How to Find an Empirical Formula from Percentage Composition

An unknown iron oxide contains 69.9% iron and 30.1% oxygen by mass. Would you predict that it contains more iron atoms than oxygen atoms?

It actually contains more oxygen atoms. Iron atoms are much heavier, so a larger mass of iron doesn’t necessarily mean a larger number of atoms.

To find an empirical formula from percentage composition, use this method:

1. Assume you have 100 g of the compound, so each percentage becomes a mass in grams.
2. Convert each element’s mass into moles using its atomic molar mass.
3. Divide every mole amount by the smallest mole amount.
4. If the resulting ratio contains fractions, multiply every part by the same small whole number until you obtain the simplest whole-number ratio.
5. Write the empirical formula using those numbers as subscripts.

The key is that percentages describe mass, while an empirical formula describes the simplest ratio of atoms. Moles connect the two.

01What does an experiment actually measure?

Suppose a chemist wants to determine the composition of an oxide formed when iron reacts with oxygen.

A simple approach would be to weigh some iron, heat it in air, and weigh the product. If oxygen combines with the iron, what should happen to the mass?

It should increase. Oxygen atoms from the surrounding air become part of the solid, adding mass to the original iron.

The apparatus and measurement

Imagine this experimental arrangement:

• A dry, preweighed ceramic crucible holds a known mass of iron powder.
• The crucible sits on a clay triangle above a Bunsen burner.
• A lid rests slightly ajar, allowing oxygen from the air to reach the heated powder while helping prevent solid material from escaping.
• After heating, the crucible is allowed to cool and is weighed on a balance.
• Heating, cooling, and weighing are repeated until the mass becomes approximately constant.

Why not simply heat the powder once in an uncovered dish?

That first design has two problems. Some powder might escape during heating, and some iron might remain unreacted. Either problem would distort the calculated oxygen content.

A partially covered crucible reduces material loss. Repeated heating and weighing help establish whether the reaction has stopped producing a measurable mass change.

Neither improvement proves that the solid contains only one iron oxide. Different oxides can form under different conditions, and constant mass alone does not establish chemical purity.

An illustrative set of measurements

Suppose the following readings are obtained. These are invented values for demonstrating the method, not results from a performed experiment.

Assuming no iron escaped, the product contains only iron and oxygen, and the entire mass gain came from oxygen:

\[
\begin{aligned}
m_{\mathrm O}
&=2.000-1.398\\
&=0.602\ \mathrm{g}
\end{aligned}
\]

Here, \(m_{\mathrm O}\) represents the mass of oxygen incorporated into the solid.

The percentage composition of the product is:

\[
\begin{aligned}
\%\mathrm{Fe}
&=\frac{1.398}{2.000}\times100\\
&=69.9\%\\[4pt]
\%\mathrm O
&=\frac{0.602}{2.000}\times100\\
&=30.1\%
\end{aligned}
\]

Both numerator and denominator are masses in grams, so the units cancel.

The balance tells us how much mass was gained. It doesn’t tell us directly how many oxygen atoms joined each iron atom.

That’s the next calculation.

02From percentage composition to an empirical formula

03How do you know when to round a mole ratio?

The hardest part of many empirical formula questions is deciding what to do with a decimal.

Not every decimal should be rounded to the nearest whole number.

Consider these results:

Why multiplying works but ordinary rounding can fail

Imagine a ratio of 1 iron atom for every 1.5 oxygen atoms.

One and a half atoms isn’t a valid atom count for a formula unit. But the ratio can be expressed using whole atoms by doubling both parts.

Two iron atoms for every three oxygen atoms preserve exactly the same proportion.

Rounding 1.5 to 2 would not preserve it.

The same logic applies to thirds:

\[
1:1.333\ldots
\quad\longrightarrow\quad
3:4
\]

How close is close enough?

Suppose your calculated ratio is \(1:1.503\).

The intended value of 1.5 is strongly supported, especially when the original percentages are given to one decimal place.

