How to Calculate the Mass of Metal Recovered from an Oxide

A ceramic boat contains 7.95 g of black copper(II) oxide. After the oxide reacts with hydrogen, the solid is 1.60 g lighter.

Has the reaction produced 1.60 g of copper?

No. The 1.60 g is oxygen that has been removed from the oxide. Assuming the reaction is complete and no solid is lost, the mass of copper recovered is:

\[ 7.95-1.60=6.35\ \mathrm{g} \]

To find the mass of metal recovered from any metal oxide, use the same four-stage method:

Mass of oxide → moles of oxide → moles of metal → mass of metal.

The balanced chemical equation provides the mole ratio between the oxide and the metal. It doesn’t directly provide a mass ratio.

This distinction becomes especially important when the oxide contains more than one metal atom per formula unit, the starting material is impure, or the reaction doesn’t go to completion.

01How does the experiment separate metal from oxygen?

Consider the reduction of copper(II) oxide using hydrogen.

The balanced equation is:

\[ \ce{CuO(s) + H2(g) -> Cu(s) + H2O(g)} \]

Copper(II) oxide contains copper and oxygen. During the reaction, hydrogen removes oxygen from the oxide, producing water vapour and leaving copper metal behind.

The copper doesn’t disappear into the gas. Its atoms remain in the solid product.

Picture the apparatus

Imagine a horizontal heat-resistant glass tube supported above a heat source.

From left to right:

1. A controlled supply of dry hydrogen enters the left end of the tube.
2. In the centre, a ceramic boat holds a measured sample of black copper(II) oxide.
3. The oxide is heated while hydrogen passes over it.
4. Water vapour produced by the reaction travels towards the right-hand outlet.
5. A cooled collection trap can capture the water, while remaining gas passes to an appropriately managed outlet.

The ceramic boat can be weighed separately before and after the reaction, once the solid is dry and cool.

This is a modelled investigation with illustrative measurements, not a record of an experiment performed by tutorgum. Hydrogen is flammable, so any practical reduction requires a teacher-controlled laboratory setup and appropriate safety measures.

What should the experiment show?

Before seeing the result, predict what happens to the mass of the solid.

The oxide contains oxygen, which leaves the boat as part of water vapour. Copper remains behind. Therefore, the dry solid should become lighter.

For the illustrative 7.95 g sample of pure copper(II) oxide, complete reduction predicts:

The last two masses are different. That’s worth understanding.

The removed oxygen contributes 1.60 g to the water. Another 0.20 g comes from hydrogen supplied to the reaction.

Therefore:

\[ 1.60+0.20=1.80\ \mathrm{g} \]

The solid loses oxygen, but the water contains oxygen and hydrogen.

A student who subtracts the full mass of water from the starting oxide would get the wrong mass of copper.

02Why the balanced equation uses moles, not grams

A balanced equation counts particles in fixed proportions.

For copper(II) oxide:

\[ \ce{CuO + H2 -> Cu + H2O} \]

One formula unit of copper(II) oxide contains one copper atom. Therefore, one mole of CuO can produce one mole of Cu.

The ratio is:

\[ n(\ce{CuO}):n(\ce{Cu})=1:1 \]

Here, \(n\) represents the amount of substance, measured in moles (mol).

But one mole of copper(II) oxide does not have the same mass as one mole of copper.

Using the following approximate atomic molar masses:

• Copper: \(63.5\ \mathrm{g\,mol^{-1}}\)
• Oxygen: \(16.0\ \mathrm{g\,mol^{-1}}\)
• Hydrogen: \(1.0\ \mathrm{g\,mol^{-1}}\)

The molar mass of copper(II) oxide is:

\[ M(\ce{CuO})=63.5+16.0 =79.5\ \mathrm{g\,mol^{-1}} \]

Here, \(M\) is molar mass: the mass of one mole of a substance.

One mole of CuO has a mass of 79.5 g, but it contains only 63.5 g of copper.

