How to Calculate Element Percentages from Chemical Formulas

Two compounds have the formulas \(\ce{CH2O}\) and \(\ce{C2H4O2}\). The second has twice as many carbon atoms per molecule. Would you expect it to have twice the percentage of carbon by mass?

It doesn’t. Both have exactly the same elemental percentages because their carbon, hydrogen, and oxygen counts increase in the same proportion. This is the key to calculating percentages from chemical formulas: you’re interested in how much each element contributes to the total mass, not simply how many atoms are present.

To calculate the percentage by mass of an element in a pure compound:

1. Count the atoms of that element in the chemical formula.
2. Multiply that count by the element’s atomic molar mass.
3. Calculate the molar mass of the entire formula.
4. Divide the element’s mass contribution by the total, then multiply by 100.

The calculation is:

\[
\%\mathrm{E}=\frac{n_{\mathrm{E}}M_{\mathrm{E}}}{M}\times100
\]

Here:

• \(\mathrm{E}\) is the element being investigated.
• \(n_{\mathrm{E}}\) is the number of its atoms in one formula unit.
• \(M_{\mathrm{E}}\) is its atomic molar mass, in \(\mathrm{g\,mol^{-1}}\).
• \(M\) is the molar mass of the entire compound, also in \(\mathrm{g\,mol^{-1}}\).

The molar-mass units cancel, leaving a dimensionless percentage. This is also called elemental percentage composition by mass.

01Why the mole ratio determines the percentage

Consider \(\ce{C2H4O2}\), the molecular formula of ethanoic acid.

One molecule contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. That means one mole of its molecules contains:

• 2 mol of carbon atoms
• 4 mol of hydrogen atoms
• 2 mol of oxygen atoms

A mole is a counting unit, just like a dozen, except that one mole contains approximately \(6.022\times10^{23}\) particles.

The useful part is that mole ratios let us turn the subscripts in a chemical formula into masses.

Using these atomic molar masses:

For carbon:

\[
\%\mathrm{C}=\frac{24.02}{60.052}\times100
=40.0\%
\]

For hydrogen:

\[
\%\mathrm{H}=\frac{4.032}{60.052}\times100
=6.71\%
\]

For oxygen:

\[
\%\mathrm{O}=\frac{32.00}{60.052}\times100
=53.3\%
\]

These percentages describe the mass of each element in pure \(\ce{C2H4O2}\). They do not describe the percentage of atoms.

The surprising comparison: CH2O versus C2H4O2

Now compare the carbon percentages.

Doubling the formula doubles the carbon mass contribution and the total molar mass. Their ratio stays unchanged.

Both formulas reduce to the same simplest atom ratio:

\[
\mathrm{C:H:O}=1:2:1
\]

That simplest whole-number ratio is the empirical formula, \(\ce{CH2O}\).

But there’s an important limit to what the ratio tells us.

• \(\ce{CH2O}\) can represent the molecular formula of methanal.
• \(\ce{C2H4O2}\) can represent the molecular formula of ethanoic acid.
• Other compounds can share the same molecular formula while having different atomic arrangements.

Elemental percentages alone cannot distinguish them.

The same caution applies to ionic compounds. A formula such as \(\ce{Mg3(PO4)2}\) describes the simplest ratio of ions in a crystal, not a separate molecule containing exactly three magnesium ions and two phosphate ions.

03How could an experiment test a calculated percentage?

A chemical formula predicts the composition of a pure compound. But suppose the formula is unknown.

How could a chemist measure the elemental percentages and work backwards?

Imagine a small sample containing only carbon, hydrogen, and oxygen.

A first attempt might be to weigh the sample, burn it in an open dish, and weigh whatever remains.

What should the mass loss tell us?

Unfortunately, not enough. Carbon and hydrogen can leave as gaseous products, oxygen from the air enters the reaction, and combustion might be incomplete. A change in the sample’s mass cannot separately identify its original carbon, hydrogen, and oxygen contents.

A better design captures the combustion products separately.

A controlled combustion apparatus

The apparatus works in this order.

