Atoms, Ions & Nomenclature

CHEM 50 — General Chemistry I · Dr. Karmach

build Jul 13 · 09:43 · CC BY-NC-SA 4.0 · OpenStax Chemistry 2e where noted
Dr. Karmach

By the end of this unit, you can…

  • Locate protons, neutrons, and electrons in the atom and state what each count determines
  • Read and write isotope symbols, converting between mass number, atomic number, and particle counts
  • Calculate the average atomic mass of an element from isotope masses and abundances
  • Predict the charge an atom takes when it forms an ion, and count the particles in that ion
  • Name ionic and molecular compounds and acids, and write formulas from names
  • Recognize the common polyatomic ions and build formulas that contain them
Dr. Karmach

Today's route 🗺️

  1. Atomic Structure
  2. Isotope Notation
  3. Average Atomic Mass
  4. Ions
  5. Naming Ionic Compounds
  6. Naming Molecular Compounds and Acids
  7. Polyatomic Ions
Dr. Karmach

1 · Atomic Structure

Count the protons, neutrons, and electrons in any neutral atom, and name the element from its proton count.

Dr. Karmach

About 90 kinds of atoms

A gold ring, a copper wire, a diamond: each is a single kind of atom. One count inside the atom sets the kinds apart.

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Protons name the element

All matter is atoms, and a single count separates the elements: the protons. An atom with six protons is carbon, every time. Change the count and the atom is a different element.

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Three particles build every atom

proton p⁺ · charge 1+ · relative mass 1
in the nucleus
neutron n⁰ · charge 0 · relative mass ≈ 1
in the nucleus
electron e⁻ · charge 1− · relative mass ≈ 1/2000
outside the nucleus

The nucleus is the dense center; electrons fill the space around it.

Dr. Karmach

Where the mass sits

Protons and neutrons give the atom nearly all its mass, packed into the tiny nucleus. If the nucleus were a marble, the atom would be a stadium.

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Two numbers describe an atom

atomic number Z = protons
Z names the element — every carbon atom has Z = 6
mass number A = protons + neutrons
A counts the particles in the nucleus of one atom

Z is the same for every atom of an element. In a neutral atom, electrons match protons, so the charges cancel.

Dr. Karmach

Reading the periodic table

The table lists the elements in order of Z. A tile carries the atomic number and the average mass of the element's atoms. Metals fill the left and center; nonmetals sit to the upper right.

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The method

  1. Match protons and Z. The atomic number is the proton count; it names the element.
  2. Use A = protons + neutrons. Subtract to find whichever count is missing.
  3. Match electrons to protons. A neutral atom holds equal numbers.
Dr. Karmach

Worked example 1 — silicon

silicon — Z = 14 · mass number A = 29
given: one neutral atom · wanted: protons, neutrons, electrons

A neutral silicon atom has mass number 29. Count its protons, neutrons, and electrons.

Dr. Karmach

Worked example 1 — solution

silicon — Z = 14 · mass number A = 29
given: one neutral atom · wanted: protons, neutrons, electrons

Step 1 · Match protons and Z

Z = 14, so the atom holds 14 protons — and 14 protons is what makes it silicon.

Dr. Karmach

Worked example 1 — solution

silicon — Z = 14 · mass number A = 29
given: one neutral atom · wanted: protons, neutrons, electrons
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons

The mass number counts protons and neutrons together: neutrons = 29 − 14 = 15.

Dr. Karmach

Worked example 1 — solution

silicon — Z = 14 · mass number A = 29
given: one neutral atom · wanted: protons, neutrons, electrons
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons Step 3 · Match electrons to protons
silicon → 14 p⁺ · 15 n⁰ · 14 e⁻
protons = Z = 14 · neutrons = 29 − 14 = 15 · electrons = 14 — neutral
Dr. Karmach

Worked example 1 — solution

silicon — Z = 14 · mass number A = 29
given: one neutral atom · wanted: protons, neutrons, electrons
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons Step 3 · Match electrons to protons
silicon → 14 p⁺ · 15 n⁰ · 14 e⁻
protons = Z = 14 · neutrons = 29 − 14 = 15 · electrons = 14 — neutral
Rebuild the mass number: 14 + 15 = 29, the given A. ✓
Dr. Karmach

Worked example 2 — copper

periodic-table tile: 29 · Cu · 63.55
given: one neutral atom, mass number A = 65 · wanted: protons, neutrons, electrons

A neutral copper atom has mass number 65. The tile supplies the atomic number. Count the protons, neutrons, and electrons.

Dr. Karmach

Worked example 2 — solution

periodic-table tile: 29 · Cu · 63.55
given: one neutral atom, mass number A = 65

Step 1 · Match protons and Z

The tile gives Z = 29: the atom holds 29 protons, and 29 protons is copper.

Dr. Karmach

Worked example 2 — solution

periodic-table tile: 29 · Cu · 63.55
given: one neutral atom, mass number A = 65
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons

Neutrons = 65 − 29 = 36. The tile's 63.55 is an average mass of many atoms — the mass number 65 belongs to this one atom.

Dr. Karmach

Worked example 2 — solution

periodic-table tile: 29 · Cu · 63.55
given: one neutral atom, mass number A = 65
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons Step 3 · Match electrons to protons
copper → 29 p⁺ · 36 n⁰ · 29 e⁻
protons = Z = 29 · neutrons = 65 − 29 = 36 · electrons = 29 — neutral
Dr. Karmach

Worked example 2 — solution

periodic-table tile: 29 · Cu · 63.55
given: one neutral atom, mass number A = 65
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons Step 3 · Match electrons to protons
copper → 29 p⁺ · 36 n⁰ · 29 e⁻
protons = Z = 29 · neutrons = 65 − 29 = 36 · electrons = 29 — neutral
29 + 36 rebuilds the given mass number, 65. The tile's 63.55 never entered a particle count. ✓
Dr. Karmach

Your turn — fluorine

fluorine — Z = 9 · mass number A = 19
one neutral atom
step question answer
1 · match protons and Z protons?
2 · use A = protons + neutrons 19 − 9 = ? neutrons
3 · match electrons to protons electrons?

Complete the three counts.

Dr. Karmach

Your turn — fluorine

fluorine — Z = 9 · mass number A = 19
one neutral atom
step question answer
1 · match protons and Z protons?
2 · use A = protons + neutrons 19 − 9 = ? neutrons
3 · match electrons to protons electrons?

Complete the three counts.

fluorine → 9 p⁺ · 10 n⁰ · 9 e⁻
protons = Z = 9 · neutrons = 19 − 9 = 10 · electrons = 9 — neutral
Dr. Karmach

Where this goes wrong

Naming the element from the electron count. A neutral atom holds equal electrons and protons, so the two counts agree. Electrons are gained and lost in chemical changes; the proton count is the one that names the element.
Reading the tile's decimal as a mass number. Chlorine's 35.45 is an average over many atoms, not a count. A mass number is a whole number and belongs to one specific atom.
Taking the mass number as the neutron count. A = 29 does not mean 29 neutrons. The mass number counts protons and neutrons together: for silicon, neutrons = 29 − 14 = 15.
Dr. Karmach

Practice 1

zinc — Z = 30 · mass number A = 66
one neutral atom

A neutral zinc atom has mass number 66. Which row counts its particles?

  1. 30 p⁺ · 66 n⁰ · 30 e⁻
  2. 30 p⁺ · 96 n⁰ · 30 e⁻
  3. 30 p⁺ · 36 n⁰ · 30 e⁻
  4. 36 p⁺ · 30 n⁰ · 36 e⁻
Dr. Karmach

Practice 1 — answer: C

zinc, Z = 30, A = 66 → 30 p⁺ · 36 n⁰ · 30 e⁻ — answer C
protons = Z = 30 · neutrons = 66 − 30 = 36 · electrons = 30 — neutral

A read the mass number as the neutron count: 66. B added instead of subtracting: 66 + 30 = 96. D swapped protons and neutrons: 36 protons is a different element — krypton, not zinc.

30 protons + 36 neutrons returns the mass number, 66. ✓
Dr. Karmach

Worked example 3 — the element from the counts

one neutral atom — 24 p⁺ · 28 n⁰ · 24 e⁻
wanted: the element and its mass number

An atom holds 24 protons, 28 neutrons, and 24 electrons. Identify the element and give the mass number.

A common first attempt: A = protons + electrons = 24 + 24 = 48. Test it against what the mass number counts.

Dr. Karmach

Worked example 3 — solution

A common first attempt

24 p⁺ · 28 n⁰ · 24 e⁻ — attempt: A = 24 + 24 = 48?
electrons sit outside the nucleus and weigh ≈ 1/2000 of a proton ✗

The mass number counts the nucleus. Electrons never enter it.

Dr. Karmach

Worked example 3 — solution

A common first attempt

24 p⁺ · 28 n⁰ · 24 e⁻ — attempt: A = 24 + 24 = 48?
electrons sit outside the nucleus and weigh ≈ 1/2000 of a proton ✗
Step 1 · Match protons and Z

24 protons means Z = 24: the element is chromium. The neutron and electron counts have no part in the identity.

Dr. Karmach

Worked example 3 — solution

A common first attempt

24 p⁺ · 28 n⁰ · 24 e⁻ — attempt: A = 24 + 24 = 48?
electrons sit outside the nucleus and weigh ≈ 1/2000 of a proton ✗
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons

Only the nucleus counts: A = 24 + 28 = 52.

Dr. Karmach

Worked example 3 — solution

A common first attempt

24 p⁺ · 28 n⁰ · 24 e⁻ — attempt: A = 24 + 24 = 48?
electrons sit outside the nucleus and weigh ≈ 1/2000 of a proton ✗
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons Step 3 · Match electrons to protons
chromium — Z = 24 · A = 52
24 p⁺ · 28 n⁰ · 24 e⁻ — electrons equal protons: neutral ✓
Dr. Karmach

Worked example 3 — solution

A common first attempt

24 p⁺ · 28 n⁰ · 24 e⁻ — attempt: A = 24 + 24 = 48?
electrons sit outside the nucleus and weigh ≈ 1/2000 of a proton ✗
Step 1 · Match protons and Z Step 2 · Use A = protons + neutrons Step 3 · Match electrons to protons
chromium — Z = 24 · A = 52
24 p⁺ · 28 n⁰ · 24 e⁻ — electrons equal protons: neutral ✓
A comes from the nucleus alone: 24 + 28 = 52. The electrons never entered the sum.
Dr. Karmach

Take-home: the mass number counts the nucleus

A = protons + neutrons
24 + 28 = 52 ✓ — never protons + electrons: 24 + 24 = 48 ✗

Electrons balance the charge and fill the atom's volume. At 1/2000 the mass of a proton, they never enter the mass number.

Dr. Karmach

Practice 2

one neutral atom — 34 p⁺ · 44 n⁰ · 34 e⁻
wanted: the element and its mass number

A neutral atom holds 34 protons, 44 neutrons, and 34 electrons. Which identification is correct?

  1. Selenium, A = 68
  2. Selenium, A = 78
  3. Ruthenium, A = 78
  4. Platinum, A = 78
Dr. Karmach

Practice 2 — answer: B

34 p⁺ · 44 n⁰ · 34 e⁻ → selenium, A = 78 — answer B
protons = Z = 34 = selenium · A = 34 + 44 = 78 · electrons match the protons — neutral

A put the electrons into the mass number: 34 + 34 = 68 — electrons carry almost no mass. C matched the element to the neutron count: Z = 44 is ruthenium, but neutrons never set the identity. D read the mass number as an atomic number: Z = 78 is platinum, but 78 counts this atom's whole nucleus.