That is why equal mole amounts don’t mean equal masses.

The metal oxide formula can change the ratio

Now compare copper(II) oxide, \(\ce{CuO}\), with copper(I) oxide, \(\ce{Cu2O}\).

Copper(I) oxide contains two copper atoms for every oxygen atom.

Its reduction equation is:

\[ \ce{Cu2O + H2 -> 2Cu + H2O} \]

One mole of \(\ce{Cu2O}\) produces two moles of copper.

This gives us three useful reference cases.

For iron, the table uses an atomic molar mass of \(55.8\ \mathrm{g\,mol^{-1}}\).

Notice that the ratio is determined by the oxide’s formula and the balanced equation, not simply by the fact that the reactant is a metal oxide.

03The mass-to-moles-to-moles-to-mass method

Suppose an exam question gives the mass of a metal oxide and asks you to calculate the theoretical mass of metal produced.

Follow these four steps.

Step 1: Write the balanced reduction equation

Identify the metal oxide and the metal being produced.

For copper(II) oxide:

\[ \ce{CuO + H2 -> Cu + H2O} \]

For iron(III) oxide reduced by hydrogen:

\[ \ce{Fe2O3 + 3H2 -> 2Fe + 3H2O} \]

In the second equation, one mole of iron(III) oxide produces two moles of iron.

Step 2: Convert the oxide mass into moles

Use:

\[ n=\frac{m}{M} \]

Where:

• \(n\) is the amount of substance in mol.
• \(m\) is the mass in g.
• \(M\) is the molar mass in \(\mathrm{g\,mol^{-1}}\).

Make sure you’re dividing by the molar mass of the entire oxide, not just the metal.

Step 3: Apply the mole ratio

Read the coefficients from the balanced equation.

For example:

\[ \ce{Fe2O3 -> 2Fe} \]

This is the oxide-to-metal relationship taken from the full balanced reaction.

Therefore:

\[ n(\ce{Fe})=2n(\ce{Fe2O3}) \]

The amount of iron is twice the amount of iron(III) oxide.

Step 4: Convert metal moles into grams

Use:

\[ m=nM \]

This time, use the molar mass of the metal itself.

The result is the theoretical mass of metal, assuming the stated amount of oxide can react completely, the reducing agent is sufficient, and no metal is lost.

04Why a smaller solid mass doesn’t always prove that all the metal has been recovered

A balance measures mass. It doesn’t tell you exactly which substances are present.

Imagine the first experimental design uses only a before-and-after weighing of the ceramic boat.

After heating, the solid is lighter.

That observation supports the idea that oxygen has been removed. But it doesn’t prove that the final solid is pure metal.

Three problems could affect the interpretation:

• Some metal oxide may remain unreduced.
• The recovered metal may react with oxygen again while cooling.
• Material may be lost from the boat, or moisture may affect the measurement.

A mass change caused by spilled powder isn’t the same as a mass change caused by chemical reduction.

Improving the experiment

A more informative design would use a pre-weighed ceramic boat, a dry gas supply, a separate water collection trap, and a controlled way of cooling the product without exposing it unnecessarily to oxygen.

The investigator would compare three pieces of evidence: the initial oxide mass, the final dry solid mass, and the water produced.

If copper(II) oxide is completely reduced to copper, each mole of CuO should lose one mole of oxygen atoms and produce one mole of water.

For the 15.9 g sample in the worked example, the ideal predictions are:

These predictions assume dry incoming hydrogen, complete water collection, no solid losses, and no side reactions.

Matching results would support the proposed reaction. They wouldn’t automatically prove the product’s purity, because some partially reduced oxides can also produce water.

The revised design improves the evidence, but identifying the final solid remains important.

05A contrasting case: Starting with the same oxide mass can give a different metal mass

Suppose two samples each have a mass of 7.95 g.

Sample A is pure \(\ce{CuO}\). Sample B is pure \(\ce{Cu2O}\).

Both are reduced completely using excess hydrogen.

Would they produce equal masses of copper?

