Some Basic Concepts of Chemistry · Lesson 8 of 11

Percentage Composition and Chemical Formulas

“Translate between element mass percentages, simplest atom ratios, and molecular formulas.”

Learning Objectives

• Calculate element mass percentages from a formula. • Distinguish empirical formula from molecular formula. • Use composition data to find the simplest whole-number atom ratio. • Use an independently known molar mass to determine a molecular formula. • Explain how combustion products can reveal carbon and hydrogen amounts.

Two directions of the composition problem

A formula tells us how many atoms of each element occur in a chemical unit. A chemical analysis often supplies masses instead. We need a way to move from a formula to mass information and, in the reverse direction, from measured masses to an atom ratio.

Mass percentage expresses the mass contribution of one element as a fraction of the total compound mass, multiplied by 100. Use the same basis in numerator and denominator: for example, the mass of that element in one mole of compound divided by the mass of that mole of compound.

Elemental mass percentageLaTeX

Mass % of element = \frac{a A}{M} \times 100

a

is the number of atoms of the element in the formula,

A

its atomic molar mass, and

M

the compound molar mass. Numerator and denominator must use matching mass units and the same formula basis.

Example — Water composition

Problem
Find the hydrogen and oxygen mass percentages in $H_{2} O$ using H = 1.008 and O = 16.00.

1.One mole has mass $2 (1.008) + 16.00 = 18.016 g$ . Retain this working value during the percentage calculation.
2.Hydrogen: $(2.016/18.016) 100 = 11.19%$ . Oxygen: $(16.00/18.016) 100 = 88.81%$ to two decimal places.
3.The percentages add to 100.00%. Small differences from calculations that first round water mass to 18.02 arise from rounding, not from a change in chemical composition.

Example — Ethanol composition

Problem
Find the elemental percentages in $C_{2} H_{5} OH$ using C = 12.01, H = 1.008, O = 16.00.

1.Count the hydrogen in both parts of the formula: $C_{2} H_{5} OH$ has six hydrogens. Its molar mass is $2 (12.01) + 6 (1.008) + 16.00 = 46.068 g mo l^{- 1}$ .
2.The mass contributions per mole are 24.02 g C, 6.048 g H, and 16.00 g O.
3.Dividing each by 46.068 g and multiplying by 100 gives approximately 52.14% C, 13.13% H, and 34.73% O. Their sum provides a useful check.

Composition analysis can reveal that a sample is inconsistent with the expected pure compound. Agreement is evidence consistent with purity, but it is not an absolute guarantee: a mixture could sometimes give the same overall elemental percentages. Chemical identification may need more information than percentages alone.

The simplest ratio is not always the actual molecule

The ratio of atom numbers can be reduced like an ordinary whole-number ratio. This gives the empirical formula. It preserves the proportions of the elements, but not necessarily the number of atoms in one molecule.

Definition

Empirical formula

A formula showing the simplest whole-number ratio of atoms of the constituent elements.

Definition

Molecular formula

A formula showing the actual number of atoms of each constituent element in one molecule.

The molecular formula $C_{6} H_{12} O_{6}$ has ratio $6 : 12 : 6$ . Dividing by 6 gives $1 : 2 : 1$ , so its empirical formula is $C H_{2} O$ . The formulas $C H_{2} O$ and $C_{6} H_{12} O_{6}$ have the same elemental mass percentages but describe different molecule sizes. For an ionic solid, an empirical formula usually expresses an ion or atom ratio rather than a molecule.

From measured composition to a formula— Mass ratios must first become mole ratios. Atom ratios follow mole ratios, not mass ratios.

Why composition must be converted into moles

Equal masses of different elements contain different atom counts because the atoms have different masses. Dividing each element mass by its atomic molar mass converts the data into amounts. Their ratio then gives the ratio of atom numbers because every mole contains the same number of atoms.

Example — From percentages to two formulas

Problem
A compound contains 24.27% C, 4.07% H, and 71.65% Cl. Its molar mass is $98.96 g mo l^{- 1}$ . Find empirical and molecular formulas using C = 12.01, H = 1.008, Cl = 35.453.

1.Choose a 100 g basis. It contains 24.27 g C, 4.07 g H, and 71.65 g Cl. The slightly non-100 total of the percentages reflects rounding.
2.Convert masses to moles: C gives $24.27/12.01 = 2.021 mol$ ; H gives $4.07/1.008 = 4.038 mol$ ; Cl gives $71.65/35.453 = 2.021 mol$ .
3.Divide by the smallest value, about 2.021: C:H:Cl is approximately $1 : 2 : 1$ . The empirical formula is $C H_{2} Cl$ .
4.Its empirical formula molar mass is $12.01 + 2 (1.008) + 35.453 = 49.479 g mo l^{- 1}$ .
5.The multiplier is $98.96/49.479 \approx 2$ . Multiply every empirical subscript by 2 to obtain $C_{2} H_{4} C l_{2}$ .
6.Check: the molecular formula has twice the empirical formula mass and preserves the same atom ratios.