Z = 34 and A = 78 rebuild the given counts: 78 − 34 = 44 neutrons. ✓
Dr. Karmach

Check yourself

  1. Which count names the element, and what happens to an atom's identity when that count changes?
  2. A neutral manganese atom (Z = 25) has mass number 55. Count its protons, neutrons, and electrons.

Chemists record all three counts in one symbol: the element symbol with its atomic number and mass number attached — isotope notation. Atoms of one element can differ in neutron count; the notation tells those atoms apart.

Dr. Karmach

2 · Isotope Notation

Read and write isotope symbols in both notations and count the protons, neutrons, and electrons in a neutral atom of any isotope.

Dr. Karmach

Heavy water

Heavy water looks, pours, and reacts like ordinary water, but a liter of it weighs about a tenth more. The extra mass sits inside its hydrogen atoms.

Dr. Karmach

Same element, different mass

The proton count fixes the element: one proton means hydrogen. Neutrons add mass and never change the element. Atoms with the same protons but different neutrons are isotopes of one element.

Dr. Karmach

Two numbers label the nucleus

atomic number Z = protons
Z names the element — every chlorine atom has Z = 17
mass number A = protons + neutrons
a chlorine atom with 20 neutrons: A = 17 + 20 = 37

Both are whole-number counts of the particles in one atom. Rearranged, A − Z isolates the neutrons.

Dr. Karmach

Writing an isotope down

The nuclide symbol carries both counts: the mass number upper left, the atomic number lower left. Hyphen notation writes the name, then A. The name already fixes Z, so the hyphen form drops it.

Dr. Karmach

Isotopes of an element share chemistry

³⁵Cl: 17 p⁺ · 18 n · 17 e⁻   ·   ³⁷Cl: 17 p⁺ · 20 n · 17 e⁻
same electron count → same bonds · 18 vs 20 neutrons → different mass only

Electrons make the chemistry, and a neutral atom holds electrons equal to its protons. Isotopes share the proton count, so both chlorine isotopes form the same compounds.

Dr. Karmach

The method

  1. Find Z. The element and its atomic number fix each other on the periodic table.
  2. Apply A = Z + N. Any two of the three counts give the third.
  3. Count electrons. Neutral atom: electrons = protons.
Dr. Karmach

Worked example 1 — reading a symbol

⁶⁵₂₉Cu — one neutral atom
given: A = 65 · Z = 29 · wanted: protons, neutrons, electrons

About a third of natural copper atoms carry this symbol. Count the protons, neutrons, and electrons in one of them.

Dr. Karmach

Worked example 1 — solution

⁶⁵₂₉Cu — one neutral atom
given: A = 65 · Z = 29 · wanted: protons, neutrons, electrons

Step 1 · Find Z

The lower-left number is the atomic number: Z = 29, so 29 protons. The periodic table agrees: element 29 is copper.

Dr. Karmach

Worked example 1 — solution

⁶⁵₂₉Cu — one neutral atom
given: A = 65 · Z = 29 · wanted: protons, neutrons, electrons
Step 1 · Find Z Step 2 · Apply A = Z + N
neutrons = A − Z = 65 − 29 = 36
65 heavy particles − 29 protons = 36 neutrons
Dr. Karmach

Worked example 1 — solution

⁶⁵₂₉Cu — one neutral atom
given: A = 65 · Z = 29 · wanted: protons, neutrons, electrons
Step 1 · Find Z Step 2 · Apply A = Z + N
neutrons = A − Z = 65 − 29 = 36
65 heavy particles − 29 protons = 36 neutrons
Step 3 · Count electrons
⁶⁵₂₉Cu → 29 p⁺ · 36 n · 29 e⁻
neutral atom: electrons = protons = 29 · check: 29 + 36 = 65 = A ✓
Dr. Karmach

Worked example 1 — solution

⁶⁵₂₉Cu — one neutral atom
given: A = 65 · Z = 29 · wanted: protons, neutrons, electrons
Step 1 · Find Z Step 2 · Apply A = Z + N
neutrons = A − Z = 65 − 29 = 36
65 heavy particles − 29 protons = 36 neutrons
Step 3 · Count electrons
⁶⁵₂₉Cu → 29 p⁺ · 36 n · 29 e⁻
neutral atom: electrons = protons = 29 · check: 29 + 36 = 65 = A ✓
Neutrons outnumber protons, 36 to 29 — typical beyond the lightest elements. The sum rebuilds A: 29 + 36 = 65. ✓
Dr. Karmach

Worked example 2 — hyphen notation

carbon-14 — one neutral atom
given: the name carries A = 14 · wanted: protons, neutrons, electrons

Living wood holds a trace of carbon-14; the amount left in an artifact dates it. Count the particles in one neutral atom.

A common first attempt: carbon-14 holds 14 neutrons. Test it.

Dr. Karmach

Worked example 2 — the first attempt

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it

A common first attempt

carbon-14 → 14 neutrons?
14 counts protons and neutrons together — the protons are still inside ✗
Dr. Karmach

Worked example 2 — the first attempt

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it

A common first attempt

carbon-14 → 14 neutrons?
14 counts protons and neutrons together — the protons are still inside ✗
The 14 is the mass number: every heavy particle in the nucleus. Some of those particles are protons, so the neutron count must be smaller than 14.
Dr. Karmach

Worked example 2 — the first attempt

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it

A common first attempt

carbon-14 → 14 neutrons?
14 counts protons and neutrons together — the protons are still inside ✗
The 14 is the mass number: every heavy particle in the nucleus. Some of those particles are protons, so the neutron count must be smaller than 14.
A is a total; the protons take part of it. Finding neutrons needs a subtraction, not a copy of A.
Dr. Karmach

Worked example 2 — solution

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it

Step 1 · Find Z

Carbon is element 6 on the periodic table: Z = 6, so 6 protons.

Dr. Karmach

Worked example 2 — solution

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it
Step 1 · Find Z Step 2 · Apply A = Z + N
neutrons = A − Z = 14 − 6 = 8
of the 14 heavy particles, 6 are protons and 8 are neutrons
Dr. Karmach

Worked example 2 — solution

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it
Step 1 · Find Z Step 2 · Apply A = Z + N
neutrons = A − Z = 14 − 6 = 8
of the 14 heavy particles, 6 are protons and 8 are neutrons
Step 3 · Count electrons
carbon-14 → 6 p⁺ · 8 n · 6 e⁻
neutral atom: electrons = protons = 6 · check: 6 + 8 = 14 ✓
Dr. Karmach

Worked example 2 — solution

carbon-14 — one neutral atom
given: A = 14 · Z not written — the periodic table supplies it
Step 1 · Find Z Step 2 · Apply A = Z + N
neutrons = A − Z = 14 − 6 = 8
of the 14 heavy particles, 6 are protons and 8 are neutrons
Step 3 · Count electrons
carbon-14 → 6 p⁺ · 8 n · 6 e⁻
neutral atom: electrons = protons = 6 · check: 6 + 8 = 14 ✓
8 of the 14 heavy particles are neutrons; the other 6 are the protons. The sum rebuilds A: 6 + 8 = 14. ✓
Dr. Karmach

Take-home: the mass number is a total

carbon-14: A = 14 = 6 protons + 8 neutrons
the total already includes the protons
neutrons = A − Z = 14 − 6 = 8
subtracting the protons out leaves the neutrons

A counts every heavy particle in the nucleus, protons included. Reading A as a neutron count counts the protons twice; subtracting Z removes them.

Dr. Karmach

Your turn — sulfur-34

³⁴₁₆S — one neutral atom
A = 34 upper left · Z = 16 lower left
step reading count
1 · Find Z lower left: Z = 16 protons =
2 · Apply A = Z + N N = 34 − 16 neutrons =
3 · Count electrons neutral atom electrons =

Complete the three counts.

Dr. Karmach

Your turn — sulfur-34

³⁴₁₆S — one neutral atom
A = 34 upper left · Z = 16 lower left
step reading count
1 · Find Z lower left: Z = 16 protons =
2 · Apply A = Z + N N = 34 − 16 neutrons =
3 · Count electrons neutral atom electrons =

Complete the three counts.

³⁴₁₆S → 16 p⁺ · 18 n · 16 e⁻
neutrons: 34 − 16 = 18 · check: 16 + 18 = 34 ✓
Dr. Karmach

Where this goes wrong

³⁷₁₇Cl — chlorine-37
A = 37 · Z = 17 · neutrons = 37 − 17 = 20
Reading A as the neutron count. "37 neutrons" reads a total as one of its parts. The 37 counts protons and neutrons together; subtract: 37 − 17 = 20 neutrons.
Adding A and Z. 37 + 17 = 54 counts the 17 protons twice — A already includes them. Neutrons = A − Z, never A + Z.
Reading Z as the neutron count. The 17 counts protons. The neutron count never appears in the symbol; only the subtraction produces it.
Expecting A on the periodic table. The table lists 35.45 for chlorine, and that is not a mass number. A is a whole-number count for one specific atom.
Dr. Karmach

Practice 1

strontium-88 — ⁸⁸Sr · atomic number 38
the most common strontium atom in nature

How many neutrons are in one atom of strontium-88?

  1. 88
  2. 50
  3. 126
  4. 38
Dr. Karmach

Practice 1 — answer: B

⁸⁸Sr: neutrons = A − Z = 88 − 38 = 50 — answer B
38 p⁺ · 50 n · 38 e⁻ · check: 38 + 50 = 88 ✓

A read the mass number as the neutron count; 88 counts protons and neutrons together. C added the two numbers: 88 + 38 = 126, counting the protons twice. D read the atomic number; 38 counts the protons.

Beyond the lightest elements, neutrons outnumber protons: 50 > 38 fits. ✓
Dr. Karmach

Practice 2

cadmium-114 — hyphen notation
the periodic table: cadmium is element 48

Cadmium-114 is the most common cadmium atom. How many neutrons does one neutral atom hold?

  1. 114
  2. 48
  3. 66
  4. 162
Dr. Karmach

Practice 2 — answer: C

cadmium-114: neutrons = A − Z = 114 − 48 = 66 — answer C
48 p⁺ · 66 n · 48 e⁻ · check: 48 + 66 = 114 ✓

A read the mass number as the neutron count; 114 counts protons and neutrons together. B read the atomic number; 48 counts the protons. D added the two numbers: 114 + 48 = 162, counting the protons twice.

The neutron excess grows with heavier elements: 66 neutrons to 48 protons. ✓
Dr. Karmach

Worked example 3 — writing the symbol

a neutral gallium atom with 38 neutrons
given: the element · N = 38 · wanted: nuclide symbol and hyphen notation

This time the particle counts are given and the symbol is wanted. Write both notations for this atom. The same steps apply.

Dr. Karmach

Worked example 3 — solution

a neutral gallium atom with 38 neutrons
given: the element · N = 38 · wanted: nuclide symbol and hyphen notation

Step 1 · Find Z

The periodic table places gallium at element 31: Z = 31, so 31 protons.

