How to Find Molecular Formula from Empirical Formula and Molar Mass

A compound has the empirical formula \(\ce{CH}\), but its measured molar mass is \(78.1\ \mathrm{g\,mol^{-1}}\). Would you write \(\ce{CH}\) as its molecular formula?

No. The atoms in \(\ce{CH}\) account for only \(13.02\ \mathrm{g\,mol^{-1}}\). The actual molecule is six times as massive, so its molecular formula is \(\ce{C6H6}\).

That’s the method for finding a molecular formula from an empirical formula and molar mass: divide the measured molar mass by the empirical formula mass, then multiply every subscript by the resulting whole number.

\[ k=\frac{M}{M_{\mathrm{emp}}} \]

Here:

• \(k\) is the number of empirical-formula units in one molecule. It must be a positive whole number for a pure molecular compound.
• \(M\) is the independently measured molar mass of the compound, in \(\mathrm{g\,mol^{-1}}\).
• \(M_{\mathrm{emp}}\) is the molar mass calculated from the empirical formula, also in \(\mathrm{g\,mol^{-1}}\).

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of each type of atom in one molecule.

The important question is where these two pieces of information come from. An experiment that measures elemental composition does not necessarily tell us how many atoms belong to each molecule.

01Why elemental analysis gives a ratio, not a molecular formula

Suppose a chemist receives an unknown pure compound containing only carbon, hydrogen, and oxygen.

The first problem is finding how much of each element is present. One way to investigate this is combustion analysis.

How the combustion apparatus works

Imagine a horizontal combustion tube.

At the left, a controlled supply of oxygen enters the tube. A weighed sample of the unknown compound sits in a small sample boat inside a heated section.

As the sample burns completely, its carbon forms carbon dioxide, and its hydrogen forms water.

The gas stream then passes through two separate collection tubes:

1. The first contains a substance that selectively absorbs water.
2. The second contains a substance that selectively absorbs carbon dioxide.

Both collection tubes are weighed before and after the experiment. Their increases in mass reveal how much water and carbon dioxide were produced.

Consider these illustrative measurements. They are invented teaching data, not a record of a particular historical experiment.

The collected products have a combined mass of \(0.620\ \mathrm{g}\), more than twice the original sample mass.

Does that violate conservation of mass?

No. Oxygen from the external supply becomes part of the combustion products. The original sample is not the only source of atoms.

This also explains why we cannot calculate the unknown compound’s oxygen content by simply adding up the oxygen in the carbon dioxide and water.

What the measurements actually reveal

Every mole of \(\ce{CO2}\) contains one mole of carbon atoms. Every mole of \(\ce{H2O}\) contains two moles of hydrogen atoms.

Using \(M(\ce{CO2})=44.01\ \mathrm{g\,mol^{-1}}\) and \(M(\ce{H2O})=18.02\ \mathrm{g\,mol^{-1}}\):

\[ \begin{aligned} n(\mathrm C)&=\frac{0.440}{44.01}\\ &=0.0100\ \mathrm{mol}\\[4pt] n(\mathrm H)&=2\left(\frac{0.180}{18.02}\right)\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

Here, \(n\) means amount in moles. The numerators are measured masses in grams, and the denominators are molar masses in \(\mathrm{g\,mol^{-1}}\).

Because the original compound contains only carbon, hydrogen, and oxygen, we can find its oxygen mass by subtraction.

Using atomic molar masses of \(12.01\ \mathrm{g\,mol^{-1}}\) for carbon, \(1.008\ \mathrm{g\,mol^{-1}}\) for hydrogen, and \(16.00\ \mathrm{g\,mol^{-1}}\) for oxygen:

\[ \begin{aligned} m(\mathrm O)&=0.300-0.120-0.0201\\ &=0.160\ \mathrm g\\[4pt] n(\mathrm O)&=\frac{0.160}{16.00}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The symbol \(m\) represents mass. Values are rounded to the precision of the measurements.

Our carbon-to-hydrogen-to-oxygen mole ratio is therefore approximately:

\[ 0.0100:0.0200:0.0100 \]

Dividing all three quantities by \(0.0100\) gives:

\[ 1:2:1 \]

The empirical formula is \(\ce{CH2O}\).

