Worked example: solution

Apply the law

The totals must match: 15.9 g of reactant becomes 8.3 g of solid plus every gram of escaped gas.

Worked example: solution

Worked example: solution

Guided example: solution

Step 1 · Read the case

A common first attempt reads C and u apart. No symbol starts with a lowercase letter, so Cu is one symbol.

Guided example: solution

Step 1 · Read the case

Guided example: solution

Step 1 · Read the case

Guided example: the route on the map

Step 4 · Count

Practice 1 · answer: B

B is the wrong pairing: two capitals mean two elements, so CO is a compound of carbon and oxygen. A, C, and D are correct Latin pairs: ferrum gives Fe, natrium gives Na, stannum gives Sn.

Practice 1: the route on the map

Practice 2 · answer: A

B swapped the two precious metals: gold is Au, from aurum. C matched Ag to an English name that starts with A: argon is Ar. D dropped the lowercase r: boron is B alone, and the r joins B in Br, bromine.

Practice 2: the route on the map

Practice 3 · answer: D

A took potassium's first letter: P is phosphorus. B capitalized chlorine's second letter: CL reads as two symbols, C and L, and no element is L. C took the first two letters: Po is polonium.

Practice 3: the route on the map

Practice 4 · answer: C

A read CO as Co, cobalt: C, H, Co, O makes 4. B counted atoms, not elements: 2 C + 4 H + 2 O = 8. D counted every symbol as written, repeats included: 6.

Practice 4: the route on the map

Practice 5 · answer: D

A read Co as two symbols, C and O: a lowercase o never starts a symbol, so Co is one element, cobalt. B read CO as Co too: two capitals are two symbols, carbon and oxygen. C matched Co to copper: copper is Cu, from cuprum.

Practice 5: the route on the map

Guided example: solution

Three moves are needed, one per method step.

Step 1 · Match protons and Z

Z is the proton count. Sodium's Z = 11, so the atom holds 11 protons.

Guided example: solution

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: C

A read the tile's average mass, rounded, as the proton count: 108. B subtracted Z from that rounded mass: 108 − 47 = 61, a neutron estimate, not protons. D added the tile's two numbers: 47 + 108 = 155.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Match protons and Z

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

Worked example 1: solution

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

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Match protons and Z

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

Worked example 2: solution

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.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: A

C 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.

Practice 2: the route on the map

Worked example 3: solution

A common first attempt

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

Worked example 3: solution

A common first attempt

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

Worked example 3: solution

A common first attempt

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

Worked example 3: solution

A common first attempt

Worked example 3: solution

A common first attempt

Worked example 3: the route on the map

Practice 3 · answer: D

A read the mass number as the atomic number: Z = 27 is cobalt, and 27 protons leave no room for 14 neutrons in a nucleus of 27. B read the neutron count as the proton count: Z = 14 is silicon, but neutrons never name the element. C matched the electrons to the neutrons; a neutral atom matches them to the protons, 13.

Practice 3: the route on the map

Practice 4 · answer: E

Practice 4 · answer: E

Practice 4: the route on the map

Guided example: solution

Step 1 · Find Z

The lower-left number is Z = 54: 54 protons. Element 54 on the periodic table is xenon, so the number and the symbol agree.

Guided example: solution

Two of the three are known, A and Z. Subtract to find the third.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: D

A read the mass number as the electron count; 127 counts protons and neutrons, and neutrons carry no charge. B gave the neutron count, 127 − 53 = 74; the question asks for electrons. C added the two numbers: 127 + 53 = 180.

Practice 1: the route on the map

Worked example 1: solution

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.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: the first attempt

A common first attempt

Worked example 2: the first attempt

A common first attempt

Worked example 2: the first attempt

A common first attempt

Worked example 2: solution

Step 1 · Find Z

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

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 2 · answer: B

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.

Practice 2: the route on the map

Practice 3 · answer: C

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.