Dr. Karmach

Worked example 3 — solution

a neutral gallium atom with 38 neutrons
given: the element · N = 38 · wanted: nuclide symbol and hyphen notation
Step 1 · Find Z Step 2 · Apply A = Z + N
A = Z + N = 31 + 38 = 69
31 protons + 38 neutrons = 69 heavy particles
Dr. Karmach

Worked example 3 — solution

a neutral gallium atom with 38 neutrons
given: the element · N = 38 · wanted: nuclide symbol and hyphen notation
Step 1 · Find Z Step 2 · Apply A = Z + N
A = Z + N = 31 + 38 = 69
31 protons + 38 neutrons = 69 heavy particles
Assemble the notation
⁶⁹₃₁Ga — gallium-69
A = 69 upper left · Z = 31 lower left · the hyphen form keeps only A
Dr. Karmach

Worked example 3 — solution

a neutral gallium atom with 38 neutrons
given: the element · N = 38 · wanted: nuclide symbol and hyphen notation
Step 1 · Find Z Step 2 · Apply A = Z + N
A = Z + N = 31 + 38 = 69
31 protons + 38 neutrons = 69 heavy particles
Assemble the notation
⁶⁹₃₁Ga — gallium-69
A = 69 upper left · Z = 31 lower left · the hyphen form keeps only A
Subtraction runs the check in reverse: 69 − 31 = 38, the given neutron count. ✓
Dr. Karmach

Practice 3

a neutral atom: 34 protons · 46 neutrons
wanted: the nuclide symbol

Which nuclide symbol represents this atom?

  1. ⁴⁶₃₄Se
  2. ⁸⁰₄₆Pd
  3. ³⁴₈₀Se
  4. ⁸⁰₃₄Se
Dr. Karmach

Practice 3 — answer: D

34 protons → selenium · A = 34 + 46 = 80 → ⁸⁰₃₄Se — answer D
A = 80 upper left · Z = 34 lower left · check: 80 − 34 = 46 neutrons ✓

A wrote the neutron count in the mass-number spot; the mass number is the sum, 34 + 46 = 80. B matched the element to the neutron count; protons alone name the element, and element 34 is selenium. C swapped the two positions; the mass number sits upper left.

Upper minus lower must give back the given neutrons: 80 − 34 = 46. ✓
Dr. Karmach

Check yourself

  1. A neutral atom is written ⁵⁹₂₇Co. Work the counts: protons, neutrons, electrons.
  2. Two neutral atoms each hold 20 protons; one holds 20 neutrons, the other 24. Name the element, and name what differs between the atoms.

The periodic table lists chlorine at 35.45 — neither 35 nor 37. Natural chlorine is a mixture of both isotopes, and the table's number is the abundance-weighted average atomic mass of that mixture.

Dr. Karmach

3 · Average Atomic Mass

Calculate an element's average atomic mass from isotopic masses and percent abundances, and check that the answer lands between the isotope masses, closer to the more abundant one.

Dr. Karmach

The number under every symbol

A chlorine atom weighs 34.97 or 36.97 amu, never 35.45. Every periodic table lists 35.45: the average of the natural mix.

Dr. Karmach

A natural sample is a fixed mix of isotopes

Natural chlorine is always the same mixture: 75.76% Cl-35 and 24.24% Cl-37, in every bottle. The mass that describes chlorine is the sample's average, weighted by those fixed proportions.

Dr. Karmach

The weighted average

average atomic mass = (mass₁ × fraction₁) + (mass₂ × fraction₂) + …
one term per isotope · fraction = percent ÷ 100 · masses in amu — 1 amu = 1/12 of one carbon-12 atom

Each isotope contributes its mass in proportion to its share of the sample. Convert every percent to a fraction first: 75.76% → 0.7576. Mass spectrometry measures the masses and the abundances.

Dr. Karmach

Mass number and atomic mass

mass number A = 35
a whole-number count: 17 protons + 18 neutrons in one Cl-35 atom
atomic mass = 35.45 amu
a weighted average over the natural sample — the periodic-table entry

A mass number counts particles in one atom, so it is whole. The periodic-table mass is an average over the sample; it is not whole, and it matches no single atom.

Dr. Karmach

Where the average lands

A weighted average lands between the lightest and heaviest masses, closer to the more abundant isotope. Check every answer against that range before trusting the arithmetic.

Dr. Karmach

The method

  1. Percents → fractions: divide each percent abundance by 100.
  2. Mass × fraction: multiply each isotopic mass by its fraction of the sample.
  3. Add the contributions: the sum is the average atomic mass.
Dr. Karmach

Worked example 1 — boron

B-10 · 10.0129 amu · 19.9%   ·   B-11 · 11.0093 amu · 80.1%
given: two isotopic masses with abundances · wanted: average atomic mass

Natural boron is the two isotopes above. Calculate the average atomic mass of boron.

Dr. Karmach

Worked example 1 — solution

B-10 · 10.0129 amu · 19.9%   ·   B-11 · 11.0093 amu · 80.1%
wanted: average atomic mass of boron

Two products are needed, then one sum.

Step 1 · Percents → fractions

19.9 ÷ 100 = 0.199 and 80.1 ÷ 100 = 0.801. The fractions cover the whole sample: 0.199 + 0.801 = 1.

Dr. Karmach

Worked example 1 — solution

B-10 · 10.0129 amu · 19.9%   ·   B-11 · 11.0093 amu · 80.1%
wanted: average atomic mass of boron
Step 1 · Percents → fractions Step 2 · Mass × fraction
10.0129 × 0.199 = 1.9926  ·  11.0093 × 0.801 = 8.8184
one contribution per isotope, in amu
Dr. Karmach

Worked example 1 — solution

B-10 · 10.0129 amu · 19.9%   ·   B-11 · 11.0093 amu · 80.1%
wanted: average atomic mass of boron
Step 1 · Percents → fractions Step 2 · Mass × fraction
10.0129 × 0.199 = 1.9926  ·  11.0093 × 0.801 = 8.8184
one contribution per isotope, in amu
Step 3 · Add the contributions
1.9926 + 8.8184 = 10.81 amu
the periodic-table entry for boron
Dr. Karmach

Worked example 1 — solution

B-10 · 10.0129 amu · 19.9%   ·   B-11 · 11.0093 amu · 80.1%
wanted: average atomic mass of boron
Step 1 · Percents → fractions Step 2 · Mass × fraction
10.0129 × 0.199 = 1.9926  ·  11.0093 × 0.801 = 8.8184
one contribution per isotope, in amu
Step 3 · Add the contributions
1.9926 + 8.8184 = 10.81 amu
the periodic-table entry for boron
10.81 lies between 10.0129 and 11.0093, close to B-11 — the isotope carrying 80.1% of the sample. ✓
Dr. Karmach

Worked example 2 — chlorine

Cl-35 · 34.97 amu · 75.76%   ·   Cl-37 · 36.97 amu · 24.24%
given: two isotopic masses with abundances · wanted: average atomic mass

Chlorine disinfects drinking water. Calculate its average atomic mass.

A common first attempt: add the two masses and divide by two. Test it.

Dr. Karmach

Worked example 2 — solution

Cl-35 · 34.97 amu · 75.76%   ·   Cl-37 · 36.97 amu · 24.24%
wanted: average atomic mass of chlorine

A common first attempt

(34.97 + 36.97) ÷ 2 = 35.97 amu
every periodic table prints 35.45 — the even split fails ✗

Dividing by two weights each isotope equally. This sample is not an even split: 75.76% of its atoms carry the lighter mass.

Dr. Karmach

Worked example 2 — solution

Cl-35 · 34.97 amu · 75.76%   ·   Cl-37 · 36.97 amu · 24.24%
wanted: average atomic mass of chlorine
A common first attempt
(34.97 + 36.97) ÷ 2 = 35.97 amu
every periodic table prints 35.45 — the even split fails ✗
Step 1 · Percents → fractions

Two products are needed, then one sum. First the fractions: 75.76 ÷ 100 = 0.7576 and 24.24 ÷ 100 = 0.2424.

Dr. Karmach

Worked example 2 — solution

Cl-35 · 34.97 amu · 75.76%   ·   Cl-37 · 36.97 amu · 24.24%
wanted: average atomic mass of chlorine
A common first attempt
(34.97 + 36.97) ÷ 2 = 35.97 amu
every periodic table prints 35.45 — the even split fails ✗
Step 1 · Percents → fractions Step 2 · Mass × fraction Step 3 · Add the contributions
34.97 × 0.7576 + 36.97 × 0.2424 = 26.4933 + 8.9616 = 35.45 amu
two contributions, one sum — the periodic-table entry ✓
Dr. Karmach

Worked example 2 — solution

Cl-35 · 34.97 amu · 75.76%   ·   Cl-37 · 36.97 amu · 24.24%
wanted: average atomic mass of chlorine
A common first attempt
(34.97 + 36.97) ÷ 2 = 35.97 amu
every periodic table prints 35.45 — the even split fails ✗
Step 1 · Percents → fractions Step 2 · Mass × fraction Step 3 · Add the contributions
34.97 × 0.7576 + 36.97 × 0.2424 = 26.4933 + 8.9616 = 35.45 amu
two contributions, one sum — the periodic-table entry ✓
35.45 lies between 34.97 and 36.97, closer to Cl-35, the isotope in three quarters of the sample. The even split lands at 35.97 because it ignores which isotope is common. ✓
Dr. Karmach

Take-home: abundance weights the average

(34.97 + 36.97) ÷ 2 = 35.97 amu
treats a 76 : 24 sample as an even split ✗
34.97 × 0.7576 + 36.97 × 0.2424 = 35.45 amu
weights each mass by its share of the sample ✓

A simple average is correct only when the abundances are equal. Natural abundances rarely are. Weight each isotopic mass by its fraction of the sample.

Dr. Karmach

Your turn — copper

Cu-63 · 62.9296 amu · 69.15%   ·   Cu-65 · 64.9278 amu · 30.85%
wanted: average atomic mass of copper
62.9296 × + 64.9278 × 0.3085 = + 20.0302 = amu
fractions from 69.15% and 30.85% · one contribution per isotope · then the sum

Fill the missing fraction, then complete the first contribution and the sum.

Dr. Karmach

Your turn — copper

Cu-63 · 62.9296 amu · 69.15%   ·   Cu-65 · 64.9278 amu · 30.85%
wanted: average atomic mass of copper
62.9296 × + 64.9278 × 0.3085 = + 20.0302 = amu
fractions from 69.15% and 30.85% · one contribution per isotope · then the sum

Fill the missing fraction, then complete the first contribution and the sum.

62.9296 × 0.6915 + 64.9278 × 0.3085 = 43.5158 + 20.0302 = 63.55 amu
between 62.9296 and 64.9278, closer to Cu-63 — 69.15% of the sample ✓
Dr. Karmach

Where this goes wrong

Cu-63 · 62.9296 amu · 69.15%   ·   Cu-65 · 64.9278 amu · 30.85%
average atomic mass: 63.55 amu
Using whole percents. 62.9296 × 69.15 + 64.9278 × 30.85 = 6354.6 amu — one hundred times too heavy. A share of a sample is a fraction: divide each percent by 100 first. 0.6915 and 0.3085 give 63.55 amu.
Attaching the abundances to the wrong isotopes. 62.9296 × 0.3085 + 64.9278 × 0.6915 = 64.31 amu — closer to Cu-65, the rarer isotope. The average must sit closer to the 69.15% isotope: 63.55 amu.
Weighting mass numbers instead of isotopic masses. 63 × 0.6915 + 65 × 0.3085 = 63.62 amu, not 63.55. Mass numbers count protons and neutrons; the average takes the measured masses, 62.9296 and 64.9278.
Dr. Karmach

Practice 1

Ga-69 · 68.9256 amu · 60.108%   ·   Ga-71 · 70.9247 amu · 39.892%
wanted: average atomic mass of gallium

Gallium nitride makes the blue light in LED bulbs. Natural gallium is the two isotopes above.

What is the average atomic mass of gallium?

