Empirical and molecular formulae — Chemistry

Before this

This leads to

Stoichiometry — mole ratios in equations

Three substances, one recipe

Here are three substances. Methanal is a gas with a sharp smell, used to preserve specimens. Ethanoic acid is the acid in vinegar. Glucose is the sugar your cells run on. They look, smell and behave nothing alike. Yet if you analyse each one, you find exactly the same result: 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.

How can three different substances have the same composition? Look at their formulae: methanal is CH₂O, ethanoic acid is C₂H₄O₂, and glucose is C₆H₁₂O₆. In all three, there are twice as many hydrogen atoms as carbon atoms, and as many oxygen atoms as carbon atoms. The ratio is the same, CH₂O. The numbers are different.

This is the problem this lecture solves. An analysis in a laboratory gives you masses. From masses, you can find the ratio of atoms. But the ratio alone cannot tell glucose from vinegar. You need one more piece of information to find the real formula.

Two kinds of formula

Empirical and molecular formulae

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

The molecular formula is the actual number of atoms of each element in one molecule. It is always a whole-number multiple of the empirical formula.

Sometimes the two are the same. Water, H₂O, cannot be simplified, so its empirical and molecular formulae are both H₂O. Sometimes they differ. Ethane is C₂H₆, and dividing by 2 gives the empirical formula CH₃. Benzene is C₆H₆, with empirical formula CH. Hydrogen peroxide is H₂O₂, with empirical formula HO.

The word "empirical" means "found by experiment". That is the point of the empirical formula. It is what an experiment gives you directly, before you know anything else about the molecule.

For ionic compounds, such as magnesium oxide or iron oxide, there are no molecules. A crystal is a huge lattice of ions. So for ionic compounds we only ever write the empirical formula. MgO means "magnesium and oxide ions in a 1 : 1 ratio".

From masses to a ratio of atoms

A laboratory finds that magnesium oxide is 60.3% magnesium and 39.7% oxygen by mass. What is its formula?

Start by imagining exactly 100 g of the compound. This is only a convenience: percentages then become grams directly. So 100 g of magnesium oxide contains 60.3 g of magnesium and 39.7 g of oxygen.

Stop and think

The mass ratio is 60.3 : 39.7, which is about 3 : 2. So is the formula Mg₃O₂?

Show the answer

No. A formula counts atoms, not grams. A magnesium atom is heavier than an oxygen atom (relative masses 24 and 16), so a gram of magnesium contains fewer atoms than a gram of oxygen. To compare numbers of atoms, you must turn each mass into moles first. When you do, the ratio becomes 1 : 1, and the formula is MgO. Comparing masses directly is like comparing a bag of oranges with a bag of grapes by weight and calling it a count of fruit.

The last lecture called the mole the hub of chemistry calculations, and promised that finding a formula from data is "a trip through the hub". Here it is. Divide each mass by its relative atomic mass:

n (Mg) = \frac{60.3}{24} \approx 2.51 mol

n (O) = \frac{39.7}{16} \approx 2.48 mol

Now the amounts are counts of atoms (in units of $6.02 \times 1 0^{23}$ ). To see the simplest ratio, divide both by the smaller one, 2.48:

Mg : O = \frac{2.51}{2.48} : \frac{2.48}{2.48} \approx 1.01 : 1

That is 1 : 1, so the empirical formula is MgO. The 1.01 is not exactly 1 only because the percentages were rounded.

The same 100 g of magnesium oxide. By mass, magnesium is about 3 parts in 5. Counted in moles, there is one magnesium atom for every oxygen atom.

That is it. That is the whole idea. Masses do not count atoms. Moles do. Turn every mass into moles, then compare. Everything else in this lecture is organisation.

The method, with the reason for each step

Write the working as a table, with one column for each element. Each row is one step:

Step 1 — Mass. Write the mass of each element in grams. From percentages, imagine 100 g. Reason: we need amounts we can convert.
Step 2 — Divide by $A_{r}$ . This gives the amount of each element in moles. Reason: moles count atoms; grams do not.
Step 3 — Divide by the smallest amount. Reason: this scales the counts so that the rarest element has 1, which shows the ratio.
Step 4 — Make whole numbers. If every value is within about 0.05 of a whole number, round. If not, multiply all of them by the same small number until they are whole. Reason: a formula counts whole atoms.