If you need to refresh the conversion between grams and moles, see Moles from Mass and Molar Mass: HSC Chemistry Guide.

The problem this experiment cannot solve

We know the atoms occur in the simplest ratio \(1:2:1\). But what if each molecule contains twice or three times those numbers?

All these molecular formulas have the same empirical formula:

Would a larger molecule produce more carbon dioxide when it burns?

Per molecule, yes. For example:

\[ \ce{CH2O + O2 -> CO2 + H2O} \]

Compared with:

\[ \ce{C3H6O3 + 3O2 -> 3CO2 + 3H2O} \]

But a fixed mass of the larger molecules contains fewer molecules.

For instance, \(0.300\ \mathrm{g}\) of \(\ce{CH2O}\) contains about \(0.0100\ \mathrm{mol}\) of molecules. The same mass of \(\ce{C3H6O3}\) contains about \(0.00333\ \mathrm{mol}\) of molecules, with three times as many carbon and hydrogen atoms per molecule.

Both samples would produce approximately \(0.0100\ \mathrm{mol}\) of carbon dioxide and \(0.0100\ \mathrm{mol}\) of water.

The combustion measurements cannot distinguish them.

This is the limitation of the first measurement: it reveals the proportions of atoms in the material, not how many atoms are packaged into each molecule.

02The second measurement: finding the size of the molecule

To determine a molecular formula, we need another piece of evidence: the molar mass of the molecular substance.

One instrumental method that can help establish molecular mass is mass spectrometry.

A sample enters an ion source, where particles are converted into charged ions. Those ions travel into a mass analyser, which separates them according to their mass-to-charge ratios. A detector then records the ions reaching it.

If an intact molecular ion can be correctly identified, its mass provides evidence about the original molecule.

There is an important limitation. Some molecules fragment readily during ionisation, producing smaller ions. The strongest signal is not necessarily the intact molecular ion. Ion charge and isotope patterns also matter.

The molecular mass therefore needs to come from an appropriate, correctly interpreted measurement. For the following example, we will assume the stated molar mass has been independently established.

That second measurement answers the question combustion analysis left open: how much does one mole of complete molecules weigh?

03How to use mole ratios and round correctly

The arithmetic is straightforward when the empirical formula is supplied. The trickier questions ask you to find that formula first.

There are two different places where rounding can cause mistakes.

The first is when converting experimental mole amounts into the simplest whole-number ratio. The second is when deciding whether the molecular multiplier \(k\) is genuinely an integer.

These need different checks.

Rounding elemental mole ratios

After converting each element’s mass into moles, divide every amount by the smallest non-zero amount.

You might obtain a ratio such as:

\[ 1.50:3.00:1.00 \]

Don’t round \(1.50\) to \(2\). The ratio tells you that the first quantity is one and a half times the smallest quantity, not approximately twice it.

Multiply all parts by two instead:

\[ 3:6:2 \]

The corresponding empirical formula is \(\ce{C3H6O2}\).

Experimental ratios may also be slightly different from neat fractions because measurements have limited precision. For example, \(1.499:2.996:1.000\) is strong evidence for \(1.5:3:1\).

The appropriate question is whether the proposed simple ratio agrees with the measured values to a reasonable degree of precision.

04When you should not round the molecular multiplier

Suppose the empirical formula is \(\ce{CH2O}\), with a formula mass of approximately \(30.03\ \mathrm{g\,mol^{-1}}\).

Different independent molar-mass measurements would lead to different conclusions.

Why is \(k=3.16\) different from \(k=3.00\)?

The ratio is not a free choice. If \(k=3\), the predicted molecular molar mass is approximately \(90.09\ \mathrm{g\,mol^{-1}}\), not \(95.0\ \mathrm{g\,mol^{-1}}\).

A difference of nearly \(5\ \mathrm{g\,mol^{-1}}\) is too large to dismiss merely because three is the nearest integer. Whether it could be explained by experimental uncertainty depends on the measurement method and its stated uncertainty.

If the measurements are incompatible, investigate the assumptions before reporting a formula. Possible problems include incorrect elemental analysis, an incorrect molar-mass measurement, an impure sample, or a mixture of molecular species.

