What are molecular and empirical formulae?

The molecular formula gives the actual atoms in one molecule; the empirical formula gives their simplest whole-number ratio. Glucose is C6H12O6. Dividing by 6 gives its empirical formula, CH2O. The molecular formula is always a whole-number multiple of the empirical formula.

Formulae, Functional Groups and Terminology Notes

Cambridge IGCSE Chemistry: Revision notes

Key facts

The empirical formula is the simplest whole-number ratio of atoms; the molecular formula shows the actual atoms.
Balance equations with coefficients only, never by changing subscripts.
Ionic formulae come from the charges: swap the numbers (criss-cross) and simplify.
A functional group decides the chemical properties; a homologous series differs by $\ce{CH2}$ .
Net ionic equations leave out spectator ions.

Molecular and empirical formulae

The molecular formula gives the actual atoms in one molecule; the empirical formula gives their simplest whole-number ratio.

Glucose is $\ce{C6H12O6}$ . Dividing by 6 gives its empirical formula, $\ce{CH2O}$ . The molecular formula is always a whole-number multiple of the empirical formula.

To find an empirical formula:

Divide each mass by $A_r$ to get moles.
Divide by the smallest number.
Multiply up if needed to get whole numbers.

1
Moles
Divide each mass by Ar
2
Smallest number
Divide all the moles by the smallest
3
Whole numbers
Multiply up if needed: this is the empirical formula
4
Molecular formula
A whole-number multiple n of the empirical formula

Finding an empirical formula, then the molecular formula (glucose, C₆H₁₂O₆, has empirical formula CH₂O).

Molecular formula= n × empirical formula

Worked example

A compound contains 36 g of carbon, 6 g of hydrogen and 32 g of oxygen. Find its empirical formula.

1
Moles: C = 36 ÷ 12 = 3 mol; H = 6 ÷ 1 = 6 mol; O = 32 ÷ 16 = 2 mol.
2
Divide by the smallest (2): C = 1.5, H = 3, O = 1.
3
Multiply by 2 to get whole numbers: C = 3, H = 6, O = 2.

Molecular

Actual atoms in one molecule
Ethene: $\ce{C2H4}$
Glucose: $\ce{C6H12O6}$

Empirical

Simplest whole-number ratio
Ethene: $\ce{CH2}$
Glucose: $\ce{CH2O}$

What is the empirical formula of $\ce{C2H4}$ ?

Balancing equations

A balanced symbol equation has equal numbers of each atom on both sides, adjusted using coefficients only.

Word equations use names: magnesium + oxygen → magnesium oxide.

Symbol equations use formulae and must be balanced. Add coefficients in front of formulae and never change subscripts. State symbols are (s) solid, (l) liquid, (g) gas and (aq) aqueous.

$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$

1
Write the formulae
correct formulae for every substance
2
Count atoms
each element on each side
3
Add coefficients
numbers in front of formulae
4
Check
atoms equal on both sides

Balancing an equation

Worked example

Balance: $\ce{Fe + Cl2 -> FeCl3}$

1
Left: 1 Fe, 2 Cl. Right: 1 Fe, 3 Cl.
2
Chlorine needs a common multiple of 2 and 3, which is 6: use 3 Cl₂ and 2 FeCl₃.
3
Now Fe needs a 2 to match 2 FeCl₃.
4
Check: left 2 Fe and 6 Cl; right 2 Fe and 6 Cl.

Which numbers balance $\ce{Mg + O2 -> MgO}$ ?

Formulae of ionic compounds

The charges on the ions decide the formula, so the total charge of the compound is zero.

Use the criss-cross method: write each ion's charge, swap the numbers (ignoring signs) to become subscripts, then simplify if there is a common factor. Use brackets round a polyatomic ion when its subscript is more than 1.

$\ce{Al^3+}$ and $\ce{Cl-}$ give $\ce{AlCl3}$ . $\ce{Ca^2+}$ and $\ce{O^2-}$ give $\ce{CaO}$ . $\ce{Ca^2+}$ and $\ce{NO3-}$ give $\ce{Ca(NO3)2}$ .

