1.4.4—Empirical and molecular formulas

An empirical formula gives the simplest whole-number ratio of atoms. A molecular formula gives the actual number of each atom in a molecule.

Step	Operation
1	Convert each composition value to moles
2	Divide all mole values by the smallest
3	Multiply to reach the simplest whole-number ratio
4	Write the empirical formula
5	Divide molecular molar mass by empirical-formula mass and multiply every subscript by that integer

Keep the empirical ratio simplest, and make sure the molecular-formula multiplier is a whole number consistent with the given molar mass.

A composition of 40.0% C, 6.7% H and 53.3% O gives the simplest ratio CH₂O after division by atomic masses. If the molar mass is 180 g mol⁻¹, compare it with the empirical-formula mass 30 to obtain the multiplier 6 and molecular formula C₆H₁₂O₆.

Do not round a ratio such as 1 : 1.50 : 1 to 1 : 2 : 1. Preserve the calculated values and multiply every ratio by the same small integer: ×2 converts halves, while ×3 can resolve values close to thirds such as 1.33 or 1.67. Round only after the common multiplier produces values consistent with the data precision.

How it is assessed

Questions derive an empirical formula from composition or scale an empirical formula to the molecular formula using molar mass.

Command terms

determine

What earns marks

Show conversion to moles, the simplest whole-number ratio, and the final formula; for molecular formula, use the integer molar-mass multiplier on every subscript.

Watch for

Rounding mole ratios before reaching whole numbers, or multiplying only one subscript when converting to the molecular formula.

Representative question

[Maximum number: 3]

4.32 g of the compound was combusted completely in oxygen and produced 9.49 g of $\mathrm{CO}_{2}$ and 5.18 g of $\mathrm{H}_{2} \mathrm{O}$ .
Determine the empirical formula of the compound, using sections 1 and 7 of the data booklet.

（a）

$n(\mathrm{C}) \ll=n\left(\mathrm{CO}_{2}\right)=\frac{9.49 \mathrm{~g}}{44.01 \mathrm{~g} \mathrm{~mol}^{-1}}>=0.216<\mathrm{mol}>$

AND

$n(\mathrm{H}) ~ « 2 n\left(\mathrm{H}_{2} \mathrm{O}\right)=2 \times \frac{5.18 \mathrm{~g}}{18.02 \mathrm{~g} \mathrm{~mol}^{-1}} »=0.575<\mathrm{mol} \gg$

OR
$H=0.581 \text { 《g》 }$

AND
$C=2.59 \text { «g»v }$ $\begin{aligned} & \text { 《 } m(\mathrm{O})=4.32-(m(\mathrm{C})+m(\mathrm{H})) & =4.32-(2.59+0.581)=1.15 \mathrm{~g} \| \end{aligned}$ $n(\mathrm{O})=\frac{1.15 \mathrm{~g}}{16.00 \mathrm{~g} \mathrm{~mol}^{-1}}$

／ 0.0718 «mol»
$<n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{O})=0.216: 0.575: 0.0718=3: 8: 1>$ $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$

M2 for finding mass of oxygen

M3 for finding empirical formula Award［3］for correct final answer．

Retrieve the quantitative chain: count entities with mN_A, sum relative masses from formulae, convert mass with n=m/M, derive formula ratios, use n=VC for solutions, and apply gas-volume ratios at the same temperature and pressure.

Before finalising, check the requested entity, formula subscripts, units, dm³ conversion, balanced-equation coefficients, and any limiting reactant.

1.4.4—Empirical and molecular formulas

Empirical and Molecular Formulae

Deriving Chemical Formulae

Assessment in practice

The Mole Summary

Question 1