  1. 69.93 amu
  2. 69.72 amu
  3. 70.13 amu
  4. 6972 amu
Dr. Karmach

Practice 1 — answer: B

Ga-69 · 68.9256 amu · 60.108%   ·   Ga-71 · 70.9247 amu · 39.892%
wanted: average atomic mass of gallium
68.9256 × 0.60108 + 70.9247 × 0.39892 = 41.4298 + 28.2933 = 69.72 amu — answer B
each mass weighted by its fraction of the sample

A averaged the masses equally: (68.9256 + 70.9247) ÷ 2 = 69.93. C attached the abundances to the wrong isotopes: 68.9256 × 0.39892 + 70.9247 × 0.60108 = 70.13. D used whole percents: 68.9256 × 60.108 + 70.9247 × 39.892 = 6972.

D is one hundred times too heavy. A and C sit inside the range, but only 69.72 leans toward Ga-69, the isotope in 60.108% of the sample. ✓
Dr. Karmach

Worked example 3 — magnesium

Mg-24 · 23.9850 amu · 78.99%  ·  Mg-25 · 24.9858 amu · 10.00%  ·  Mg-26 · 25.9826 amu · 11.01%
abundances: 78.99 + 10.00 + 11.01 = 100.00 — the whole sample

Magnesium has three natural isotopes. The method does not change: one term per isotope. Calculate the average atomic mass.

Dr. Karmach

Worked example 3 — fractions and contributions

Mg-24 · 23.9850 amu · 78.99%  ·  Mg-25 · 24.9858 amu · 10.00%  ·  Mg-26 · 25.9826 amu · 11.01%
wanted: average atomic mass of magnesium

Three products are needed, then one sum.

Step 1 · Percents → fractions

78.99 ÷ 100 = 0.7899, 10.00 ÷ 100 = 0.1000, 11.01 ÷ 100 = 0.1101. Together: 0.7899 + 0.1000 + 0.1101 = 1.

Dr. Karmach

Worked example 3 — fractions and contributions

Mg-24 · 23.9850 amu · 78.99%  ·  Mg-25 · 24.9858 amu · 10.00%  ·  Mg-26 · 25.9826 amu · 11.01%
wanted: average atomic mass of magnesium
Step 1 · Percents → fractions Step 2 · Mass × fraction
23.9850 × 0.7899 = 18.9458 amu
24.9858 × 0.1000 = 2.4986 amu
25.9826 × 0.1101 = 2.8607 amu
one contribution per isotope — the largest from the most abundant
Dr. Karmach

Worked example 3 — fractions and contributions

Mg-24 · 23.9850 amu · 78.99%  ·  Mg-25 · 24.9858 amu · 10.00%  ·  Mg-26 · 25.9826 amu · 11.01%
wanted: average atomic mass of magnesium
Step 1 · Percents → fractions Step 2 · Mass × fraction
23.9850 × 0.7899 = 18.9458 amu
24.9858 × 0.1000 = 2.4986 amu
25.9826 × 0.1101 = 2.8607 amu
one contribution per isotope — the largest from the most abundant
Mg-24 supplies 18.9458 of the total; nearly four fifths of the sample is Mg-24. ✓
Dr. Karmach

Worked example 3 — the sum

contributions: Mg-24 → 18.9458 amu · Mg-25 → 2.4986 amu · Mg-26 → 2.8607 amu
from 23.9850 × 0.7899 · 24.9858 × 0.1000 · 25.9826 × 0.1101

Step 3 · Add the contributions

18.9458 + 2.4986 + 2.8607 = 24.305 amu
the periodic-table entry for magnesium
Dr. Karmach

Worked example 3 — the sum

contributions: Mg-24 → 18.9458 amu · Mg-25 → 2.4986 amu · Mg-26 → 2.8607 amu
from 23.9850 × 0.7899 · 24.9858 × 0.1000 · 25.9826 × 0.1101

Step 3 · Add the contributions

18.9458 + 2.4986 + 2.8607 = 24.305 amu
the periodic-table entry for magnesium
24.305 lies between 23.9850 and 25.9826, close to Mg-24 — 78.99% of the sample. The two heavier isotopes hold 21.01% between them and raise the average only slightly. ✓
Dr. Karmach

Practice 2

In-113 · 112.9041 amu · 4.29%   ·   In-115 · 114.9039 amu · 95.71%
wanted: average atomic mass of indium

Indium tin oxide is the transparent conductor in touchscreens. Natural indium is the two isotopes above.

What is the average atomic mass of indium?

  1. 112.99 amu
  2. 113.90 amu
  3. 114.82 amu
  4. 11482 amu
Dr. Karmach

Practice 2 — answer: C

In-113 · 112.9041 amu · 4.29%   ·   In-115 · 114.9039 amu · 95.71%
wanted: average atomic mass of indium
112.9041 × 0.0429 + 114.9039 × 0.9571 = 4.8436 + 109.9745 = 114.82 amu — answer C
each mass weighted by its fraction of the sample

A attached the abundances to the wrong isotopes: 112.9041 × 0.9571 + 114.9039 × 0.0429 = 112.99. B averaged the masses equally: (112.9041 + 114.9039) ÷ 2 = 113.90, treating a 4 : 96 split as even. D used whole percents: 112.9041 × 4.29 + 114.9039 × 95.71 = 11482.

114.82 sits between the two masses and nearly on In-115 — 95.71% of the sample. ✓
Dr. Karmach

Check yourself

  1. Lithium is 7.59% Li-6 (6.0151 amu) and 92.41% Li-7 (7.0160 amu). Calculate the average atomic mass, then check it against a periodic table.
  2. Chlorine's mass number 35 is a whole number; its atomic mass, 35.45 amu, is not. Explain why the mass number is whole and the atomic mass is not.

The periodic-table mass reads two ways. In amu it is the average mass of one atom. In grams it is the mass of one mole of atoms — the counting unit that turns balanced equations into weighable amounts.

Dr. Karmach

4 · Ions

Predict the charge a main-group atom takes when it forms an ion, count the particles in the ion, and write its symbol.

Dr. Karmach

Salt water conducts electricity

Pure water barely conducts electricity. Stir in table salt and the same water lights a bulb. Dissolved salt releases charged particles, and moving charges are an electric current.

Dr. Karmach

Ions form by losing or gaining electrons

A neutral atom holds equal protons and electrons. Losing or gaining electrons turns it into an ion: charge = protons − electrons. The protons never change — they name the element.

Dr. Karmach

Cation or anion

Na → Na⁺ + e⁻
cation: 11 p⁺ · 10 e⁻ — charge 11 − 10 = 1+
Cl + e⁻ → Cl⁻
anion: 17 p⁺ · 18 e⁻ — charge 17 − 18 = 1−

Losing electrons removes negative charge: a positive ion, a cation. Gaining electrons adds negative charge: a negative ion, an anion. Metals lose; nonmetals gain.

Dr. Karmach

Predicting the charge

Main-group atoms lose or gain electrons to reach the stable count of the nearest noble gas. Groups 1, 2, 13 lose: 1+, 2+, 3+. Groups 15, 16, 17 gain: 3−, 2−, 1−.

Dr. Karmach

Writing the ion symbol

Ca²⁺ · Al³⁺ · S²⁻
charge upper-right · number before sign: 2+, never +2

The charge sits at the upper right of the element symbol, number before sign. A charge of one shows the sign alone: Na⁺, Cl⁻.

Dr. Karmach

The method

  1. Count the protons. The atomic number; it never changes.
  2. Count the electrons. Neutral = protons; subtract lost, add gained.
  3. Compute the charge. Charge = protons − electrons.
  4. Write the symbol. Charge upper-right, number before sign.
Dr. Karmach

Worked example 1 — magnesium

²⁴Mg loses 2 e⁻ → ?
neutral atom: 12 p⁺ · 12 n · 12 e⁻ · wanted: particle counts and symbol

A magnesium atom loses two electrons. Count each particle in the ion and write its symbol.

Dr. Karmach

Worked example 1 — solution

²⁴Mg loses 2 e⁻ → ?
neutral atom: 12 p⁺ · 12 n · 12 e⁻

Step 1 · Count the protons

12 protons before, 12 after: the ion is still magnesium. The neutrons also stay at 12.

Dr. Karmach

Worked example 1 — solution

²⁴Mg loses 2 e⁻ → ?
neutral atom: 12 p⁺ · 12 n · 12 e⁻
Step 1 · Count the protons Step 2 · Count the electrons

Neutral means 12 electrons. Two are lost: 12 − 2 = 10 e⁻.

Dr. Karmach

Worked example 1 — solution

²⁴Mg loses 2 e⁻ → ?
neutral atom: 12 p⁺ · 12 n · 12 e⁻
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge Step 4 · Write the symbol
Mg²⁺
12 p⁺ · 12 n · 10 e⁻ — charge 12 − 10 = 2+
Dr. Karmach

Worked example 1 — solution

²⁴Mg loses 2 e⁻ → ?
neutral atom: 12 p⁺ · 12 n · 12 e⁻
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge Step 4 · Write the symbol
Mg²⁺
12 p⁺ · 12 n · 10 e⁻ — charge 12 − 10 = 2+
Twelve positive protons against ten negative electrons: two positives are unmatched, so the ion carries 2+. ✓
Dr. Karmach

Worked example 2 — sulfur

³²S → ?
neutral atom: 16 p⁺ · 16 n · 16 e⁻ · group 16 · wanted: charge, counts, symbol

Sulfur forms an ion. Predict how many electrons move, count the particles, and write the symbol.

A common first attempt: an ion that gains electrons gains particles, so its charge comes out positive. Test it.

Dr. Karmach

Worked example 2 — solution

³²S → ?
neutral atom: 16 p⁺ · 16 n · 16 e⁻ · group 16

A common first attempt

S gains 2 e⁻ → S²⁺?
16 p⁺ · 18 e⁻ — charge 16 − 18 = 2−, not 2+ ✗

Each electron carries one negative charge. Adding electrons can only push the total negative.

Dr. Karmach

Worked example 2 — solution

³²S → ?
neutral atom: 16 p⁺ · 16 n · 16 e⁻ · group 16
A common first attempt
S gains 2 e⁻ → S²⁺?
16 p⁺ · 18 e⁻ — charge 16 − 18 = 2−, not 2+ ✗
Step 1 · Count the protons

16 protons, unchanged: still sulfur. The neutrons stay at 16.

Dr. Karmach

Worked example 2 — solution

³²S → ?
neutral atom: 16 p⁺ · 16 n · 16 e⁻ · group 16
A common first attempt
S gains 2 e⁻ → S²⁺?
16 p⁺ · 18 e⁻ — charge 16 − 18 = 2−, not 2+ ✗
Step 1 · Count the protons Step 2 · Count the electrons

The nearest noble gas is argon, 18 electrons. Sulfur holds 16 and gains two: 16 + 2 = 18 e⁻.

Dr. Karmach

Worked example 2 — solution

³²S → ?
neutral atom: 16 p⁺ · 16 n · 16 e⁻ · group 16
A common first attempt
S gains 2 e⁻ → S²⁺?
16 p⁺ · 18 e⁻ — charge 16 − 18 = 2−, not 2+ ✗
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge Step 4 · Write the symbol
S²⁻
16 p⁺ · 16 n · 18 e⁻ — charge 16 − 18 = 2−
Dr. Karmach

Worked example 2 — solution

³²S → ?
neutral atom: 16 p⁺ · 16 n · 16 e⁻ · group 16
A common first attempt
S gains 2 e⁻ → S²⁺?
16 p⁺ · 18 e⁻ — charge 16 − 18 = 2−, not 2+ ✗
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge Step 4 · Write the symbol
S²⁻
16 p⁺ · 16 n · 18 e⁻ — charge 16 − 18 = 2−
Two electrons beyond the protons, each carrying one negative charge: 2−. Gained electrons always land the charge negative. ✓
Dr. Karmach

Take-home: the sign follows the electrons

lose e⁻ → positive ion
Mg²⁺: 12 p⁺ · 10 e⁻ — protons outnumber electrons: 12 − 10 = 2+
gain e⁻ → negative ion
S²⁻: 16 p⁺ · 18 e⁻ — electrons outnumber protons: 16 − 18 = 2−

Losing negative particles leaves a positive ion. Gaining negative particles makes a negative ion. To check a sign, compute charge = protons − electrons.

