Percentage Composition from a Formula

Introduction

A chemical formula fixes not only atom ratios but also the share of a pure compound's mass contributed by each element. Percentage composition uses that formula and atomic masses. It is a property of the specified compound formula, distinct from how pure a particular sample is or how much product a reaction yields.

Core explanation

For a compound with molar mass M, the mass percentage of element E is [number of E atoms in one formula × atomic molar mass of E]/M × 100%. The numerator and denominator can both be expressed as grams per mole of formula units, so their units cancel. One could perform the calculation with formula masses in atomic mass units instead and obtain the same numerical percentages. Every element's percentages should add to about 100%, allowing for rounding.

For CO₂, using C 12.01 and O 16.00 gives M = 12.01 + 2(16.00) = 44.01 g mol⁻¹. Carbon contributes 12.01/44.01 × 100 ≈ 27.3% by mass; oxygen contributes 32.00/44.01 × 100 ≈ 72.7%. The atom count is one C and two O, or 33.3% carbon atoms by number, but carbon's mass fraction is only 27.3%. Mass percent and atom percent answer different questions because C and O atoms have different masses.

Parentheses must be expanded before calculating. In Al₂(SO₄)₃, the formula contains two Al, three S and twelve O atoms per formula unit. With Al 26.98, S 32.06 and O 16.00, M = 342.14 g mol⁻¹. Oxygen contributes 12 × 16.00 = 192.00 g per mole of formula units, so its mass percentage is 192.00/342.14 × 100 ≈ 56.1%. Counting only four oxygen atoms because sulfate is in parentheses would produce a false result and percentages that fail the sum check.

Hydrates add the water atoms and mass shown after the dot. CuSO₄·5H₂O has five water units per displayed hydrate formula. A percentage of water of crystallization can be calculated as five water molar masses divided by the full hydrate molar mass, while elemental hydrogen percentage uses ten H atoms per hydrate unit. These are related but different percentages. Specify whether “water percent” or “hydrogen percent” is requested.

Percentage composition can support formula inference. If experimental elemental mass percentages are known, treat a hypothetical 100 g pure sample so each percentage becomes grams, convert each element's grams to moles, and reduce to the simplest ratio. The next pages develop that reverse path. A composition result calculated from a known formula, however, does not prove that a laboratory sample has that formula; experimental analysis is needed to test identity and purity.

Purity has a different denominator. A jar might contain 80% CaCO₃ by total sample mass, while pure CaCO₃ itself has a fixed calcium mass percentage from its formula. Multiplying the two may estimate calcium's share of the jar only if the other 20% contains no calcium. The chemical formula's elemental percentages describe the pure compound, not necessarily a mixed material.

Use sensible atomic masses and rounding. Different data tables may give small last-digit differences. A mass percentage slightly above 100 when summed from individually rounded values can be a rounding artifact, but a large discrepancy suggests formula expansion or arithmetic error. Keep guard digits through the element contributions and round final percentages together.

Step-by-step reasoning

1. Verify the compound formula, including parentheses and waters of hydration. 2. Count each element's atoms per molecule or formula unit. 3. Multiply each count by its atomic molar mass and sum to obtain M. 4. Divide each element's contribution by M and multiply by 100%. 5. Check that element percentages sum to about 100% and label the mass basis.

Visual explanation

Draw a stacked 44.01 g bar representing one mole of CO₂. Shade 12.01 g for carbon and 32.00 g for oxygen. Label their fractions 27.3% and 72.7%. Beside it, draw three atom icons, one C and two O, to show why the atom-number fraction differs from mass fraction.

Real-world analogy

A package can contain one heavy item and two lighter items. The item counts are one-third and two-thirds, but the mass shares depend on individual weights. A chemical formula gives atom counts; atomic masses convert those counts into shares of total compound mass.

Real-world example

In Fe₂O₃, two Fe atoms contribute 111.70 g per mole of formula units and three O atoms contribute 48.00 g, for M = 159.70 g mol⁻¹. Iron is about 69.9% of pure Fe₂O₃ by mass. A 10.0 g sample of pure Fe₂O₃ therefore contains about 6.99 g Fe atoms as part of the compound, though obtaining elemental iron would require a chemical reduction and may have a separate yield.

Why?

Why does a formula determine mass percentage? Each formula unit has a fixed count of each element. Multiplying those counts by atomic masses gives a fixed element-mass contribution for every unit, so scaling to one mole or a larger pure sample preserves the same fractions.

Common misconception

“Two oxygen atoms in CO₂ mean oxygen is exactly two-thirds of its mass.” Two-thirds is the fraction of atoms that are oxygen. Oxygen's mass share is about 72.7% because an O atom is heavier than a C atom.

Worked example

Find percentage composition of water using H 1.008 and O 16.00. M(H₂O) = 2(1.008) + 16.00 = 18.016 g mol⁻¹. Hydrogen contributes 2.016 g mol⁻¹, so %H = 2.016/18.016 × 100 = 11.19%. Oxygen contributes 16.00 g mol⁻¹, so %O = 16.00/18.016 × 100 = 88.81%. The percentages add to 100.00% with the shown rounding. This applies to pure H₂O regardless of whether the sample contains 1 mol or 0.001 mol; it is not the percentage of water in a wet mixture.

Quick check

1. Why is carbon about 27.3% by mass of CO₂ even though it is one of three atoms? Answer: Mass percentages use atomic masses, and carbon's 12.01 contribution is divided by the full 44.01 formula mass.

Exam focus

Write an element-contribution table for grouped formulas. State “by mass” and use the full formula as denominator. Check the sum of all elemental percentages; do not confuse pure-compound composition with the purity of a mixed sample.

Advanced insight

For a molecular formula that is an integer multiple of an empirical formula, elemental mass percentages are identical because both numerator and denominator scale by the same integer. Thus percentage composition alone can determine the simplest ratio but cannot distinguish, for example, CH₂O from C₂H₄O₂ without additional molar-mass information.

Summary

Element mass percentage equals its formula-based mass contribution divided by the compound's full molar mass, times 100. Subscripts and parentheses fix atom counts; atomic masses set mass shares. The percentages characterize a pure formula and differ from atom percentages, sample purity and reaction yield.

Practice questions

1. What is carbon's approximate mass percentage in CO₂ using C 12.01 and O 16.00? Answer: 27.3% by mass. 2. What is oxygen's corresponding percentage in CO₂? Answer: 72.7% by mass, complementing the carbon fraction. 3. How many oxygen atoms contribute to the mass of Al₂(SO₄)₃? Answer: Twelve per formula unit, because three sulfate groups each contain four. 4. Is oxygen two-thirds of CO₂ by mass because two of its three atoms are O? Answer: No. Two-thirds is atom-number fraction, while oxygen is about 72.7% by mass. 5. Can percent composition alone distinguish CH₂O from C₂H₄O₂? Answer: No. The second formula is an integer multiple and has the same elemental mass percentages.