A useful final check is to calculate the molar mass from your proposed molecular formula. It must agree with the independently measured value within the relevant precision.

If calculating formula masses is the step you need to revise, see Molar Mass of Elements and Compounds: HSC Chemistry Guide.

05Molecular formula is not the same as molecular identity

Suppose two compounds both have the empirical formula \(\ce{CH2O}\) and measured molar masses close to \(60.05\ \mathrm{g\,mol^{-1}}\).

Both therefore have molecular formula \(\ce{C2H4O2}\).

Are they necessarily the same chemical?

No. Consider ethanoic acid and methyl methanoate.

Ethanoic acid has the structural formula \(\ce{CH3COOH}\). Methyl methanoate has the structural formula \(\ce{HCOOCH3}\).

Both contain two carbon atoms, four hydrogen atoms, and two oxygen atoms. Their molecular formulas are identical, but the atoms are connected differently.

This is structural isomerism. Compounds with the same molecular formula can have different functional groups, properties, and reactions.

Elemental composition and molar mass establish atom counts, not bonding arrangements. A structural formula requires additional evidence, such as characteristic chemical reactions or appropriately interpreted spectroscopic data.

There’s another assumption worth keeping in mind: the multiplier method applies to a pure compound made of discrete molecules. An ionic solid such as \(\ce{NaCl}\) is described using formula units, not individual \(\ce{NaCl}\) molecules. A mixture also cannot automatically be assigned one molecular formula.

That last distinction becomes important when measurements describe an entire sample rather than a single molecular species.

06Questions and solutions

The questions progress from a standard HSC calculation to situations where different experimental measurements appear to support competing conclusions.

Use the following relative atomic masses throughout:

\(\mathrm C=12.01\), \(\mathrm H=1.008\), \(\mathrm O=16.00\), and \(\mathrm N=14.01\).

All atomic molar masses are in \(\mathrm{g\,mol^{-1}}\). Assume measurements are valid to their stated precision unless a question specifies an uncertainty.

Question 1

A pure molecular compound has the empirical formula \(\ce{CH}\) and a measured molar mass of \(78.1\ \mathrm{g\,mol^{-1}}\).

Determine its molecular formula. Explain why the empirical formula cannot be reported as the molecular formula.

Solution 1

The molecular formula is \(\ce{C6H6}\). The empirical formula \(\ce{CH}\) gives only the simplest carbon-to-hydrogen atom ratio.

First, calculate the empirical formula mass.

\[ \begin{aligned} M_{\mathrm{emp}}&=12.01+1.008\\ &=13.018\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Divide the measured molar mass by this value.

\[ \begin{aligned} k&=\frac{78.1}{13.018}\\ &=6.00 \end{aligned} \]

Multiply both empirical subscripts by six.

\[ \ce{CH}\longrightarrow\ce{C6H6} \]

Checking the result:

\[ 6(13.018)=78.108\ \mathrm{g\,mol^{-1}} \]

This agrees with the measured molar mass.

Reporting \(\ce{CH}\) would account for only one-sixth of the molecular molar mass. The correct answer is \(\ce{C6H6}\), although this alone does not establish which structural isomer is present.

Question 2

An unknown pure compound contains only carbon, hydrogen, and oxygen.

A \(0.580\ \mathrm g\) sample is burned completely in excess oxygen. The products are collected separately.

The following illustrative data are obtained:

• Carbon dioxide produced: \(1.320\ \mathrm g\)
• Water produced: \(0.540\ \mathrm g\)
• Independently measured molar mass: \(116.2\ \mathrm{g\,mol^{-1}}\)

Assume all carbon forms \(\ce{CO2}\), all hydrogen forms \(\ce{H2O}\), both products are completely collected, and no other elements are present.

Determine the empirical formula and molecular formula.

A student suggests calculating the original oxygen content directly from all the oxygen atoms in the collected combustion products. Explain why that method fails.

Solution 2

The empirical formula is \(\ce{C3H6O}\), and the molecular formula is \(\ce{C6H12O2}\). The compound’s oxygen content must be calculated from the original sample mass, not directly from all the oxygen in the combustion products.

