How to Find a Metal Oxide Formula from Mass Gain

A metal sample is heated in air. Before heating, it has a mass of 0.486 g. After heating and cooling, the solid product has a mass of 0.806 g.

The sample has gained 0.320 g. Where did that extra mass come from?

Oxygen from the air has combined with the metal to form a metal oxide. More importantly, that mass gain gives us a way to work backwards and determine the oxide’s formula.

Before calculating, consider this: is the 0.320 g gain enough to determine the formula by itself? No. You also need the original mass of metal. A chemical formula describes a ratio of atoms, so we need to compare the number of metal atoms with the number of oxygen atoms.

The method is straightforward:

1. Calculate the mass of oxygen gained.
2. Convert the metal mass and oxygen mass into moles.
3. Divide both mole quantities by the smaller value.
4. Convert the resulting ratio into the simplest whole-number ratio.

This gives the empirical formula of the oxide, assuming the product is a single, pure oxide and the metal has reacted completely.

01What does the mass gain actually measure?

Imagine heating a small piece of metal in a crucible.

Before heating, you weigh the metal. During heating, oxygen from the surrounding air reacts with the metal. After cooling, you weigh the solid product again.

If oxygen is the only substance entering the sample, and no solid is lost, the increase in mass must be the mass of oxygen that has combined with the metal.

The relationship is:

\[ m_{\mathrm{oxygen}}=\Delta m \]

where:

\[ \Delta m=m_{\mathrm{after}}-m_{\mathrm{before}} \]

Here:

• \(m_{\mathrm{oxygen}}\) is the mass of oxygen incorporated into the oxide, in grams.
• \(m_{\mathrm{after}}\) is the mass of the solid after heating, in grams.
• \(m_{\mathrm{before}}\) is the original mass of the metal, in grams.
• \(\Delta m\) is the change in mass, in grams.

A positive mass change indicates that the sample has gained mass.

For example, if a metal starts at 1.20 g and its oxide product weighs 1.60 g:

\[ m_{\mathrm{oxygen}}=1.60-1.20 \]

\[ =0.40\ \mathrm{g} \]

The metal contributed 1.20 g, and oxygen contributed the remaining 0.40 g.

An important distinction: the final mass is the mass of the entire oxide, not the mass of oxygen alone. You must subtract the original metal mass.

How the experiment is arranged

A typical arrangement uses a ceramic crucible with a lid. The metal sample sits inside the crucible, which rests on a clay triangle above a heat source.

The lid is kept slightly open during heating.

This small gap has two jobs:

• It allows oxygen from the air to enter and react with the metal.
• It helps prevent fine solid oxide particles from escaping.

An uncovered crucible allows oxygen in, but some product may be lost. A completely closed crucible limits the supply of oxygen, potentially preventing complete oxidation.

The partially covered crucible is a compromise between these problems.

The experiment relies on a few important assumptions:

• The original metal sample is pure.
• The metal reacts completely.
• The final solid contains only the metal and oxygen.
• No metal or oxide is lost during heating.
• The crucible and lid do not undergo a chemical reaction or change in mass.
• The product is dry when weighed.

These assumptions matter. If oxide powder escapes, the measured mass gain will be too small. If moisture or another substance enters the sample, the gain may be too large.

The calculation is only as reliable as the experiment.

02Convert the masses into moles

Now we can work out the number of metal and oxygen atoms involved.

Remember that a chemical formula counts atoms, not grams. We therefore need to convert each element’s mass into moles.

For the metal:

\[ n(\mathrm{M})=\frac{m(\mathrm{M})}{M(\mathrm{M})} \]

For oxygen:

\[ n(\mathrm{O})=\frac{m(\mathrm{O})}{16.00} \]

Here:

• \(n(\mathrm{M})\) is the amount of metal, in moles.
• \(m(\mathrm{M})\) is the original metal mass, in grams.
• \(M(\mathrm{M})\) is the metal’s molar mass, in \(\mathrm{g\,mol^{-1}}\).
• \(n(\mathrm{O})\) is the amount of oxygen atoms, in moles.
• \(m(\mathrm{O})\) is the measured oxygen mass gain, in grams.
• \(16.00\ \mathrm{g\,mol^{-1}}\) is the molar mass of oxygen atoms.