It’s tempting to say yes because they start with the same mass and contain the same metal.

But \(\ce{Cu2O}\) contains less oxygen relative to copper. A greater fraction of its starting mass is already copper.

For Sample A, the calculation gives:

\[ m(\ce{Cu})=6.35\ \mathrm{g} \]

For Sample B:

\[ \ce{Cu2O + H2 -> 2Cu + H2O} \]

\[ \begin{aligned} n(\ce{Cu2O})&=\frac{7.95}{143.0}\\ &=0.0556\ \mathrm{mol}\\ n(\ce{Cu})&=2(0.0556)\\ &=0.111\ \mathrm{mol}\\ m(\ce{Cu})&=0.11119\times63.5\\ &=7.06\ \mathrm{g} \end{aligned} \]

The final calculation uses the unrounded copper amount.

So the results are:

The copper(I) oxide produces more copper from the same starting mass because it contains a larger proportion of copper by mass.

This exposes two common errors: assuming all metal oxides have the same metal percentage, and treating equation coefficients as mass ratios.

06What if the oxide is part of an impure ore?

An ore sample is rarely just pure metal oxide. It may contain minerals that don’t contribute to the amount of metal being calculated.

The mass-to-moles method still works, but you must first find how much actual oxide is present.

If the oxide represents \(p\%\) of an ore sample:

\[ m(\text{oxide})= m(\text{ore})\times\frac{p}{100} \]

A second complication is recovery yield.

Theoretical yield is the maximum predicted mass of metal under the stated reaction assumptions. Actual recovered mass may be smaller because of incomplete conversion, physical losses, or other processes.

If the percentage recovery is known:

\[ m_{\text{actual}}= m_{\text{theoretical}}\times\frac{Y}{100} \]

Here, \(Y\) is the percentage recovery, and both masses are measured in grams.

07Questions and solutions

Use the following approximate atomic molar masses throughout:

\[ \begin{aligned} M(\ce{Cu})&=63.5\ \mathrm{g\,mol^{-1}}\\ M(\ce{Fe})&=55.8\ \mathrm{g\,mol^{-1}}\\ M(\ce{O})&=16.0\ \mathrm{g\,mol^{-1}}\\ M(\ce{H})&=1.0\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Unless a question says otherwise, assume the stated reactions occur without side reactions and that the materials are dry.

Question 1

A 7.95 g sample of pure copper(II) oxide is completely reduced using excess hydrogen.

The equation is:

\[ \ce{CuO + H2 -> Cu + H2O} \]

Calculate the theoretical mass of copper recovered.

Solution 1

The theoretical mass of copper recovered is 6.35 g.

The balanced equation gives a \(1:1\) mole ratio between copper(II) oxide and copper.

First, calculate the amount of CuO:

\[ \begin{aligned} n(\ce{CuO})&=\frac{7.95}{79.5}\\ &=0.100\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ n(\ce{Cu})=0.100\ \mathrm{mol} \]

Convert to mass:

\[ \begin{aligned} m(\ce{Cu})&=0.100\times63.5\\ &=6.35\ \mathrm{g} \end{aligned} \]

The remaining 1.60 g of the original oxide mass was oxygen.

The tempting mistake is to assume that a \(1:1\) mole ratio means 7.95 g of CuO produces 7.95 g of copper. It doesn’t. The oxygen contributes mass to the oxide but isn’t part of the final copper metal.

Question 2

A student places 12.0 g of pure copper(II) oxide in a ceramic boat and heats it in hydrogen.

After cooling, the dry solid remaining in the boat has a mass of 10.4 g.

Analysis shows that the final solid contains only copper and unreacted copper(II) oxide. No solid was lost, and no reoxidation occurred.

Calculate:

a. The mass of copper produced.

b. The mass of unreacted copper(II) oxide.

c. The percentage of the original copper(II) oxide that was reduced.

Solution 2

The reaction produced 6.35 g of copper, left 4.05 g of CuO unreacted, and reduced approximately 66.3% of the original oxide.