But a ratio of \(1:1.47\) needs more care.

It might reflect experimental uncertainty, contamination, or incorrect assumptions about the sample. It might also indicate that the proposed formula is wrong.

There is no universal rule that says to round anything within a particular decimal distance.

A more reliable approach is to calculate the percentage composition predicted by your proposed formula and compare it with the reported percentages.

If the data represent exact theoretical values rounded to one decimal place, each predicted percentage should agree when rounded to one decimal place. If the values are experimental, consider the measurement uncertainty as well.

Don’t invent precision the measurements don’t provide.

05What an empirical formula can and cannot identify

Suppose two different molecular substances have the formulas \(\ce{C2H2}\) and \(\ce{C6H6}\).

Both simplify to:

\[
\ce{CH}
\]

They have identical percentage compositions by mass. Yet they are different substances with different structures and molar masses.

The first could be ethyne, with a carbon-carbon triple bond. The second could be benzene, with a six-carbon ring.

This is why an empirical formula should not automatically be called a molecular formula.

For a pure molecular substance, an independently determined molar mass can help find the molecular formula:

\[
k=\frac{M_{\mathrm{molecule}}}
{M_{\mathrm{empirical}}}
\]

Here:

• \(k\) is the positive whole-number multiplier.
• \(M_{\mathrm{molecule}}\) is the molecular molar mass in \(\mathrm{g\,mol^{-1}}\).
• \(M_{\mathrm{empirical}}\) is the molar mass calculated from the empirical formula, in the same units.

You multiply every subscript in the empirical formula by \(k\).

However, even the molecular formula may not uniquely identify a substance. Structural isomers can have exactly the same molecular formula.

For ionic solids, the empirical formula normally represents the simplest ratio of ions in the crystal. There may be no separate molecules to identify.

There is one more important limitation: the calculation assumes the percentage composition belongs to a single pure compound.

A mixture can sometimes produce a perfectly neat whole-number elemental ratio. That ratio describes the overall sample, but it does not prove that one compound with that formula is present.

06Questions and solutions

These questions use hypothetical samples and measurements. Assume the stated compositions are by mass. Use the atomic molar masses supplied, and retain extra digits during calculations before rounding your results.

Question 1

A pure zinc oxide contains 80.3% zinc and 19.7% oxygen by mass.

Determine its empirical formula.

Use:

• Zn = \(65.38\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)

Solution 1

The empirical formula is \(\ce{ZnO}\), because the calculated amounts of zinc and oxygen are approximately equal.

Assume a 100 g sample. It contains 80.3 g Zn and 19.7 g O.

Using \(n=m/M\):

\[
\begin{aligned}
n_{\mathrm{Zn}}
&=\frac{80.3}{65.38}\\
&=1.228\ \mathrm{mol}\\[4pt]
n_{\mathrm O}
&=\frac{19.7}{16.00}\\
&=1.231\ \mathrm{mol}
\end{aligned}
\]

The masses are in grams and the molar masses are in \(\mathrm{g\,mol^{-1}}\).

Divide both mole amounts by \(1.228\ \mathrm{mol}\):

\[
\mathrm{Zn:O}\approx1.000:1.002
\]

This is effectively \(1:1\).

The result means there is one zinc ion for every oxide ion in the simplest formula-unit ratio.

The tempting mistake is to use \(80.3:19.7\) directly as an atom ratio. That is a mass ratio, not a mole ratio.

Question 2

A 5.00 g sample of an unknown pure tin oxide, labelled oxide A, contains:

• 88.1% tin
• 11.9% oxygen

The sample is heated in excess oxygen until it forms another single pure tin oxide, oxide B.

The final solid has a mass of 5.59 g.

Assume that no tin is lost, and oxygen is the only substance added.

Determine:

a) The empirical formula of oxide A.

b) The empirical formula of oxide B.

c) The percentage by mass of oxygen in oxide B.

d) Why adding the original oxygen percentage to the percentage mass gain would give an incorrect answer.