Why do we use 16.00 rather than 32.00 for oxygen?

This is a common source of mistakes.

Oxygen enters the reaction from the air as diatomic oxygen molecules, \(\ce{O2}\). Each molecule contains two oxygen atoms.

However, the chemical formula of an oxide counts individual oxygen atoms.

One mole of \(\ce{O2}\) molecules contains two moles of oxygen atoms.

Therefore:

\[ M(\ce{O2})=32.00\ \mathrm{g\,mol^{-1}} \]

but:

\[ M(\mathrm{O})=16.00\ \mathrm{g\,mol^{-1}} \]

You can use either approach, provided you account for the two atoms in each oxygen molecule.

The simplest approach is to divide the oxygen mass gain by 16.00.

Finding the simplest mole ratio

Once you have calculated the moles of each element, divide both quantities by the smaller number.

Suppose your calculation gives:

\[ \mathrm{M:O}=0.0200:0.0300 \]

Divide both values by 0.0200:

\[ \mathrm{M:O}=1:1.5 \]

We still don’t have a whole-number ratio.

To remove the decimal, multiply both sides by 2:

\[ \mathrm{M:O}=2:3 \]

The empirical formula is therefore:

\[ \ce{M2O3} \]

This is the key step. You are looking for the simplest whole-number ratio that is consistent with the experimental data.

Some common ratios are:

Do not round every decimal to the nearest integer immediately.

For example, 1.33 is not close enough to 1 to justify rounding it to 1. Instead, recognise that 1.33 is approximately \(4/3\), and multiply the entire ratio by 3.

Rounding is justified when the resulting whole-number ratio is consistent with the precision of the measurements.

If your calculated ratio is 1 : 1.42, don’t automatically force it into a familiar formula. Check your arithmetic, experimental assumptions, and measurement precision first.

03Worked examples

All measurements in the following examples are illustrative data.

04The common mistake: Comparing masses instead of moles

Consider a copper sample with an initial mass of 2.542 g. After heating, its oxide product weighs 2.862 g.

The oxygen mass gain is:

\[ m(\mathrm{O})=2.862-2.542 \]

\[ =0.320\ \mathrm{g} \]

A student might compare the two masses directly:

\[ \mathrm{Cu:O}=2.542:0.320 \]

\[ \approx 7.94:1 \]

They might conclude that the formula is approximately \(\ce{Cu8O}\).

That is incorrect.

The problem is that copper atoms are much heavier than oxygen atoms. Comparing their masses does not tell us how many atoms of each element are present.

Let’s convert the masses into moles instead.

Use:

• Cu = \(63.55\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)

For copper:

\[ n(\mathrm{Cu})=\frac{2.542}{63.55} \]

\[ =0.04000\ \mathrm{mol} \]

For oxygen:

\[ n(\mathrm{O})=\frac{0.320}{16.00} \]

\[ =0.02000\ \mathrm{mol} \]

The mole ratio is:

\[ \mathrm{Cu:O}=0.04000:0.02000 \]

\[ =2:1 \]

Therefore:

\[ \boxed{\ce{Cu2O}} \]

The correct empirical formula is \(\ce{Cu2O}\).

The important distinction is:

• Mass ratios compare how heavy the elements are.
• Mole ratios compare how many atoms of each element are present.

Chemical formulas require the second.

Also remember that the mass gain alone is insufficient. If you only knew that the sample gained 0.320 g, you could calculate the amount of oxygen, but you would have no information about how much metal was present.

05Does the empirical formula tell you everything?

Not quite.

A mass-gain experiment tells you the simplest ratio of atoms in the sample, provided the sample is a single pure compound.

This is called the empirical formula.

For most metal oxides, the solid has an extended ionic lattice. Its formula represents the simplest ratio of metal ions to oxide ions, rather than the number of atoms in a separate molecule.

For example, \(\ce{MgO}\) tells us that magnesium and oxygen occur in a 1:1 ratio throughout the solid.

It does not mean that a piece of magnesium oxide consists of individual, isolated MgO molecules.

