{"id":5276,"date":"2026-06-19T04:55:02","date_gmt":"2026-06-19T03:55:02","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/empirical-formula\/"},"modified":"2026-06-19T05:45:29","modified_gmt":"2026-06-19T04:45:29","slug":"empirical-formula","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/empirical-formula\/","title":{"rendered":"Empirical Formula"},"content":{"rendered":"\n<!--\n===============================================================================\nONLINE LEARNING SYSTEM COPYRIGHT NOTICE\n\u00a9 Online Learning System. All rights reserved.\n\nThis WordPress revision page HTML, CSS, content sequence, educational wording,\nlayout structure, schema structure and embedded design logic are protected\nintellectual property of Online Learning System.\n\nUnauthorised copying, redistribution, resale, modification, republication,\nscraping, extraction, derivative reuse, automated harvesting or removal of\ncopyright notices is strictly prohibited.\n\nBackend copyright marker:\nOLS-AQA-312-EMPIRICAL-FORMULA-REVISION-PAGE-7405-2026\n\nPage:\nEmpirical Formula\nAQA A Level Chemistry\nPaper 1 and Paper 2\n3.1.2 Amount of Substance\n7405\/1 and 7405\/2\n\nCopyright enforcement notes:\n- The visible student page is a free OLS revision resource.\n- The backend HTML structure, CSS architecture, card sequencing, schema graph,\n  revision wording, responsive table behaviour and embedded learning pathway are\n  proprietary OLS production assets.\n- Do not remove this notice.\n===============================================================================\n-->\n\n<section class=\"ols-revision-page ols-empirical-formula-aqa-page\" data-owner=\"Online Learning System\" data-copyright=\"\u00a9 Online Learning System. 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OLS protected content block.\n    Page: Empirical Formula.\n    This content is authored for Online Learning System and must not be scraped,\n    cloned, republished, resold, reformatted into derivative notes, or reused in\n    competing resources without written permission.\n  -->\n\n  <div class=\"ols-revision-layout\">\n    <aside class=\"ols-sidebar\">\r\n  <div class=\"ols-sidebar-header\">\r\n    <h3>Revision Notes<\/h3>\r\n    <p>AQA A Level Chemistry<\/p>\r\n  <\/div>\r\n\r\n  <div class=\"ols-topic-group\">\r\n    <h4>3.1.2 Amount of Substance<\/h4>\r\n\r\n    <ul class=\"ols-topic-list\">\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/relative-masses\/\">Relative Masses<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/the-mole-and-avogadro-constant\/\">The Mole and Avogadro Constant<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/\">Empirical and Molecular Formulae<\/a>\r\n\r\n        <ul class=\"ols-subtopic-list\">\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/empirical-formula\/\">Empirical Formula<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/molecular-formula\/\">Molecular Formula<\/a>\r\n          <\/li>\r\n        <\/ul>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/\">Chemical Equations and Reacting Masses<\/a>\r\n\r\n        <ul class=\"ols-subtopic-list\">\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/writing-chemical-equations\/\">Writing Chemical Equations<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/ionic-equations\/\">Ionic Equations<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/calculations-using-reacting-masses\/\">Calculations Using Reacting Masses<\/a>\r\n          <\/li>\r\n        <\/ul>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/\">Concentration and Titration Calculations<\/a>\r\n\r\n        <ul class=\"ols-subtopic-list\">\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/concentrations-of-solutions\/\">Concentrations of Solutions<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/titration-calculations\/\">Titration Calculations<\/a>\r\n          <\/li>\r\n        <\/ul>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/gas-volumes-and-the-ideal-gas-equation\/\">Gas Volumes and the Ideal Gas Equation<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/percentage-yield\/\">Percentage Yield<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/atom-economy\/\">Atom Economy<\/a>\r\n      <\/li>\r\n    <\/ul>\r\n  <\/div>\r\n\r\n  <div class=\"ols-topic-group\">\r\n    <h4>Adjacent Topics<\/h4>\r\n\r\n    <ul class=\"ols-topic-list\">\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-1-atomicstructure\/\">3.1.1 Atomic Structure<\/a>\r\n      <\/li>\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-3-bonding\/\">3.1.3 Bonding<\/a>\r\n      <\/li>\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-4-energetics\/\">3.1.4 Energetics<\/a>\r\n      <\/li>\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-2-1-periodicity\/\">3.2.1 Periodicity<\/a>\r\n      <\/li>\r\n    <\/ul>\r\n  <\/div>\r\n<\/aside>\r\n\r\n<script>\r\n(function() {\r\n  function normalisePath(path) {\r\n    return String(path || '')\r\n      .split('?')