1. Dry oxygen enters a heated combustion tube containing the weighed sample.
2. The sample burns, ideally converting all its carbon into \(\ce{CO2}\) and all its hydrogen into \(\ce{H2O}\).
3. The gases pass through a water-absorbing trap. Its increase in mass measures the water collected.
4. The remaining gases pass through a carbon dioxide absorber. Its increase in mass measures the carbon dioxide collected.
5. The gases leave through the outlet.

Both absorbers are weighed before and after the experiment.

Why collect water first? Because the downstream carbon dioxide absorbent may also retain water. If water reaches that absorber, its mass gain no longer represents carbon dioxide alone.

Dry incoming oxygen, complete combustion, leak-free connections, and complete collection are necessary assumptions. These are experimental controls, not facts guaranteed simply by drawing the apparatus.

The gas products also contain oxygen supplied from outside the original sample. We must account for that when interpreting their masses.

04Where formula calculations and experiments can disagree

If a sample is genuinely pure and its formula is correct, its measured elemental composition should agree with the calculated values within experimental uncertainty.

When they don’t agree, don’t immediately rewrite the molecular formula.

Consider what the experiment actually measured.

The formula method assumes a known, pure chemical composition. The experimental method depends on the reliability of the measurements and assumptions.

Those are two different sources of uncertainty.

05How much rounding is appropriate?

In HSC calculations, premature rounding can turn a straightforward mole ratio into something confusing.

A useful rule is to keep several extra digits during the working, then round only the final percentage to the precision justified by the question.

For example:

\[
\frac{128.00}{262.87}\times100
=48.693270\ldots\%
\]

Reporting 48.7% to three significant figures is sensible when the given data support that precision.

For the combustion example, the \(0.600\ \mathrm{g}\), \(0.360\ \mathrm{g}\), and \(0.880\ \mathrm{g}\) measurements have three significant figures. Reporting the elemental percentages to three significant figures is therefore appropriate.

You might notice:

\[
40.0+6.71+53.3=100.01\%
\]

That tiny difference comes from rounding. It isn’t evidence of an extra element or an error in mass conservation.

Don’t adjust one percentage arbitrarily to force an exact total. Keep unrounded numbers for subsequent calculations.

For mole ratios, especially when finding an empirical formula, retain enough precision to recognise small whole-number ratios. A measured ratio slightly different from \(1:2:1\) may reflect experimental uncertainty rather than a genuinely complicated formula.

For a broader explanation of composition calculations involving pure compounds and mixtures, see Percentage Composition by Mass for HSC Chemistry.

06Questions and solutions

These questions move from applying the formula directly to interpreting experimental evidence, competing explanations, and the limits of what percentage composition can prove.

Unless otherwise stated, assume the substances are pure when described as pure, use the atomic molar masses supplied, and round calculated percentages appropriately.

Question 1

Calculate the percentage by mass of aluminium in aluminium oxide, \(\ce{Al2O3}\), to three significant figures.

Use:

• Al = \(26.98\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)

Solution 1

Aluminium makes up 52.9% by mass of pure aluminium oxide.

There are two aluminium atoms and three oxygen atoms per formula unit.

The total molar mass is:

\[
\begin{aligned}
M&=2(26.98)+3(16.00)\\
&=101.96\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The aluminium contribution is:

\[
2(26.98)=53.96\ \mathrm{g\,mol^{-1}}
\]

Therefore:

\[
\begin{aligned}
\%\mathrm{Al}&=\frac{53.96}{101.96}\times100\\
&=52.9\%
\end{aligned}
\]

Two of the five atoms in the formula-unit ratio are aluminium, but that does not make aluminium 40% of the mass.

The calculation uses atomic masses, not atom counts alone.

Question 2

A student heats \(5.000\ \mathrm{g}\) of pure hydrated iron(II) sulfate, \(\ce{FeSO4*7H2O}\).

After the first heating, the residue has a mass of \(3.000\ \mathrm{g}\).

Assume that:

• Only water is lost during heating.
• No iron(II) sulfate decomposes or escapes.
• The residue contains iron(II) sulfate and any water not yet removed.

