Calculating Metal from Ore Mass

Introduction

An ore mass does not directly state a metal mass. A calculation may need three separate filters: what fraction of ore is valuable mineral, what fraction of that mineral is the target element, and what fraction of contained metal is recovered. Writing these stages in order makes the units and assumptions visible.

Core explanation

Consider a 1000 kg ore that is 50.0% hematite, Fe₂O₃, by dry mass. The feed contains 500 kg hematite. Using Fe ≈ 55.85 and O ≈ 16.00 g mol⁻¹, Fe₂O₃ formula mass is 159.7 g mol⁻¹ and iron contributes 111.7 g. Iron fraction of pure hematite is 111.7/159.7 ≈ 0.699. The ore therefore contains approximately 500 × 0.699 = 350 kg iron. This is a theoretical contained amount, not the mass of pig iron or steel product.

If extraction recovers 80.0% of the iron, recovered elemental iron content is about 350 × 0.800 = 280 kg. A crude metal stream may weigh more than 280 kg if it contains carbon and impurities. It may weigh less if only a portion of recovered metal is being reported at a particular stage. Be precise about whether the requested answer is iron content, crude product mass or a specified pure metal mass.

Some questions supply elemental grade directly. If the same 1000 kg ore is stated to assay 35.0% Fe by mass, contained iron is 350 kg immediately. Do not multiply by the hematite iron fraction again; the oxygen has already been accounted for in the elemental assay. The word “grade” without a basis can be ambiguous, so read the exact label or ask for clarification when necessary.

Other mineral formulas require other fractions. Pure CuO has copper fraction about 63.55/(63.55+16.00) = 0.799. Pure ZnS has zinc fraction about 65.38/(65.38+32.06) = 0.671. Even if two ores each contain 40.0% of their named mineral, their elemental metal grades differ. Molar masses translate formula composition into mass fractions before recovery is considered.

The mole pathway provides an equivalent method: ore mass → pure mineral mass → mineral moles → metal atom moles using formula subscripts → metal mass. The fraction pathway is algebraically shorter, but the mole pathway is valuable when the next question asks for reagent demand or gas output. Both should agree if formulas and units are correct.

Recovery may cover several steps. If separate concentration and reduction recoveries are given, multiply them only when both refer to the metal entering each respective step. A statement of “overall 80% recovery” already includes stage losses and should be applied once. Avoid treating product purity as another recovery unless its definition says so.

An ore may contain more than one mineral with the same target metal. Then calculate contained metal from each mineral and add, or use a direct elemental assay. If a mineral contains multiple useful metals, calculate each element's fraction separately. The result depends on accurate mineralogical or assay data, not just total rock mass.

As a check, recovered pure metal cannot exceed contained metal from the stated feed unless another metal source enters. A result above that bound usually indicates a percentage denominator, molar mass or unit conversion error. Keep kilograms, grams and tonnes consistent through the chain.

Step-by-step reasoning

1. Determine whether grade is mineral fraction or elemental metal fraction. 2. If mineral-based, multiply ore mass by mineral fraction. 3. Calculate metal fraction from the mineral formula and atomic masses. 4. Multiply to get contained metal, then apply correctly based recovery. 5. Check requested product identity and compare recovered metal with contained maximum.

Visual explanation

Draw a bar for 1000 kg ore. Shade 500 kg as Fe₂O₃, then shade 69.9% of that segment as iron atoms. A final arrow sends 80.0% of the iron segment to recovered iron. Label the three masses 1000, 350 and 280 kg.

Real-world analogy

A box may be 50% wrapped candies, and each wrapped candy may be 70% edible candy by mass. If sorting recovers 80% of edible content, all three percentages are needed. Applying 50% alone counts wrappers as food; applying 80% before identifying food obscures the basis.

Real-world example

An iron-ore planner compares a reported hematite percentage with an elemental iron assay. They should imply compatible theoretical iron contents if they describe the same representative sample. The expected furnace iron output is lower after practical recovery and may carry carbon in crude product.

Why?

Why does hematite contain less iron mass than its total mass? Oxygen atoms contribute 48.0 g to every approximately 159.7 g formula amount of Fe₂O₃. Reduction moves oxygen into other products; it does not turn oxygen mass into extra iron.

Common misconception

“A 50% hematite ore contains 50% iron.” Hematite is a compound. Iron fraction in ore is approximately 0.50 × 0.699 = 34.9% if hematite is the only iron source.

Worked example

A 250 kg feed is 60.0% pure CuO, with the rest inert. CuO mass is 150 kg. Copper fraction is 63.55/79.55 ≈ 0.799, so contained copper is about 120 kg. At 75.0% overall copper recovery, recovered elemental copper is about 90.0 kg. Reporting 150 kg copper would count oxygen as copper; reporting 112.5 kg would apply recovery to CuO mass without first removing oxygen's fraction.

Quick check

1. How much hematite is present in 200 kg ore at 40.0% Fe₂O₃ grade? Answer: 80.0 kg Fe₂O₃ before formula-fraction calculation.

Exam focus

Write a labeled mass pathway and distinguish elemental assay from mineral grade. Use formula masses before recovery when the grade names a compound. State whether final mass means pure element or an impure metal stream.

Advanced insight

Representative sampling is essential because ore grade varies spatially. An analytically precise assay of an unrepresentative sample can still predict plant output poorly. Ore-to-metal calculations are therefore sensitive to geological sampling as well as chemical arithmetic.

Summary

Metal from ore mass is calculated by applying the mineral fraction, the element's formula mass fraction and process recovery on their proper bases. Direct elemental grade skips the formula step. Recovered pure metal cannot exceed contained metal without another input.

Practice questions

1. What iron fraction is in pure Fe₂O₃ using 111.7/159.7? Answer: About 0.699, or 69.9%. 2. How much iron is contained in 100 kg pure Fe₂O₃? Answer: About 69.9 kg. 3. What is recovered at 90.0% iron recovery? Answer: About 62.9 kg elemental iron content. 4. When is the formula fraction unnecessary? Answer: When grade is already reported as elemental metal fraction by mass.