Chemistry: Atoms First 2e · Representative Metals, Metalloids, and Nonmetals

Occurrence and Preparation of the Representative Metals

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In 30 seconds
Why this matters
The college version
Eli explains
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

The representative metals are the s-block metals (Groups 1 and 2) plus the p-block metals: aluminum, gallium, indium, thallium, tin, lead, and bismuth. Except for a few "noble" metals, none are found free in nature: over geological time, oxygen, sulfur, carbon dioxide, and water oxidized them into the ores we mine today — oxides, sulfides, carbonates, and chlorides.

is the science of getting a metal back out of its , and it rests on one matching game: the more reactive the metal, the harder it is to free, and the more powerful the reducing agent must be. Gold, silver, and platinum occur free and need almost no chemistry; sodium, potassium, calcium, magnesium, and aluminum bond so strongly that only of molten compounds can reduce them; in between, iron, zinc, tin, and lead are won by heating their oxides with carbon or carbon monoxide.

Why this matters

Metals built your world: structural steel, aircraft and cans (aluminum), wiring, and batteries all start as ores that must be reduced.
Biology runs on representative metals: Na⁺ and K⁺ drive nerve impulses, Ca²⁺ builds bone and triggers muscle contraction, and Mg²⁺ sits at the center of chlorophyll.
Extraction is expensive and environmental: aluminum production consumes a large share of industrial electricity precisely because Al–O bonds are so strong; recycling aluminum saves roughly 95% of that energy.
Exam value: pairing a metal with its extraction method (electrolysis vs. carbon reduction vs. native) tests whether you understand the reactivity series instead of memorizing lists.

The college version

Core Concepts

Where the representative metals are found

Earth's crust is about 46.6% oxygen and 27.7% silicon by mass. The most abundant metals follow: aluminum (~8.1%), iron (~5.0%), calcium (~3.6%), sodium (~2.8%), potassium (~2.6%), and magnesium (~2.1%). Seawater adds a vast reservoir of sodium and magnesium chlorides. Ore minerals belong to a few families:

Oxides: bauxite (extAl₂ extO₃ · x extH₂ extO) → aluminum; hematite (extFe₂ extO₃) → iron; cassiterite (extSnO₂) → tin.
Sulfides: sphalerite (ZnS) → zinc; galena (PbS) → lead; these are roasted to oxides first.
Carbonates and halides: limestone (CaCO₃), magnesite (MgCO₃), halite (NaCl).
Silicates: the most common rock-forming minerals, but too stable and complex to be economical ore sources.

Only the least reactive metals — gold, silver, platinum, occasionally copper and mercury — occur native (uncombined): their oxides are so easily reduced that weak natural agents kept them metallic, and physical methods (panning, sluicing) concentrate the grains.

The matching game: reactivity decides the method

A metal is obtained by reducing its compounds, and the reducing agent must be stronger than the metal itself. Carbon, CO, hydrogen, a more reactive metal, or an electric current can serve — if strong enough.

Electrolysis — the most reactive metals (Na, K, Ca, Mg, Al). No chemical reducing agent works, so an electric current supplies electrons directly.

Sodium (Downs cell): molten NaCl, with CaCl₂ added to lower the melting point from ~801 °C to ~600 °C:

$2 NaCl (l) ⟶ 2 Na (l) + Cl 2 (g)$

Magnesium (Dow process): Mg(OH)₂ precipitated from seawater is converted to extMgCl₂, dried, and electrolyzed molten. The Pidgeon process instead reduces MgO with ferrosilicon under vacuum.
Aluminum (Hall–Héroult process): purified extAl₂ extO₃ is dissolved in molten cryolite (extNa₃ extAlF₆) at 950–1000 °C. At the cathode, extAl³⁺ + 3e^- ⟶ Al; at the carbon anodes, oxide ions are oxidized to O₂, which slowly burns the anodes away.

