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D. Quantities in Chemical Reactions · Quantities in Chemical Reactions

Molecular Formulas

D2.4 D3.3 D2.1About 70 minutes
Learning goals
  • I can explain the difference between an empirical formula and a molecular formula.
  • I can find a molecular formula from an empirical formula and a molar mass.
  • I can find a molecular formula starting from percentage composition or combustion data.
  • I can explain why ionic compounds are written with empirical formulas only.

Words to know

molecular formula
The actual number of atoms of each element in one molecule, for example C₆H₁₂O₆.
empirical formula
The simplest whole-number ratio of atoms, for example CH₂O.
multiplier (n)
The whole number that the empirical formula is multiplied by: n = M(molecular) ÷ M(empirical).
mass spectrometer
An instrument that measures the masses of molecules (and their fragments), often used to find molar mass.

Sweet, sour and sharp-smelling

Glucose is the sugar your cells burn for energy. Acetic acid gives vinegar its sour taste. Formaldehyde is a sharp-smelling gas used to make some glues and resins. They could not be more different, yet a percentage-composition analysis of each one gives exactly the same result: 40.0% C, 6.7% H, 53.3% O. All three have the empirical formula CH₂O.

So the empirical formula is not the whole story. To know which compound you have, you also need its molar mass.

Empirical vs molecular formulas

  • The empirical formula gives the simplest ratio of atoms.
  • The molecular formula gives the actual number of atoms in one molecule.

The molecular formula is always a whole-number multiple of the empirical formula:

molecular formula = (empirical formula) × n, where n = M(molecular) ÷ M(empirical)

Sometimes n = 1 and the two formulas are the same: water, H₂O, and carbon dioxide, CO₂, are both.

Ionic compounds such as NaCl or CaCl₂ do not form separate molecules; they are giant lattices of ions. So they are always written with the simplest ratio, the empirical formula (one formula unit).

empirical formulaCH₂O (30.03 g/mol)formaldehydeCH₂O30.03 g/mol× 1acetic acid (in vinegar)C₂H₄O₂60.06 g/mol× 2glucoseC₆H₁₂O₆180.18 g/mol× 6
One empirical formula, three different compounds. The multiplier n comes from the molar mass of each compound.
Worked example

Example 1: Empirical formula and molar mass given

A compound has the empirical formula CH₂O and a molar mass of 180.18 g/mol. Find its molecular formula.

  1. M(CH₂O) = 12.01 + 2(1.01) + 16.00 = 30.03 g/mol.
  2. n = 180.18 ÷ 30.03 = 6.00.
  3. Multiply every subscript by 6: C (1 × 6 = 6), H (2 × 6 = 12), O (1 × 6 = 6).

The molecular formula is C₆H₁₂O₆ (glucose).

CH₂OCH₂OCH₂OCH₂OCH₂OCH₂Oglucose, C₆H₁₂O₆: 180.18 g/mol30.0330.0330.0330.0330.0330.03n = 180.18 ÷ 30.03 = 6.00, so molecular formula = (CH₂O)₆ = C₆H₁₂O₆
The molar mass of glucose is made of exactly six "blocks" of CH₂O. That is why n must be a whole number.
Worked example

Example 2: Starting from percentages

A hydrocarbon is 85.6% C and 14.4% H. Its molar mass is 56.12 g/mol. Find its molecular formula.

  1. Empirical formula (last lesson): C: 85.6 ÷ 12.01 = 7.127; H: 14.4 ÷ 1.01 = 14.26. Ratio 1 : 2. Empirical formula CH₂.
  2. M(CH₂) = 12.01 + 2(1.01) = 14.03 g/mol.
  3. n = 56.12 ÷ 14.03 = 4.00.

Molecular formula C₄H₈.

Worked example

Example 3: Same empirical formula, different molecules

Benzene is 92.24% C and 7.76% H, with a molar mass of 78.12 g/mol. Find its molecular formula.

  1. C: 92.24 ÷ 12.01 = 7.680 mol; H: 7.76 ÷ 1.01 = 7.683 mol. Ratio 1 : 1. Empirical formula CH.
  2. M(CH) = 13.02 g/mol. n = 78.12 ÷ 13.02 = 6.00.

Molecular formula C₆H₆. Ethyne (acetylene), C₂H₂, the gas in welding torches, has the same empirical formula, CH.

HCCHethyne (acetylene), C₂H₂26.04 g/molHHHHHHbenzene, C₆H₆ (a ring of 6 C)78.12 g/molBoth have the empirical formula CH (13.02 g/mol): n = 2 and n = 6.
Ethyne and benzene have the same empirical formula, CH, and the same percentage composition, but different molecular formulas and very different structures.

Where does the molar mass come from?

In research labs, molar mass usually comes from a mass spectrometer. In a school lab it can come from measuring the mass and volume of a gas sample (you will learn the gas laws in the unit Gases and Atmospheric Chemistry) or from how much a substance lowers the freezing point of a solvent. For problems in this unit, the molar mass is given.

Worked example

Example 4: Combustion analysis to molecular formula

A 0.600 g sample of a compound containing C, H and O burns completely to give 0.880 g of CO₂ and 0.360 g of H₂O. Its molar mass is 60.06 g/mol. Find the molecular formula.

