How to Calculate the Percentage of Metal in an Ore

An ore sample weighs 2.000 g. After adding a chemical reagent, you collect 0.510 g of a rusty-brown precipitate. Does that mean the ore contains 25.5% iron?

Not necessarily. The precipitate is not pure iron. It contains iron combined with other elements, and the wet solid also contains water and potentially dissolved impurities.

To calculate the percentage of metal in an ore, you need to work backwards from the mass of a known, pure compound containing that metal. Use its chemical formula and mole ratio to calculate the mass of metal, account for any dilution or aliquot, and divide by the original ore sample mass.

This is the key calculation:

\[ \%\text{metal}= \frac{\text{mass of metal in original ore sample}} {\text{mass of original ore sample}} \times100 \]

The challenging part is obtaining a precipitate whose measured mass genuinely represents a known amount of metal.

We’ll follow an iron ore analysis from the original sample through dissolution, precipitation, filtration, and heating. The same reasoning applies to other metals, provided you use the correct reaction and stoichiometric relationship.

01Start with a representative ore sample

Before doing any calculations, consider where the ore sample came from.

An ore is not necessarily a uniform mixture. One piece might contain mostly iron-rich minerals, while another contains much more rock and silicate material.

If you take a small handful from one part of a large ore deposit, your sample might not represent the whole material.

This introduces an important distinction:

• Sampling determines whether the material you analyse represents the bulk ore.
• Chemical analysis determines how much metal is present in the sample you actually collected.

A very precise chemical analysis cannot correct a badly chosen sample.

A suitable preparation process involves:

1. Collecting portions from different locations within the bulk ore.
2. Crushing and grinding the material to reduce particle size.
3. Mixing the resulting powder thoroughly.
4. Using a suitable splitting method to obtain a representative subsample.
5. Drying the sample if a dry-basis percentage is required.
6. Accurately measuring the mass of the subsample.

Suppose the representative dry sample has a mass of 2.500 g.

This is the denominator in your final percentage calculation. Do not replace it with the mass of the dissolved solution or the precipitate.

02Following the metal through precipitation

Iron in the ore cannot simply be placed on a balance and weighed separately from the other substances.

Instead, chemists use a chemical reaction to convert dissolved iron ions into an insoluble solid.

The substance being measured is called the analyte. In this example, the analyte is iron.

Step 1: Dissolve the ore

The representative ore sample is treated with a suitable acid to dissolve the iron-containing minerals.

The aim is to transfer all the iron into solution as iron ions.

Any insoluble gangue can be separated by filtration, provided the chosen digestion method has dissolved all the iron.

If necessary, iron(II) ions are quantitatively oxidised to iron(III) ions before precipitation.

The resulting solution is then transferred quantitatively into a volumetric flask and diluted to a known volume.

For example:

• Original ore sample: 2.500 g
• Final digest volume: 250.0 mL
• Iron present in the digest: assumed to be entirely dissolved as \(\mathrm{Fe^{3+}}\)

The known solution volume will become important if you analyse only a portion of the digest.

Step 2: Precipitate the iron ions

A suitable base is added to the iron-containing solution.

Iron(III) ions react with hydroxide ions to form iron(III) hydroxide, a brown solid.

The balanced ionic equation is:

\[ \ce{Fe^3+(aq) + 3OH-(aq) -> Fe(OH)3(s)} \]

The solid formed is called a precipitate.

What should happen if the precipitation is complete?

Ideally, all the dissolved iron ions should have been converted into solid iron(III) hydroxide. The remaining solution should contain negligible iron.

A useful experimental check is to add more precipitating reagent to a portion of the filtrate. If additional brown precipitate forms, some iron remained in solution after the original precipitation.

A solution that looks clear does not necessarily contain zero dissolved iron.

Step 3: Filter and wash the precipitate

The mixture is passed through filter paper in a funnel.

The solid iron(III) hydroxide remains on the filter paper, while the remaining solution passes through into a receiving flask.