There is another important limitation: a bulk mass ratio does not necessarily prove that the product is a single compound.

A sample containing two different metal oxides can have an overall elemental ratio that looks like the empirical formula of a different compound.

We’ll explore that problem in Question 5.

If you want to reverse the calculation and determine the percentage by mass of each element once you know a formula, see How to Calculate Element Percentages from Chemical Formulas.

06Experimental errors and how they affect the formula

A calculated ratio is only meaningful if the experiment has measured the correct mass gain.

Different experimental problems can push the result in different directions.

Why repeated heating can help

Suppose a sample is heated, cooled, and weighed. It is then heated again and weighed after cooling.

If the mass increases again, the original reaction was probably incomplete. More oxygen has combined with the remaining metal.

Repeating this process until the mass stops changing is called heating to constant mass.

It provides evidence that no further detectable mass change is occurring under the experimental conditions.

However, constant mass does not prove that the sample is a single pure oxide. It also cannot rule out solid product that escaped during an earlier heating.

For magnesium, there is an additional complication: magnesium can react with nitrogen in air to form magnesium nitride. A real experimental method must account for this possibility if the aim is to determine the composition of pure magnesium oxide.

The simple mass-gain calculation assumes that the only reaction contributing to the mass increase is the formation of the oxide.

07Questions and solutions

The following questions progress from straightforward mass-gain calculations to experimental evaluation and more demanding stoichiometric reasoning.

Question 1

A 1.308 g sample of zinc is completely converted into a single metal oxide. After heating, the final solid has a mass of 1.628 g.

Calculate the empirical formula of the oxide.

Use:

• Zn = \(65.4\ \mathrm{g\,mol^{-1}}\)
• O = \(16.0\ \mathrm{g\,mol^{-1}}\)

Solution 1

The empirical formula is \(\ce{ZnO}\).

The key is to calculate the oxygen mass from the difference in mass, then compare the moles of zinc and oxygen.

Step 1: Calculate the oxygen mass gain.

\[ m(\mathrm{O})=1.628-1.308 \]

\[ =0.320\ \mathrm{g} \]

Step 2: Calculate the moles of each element.

For zinc:

\[ n(\mathrm{Zn})=\frac{1.308}{65.4} \]

\[ =0.0200\ \mathrm{mol} \]

For oxygen:

\[ n(\mathrm{O})=\frac{0.320}{16.0} \]

\[ =0.0200\ \mathrm{mol} \]

Step 3: Determine the simplest ratio.

\[ \mathrm{Zn:O}=0.0200:0.0200 \]

\[ =1:1 \]

Therefore, the empirical formula is:

\[ \boxed{\ce{ZnO}} \]

There is one zinc atom for every oxygen atom in the oxide.

Question 2

A 2.374 g sample of tin is completely converted into a single oxide. The final product has a mass of 3.014 g.

Determine the empirical formula of the oxide.

Use:

• Sn = \(118.7\ \mathrm{g\,mol^{-1}}\)
• O = \(16.0\ \mathrm{g\,mol^{-1}}\)

A student argues that oxygen should be divided by 32.0 rather than 16.0 because oxygen enters the reaction as \(\ce{O2}\).

Explain whether this approach is necessarily incorrect.

Solution 2

The empirical formula is \(\ce{SnO2}\).

The student’s use of 32.0 is not necessarily incorrect, provided they remember that each oxygen molecule contains two oxygen atoms.

Step 1: Calculate the oxygen mass gain.

\[ m(\mathrm{O})=3.014-2.374 \]

\[ =0.640\ \mathrm{g} \]

Step 2: Calculate the moles of tin.

\[ n(\mathrm{Sn})=\frac{2.374}{118.7} \]

\[ =0.02000\ \mathrm{mol} \]

Step 3: Calculate the moles of oxygen atoms.

Using the atomic molar mass:

\[ n(\mathrm{O})=\frac{0.640}{16.0} \]

\[ =0.04000\ \mathrm{mol} \]

Step 4: Find the simplest ratio.

\[ \mathrm{Sn:O}=0.02000:0.04000 \]

\[ =1:2 \]

Therefore:

\[ \boxed{\ce{SnO2}} \]

Why can the student also use 32.0?