Dr. Karmach

Your turn — potassium

³⁹K → ?
neutral atom: 19 p⁺ · 20 n · 19 e⁻ · group 1
step count
1 · protons 19, unchanged
2 · electrons argon holds 18, so one electron leaves: 19 − 1 =
3 · charge 19 − 18 =
4 · symbol

Complete the counts and write the symbol.

Dr. Karmach

Your turn — potassium

³⁹K → ?
neutral atom: 19 p⁺ · 20 n · 19 e⁻ · group 1
step count
1 · protons 19, unchanged
2 · electrons argon holds 18, so one electron leaves: 19 − 1 =
3 · charge 19 − 18 =
4 · symbol

Complete the counts and write the symbol.

K⁺
19 p⁺ · 20 n · 18 e⁻ — charge 19 − 18 = 1+
Dr. Karmach

Where this goes wrong

Reading "lost" as negative. Mg loses 2 e⁻, leaving 12 p⁺ and 10 e⁻: charge 12 − 10 = 2+. The particles lost were the negative ones. Losing electrons always leaves a positive ion.
Charging the nucleus. A 2+ charge never comes from added protons: 12 protons is magnesium, 14 is silicon. Ion formation moves electrons only.
Naming the element from the electrons. Na⁺, Mg²⁺, and O²⁻ each hold 10 electrons, and none is neon. Protons identify the element.
Counting electrons into the mass number. Only nucleus particles count: ²⁴Mg²⁺ keeps mass number 12 + 12 = 24, with 12 electrons or with 10.
Dr. Karmach

Practice 1

Ba → Ba²⁺
neutral atom: 56 p⁺ · 56 e⁻ · group 2

A neutral barium atom becomes a Ba²⁺ ion. Which statement describes what happens?

  1. The atom gains two electrons; the extra particles give it the 2+ charge
  2. The nucleus gains two protons, which makes the atom 2+
  3. The atom loses two electrons; with more protons than electrons left, it carries the 2+ charge
  4. The atom loses two protons from its nucleus, leaving a 2+ charge
Dr. Karmach

Practice 1 — answer: C

Ba → Ba²⁺ + 2 e⁻ — answer C
56 p⁺ · 54 e⁻ — charge 56 − 54 = 2+

A: gaining two electrons computes 56 − 58 = 2−, an anion. B: 58 protons is no longer barium; protons never change in chemistry. D: losing two protons changes the element too, and 54 p⁺ against 56 e⁻ computes 54 − 56 = 2−.

Barium sits in group 2: it loses two electrons, and 54 electrons is xenon's count, the nearest noble gas. ✓
Dr. Karmach

Worked example 3 — identifying an unknown ion

? — 34 p⁺ · 46 n · 36 e⁻
given: the particle counts · wanted: element, mass number, charge, full symbol

A particle holds 34 protons, 46 neutrons, and 36 electrons. Identify it and write its full symbol.

Dr. Karmach

Worked example 3 — solution

? — 34 p⁺ · 46 n · 36 e⁻
given: the particle counts · wanted: element, mass number, charge, full symbol

Step 1 · Count the protons

34 protons: selenium. The 36 electrons match krypton's count, but electrons come and go; protons name the element.

Dr. Karmach

Worked example 3 — solution

? — 34 p⁺ · 46 n · 36 e⁻
given: the particle counts · wanted: element, mass number, charge, full symbol
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge

36 electrons are given, two more than the protons: charge 34 − 36 = 2−.

Dr. Karmach

Worked example 3 — solution

? — 34 p⁺ · 46 n · 36 e⁻
given: the particle counts · wanted: element, mass number, charge, full symbol
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge The mass number

Protons plus neutrons: 34 + 46 = 80. Electrons never enter the mass number.

Dr. Karmach

Worked example 3 — solution

? — 34 p⁺ · 46 n · 36 e⁻
given: the particle counts · wanted: element, mass number, charge, full symbol
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge The mass number Step 4 · Write the symbol
⁸⁰Se²⁻
34 p⁺ · 46 n · 36 e⁻ — mass number 34 + 46 = 80 · charge 34 − 36 = 2−
Dr. Karmach

Worked example 3 — solution

? — 34 p⁺ · 46 n · 36 e⁻
given: the particle counts · wanted: element, mass number, charge, full symbol
Step 1 · Count the protons Step 2 · Count the electrons Step 3 · Compute the charge The mass number Step 4 · Write the symbol
⁸⁰Se²⁻
34 p⁺ · 46 n · 36 e⁻ — mass number 34 + 46 = 80 · charge 34 − 36 = 2−
Selenium sits in group 16, two electrons short of krypton, and a 2− anion is exactly what the periodic table predicts. ✓
Dr. Karmach

Practice 2

I → I⁻
neutral atom: 53 p⁺ · 53 e⁻ · group 17

A neutral iodine atom becomes an I⁻ ion. Which statement describes what happens?

  1. The nucleus loses one proton, which makes the atom 1−
  2. The atom gains one electron; with more electrons than protons, it carries the 1− charge
  3. The atom loses one electron, and losing a particle leaves the 1− charge
  4. The nucleus gains one proton; the extra particle gives it the 1− charge
Dr. Karmach

Practice 2 — answer: B

I + e⁻ → I⁻ — answer B
53 p⁺ · 54 e⁻ — charge 53 − 54 = 1−

A: losing a proton computes 52 − 53 = 1−, but 52 protons is tellurium, a different element; the nucleus never changes. C: losing an electron computes 53 − 52 = 1+, a cation. D: a gained proton computes 54 − 53 = 1+ and changes the element as well.

54 electrons is xenon's count: iodine sits one electron short of the nearest noble gas, so it gains exactly one. ✓
Dr. Karmach

Check yourself

  1. Strontium (38 protons) sits in group 2. How many electrons does its ion hold, what is the charge, and what is the symbol?
  2. A particle holds 30 protons, 34 neutrons, and 28 electrons. Which element is it, and what is its full symbol?

Cations and anions attract into ionic compounds, and the charges must cancel: Na⁺ pairs one-to-one with Cl⁻, while Ca²⁺ takes two F⁻. These predicted charges fix the formula and the name of every ionic compound.

Dr. Karmach

5 · Naming Ionic Compounds

Name any binary ionic compound from its formula and write its formula from its name, letting charge balance set every subscript and every Roman numeral.

Dr. Karmach

What a chemical name is for

Rust is iron combined with oxygen from the air — two iron for every three oxygen, in every flake. A chemical name reports exactly what a compound contains.

Dr. Karmach

Every ionic compound is neutral

Cations and anions carry charge; the compound carries none. Total positive charge cancels total negative charge. That balance fixes the ratio of ions — every name and every formula follows from it.

Dr. Karmach

Recognizing an ionic compound

A metal with a nonmetal is ionic: the metal's atoms become cations, the nonmetal's become anions. Periodic position gives each ion its charge.

Dr. Karmach

The name: cation, then anion

NaCl → sodium chloride
1(+1) + 1(−1) = 0 ✓ · the cation keeps its element name
MgBr₂ → magnesium bromide
1(+2) + 2(−1) = 0 ✓ · brom- + -ide · the subscript is never spoken

The cation is named first, unchanged. The anion takes its element's stem plus -ide. Subscripts come from charge balance, so the name does not repeat them.

Dr. Karmach

Fixed-charge and variable-charge metals

one possible charge: Group 1 → 1+ · Group 2 → 2+ · Al³⁺ · Zn²⁺ · Ag⁺
the name never carries a numeral — NaCl is sodium chloride
more than one: Fe²⁺/Fe³⁺ · Cu⁺/Cu²⁺ · Sn²⁺/Sn⁴⁺ · Pb²⁺/Pb⁴⁺
iron(II) = Fe²⁺ · iron(III) = Fe³⁺ — the Roman numeral states the cation's charge

A fixed-charge metal forms one cation; its name needs no numeral. A variable-charge metal forms more than one, so its name carries a Roman numeral stating the cation's charge.

Dr. Karmach

The method

  1. Classify the compound. Metal + nonmetal → ionic.
  2. Identify the ions. Charges from periodic position.
  3. Balance the charges to zero. Subscripts count ions.
  4. Assemble the answer. Name: cation, anion stem + -ide. Formula: smallest whole-number ratio.
Dr. Karmach

Worked example 1 — K₂S

Step 1 · Classify the compound

K₂S
potassium, a metal · sulfur, a nonmetal → ionic · wanted: the name

Black powder burns to a residue containing K₂S. Name the compound.

Dr. Karmach

Worked example 1 — solution

K₂S
potassium, a metal · sulfur, a nonmetal → ionic · wanted: the name

Step 2 · Identify the ions

K⁺ and S²⁻
K: Group 1 → 1+, fixed · S: Group 6A → 8 − 6 = 2 → 2−

Potassium has one possible charge, so no Roman numeral will appear.

Dr. Karmach

Worked example 1 — solution

K₂S
potassium, a metal · sulfur, a nonmetal → ionic · wanted: the name
Step 2 · Identify the ions
K⁺ and S²⁻
K: Group 1 → 1+, fixed · S: Group 6A → 8 − 6 = 2 → 2−
Step 3 · Balance the charges to zero Step 4 · Assemble the answer

The subscript 2 records the balance already:

K₂S → potassium sulfide
2(+1) + 1(−2) = 0 ✓ · sulf- + -ide · the 2 is not spoken
Dr. Karmach

Worked example 1 — solution

K₂S
potassium, a metal · sulfur, a nonmetal → ionic · wanted: the name
Step 2 · Identify the ions
K⁺ and S²⁻
K: Group 1 → 1+, fixed · S: Group 6A → 8 − 6 = 2 → 2−
Step 3 · Balance the charges to zero Step 4 · Assemble the answer
K₂S → potassium sulfide
2(+1) + 1(−2) = 0 ✓ · sulf- + -ide · the 2 is not spoken
No numbers appear in the name, and none are needed: K⁺ and S²⁻ reach zero charge only in a 2 : 1 ratio.
Dr. Karmach

Worked example 2 — magnesium nitride

Step 1 · Classify the compound

magnesium nitride
magnesium, a metal · nitride, a nonmetal anion → ionic · wanted: the formula

Magnesium burning in air combines with nitrogen as well as oxygen. Write the formula for magnesium nitride.

Dr. Karmach

Worked example 2 — solution

magnesium nitride
a metal with a nonmetal → ionic · wanted: the formula

Step 2 · Identify the ions

Mg²⁺ and N³⁻
Mg: Group 2 → 2+ · N: Group 5A → 8 − 5 = 3 → 3−
Dr. Karmach

Worked example 2 — solution

magnesium nitride
a metal with a nonmetal → ionic · wanted: the formula
Step 2 · Identify the ions
Mg²⁺ and N³⁻
Mg: Group 2 → 2+ · N: Group 5A → 8 − 5 = 3 → 3−
Step 3 · Balance the charges to zero Step 4 · Assemble the answer

Neither charge cancels the other one-for-one. The smallest totals that cancel are +6 and −6: three Mg²⁺ with two N³⁻. The criss-cross shortcut writes each ion's charge as the other ion's subscript.

magnesium nitride → Mg₃N₂
3(+2) + 2(−3) = +6 − 6 = 0 ✓ · 3 and 2 share no common factor
Dr. Karmach

Worked example 2 — solution

magnesium nitride
a metal with a nonmetal → ionic · wanted: the formula
Step 2 · Identify the ions
Mg²⁺ and N³⁻
Mg: Group 2 → 2+ · N: Group 5A → 8 − 5 = 3 → 3−
Step 3 · Balance the charges to zero Step 4 · Assemble the answer
magnesium nitride → Mg₃N₂
3(+2) + 2(−3) = +6 − 6 = 0 ✓ · 3 and 2 share no common factor
The formula sums to zero charge, in the smallest whole numbers that do it.
Dr. Karmach

Your turn — calcium sulfide

calcium sulfide
wanted: the formula
step question answer
1 · classify the compound metal + nonmetal? ionic
2 · identify the ions charges from periodic position Ca and S
3 · balance the charges to zero the criss-cross gives Ca₂S₂ — is that the smallest ratio?
4 · assemble the answer

Complete the formula.