The collected solid must then be washed with distilled water.

Why?

The precipitate can trap small amounts of the original solution. That solution may contain dissolved salts and excess reagent.

If these remain attached to the solid, they can contribute to the measured mass.

Washing removes these soluble impurities.

Step 4: Convert the precipitate into a known compound

Here is an important detail.

You should not simply weigh the wet iron(III) hydroxide and assume that its mass represents iron.

The precipitate contains water, and freshly formed iron(III) hydroxide can also have variable amounts of associated water.

Instead, the collected solid is heated strongly to convert it into iron(III) oxide.

The reaction is:

\[ \ce{2Fe(OH)3(s) -> Fe2O3(s) + 3H2O(g)} \]

The iron remains in the solid, but water is released.

The final product is iron(III) oxide, \(\ce{Fe2O3}\), which has a known chemical formula.

The solid is heated, cooled, and weighed repeatedly until a constant mass is obtained.

This means successive heating, cooling, and weighing cycles produce no further significant change in mass.

The measurement is now much more useful.

You have a known mass of a compound with a known proportion of iron.

That allows you to calculate the original iron content.

03Wet precipitate versus pure dry mass

This is one of the most important distinctions in ore analysis.

Imagine weighing a wet towel. The balance measures the towel and the water together.

A wet precipitate creates the same problem: the balance cannot distinguish the chemical solid from the water it contains.

The analogy has a limit. In chemical analysis, some water may also be associated with the precipitate itself, and heating can change the chemical composition of the solid.

Consider these two measurements:

The difference can be substantial.

For example, suppose a wet precipitate has a mass of 0.510 g, but after washing, ignition, and cooling, the final iron(III) oxide has a mass of only 0.164 g.

These are not interchangeable measurements.

Using 0.510 g as though it were pure \(\ce{Fe2O3}\) would produce a seriously inflated estimate of the iron content.

There are several possible sources of error:

Notice that constant mass does not automatically mean pure precipitate.

A contaminant that remains after heating can still contribute to the final mass. Chemical selectivity must also be considered.

04The calculation: converting iron oxide mass into iron mass

Now we can connect the experimental measurement to the percentage of iron in the original ore.

The key is to remember that the balance measures the entire \(\ce{Fe2O3}\) compound, not just its iron atoms.

Calculate the molar mass of iron(III) oxide

Using:

• \(A_r(\mathrm{Fe})=55.85\)
• \(A_r(\mathrm{O})=16.00\)

The molar mass is:

\[ \begin{aligned} M(\ce{Fe2O3}) &=2(55.85)+3(16.00)\\ &=159.70\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The iron contribution to that molar mass is:

\[ 2(55.85)=111.70\ \mathrm{g\,mol^{-1}} \]

Therefore, the fraction of the oxide’s mass that comes from iron is:

\[ \frac{111.70}{159.70}=0.6994 \]

This means that 69.94% of pure iron(III) oxide’s mass is iron.

So:

\[ m(\mathrm{Fe})= m(\ce{Fe2O3})\times0.6994 \]

This conversion factor is extremely useful in calculations involving iron precipitated and weighed as iron(III) oxide.

What if only part of the solution was analysed?

You must also account for the fraction of the original digest that was used.

An aliquot is a measured portion of a larger solution.

Suppose you dissolve the original ore sample and make the solution up to 250.0 mL. You then use a pipette to transfer 25.00 mL into a beaker for precipitation.

Only one-tenth of the original solution has been analysed.

Therefore:

\[ \text{Aliquot factor} = \frac{250.0}{25.00}=10.00 \]

The amount of iron in the complete digest is ten times the amount found in the aliquot.