Molecular formula multiplierLaTeX

k = \frac{M}{M _{emp}} molecular formula = (empirical formula)_{k}

M

is molecular molar mass,

M_{emp}

is the molar mass of the empirical-formula composition, and

k

should be a positive integer within experimental precision. The notation means multiply every subscript by

k

After division by the smallest amount, a ratio may contain a simple fraction such as 1.5, 1.333, or 1.25. Multiply all terms by 2, 3, or 4 respectively to obtain small whole numbers. Do not round a genuine 1.5 to 2. Small deviations such as 1.998 from 2.000 can arise from measurement precision.

Example — A fractional ratio

Problem
An iron oxide contains 69.9% Fe and 30.1% O. Find its empirical formula using Fe = 55.85 and O = 16.00.

1.For 100 g, the amounts are $69.9/55.85 = 1.252 mol$ Fe and $30.1/16.00 = 1.881 mol$ O.
2.Divide by 1.252 to obtain about $1 : 1.503$ , recognised as $1 : 1.5$ at the data precision.
3.Multiply both terms by 2 to get $2 : 3$ . The empirical formula is $F e_{2} O_{3}$ .
4.These percentages alone do not establish a distinct molecular size. For this ionic oxide the formula describes the composition ratio rather than an isolated molecule.

Composition alone does not determine molecule size

Several molecular formulas can share one empirical formula. An independent molar mass or equivalent structural information is required to determine the molecular multiplier. If a question asks for a molecular formula from percentages alone, identify the missing information instead of guessing.

Using reaction products as composition evidence

Combustion analysis is another way to obtain element amounts. If a compound contains only carbon and hydrogen and burns completely to carbon dioxide and water, every carbon atom is accounted for in the carbon dioxide and every hydrogen atom in the water. Oxygen in those products came from the oxygen supply as well, so it must not be assigned to the original hydrocarbon.

Example — Reading composition from combustion products

Problem
A hydrocarbon produces $4.40 g$ carbon dioxide and $1.80 g$ water on complete combustion. Find its empirical formula using molar masses 44.0 and 18.0.

1.Carbon dioxide amount is $4.40/44.0 = 0.100 mol$ . Each molecule contains one C atom, so the fuel contained $0.100 mol$ C atoms.
2.Water amount is $1.80/18.0 = 0.100 mol$ . Each molecule contains two H atoms, so the fuel contained $0.200 mol$ H atoms.
3.The atom ratio C:H is $1 : 2$ , giving empirical formula $C H_{2}$ . A molecular molar mass would be needed to choose among its molecular multiples.

Quick check

Why divide element masses by their atomic molar masses?

Quick check

What is the empirical formula of $C_{6} H_{12} O_{6}$ ?

Quick check

A mole ratio is 1:1.5. What is the simplest whole-number ratio?

Quick check

An empirical formula has molar mass $30 g mo l^{- 1}$ and the molecular molar mass is $180 g mo l^{- 1}$ . What is the multiplier?

Quick check

Which statement about percentage composition is correct?

Practice Problems

Find the mass percentages in $N a_{2} S O_{4}$ using Na = 23.0, S = 32.0, O = 16.0.
Find the empirical formula for 40.0% C, 6.7% H, and 53.3% O using C = 12.0, H = 1.0, O = 16.0.
A compound has empirical formula $C H_{2} O$ and molar mass $180 g mo l^{- 1}$ . Find its molecular formula using empirical-formula molar mass 30.0.
Why does a ratio $1 : 1.5$ require multiplication rather than ordinary rounding?
How much copper is represented in $100 g$ pure anhydrous $CuS O_{4}$ ? Use Cu = 63.5, S = 32.0, O = 16.0.
A hydrocarbon produces $8.80 g$ carbon dioxide and $5.40 g$ water. Find its empirical formula using molar masses 44.0 and 18.0.
Explain what extra information is needed if only an empirical formula is known.

Key Takeaways

Mass percentages compare element mass with total compound mass on the same basis. Empirical formulas show simplest atom ratios; molecular formulas show actual molecule composition. Convert element masses to moles before finding a ratio. Preserve simple fractional ratios by multiplying all terms to whole numbers. Molar mass determines the integer molecular multiplier; percentages alone cannot.

Previous · Lesson 7

The Mole and Molar Mass

Next · Lesson 9

Chemical Equations, Stoichiometry, and Limiting Reagents

Two directions of the composition problem

The simplest ratio is not always the actual molecule

Why composition must be converted into moles

Using reaction products as composition evidence

Quiz

Practice Problems

Key Takeaways