Dr. Karmach

Your turn — calcium sulfide

calcium sulfide
wanted: the formula
step question answer
1 · classify the compound metal + nonmetal? ionic
2 · identify the ions charges from periodic position Ca and S
3 · balance the charges to zero the criss-cross gives Ca₂S₂ — is that the smallest ratio?
4 · assemble the answer

Complete the formula.

calcium sulfide → CaS
1(+2) + 1(−2) = 0 ✓ — Ca₂S₂ reduces: a formula unit is the smallest ratio of ions, not a molecule
Dr. Karmach

Where this goes wrong

Greek prefixes on an ionic compound. CaCl₂ is calcium chloride, never calcium dichloride. Prefixes count atoms in molecular compounds. An ionic subscript comes from charge balance and is not spoken.
A Roman numeral on a fixed-charge metal. Sodium(I) chloride and calcium(II) bromide are never written. The numeral appears only when the metal has more than one possible charge.
Writing each ion's own charge as its own subscript. Al³⁺ with Cl⁻ is not Al₃Cl: 3(+3) + 1(−1) = +8, not neutral. Cross the charges instead: AlCl₃, 1(+3) + 3(−1) = 0.
Stopping before the smallest ratio. The criss-cross on Mg²⁺ and O²⁻ gives Mg₂O₂, and 2(+2) + 2(−2) = 0 balances. A formula unit is the smallest whole-number ratio: MgO.
Dr. Karmach

Practice 1

Na₂O
sodium, a metal · oxygen, a nonmetal → ionic

Window glass is made from a melt containing Na₂O. What is the correct name for Na₂O?

  1. sodium(II) oxide
  2. disodium monoxide
  3. sodium oxide
  4. sodium(I) oxide
Dr. Karmach

Practice 1 — answer: C

Na₂O → sodium oxide — answer C
2(+1) + 1(−2) = 0 ✓ · Na: Group 1 → 1+, fixed

A read the subscript as a charge: sodium 2+ would give 2(+2) + 1(−2) = +2, not neutral — the 2 counts Na⁺ ions. B counts atoms with Greek prefixes, the naming system for molecular compounds. D writes a numeral for a metal with only one possible charge; numerals mark variable-charge metals only.

Na⁺ and O²⁻ reach zero charge only as 2 : 1, so the name needs no number.
Dr. Karmach

Worked example 3 — Cu₂O

Step 1 · Classify the compound

Cu₂O
copper, a metal · oxygen, a nonmetal → ionic · wanted: the name

Cu₂O is the red pigment in antifouling boat paint. Copper is a variable-charge metal, so the name needs a Roman numeral.

A common first attempt: read the subscript 2 as copper's charge — copper(II) oxide. Test it.

Dr. Karmach

Worked example 3 — testing the first attempt

Cu₂O
copper, a variable-charge metal · oxygen, a nonmetal → ionic · wanted: the name

A common first attempt

copper(II) oxide → each Cu would be 2+
2(+2) + 1(−2) = +2 ✗ — not neutral
Dr. Karmach

Worked example 3 — testing the first attempt

Cu₂O
copper, a variable-charge metal · oxygen, a nonmetal → ionic · wanted: the name

A common first attempt

copper(II) oxide → each Cu would be 2+
2(+2) + 1(−2) = +2 ✗ — not neutral
The subscript 2 counts copper ions. It is not a charge.
Cu₂O is neutral, so any name for it must balance to zero.
Dr. Karmach

Worked example 3 — solution

Cu₂O
copper, a variable-charge metal · oxygen, a nonmetal → ionic · wanted: the name

Step 2 · Identify the ions Step 3 · Balance the charges to zero

Oxide is fixed at 2−. Copper's charge must come from this formula: one O²⁻ contributes 2−, so the two Cu contribute +2 in total.

each Cu = +2 ÷ 2 = 1+
2(+1) + 1(−2) = 0 ✓
Dr. Karmach

Worked example 3 — solution

Cu₂O
copper, a variable-charge metal · oxygen, a nonmetal → ionic · wanted: the name
Step 2 · Identify the ions Step 3 · Balance the charges to zero
each Cu = +2 ÷ 2 = 1+
2(+1) + 1(−2) = 0 ✓
Step 4 · Assemble the answer
Cu₂O → copper(I) oxide
the numeral reports the charge on each Cu, 1+ · the subscript already counts the ions
Dr. Karmach

Worked example 3 — solution

Cu₂O
copper, a variable-charge metal · oxygen, a nonmetal → ionic · wanted: the name
Step 2 · Identify the ions Step 3 · Balance the charges to zero
each Cu = +2 ÷ 2 = 1+
2(+1) + 1(−2) = 0 ✓
Step 4 · Assemble the answer
Cu₂O → copper(I) oxide
the numeral reports the charge on each Cu, 1+ · the subscript already counts the ions
Copper(II) oxide exists, and it is a different compound: CuO, where 1(+2) + 1(−2) = 0. One numeral, one formula.
Dr. Karmach

Take-home: the Roman numeral is a charge, not a count

Cu₂O → copper(I) oxide
two Cu ions, each 1+ · 2(+1) + 1(−2) = 0 ✓
CuO → copper(II) oxide
one Cu ion at 2+ · 1(+2) + 1(−2) = 0 ✓

The numeral states the charge on the cation, found by balancing the anion's total. Subscripts count ions; the numeral never does.

Dr. Karmach

Practice 2

FeO
iron, a variable-charge metal · oxygen, a nonmetal → ionic

FeO gives green bottle glass its tint. What is the correct name for FeO?

  1. iron(II) oxide
  2. iron(III) oxide
  3. iron oxide
  4. iron monoxide
Dr. Karmach

Practice 2 — answer: A

FeO → iron(II) oxide — answer A
1(+2) + 1(−2) = 0 ✓ · one O²⁻ demands 2+ from one Fe

B recycles the 3+ from rust: here 1(+3) + 1(−2) = +1, not neutral — the numeral must balance this formula. C omits the numeral: iron has more than one possible charge, so "iron oxide" cannot separate FeO from Fe₂O₃. D counts atoms with a Greek prefix, the system for molecular compounds.

The numeral is settled one compound at a time: FeO holds Fe²⁺, and Fe₂O₃ holds Fe³⁺.
Dr. Karmach

Check yourself

  1. Name CrCl₃. Chromium is a variable-charge metal — where does its Roman numeral come from?
  2. Write the formula for barium nitride. Does your formula sum to zero charge?

A compound of two nonmetals contains no ions. Those molecular compounds are named with Greek prefixes: CO₂ is carbon dioxide. And some ions are charged groups of atoms; they carry their own names and follow the same charge-balance rule.

Dr. Karmach

6 · Naming Molecular Compounds and Acids

Decide whether a compound is ionic, molecular, or an acid, then build its name with that system's rules — or rebuild the formula from the name.

Dr. Karmach

One atom apart

A faulty furnace releases a deadly gas; homes carry an alarm for it. The fizz in soda is a different gas made of the same two elements.

Dr. Karmach

The compound's type decides the name

Chemistry has three naming systems: ionic, molecular, acid. A compound's type decides which one applies; the systems never mix. Naming starts with a classification.

Dr. Karmach

Molecular compounds: prefixes count atoms

1 mono- · 2 di- · 3 tri- · 4 tetra- · 5 penta- · 6 hexa- · 7 hepta- · 8 octa- · 9 nona- · 10 deca-
mono- is dropped on the first element only · a prefix's final a or o drops before oxide: mono- + oxide → monoxide

Two nonmetals form a molecular compound, and several ratios are often possible. The name carries the formula: a Greek prefix counts each element's atoms.

Dr. Karmach

Acids: a category of their own

HCl(g) — hydrogen chloride, a gas · HCl(aq) — an acid
the same molecule; dissolved in water it releases H⁺

An acid is a compound that releases H⁺ when dissolved in water. Its formula starts with H and carries (aq). Acids get their own names, under their own rules.

Dr. Karmach

The method

  1. Classify the compound. H first, dissolved in water → acid. Metal present → ionic. Two nonmetals → molecular.
  2. Apply that system's rules. One system per compound; rules never mix.
  3. Read the name back. A correct name rebuilds the formula.
Dr. Karmach

Worked example 1 — CO and CO₂

CO · CO₂
the furnace gas: 1 C + 1 O · the soda gas: 1 C + 2 O

Both gases pair carbon with oxygen, one oxygen atom apart. Name each compound.

A common first attempt: name the elements and stop — carbon oxide. Test it.

Dr. Karmach

Worked example 1 — solution

CO · CO₂
the furnace gas: 1 C + 1 O · the soda gas: 1 C + 2 O

A common first attempt

carbon oxide
fits CO and CO₂ equally — the name cannot rebuild either formula ✗

In an ionic name two element names are enough, because charges fix the ratio. Carbon and oxygen carry no charges to fix it.

Dr. Karmach

Worked example 1 — solution

CO · CO₂
the furnace gas: 1 C + 1 O · the soda gas: 1 C + 2 O
A common first attempt
carbon oxide
fits CO and CO₂ equally — the name cannot rebuild either formula ✗
Step 1 · Classify the compound

No leading H, no metal: two nonmetals. A molecular compound, so prefixes count the atoms.

Dr. Karmach

Worked example 1 — solution

CO · CO₂
the furnace gas: 1 C + 1 O · the soda gas: 1 C + 2 O
A common first attempt
carbon oxide
fits CO and CO₂ equally — the name cannot rebuild either formula ✗
Step 1 · Classify the compound Step 2 · Apply that system's rules
CO → carbon monoxide · CO₂ → carbon dioxide
1 C — mono- dropped on the first element · mono- + oxide → monoxide · 2 O → dioxide
Dr. Karmach

Worked example 1 — solution

CO · CO₂
the furnace gas: 1 C + 1 O · the soda gas: 1 C + 2 O
A common first attempt
carbon oxide
fits CO and CO₂ equally — the name cannot rebuild either formula ✗
Step 1 · Classify the compound Step 2 · Apply that system's rules
CO → carbon monoxide · CO₂ → carbon dioxide
1 C — mono- dropped on the first element · mono- + oxide → monoxide · 2 O → dioxide
Step 3 · Read the name back

Monoxide rebuilds exactly one O; dioxide rebuilds exactly two. Each name recovers its own formula.

Dr. Karmach

Worked example 1 — solution

CO · CO₂
the furnace gas: 1 C + 1 O · the soda gas: 1 C + 2 O
A common first attempt
carbon oxide
fits CO and CO₂ equally — the name cannot rebuild either formula ✗
Step 1 · Classify the compound Step 2 · Apply that system's rules
CO → carbon monoxide · CO₂ → carbon dioxide
1 C — mono- dropped on the first element · mono- + oxide → monoxide · 2 O → dioxide
Step 3 · Read the name back
Two different gases, two different names. The prefix is the part of the name that keeps them apart.
Dr. Karmach

Take-home: prefixes carry the formula

nitrogen + oxygen: NO · NO₂ · N₂O · N₂O₄
four different compounds — "nitrogen oxide" fits every one ✗

Two nonmetals often combine in several ratios. A molecular name without prefixes loses the formula. Ionic names never need prefixes; molecular names always do.