Calcium ion

2,8,8

Oxide ion

2,8

Calcium oxide: Ca²⁺ (2, 8, 8) and O²⁻ (2, 8) have equal and opposite charges, so the formula is CaO and the total charge is zero.

Cations

$\ce{Na+}$
$\ce{Mg^2+}$
$\ce{Ca^2+}$
$\ce{Al^3+}$
$\ce{NH4+}$

Anions

$\ce{Cl-}$
$\ce{O^2-}$
$\ce{NO3-}$
$\ce{SO4^2-}$
$\ce{CO3^2-}$

Worked example

Deduce the formula of the compound formed between $\ce{Al^3+}$ and $\ce{SO4^2-}$ .

1
Charges: Al is 3+, sulfate is 2−.
2
Criss-cross: Al takes the 2 from sulfate, and sulfate takes the 3 from Al.
3
Use brackets round the polyatomic ion: $\ce{Al2(SO4)3}$ .
4
Check the total charge: (2 × +3) + (3 × −2) = +6 − 6 = 0.

What is the formula of calcium nitrate?

Functional groups

A functional group is the atom or group that decides how a compound reacts, so members of a series behave alike.

A functional group determines the chemical properties of an organic compound. A homologous series has the same functional group and general formula, differs by $\ce{CH2}$ and has similar chemical properties. Alkanes are $C_nH_{2n+2}$ and alkenes $C_nH_{2n}$ .

Saturated compounds have only single C–C bonds; unsaturated ones contain C=C. Isomers share a molecular formula but have different structures: butane and methylpropane are both $\ce{C4H10}$ .

1
Alkane
C–C single bonds only; -ane; methane CH₄
2
Alkene
C=C; -ene; ethene C₂H₄
3
Alcohol
–OH; -ol; methanol CH₃OH
4
Carboxylic acid
–COOH; -oic acid; ethanoic acid CH₃COOH

Functional groups, their name endings and an example of each.

\ce{CH4}

	Alkane	Alkene	Alcohol	Carboxylic acid
Functional group	C–C single bonds only	C=C	–OH	–COOH
Name ending	-ane	-ene	-ol	-oic acid
Example	$\ce{CH4}$ methane	$\ce{C2H4}$ ethene	$\ce{CH3OH}$ methanol	$\ce{CH3COOH}$ ethanoic acid

Alkane

Functional group:: C–C single bonds only
Name ending:: -ane
Example:: $\ce{CH4}$ methane

Alkene

Functional group:: C=C
Name ending:: -ene
Example:: $\ce{C2H4}$ ethene

Alcohol

Functional group:: –OH
Name ending:: -ol
Example:: $\ce{CH3OH}$ methanol

Carboxylic acid

Functional group:: –COOH
Name ending:: -oic acid
Example:: $\ce{CH3COOH}$ ethanoic acid

Other series you may meet: aldehydes (–CHO, ending -al, e.g. methanal) and ketones (–CO–, ending -one, e.g. propanone).

What is a homologous series?

Ionic equations

An ionic equation shows only the particles that change; spectator ions are left out.

Write the balanced equation with state symbols, split soluble ionic compounds into ions, keep solids, gases and covalent compounds such as water whole, then cancel the spectator ions to leave the net ionic equation.

$\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$

In $\ce{AgNO3 + NaCl}$ the spectators are $\ce{NO3-}$ and $\ce{Na+}$ .

1
Full equation
balanced, with state symbols
2
Split into ions
soluble ionic compounds only
3
Spot spectators
identical on both sides
4
Remove them
what is left is the net ionic equation

Writing a net ionic equation

Worked example

Write the net ionic equation for $\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$ .

1
Complete ionic: $\ce{H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + Cl-(aq) + H2O(l)}$ .
2
Water stays as a whole formula because it is covalent.
3
Spectator ions are $\ce{Na+}$ and $\ce{Cl-}$ because they appear unchanged on both sides.

What is a spectator ion?

Try an exam question

Write a balanced equation with state symbols for the reaction of silver nitrate solution with sodium chloride solution, and then write the ionic equation for the reaction.

[4 marks]

That's the notes covered.

Carry on to the next subtopic.