The critical clue is that the 1.60 g decrease represents oxygen removed, not copper produced.

Part a: Copper produced

The solid mass decrease is:

\[ 12.0-10.4=1.60\ \mathrm{g} \]

Each mole of CuO that reacts loses one mole of oxygen atoms from the solid.

Therefore, the amount of oxygen removed is:

\[ \begin{aligned} n(\ce{O})&=\frac{1.60}{16.0}\\ &=0.100\ \mathrm{mol} \end{aligned} \]

From the reaction:

\[ \ce{CuO + H2 -> Cu + H2O} \]

One mole of CuO reduced produces one mole of copper.

Therefore, \(0.100\ \mathrm{mol}\) of copper formed.

\[ \begin{aligned} m(\ce{Cu})&=0.100\times63.5\\ &=6.35\ \mathrm{g} \end{aligned} \]

Part b: Unreacted copper(II) oxide

Since the final solid contains only Cu and CuO:

\[ \begin{aligned} m(\ce{CuO\ remaining})&=10.4-6.35\\ &=4.05\ \mathrm{g} \end{aligned} \]

Part c: Percentage reduced

The amount of CuO consumed was \(0.100\ \mathrm{mol}\), equivalent to:

\[ m(\ce{CuO\ reacted})=0.100\times79.5 =7.95\ \mathrm{g} \]

Therefore:

\[ \begin{aligned} \%\text{ reduced} &=\frac{7.95}{12.0}\times100\\ &=66.3\% \end{aligned} \]

The balance shows that some reduction occurred, but the final solid is not entirely metal.

A particularly tempting error would be to call 10.4 g the mass of copper recovered. That ignores the explicit evidence that unreacted oxide remains in the boat.

Question 3

A 24.0 g mineral sample contains 75.0% iron(III) oxide by mass. The remainder is inert and stays in the reaction vessel.

The sample is exposed to exactly 0.240 mol of hydrogen available for reaction, with no additional hydrogen supplied.

The reaction is:

\[ \ce{Fe2O3 + 3H2 -> 2Fe + 3H2O} \]

Assume that the available limiting reactant is completely consumed, all produced iron stays in the vessel, and the water leaves as vapour.

Determine:

a. The limiting reactant.

b. The mass of iron produced.

c. The total mass of dry solid remaining in the vessel, including any unreacted iron oxide and inert material.

Solution 3

Hydrogen is the limiting reactant. The reaction produces 8.93 g of iron and leaves approximately 20.2 g of dry solid.

The important constraint is that the ore contains impurities and the hydrogen supply is insufficient to reduce all the iron(III) oxide.

Part a: Identify the limiting reactant

First, calculate the oxide mass:

\[ \begin{aligned} m(\ce{Fe2O3})&=24.0\times0.750\\ &=18.0\ \mathrm{g} \end{aligned} \]

Its amount is:

\[ \begin{aligned} n(\ce{Fe2O3})&=\frac{18.0}{159.6}\\ &=0.1128\ \mathrm{mol} \end{aligned} \]

The equation requires three moles of hydrogen per mole of iron(III) oxide.

To reduce all the oxide:

\[ \begin{aligned} n(\ce{H2\ required})&=3(0.1128)\\ &=0.338\ \mathrm{mol} \end{aligned} \]

Only 0.240 mol is available.

Therefore, hydrogen is limiting, and some iron(III) oxide must remain.