Use Sn = \(118.71\ \mathrm{g\,mol^{-1}}\) and O = \(16.00\ \mathrm{g\,mol^{-1}}\).

Solution 2

Oxide A is \(\ce{SnO}\), oxide B is \(\ce{SnO2}\), and oxide B contains approximately 21.2% oxygen by mass.

The key is that the amount of tin remains constant while oxygen is added. The total mass also changes, so the oxygen percentage must be recalculated using the final mass.

First, calculate the original elemental masses:

\[
\begin{aligned}
m_{\mathrm{Sn}}
&=5.00(0.881)\\
&=4.405\ \mathrm{g}\\[4pt]
m_{\mathrm{O,A}}
&=5.00(0.119)\\
&=0.595\ \mathrm{g}
\end{aligned}
\]

Convert to moles:

\[
\begin{aligned}
n_{\mathrm{Sn}}
&=\frac{4.405}{118.71}\\
&=0.03711\ \mathrm{mol}\\[4pt]
n_{\mathrm{O,A}}
&=\frac{0.595}{16.00}\\
&=0.03719\ \mathrm{mol}
\end{aligned}
\]

The ratio is approximately \(1:1\), so oxide A has empirical formula \(\ce{SnO}\).

Next, calculate the mass of oxygen added during heating:

\[
\begin{aligned}
m_{\mathrm{added}}
&=5.59-5.00\\
&=0.59\ \mathrm{g}
\end{aligned}
\]

The mass of oxygen in oxide B is:

\[
\begin{aligned}
m_{\mathrm{O,B}}
&=0.595+0.590\\
&=1.185\ \mathrm{g}
\end{aligned}
\]

The number of moles of tin is unchanged at \(0.03711\ \mathrm{mol}\).

The new oxygen amount is:

\[
\begin{aligned}
n_{\mathrm{O,B}}
&=\frac{1.185}{16.00}\\
&=0.07406\ \mathrm{mol}
\end{aligned}
\]

The final mole ratio is:

\[
\begin{aligned}
\mathrm{Sn:O}
&=0.03711:0.07406\\
&\approx1:2
\end{aligned}
\]

Therefore, oxide B is \(\ce{SnO2}\).

Its oxygen percentage is:

\[
\begin{aligned}
\%\mathrm{O}
&=\frac{1.185}{5.59}\times100\\
&=21.2\%
\end{aligned}
\]

For part (d), adding \(11.9\%\) to the percentage mass gain is incorrect because those percentages use different starting totals.

The original 11.9% refers to a 5.00 g sample. The final oxygen percentage refers to a 5.59 g sample.

The change in total mass changes the denominator.

This is also why an oxidation experiment can increase the oxygen mass without increasing its percentage by the same number of percentage points.

Question 3

A pure gaseous hydrocarbon contains 85.6% carbon and 14.4% hydrogen by mass.

Its independently measured molar mass is \(42.1\ \mathrm{g\,mol^{-1}}\).

Two possible identities are proposed:

• Propene, which contains a carbon-carbon double bond.
• Cyclopropane, which contains a three-carbon ring with only single carbon-carbon bonds.

a) Determine the empirical formula.

b) Determine the molecular formula.

c) Decide whether the percentage composition and molar mass distinguish the two proposed substances.

d) State what additional structural information would resolve the identity.

Use C = \(12.01\ \mathrm{g\,mol^{-1}}\) and H = \(1.008\ \mathrm{g\,mol^{-1}}\).

Solution 3

The empirical formula is \(\ce{CH2}\), and the molecular formula is \(\ce{C3H6}\). Both propene and cyclopropane satisfy these results, so the measurements do not uniquely identify the substance.

Assume a 100 g sample.

\[
\begin{aligned}
n_{\mathrm C}
&=\frac{85.6}{12.01}\\
&=7.127\ \mathrm{mol}\\[4pt]
n_{\mathrm H}
&=\frac{14.4}{1.008}\\
&=14.286\ \mathrm{mol}
\end{aligned}
\]

Dividing by \(7.127\ \mathrm{mol}\):

\[
\mathrm{C:H}\approx1:2.004
\]

This gives the simplest ratio \(1:2\), so the empirical formula is \(\ce{CH2}\).