Practice 3: the route on the map

Worked example 3: solution

Step 1 · Find Z

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

Worked example 3: solution

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 4 · answer: A

B matched the mass numbers: atom 1 has A = 26 + 32 = 58, like nickel-58, but 26 and 28 protons are two elements. C matched the neutron counts: nickel-58 holds 58 − 28 = 30 neutrons, like atom 3, but neutrons never name the element. D read atom 3's 30 neutrons as its proton count, zinc; subtract instead: 56 − 30 = 26.

Practice 4: the route on the map

Practice 5 · answer: D

A subtracted the mass numbers: 40 − 39 = 1. That shortcut works only for isotopes, which share Z. B stopped at argon-40's own count, 22. C added A and Z for each atom: (40 + 18) − (39 + 19) = 0. E gave potassium-39's own count, 20.

Practice 5: the route on the map

Guided example: solution

Set up the abundances

A natural sample with both percents given: 72.17 + 27.83 = 100.00. Two isotopes, so two terms.

Guided example: solution

72.17 ÷ 100 = 0.7217 and 27.83 ÷ 100 = 0.2783.

Guided example: solution

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: C

A attached Sb-121's abundance to Sb-123: 122.9042 × 0.5721 = 70.31. B treated the sample as an even split: 122.9042 × 0.50 = 61.45. D used the whole percent: 122.9042 × 42.79 = 5259.

Practice 1: the route on the map

Worked example 1: solution

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.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

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

Worked example 2: solution

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

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: B

C averaged the masses equally: (14.0031 + 15.0001) ÷ 2 = 14.50. A attached the abundances to the wrong isotopes: 14.0031 × 0.00364 + 15.0001 × 0.99636 = 15.00. D used whole percents: 14.0031 × 99.636 + 15.0001 × 0.364 = 1401.

Practice 2: the route on the map

Worked example 3: fractions and contributions

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.

Worked example 3: fractions and contributions

Worked example 3: fractions and contributions

Worked example 3: the sum

Step 3 · Add the contributions

Worked example 3: the sum

Step 3 · Add the contributions

Worked example 3: the route on the map

Practice 3 · answer: E

Practice 3 · answer: E

Practice 3: the route on the map

Practice 4 · answer: D

Three products are needed, then one sum.

Practice 4 · answer: D

Practice 4: the route on the map

Guided example: solution

Step 1 · Find the period

Xenon closes the fifth row, which runs from rubidium (Z = 37) to xenon (Z = 54): period 5.

Guided example: solution

Xenon sits in the last column on the right: group 18, labeled 8A.

Guided example: solution

Group 18 is the noble gases: helium, neon, argon, krypton, xenon, and radon.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: A

B stretched the staircase: aluminum touches the line but is a metal, an excellent conductor. C sits one column right of arsenic, past the line: selenium is a nonmetal. D carried germanium's class down group 14: tin, below it, is a metal.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Find the period

Chlorine sits in the third row: period 3. That row runs from sodium (Z = 11) to argon (Z = 18): 18 − 11 + 1 = 8 elements.

Worked example 1: solution

Counting columns from the left edge, chlorine lands in column 17: group 17.

Worked example 1: solution

Group 17 is the halogens: fluorine, chlorine, bromine, and iodine.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Thirty-eight protons against twenty is a real difference, and mass roughly doubles. But behavior follows the column: 38 − 20 = 18, exactly one full row, so strontium lands directly under calcium.

Worked example 2: solution

One row below calcium's period 4: period 5. Same column: group 2.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Your turn: the route on the map

Practice 2 · answer: B

A counted rows inside the halogen column: fluorine's row is period 2, so the fourth halogen down is iodine, period 5. C read group 17 as group 7 and landed in the transition block: manganese sits in period 4, group 7. D overshot by one column: krypton is group 18, a noble gas.

Practice 2: the route on the map

Worked example 3: solution

The ordering rule

By mass, iodine should come first. The table disagrees: position follows atomic number, protons only, and 52 comes before 53. Extra neutrons make tellurium heavier; position ignores them.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 · answer: C

A misreads the geometry: sodium and potassium share a group, a column; period-mates differ. B fails on its own numbers: 39.10 − 22.99 = 16.11, the masses are not close, and mass does not set behavior. D misfiles them: group 1 is a main-group column; the transition block starts at group 3.