Use Fe = 55.85, S = 32.06, O = 16.00, and H = 1.008, with atomic molar masses in \(\mathrm{g\,mol^{-1}}\).

(a) Calculate the mass of completely anhydrous \(\ce{FeSO4}\) that should remain after all water has been removed.

(b) Calculate the mass of water still present after the first heating.

(c) Determine the percentage by mass of iron in the \(3.000\ \mathrm{g}\) residue.

(d) A student claims that the residue must be a pure compound with the formula \(\ce{FeSO4*0.83H2O}\). Explain why the mass measurements do not establish this.

Solution 2

The completely dry residue should weigh 2.732 g, so the first heating has left 0.268 g of water. Iron makes up 33.5% of the measured residue. The data do not prove that the residue has a single, uniform hydrate formula.

Part (a): Calculate the theoretical anhydrous mass.

One mole of hydrated iron(II) sulfate contains one mole of \(\ce{FeSO4}\) and seven moles of water.

The molar mass of anhydrous \(\ce{FeSO4}\) is:

\[
M(\ce{FeSO4})=151.91\ \mathrm{g\,mol^{-1}}
\]

The molar mass of one water unit is \(18.016\ \mathrm{g\,mol^{-1}}\).

Therefore:

\[
\begin{aligned}
M_{\mathrm{hydrate}}&=151.91+7(18.016)\\
&=278.022\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Since the amount of iron(II) sulfate is conserved:

\[
\begin{aligned}
m_{\mathrm{dry}}
&=5.000\times\frac{151.91}{278.022}\\
&=2.732\ \mathrm{g}
\end{aligned}
\]

Part (b): Calculate the water remaining.

The measured residue is heavier than the theoretical anhydrous residue.

\[
\begin{aligned}
m_{\mathrm{water}}
&=3.000-2.732\\
&=0.268\ \mathrm{g}
\end{aligned}
\]

The first heating was incomplete under the stated assumptions.

Part (c): Find the percentage of iron in the measured residue.

All the original iron remains.

Its mass can be calculated from the starting hydrate:

\[
\begin{aligned}
m_{\mathrm{Fe}}
&=5.000\times\frac{55.85}{278.022}\\
&=1.0044\ \mathrm{g}
\end{aligned}
\]

The denominator for the requested percentage is the measured residue, not the original hydrate.

\[
\begin{aligned}
\%\mathrm{Fe}
&=\frac{1.0044}{3.000}\times100\\
&=33.5\%
\end{aligned}
\]

Heating has increased the iron mass percentage even though no iron has been created or added. The total mass decreased as water escaped.

Part (d): Interpret the supposed hydrate formula.

There are:

\[
\frac{5.000}{278.022}=0.01798\ \mathrm{mol}
\]

of iron(II) sulfate formula units in the original sample.

The remaining water corresponds to:

\[
\frac{0.2680}{18.016}=0.01488\ \mathrm{mol}
\]

The average ratio of water to iron(II) sulfate is:

\[
\frac{0.01488}{0.01798}\approx0.827
\]

That is approximately 0.83 water units per iron(II) sulfate formula unit on average.

However, an average composition is not proof of a pure crystalline hydrate with that formula. Different partially dehydrated regions or a mixture of hydrated and anhydrous material could give the same overall ratio.

The tempting mistake is treating an experimentally calculated average ratio as an independently established chemical identity.

Question 3

An analytical chemist combusts \(0.600\ \mathrm{g}\) of pure ethanoic acid, \(\ce{C2H4O2}\), in dry oxygen.

The apparatus places a water absorber before a carbon dioxide absorber.

During one run, the following mass gains are recorded:

Assume combustion is complete, all gaseous products are collected, and the carbon dioxide absorber captures all the \(\ce{CO2}\) reaching it. However, the carbon dioxide absorber will also retain any water that escapes the first absorber.

Use C = 12.01, H = 1.008, and O = 16.00, with atomic molar masses in \(\mathrm{g\,mol^{-1}}\).

(a) Calculate the theoretical masses of \(\ce{CO2}\) and \(\ce{H2O}\) produced.