Now find its empirical formula mass:

\[
\begin{aligned}
M_{\mathrm{empirical}}
&=12.01+2(1.008)\\
&=14.026\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The multiplier is:

\[
\begin{aligned}
k&=\frac{42.1}{14.026}\\
&\approx3.00
\end{aligned}
\]

Multiplying both empirical subscripts by 3 gives:

\[
\ce{C3H6}
\]

That answers parts (a) and (b).

For part (c), both proposed substances have the molecular formula \(\ce{C3H6}\), despite their different bonding arrangements.

The tempting conclusion is that knowing the molecular formula establishes the molecular identity. It doesn’t.

For part (d), an additional investigation needs to distinguish a carbon-carbon double bond from a three-membered carbon ring.

Evidence establishing the presence of a C=C bond would support propene. Evidence showing a three-carbon ring containing only single C-C bonds would support cyclopropane.

Question 4

A hydrated sample of copper(II) sulfate contains 25.5% copper by mass.

A 2.50 g sample is heated under conditions known to remove all water of crystallisation without decomposing the salt.

After heating, 1.60 g of pure anhydrous \(\ce{CuSO4}\) remains.

a) Determine the formula of the original hydrate, written \(\ce{CuSO4}\cdot x\ce{H2O}\), where \(x\) is a positive whole number.

b) Calculate the copper percentage predicted by your formula and compare it with the reported value.

c) Explain why elemental percentages alone do not directly establish that hydrogen and oxygen occur as water of crystallisation.

Use:

• Cu = \(63.55\ \mathrm{g\,mol^{-1}}\)
• S = \(32.06\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)
• H = \(1.008\ \mathrm{g\,mol^{-1}}\)

Solution 4

The original hydrate is \(\ce{CuSO4}\cdot5\ce{H2O}\). Its predicted copper composition is approximately 25.5%, consistent with the independent percentage measurement.

First, determine the mass of water lost:

\[
\begin{aligned}
m_{\mathrm{water}}
&=2.50-1.60\\
&=0.90\ \mathrm{g}
\end{aligned}
\]

The molar mass of anhydrous copper(II) sulfate is:

\[
\begin{aligned}
M(\ce{CuSO4})
&=63.55+32.06+4(16.00)\\
&=159.61\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The amount of anhydrous salt remaining is:

\[
\begin{aligned}
n(\ce{CuSO4})
&=\frac{1.60}{159.61}\\
&=0.01002\ \mathrm{mol}
\end{aligned}
\]

The molar mass of water is:

\[
\begin{aligned}
M(\ce{H2O})
&=2(1.008)+16.00\\
&=18.016\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
n(\ce{H2O})
&=\frac{0.90}{18.016}\\
&=0.04996\ \mathrm{mol}
\end{aligned}
\]

Divide the water amount by the anhydrous salt amount:

\[
\begin{aligned}
x&=\frac{0.04996}{0.01002}\\
&\approx4.98\\
&\approx5
\end{aligned}
\]

The slight departure from 5 comes from the precision of the supplied masses.

The hydrate formula is:

\[
\ce{CuSO4}\cdot5\ce{H2O}
\]

For part (b), calculate its molar mass:

\[
\begin{aligned}
M_{\mathrm{hydrate}}
&=159.61+5(18.016)\\
&=249.69\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The percentage of copper is:

\[
\begin{aligned}
\%\mathrm{Cu}
&=\frac{63.55}{249.69}\times100\\
&=25.45\%\\
&\approx25.5\%
\end{aligned}
\]

This agrees with the reported 25.5%.

For part (c), elemental percentages reveal how much copper, sulfur, oxygen, and hydrogen are present. They don’t directly establish which atoms belong to sulfate ions and which belong to water molecules.