Combining the two conversions gives:

\[ \boxed{ m(\mathrm{Fe})= m(\ce{Fe2O3}) \times0.6994 \times \frac{V_{\mathrm{digest}}}{V_{\mathrm{aliquot}}} } \]

Here:

• \(m(\mathrm{Fe})\) is the total mass of iron in the original ore sample, in grams.
• \(m(\ce{Fe2O3})\) is the measured mass of pure iron(III) oxide from the aliquot, in grams.
• \(V_{\mathrm{digest}}\) is the total volume of the prepared solution.
• \(V_{\mathrm{aliquot}}\) is the volume analysed.

The two volumes must use the same units.

Finally:

\[ \%\mathrm{Fe}= \frac{m(\mathrm{Fe})}{m(\mathrm{ore})}\times100 \]

This is the complete calculation pathway.

05A quick method for exam questions

When you see a question asking you to calculate the percentage of metal in an ore from precipitation data, follow this sequence.

A useful way to check your work is to ask:

• Have I used the mass of the final compound rather than the wet precipitate?
• Have I accounted for the atoms other than the metal in that compound?
• Have I used the correct mole ratio?
• Have I accounted for the aliquot?
• Have I divided by the original ore sample mass?

One further complication deserves attention: a precipitate can have a constant mass and still contain more than one compound. This matters when other metal ions react with the same reagent.

06Questions and solutions

The following questions build from a straightforward percentage calculation towards experimental evaluation and the problem of competing precipitates.

Use \(A_r(\mathrm{Fe})=55.85\), \(A_r(\mathrm{O})=16.00\), and assume ideal quantitative recovery unless the question states otherwise.

Question 1

A 1.000 g sample of dry iron ore is analysed. All the iron is converted into pure iron(III) oxide, producing 0.250 g of \(\ce{Fe2O3}\).

Calculate the percentage by mass of iron in the original ore.

Solution 1

The ore contains 17.5% iron by mass.

The important point is that the measured 0.250 g represents iron(III) oxide, not elemental iron.

First, calculate the molar mass of iron(III) oxide:

\[ M(\ce{Fe2O3})=159.70\ \mathrm{g\,mol^{-1}} \]

The mass fraction of iron is:

\[ \frac{2(55.85)}{159.70}=0.6994 \]

Therefore, the mass of iron is:

\[ m(\mathrm{Fe})=0.250\times0.6994 \]

\[ =0.1749\ \mathrm{g} \]

Now calculate the percentage:

\[ \%\mathrm{Fe}= \frac{0.1749}{1.000}\times100 \]

\[ =17.5\% \]

The answer is 17.5% iron by mass.

A tempting mistake is to calculate \(0.250/1.000\times100\) and report 25.0%.

That would give the percentage of iron(III) oxide, not the percentage of iron. The oxygen must be excluded from the metal mass.

Question 2

A 2.000 g sample of iron ore is completely dissolved and diluted to 250.0 mL.

A 25.00 mL aliquot is transferred into a beaker. Iron is precipitated, filtered, washed, and ignited to produce 0.145 g of pure \(\ce{Fe2O3}\).

Calculate the percentage by mass of iron in the original ore.

Solution 2

The original ore contains 50.7% iron by mass.

There are two separate conversions to make: from iron oxide to iron, and from the aliquot to the complete digest.

First, calculate the iron mass fraction in \(\ce{Fe2O3}\):

\[ \frac{111.70}{159.70}=0.6994 \]

The iron mass in the analysed aliquot is:

\[ m(\mathrm{Fe})= 0.145\times0.6994 \]

\[ =0.1014\ \mathrm{g} \]

However, this is only the iron in 25.00 mL.

The original solution volume was 250.0 mL, so:

\[ \text{Aliquot factor} = \frac{250.0}{25.00}=10.00 \]

The total iron mass is:

\[ m(\mathrm{Fe,\ total}) = 0.1014\times10.00 \]

\[ =1.014\ \mathrm{g} \]

Finally:

\[ \%\mathrm{Fe}= \frac{1.014}{2.000}\times100 \]

\[ =50.7\% \]

The ore contains 50.7% iron by mass.