(b) Determine approximately how much water escaped the first absorber.

(c) Calculate the incorrect carbon percentage that would be obtained by treating the entire \(1.049\ \mathrm{g}\) downstream mass gain as carbon dioxide.

(d) Explain why the nearly correct combined mass gain does not establish that the experiment measured both products correctly.

Solution 3

Approximately 0.170 g of water bypassed the first absorber, causing an incorrect apparent carbon percentage of 47.7% rather than 40.0%. The combined mass gain can be almost correct while the individual product measurements are wrong.

Part (a): Calculate the theoretical products.

Complete combustion follows:

\[
\ce{C2H4O2 + 2O2 -> 2CO2 + 2H2O}
\]

The molar mass of ethanoic acid is:

\[
M=60.052\ \mathrm{g\,mol^{-1}}
\]

Therefore:

\[
n=\frac{0.600}{60.052}
=0.009991\ \mathrm{mol}
\]

Each mole of ethanoic acid produces two moles of carbon dioxide and two moles of water.

The predicted carbon dioxide mass is:

\[
\begin{aligned}
m_{\ce{CO2}}&=2(0.009991)(44.01)\\
&=0.879\ \mathrm{g}
\end{aligned}
\]

The predicted water mass is:

\[
\begin{aligned}
m_{\ce{H2O}}&=2(0.009991)(18.016)\\
&=0.360\ \mathrm{g}
\end{aligned}
\]

Part (b): Determine the water that escaped its absorber.

The water absorber collected only \(0.190\ \mathrm{g}\).

\[
\begin{aligned}
m_{\mathrm{bypass}}
&=0.360-0.190\\
&=0.170\ \mathrm{g}
\end{aligned}
\]

That water was captured downstream, adding to the carbon dioxide absorber’s mass gain.

Indeed:

\[
0.879+0.170\approx1.049\ \mathrm{g}
\]

The readings are consistent with the proposed fault, within rounding.

Part (c): Find the incorrect carbon percentage.

A student who assumes the entire downstream mass gain is \(\ce{CO2}\) would calculate:

\[
\begin{aligned}
m_{\mathrm C,app}
&=1.049\times\frac{12.01}{44.01}\\
&=0.2863\ \mathrm{g}
\end{aligned}
\]

Here, \(m_{\mathrm C,app}\) is the apparent carbon mass.

Therefore:

\[
\begin{aligned}
\%\mathrm{C}_{\mathrm{app}}
&=\frac{0.2863}{0.600}\times100\\
&=47.7\%
\end{aligned}
\]

But the correct carbon percentage from the formula is approximately 40.0%.

Part (d): Explain the misleading agreement.

The measured total trap gain is:

\[
0.190+1.049=1.239\ \mathrm{g}
\]

The theoretical total is:

\[
0.360+0.879=1.239\ \mathrm{g}
\]

Those totals agree, but water was assigned to the wrong product.

The error affects the calculated elemental percentages without significantly affecting the total mass collected.

A suitable revision is to improve or replace the water absorbent, ensure that it has sufficient capacity, and verify that water is removed before the gases enter the carbon dioxide trap.

This is a useful experimental lesson: agreement with one conservation rule does not guarantee that every measurement within the system is correct.

Question 4

A \(5.00\ \mathrm{g}\) sample of pure copper(II) carbonate, \(\ce{CuCO3}\), is heated.

The only reaction is:

\[
\ce{CuCO3 -> CuO + CO2}
\]

Heating is stopped when the solid residue has a mass of \(4.00\ \mathrm{g}\).

Assume all carbon dioxide escapes, no solid is lost, and both copper-containing solids remain stable under the conditions.

Use Cu = 63.55, C = 12.01, and O = 16.00, with atomic molar masses in \(\mathrm{g\,mol^{-1}}\).

(a) Calculate the percentage of the original copper(II) carbonate that has decomposed, expressed as a percentage of its initial amount in moles.

(b) Determine the percentage by mass of copper in the \(4.00\ \mathrm{g}\) residue.