Practice 3: the route on the map

Worked example 4: solution

Family → group

The family name fixes the column: alkaline earth metals are group 2.

Worked example 4: solution

The period fixes the row: period 3 runs from sodium (Z = 11) to argon (Z = 18).

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 · answer: C

A fits the bench tests, but potassium sits in group 1: one outer electron, so it loses one, never two. B fails at the bench: silicon is a metalloid on the staircase, a semiconductor, not a full conductor. D reverses the electron move: sulfur is a brittle nonmetal, and nonmetals gain electrons.

Practice 4: the route on the map

Practice 5 · answer: B

A read 5A as group 5: period 6, group 5 is tantalum, a transition metal; an A label always marks a main-group column. C swapped the two numbers: period 5, group 6A (16) is tellurium, a metalloid on the staircase. D carried antimony's class one row down: bismuth sits left of the staircase, a metal.

Practice 5: the route on the map

Extra practice 1 · answer: D

A carried the class down the column: class shifts down group 15, from nonmetal (N, P) to metalloid (As, Sb) to metal (Bi). B treated row-mates as alike: rubidium and strontium sit far left of the staircase; antimony sits on it. C judged by one bench property: metalloids look metallic, and antimony is only a modest conductor.

Extra practice 2 · answer: A

B read the mass number as the atomic number: element 85 is astatine, period 6, group 17. C found rubidium but swapped the address: period 1 holds only H and He, so no group 5 seat exists there. D took the neutron count as the atomic number: 85 − 37 = 48 is cadmium, and 48 − 37 + 1 = 12 puts it in period 5, group 12.

Extra practice 3 · answer: B

A reversed the electron move: cesium is a metal, and metals lose electrons; gaining one is the halogen move. C carried group 17's fade into group 1: down the alkali column the water reaction escalates. D let the proton gap override the column: 36 = 18 + 18 is exactly two full rows, so cesium stays in group 1 with one outer electron.

Guided example: solution

Step 1 · Count the protons

Oxygen is element 8 on the periodic table: 8 protons.

Guided example: solution

A neutral oxygen atom holds 8 electrons. A negative charge means electrons were gained, two of them: 8 + 2 = 10 e⁻.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 · answer: A

B ignored the charge: 13 electrons is the neutral atom. C counted neutrons: 27 − 13 = 14. D added electrons for a positive charge: 13 + 3 = 16 computes 13 − 16 = 3−. E took the charge off the mass number: 27 − 3 = 24.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Count the protons

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

Worked example 1: solution

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

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

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

Worked example 2: solution

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

Worked example 2: solution

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

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 · answer: A

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

Practice 2: the route on the map

Practice 3 · answer: E

A gave nitrogen a metal's move: losing three leaves 7 − 3 = 4 e⁻ and a 3+ charge. B lost all five outer electrons, 7 − 5 = 2: the long way to a noble-gas count. C used group 17's charge: 7 + 1 = 8 e⁻, two short of neon. D kept the 3− label but subtracted: 7 − 3 = 4 e⁻ computes 7 − 4 = 3+.

Practice 3: the route on the map

Worked example 3: solution

Step 1 · Count the protons

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

Worked example 3: solution

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

Worked example 3: solution

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

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 4 · answer: D

A accepted statement 1: electrons never enter the mass number, so ⁴⁰Ca²⁺ keeps A = 20 + 20 = 40, not 40 − 2 = 38. B read 3− as three electrons lost, 15 − 3 = 12, and rejected statement 2; a negative ion gained them. C found neutrons by subtracting electrons, 85 − 36 = 49, and rejected statement 3; neutrons = A − protons = 48.

Practice 4: the route on the map

Practice 5 · answer: C

A named the element from the electrons: 54 is xenon's count, and 54 + 76 = 130. B ran the charge backward: 54 + 2 = 56 protons is barium, and 56 − 54 computes 2+. D added the two gained electrons to the mass number: 52 + 76 + 2 = 130. E added all 54: 52 + 76 + 54 = 182.

Practice 5: the route on the map