Dr. Karmach

Worked example 2 — formula from the name

tetraphosphorus decoxide
given: the name · wanted: the formula

Tetraphosphorus decoxide is a laboratory drying agent, sold as a white powder. Write its formula.

Dr. Karmach

Worked example 2 — solution

tetraphosphorus decoxide
given: the name · wanted: the formula

Step 1 · Classify the compound

Counting prefixes appear only in molecular names: this is a molecular compound of phosphorus and oxygen.

Dr. Karmach

Worked example 2 — solution

tetraphosphorus decoxide
given: the name · wanted: the formula
Step 1 · Classify the compound Step 2 · Apply that system's rules

Each prefix sets its own element's subscript:

tetraphosphorus decoxide → P₄O₁₀
tetra- → 4 P · dec(a)- → 10 O — the a dropped before oxide
Dr. Karmach

Worked example 2 — solution

tetraphosphorus decoxide
given: the name · wanted: the formula
Step 1 · Classify the compound Step 2 · Apply that system's rules
tetraphosphorus decoxide → P₄O₁₀
tetra- → 4 P · dec(a)- → 10 O — the a dropped before oxide
Step 3 · Read the name back

P₄O₁₀ reads back to the same name — the subscripts stay 4 and 10, not a reduced ratio.

Dr. Karmach

Worked example 2 — solution

tetraphosphorus decoxide
given: the name · wanted: the formula
Step 1 · Classify the compound Step 2 · Apply that system's rules
tetraphosphorus decoxide → P₄O₁₀
tetra- → 4 P · dec(a)- → 10 O — the a dropped before oxide
Step 3 · Read the name back
Ten O for four P: the name counted every atom, so the formula keeps exactly those counts.
Dr. Karmach

Your turn — Cl₂O₇

Cl₂O₇
chlorine and oxygen — two nonmetals, no leading H
step question answer
1 · classify acid, ionic, or molecular? molecular → prefixes
2 · apply 2 Cl · 7 O — which prefixes? chlorine oxide
3 · read back does the name rebuild Cl₂O₇?

Complete the name.

Dr. Karmach

Your turn — Cl₂O₇

Cl₂O₇
chlorine and oxygen — two nonmetals, no leading H
step question answer
1 · classify acid, ionic, or molecular? molecular → prefixes
2 · apply 2 Cl · 7 O — which prefixes? chlorine oxide
3 · read back does the name rebuild Cl₂O₇?

Complete the name.

Cl₂O₇ → dichlorine heptoxide
di- → 2 Cl · hept(a)- → 7 O — the a drops before oxide · reads back to Cl₂O₇ ✓
Dr. Karmach

Practice 1

N₂O₃
nitrogen and oxygen — two nonmetals, no leading H

N₂O₃ is one of several oxides of nitrogen found in polluted air. What is the correct name for N₂O₃?

  1. nitrogen oxide
  2. dinitrogen trioxide
  3. trinitrogen dioxide
  4. nitrous acid
Dr. Karmach

Practice 1 — answer: B

N₂O₃ → dinitrogen trioxide — answer B
two nonmetals → molecular · di- → 2 N · tri- → 3 O

A drops the prefixes, and NO, NO₂, N₂O, and N₂O₃ would all share that name — the formula is lost. C swaps the prefixes: trinitrogen dioxide rebuilds N₃O₂, a different compound. D uses an acid name, but nitrous acid is HNO₂ dissolved in water, and N₂O₃ contains no hydrogen.

Read the name back: di- and tri- rebuild N₂O₃ ✓. Each prefix counts its own element.
Dr. Karmach

Acid names come from the anion

Remove the H and look at the anion. No oxygen: hydro- + root + -ic acid. Oxygen present: the anion's -ate becomes -ic, -ite becomes -ous, and hydro- never appears.

Dr. Karmach

Worked example 3 — two acids

HCl(aq) · HNO₃(aq)
both start with H · both dissolved in water

HCl(aq) cleans concrete; HNO₃(aq) is used to make fertilizer. Name each compound.

A common first attempt: hydro- on both — hydrochloric acid and hydronitric acid. Test it.

Dr. Karmach

Worked example 3 — HCl(aq)

HCl(aq) · HNO₃(aq)
both start with H · both dissolved in water

Step 1 · Classify the compound

H first and dissolved in water: both are acids. Acid rules, not counting prefixes.

Dr. Karmach

Worked example 3 — HCl(aq)

HCl(aq) · HNO₃(aq)
both start with H · both dissolved in water
Step 1 · Classify the compound Step 2 · Apply that system's rules

Remove the H from HCl: the anion is chloride, Cl⁻, with no oxygen.

HCl(aq) → hydrochloric acid
anion: chloride — no oxygen → hydro- + chlor- + -ic acid
Dr. Karmach

Worked example 3 — HCl(aq)

HCl(aq) · HNO₃(aq)
both start with H · both dissolved in water
Step 1 · Classify the compound Step 2 · Apply that system's rules
HCl(aq) → hydrochloric acid
anion: chloride — no oxygen → hydro- + chlor- + -ic acid
Without the water it is hydrogen chloride, a gas. The (aq) is what the acid name records.
Dr. Karmach

Worked example 3 — HNO₃(aq)

HNO₃(aq)
an acid · anion: NO₃⁻, called nitrate — it holds three O

A common first attempt

hydronitric acid?
hydro- reads back to an anion with no oxygen — NO₃⁻ holds three ✗

Hydro- means the anion holds no oxygen. Nitrate holds three.

Dr. Karmach

Worked example 3 — HNO₃(aq)

HNO₃(aq)
an acid · anion: NO₃⁻, called nitrate — it holds three O
A common first attempt
hydronitric acid?
hydro- reads back to an anion with no oxygen — NO₃⁻ holds three ✗
Step 2 · Apply that system's rules

Oxygen present, so the anion's suffix maps: -ate → -ic acid.

HNO₃(aq) → nitric acid
anion: nitrate → -ate becomes -ic · no hydro-
Dr. Karmach

Worked example 3 — HNO₃(aq)

HNO₃(aq)
an acid · anion: NO₃⁻, called nitrate — it holds three O
A common first attempt
hydronitric acid?
hydro- reads back to an anion with no oxygen — NO₃⁻ holds three ✗
Step 2 · Apply that system's rules
HNO₃(aq) → nitric acid
anion: nitrate → -ate becomes -ic · no hydro-
Step 3 · Read the name back

Nitric acid reads back through nitrate to HNO₃(aq); hydrochloric acid reads back through chloride to HCl(aq).

Dr. Karmach

Worked example 3 — HNO₃(aq)

HNO₃(aq)
an acid · anion: NO₃⁻, called nitrate — it holds three O
A common first attempt
hydronitric acid?
hydro- reads back to an anion with no oxygen — NO₃⁻ holds three ✗
Step 2 · Apply that system's rules
HNO₃(aq) → nitric acid
anion: nitrate → -ate becomes -ic · no hydro-
Step 3 · Read the name back
Oxygen in the anion picks the rule. Nitrate holds three O, so the name takes no hydro-.
Dr. Karmach

Take-home: hydro- means no oxygen

H₂S(aq) → hydrosulfuric acid
anion: sulfide, S²⁻ — no oxygen → hydro- + -ic
H₂SO₄(aq) → sulfuric acid · H₂SO₃(aq) → sulfurous acid
sulfate SO₄²⁻ → -ic · sulfite SO₃²⁻ → -ous · no hydro- on either

Hydro- appears only when the anion has no oxygen. With oxygen, the anion's suffix sets the acid's suffix: -ate → -ic, -ite → -ous.

Dr. Karmach

Where this goes wrong

Prefixes on an ionic compound. MgCl₂ is magnesium chloride, never magnesium dichloride. A metal is present, so charges fix the ratio; prefixes belong to molecular names.
Swapped prefixes. Each prefix counts its own element's atoms. Tetranitrogen dioxide rebuilds N₄O₂; the compound N₂O₄ is dinitrogen tetroxide.
An acid name without hydrogen. SO₃ is not sulfuric acid — sulfuric acid is H₂SO₄(aq). No leading H and no water: SO₃ is the molecular compound sulfur trioxide.
-ate mapped to -ous. Sulfate → sulfuric, sulfite → sulfurous. Naming H₂SO₄(aq) "sulfurous acid" points at the wrong compound: sulfurous acid is H₂SO₃(aq), built on sulfite.
Dr. Karmach

Practice 2

HF(aq)
H first · dissolved in water

HF dissolved in water etches patterns into glass. What is the correct name for HF(aq)?

  1. hydrogen fluoride
  2. fluoric acid
  3. hydrofluoric acid
  4. hydrogen monofluoride
Dr. Karmach

Practice 2 — answer: C

HF(aq) → hydrofluoric acid — answer C
an acid · anion: fluoride, F⁻ — no oxygen → hydro- + fluor- + -ic acid

A names the pure gas, HF(g); the (aq) marks a dissolved acid with its own name. B drops hydro-: without it the name reads as an oxyacid, and fluoride holds no oxygen. D uses counting prefixes, and prefixes never appear in acid names.

Read the name back: hydro- marks a no-oxygen anion, fluoride ✓. The name rebuilds HF(aq).
Dr. Karmach

Check yourself

  1. Sulfur and fluorine form SF₆, a gas used to insulate electrical equipment. Classify the compound and name it.
  2. NO₂⁻ is nitrite and NO₃⁻ is nitrate. Name HNO₂(aq) and HNO₃(aq).

Sulfate, sulfite, nitrate, and nitrite belong to a larger family of polyatomic ions — charged groups of atoms that act as a unit. Their names and charges build ionic formulas such as Ca(NO₃)₂, where parentheses group the ion.

Dr. Karmach

7 · Polyatomic Ions

Recognize the common polyatomic ions, name compounds that contain them, and build formulas with parentheses wherever a group is multiplied.

Dr. Karmach

The names on the shelf

Baking soda's ingredient label reads sodium hydrogen carbonate. Household bleach lists sodium hypochlorite. Garden fertilizer lists ammonium nitrate. Everyday products; the labels name the chemistry inside.

Dr. Karmach

One group, one charge

A polyatomic ion is a bonded group of atoms carrying one overall charge. The charge belongs to the whole group, and the group travels through reactions and formulas as a single unit.

Dr. Karmach

The common polyatomic ions

Each formula is memorized with its charge. Together they are the ion's identity.

OH⁻ hydroxide · NO₃⁻ nitrate · NO₂⁻ nitrite · ClO₃⁻ chlorate · HCO₃⁻ hydrogen carbonate
charge 1− · the largest family
SO₄²⁻ sulfate · SO₃²⁻ sulfite · CO₃²⁻ carbonate · CrO₄²⁻ chromate
charge 2−
PO₄³⁻ phosphate · NH₄⁺ ammonium
phosphate: charge 3− · ammonium: 1+, the one common polyatomic cation
Dr. Karmach

-ate and -ite: the oxygen count

Suffixes and prefixes report oxygen count, never charge. -ate marks the higher count, -ite one fewer; per- sits one above -ate, hypo- one below -ite. The whole chlorine series carries 1−.