(c) A student calculates that copper makes up approximately 79.9% of pure \(\ce{CuO}\) and concludes that the residue must also contain 79.9% copper. Explain the error and determine the actual solid composition.

Solution 4

Approximately 56.2% of the copper(II) carbonate decomposed. The remaining solid contains 64.3% copper by mass, and consists of both \(\ce{CuCO3}\) and \(\ce{CuO}\), not pure copper(II) oxide.

Part (a): Calculate the amount decomposed.

The molar masses are:

\[
\begin{aligned}
M(\ce{CuCO3})&=123.56\ \mathrm{g\,mol^{-1}}\\
M(\ce{CuO})&=79.55\ \mathrm{g\,mol^{-1}}\\
M(\ce{CO2})&=44.01\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The original amount of copper(II) carbonate was:

\[
n_0=\frac{5.00}{123.56}
=0.04047\ \mathrm{mol}
\]

The sample lost \(1.00\ \mathrm{g}\) as carbon dioxide.

Since one mole of carbonate produces one mole of \(\ce{CO2}\), the amount decomposed is:

\[
n_{\mathrm{dec}}=\frac{1.00}{44.01}
=0.02272\ \mathrm{mol}
\]

Therefore:

\[
\begin{aligned}
\%\mathrm{decomposed}
&=\frac{0.02272}{0.04047}\times100\\
&=56.2\%
\end{aligned}
\]

Part (b): Calculate the copper percentage.

Copper stays in the solid throughout the reaction.

The original sample contained:

\[
\begin{aligned}
m_{\mathrm{Cu}}
&=5.00\times\frac{63.55}{123.56}\\
&=2.572\ \mathrm{g}
\end{aligned}
\]

That copper is now contained within the \(4.00\ \mathrm{g}\) residue.

\[
\begin{aligned}
\%\mathrm{Cu}
&=\frac{2.572}{4.00}\times100\\
&=64.3\%
\end{aligned}
\]

Part (c): Determine the actual composition.

The amount of \(\ce{CuO}\) formed equals the amount of carbonate decomposed:

\[
\begin{aligned}
m_{\ce{CuO}}
&=0.02272\times79.55\\
&=1.81\ \mathrm{g}
\end{aligned}
\]

The remaining copper(II) carbonate mass is:

\[
4.00-1.81=2.19\ \mathrm{g}
\]

The residue therefore contains approximately:

• \(1.81\ \mathrm{g}\) of \(\ce{CuO}\)
• \(2.19\ \mathrm{g}\) of \(\ce{CuCO3}\)

The tempting 79.9% answer assumes the residue is pure \(\ce{CuO}\).

But complete decomposition would have released approximately \(1.78\ \mathrm{g}\) of carbon dioxide, leaving only \(3.22\ \mathrm{g}\) of solid. The measured residue is heavier than that.

The mass-loss evidence shows that the reaction stopped before completion.

The copper mass stays constant, but its percentage rises because part of the original sample leaves as carbon dioxide.

This is an important distinction: elemental mass percentage can change during a reaction even when none of that element is lost.

Question 5

An analytical laboratory receives a \(6.0052\ \mathrm{g}\) sample containing only carbon, hydrogen, and oxygen.

The sample is known to be exactly one of the following:

Candidate A: Pure ethanoic acid, \(\ce{C2H4O2}\).

Candidate B: A mixture containing methanal, \(\ce{CH2O}\), and lactic acid, \(\ce{C3H6O3}\), in an unknown proportion.

The following measurements have been made:

• Elemental analysis: 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen by mass.
• An independent measurement establishes that the sample contains \(0.1000\ \mathrm{mol}\) of individual molecules in total.
• A separate portion equivalent to the entire sample reacts with excess sodium hydrogen carbonate, \(\ce{NaHCO3}\), producing \(2.20\ \mathrm{g}\) of carbon dioxide.

Additional information:

• Ethanoic acid and lactic acid are monoprotic acids. Each mole reacts with excess sodium hydrogen carbonate to produce one mole of \(\ce{CO2}\).
• Methanal produces no carbon dioxide with sodium hydrogen carbonate under the test conditions.
• There are no other reactions producing carbon dioxide.
• The molecular-count measurement is reliable and counts individual molecules rather than associated or dissociated particles.