Part b: Calculate iron produced

From the balanced equation:

\[ 3\ \mathrm{mol\ H_2}\rightarrow2\ \mathrm{mol\ Fe} \]

Therefore:

\[ \begin{aligned} n(\ce{Fe})&=0.240\times\frac{2}{3}\\ &=0.160\ \mathrm{mol}\\ m(\ce{Fe})&=0.160\times55.8\\ &=8.93\ \mathrm{g} \end{aligned} \]

Part c: Calculate the remaining dry solid

The amount of oxide reduced is:

\[ \begin{aligned} n(\ce{Fe2O3\ reacted}) &=\frac{0.240}{3}\\ &=0.0800\ \mathrm{mol} \end{aligned} \]

Its mass is:

\[ m(\ce{Fe2O3\ reacted}) =0.0800\times159.6 =12.768\ \mathrm{g} \]

Unreacted oxide:

\[ 18.0-12.768=5.232\ \mathrm{g} \]

Inert material:

\[ 24.0-18.0=6.00\ \mathrm{g} \]

Add the solid components:

\[ \begin{aligned} m_{\text{final solid}} &=8.928+5.232+6.00\\ &=20.16\ \mathrm{g}\\ &\approx20.2\ \mathrm{g} \end{aligned} \]

The final dry solid contains iron, unreacted iron(III) oxide, and inert material.

The trap is treating the 18.0 g of oxide as though it can all react simply because hydrogen is present. Checking the mole requirement first changes the entire calculation.

Notice also that the 4.32 g of water produced isn’t the amount lost from the dry solid. The water includes 0.480 g of hydrogen that entered from outside the original ore sample.

Question 4

A 22.25 g solid sample contains only copper(II) oxide, \(\ce{CuO}\), and copper(I) oxide, \(\ce{Cu2O}\).

Both oxides are completely reduced using excess dry hydrogen. All the resulting copper is recovered, with no loss or reoxidation.

The final mass of copper is 19.05 g.

The reactions are:

\[ \ce{CuO + H2 -> Cu + H2O} \]

\[ \ce{Cu2O + H2 -> 2Cu + H2O} \]

Calculate:

a. The initial mass of each oxide.

b. The theoretical mass of water produced.

c. Explain why the amount of collected water cannot simply be subtracted from the original oxide mass to calculate copper recovered.

Solution 4

The starting mixture contained 7.95 g of CuO and 14.30 g of Cu2O. Complete reduction produces 3.60 g of water.

The key is to use copper and oxygen conservation together. A single mass difference tells us how much oxygen was removed, but the final copper mass also tells us how many copper atoms were present.

Part a: Determine the oxide masses

The total oxygen mass initially present was:

\[ \begin{aligned} m(\ce{O})&=22.25-19.05\\ &=3.20\ \mathrm{g} \end{aligned} \]

Therefore:

\[ n(\ce{O})=\frac{3.20}{16.0} =0.200\ \mathrm{mol} \]

The total amount of copper was:

\[ n(\ce{Cu})=\frac{19.05}{63.5} =0.300\ \mathrm{mol} \]

Let:

• \(x\) be the initial amount of CuO in mol.
• \(y\) be the initial amount of Cu2O in mol.

Each formula unit of either oxide contains one oxygen atom.

Therefore:

\[ x+y=0.200 \]

But CuO contains one copper atom per formula unit, while Cu2O contains two.

Therefore:

\[ x+2y=0.300 \]

Subtracting the first equation from the second:

\[ \begin{aligned} y&=0.100\ \mathrm{mol}\\ x&=0.100\ \mathrm{mol} \end{aligned} \]

Now convert each to mass.

For CuO:

\[ m(\ce{CuO})=0.100\times79.5 =7.95\ \mathrm{g} \]

For Cu2O:

\[ m(\ce{Cu2O})=0.100\times143.0 =14.30\ \mathrm{g} \]

Together, these give the required 22.25 g.

Part b: Calculate water produced

Each mole of either oxide forms one mole of water on complete reduction.

Therefore:

\[ \begin{aligned} n(\ce{H2O})&=x+y\\ &=0.200\ \mathrm{mol} \end{aligned} \]

\[ \begin{aligned} m(\ce{H2O})&=0.200\times18.0\\ &=3.60\ \mathrm{g} \end{aligned} \]

Part c: Explain the mass difference

Subtracting the water mass from the starting oxide mass would give:

\[ 22.25-3.60=18.65\ \mathrm{g} \]

That is incorrect.

