Guided example: solution

Step 2 · Identify the ions

Both charges come from periodic position.

Guided example: solution

Guided example: solution

Guided example: the name

Step 4 · Assemble the answer

Name the cation first, unchanged. Then the anion: its element's stem plus -ide.

Guided example: the name

Guided example: the name

Guided example: the route on the map

Practice 1 · answer: B

A wrote each ion's own charge as its own subscript: 2(+2) + 1(−1) = +3, not neutral. C pairs one of each: 1(+2) + 1(−1) = +1. D has the right counts in the wrong order; the cation is written first.

Practice 1: the route on the map

Worked example 1: solution

Step 2 · Identify the ions

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

Worked example 1: solution

The subscript 2 records the balance already:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 2 · Identify the ions

Worked example 2: solution

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.

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: C

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.

Practice 2: the route on the map

Worked example 3: testing the first attempt

A common first attempt

Worked example 3: testing the first attempt

A common first attempt

Worked example 3: solution

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.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 · answer: A

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.

Practice 3: the route on the map

Worked example 4: solution

Step 2 · Identify the ions

The numeral states tin's charge directly: Sn⁴⁺. Oxide is O²⁻ from periodic position.

Worked example 4: solution

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 · answer: D

B balances a different cation: 1(+2) + 2(−1) = 0 holds for Sn²⁺, so SnF₂ is tin(II) fluoride, the stannous fluoride on toothpaste labels. C pairs one of each ion: 1(+4) + 1(−1) = +3, not neutral. A writes the numeral as a count of tin atoms: 4(+4) + 1(−1) = +15.

Practice 4: the route on the map

Practice 5 · answer: A

B is correct: three S²⁻ carry 3(−2) = −6, split over two Fe, 6 ÷ 2 = 3+ each; reading the subscript 2 as the charge gives the wrong numeral. C is correct: aluminum is fixed at 3+, 1(+3) + 1(−3) = 0, so no numeral. D is correct: 1(+2) + 2(−1) = 0, and strontium diiodide would borrow molecular prefixes.

Practice 5: the route on the map

Guided example: solution

Three moves are needed, one per method step.

Step 1 · Identify the ions

Sodium sits in Group 1: Na⁺. Nitrite holds one O fewer than the given nitrate, with the same charge: NO₂⁻.

Guided example: solution

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: A

Practice 1 · answer: A

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Identify the ions

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

Worked example 1: solution

Worked example 1: solution

Cation first, then the anion:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: balancing the charges

A common first attempt

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

Worked example 2: balancing the charges

Magnesium forms Mg²⁺. Nitrate is a given -ate: NO₃⁻.

Worked example 2: balancing the charges

Worked example 2: balancing the charges

Worked example 2: writing the formula

Step 3 · Write the formula or the name

Cation first, and the repeated polyatomic goes in parentheses:

Worked example 2: writing the formula

Worked example 2: the route on the map

Practice 2 · answer: D

Four oxygens is one above the -ate member of the chlorine series: per- + chlorate. C, chlorate, is ClO₃⁻, one oxygen fewer. A, chlorite, is ClO₂⁻, two fewer. B, chloride, is Cl⁻, a monatomic ion with no oxygen at all. E puts a numeral on potassium, a fixed-charge metal: K⁺ never takes one.

Practice 2: the route on the map

Worked example 3: solution

Step 1 · Identify the ions

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

Worked example 3: solution

Worked example 3: solution

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

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 · answer: B

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.

Practice 3: the route on the map

Practice 4 · answer: C

Practice 4 · answer: C

Practice 4: the route on the map

Worked example 4: testing the first attempt

A common first attempt

Worked example 4: testing the first attempt

A common first attempt

Worked example 4: solution

Step 1 · Identify the ions

Aluminum forms Al³⁺. Sulfate is a given -ate: SO₄²⁻.

Worked example 4: solution

Worked example 4: solution

The crossed 3 lands outside the parentheses; the 4 inside never changes.

Worked example 4: solution

Worked example 4: the route on the map

Practice 6 · answer: D

A kept the -ate: Fe₂(SO₄)₃ is iron(III) sulfate. B pairs one of each: 1(+3) + 1(−2) = +1; FeSO₃ balances only with Fe²⁺, iron(II) sulfite. C writes each ion's own charge as its own subscript: 3(+3) + 2(−2) = +5. E multiplies the 3 through: Fe₂S₃O₉ has the atoms but no sulfite group left.

Practice 6: the route on the map

Guided example: solution

Three moves are needed, one per method step.

Step 1 · Classify the compound

Guided example: solution

Guided example: solution

Carbon tetrachloride reads back to 1 C and 4 Cl: CCl₄ ✓.

