From Composition Data to Molecular Formulae

Welcome back. Last lesson established the central calculation route:

$$\text{mass}⟶\text{moles}⟶\text{particles}$$

This lesson uses the same first conversion—mass to moles—but for a different purpose. Rather than finding an amount of substance, you will compare the mole amounts of elements to find their simplest whole-number ratio: the empirical formula. You will then use $M_r$ to scale that ratio up to the compound’s molecular formula.

These are high-value A-level calculation marks. The reliable method is more important than trying to spot ratios by inspection.

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Empirical versus molecular formula

An empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.

For example:

$${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$$

has the atom ratio $6:12:6$. Dividing each number by 6 gives $1:2:1$, so its empirical formula is:

$$\mathrm{C}{\mathrm{H}}_2\mathrm{O}$$

A molecular formula gives the actual number of atoms in one molecule. Therefore, a molecular formula is always a whole-number multiple of the empirical formula:

$$\text{molecular formula}=(\text{empirical formula})\times k$$

where $k$ is a positive integer.

For glucose:

$$\mathrm{C}{\mathrm{H}}_2\mathrm{O}\times 6={\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$$

Do not assume that the empirical formula is the molecular formula. They are sometimes identical—for example, ${\mathrm{H}}_2\mathrm{O}$—but often they are not.

For ionic compounds, we normally use the term empirical formula or formula unit, because an ionic lattice does not consist of separate molecules. For example, $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$ is already the simplest ratio of aluminium ions to oxide ions.

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The empirical-formula method: convert first, compare second

A mass tells you how heavy a sample is, but it does not tell you directly how many atoms are present. Different elements have different atomic masses, so you must convert each mass into moles before comparing.

The method is always:

1. Convert each element’s mass to moles.
2. Divide every mole value by the smallest mole value.
3. Convert the resulting ratio to the smallest whole numbers.
4. Write the empirical formula.

The reason for dividing by the smallest value is that it sets the least abundant element’s ratio to $1$. Every other value then tells you how many times as many moles—and therefore atoms—are present relative to that element.

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Worked example: empirical formula from masses

A compound contains $2.70\mathrm{g}$ of aluminium and $2.40\mathrm{g}$ of oxygen. Determine its empirical formula.

1. Convert each mass to moles

Use:

$$n=\frac{m}{M}$$ $$n(\mathrm{Al})=\frac{2.70}{27.0}=0.100\mathrm{mol}$$ $$n(\mathrm{O})=\frac{2.40}{16.0}=0.150\mathrm{mol}$$

At this point, do not write $\mathrm{A}{\mathrm{l}}_{0.100}{\mathrm{O}}_{0.150}$. These are mole amounts, not the atom ratio.

2. Divide by the smallest mole amount

The smallest value is $0.100$.

$$\mathrm{Al}:\frac{0.100}{0.100}=1$$ $$\mathrm{O}:\frac{0.150}{0.100}=1.5$$

So the ratio is:

$$\mathrm{Al}:\mathrm{O}=1:1.5$$

3. Turn the ratio into whole numbers

A formula cannot contain half an atom, so multiply every part of the ratio by 2:

$$1:1.5⟶2:3$$

The empirical formula is therefore:

$$\boxed{\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3}$$

A clear exam layout is usually a small table:

Element	Mass / $\mathrm{g}$	Moles	Divide by smallest	Final ratio
Al	2.70	0.100	$0.100/0.100$	2
O	2.40	0.150	$0.150/0.100$	3

You do not have to include every column if your workings are clear, but this format prevents the common error of comparing masses instead of moles.

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Recognising ratios that need scaling

After dividing by the smallest mole amount, values close to whole numbers can normally be rounded:

$$0.999\approx 1$$ $$1.998\approx 2$$

But values close to simple fractions must be scaled up rather than rounded.

Mole ratio obtained	Multiply all ratios by	Resulting whole-number ratio
$1:1.5$	2	$2:3$
$1:1.33$ or $1:1.67$	3	$3:4$ or $3:5$
$1:1.25$ or $1:1.75$	4	$4:5$ or $4:7$
$1:1.2$, $1:1.4$, $1:1.6$, $1:1.8$	5	$5:6$, $5:7$, $5:8$, $5:9$

The key rule is:

If you multiply one ratio, multiply every ratio by the same number.

For instance, a ratio of $1:1.5:2$ becomes:

$$2:3:4$$

not $1:3:2$.

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A visual walkthrough and a second method check

The Chemistry Tutor video gives a useful visual explanation of direct mass data, percentage composition, awkward ratios, and the link to molecular formulae.

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Percentage composition: assume a 100 g sample

If a question gives percentage composition by mass, assume you have exactly $100\mathrm{g}$ of the compound. That makes each percentage numerically equal to a mass in grams.

For example:

$$40.0\%\text{ carbon}=40.0\mathrm{g}\mathrm{carbon}$$ $$6.7\%\text{ hydrogen}=6.7\mathrm{g}\mathrm{hydrogen}$$

This is not a shortcut without justification: “percent by mass” literally means grams per $100\mathrm{g}$.