Only 3.20 g of the water’s mass came from the original oxide. The remaining 0.40 g came from hydrogen supplied during reduction.

The correct solid mass calculation is:

\[ 22.25-3.20=19.05\ \mathrm{g} \]

This question also shows why oxide composition matters. The mixture contains equal amounts of the two oxides in moles, but not equal masses, and not equal amounts of copper.

Question 5

A researcher wants to determine how much metallic copper remains after a reduction experiment.

Initially, a ceramic boat contains 15.90 g of pure copper(II) oxide.

The oxide undergoes complete reduction using dry hydrogen:

\[ \ce{CuO + H2 -> Cu + H2O} \]

All the copper remains in the boat, and a water trap collects 3.60 g of water, consistent with complete reduction.

The trap is then isolated from the reaction tube.

During cooling, a small air leak allows oxygen to react with some of the freshly produced copper. No material is physically lost, and no further water forms.

The final dry solid weighs 13.50 g.

The researcher knows that the cooling stage formed only one type of oxide, alongside any remaining metallic copper, but initially doesn’t know whether that oxide is CuO or Cu2O.

Consider the following oxidation reactions:

\[ \ce{2Cu + O2 -> 2CuO} \]

\[ \ce{4Cu + O2 -> 2Cu2O} \]

Answer the following.

a. Calculate the mass of metallic copper present immediately after complete reduction, before the air leak.

b. Calculate the mass of oxygen absorbed during cooling.

c. Determine the final mass of metallic copper if the oxide formed during cooling is exclusively CuO.

d. Determine the final mass of metallic copper if the oxide formed during cooling is exclusively Cu2O.

e. Can the measured water mass and final solid mass alone distinguish between these two possibilities?

A subsequent composition analysis establishes that Cu2O is the only oxide present in the cooled solid. State the actual final mass of metallic copper and explain the surprising relationship between oxide composition and how much metal remains.

Solution 5

The copper mass immediately after reduction was 12.70 g. The air leak added 0.80 g of oxygen to the solid. Depending on the oxide formed, the final metallic copper mass could be 9.53 g or 6.35 g.

The mass measurements alone cannot distinguish these outcomes. Once the additional analysis identifies Cu2O, the final answer is 6.35 g of metallic copper.

The central complication is that the experiment has two separate chemical stages: reduction followed by partial reoxidation.

Part a: Copper immediately after reduction

The initial amount of CuO was:

\[ \begin{aligned} n(\ce{CuO})&=\frac{15.90}{79.5}\\ &=0.200\ \mathrm{mol} \end{aligned} \]

Complete reduction produces \(0.200\ \mathrm{mol}\) of copper.

\[ \begin{aligned} m(\ce{Cu})&=0.200\times63.5\\ &=12.70\ \mathrm{g} \end{aligned} \]

This agrees with the water measurement:

\[ \begin{aligned} n(\ce{H2O})&=0.200\ \mathrm{mol}\\ m(\ce{H2O})&=0.200\times18.0\\ &=3.60\ \mathrm{g} \end{aligned} \]

So the first stage is consistent with complete reduction.

Part b: Oxygen absorbed during cooling

Immediately after reduction, the solid consisted of 12.70 g of copper.

After the air leak, it weighed 13.50 g.

Since the only substance added to the solid was oxygen:

\[ \begin{aligned} m(\ce{O\ absorbed})&=13.50-12.70\\ &=0.80\ \mathrm{g} \end{aligned} \]

In moles of oxygen atoms:

\[ \begin{aligned} n(\ce{O})&=\frac{0.80}{16.0}\\ &=0.0500\ \mathrm{mol} \end{aligned} \]

The oxygen has increased the solid mass. But that doesn’t tell us how many copper atoms have become part of an oxide until we know which oxide formed.

Part c: If CuO forms

For copper(II) oxide:

\[ \ce{2Cu + O2 -> 2CuO} \]

Each oxygen atom combines with one copper atom.