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: D

A borrows the ionic numeral: two nonmetals share electrons, and a numeral names a metal's charge. B drops the prefixes, and that name fits NF₃ and N₂F₄ alike. C puts tri- on the wrong element and rebuilds N₃F.

Practice 1: the route on the map

Worked example 1: solution

A common first attempt

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

Worked example 1: solution

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

Worked example 1: solution

Worked example 1: solution

Monoxide rebuilds one O, dioxide two: each name recovers its own formula.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Classify the compound

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

Worked example 2: solution

Each prefix sets its own element's subscript:

Worked example 2: solution

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

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: B

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.

Practice 2: the route on the map

Worked example 3: HCl(aq)

Step 1 · Classify the compound

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

Worked example 3: HCl(aq)

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

Worked example 3: HCl(aq)

Worked example 3: HNO₃(aq)

A common first attempt

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

Worked example 3: HNO₃(aq)

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

Worked example 3: HNO₃(aq)

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

Worked example 3: HNO₃(aq)

Worked example 3: the route on the map

Practice 3 · answer: C

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.

Practice 3: the route on the map

Worked example 4: solution

Step 1 · Classify the compound

An acid name, so the formula starts with H and carries (aq).

Worked example 4: solution

Run the suffix map backward: -ous came from -ite. The anion is sulfite, SO₃²⁻.

Worked example 4: solution

Worked example 4: solution

H₂SO₃ minus its hydrogens is sulfite, and -ite returns -ous: sulfurous acid ✓.

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 · answer: D

B writes one H: 1(+1) + 1(−3) = −2, not neutral. C writes two: 2(+1) + 1(−3) = −1. A balances three H on the wrong rung: PO₃³⁻ is phosphite, and its acid is phosphorous acid, one rung down the ladder.

Practice 4: the route on the map

Worked example 5: solution

Step 1 · Classify the compound

A metal with a polyatomic anion: an ionic salt, carrying attached water. A hydrate.

Worked example 5: solution

Name the salt by the ionic rules, then count the waters with a prefix.

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the map

Practice 5 · answer: C

A is correct: two nonmetals take prefixes, di- → 2 N, pent(a)- → 5 O; nitrogen(V) oxide would borrow the ionic numeral. B is correct: iodide holds no oxygen, so hydro- + iod- + -ic. D is correct: three Cl⁻ demand Fe³⁺, 1(+3) + 3(−1) = 0, and hexa- counts the 6 waters.

Practice 5: the route on the map

Guided example: solution

Step 1 · Classify the compound

Guided example: solution

A common first attempt writes calcium(II) chloride. A numeral appears only for a metal with more than one possible charge.

Guided example: solution

Guided example: the route on the map

Step 3 · Read the name back

Worked example 1: testing the first attempt

A common first attempt

Worked example 1: testing the first attempt

A common first attempt

Worked example 1: solution

Step 1 · Classify the compound

No leading H. A metal is present: ionic, and copper is a variable-charge metal.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Classify the compound

H first and dissolved in water: an acid. The metal test and the prefix test never run.

Worked example 2: solution

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Your turn: the route on the map

Practice 1 · answer: A

B holds sodium, a metal: Na₂O is sodium oxide, and its 2 comes from charge balance. C is all nonmetals but holds NH₄⁺, which question 2 catches: ammonium nitrate is ionic. D has two nonmetals, but H first in water answers question 1: hydroiodic acid.

Practice 1: the route on the map

Practice 2 · answer: B

A borrows the ionic system's numeral, but a numeral reports an ion's charge and ICl₃ holds no ions. C is the fixed-charge ionic style, and without a prefix the name cannot separate ICl₃ from ICl. D swaps the prefix onto iodine: triiodine monochloride rebuilds I₃Cl.

Practice 2: the route on the map

Practice 3 · answer: D

A writes a numeral for a metal with one possible charge; Group 2 is always 2+, so no numeral appears. B counts atoms with molecular prefixes, but the metal makes this ionic. C names the monatomic anion P³⁻; this compound holds the polyatomic group phosphate, PO₄³⁻.

Practice 3: the route on the map

Practice 4 · answer: A

B is the ionic-style read-back, and hydrogen carbonate already names a different species, the ion HCO₃⁻. C adds hydro-, which claims a no-oxygen anion; carbonate holds three. D counts atoms with a prefix, and prefixes never appear in acid names.

Practice 4: the route on the map

Practice 5 · answer: B

A names ClO₃⁻, one oxygen short: copper(II) chlorate rebuilds Cu(ClO₃)₂. C drops the numeral, but copper has more than one possible charge. D counts the anions with a prefix, a molecular rule on a compound with a metal. E names the one-atom ion Cl⁻: copper(II) chloride rebuilds CuCl₂, with no oxygen.

Practice 5: the route on the map