If the student calculates the amount of oxygen molecules:

\[ n(\ce{O2})=\frac{0.640}{32.0} \]

\[ =0.02000\ \mathrm{mol} \]

Each oxygen molecule contains two oxygen atoms, so:

\[ n(\mathrm{O})=2(0.02000) \]

\[ =0.04000\ \mathrm{mol} \]

This gives exactly the same result.

The mistake would be treating 0.02000 mol of \(\ce{O2}\) molecules as 0.02000 mol of oxygen atoms. That would halve the oxygen quantity and produce an incorrect formula.

Question 3

A student heats a sample of chromium in a ceramic crucible with a lid.

The following measurements are recorded:

Assume the chromium reacts completely to form one pure oxide, with no loss of solid and no change in the mass of the crucible.

Determine the empirical formula of the oxide.

Use:

• Cr = \(52.0\ \mathrm{g\,mol^{-1}}\)
• O = \(16.0\ \mathrm{g\,mol^{-1}}\)

Explain why the empty crucible mass must be accounted for when calculating the masses of the elements.

Solution 3

The empirical formula is \(\ce{Cr2O3}\).

The empty crucible and lid contribute to both initial and final measurements. Their mass must be removed before calculating the individual sample masses.

Step 1: Calculate the original chromium mass.

\[ m(\mathrm{Cr})=29.680-28.640 \]

\[ =1.040\ \mathrm{g} \]

Step 2: Calculate the final oxide mass.

\[ m(\mathrm{oxide})=30.160-28.640 \]

\[ =1.520\ \mathrm{g} \]

Step 3: Calculate the oxygen mass gained.

\[ m(\mathrm{O})=1.520-1.040 \]

\[ =0.480\ \mathrm{g} \]

Notice that we could also calculate the gain directly from the two total assembly readings:

\[ 30.160-29.680=0.480\ \mathrm{g} \]

The empty crucible mass cancels because it is unchanged.

Step 4: Convert both masses into moles.

For chromium:

\[ n(\mathrm{Cr})=\frac{1.040}{52.0} \]

\[ =0.0200\ \mathrm{mol} \]

For oxygen:

\[ n(\mathrm{O})=\frac{0.480}{16.0} \]

\[ =0.0300\ \mathrm{mol} \]

Step 5: Determine the simplest ratio.

\[ \mathrm{Cr:O}=0.0200:0.0300 \]

Dividing both values by 0.0100 gives:

\[ \mathrm{Cr:O}=2:3 \]

Therefore:

\[ \boxed{\ce{Cr2O3}} \]

The empirical formula is chromium(III) oxide, \(\ce{Cr2O3}\).

The important experimental point is that the crucible is not part of the reacting sample. Its mass must not be mistaken for the mass of the metal or oxide.

Question 4

Two students attempt to determine the empirical formula of a metal oxide.

The first student heats the metal in a completely open dish. Oxygen has easy access to the metal, but some fine oxide powder escapes during heating.

The second student heats the metal in a completely sealed crucible. No solid escapes, but the supply of oxygen inside the crucible is limited.

Both students calculate the formula using their measured mass gains.

Evaluate the reliability of the two methods.

Propose a revised experimental arrangement that addresses both problems. Explain one useful check that could be performed to assess whether oxidation is complete, and identify one limitation of that check.

Solution 4

Neither student’s result is automatically reliable. The first arrangement risks losing product, while the second risks incomplete oxidation.

The two problems affect the experiment in different ways.

The first student: open dish

Oxygen can freely enter the dish, which helps the reaction proceed.

However, fine oxide particles can escape during heating. These particles contain both metal and oxygen, so losing them decreases the final measured mass.

The measured mass gain is therefore smaller than the actual oxygen uptake.

This leads to an underestimate of the oxygen content and an incorrect mole ratio.

The second student: completely sealed crucible

The sealed crucible prevents solid product from escaping.

However, it also limits the supply of oxygen. Once the oxygen trapped inside has reacted, further oxidation may stop even if unreacted metal remains.

The final mass gain may therefore be smaller than expected.

The student could incorrectly interpret a partially oxidised sample as a pure oxide with a different metal-to-oxygen ratio.