[0]\r\n      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Notes<\/a> \/\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/\">A Level Chemistry<\/a> \/\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/\">AQA<\/a> \/\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/\">3.1.2 Amount of Substance<\/a> \/\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/\">Empirical and Molecular Formulae<\/a> \/\n        <span>Empirical Formula<\/span>\n      <\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Empirical Formula<\/h1>\n        <p class=\"ols-page-intro\">A focused AQA revision guide to empirical formula. This page covers the definition, the mole-ratio method, worked examples using mass data and percentage composition, combustion analysis and the link between empirical and molecular formulae.<\/p>\n\n        <div class=\"ols-badges\">\n          <div class=\"ols-badge\">Paper 1 and Paper 2<\/div>\n          <div class=\"ols-badge\">AQA<\/div>\n          <div class=\"ols-badge\">3.1.2 Amount of Substance<\/div>\n          <div class=\"ols-badge\">7405\/1 and 7405\/2<\/div>\n        <\/div>\n\n\n        <div class=\"ols-author\">\n          <img decoding=\"async\" class=\"ols-author-avatar-img\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Author-Profile.jpeg\" alt=\"Dr. Mohammed Al-Fatah\">\n          <div class=\"ols-author-content\">\n            <h2 class=\"ols-author-title\">Written by:<br><span>Dr. Mohammed Al-Fatah<\/span><\/h2>\n            <p class=\"ols-author-description\">Chemistry specialist revision notes for A Level Chemistry.<\/p>\n            <a class=\"ols-linkedin-pill\" href=\"https:\/\/www.linkedin.com\/in\/doctormohammedfatah\/\" target=\"_blank\" rel=\"noopener noreferrer\">\n              <svg class=\"ols-linkedin-icon\" viewBox=\"0 0 24 24\" fill=\"currentColor\" aria-hidden=\"true\"><path d=\"M4.98 3.5C4.98 4.88 3.86 6 2.48 6S0 4.88 0 3.5 1.12 1 2.48 1s2.5 1.12 2.5 2.5zM.5 8h4V24h-4V8zm7 0h3.8v2.2h.1c.5-.9 1.8-2.2 3.9-2.2 4.2 0 5 2.8 5 6.4V24h-4v-7.6c0-1.8 0-4.2-2.6-4.2s-3 2-3 4v7.8h-4V8z\"><\/path><\/svg>\n              View LinkedIn Profile\n            <\/a>\n          <\/div>\n        <\/div>\n      <\/header>\n\n      <!-- Card 1: What is an Empirical Formula -->\n      <article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">1<\/div><h2>What Is an Empirical Formula?<\/h2><\/div>\n        <p>The <strong>empirical formula<\/strong> of a compound is the simplest whole-number ratio of the atoms of each element present. It does not show the actual number of atoms in one molecule; it shows the ratio in its most reduced form.<\/p>\n        <p>For example, glucose has the molecular formula C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, but its empirical formula is <strong>CH<sub>2<\/sub>O<\/strong> because the ratio of C : H : O reduces to 1 : 2 : 1.<\/p>\n        <div class=\"ols-definition-box\">\n          <p><strong>Definition:<\/strong> The empirical formula is the <strong>simplest ratio<\/strong> of atoms of each <strong>element<\/strong> in the compound.<\/p>\n        <\/div>\n        <div class=\"ols-key-box\">\n          <p><strong>Key distinction:<\/strong> the empirical formula and the molecular formula can be the same (for example, water is H<sub>2<\/sub>O in both) or different (for example, hydrogen peroxide has the molecular formula H<sub>2<\/sub>O<sub>2<\/sub> but the empirical formula HO).<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Image 1: general method \/ explained -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Finding-the-empirical-formula-explained.webp\" alt=\"Finding the empirical formula explained step by step for A Level Chemistry\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Dividing by atomic mass converts mass into moles, and dividing by the smallest mole value gives the simplest whole-number ratio.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 2: The General Method -->\n      <article class=\"ols-note-card\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">2<\/div><h2>The General Method<\/h2><\/div>\n        <p>The same three-step method applies whether you are given the <strong>mass of each element<\/strong> or the <strong>percentage mass of each element<\/strong>. Percentage values can be treated directly as if they were masses in grams.<\/p>\n\n        <div class=\"ols-rule-list\">\n          <div class=\"ols-rule-item\">\n            <h3>Step 1 &#8211; Divide by atomic mass<\/h3>\n            <p>Divide the mass (or percentage mass) of each element by its relative atomic mass. This converts the value into an amount of substance, n, in moles.<\/p>\n          <\/div>\n          <div class=\"ols-rule-item\">\n            <h3>Step 2 &#8211; Divide by the smallest<\/h3>\n            <p>Divide each mole value from Step 1 by the smallest mole value obtained. This scales the amount-of-substance ratio so the smallest value becomes 1.<\/p>\n          <\/div>\n          <div class=\"ols-rule-item\">\n            <h3>Step 3 &#8211; Scale to whole numbers if needed<\/h3>\n            <p>If the ratios are not whole numbers, multiply all values by the smallest integer that converts them all to whole numbers. Common examples: multiply by 2 if you get 0.5, or by 3 if you get 0.33.<\/p>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Data types:<\/strong> the AQA method works for (1) the mass of each element in a sample, and (2) the percentage mass of each element in the compound. The steps are identical for both because both are converted into amount of substance ratios.