First, calculate the amount of carbon from the carbon dioxide.

\[ \begin{aligned} n(\mathrm C)&=\frac{1.320}{44.01}\\ &=0.02999\ \mathrm{mol} \end{aligned} \]

Next, calculate the hydrogen from the water.

\[ \begin{aligned} n(\mathrm H)&=2\left(\frac{0.540}{18.02}\right)\\ &=0.05993\ \mathrm{mol} \end{aligned} \]

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

Calculate the carbon and hydrogen masses in the original sample.

\[ \begin{aligned} m(\mathrm C)&=0.02999(12.01)\\ &\approx0.3602\ \mathrm g\\[4pt] m(\mathrm H)&=0.05993(1.008)\\ &\approx0.0604\ \mathrm g \end{aligned} \]

Because the original compound contains only carbon, hydrogen, and oxygen:

\[ \begin{aligned} m(\mathrm O)&=0.580-0.3602-0.0604\\ &=0.1594\ \mathrm g \end{aligned} \]

Therefore:

\[ \begin{aligned} n(\mathrm O)&=\frac{0.1594}{16.00}\\ &=0.00996\ \mathrm{mol} \end{aligned} \]

Divide the three mole amounts by the smallest, \(0.00996\ \mathrm{mol}\).

\[ \mathrm{C:H:O}\approx3.01:6.02:1 \]

The small deviations are consistent with measurement rounding, giving the simplest whole-number ratio \(3:6:1\).

The empirical formula is \(\ce{C3H6O}\).

Its molar mass is:

\[ \begin{aligned} M_{\mathrm{emp}}&=3(12.01)+6(1.008)+16.00\\ &=58.078\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Now calculate the multiplier.

\[ \begin{aligned} k&=\frac{116.2}{58.078}\\ &\approx2.00 \end{aligned} \]

The molecular formula is therefore \(\ce{C6H12O2}\).

Its predicted molar mass is approximately \(116.16\ \mathrm{g\,mol^{-1}}\), consistent with the measured value.

The proposed oxygen method fails because the oxygen supplied to the combustion tube contributes oxygen atoms to both \(\ce{CO2}\) and \(\ce{H2O}\).

In fact, the collected products contain more oxygen by mass than the entire original \(0.580\ \mathrm g\) sample. That oxygen could not all have come from the unknown compound.

The oxygen-by-difference calculation is valid here because the sample is assumed pure, contains only C, H, and O, and undergoes complete combustion with quantitative product collection.

Question 3

A pure compound contains only carbon, hydrogen, and oxygen.

Elemental analysis gives the following mole ratio, normalised so that carbon equals \(1.00\):

\[ \mathrm{C:H:O}=1.00:2.49:0.50 \]

The uncertainty in each reported relative value is approximately \(\pm0.02\).

An independent experiment measures the molar mass as:

\[ 90.1\pm0.2\ \mathrm{g\,mol^{-1}} \]

Two students propose different answers.

Student A rounds each elemental ratio separately to obtain the empirical formula \(\ce{CH2O}\), giving molecular formula \(\ce{C3H6O3}\).

Student B keeps the fractional ratio and proposes \(\ce{C4H10O2}\).

Determine which molecular formula is supported by the measurements.

Calculate the molar masses of both proposed molecular formulas and explain why the molar-mass measurement alone is insufficient to decide between them.

Solution 3

Student B’s molecular formula, \(\ce{C4H10O2}\), is supported by the evidence. Both candidates have molar masses close to \(90.1\ \mathrm{g\,mol^{-1}}\), but only \(\ce{C4H10O2}\) matches the measured elemental ratio.

Start with the ratio:

\[ 1.00:2.49:0.50 \]

The hydrogen ratio is close to \(2.50\), and the oxygen ratio is close to \(0.50\), both within the stated uncertainty.

Multiply all three values by two.

\[ 2.00:4.98:1.00 \]

Within the measurement precision, this gives \(2:5:1\).

The empirical formula is therefore \(\ce{C2H5O}\).

Calculate its formula mass.

\[ \begin{aligned} M_{\mathrm{emp}}&=2(12.01)+5(1.008)+16.00\\ &=45.06\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The molecular multiplier is:

\[ \begin{aligned} k&=\frac{90.1}{45.06}\\ &\approx2.00 \end{aligned} \]

Hence the molecular formula is \(\ce{C4H10O2}\).