Chemical reduction — moderately reactive metals (Fe, Zn, Sn, Pb). Sulfide ores are first roasted to oxides, then the oxide is heated with carbon or CO:

$Fe 2 O 3 (s) + 3 CO (g) ⟶ 2 Fe (l) + 3 CO 2 (g) (blast furnace)$

$SnO 2 (s) + C (s) ⟶ Sn (l) + CO 2 (g)$

Zinc is roasted (ZnS → ZnO, releasing SO₂ that is captured for sulfuric acid), then reduced: ZnO(s) + C(s) ⟶ Zn(g) + CO(g).

Found native — noble metals (Au, Ag, Pt). No reduction needed; the metal is separated from rock by density (panning) or, industrially, dissolved with cyanide and re-precipitated. Cyanide leaching is strictly industrial — cyanide salts are deadly and never used in student labs.

Why carbon works for iron but not aluminum

The more negative the free-energy change for forming a metal oxide, the more stable the oxide and the harder it is to reduce. At blast-furnace temperatures, CO binds oxygen strongly enough to win it from iron; aluminum oxide is so stable that carbon cannot compete at any practical temperature. Only electrolysis — not limited by chemical reducing strength — can pull aluminum out. The activity series summarizes this: metals above carbon (Li through Al) need electrolysis; metals below carbon (Zn through Cu) can be reduced by carbon.

Examples: stoichiometry of extraction

Example 1 — Aluminum from a ton of alumina

The cell reaction is 2Al₂O₃ ⟶ 4Al + 3O₂, so 2 mol of Al form per mole of Al₂O₃. How much aluminum comes from 1.00 × 10³ kg of pure extAl₂ extO₃? Molar mass:

$M(Al 2 O 3) = 2(26.98) + 3(16.00) = 101.96 g/mol$

$n(Al 2 O 3) = 1.00 \times 106 g 101.96 g/mol = 9.81 \times 103 mol$

Convert with the mole ratio, then to mass:

$m(Al) = 9.81 \times 103 mol Al 2 O 3 \times 2 mol Al 1 mol Al 2 O 3 \times 26.98 g 1 mol Al = 5.29 \times 105 g \approx 529 kg$

One ton of alumina yields about 529 kg of aluminum — the rest leaves as oxygen at the anodes.

Example 2 — Iron from impure hematite

A furnace charges 2.00 × 10³ kg of ore that is 85.0% extFe₂ extO₃ by mass. Using extFe₂ extO₃ + 3CO ⟶ 2Fe + 3CO₂, find the mass of iron. First isolate the hematite:

$m(Fe 2 O 3) = 0.850 \times 2.00 \times 106 g = 1.70 \times 106 g$

With M(Fe₂O₃) = 2(55.85) + 3(16.00) = 159.70 g/mol:

$m(Fe) = 1.70 \times 106 g 159.70 g/mol \times 2 mol Fe 1 mol Fe 2 O 3 \times 55.85 g 1 mol Fe = 1.19 \times 106 g \approx 1.19 \times 103 kg$

The 15% never reacts and ends up in the slag.

Example 3 — Sodium by the Downs cell

A Downs cell runs at 5.00 × 10⁴ A for 2.00 h. Each Na⁺ needs one electron: Na⁺ + e^- ⟶ Na. Find the charge first:

$q = I \times t = (5.00 \times 104 A)(7200 s) = 3.60 \times 108 C$

Convert to moles of electrons with the Faraday constant F = 96,485 C/mol e^-:

$m(Na) = 3.60 \times 108 C 96, 485 C/mol e - \times 1 mol Na 1 mol e - \times 22.99 g 1 mol Na = 8.58 \times 104 g \approx 85.8 kg$

Two hours at industrial current deliver roughly 86 kg of sodium, plus a stoichiometric amount of Cl₂ at the anode.