The heating experiment supplies different information: under the stated conditions, the lost mass is known to be water, and the residue is known to be anhydrous copper(II) sulfate.

Together, those observations support the hydrate formula.

If heating had also decomposed the salt, treating the entire mass loss as water would be unjustified. Reaching constant mass would not fix that mistaken assumption.

Question 5

An unknown iron oxide powder contains:

• 72.36% iron
• 27.64% oxygen

The sample contains no elements other than iron and oxygen.

A chemist calculates the empirical formula \(\ce{Fe3O4}\) and concludes that the sample is pure magnetite.

Another chemist proposes a different explanation: the powder could be a physical mixture of two separate crystalline substances, iron(II) oxide (\(\ce{FeO}\)) and iron(III) oxide (\(\ce{Fe2O3}\)).

To investigate, a 100.00 g portion is heated in excess oxygen until all its iron has been converted into pure \(\ce{Fe2O3}\). The final mass is 103.45 g.

Assume no material is lost, treat \(\ce{FeO}\) as having the ideal formula shown, and use:

• Fe = \(55.85\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)

Answer the following:

a) Calculate the overall simplest iron-to-oxygen mole ratio.

b) Determine whether a mixture of \(\ce{FeO}\) and \(\ce{Fe2O3}\) could produce exactly that ratio. If so, find the required mole ratio of the two oxides.

c) Calculate the expected mass after complete oxidation to \(\ce{Fe2O3}\). Does the measured mass of 103.45 g distinguish the competing models?

d) In the usual oxidation-state description of magnetite, one iron ion is \(\mathrm{Fe}^{2+}\) and two are \(\mathrm{Fe}^{3+}\) for every \(\ce{Fe3O4}\) formula unit. Would measuring the overall proportion of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) settle the disagreement?

e) What type of additional evidence would actually distinguish the two proposed models?

Solution 5

Neither the elemental percentages nor the measured oxygen uptake prove that the powder is pure magnetite. A particular mixture of idealised \(\ce{FeO}\) and \(\ce{Fe2O3}\) gives the same bulk composition, the same final mass after oxidation, and even the same overall proportions of iron oxidation states.

The decisive distinction is between the composition of the entire sample and the identity of the crystalline phase or phases within it.

Part (a): Determine the overall atom ratio.

Assume a 100 g sample.

\[
\begin{aligned}
n_{\mathrm{Fe}}
&=\frac{72.36}{55.85}\\
&=1.2956\ \mathrm{mol}\\[4pt]
n_{\mathrm O}
&=\frac{27.64}{16.00}\\
&=1.7275\ \mathrm{mol}
\end{aligned}
\]

Dividing by the smaller amount:

\[
\mathrm{Fe:O}\approx1:1.333
\]

Multiplying both parts by 3 gives:

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

The overall elemental ratio is consistent with \(\ce{Fe3O4}\).

But that alone does not establish the presence of a single pure \(\ce{Fe3O4}\) phase.

Part (b): Test the mixture model.

Let \(a\) be the amount of \(\ce{FeO}\), and \(b\) the amount of \(\ce{Fe2O3}\), each in moles of formula units.

Each mole of \(\ce{FeO}\) contributes 1 mol Fe atoms and 1 mol O atoms.

Each mole of \(\ce{Fe2O3}\) contributes 2 mol Fe atoms and 3 mol O atoms.

The total amounts of atoms are therefore:

\[
\begin{aligned}
n_{\mathrm{Fe}}&=a+2b\\
n_{\mathrm O}&=a+3b
\end{aligned}
\]

For a \(3:4\) iron-to-oxygen ratio:

\[
\frac{a+2b}{a+3b}=\frac34
\]

Cross-multiplying:

\[
\begin{aligned}
4a+8b&=3a+9b\\
a&=b
\end{aligned}
\]

So an equimolar mixture of \(\ce{FeO}\) and \(\ce{Fe2O3}\) produces exactly the same overall atom ratio as \(\ce{Fe3O4}\).