Its calculated molar mass is:

\[ \begin{aligned} M(\ce{C4H10O2})&=4(12.01)+10(1.008)\\ &\quad+2(16.00)\\ &=90.12\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Now examine Student A’s proposed formula.

\[ \begin{aligned} M(\ce{C3H6O3})&=3(12.01)+6(1.008)\\ &\quad+3(16.00)\\ &=90.078\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Both \(90.12\) and \(90.078\ \mathrm{g\,mol^{-1}}\) agree with \(90.1\pm0.2\ \mathrm{g\,mol^{-1}}\).

The molar-mass measurement cannot resolve these two candidates.

The deciding evidence is the elemental mole ratio.

For \(\ce{C4H10O2}\):

\[ 4:10:2=1:2.5:0.5 \]

For \(\ce{C3H6O3}\):

\[ 3:6:3=1:2:1 \]

The latter disagrees strongly with the measured values \(1.00:2.49:0.50\). Differences of approximately \(0.49\) and \(0.50\) are far larger than the stated uncertainty of \(0.02\).

Student A’s mistake was rounding each mole ratio independently before removing the fractional common ratio.

The deeper lesson is that matching the measured molar mass does not automatically validate a proposed formula. Its elemental composition must also agree with the independent evidence.

Question 4

Nitrogen dioxide can reversibly combine to form dinitrogen tetroxide:

\[ \ce{2NO2(g) <=> N2O4(g)} \]

A sealed, rigid container initially holds \(0.100\ \mathrm{mol}\) of pure \(\ce{NO2}\).

The system reaches equilibrium at a temperature where both substances remain gaseous.

An ideal-gas measurement of the equilibrium pressure indicates that the container now holds a total of \(0.0750\ \mathrm{mol}\) of gas particles.

Assume no gas escapes, no other reactions occur, and the two gases behave ideally.

Determine:

(a) The equilibrium amounts of \(\ce{NO2}\) and \(\ce{N2O4}\).

(b) The total gas mass and the apparent average molar mass obtained by dividing total mass by total moles of gas particles.

(c) Why the ratio of this average molar mass to the empirical formula mass is not a valid molecular multiplier.

Solution 4

At equilibrium, the container holds \(0.0500\ \mathrm{mol}\) of \(\ce{NO2}\) and \(0.0250\ \mathrm{mol}\) of \(\ce{N2O4}\). The apparent molar mass is approximately \(61.3\ \mathrm{g\,mol^{-1}}\), but it describes a mixture, not one molecular species.

(a) Calculate the equilibrium amounts.

Let \(x\) be the number of moles of \(\ce{N2O4}\) formed.

Every mole of \(\ce{N2O4}\) consumes two moles of \(\ce{NO2}\).

Therefore:

\[ \begin{aligned} n(\ce{NO2})&=0.100-2x\\ n(\ce{N2O4})&=x \end{aligned} \]

The total equilibrium amount is:

\[ \begin{aligned} 0.0750&=(0.100-2x)+x\\ 0.0750&=0.100-x\\ x&=0.0250\ \mathrm{mol} \end{aligned} \]

Hence:

\[ \begin{aligned} n(\ce{NO2})&=0.100-2(0.0250)\\ &=0.0500\ \mathrm{mol}\\[4pt] n(\ce{N2O4})&=0.0250\ \mathrm{mol} \end{aligned} \]

The dimerisation reaction reduces the number of gas particles without removing nitrogen or oxygen atoms from the vessel.

(b) Calculate total mass and average molar mass.

The molar mass of \(\ce{NO2}\) is:

\[ \begin{aligned} M(\ce{NO2})&=14.01+2(16.00)\\ &=46.01\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The initial gas mass is:

\[ \begin{aligned} m&=0.100(46.01)\\ &=4.601\ \mathrm g \end{aligned} \]

The vessel is closed, so its total gas mass remains \(4.601\ \mathrm g\).

At equilibrium, the apparent average molar mass is:

\[ \begin{aligned} M_{\mathrm{avg}}&=\frac{4.601\ \mathrm g}{0.0750\ \mathrm{mol}}\\ &=61.35\ \mathrm{g\,mol^{-1}} \end{aligned} \]

This is approximately \(61.3\ \mathrm{g\,mol^{-1}}\).