Common Confusions

Do Not Confuse	With	Difference
Native metal	Refined metal	Native = found free in nature (Au, Ag, Pt); refined = extracted and purified by humans.
Roasting	Smelting	Roasting converts sulfide ores to oxides in air; smelting reduces oxides with C/CO to metal.
Electrolysis	Smelting	Electrolysis uses electric current (very reactive metals); smelting uses chemical reducing agents (moderately reactive metals).
Abundance in crust	Ease of extraction	Aluminum is the most abundant metal yet among the hardest to obtain — abundance says nothing about reactivity.
Metal at the anode	Metal at the cathode	Reduction (metal deposition) always occurs at the cathode; oxidation (Cl₂, O₂ evolution) at the anode.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Most metals hide inside rocks, locked up with oxygen, because they are "friendly" — they react easily. To free them you must win a tug-of-war for the oxygen. For super-friendly metals like aluminum you need a giant electric plug. For less friendly metals like iron, heating with charcoal is enough. A few metals like gold never made friends with oxygen, so you can just pick them out of the dirt.

Key takeaways

Most representative metals occur combined, not free; only Au, Ag, Pt (and sometimes Cu, Hg) occur native.
Reactivity controls the method: electrolysis for Na, K, Ca, Mg, Al; carbon/CO reduction for Fe, Zn, Sn, Pb; physical separation for noble metals.
Name the processes: Downs cell (molten NaCl + CaCl₂) → Na; Hall–Héroult (Al₂O₃ in molten cryolite) → Al; blast furnace (Fe₂O₃ + 3CO → 2Fe + 3CO₂) → Fe.
Sulfide ores are roasted to oxides first, releasing SO₂ that must be captured.
Aluminum is the most abundant crustal metal (~8%) yet was isolated only in the 1800s: abundant ≠ easy to obtain.
Metal forms at the cathode (reduction); Cl₂ or O₂ forms at the anode (oxidation).

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Why are almost no representative metals found free in nature?
Show answer
Because they are reactive: over geological time they combined with oxygen, sulfur, carbon dioxide, and water. Only very unreactive metals (Au, Ag, Pt) survive uncombined.
Match each metal to its preparation: (a) Na, (b) Fe, (c) Au, (d) Al.
Show answer
(a) Na — electrolysis of molten NaCl (Downs cell); (b) Fe — reduction of Fe₂O₃ with CO in the blast furnace; (c) Au — found native; (d) Al — electrolysis of Al₂O₃ in molten cryolite (Hall–Héroult).
In the Downs cell, why is CaCl₂ added to the molten NaCl?
Show answer
CaCl₂ lowers the melting point from ~801 °C to ~600 °C, saving energy and limiting sodium vaporization losses.
How many kilograms of aluminum can be produced from 2.00 × 10³ kg of pure Al₂O₃? (M: Al₂O₃ = 101.96, Al = 26.98 g/mol.)
Show answer
Moles Al₂O₃ = 2.00×10⁶ g ÷ 101.96 g/mol = 1.96×10⁴ mol; moles Al = 2 × 1.96×10⁴ = 3.92×10⁴ mol; mass = 3.92×10⁴ × 26.98 g/mol = 1.06×10⁶ g ≈ 1.06×10³ kg.
A current of 1.93 × 10⁵ A runs for exactly 1.00 h in a Downs cell. How many moles of electrons pass? (F = 96,485 C/mol.)
Show answer
q = (1.93×10⁵ A)(3600 s) = 6.95×10⁸ C; n(e⁻) = 6.95×10⁸ C ÷ 96,485 C/mol = 7.20×10³ mol.
Why must sulfide ores like ZnS be roasted before carbon reduction?
Show answer
Carbon cannot easily reduce sulfide minerals directly; roasting converts ZnS to ZnO so the oxide can be reduced by carbon (and the SO₂ is captured).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Ore: A mineral deposit from which a metal can be extracted at a profit.
Native metal: A metal found uncombined in nature (Au, Ag, Pt).
Metallurgy: The science of extracting and refining metals from ores.
Roasting: Heating a sulfide ore in air to convert it to an oxide.
Smelting: Heating an ore with a reducing agent (C or CO) to get the metal.
Electrolysis: Using electric current to force a nonspontaneous reduction.
Gangue: Worthless rock and mineral impurities in an ore.

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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