A better experimental arrangement

A crucible with a lid kept slightly ajar addresses both problems.

Oxygen can enter through the gap, while the lid helps retain solid product.

A useful check

Heat the sample, allow it to cool, and weigh it. Repeat the heating and weighing process.

If the mass increases again, additional oxidation has occurred. The original reaction was incomplete.

Continue until the mass remains constant under the same conditions.

However, constant mass does not prove that the product is a single pure oxide. It also cannot establish that no powder escaped during an earlier heating.

Additional experimental evidence may be needed to establish the product’s actual composition.

Question 5

A 2.542 g sample of copper is heated to produce a solid oxide product with a final mass of 3.022 g.

The mass gain is therefore 0.480 g.

Use:

• Cu = \(63.55\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)

An independent analysis establishes that the final solid contains exactly two compounds:

• Copper(I) oxide, \(\ce{Cu2O}\)
• Copper(II) oxide, \(\ce{CuO}\)

There is no unreacted copper and no other material in the sample.

A student calculates the overall copper-to-oxygen atom ratio and obtains 4:3. They conclude that the product must be a single compound with the formula \(\ce{Cu4O3}\).

Determine:

(a) The total moles of copper and oxygen atoms in the sample.

(b) The number of moles of \(\ce{CuO}\) and \(\ce{Cu2O}\) formula units present.

(c) The ratio of \(\ce{CuO}\) to \(\ce{Cu2O}\) formula units.

(d) The fraction of the total copper atoms contained in each oxide.

(e) Explain why the overall empirical ratio does not establish that \(\ce{Cu4O3}\) is a single compound.

Solution 5

The sample contains 0.02000 mol of \(\ce{CuO}\) and 0.01000 mol of \(\ce{Cu2O}\), giving a formula-unit ratio of 2:1.

The tempting answer, \(\ce{Cu4O3}\), describes the overall atom ratio, but it does not describe the identity of either compound in the mixture.

This question requires us to distinguish the composition of the entire sample from the composition of each individual oxide.

Part (a): Calculate the total moles of copper and oxygen atoms.

The original copper mass is 2.542 g.

\[ n(\mathrm{Cu})=\frac{2.542}{63.55} \]

\[ =0.04000\ \mathrm{mol} \]

The oxygen mass gain is:

\[ m(\mathrm{O})=3.022-2.542 \]

\[ =0.480\ \mathrm{g} \]

Therefore:

\[ n(\mathrm{O})=\frac{0.480}{16.00} \]

\[ =0.03000\ \mathrm{mol} \]

The overall mole ratio is:

\[ \mathrm{Cu:O}=0.04000:0.03000 \]

Dividing by 0.01000:

\[ \mathrm{Cu:O}=4:3 \]

If the sample were known to contain only one pure compound, this would give the empirical formula \(\ce{Cu4O3}\).

But we have additional information: the sample contains a mixture of \(\ce{CuO}\) and \(\ce{Cu2O}\).

We must use that information.

Part (b): Calculate the moles of each oxide.

Let:

• \(x\) = moles of \(\ce{CuO}\) formula units.
• \(y\) = moles of \(\ce{Cu2O}\) formula units.

Now consider how many atoms each formula unit contains.

One mole of \(\ce{CuO}\) contains:

• 1 mol of copper atoms.
• 1 mol of oxygen atoms.

One mole of \(\ce{Cu2O}\) contains:

• 2 mol of copper atoms.
• 1 mol of oxygen atoms.

We can therefore construct two equations.

From the total copper content:

\[ x+2y=0.04000 \]

From the total oxygen content:

\[ x+y=0.03000 \]

Subtract the second equation from the first:

\[ (x+2y)-(x+y) \]

\[ 0.04000-0.03000=y \]

Therefore:

\[ y=0.01000\ \mathrm{mol} \]

Substitute this into the oxygen equation:

\[ x+0.01000=0.03000 \]

\[ x=0.02000\ \mathrm{mol} \]

The sample contains:

\[ \boxed{0.02000\ \mathrm{mol\ CuO}} \]

and:

\[ \boxed{0.01000\ \mathrm{mol\ Cu2O}} \]

Part (c): Determine the formula-unit ratio.