(c) Explain why the multiplier method fails.

Both \(\ce{NO2}\) and \(\ce{N2O4}\) have the same simplest nitrogen-to-oxygen ratio, \(1:2\).

The empirical formula mass corresponding to that ratio is \(46.01\ \mathrm{g\,mol^{-1}}\).

Dividing the apparent average molar mass by this value gives:

\[ \begin{aligned} k&=\frac{61.35}{46.01}\\ &\approx1.333 \end{aligned} \]

A student might try rounding this value to one or conclude that the experimental data must be wrong.

Neither conclusion is justified.

The problem is that the measured \(61.35\ \mathrm{g\,mol^{-1}}\) is an average over two molecular species.

Some gas particles are \(\ce{NO2}\), with molar mass \(46.01\ \mathrm{g\,mol^{-1}}\). Others are \(\ce{N2O4}\), with molar mass \(92.02\ \mathrm{g\,mol^{-1}}\).

The average lies between those values.

It does not imply a molecule containing a fractional number of empirical units.

The molecular multiplier must be a positive integer only when the independently measured molar mass belongs to one pure molecular species.

The surprising result arises because the reaction changes the number of molecules while conserving every atom in the vessel.

Question 5

An unknown liquid contains only carbon, hydrogen, and oxygen.

A careful elemental analysis finds that the overall atom mole ratio in the sample is exactly:

\[ \mathrm{C:H:O}=2:6:1 \]

A separate measurement gives the following illustrative results:

• Sample mass: \(9.21\ \mathrm g\)
• Total amount of molecules in the sample: \(0.200\ \mathrm{mol}\)

Assume the molecule-counting method is valid for this sample, with no molecular association, dissociation, or chemical reaction during the measurement.

A student calculates an apparent molar mass of approximately \(46.1\ \mathrm{g\,mol^{-1}}\).

The student concludes that the sample is pure ethanol, \(\ce{C2H6O}\), because its empirical formula mass and measured molar mass agree.

However, a later separation shows that the sample contains only two different molecular substances:

• Substance A has molecular formula \(\ce{CH4O}\).
• Substance B has molecular formula \(\ce{C3H8O}\).

The substances do not react with each other. Assume the separation identifies both species correctly and confirms that no other components are present.

(a) Determine the amount in moles of substances A and B in the original sample.

(b) Prove that this mixture reproduces both the measured overall elemental ratio and the apparent molar mass.

(c) Could the original elemental analysis and total-molecule measurement, even if perfectly accurate, distinguish this mixture from a pure compound with molecular formula \(\ce{C2H6O}\)? Explain what additional type of experimental evidence is needed.

Solution 5

The original sample contains approximately \(0.100\ \mathrm{mol}\) of \(\ce{CH4O}\) and \(0.100\ \mathrm{mol}\) of \(\ce{C3H8O}\). This equimolar mixture reproduces both original measurements, so those measurements alone cannot establish that the sample is a pure \(\ce{C2H6O}\) compound.

The key is that elemental analysis measures total atom proportions, while dividing a mixture’s mass by its total amount of molecules gives an average molar mass.

Both quantities can accidentally match those of a pure molecular substance.

(a) Determine the amounts of the two compounds.

First, calculate their molar masses.

For substance A:

\[ \begin{aligned} M_A&=12.01+4(1.008)+16.00\\ &=32.042\ \mathrm{g\,mol^{-1}} \end{aligned} \]

For substance B:

\[ \begin{aligned} M_B&=3(12.01)+8(1.008)+16.00\\ &=60.094\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Let \(n_A\) and \(n_B\) be the amounts in moles of substances A and B.

The total amount is:

\[ n_A+n_B=0.200 \]

The total mass is the sum of the two component masses:

\[ M_A n_A+M_B n_B=9.21\ \mathrm g \]

Substitute \(n_A=0.200-n_B\) and the molar masses in \(\mathrm{g\,mol^{-1}}\).

\[ \begin{aligned} 32.042(0.200-n_B)&\\ +60.094n_B&=9.21 \end{aligned} \]

Simplifying, with the mass in grams:

\[ \begin{aligned} 6.4084+28.052n_B&=9.21\\ n_B&\approx0.0999\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ n_A\approx0.100\ \mathrm{mol} \]

To the precision of the data, the mixture contains equal numbers of molecules of A and B.