Step 4 needs care. A value of 1.01 or 2.98 is a whole number spoiled by rounding. A value of 1.5 is not. It means that for every 2 atoms of one element there are 3 of the other. Learn to recognise the common fractions:

Value after Step 3 ends in about	It is	Multiply everything by
.5	a half	2
.33 or .67	a third or two thirds	3
.25 or .75	a quarter or three quarters	4

The most common mistake in this whole topic is rounding 1.5 to 2. Find it in the working below.

Spot the error — an iron oxide

Tunde analyses an iron oxide: 70.0% iron and 30.0% oxygen by mass. One line of his working is wrong. Tap it.

Tunde's working for an iron oxide with 70.0% iron and 30.0% oxygen, with the error marked:

In 100 g: 70.0 g Fe and 30.0 g O. Correct.
Moles: Fe $70.0 \div 56 = 1.25$ ; O $30.0 \div 16 = 1.875$ . Correct.
Divide by the smaller, 1.25: Fe 1, O 1.5. Correct.
Round to whole numbers: Fe 1, O 2, so FeO₂. Wrong: 1.5 is not a rounding error. It is a half.
Correct: multiply both by 2. Fe 2, O 3, so the empirical formula is Fe₂O₃.

Notice the check in the last line. Any formula can be tested by working the percentage backwards. Fe₂O₃ has $M_{r} = 2 \times 56 + 3 \times 16 = 160$ , and $112 \div 160 = 70.0%$ iron, which matches the data. FeO₂ gives 63.6%, which does not.

From the empirical formula to the molecular formula

Back to the three substances with the same recipe. The analysis gives CH₂O for all of them. To tell them apart, you need one more measurement: the relative molecular mass, $M_{r}$ , of the substance. Today this usually comes from a mass spectrometer, which measures the mass of whole molecules.

The empirical formula CH₂O has a mass of $12 + 2 \times 1 + 16 = 30$ . We call this the empirical formula mass. The molecule must contain a whole number $n$ of these units, so

n = \frac{M _{r} of the compound}{empirical formula mass}

For methanal, $M_{r} = 30$ , so $n = 1$ and the formula is CH₂O. For ethanoic acid, $M_{r} = 60$ , so $n = 2$ : C₂H₄O₂. For glucose, $M_{r} = 180$ , so $n = 6$ : C₆H₁₂O₆. One analysis plus one mass gives each answer.

Combustion analysis: burning a compound to read its formula

Most compounds of carbon do not come with percentages printed on them. For two centuries, chemists found their formulae by burning them. When a compound of carbon, hydrogen and oxygen burns completely in oxygen, every carbon atom ends up in carbon dioxide and every hydrogen atom ends up in water. Catch and weigh the carbon dioxide and the water, and you know how much carbon and hydrogen were in the sample.

A simple combustion train. The sample burns in a stream of oxygen. Calcium chloride absorbs the water, and potassium hydroxide absorbs the carbon dioxide. The gain in mass of each absorber is the mass of water or carbon dioxide formed.

In 1831, Justus von Liebig made this into a reliable routine: calcium chloride to absorb water, potassium hydroxide to absorb carbon dioxide (LibreTexts, Physical Methods in Chemistry). Oxygen cannot be found the same way, because the compound burns in a stream of oxygen. So oxygen is found last, by subtraction. Step through a full example below.

Step through — a liquid burned in oxygen

2.30 g of a liquid containing only carbon, hydrogen and oxygen burns completely. It forms 4.40 g of carbon dioxide and 2.70 g of water. Press Next to add one step.

2.30 g of a liquid containing only C, H and O burns completely, forming 4.40 g of CO₂ and 2.70 g of H₂O. Each step and its reason:

Step 1. $n (CO_{2}) = 4.40 \div 44 = 0.100$ mol, so $n (C) = 0.100$ mol. Each CO₂ holds one C atom, and all the carbon ends up in CO₂.
Step 2. Mass of C $= 0.100 \times 12 = 1.20$ g. Needed for the subtraction later.
Step 3. $n (H_{2} O) = 2.70 \div 18 = 0.150$ mol, so $n (H) = 2 \times 0.150 = 0.300$ mol. Each H₂O holds two H atoms.
Step 4. Mass of H $= 0.300 \times 1 = 0.300$ g.
Step 5. Mass of O $= 2.30 - 1.20 - 0.300 = 0.800$ g, so $n (O) = 0.800 \div 16 = 0.0500$ mol. The rest of the sample's mass must be oxygen.
Step 6. Divide by 0.0500: C 2, H 6, O 1. The smallest amount becomes 1.