For every 1 mol of each oxide, the total is 3 mol Fe atoms and 4 mol O atoms.

The tempting conclusion was that obtaining the exact ratio \(3:4\) establishes magnetite. The mixture calculation shows why that conclusion is not justified.

Part (c): Test the mass gain.

After complete oxidation, every two iron atoms correspond to three oxygen atoms in \(\ce{Fe2O3}\).

The final oxygen amount must therefore be 1.5 times the amount of iron atoms:

\[
\begin{aligned}
n_{\mathrm{O,final}}
&=1.5(1.2956)\\
&=1.9434\ \mathrm{mol}
\end{aligned}
\]

The original oxygen amount was \(1.7275\ \mathrm{mol}\).

The extra oxygen required is:

\[
\begin{aligned}
\Delta n_{\mathrm O}
&=1.9434-1.7275\\
&=0.2159\ \mathrm{mol}
\end{aligned}
\]

Converting this to mass:

\[
\begin{aligned}
\Delta m
&=0.2159(16.00)\\
&\approx3.45\ \mathrm{g}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
m_{\mathrm{final}}
&=100.00+3.45\\
&=103.45\ \mathrm{g}
\end{aligned}
\]

This matches the observed final mass.

But the result does not distinguish the two models.

Both start with the same total amounts of iron and oxygen. Both finish with those iron atoms incorporated into pure \(\ce{Fe2O3}\).

Conservation of mass requires the same amount of additional oxygen, regardless of how the original atoms were distributed between crystalline phases.

The mass-gain measurement looks like a second confirmation, but it is actually another consequence of the same overall elemental composition.

Part (d): Would oxidation-state measurements help?

Not necessarily.

In the idealised ionic description of pure magnetite, the ratio is:

\[
\mathrm{Fe}^{2+}:\mathrm{Fe}^{3+}=1:2
\]

Now consider the equimolar mixture.

• One mole of \(\ce{FeO}\) contributes 1 mol of \(\mathrm{Fe}^{2+}\).
• One mole of \(\ce{Fe2O3}\) contributes 2 mol of \(\mathrm{Fe}^{3+}\).

The mixture also has:

\[
\mathrm{Fe}^{2+}:\mathrm{Fe}^{3+}=1:2
\]

Even a perfectly accurate measurement of the overall oxidation-state proportions would give the same answer for both models.

This is the deeper problem. The samples can agree in elemental proportions, oxidation-state proportions, and total oxygen uptake while still containing different solid phases.

Part (e): What would settle the disagreement?

We need information about how the atoms are arranged within the solid, rather than another measurement of their overall amounts.

For example, powder X-ray diffraction is a phase-sensitive technique (an extension beyond the calculation required here). Under suitable conditions, it can distinguish diffraction patterns associated with a single magnetite phase from patterns containing separate iron(II) oxide and iron(III) oxide phases.

The exact technique is less important than the information it provides.

Elemental composition measures how much of each element exists in the sample. Phase-sensitive evidence tests whether those atoms belong to one crystalline substance or multiple substances.

The answer to the original problem is therefore precise: the sample has an overall Fe:O ratio of \(3:4\), but the available bulk measurements do not uniquely establish its chemical phase identity.

07What to do after finding the empirical formula

Once you obtain a simplest whole-number ratio, make two decisions before treating the problem as finished.

First, check whether your formula reproduces the reported percentage composition to the precision supported by the data. Working backwards from a proposed formula is covered in How to Calculate Element Percentages from Chemical Formulas.

Second, check what the question actually asks you to identify. If it asks only for the empirical formula of a pure compound, the simplest atom ratio is the required result. If it asks for the molecular formula, you may need an independently measured molar mass. If it asks for the identity of an unknown solid, you may also need evidence about purity, ions, bonding, or crystal structure.

For a broader treatment of the relationship between these calculations, see Percentage Composition and Empirical Formulae: HSC Chemistry.

The calculation converts mass information into an atom ratio. Knowing what that ratio can prove, and what it cannot, is what makes the result chemically meaningful.