little energy to spare.
Left = held tightly, close to the nucleus.
\(\mathrm{2p^6}\) is three times as tall as \(\mathrm{2s^2}\).
Na has four: 1s, 2s, 2p, 3s.
Ar's 1s is far left of Na's 1s.
Photoelectron spectrometer: Sodium
?Peak heights 2, 2, 6, 2, 1, from high to low binding energy. Which element?Think, then tap to check
\(\mathrm{1s^2\,2s^2\,2p^6\,3s^2\,3p^1}\) = 13 electrons: aluminum. Pick Al in the simulation to confirm.
Periodic trends
Across a period: stronger pull.
More protons, same shell, same shielding. Atoms shrink and grip their electrons harder.
Down a group: farther away.
Each row adds a shell. Valence electrons sit farther out, so atoms grow and let go more easily.
First ionization energy
Swipe the table sideways to see every element.
First ionization energy (kJ/mol) vs atomic number. Click the graph or the table to pick an element.
Orange line = pull on the valence electron. It grows with net charge and shrinks with distance.
Successive ionization energies
Ionization energy in kJ/mol on a log scale: each gridline is ten times the one below.
Explain why the first ionization energy of Mg is greater than that of Na.
- Both remove a 3s electron: same shell (\(n = 3\)), shielded by the same 10 core electrons.
- Mg has 12 protons, Na has 11, so Mg's valence electron feels a greater net nuclear charge.
- By Coulomb's law the attraction is stronger, so more energy is needed to remove it.
?Rank by size, smallest first: \(\ce{O^2-}\), \(\ce{F-}\), \(\ce{Na+}\), \(\ce{Mg^2+}\).Think, then tap to check
All have 10 electrons. More protons pull them in tighter: \(\ce{Mg^2+} < \ce{Na+} < \ce{F-} < \ce{O^2-}\).
Two dips on the graph: B is below Be (2p is higher in energy than 2s) and O is below N (the paired 2p electron is repelled). Find them above.
Ions and formulas
Atoms gain or lose electrons to look like a noble gas.
Group 1 → +1. Group 2 → +2. Al → +3. Group 15 → −3. Group 16 → −2. Group 17 → −1.
An ionic formula is whatever ratio makes the charges cancel.
Balance the charges
?Element X forms \(\ce{XCl2}\). What is the formula of its oxide?Think, then tap to check
Two \(\ce{Cl-}\) balance one X, so the ion is \(\ce{X^2+}\). One \(\ce{O^2-}\) balances it: \(\ce{XO}\).
Everything on one screen
\(n = \dfrac{m}{M}\)moles from grams\(N = n \times 6.022 \times 10^{23}\)particles from moles\(\text{avg} = \sum (\text{mass} \times \text{fraction})\)average atomic mass\(\text{neutrons} = A - Z\)mass number minus protons\(\% = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100\)percent composition\(F \propto \dfrac{q_1 q_2}{r^2}\)Coulomb's law: every trendReady to test yourself?
36 exam-style questions with a worked explanation after every answer. Create a free account to save your score.
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