Dr. Karmach

The method

  1. Identify the ions. Recall each ion's atoms and charge from the memorized list.
  2. Balance the charges to zero. Smallest counts that sum to zero.
  3. Write the formula or the name. Cation first; parentheses around a repeated polyatomic; no prefixes.
Dr. Karmach

Worked example 1 — naming K₂CO₃

K₂CO₃ — potash, a traditional glassmaking ingredient
given: the formula · wanted: the name

Potash lowers the melting point of the sand in a glass furnace. Name the compound.

Dr. Karmach

Worked example 1 — solution

K₂CO₃
given: the formula · wanted: the name

Step 1 · Identify the ions

The group CO₃ with its charge is on the memorized list: carbonate, CO₃²⁻. The rest is potassium, K⁺.

Dr. Karmach

Worked example 1 — solution

K₂CO₃
given: the formula · wanted: the name
Step 1 · Identify the ions Step 2 · Balance the charges to zero
K₂CO₃ — 2 K⁺ and 1 CO₃²⁻
charge: 2(+1) + 1(−2) = 0 ✓ — the split is consistent
Dr. Karmach

Worked example 1 — solution

K₂CO₃
given: the formula · wanted: the name
Step 1 · Identify the ions Step 2 · Balance the charges to zero
K₂CO₃ — 2 K⁺ and 1 CO₃²⁻
charge: 2(+1) + 1(−2) = 0 ✓ — the split is consistent
Step 3 · Write the formula or the name

Cation first, then the anion:

K₂CO₃ → potassium carbonate
not "dipotassium carbonate" — charge balance already fixes the counts
Dr. Karmach

Worked example 1 — solution

K₂CO₃
given: the formula · wanted: the name
Step 1 · Identify the ions Step 2 · Balance the charges to zero
K₂CO₃ — 2 K⁺ and 1 CO₃²⁻
charge: 2(+1) + 1(−2) = 0 ✓ — the split is consistent
Step 3 · Write the formula or the name
K₂CO₃ → potassium carbonate
not "dipotassium carbonate" — charge balance already fixes the counts
The name carries no numbers. The memorized charges rebuild them: reaching zero requires two K⁺ for one CO₃²⁻.
Dr. Karmach

Worked example 2 — magnesium nitrate

magnesium nitrate — a nitrogen source in fertilizers
given: the name · wanted: the formula

Write the formula. A common first attempt: MgNO₃₂. Test it.

Dr. Karmach

Worked example 2 — balancing the charges

magnesium nitrate
given: the name · wanted: the formula

A common first attempt

MgNO₃₂
reads as one N and one O₃₂ — a 32-oxygen subscript, no nitrate group left ✗

Two nitrate ions were intended. Written without parentheses, the subscripts run together and the group disappears.

Dr. Karmach

Worked example 2 — balancing the charges

magnesium nitrate
given: the name · wanted: the formula
A common first attempt
MgNO₃₂
reads as one N and one O₃₂ — a 32-oxygen subscript, no nitrate group left ✗
Step 1 · Identify the ions

Magnesium forms Mg²⁺. Nitrate is on the memorized list: NO₃⁻.

Dr. Karmach

Worked example 2 — balancing the charges

magnesium nitrate
given: the name · wanted: the formula
A common first attempt
MgNO₃₂
reads as one N and one O₃₂ — a 32-oxygen subscript, no nitrate group left ✗
Step 1 · Identify the ions Step 2 · Balance the charges to zero
1(+2) + 1(−1) = +1 ✗  ·  1(+2) + 2(−1) = 0 ✓
one nitrate leaves +1 — one Mg²⁺ needs two NO₃⁻
Dr. Karmach

Worked example 2 — balancing the charges

magnesium nitrate
given: the name · wanted: the formula
A common first attempt
MgNO₃₂
reads as one N and one O₃₂ — a 32-oxygen subscript, no nitrate group left ✗
Step 1 · Identify the ions Step 2 · Balance the charges to zero
1(+2) + 1(−1) = +1 ✗  ·  1(+2) + 2(−1) = 0 ✓
one nitrate leaves +1 — one Mg²⁺ needs two NO₃⁻
The count of each ion may change; the charge on each ion may not. Two nitrates cancel one Mg²⁺.
Dr. Karmach

Worked example 2 — writing the formula

magnesium nitrate — Mg²⁺ with two NO₃⁻
charge: 1(+2) + 2(−1) = 0 ✓

Step 3 · Write the formula or the name

Cation first, and the repeated polyatomic goes in parentheses:

Mg(NO₃)₂
atoms: 1 Mg · 2 N · 2 × 3 = 6 O — the 2 multiplies everything inside
Dr. Karmach

Worked example 2 — writing the formula

magnesium nitrate — Mg²⁺ with two NO₃⁻
charge: 1(+2) + 2(−1) = 0 ✓
Step 3 · Write the formula or the name
Mg(NO₃)₂
atoms: 1 Mg · 2 N · 2 × 3 = 6 O — the 2 multiplies everything inside
The charges sum to zero, and each nitrate stays whole inside its parentheses.
Dr. Karmach

Take-home: parentheses keep the group whole

A subscript outside parentheses multiplies everything inside. Mg(NO₃)₂ holds 1 Mg, 2 N, and 6 O. A polyatomic ion taken more than once is always written in parentheses.

Dr. Karmach

Your turn — calcium hydroxide

calcium hydroxide — slaked lime, the base in mortar and plaster
given: the name · wanted: the formula
step question answer
1 · identify the ions cation and anion? Ca²⁺ and
2 · balance the charges to zero 1(+2) + (−1) = 0 hydroxides
3 · write the formula or the name parentheses needed?

Complete the three steps.

Dr. Karmach

Your turn — calcium hydroxide

calcium hydroxide — slaked lime, the base in mortar and plaster
given: the name · wanted: the formula
step question answer
1 · identify the ions cation and anion? Ca²⁺ and
2 · balance the charges to zero 1(+2) + (−1) = 0 hydroxides
3 · write the formula or the name parentheses needed?

Complete the three steps.

Ca(OH)₂
charge: 1(+2) + 2(−1) = 0 ✓ · atoms: 1 Ca · 2 O · 2 H
Hydroxide is 1−, so two groups balance one Ca²⁺. The parentheses keep each OH whole.
Dr. Karmach

Where this goes wrong

Reading -ite as a different charge. Sulfate SO₄²⁻ and sulfite SO₃²⁻ both carry 2−. The suffix changes the oxygen count, never the charge.
Borrowing another ion's charge. Nitrate is NO₃⁻, never NO₃²⁻ — the 2− belongs to carbonate and sulfate. The charge is part of each ion's memorized identity.
Dropping the parentheses. CaOH₂ shows 1 O and 2 H. Calcium hydroxide holds two whole OH⁻ groups: Ca(OH)₂, with 2 O and 2 H.
Adding counting prefixes to the name. Mg(NO₃)₂ is magnesium nitrate, never magnesium dinitrate. Charge balance already fixes the counts; prefixes belong to molecular compounds.
Dr. Karmach

Practice 1

KClO₃ — K⁺ with one ion from the chlorine series
given: the formula · wanted: the name

Match heads carry KClO₃ as their oxygen supply. Name the compound.

  1. potassium chlorite
  2. potassium chlorate
  3. potassium perchlorate
  4. potassium chloride
Dr. Karmach

Practice 1 — answer: B

KClO₃ → potassium chlorate — answer B
K⁺ and ClO₃⁻ · charge: 1(+1) + 1(−1) = 0 ✓

Three oxygens is the -ate member of the chlorine series. A, chlorite, is ClO₂⁻ — one oxygen fewer. C, perchlorate, is ClO₄⁻ — one oxygen more. D, chloride, is Cl⁻ — a monatomic ion with no oxygen at all.

Only the oxygen count separates these four names. Every choice pairs one K⁺ with one 1− anion, so the charge test cannot pick the name; the memorized series does.
Dr. Karmach

Worked example 3 — ammonium phosphate

ammonium phosphate — a fertilizer supplying nitrogen and phosphorus at once
given: the name · wanted: the formula · both ions polyatomic

Write the formula. Both ions come from the memorized list.

Dr. Karmach

Worked example 3 — solution

ammonium phosphate
given: the name · wanted: the formula

Step 1 · Identify the ions

Ammonium, the one common polyatomic cation: NH₄⁺. Phosphate: PO₄³⁻.

Dr. Karmach

Worked example 3 — solution

ammonium phosphate
given: the name · wanted: the formula
Step 1 · Identify the ions Step 2 · Balance the charges to zero
1(+1) + 1(−3) = −2 ✗  ·  3(+1) + 1(−3) = 0 ✓
one PO₄³⁻ needs three NH₄⁺
Dr. Karmach

Worked example 3 — solution

ammonium phosphate
given: the name · wanted: the formula
Step 1 · Identify the ions Step 2 · Balance the charges to zero
1(+1) + 1(−3) = −2 ✗  ·  3(+1) + 1(−3) = 0 ✓
one PO₄³⁻ needs three NH₄⁺
Step 3 · Write the formula or the name

The repeated ion is polyatomic, so it takes the parentheses. Phosphate appears once and needs none.

(NH₄)₃PO₄
atoms: 3 × 1 = 3 N · 3 × 4 = 12 H · 1 P · 1 × 4 = 4 O
Dr. Karmach

Worked example 3 — solution

ammonium phosphate
given: the name · wanted: the formula
Step 1 · Identify the ions Step 2 · Balance the charges to zero
1(+1) + 1(−3) = −2 ✗  ·  3(+1) + 1(−3) = 0 ✓
one PO₄³⁻ needs three NH₄⁺
Step 3 · Write the formula or the name
(NH₄)₃PO₄
atoms: 3 × 1 = 3 N · 3 × 4 = 12 H · 1 P · 1 × 4 = 4 O
The charges sum to zero, and both groups stay intact on the page: three whole ammoniums, one whole phosphate.
Dr. Karmach

Practice 2

ammonium sulfate — a lawn fertilizer
NH₄⁺, charge 1+ · SO₄²⁻, charge 2− · wanted: the formula

Write the formula for ammonium sulfate.

  1. NH₄SO₄
  2. (NH₄)₂SO₄
  3. (NH₄)₂SO₃
  4. NH₄(SO₄)₂
Dr. Karmach

Practice 2 — answer: B

(NH₄)₂SO₄ — answer B
charge: 2(+1) + 1(−2) = 0 ✓ · atoms: 2 N · 2 × 4 = 8 H · 1 S · 4 O

A stops at one of each: 1(+1) + 1(−2) = −1, not zero. C balances its charges, 2(+1) + 1(−2) = 0, but holds sulfite, SO₃²⁻ — the -ite ion, one oxygen fewer. D doubles the wrong ion: 1(+1) + 2(−2) = −3.

Two 1+ cations cancel one 2− anion, and the repeated polyatomic, ammonium, takes the parentheses.
Dr. Karmach

Check yourself

  1. Sodium carbonate contains Na⁺ and CO₃²⁻. Write the formula, and show the charge sum that makes it neutral.
  2. KClO₄ and KClO₃ — name both. Which piece of each name reports the oxygen count?

In a chemical equation, a polyatomic ion that passes through a reaction unchanged is balanced as one unit: count nitrate as nitrate, not as separate N and O atoms.

Dr. Karmach

Can you…?

  • ☐ locate protons, neutrons, and electrons in the atom and state what each count determines?
  • ☐ read and write isotope symbols, converting between mass number, atomic number, and particle counts?
  • ☐ calculate the average atomic mass of an element from isotope masses and abundances?
  • ☐ predict the charge an atom takes when it forms an ion, and count the particles in that ion?
  • ☐ name ionic and molecular compounds and acids, and write formulas from names?
  • ☐ recognize the common polyatomic ions and build formulas that contain them?

If any box stays empty, the practice site has a drill for it. 🧪

Dr. Karmach