Worked example: empirical formula from percentages

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Determine its empirical formula.

Assume a $100\mathrm{g}$ sample.

Element	Mass in assumed sample / $\mathrm{g}$	Moles
C	40.0	$40.0/12.0=3.33$
H	6.7	$6.7/1.0=6.7$
O	53.3	$53.3/16.0=3.33$

The smallest mole value is $3.33$.

$$\mathrm{C}:\frac{3.33}{3.33}=1$$ $$\mathrm{H}:\frac{6.7}{3.33}\approx 2$$ $$\mathrm{O}:\frac{3.33}{3.33}=1$$

So the ratio is:

$$\mathrm{C}:\mathrm{H}:\mathrm{O}=1:2:1$$ $$\boxed{\mathrm{C}{\mathrm{H}}_2\mathrm{O}}$$

Notice that the percentages do not add exactly to $100.0\%$:

$$40.0+6.7+53.3=100.0$$

They do here, but in many exam questions they may total $99.9\%$ or $100.1\%$ because of rounding. That is normal; use the values provided.

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Check your method against a concise worked example

The Chemrevise notes give the same general method in an exam-focused format, including a three-element example. Read this after attempting the calculations above; use it to compare the layout of your workings, rather than copying it passively.

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From empirical formula to molecular formula

Once you know the empirical formula, calculate its formula mass. Then compare it with the compound’s given $M_r$.

$$\text{multiplier}=\frac{M_r\text{ of molecular formula}}{M_r\text{ of empirical formula}}$$

Then multiply every subscript in the empirical formula by that integer.

Worked example: finding a molecular formula

A compound has empirical formula ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$ and $M_r=88.0$. Determine its molecular formula.

First calculate the mass of the empirical formula:

$$M_r({\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O})=(2\times 12.0)+(4\times 1.0)+16.0$$ $$M_r({\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O})=44.0$$

Now find how many empirical units fit into the molecular formula:

$$k=\frac{88.0}{44.0}=2$$

Multiply every subscript by 2:

$${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}\times 2={\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$$ $$\boxed{{\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2}$$

Be careful with the oxygen. In the empirical formula, $\mathrm{O}$ means one oxygen atom. That implied subscript of 1 must also be multiplied:

$$\mathrm{O}\times 2={\mathrm{O}}_2$$

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Combined example: composition data to molecular formula

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Its $M_r$ is $180.0$. Determine its molecular formula.

From the earlier calculation, its empirical formula is:

$$\mathrm{C}{\mathrm{H}}_2\mathrm{O}$$

Calculate the empirical formula mass:

$$M_r(\mathrm{C}{\mathrm{H}}_2\mathrm{O})=12.0+(2\times 1.0)+16.0=30.0$$

Calculate the multiplier:

$$k=\frac{180.0}{30.0}=6$$

Multiply all subscripts by 6:

$$\mathrm{C}{\mathrm{H}}_2\mathrm{O}\times 6={\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$$ $$\boxed{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}$$

This is a two-stage question. Do not jump from percentages directly to molecular formula:

$$\text{composition data}⟶\text{empirical formula}⟶\text{empirical formula mass}⟶\text{multiplier}⟶\text{molecular formula}$$

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Exam checks and common errors

Before committing to an answer, use these checks.

For an empirical formula

• Did you convert every mass or percentage to moles?
• Did you divide every mole amount by the same smallest value?
• Are the final subscripts the smallest possible whole numbers?
• If a value was near $1.5$, $1.33$, or $1.25$, did you scale all ratios rather than round it?

For a molecular formula

• Did you calculate the $M_r$ of the empirical formula, not of a guessed molecular formula?
• Is your multiplier a whole number?
• Did you multiply every subscript, including an implied subscript of 1?
• Does your final molecular formula have the given $M_r$?

The most frequent lost marks come from these errors:

Error	Correction
Using elemental masses as the ratio	Convert masses to moles first.
Dividing by the largest mole value	Divide all values by the smallest mole amount.
Rounding $1.5$ to $2$	Multiply the whole ratio by 2.
Multiplying only one subscript when finding molecular formula	Multiply every subscript by $k$.
Giving a multiplier of $2.4$	Recheck your arithmetic: molecular formulae are whole-number multiples of empirical formulae.

For your calculation-error log, record any issue under one of these headings: mass-to-moles, ratio simplification, fractional-ratio scaling, or molecular multiplier. That makes revision much more targeted than simply redoing every question.

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Key takeaways

An empirical formula is the simplest whole-number atom ratio. To find it:

$$\text{mass or percentage}⟶\text{moles}⟶\text{simplest ratio}$$

For percentage composition, assume a $100\mathrm{g}$ sample so that each percentage becomes a mass in grams.

To find a molecular formula:

$$k=\frac{M_r\text{ of compound}}{M_r\text{ of empirical formula}}$$

Then multiply every empirical-formula subscript by $k$.

Next, you will use mole ratios from balanced equations to calculate reacting masses and decide which reactant is limiting. The same discipline—convert to moles first, then use a ratio—will be central again.