Therefore, \(0.0500\ \mathrm{mol}\) of oxygen atoms converts \(0.0500\ \mathrm{mol}\) of copper atoms into CuO.

The mass of copper consumed by reoxidation is:

\[ \begin{aligned} m(\ce{Cu\ oxidised})&=0.0500\times63.5\\ &=3.175\ \mathrm{g} \end{aligned} \]

Metallic copper remaining:

\[ \begin{aligned} m(\ce{Cu\ metal})&=12.70-3.175\\ &=9.525\ \mathrm{g}\\ &\approx9.53\ \mathrm{g} \end{aligned} \]

The oxide formed would weigh:

\[ 3.175+0.80=3.975\ \mathrm{g} \]

The total is consistent with the measured final solid:

\[ 9.525+3.975=13.50\ \mathrm{g} \]

Part d: If Cu2O forms

For copper(I) oxide:

\[ \ce{4Cu + O2 -> 2Cu2O} \]

Each oxygen atom combines with two copper atoms.

Therefore, \(0.0500\ \mathrm{mol}\) of oxygen atoms converts \(0.100\ \mathrm{mol}\) of copper atoms into Cu2O.

Copper consumed:

\[ \begin{aligned} m(\ce{Cu\ oxidised})&=0.100\times63.5\\ &=6.35\ \mathrm{g} \end{aligned} \]

Metallic copper remaining:

\[ \begin{aligned} m(\ce{Cu\ metal})&=12.70-6.35\\ &=6.35\ \mathrm{g} \end{aligned} \]

The oxide formed weighs:

\[ 6.35+0.80=7.15\ \mathrm{g} \]

Again, the total matches:

\[ 6.35+7.15=13.50\ \mathrm{g} \]

Part e: Can the mass measurements identify the oxide?

No. Both models predict the same 13.50 g final solid mass, despite leaving different quantities of metallic copper.

The 3.60 g of water confirms the amount of oxygen removed during the first stage under the stated assumptions. It doesn’t reveal which oxide formed later.

The 0.80 g mass increase tells us how much oxygen was added during cooling, but not how many copper atoms combined with it.

The later composition analysis identifies Cu2O, so the final metallic copper mass is 6.35 g.

The surprising result is that forming Cu2O consumes more copper metal than forming CuO for the same mass of oxygen absorbed.

That happens because Cu2O has twice as many copper atoms per oxygen atom.

A student might confidently use the water trap’s evidence of complete reduction to conclude that 12.70 g of metallic copper must still be present. That conclusion ignores the subsequent oxidation stage.

Even more subtly, the two final solids have identical overall masses and identical total amounts of copper and oxygen, but different distributions of copper between metal and oxide.

Conservation of mass is necessary, but it doesn’t uniquely identify a mixture’s chemical composition.

08What to check before giving your final answer

When you calculate the mass of metal recovered from its oxide, make three decisions before writing the numerical result.

First, check the oxide formula and balanced reaction. One mole of \(\ce{CuO}\) gives one mole of copper, but one mole of \(\ce{Cu2O}\) gives two. This is the central mass-to-moles-to-moles-to-mass relationship.

Second, decide which amount can actually react. If the ore is impure, calculate the oxide mass first. If the reducing agent is limited, calculate the limiting reactant before predicting the amount of metal.

Third, distinguish theoretical metal production from the metallic material actually collected. Incomplete reduction, physical losses, and reoxidation can all affect the final result.

You can check your calculation by accounting for the metal and oxygen atoms before and after the reaction. The related guide on checking whether reaction-mass answers conserve atoms develops that method.

One useful next question is whether a particular metal oxide can be reduced by the chosen reagent at all. Stoichiometry tells you the quantity a stated reaction could produce; it doesn’t establish whether that reaction is chemically feasible under the conditions.

That connects the calculation to metal reactivity with oxygen: understanding how readily a metal forms an oxide helps explain why some oxides can be reduced relatively easily, while others require very different extraction methods.