A common mistake is to forget the aliquot factor and calculate the percentage using only 0.1014 g of iron.

That would give a result ten times smaller than the correct answer because only one-tenth of the original digest was analysed.

Question 3

A 1.500 g sample of dry iron ore is dissolved and diluted to 100.0 mL.

A 20.00 mL aliquot is analysed. After precipitation and filtration, the washed wet filter cake has a mass of 0.420 g.

After ignition and cooling to constant mass, the pure \(\ce{Fe2O3}\) residue has a mass of 0.0960 g.

A student uses the wet filter cake mass in their calculation, treating the entire 0.420 g as iron(III) oxide.

(a) Calculate the correct percentage of iron in the ore.

(b) Calculate the percentage the student would obtain using the wet mass incorrectly.

(c) Explain why the student’s result is unreliable, even if it is below 100%.

Solution 3

(a) The correct iron percentage is 22.4%.

The correct measurement is the final ignited iron(III) oxide mass of 0.0960 g.

First, convert the oxide mass into iron mass in the aliquot:

\[ m(\mathrm{Fe})= 0.0960\times\frac{111.70}{159.70} \]

\[ =0.06715\ \mathrm{g} \]

The aliquot factor is:

\[ \frac{100.0}{20.00}=5.00 \]

Therefore, the total iron mass in the original ore sample is:

\[ m(\mathrm{Fe,\ total}) = 0.06715\times5.00 \]

\[ =0.3357\ \mathrm{g} \]

Calculate the percentage:

\[ \%\mathrm{Fe}= \frac{0.3357}{1.500}\times100 \]

\[ =22.4\% \]

(b) Using the wet mass incorrectly gives 97.9%.

If the student treats 0.420 g as pure iron(III) oxide:

\[ m(\mathrm{Fe})= 0.420\times0.6994 \]

\[ =0.2938\ \mathrm{g} \]

This is the iron mass in the aliquot according to the incorrect assumption.

Scaling to the complete digest:

\[ m(\mathrm{Fe,\ total}) = 0.2938\times5.00 \]

\[ =1.469\ \mathrm{g} \]

The resulting percentage is:

\[ \%\mathrm{Fe}= \frac{1.469}{1.500}\times100 \]

\[ =97.9\% \]

(c) The result is unreliable because the wet cake contains water and is not pure \(\ce{Fe2O3}\).

The balance measures all material present, not just the iron-containing compound.

The student has incorrectly assumed that the entire wet mass has the chemical composition of iron(III) oxide.

Notice something interesting: 97.9% is below 100%, so it might initially look plausible. However, being below 100% does not establish that a result is chemically valid.

The correct method requires the known composition of the final compound, not simply a mass that looks reasonable.

Question 4

A 1.200 g sample of iron ore is dissolved and diluted to 200.0 mL.

A 20.00 mL aliquot is treated with a suitable precipitating reagent. The first precipitate is collected, washed, ignited, and found to have a mass of 0.105 g of pure \(\ce{Fe2O3}\).

The filtrate from this same aliquot is then treated with fresh precipitating reagent. A second crop of precipitate forms. After filtration, washing, and ignition, it produces another 0.0070 g of pure \(\ce{Fe2O3}\).

(a) Calculate the percentage of iron in the original ore.

(b) Calculate the percentage that would have been obtained if only the first precipitate had been measured.

(c) Explain what the second precipitate tells you about the original experiment.

A student argues that the first precipitation must have been complete because the filtrate looked clear. Evaluate this claim.

Solution 4

(a) The correct iron percentage is 65.3%.

Both precipitates came from the same original aliquot.

Therefore, both masses must be included when calculating the total iron present.