Use C = 12.01, H = 1.008, and O = 16.00, with atomic molar masses in \(\mathrm{g\,mol^{-1}}\).

(a) Explain why the elemental percentages cannot distinguish Candidate A from Candidate B, regardless of the proportions in Candidate B.

(b) Show how Candidate B could have the same total mass and total number of molecules as Candidate A. Determine the amounts, in moles, of methanal and lactic acid required.

(c) Use the reaction with sodium hydrogen carbonate to identify the sample, assuming the candidates are the only possibilities.

(d) Calculate the mass percentage of lactic acid in the identified sample.

(e) Explain why the agreement of elemental percentages and average molar mass with \(\ce{C2H4O2}\) does not establish that a sample contains molecules of \(\ce{C2H4O2}\).

Solution 5

The unknown is Candidate B, a mixture containing 0.0500 mol of methanal and 0.0500 mol of lactic acid. Lactic acid contributes 75.0% of the mixture’s mass.

The difficult part is that both candidates can produce the same elemental percentages, total mass, and total number of molecules. The acid reaction provides the additional evidence needed to distinguish them.

Part (a): Why elemental analysis cannot distinguish the candidates.

All three molecular formulas have the same simplest ratio:

\[
\begin{aligned}
\ce{CH2O}&\quad\to\quad\ce{CH2O}\\
\ce{C2H4O2}&\quad\to\quad\ce{CH2O}\\
\ce{C3H6O3}&\quad\to\quad\ce{CH2O}
\end{aligned}
\]

Each substance therefore has the same theoretical elemental percentage composition.

Using the empirical formula:

\[
M(\ce{CH2O})=30.026\ \mathrm{g\,mol^{-1}}
\]

The percentages are approximately:

\[
\begin{aligned}
\%\mathrm C&=\frac{12.01}{30.026}\times100=40.0\%\\
\%\mathrm H&=\frac{2.016}{30.026}\times100=6.71\%\\
\%\mathrm O&=\frac{16.00}{30.026}\times100=53.3\%
\end{aligned}
\]

Any mixture made from substances with identical elemental mass percentages retains those same percentages.

That remains true whether the two substances occur in equal masses, equal amounts in moles, or some other proportion.

This is not merely an experimental limitation. No improvement in the precision of elemental mass percentages could distinguish these two candidates.

Part (b): Reconcile the total mass and molecule count.

The molecular molar masses are:

\[
\begin{aligned}
M(\ce{CH2O})&=30.026\ \mathrm{g\,mol^{-1}}\\
M(\ce{C2H4O2})&=60.052\ \mathrm{g\,mol^{-1}}\\
M(\ce{C3H6O3})&=90.078\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The measured average molar mass of the unknown is:

\[
\begin{aligned}
M_{\mathrm{avg}}
&=\frac{6.0052\ \mathrm g}{0.1000\ \mathrm{mol}}\\
&=60.052\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

At first, this seems to prove the sample contains pure ethanoic acid, whose molar mass is also \(60.052\ \mathrm{g\,mol^{-1}}\).

But look at the alternative.

Methanal has a molar mass of \(30.026\ \mathrm{g\,mol^{-1}}\), while lactic acid has a molar mass of \(90.078\ \mathrm{g\,mol^{-1}}\).

Their arithmetic mean is:

\[
\frac{30.026+90.078}{2}
=60.052\ \mathrm{g\,mol^{-1}}
\]

If Candidate B contains equal amounts in moles of the two compounds, it has the same average molar mass.

To verify the amounts, let \(n_{\mathrm M}\) be the amount of methanal and \(n_{\mathrm L}\) the amount of lactic acid, both in moles.

The total amount is:

\[
n_{\mathrm M}+n_{\mathrm L}=0.1000\ \mathrm{mol}
\]

The mass constraint is:

\[
30.026n_{\mathrm M}+90.078n_{\mathrm L}=6.0052
\]

Here, the numerical coefficients are molar masses in \(\mathrm{g\,mol^{-1}}\), giving a total mass in grams.