  1. mol C = mol CO₂ = 0.880 ÷ 44.01 = 0.02000 mol; mass C = 0.02000 × 12.01 = 0.2402 g.
  2. mol H = 2 × (0.360 ÷ 18.02) = 0.03996 mol; mass H = 0.03996 × 1.01 = 0.0404 g.
  3. mass O = 0.600 − 0.2402 − 0.0404 = 0.3194 g; mol O = 0.3194 ÷ 16.00 = 0.01996 mol.
  4. Divide by 0.01996: C 1.00, H 2.00, O 1.00. Empirical formula CH₂O (30.03 g/mol).
  5. n = 60.06 ÷ 30.03 = 2.00.

Molecular formula C₂H₄O₂, acetic acid.

Just for fun

Glucose (sweet), acetic acid (sour) and formaldehyde (sharp-smelling) all share the empirical formula CH₂O. Same simplest ratio, completely different personalities: proof that you should never judge a molecule by its empirical formula.

Hands-on activity

Build a family of CH₂ molecules

You need: small marshmallows or gumdrops in two colours (or balls of modelling clay in two colours) and toothpicks. Toothpicks are sharp; younger siblings should be supervised, and do not eat the pieces after handling them.

  1. Choose one colour for carbon and one for hydrogen. Each carbon needs 4 bonds and each hydrogen 1 bond. A double bond is two toothpicks side by side.
  2. Build ethene, C₂H₄ (two carbons joined by a double bond, two H on each carbon).
  3. Build a ring of three carbons, each with two H: cyclopropane, C₃H₆.
  4. Build a ring of four carbons, each with two H: cyclobutane, C₄H₈.
  5. For each model, count atoms, write the molecular formula and the empirical formula, and find n.

What to look for: all three have the empirical formula CH₂, with n = 2, 3 and 4. Their percentage composition is identical (85.6% C), so only their molar masses (28.06, 42.09 and 56.12 g/mol) tell them apart.

Your turn

1

A gas has the empirical formula CH and a molar mass of 26.04 g/mol. Find its molecular formula.

2

A brown gas has the empirical formula NO₂ and a molar mass of 92.02 g/mol. Find its molecular formula.

3

A fuel has the empirical formula C₂H₅ and a molar mass of 58.14 g/mol. Find its molecular formula.

4

A compound is 40.0% C, 6.7% H and 53.3% O, with a molar mass of 60.06 g/mol. Find its molecular formula.

5

An oxide of phosphorus has the empirical formula P₂O₅ and a molar mass of 283.88 g/mol. Find its molecular formula.

6

Why does it make no sense to look for the "molecular formula" of sodium chloride?

More practice: warm-up

1

Lactic acid has the empirical formula CH₂O and a molar mass of 90.09 g/mol. What is n?

2

What is the empirical formula of octane, C₈H₁₈?

3

Write the formula for finding n.

4

True or false: two different compounds can have the same empirical formula.

5

What is the empirical formula of benzene, C₆H₆?

More practice: on your own

1

Caffeine is 49.5% C, 5.20% H, 28.9% N and 16.5% O. Its molar mass is 194.2 g/mol. Find its molecular formula.

2

A hydrocarbon is 82.7% C and 17.3% H, with a molar mass of 58.12 g/mol. Find its molecular formula.

3

Ethylene glycol, used in some car antifreeze, is 38.7% C, 9.74% H and 51.6% O. Its molar mass is 62.08 g/mol. Find its molecular formula.

4

A compound has the empirical formula CH₂, and 0.0250 mol of it has a mass of 2.10 g. Find its molecular formula.

5

Two unknown gases both analyse as 85.6% carbon and 14.4% hydrogen. Explain how a chemist could tell them apart.

Challenge

1

Nicotine is 74.0% C, 8.70% H and 17.3% N. Its molar mass is 162.2 g/mol. Find its molecular formula.

2

A 1.22 g sample of a compound of C, H and O burns to give 3.08 g of CO₂ and 0.540 g of H₂O. Its molar mass is 122.13 g/mol. Find its molecular formula.

3

Explain why n must always be a whole number, and what you should suspect if your calculation gives n = 2.5.

Check your understanding

Quiz

1.

A compound has empirical formula CH₂O and molar mass 120.12 g/mol. What is its molecular formula?

2.

Which pair has the same empirical formula?

3.

For which compound are the empirical and molecular formulas the same?

4.

What extra information do you need to go from an empirical formula to a molecular formula?

5.

Why are ionic compounds such as MgCl₂ written only with empirical formulas?

6.

A hydrocarbon has empirical formula C₂H₅ and molar mass 58.14 g/mol. What is its molecular formula?

For parents and teachers

Big idea: The empirical formula is the simplest ratio; the molecular formula is the real count of atoms in one molecule. n = molar mass ÷ empirical-formula mass. This lesson completes D2.4 (empirical and molecular formula problems) and D3.3 (the relationship between the two).

How to help: Ask your teen to explain why glucose and vinegar's acetic acid can share an empirical formula. Building the CH₂ models together is a quick, memorable activity (supervise toothpicks with young children).

Common difficulties:

  • Dividing the molar mass by the wrong thing (it must be the empirical-formula mass, not an atomic mass).
  • Multiplying only some subscripts by n; every subscript gets multiplied.
  • Getting a non-whole n and forcing it. A non-whole n means an earlier step went wrong.
  • In combustion analysis, forgetting to find oxygen by subtraction.