Result: the empirical formula is C₂H₆O. If $M_{r} = 46$ , then $n = 46 \div 46 = 1$ , and the molecular formula is also C₂H₆O. This is ethanol.

The order matters: carbon from carbon dioxide, hydrogen from water, oxygen by subtraction. And check the result where you can. Ethanol, C₂H₆O, has $M_{r} = 2 \times 12 + 6 + 16 = 46$ , and 2.30 g of it is exactly 0.0500 mol. Burning 0.0500 mol would make 0.100 mol of CO₂ and 0.150 mol of H₂O, which is what was measured.

Worked examples

Relative atomic masses: H 1, C 12, N 14, O 16, Mg 24, S 32, Fe 56, Cu 64.

Worked Example 1 — From a crucible experiment

A student heats 1.20 g of magnesium ribbon in a crucible until it has all reacted with oxygen from the air. The product has a mass of 2.00 g. Find the empirical formula.

Masses. Magnesium: 1.20 g. Oxygen: the gain in mass, $2.00 - 1.20 = 0.80$ g.

Moles. Mg: $1.20 \div 24 = 0.0500$ mol. O: $0.80 \div 16 = 0.0500$ mol.

Ratio. 1 : 1, so the empirical formula is MgO.

Check. MgO has $M_{r} = 40$ , and 0.0500 mol of it is $0.0500 \times 40 = 2.00$ g ✓. Notice that no percentages were needed: masses from the experiment work directly.

Worked Example 2 — A ratio ending in .33

A black iron oxide contains 72.4% iron and 27.6% oxygen by mass. Find its empirical formula.

Moles in 100 g. Fe: $72.4 \div 56 \approx 1.293$ . O: $27.6 \div 16 = 1.725$ .

Divide by the smaller. Fe 1, O $1.725 \div 1.293 \approx 1.334$ .

Whole numbers. 1.334 is a third more than 1, so multiply by 3: Fe 3, O 4.00. The empirical formula is Fe₃O₄.

Check. $M_{r} = 3 \times 56 + 4 \times 16 = 232$ , and $168 \div 232 \approx 72.4%$ iron ✓.

Worked Example 3 — The sugar

A white solid is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Find its empirical and molecular formulae.

Moles in 100 g. C: $40.0 \div 12 \approx 3.33$ . H: $6.7 \div 1 = 6.7$ . O: $53.3 \div 16 \approx 3.33$ .

Divide by 3.33. C 1.00, H 2.01, O 1.00. The 2.01 is within 0.05 of 2, so it is rounding: the empirical formula is CH₂O.

Molecular formula. Empirical formula mass $= 30$ . $n = 180 \div 30 = 6$ . So the molecular formula is C₆H₁₂O₆: glucose.

Check. $6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180$ ✓.

Three classic mistakes

Trap 1 — Rounding a real fraction

A value of 1.5, 1.33 or 1.25 after Step 3 is not a rounding error. Round only values within about 0.05 of a whole number. Otherwise multiply everything by 2, 3 or 4. A backwards check of the percentage will catch the mistake.

Trap 2 — Dividing oxygen by 32

Oxygen gas is O₂, with $M_{r} = 32$ . But in a compound you are counting oxygen atoms, so divide by 16. The same is true for nitrogen (14, not 28) and chlorine (35.5, not 71).

Trap 3 — Forgetting the 2 in water

In combustion analysis, each mole of H₂O contains two moles of hydrogen atoms. Writing $n (H) = n (H_{2} O)$ halves the hydrogen and gives a wrong formula. And never try to find oxygen from the CO₂ or H₂O: some of that oxygen came from the air. Find it by subtraction.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Relative atomic masses: H 1, C 12, N 14, O 16, S 32, Cu 64.

A gas is 75.0% carbon and 25.0% hydrogen by mass. Find its empirical formula.

Hint 1
Imagine 100 g, and divide each mass by its $A_{r}$ .

Hint 2
C: $75.0 \div 12$ . H: $25.0 \div 1$ . Then divide both by the smaller.

Full answer
C: $75.0 \div 12 = 6.25$ mol. H: $25.0 \div 1 = 25.0$ mol. Divide by 6.25: C 1, H 4. Empirical formula CH₄. Check: $12 \div 16 = 75.0%$ carbon ✓. (This gas is methane.)
An oxide of nitrogen is 30.4% nitrogen and 69.6% oxygen by mass. Its relative molecular mass is 92. Find its empirical and molecular formulae.