The combined iron(III) oxide mass is:

\[ m(\ce{Fe2O3})= 0.105+0.0070 \]

\[ =0.1120\ \mathrm{g} \]

Convert this into the mass of iron in the aliquot:

\[ m(\mathrm{Fe})= 0.1120\times\frac{111.70}{159.70} \]

\[ =0.07834\ \mathrm{g} \]

The aliquot factor is:

\[ \frac{200.0}{20.00}=10.00 \]

Therefore:

\[ m(\mathrm{Fe,\ total})= 0.07834\times10.00 \]

\[ =0.7834\ \mathrm{g} \]

Calculate the percentage:

\[ \%\mathrm{Fe}= \frac{0.7834}{1.200}\times100 \]

\[ =65.3\% \]

(b) Using only the first precipitate gives 61.2%.

If the second crop is ignored:

\[ m(\mathrm{Fe})= 0.105\times0.6994 \]

\[ =0.07344\ \mathrm{g} \]

This is the iron mass in the aliquot.

Scaling up:

\[ m(\mathrm{Fe,\ total})= 0.07344\times10.00 \]

\[ =0.7344\ \mathrm{g} \]

Therefore:

\[ \%\mathrm{Fe}= \frac{0.7344}{1.200}\times100 \]

\[ =61.2\% \]

The original calculation underestimates the iron content by approximately 4.1 percentage points.

(c) The second precipitate demonstrates that the first precipitation was incomplete.

Some dissolved iron remained in the filtrate after the first treatment.

Adding fresh reagent converted more of those remaining iron ions into an insoluble compound.

The student’s argument is flawed because a clear-looking filtrate does not prove that every iron ion has precipitated. Small concentrations of dissolved ions can remain even when no visible particles are present.

The second crop provides direct evidence that the first precipitation did not recover all the iron.

This is why checking the filtrate can be an important part of evaluating a gravimetric method.

Question 5

An iron ore sample has a mass of 2.500 g. It is dissolved and quantitatively transferred into a 250.0 mL volumetric flask.

A 25.00 mL aliquot is treated with excess hydroxide ions.

The ore contains both iron(III) ions and aluminium(III) ions. Both form insoluble hydroxides under the conditions used. After filtration and ignition, the combined solid contains \(\ce{Fe2O3}\) and \(\ce{Al2O3}\).

The total mass of the ignited mixed oxide residue is 0.220 g.

A separate, validated selective analysis of a matched 25.00 mL aliquot establishes that the aluminium would produce 0.060 g of \(\ce{Al2O3}\).

A second experiment uses a 50.00 mL aliquot from the same original digest. It produces 0.440 g of total ignited oxide.

(a) Calculate the percentage by mass of iron in the original ore.

(b) Explain why treating the entire 0.220 g residue as pure \(\ce{Fe2O3}\) would give an incorrect answer.

(c) A student argues that the second experiment proves the first precipitate was pure iron(III) oxide because doubling the aliquot volume doubled the precipitate mass. Explain why this conclusion is not justified.

(d) Suppose the separate aluminium analysis had not been performed. Could the percentage of iron still be calculated uniquely from the two total oxide masses? Demonstrate your reasoning using two possible compositions of the first residue.

Solution 5

(a) The original ore contains 44.8% iron by mass.

The first challenge is recognising that the 0.220 g residue contains two different compounds.

We cannot treat the entire mass as iron(III) oxide.

However, the separate aluminium analysis tells us that 0.060 g of the residue is aluminium oxide.

Therefore, the iron(III) oxide mass in the first aliquot is:

\[ m(\ce{Fe2O3})= 0.220-0.060 \]

\[ =0.160\ \mathrm{g} \]

Now convert this into iron mass:

\[ m(\mathrm{Fe})= 0.160\times\frac{111.70}{159.70} \]

\[ =0.1119\ \mathrm{g} \]

This is the iron mass in the 25.00 mL aliquot.

The complete digest has a volume of 250.0 mL, giving:

\[ \text{Aliquot factor} = \frac{250.0}{25.00}=10.00 \]

Therefore:

\[ m(\mathrm{Fe,\ total})= 0.1119\times10.00 \]

\[ =1.119\ \mathrm{g} \]

Finally:

\[ \%\mathrm{Fe}= \frac{1.119}{2.500}\times100 \]

\[ =44.8\% \]

The original ore contains 44.8% iron by mass.

