Percentage by Mass of an Element

Introduction

Chemical formulas show atom ratios, but many questions ask for mass percentages. Oxygen is two of the three atoms in a CO₂ molecule, yet its mass percentage is not simply two thirds. Atomic masses differ. Add the mass contributions from the formula, divide the chosen element's contribution by the whole formula mass and multiply by 100.

Core explanation

For a compound with a known formula, percentage by mass of element X is (total relative mass contributed by X in one formula unit ÷ Mᵣ of the compound) × 100%. If X appears a times, its contribution is aAᵣ(X). The ratio is dimensionless: relative mass units cancel. The same proportion would be obtained from any pure sample mass because a bigger sample scales both the element mass and total mass together.

For CO₂, use C = 12 and O = 16. One molecule has carbon contribution 12 and oxygen contribution 2(16) = 32, so Mᵣ = 44. Oxygen percentage is 32/44 × 100 ≈ 72.7%; carbon is 12/44 × 100 ≈ 27.3%. Their rounded values sum to 100.0%. The atom-count fraction for oxygen is 2/3 ≈ 66.7%, which is different because an O atom is heavier than a C atom.

For water with H = 1 and O = 16, Mᵣ(H₂O) = 2 + 16 = 18. Hydrogen contributes 2 of the 18 mass parts, so its percentage is 2/18 × 100 ≈ 11.1%. Oxygen contributes 16/18 × 100 ≈ 88.9%. A water molecule has two H atoms and one O atom, but most of its mass lies in the oxygen atom. This contrast is why subscripts alone do not give mass percentages.

For an ionic compound, interpret the formula as its simplest ratio. In CaCO₃ with Ca = 40, C = 12 and O = 16, Mᵣ = 40 + 12 + 3(16) = 100. Calcium makes up 40%, carbon 12% and oxygen 48% in the rounded model. The three contributions total 100. The numerical neatness is due to the chosen rounded atomic masses; it should not be assumed for every compound.

Brackets and hydrate dots matter. Ca(OH)₂ has two oxygen atoms and two hydrogen atoms, not one of each. CuSO₄·5H₂O has nine oxygen atoms in total: four in sulfate and five in the waters. If asked for the percentage of oxygen in the entire hydrate, include all nine O atoms in the numerator and the full hydrate Mᵣ in the denominator. If asked for percentage of water, use five complete water groups, including H, as the numerator.

Percentage composition is also a way to check a measured sample against a proposed formula. A pure compound should have the formula's expected mass ratios within measurement uncertainty. However, matching one percentage is not enough to establish a unique formula; different formulas can share or closely match one elemental fraction. Sample purity and measurement method matter.

The mass-fraction shortcut agrees with a mole calculation. In a 5.00 g CO₂ sample, oxygen mass is 5.00 × (32/44) = 3.636... g, about 3.64 g. Dividing by the total 5.00 g and multiplying by 100 recovers 72.7%. The fraction is independent of sample size because the formula composition is constant for pure CO₂.

Step-by-step reasoning

1. Write the correct complete formula and the supplied Aᵣ values. 2. Calculate the mass contribution of each element, respecting subscripts and brackets. 3. Add all contributions for Mᵣ; divide the target contribution by Mᵣ and multiply by 100. 4. Check that all element percentages sum to about 100%, allowing small rounding differences.

Visual explanation

Draw a bar of length 44 for CO₂. Colour 12 units for carbon and 32 units for oxygen. The coloured portions represent mass shares, not numbers of atoms. Mark 32/44 of the bar as about 72.7% oxygen.

Real-world analogy

A basket with two small apples and one very heavy melon has more apples by item count but may have much more melon by mass. Counting atoms and weighing their contributions answer different questions. Formula subscripts count the items; atomic masses determine the weight shares.

Real-world example

An analysis of calcium carbonate can compare measured calcium mass with the predicted 40% in the rounded model. A 10.0 g pure CaCO₃ sample should contain about 4.0 g calcium by mass. The calculation helps test composition, though a laboratory measurement has uncertainty and possible impurities.

Why?

Why use Mᵣ as the denominator? It is the combined relative mass of every atom in the formula unit. Dividing a chosen element's contribution by that whole produces its fraction of total mass; multiplying by 100 expresses that fraction as a percentage.

Common misconception

“CO₂ is 66.7% oxygen by mass because two of its three atoms are oxygen.” Two thirds is an atom-count fraction. Oxygen contributes 32 of 44 mass parts, giving about 72.7% by mass with the stated atomic masses.

Worked example

Calculate nitrogen percentage in NH₄NO₃ using N = 14, H = 1 and O = 16. There are two N atoms, four H and three O. Contributions are N: 2(14) = 28; H: 4; O: 3(16) = 48. Total Mᵣ = 28 + 4 + 48 = 80. Nitrogen percentage = 28/80 × 100 = 35%. The other shares are 5% H and 60% O, summing to 100%.

Quick check

1. What mass fraction of CO₂ is oxygen when C = 12 and O = 16? Answer: 32/44, approximately 0.727 or 72.7% by mass.

Exam focus

Display the numerator as subscript × Aᵣ and the denominator as the full Mᵣ. Use the phrase “by mass” and include a percent sign. Do not confuse the atom-number fraction with the mass fraction, especially for elements with very different atomic masses.

Advanced insight

Percentage composition can lead toward an empirical formula when measured elemental percentages are converted to moles of atoms. The reverse route divides each element's mass share by its atomic molar mass and compares the resulting mole ratios. A formula-to-percentage calculation is straightforward; percentage-to-formula needs additional ratio reasoning and careful handling of experimental uncertainty.

Summary

For an element in a compound, divide its total formula mass contribution by the complete relative formula mass and multiply by 100. Formula subscripts give atom counts, while atomic masses give weight shares. Check all elements together sum to about 100% and include hydration water where the full formula requires it.

Practice questions

1. Calculate oxygen percentage in H₂O with H = 1 and O = 16. Answer: 16/18 × 100 ≈ 88.9% oxygen by mass. 2. Calculate carbon percentage in CO₂ with C = 12 and O = 16. Answer: 12/44 × 100 ≈ 27.3% carbon by mass. 3. Find the calcium percentage in CaCO₃ with Ca = 40, C = 12 and O = 16. Answer: Mᵣ = 100 and Ca contributes 40, so calcium is 40% by mass. 4. Why can element percentages sum to 99.9% or 100.1% after calculation? Answer: Rounding individual percentages can introduce a small apparent difference from exactly 100%.