IMAT Chemistry · Stoichiometry and Reactions

Percentage Composition, Empirical and Molecular Formulas

The masses of the elements in a compound, turned into moles, give the simplest whole-number ratio of atoms; the molar mass then gives the real formula.

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Why this matters

Finding a formula from masses appears in 2015, 2019 and 2021: a hydrocarbon from its carbon content, a compound of carbon, hydrogen and oxygen, and an oxide of a metal. The options are usually formulas that differ only in their ratios.

Concept 1 of 3: Percentage by mass of an element in a compound

Every formula unit of a compound has the same make-up, so the share of the mass that belongs to one element is fixed. Work it out for one formula unit and it holds for any sample, from a milligram to a tonne.

Definition

Percentage by mass of an element: the mass of that element in one formula unit divided by the formula mass, times 100.
Multiply the element's $A_r$ by the number of its atoms in the formula.
The mass of an element in any sample is the sample mass times that fraction.
This is the law of definite proportions: a pure compound always has the same composition by mass, however it was made.

Percentage by mass

\%\,\text{element} = \frac{x \times A_r}{M_r} \times 100

$x$ number of atoms of the element in the formula
$A_r$ relative atomic mass of the element
$M_r$ relative formula mass of the compound

Worked example

Ammonium nitrate,

\mathrm{NH_4NO_3}

, is used as a fertiliser. What percentage of its mass is nitrogen? (N = 14, H = 1, O = 16)

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Share of atoms is not share of mass

Two of the five atoms in

\mathrm{Fe_2O_3}

are iron, but iron is 70% of the mass, not 40%, because an iron atom is much heavier than an oxygen atom. Always weight each atom by its

A_r

Concept 2 of 3: Empirical formula from masses or percentages

A formula is a ratio of atoms, but analysis gives a ratio of masses. Dividing each mass by its atomic mass turns masses into moles, and moles are proportional to numbers of atoms. Then you only need to tidy the ratio into small whole numbers.

Definition

The empirical formula is the simplest whole-number ratio of atoms in a compound. To find it:

Write the mass of each element (with percentages, take a 100 g sample so each percentage becomes grams).
If one element's mass is missing, it is the total minus the others.
Divide each mass by its $A_r$ to get moles.
Divide every result by the smallest one.
If a value ends in .5, double everything; if it ends in .33 or .67, triple everything.

Mole ratio of elements

\text{ratio of atoms} = \frac{m_1}{A_{r,1}} : \frac{m_2}{A_{r,2}} : \dots

$m_1, m_2$ masses of each element in the sample, in g
$A_{r,1}, A_{r,2}$ their relative atomic masses

Worked example

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. (C = 12, H = 1, O = 16)

Practice this conceptself-check · 4 quick reps

Do not round 1.5 or 1.33 to the nearest whole number

A ratio of 1 : 1.5 is 2 : 3, and 1 : 1.33 is 3 : 4. Rounding them to 1 : 1 or 1 : 2 gives a wrong formula, and IMAT lists those wrong formulas as options. Only values within about 0.1 of a whole number may be rounded.

Concept 3 of 3: Molecular formula from the empirical formula and molar mass

The empirical formula is only a ratio:

\mathrm{CH_2}

\mathrm{C_2H_4}

and

\mathrm{C_6H_{12}}

all share it. To pick the real molecule you need its molar mass. The molecular formula is a whole-number multiple of the empirical formula, and the molar mass tells you which multiple.

Definition

The molecular formula gives the actual number of each atom in one molecule.
It is $n$ times the empirical formula, where $n$ is a whole number.
$n$ is the molar mass divided by the empirical formula mass.
Without a molar mass, any multiple is possible: a question can only ask which formula could be the molecular formula. Check each option by reducing it to its simplest ratio.
Ionic compounds have no molecules, so their formula is always the empirical one (NaCl, $\mathrm{Al_2O_3}$ ).

Molecular formula multiplier

n = \frac{M_r(\text{compound})}{M_r(\text{empirical formula})}

$n$ whole number that multiplies every subscript

Worked example

A hydrocarbon has the empirical formula

\mathrm{CH_2}

and a molar mass of 84 g/mol. What is its molecular formula? (C = 12, H = 1)

Practice this conceptself-check · 3 quick reps

Different compounds can share one empirical formula

Ethyne

\mathrm{C_2H_2}

and benzene

\mathrm{C_6H_6}

both have the empirical formula CH. Analysis by mass cannot tell them apart; only the molar mass can. And sometimes the two formulas are the same, as in methane,

\mathrm{CH_4}

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Percentage by mass of an element in a compound
Percentage by mass
$\%\,\text{element} = \frac{x \times A_r}{M_r} \times 100$
Empirical formula from masses or percentages
Mole ratio of elements
$\text{ratio of atoms} = \frac{m_1}{A_{r,1}} : \frac{m_2}{A_{r,2}} : \dots$
Molecular formula from the empirical formula and molar mass
Molecular formula multiplier
$n = \frac{M_r(\text{compound})}{M_r(\text{empirical formula})}$

Watch out for (3)

Share of atoms is not share of mass
→ Percentage by mass of an element in a compound
Do not round 1.5 or 1.33 to the nearest whole number
→ Empirical formula from masses or percentages
Different compounds can share one empirical formula
→ Molecular formula from the empirical formula and molar mass

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