The ratio is:

\[ \ce{CuO:Cu2O}=0.02000:0.01000 \]

Dividing both values by 0.01000:

\[ \boxed{\ce{CuO:Cu2O}=2:1} \]

There are two formula units of copper(II) oxide for every one formula unit of copper(I) oxide, when counted on a mole basis.

Part (d): Calculate the fraction of copper atoms in each oxide.

This is where the distinction between formula units and atoms becomes important.

Each formula unit of \(\ce{CuO}\) contains one copper atom.

Therefore, the amount of copper in \(\ce{CuO}\) is:

\[ n(\mathrm{Cu\ in\ CuO})=0.02000\ \mathrm{mol} \]

Each formula unit of \(\ce{Cu2O}\) contains two copper atoms.

Therefore:

\[ n(\mathrm{Cu\ in\ Cu2O})=2(0.01000) \]

\[ =0.02000\ \mathrm{mol} \]

The total copper content is 0.04000 mol.

The fraction in \(\ce{CuO}\) is:

\[ \frac{0.02000}{0.04000}=0.50 \]

The fraction in \(\ce{Cu2O}\) is also:

\[ \frac{0.02000}{0.04000}=0.50 \]

Therefore, 50% of the copper atoms are in \(\ce{CuO}\), and 50% are in \(\ce{Cu2O}\).

Notice that the formula-unit ratio is 2:1, but the copper atom distribution is 50:50.

These are different ratios because each formula unit contains a different number of copper atoms.

Part (e): Explain why \(\ce{Cu4O3}\) is not established as a single compound.

The overall atom ratio is indeed:

\[ \mathrm{Cu:O}=4:3 \]

However, this ratio describes the entire sample, which contains two separate oxide compounds.

It does not prove that a single compound with that composition exists.

The two oxides have their own individual formulas:

\[ \ce{CuO} \]

and:

\[ \ce{Cu2O} \]

Their combined composition happens to produce an overall copper-to-oxygen ratio of 4:3.

We can verify the result using the total mass of the two oxides.

For copper(II) oxide:

\[ M(\ce{CuO})=63.55+16.00 \]

\[ =79.55\ \mathrm{g\,mol^{-1}} \]

Its mass is:

\[ m(\ce{CuO})=0.02000(79.55) \]

\[ =1.591\ \mathrm{g} \]

For copper(I) oxide:

\[ M(\ce{Cu2O})=2(63.55)+16.00 \]

\[ =143.10\ \mathrm{g\,mol^{-1}} \]

Its mass is:

\[ m(\ce{Cu2O})=0.01000(143.10) \]

\[ =1.431\ \mathrm{g} \]

The combined mass is:

\[ 1.591+1.431=3.022\ \mathrm{g} \]

This matches the measured final product mass exactly.

The important conclusion is that a bulk empirical ratio describes the overall composition, not necessarily the identity of an individual compound.

That distinction becomes essential whenever experimental data might represent a mixture rather than a single pure substance.

07What to remember when finding an oxide formula

When you encounter a metal oxide mass-gain question, follow this sequence:

1. Find the oxygen mass by subtracting the original metal mass from the final oxide mass.
2. Convert the original metal mass into moles using its molar mass.
3. Convert the oxygen mass into moles using \(16.00\ \mathrm{g\,mol^{-1}}\).
4. Divide both mole quantities by the smaller value.
5. Multiply to obtain the simplest whole-number ratio.
6. Check that your rounding is consistent with the precision of the measurements.
7. State the empirical formula, while considering whether the experimental evidence supports a single pure oxide.

The next useful step is to check whether the resulting formula is chemically consistent with the metal’s oxidation state and the charge of the oxide ion, \(\mathrm{O}^{2-}\).

For example, a metal forming \(\mathrm{M}^{2+}\) would normally combine with oxide ions in a 1:1 ratio. If your experimental result suggests a different ratio, consider whether the metal has variable oxidation states, whether the product contains multiple oxide phases, or whether the experiment has introduced an error.

The mole ratio gives you the composition. Chemical reasoning and experimental evidence help establish what substance you have actually produced.