(b) Show that both original measurements are reproduced.

Start with the total atom amounts.

Every molecule of \(\ce{CH4O}\) contains one carbon atom, four hydrogen atoms, and one oxygen atom.

Every molecule of \(\ce{C3H8O}\) contains three carbon atoms, eight hydrogen atoms, and one oxygen atom.

Using \(0.100\ \mathrm{mol}\) of each species:

Carbon atoms:

\[ \begin{aligned} n(\mathrm C)&=0.100(1)+0.100(3)\\ &=0.400\ \mathrm{mol} \end{aligned} \]

Hydrogen atoms:

\[ \begin{aligned} n(\mathrm H)&=0.100(4)+0.100(8)\\ &=1.200\ \mathrm{mol} \end{aligned} \]

Oxygen atoms:

\[ \begin{aligned} n(\mathrm O)&=0.100(1)+0.100(1)\\ &=0.200\ \mathrm{mol} \end{aligned} \]

Divide the atom amounts by \(0.200\ \mathrm{mol}\):

\[ \mathrm{C:H:O}=2:6:1 \]

The mixture reproduces the observed elemental ratio exactly.

Now consider its mass.

\[ \begin{aligned} m_A&=0.100(32.042)\\ &=3.2042\ \mathrm g\\[4pt] m_B&=0.100(60.094)\\ &=6.0094\ \mathrm g \end{aligned} \]

The total mass is:

\[ m_{\mathrm{total}}=9.2136\ \mathrm g \]

This agrees with \(9.21\ \mathrm g\) to three significant figures.

The apparent average molar mass is:

\[ \begin{aligned} M_{\mathrm{avg}}&=\frac{9.2136}{0.200}\\ &=46.068\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Now calculate the molar mass of a pure \(\ce{C2H6O}\) compound.

\[ \begin{aligned} M(\ce{C2H6O})&=2(12.01)+6(1.008)\\ &\quad+16.00\\ &=46.068\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The values are identical to the precision used.

The equimolar mixture has exactly the same overall average atom counts per molecule as a pure \(\ce{C2H6O}\) substance:

\[ \begin{aligned} \text{Average C atoms}&=2\\ \text{Average H atoms}&=6\\ \text{Average O atoms}&=1 \end{aligned} \]

But those are averages. No molecule in the mixture actually has formula \(\ce{C2H6O}\).

(c) Explain whether the original measurements can distinguish the models.

No. The original measurements alone cannot distinguish the two possibilities, even if they are perfectly accurate.

A pure \(\ce{C2H6O}\) sample and the specified equimolar mixture would give the same total elemental proportions and the same average mass per mole of molecules.

This is not a rounding error or an equipment-precision problem. The measurements simply do not contain enough independent information to determine how the atoms are distributed among individual molecules.

The student made two unsupported assumptions.

First, they assumed the sample was pure. The formula-multiplier method requires one molecular species with a definite molar mass. An average molar mass of a mixture does not satisfy that condition.

Second, even if the sample were pure \(\ce{C2H6O}\), the molecular formula would not by itself establish that it was ethanol. Dimethyl ether also has molecular formula \(\ce{C2H6O}\), but a different structure.

To distinguish the mixture from a pure compound, we need evidence that resolves the individual substances, rather than another measurement of their combined average.

For example, a suitable chromatographic separation, coupled with reliable identification of the separated components, could demonstrate that two distinct molecular species are present.

The central result is that an exact integer multiplier is necessary for a correctly identified pure molecular compound, but it is not proof that an unknown sample is pure.

07Where this method takes you next

Empirical-formula calculations establish atom ratios. An independent molar-mass measurement establishes molecular size, provided it belongs to a pure molecular substance.

Together, these give the molecular formula.

But finding \(\ce{C2H4O2}\), \(\ce{C6H6}\), or any other molecular formula does not finish the investigation. The next question is how those atoms are bonded, and whether the sample actually contains only one substance.

That is where structural formulas, isomerism, chromatography, and spectroscopy become useful: they test molecular identity rather than just elemental composition and average mass.