(b) Treating the entire residue as iron(III) oxide overestimates the iron content.

The total residue contains both iron(III) oxide and aluminium oxide.

If the student assumes all 0.220 g is \(\ce{Fe2O3}\), they would calculate:

\[ m(\mathrm{Fe,\ aliquot})= 0.220\times0.6994 \]

\[ =0.1539\ \mathrm{g} \]

Scaling up:

\[ m(\mathrm{Fe,\ total})= 0.1539\times10.00 \]

\[ =1.539\ \mathrm{g} \]

The incorrect percentage would be:

\[ \%\mathrm{Fe}= \frac{1.539}{2.500}\times100 \]

\[ =61.6\% \]

This is higher than the correct value of 44.8%.

The problem is that aluminium oxide contributes to the measured mass but contains no iron.

A mass measurement alone cannot distinguish between two different compounds.

(c) Doubling the aliquot volume does not prove that the precipitate is pure.

The second aliquot is twice the volume of the first:

\[ \frac{50.00}{25.00}=2 \]

Its total residue mass is also twice as large:

\[ \frac{0.440}{0.220}=2 \]

This shows that the total mass of precipitable solid is proportional to the aliquot volume over these measurements.

However, both iron and aluminium ions are present in the original solution.

If both precipitate in a consistent ratio, doubling the amount of solution will double the mass of both oxides.

The result is therefore consistent with the presence of two compounds, not evidence of a pure iron compound.

The tempting interpretation confuses a proportional measurement with a selective measurement.

(d) No, the iron percentage cannot be determined uniquely without additional information.

This is the most important conceptual point in the question.

The first experiment establishes only that the combined oxide mass is 0.220 g.

Consider two possible compositions:

Both compositions contain iron oxide and aluminium oxide. Both produce exactly the same measured total residue mass.

However, they imply different iron percentages.

For composition A:

\[ m(\mathrm{Fe})= 0.160\times0.6994\times10 \]

\[ =1.119\ \mathrm{g} \]

\[ \%\mathrm{Fe}=44.8\% \]

For composition B:

\[ m(\mathrm{Fe})= 0.200\times0.6994\times10 \]

\[ =1.399\ \mathrm{g} \]

\[ \%\mathrm{Fe}= \frac{1.399}{2.500}\times100 \]

\[ =56.0\% \]

Both results are compatible with the first total mass measurement.

The second experiment does not resolve the ambiguity because it measures the combined mass again. If the proportions of the two compounds remain the same, both compositions will produce a doubled total mass.

The separate selective aluminium analysis is what allows us to distinguish the two contributions.

This is a fundamental limitation of gravimetric analysis: a measured mass only identifies the analyte quantity when the composition of the weighed product is known.

07Where this connects to other HSC Chemistry analysis techniques

Calculating the percentage of metal in an ore combines several ideas:

• Representative sampling ensures that the original material is suitable for analysis.
• Precipitation converts dissolved ions into an insoluble solid.
• Filtration and washing separate the analyte-containing solid from soluble impurities.
• Heating to constant mass produces a measurable compound of known composition.
• Stoichiometry connects the compound mass to the original metal mass.
• Aliquot calculations account for the fraction of the original solution analysed.

For more practice with the general relationship between precipitate mass and analyte moles, see Gravimetric analysis: calculate analyte amounts from precipitates.

If the analyte remains in solution and is measured using light absorption rather than precipitate mass, the calculation follows a different approach. See Colourimetry: Calculate Concentration and Evaluate Data.

The key decision in any ore analysis is knowing exactly what your instrument has measured. Once you identify the actual compound, its purity, and the fraction of the original sample it represents, the percentage calculation becomes a straightforward application of stoichiometry.