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If a mining company digs up 1,000 kilograms of iron ore ($\text{Fe}_2\text{O}_3$), how much usable iron metal can they actually extract?

To find out, chemists calculate what fraction of a compound's total mass belongs to one specific element.

Every element has a **relative atomic mass** ($A_r$), which compares the mass of its atoms to one-twelfth the mass of a single carbon-12 atom.

When you add the atomic masses of every atom in a chemical formula, you get the compound's total **relative formula mass** ($M_r$).

<!DOCTYPE html>
<html lang="en">
<head>
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  <meta name="viewport" content="width=device-width, initial-scale=1.0">
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</head>
<body>

  <div class="calc-container">
    
    <div class="formula-badge">
      <span>Formula:</span>
      <span>H<sub>2</sub>O</span>
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      <div class="branch-card hydrogen">
        <div class="element-pill">2 × H</div>
        <div class="calc-step">2 × 1.0</div>
        <div class="subtotal">2.0</div>
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      <div class="branch-card oxygen">
        <div class="element-pill">1 × O</div>
        <div class="calc-step">1 × 16.0</div>
        <div class="subtotal">16.0</div>
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        <span>Total M<sub>r</sub></span>
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        <span class="sum-detail">2.0 + 16.0 =</span>
        <span>18.0</span>
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Once you know the total mass of the compound, how do you turn one element's contribution into a percentage?

Just like finding your score on a test by dividing your points by the total points, percentage by mass divides the mass of your target element by the compound's total formula mass.

$$\text{Percentage by mass} = \frac{\text{number of atoms} \times A_r}{M_r} \times 100\%$$

In 1794, French chemist Joseph Proust proved the **Law of Definite Proportions**, establishing that every pure sample of a compound always contains the exact same mass ratio of elements regardless of where it came from.

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    <div class="header-row">
      <div class="mol-title"><span id="formula-title"></span></div>
      <div class="mol-mass"><span id="mass-title"></span></div>
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    <div class="bar-container" id="barContainer">
      <div class="segment seg-h" id="segH" title="Hydrogen: 11.1%">
        <span class="segment-label">H</span>
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        <span class="segment-label">Oxygen (88.9%)</span>
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        <span><span class="dot" style="background:#0284c7;"></span>Hydrogen</span>
        <span>11.1%</span>
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        <span><span class="dot" style="background:#ea580c;"></span>Oxygen</span>
        <span>88.9%</span>
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    <div class="calc-card">
      <div class="calc-title" id="calcTitle">Percentage by Mass</div>
      <div class="math-display" id="mathTarget"></div>
    </div>
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  <script>
    const data = {
      O: {
        title: "Oxygen in H₂O",
        math: "\\%\\text{ O} = \\left( \\frac{1 \\times 16.0}{18.0} \\right) \\times 100 = 88.9\\%",
        class: "active-o"
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How do you handle larger compounds with multiple atoms and bigger atomic masses?

Let's calculate the percentage by mass of iron in iron(III) oxide ($\text{Fe}_2\text{O}_3$), using $A_r(\text{Fe}) = 56$ and $A_r(\text{O}) = 16$.

First, calculate the total formula mass: $$M_r = (2 \times 56) + (3 \times 16) = 112 + 48 = 160$$

Next, divide the total mass of the two iron atoms by the overall $M_r$: $$\text{\% Fe} = \frac{2 \times 56}{160} \times 100\% = \frac{112}{160} \times 100\% = 70\%$$

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <meta name="viewport" content="width=device-width, initial-scale=1.0">
  <meta name="aspect-ratio" content="4:3">
  <link rel="stylesheet" href="{{VENDOR_CDN}}/katex.min.css">
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  <div class="top-bar">
    <div class="molecule-pill" id="mol-label">Fe₂O₃</div>
    <div class="step-buttons">
      <button class="step-btn active" data-step="0">1. Total Mᵣ</button>
      <button class="step-btn" data-step="1">2. Iron Mass</button>
      <button class="step-btn" data-step="2">3. % Mass</button>
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  <div class="steps-grid">
    
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        <span class="step-num">Step 1</span>
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    <div class="step-card" id="card-1">
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        <span class="step-num">Step 2</span>
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      <div class="step-title">Iron Mass</div>
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  <div class="vis-container">
    <div class="bar-wrap">
      <div class="bar-fe" id="bar-fe">Iron: 112 (70%)</div>
      <div class="bar-o">Oxygen: 48 (30%)</div>
    </div>
  </div>

  
  <div class="summary-badge">
    <span>✦ 70% Iron by mass</span>
  </div>

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What traps do exam questions set when a chemical formula contains brackets or elements listed in two different spots?

A common mistake is taking only one atom's mass in the numerator instead of multiplying by the element's full atom count.

In calcium hydroxide, $\text{Ca(OH)}_2$, the subscript outside the brackets multiplies both the oxygen and hydrogen by 2, giving 2 oxygen atoms ($2 \times 16 = 32$) and 2 hydrogen atoms ($2 \times 1 = 2$).

Similarly, in ammonium nitrate ($\text{NH}_4\text{NO}_3$), nitrogen appears twice—so you must add both atoms together for a numerator of $2 \times 14 = 28$.

science

Calculating Percentage by Mass of an Element in a Compound. Calculate the percentage by mass of an element in a compound from chemical formula and relative atomic masses.

GCSE Chemistry: Relative formula mass and percentage mass calculations

Percentage Composition by Mass - GCSE Chemistry Revision Notes

Worked example: Calculating mass percent from chemical formula

Pedagogical strategy focuses on a structured 4-step algorithmic scaffolding: (1) Formula decoding and atom inventory, (2) Subtotal mass and total $M_r$ computation, (3) Proportional fraction formation, and (4) Multiplication by 100 and sanity check (sum of all elemental percentages equals 100%). Progress from simple binary compounds (e.g., $\text{CuO}$, $\text{CO}_2$) to compounds with repeated elements and polyatomic brackets (e.g., $\text{NH}_4\text{NO}_3$, $(\text{NH}_4)_2\text{SO}_4$), concluding with multistep real-world applications (fertilizers and ores).

Calculating Percentage by Mass of an Element in a Compound

The Building Blocks of Mass

The Percentage by Mass Formula