Hint 1
Divide oxygen by 16, not 32. You are counting atoms.

Hint 2
The empirical formula mass of NO₂ is $14 + 2 \times 16 = 46$ .

Full answer
N: $30.4 \div 14 \approx 2.17$ mol. O: $69.6 \div 16 = 4.35$ mol. Divide by 2.17: N 1, O 2.00. Empirical formula NO₂, with mass 46. $n = 92 \div 46 = 2$ , so the molecular formula is N₂O₄. Check: $28 \div 92 \approx 30.4%$ nitrogen ✓.
0.600 g of a compound containing only carbon and hydrogen burns completely, forming 1.76 g of carbon dioxide and 1.08 g of water. Its relative molecular mass is 30. Find its empirical and molecular formulae.

Hint 1
Carbon from CO₂ ( $M_{r} = 44$ ), hydrogen from H₂O ( $M_{r} = 18$ ). Remember the 2.

Hint 2
There is no oxygen in this compound. Check instead that the masses of C and H add up to 0.600 g.

Full answer
C: $1.76 \div 44 = 0.0400$ mol, mass $0.0400 \times 12 = 0.480$ g. H: $1.08 \div 18 = 0.0600$ mol of water, so $0.120$ mol of H atoms, mass 0.120 g. Check: $0.480 + 0.120 = 0.600$ g ✓, so there really is no oxygen. Ratio C : H = 0.0400 : 0.120 = 1 : 3. Empirical formula CH₃, mass 15. $n = 30 \div 15 = 2$ , so the molecular formula is C₂H₆ (ethane).
Blue copper sulfate crystals contain water: their formula is CuSO₄· $x$ H₂O. A student heats 2.50 g of the crystals until all the water is driven off. The white solid left, CuSO₄, has a mass of 1.60 g. Find $x$ .

Hint 1
Treat CuSO₄ and H₂O as the two "units" whose ratio you want, exactly as you treated two elements.

Hint 2
Mass of water lost $= 2.50 - 1.60$ . $M_{r}$ of CuSO₄ $= 64 + 32 + 4 \times 16 = 160$ .

Full answer
CuSO₄: $1.60 \div 160 = 0.0100$ mol. Water: $2.50 - 1.60 = 0.90$ g, and $0.90 \div 18 = 0.0500$ mol. Ratio CuSO₄ : H₂O $= 0.0100 : 0.0500 = 1 : 5$ , so $x = 5$ : CuSO₄·5H₂O. Check: $M_{r} = 160 + 5 \times 18 = 250$ , and 0.0100 mol of it is 2.50 g ✓. The same method works for any "units", not just single atoms.

The bigger idea

Combustion analysis built organic chemistry. With it, nineteenth-century chemists found the formulae of sugars, acids and many natural substances, long before anyone could see a molecule. The method kept improving. In 1923, Fritz Pregl received the Nobel Prize in Chemistry "for his invention of the method of micro-analysis of organic substances". His method needed only about 5 mg of sample or less (Nobel Prize; LibreTexts). Automated versions of the same idea still run in laboratories today, because they are cheap and reliable.

But the hook of this lecture shows the method's limit. Methanal, vinegar's acid and glucose give identical analyses. A formula from masses alone tells you the ratio, never the molecule. Even the molecular formula is not the end: different molecules can share one molecular formula. C₂H₆O is ethanol, but it is also methoxymethane, a gas, with the atoms joined in a different order. Chemistry keeps asking the same question at deeper levels: first what ratio, then how many, then how are they joined?

The next lecture uses the same move in the other direction. Instead of turning measured masses into a formula, you will use a balanced equation to turn the moles of one substance into the moles of another.

Key concepts in this lesson

The empirical formula is the simplest whole-number ratio of atoms, and the molecular formula is the actual number of atoms in one molecule, a whole-number multiple of it.
Masses do not count atoms, so every mass must be turned into moles by dividing by the relative atomic mass before any ratio is taken.
After dividing by the smallest amount, values close to whole numbers are rounded, while values ending in about .5, .33 or .25 are multiplied by 2, 3 or 4.
The molecular formula is found from $n = M_{r} \div$ empirical formula mass, using a measured relative molecular mass.
In combustion analysis, carbon is found from the carbon dioxide, hydrogen from the water (two atoms per molecule), and oxygen by subtraction.