Substituting \(n_{\mathrm M}=0.1000-n_{\mathrm L}\) gives:

\[
\begin{aligned}
3.0026+60.052n_{\mathrm L}&=6.0052\\
n_{\mathrm L}&=0.0500\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
n_{\mathrm M}=0.0500\ \mathrm{mol}
\]

Both candidates can satisfy the measured mass and molecule count exactly.

Part (c): Identify the candidate using the acid reaction.

Both ethanoic acid and lactic acid react with sodium hydrogen carbonate to release carbon dioxide.

One mole of either acid produces one mole of \(\ce{CO2}\) under the stated conditions.

For Candidate A, the \(0.1000\ \mathrm{mol}\) of ethanoic acid would produce:

\[
\begin{aligned}
m_{\ce{CO2}}
&=0.1000\times44.01\\
&=4.401\ \mathrm{g}
\end{aligned}
\]

For Candidate B, only the lactic acid produces carbon dioxide.

The \(0.0500\ \mathrm{mol}\) of lactic acid would produce:

\[
\begin{aligned}
m_{\ce{CO2}}
&=0.0500\times44.01\\
&=2.2005\ \mathrm{g}
\end{aligned}
\]

The observed mass of carbon dioxide is \(2.20\ \mathrm{g}\).

It agrees with Candidate B.

The unknown is therefore the methanal-lactic acid mixture, within the two stated possibilities.

Part (d): Calculate the lactic acid mass percentage.

The mass of lactic acid is:

\[
\begin{aligned}
m_{\mathrm L}
&=0.0500\times90.078\\
&=4.5039\ \mathrm{g}
\end{aligned}
\]

The total sample mass is \(6.0052\ \mathrm{g}\).

Therefore:

\[
\begin{aligned}
\%\mathrm{lactic\ acid}
&=\frac{4.5039}{6.0052}\times100\\
&=75.0\%
\end{aligned}
\]

This is another useful surprise: the mixture contains equal numbers of moles of methanal and lactic acid, but lactic acid contributes three times as much mass.

Equal mole amounts do not mean equal masses when molar masses differ.

Part (e): Explain the conceptual limit.

The calculated average molar mass of \(60.052\ \mathrm{g\,mol^{-1}}\) is not necessarily the molar mass of any individual molecule in a mixture.

Candidate B contains molecules with molar masses of \(30.026\) and \(90.078\ \mathrm{g\,mol^{-1}}\), not molecules of \(60.052\ \mathrm{g\,mol^{-1}}\).

Yet its average matches ethanoic acid exactly.

The apparently convincing argument would be:

The sample has the elemental ratios of C2H4O2 and an average molar mass of 60.052 g mol⁻¹, so it must be pure ethanoic acid.

The missing assumption is purity.

If the unknown were independently established to be a single pure molecular compound, its elemental ratios and molar mass could establish the molecular formula \(\ce{C2H4O2}\).

Without that purity condition, the inference fails.

And even when the molecular formula is established, identifying a particular compound may still require evidence about molecular structure or chemical behaviour.

The bicarbonate reaction supplies useful additional evidence here because it distinguishes the two specifically defined candidates. It would not, by itself, identify every possible unknown mixture or structural isomer.

07What to do with a percentage-composition result

A chemical formula gives a prediction about how a pure substance’s mass is divided among its elements.

Measured percentages can test that prediction, and converting measured elemental masses into mole ratios can reveal an empirical formula.

But the direction of reasoning matters.

Starting with a known formula, you can calculate a definite theoretical elemental composition. Starting with elemental percentages, you may have several possible formulas, structures, or mixtures.

Before assigning a molecular formula to an unknown, ask what extra evidence establishes its molar mass and whether it is pure. Before assigning a chemical identity, ask what evidence distinguishes compounds with the same formula.

That is the next useful step after calculating percentages: understanding what a measurement proves, what it only suggests, and which experiment would separate the remaining possibilities.