Occurrence of metals

Reactivity decides native versus combined. The activity series ranks metals from most reactive (potassium, sodium, calcium, aluminium) at the top to least reactive (copper, mercury, silver, gold, platinum) at the bottom. A metal high on the series reacts readily with oxygen, moisture, sulphur, or carbon dioxide in its surroundings, and over geological time it settles into whichever compound is most stable — that is why sodium, aluminium, iron, and zinc are always found combined. A metal at the very bottom barely reacts with anything it meets underground, so any of it present at the start remains as the free metal today. Gold, silver and platinum occur chiefly native; copper and mercury occur native to a smaller extent alongside their combined forms.

Combined metals fall into five families of compound, and the syllabus expects you to sort a given ore into the right one:

• Oxides — haematite, Fe₂O₃ (iron); bauxite, Al₂O₃·2H₂O (aluminium)
• Carbonates — limestone, CaCO₃; calamine, ZnCO₃ (zinc)
• Sulphides — zinc blende, ZnS (zinc); galena, PbS (lead)
• Chlorides — rock salt, NaCl (sodium)
• Sulphates — gypsum, CaSO₄·2H₂O (calcium)

Mineral versus ore is an economic distinction layered on top of a chemical one. Any naturally occurring compound of a metal is a mineral. A mineral becomes an ore only when the metal is present in a high enough proportion, by mass, that extracting it is currently profitable — a rock with a mere trace of copper is a mineral of copper but not an ore of copper. That proportion is the ore's grade, and it is nothing more than a percentage-composition calculation you already know from mole concept: how much of the compound's mass is the metal atom itself.

where: $n$ is the number of metal atoms in one formula unit of the compound, $M_{\text{metal}}$ is the atomic mass of the metal, and $M_{\text{compound}}$ is the molar (formula) mass of the whole compound. A higher percentage means a richer, more attractive ore; a lower percentage means more rock must be processed to recover the same mass of metal, which is exactly why economics — not just chemistry — decides whether a mineral earns the label "ore."

The goal is to explain, from the idea of chemical stability, why the metals near the top of the activity series are never found alone while those near the bottom almost always are — this is the reasoning your written answer should reproduce, not just the memorised rule.

Step 1. A free metal atom sitting in moist, oxygen-rich rock is not chemically at rest — it is a high-energy state compared with its compound. Whether it stays free over geological time depends on how strongly it is pulled toward reacting with what surrounds it.

Step 2. Metals high on the activity series, such as sodium and aluminium, have a strong pull toward forming compounds: their reaction with oxygen, water, or carbon dioxide releases a large amount of energy, so the compound (oxide, chloride, carbonate) is far more stable than the free metal. Given millions of years and constant exposure to air and water, essentially all the free metal converts to its compound, which is why sodium is never dug up as shiny metal — only as rock salt.

Step 3. Metals at the very bottom, such as gold and platinum, have almost no pull toward reacting with oxygen, water or the other substances present in soil and rock. Their compounds, where they exist at all, are not meaningfully more stable than the free metal, so there is no driving force pushing them into a combined state. Whatever free gold formed when the earth's crust cooled has simply stayed free ever since.

Step 4. Copper and mercury sit in between — reactive enough that most of their supply is combined (as sulphides such as copper pyrites, CuFeS₂), but unreactive enough that a measurable amount also survives as native metal. This middle position is why they are the two commonly cited exceptions to "combined metals are always combined."

Question: Haematite, Fe₂O₃, is the chief ore of iron. (a) Calculate the percentage of iron in pure haematite (atomic masses: Fe = 56, O = 16). (b) A sample of haematite rock has only 40% of its mass as pure Fe₂O₃, the rest being unwanted rock (gangue). What mass of iron can theoretically be obtained from 500 kg of this rock?

Step 1. Find the molar mass of Fe₂O₃: $M = 2(56) + 3(16) = 112 + 48 = 160$.

Step 2. Apply the percentage-metal formula from equation (1): $\%\,\text{Fe} = \dfrac{2 \times 56}{160} \times 100 = \dfrac{112}{160} \times 100 = 70\%$. So pure haematite is 70% iron by mass — a genuinely rich ore, which is exactly why it is the industrially preferred source of iron.

Step 3. Find the mass of pure Fe₂O₃ in the 500 kg rock sample: $40\%$ of $500\ \text{kg} = 0.40 \times 500 = 200\ \text{kg}$.

Step 4. Apply the 70% figure from Step 2 to this 200 kg of Fe₂O₃: mass of iron $= 0.70 \times 200 = 140\ \text{kg}$. The rock is 70% iron in its pure oxide, and 500 kg of the 40%-grade ore yields 140 kg of iron.

Competitive stretch

Using the same method, bauxite (Al₂O₃·2H₂O, atomic masses Al = 27, O = 16, H = 1) has $M = 2(27) + 3(16) + 2(18) = 54 + 48 + 36 = 138$, giving $\%\,\text{Al} = \dfrac{54}{138} \times 100 \approx 39.1\%$ — noticeably lower than haematite's 70% iron, which is one reason aluminium extraction (studied later in this unit) needs a more elaborate process than iron extraction.

• Treating "mineral" and "ore" as synonyms. Every ore is a mineral, but a mineral is called an ore only when its percentage metal content is high enough for profitable extraction — a low-grade copper trace is a mineral, not an ore.
• Assuming all metals occur combined. Gold, silver and platinum occur chiefly native; copper and mercury occur natively to a smaller degree too.
• Swapping the three prescribed ores. Bauxite is aluminium's ore, haematite is iron's, zinc blende is zinc's — mixing these up is the single most common slip.
• Confusing abundance with reactivity as the deciding factor. Aluminium is the most abundant metal in the earth's crust yet is never found native, because abundance has nothing to do with it — reactivity does.

Ask the tutor: want to work through the percentage-metal calculation for another ore, such as galena or calamine, or get quizzed on matching ores to their metals?