<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Card 3: Worked Example - Mass Data (from the image provided) -->\n      <article class=\"ols-note-card purple\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">3<\/div><h2>Worked Example: Using Mass Data<\/h2><\/div>\n        <p>Calculate the empirical formula for a compound that contains <strong>1.82 g of K<\/strong>, <strong>5.93 g of I<\/strong> and <strong>2.24 g of O<\/strong>.<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Step 1 &#8211; Divide each mass by the atomic mass<\/h3>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>K (A<sub>r<\/sub> = 39.1)<\/span>1.82 &divide; 39.1 = <strong>0.0465 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>I (A<sub>r<\/sub> = 126.9)<\/span>5.93 &divide; 126.9 = <strong>0.0467 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>O (A<sub>r<\/sub> = 16)<\/span>2.24 &divide; 16 = <strong>0.14 mol<\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-worked-example\">\n          <h3>Step 2 &#8211; Divide by the smallest (0.0465)<\/h3>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>K<\/span>0.0465 &divide; 0.0465 = <strong>1<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>I<\/span>0.0467 &divide; 0.0465 = <strong>1<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>O<\/span>0.14 &divide; 0.0465 = <strong>3<\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Empirical formula: KIO<sub>3<\/sub><\/strong> &#8211; potassium iodate. The ratio of K : I : O is 1 : 1 : 3.<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Card 4: Worked Example - Percentage Composition -->\n      <article class=\"ols-note-card orange\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">4<\/div><h2>Worked Example: Using Percentage Composition<\/h2><\/div>\n        <p>Calculate the empirical formula of a compound that contains <strong>40.0% C<\/strong>, <strong>6.7% H<\/strong> and <strong>53.3% O<\/strong> by mass.<\/p>\n        <p>Treat each percentage as a mass in grams and apply the same method.<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Step 1 &#8211; Divide each percentage by the atomic mass<\/h3>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>C (A<sub>r<\/sub> = 12)<\/span>40.0 &divide; 12 = <strong>3.33 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>H (A<sub>r<\/sub> = 1)<\/span>6.7 &divide; 1 = <strong>6.7 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>O (A<sub>r<\/sub> = 16)<\/span>53.3 &divide; 16 = <strong>3.33 mol<\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-worked-example\">\n          <h3>Step 2 &#8211; Divide by the smallest (3.33)<\/h3>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>C<\/span>3.33 &divide; 3.33 = <strong>1<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>H<\/span>6.7 &divide; 3.33 = <strong>2<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>O<\/span>3.33 &divide; 3.33 = <strong>1<\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Empirical formula: CH<sub>2<\/sub>O<\/strong> &#8211; the ratios are already whole numbers so Step 3 is not needed here.<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Image 2: percentage-based -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Finding-empirical-formulas-from-percentages.webp\" alt=\"Finding empirical formulas from percentage composition data for AQA A Level Chemistry\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Percentage values are used exactly like mass values in grams because the calculation only depends on relative proportions, not absolute amounts.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 5: When Step 3 Is Needed -->\n      <article class=\"ols-note-card\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">5<\/div><h2>When Step 3 Is Needed: Scaling to Whole Numbers<\/h2><\/div>\n        <p>After dividing by the smallest mole value, the ratios are not always whole numbers. You need to recognise when to multiply and by how much.<\/p>\n\n        <div class=\"ols-table-wrap\">\n          <table class=\"ols-table\">\n            <thead>\n              <tr>\n                <th>Ratio value after Step 2<\/th>\n                <th>What it suggests<\/th>\n                <th>Action<\/th>\n              <\/tr>\n            <\/thead>\n            <tbody>\n              <tr>\n                <td data-label=\"Ratio value after Step 2\">1.0, 2.0, 3.0 (whole number)<\/td>\n                <td data-label=\"What it suggests\">Ratios are already whole numbers.<\/td>\n                <td data-label=\"Action\">No further step needed. Write the empirical formula directly.<\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Ratio value after Step 2\">Approximately 1.5<\/td>\n                <td data-label=\"What it suggests\">Corresponds to a 3:2 relationship.<\/td>\n                <td data-label=\"Action\">Multiply all ratios by 2.<\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Ratio value after Step 2\">Approximately 1.33 or 1.67<\/td>\n                <td data-label=\"What it suggests\">Corresponds to a 4:3 or 5:3 relationship.<\/td>\n                <td data-label=\"Action\">Multiply all ratios by 3.<\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Ratio value after Step 2\">Approximately 1.25 or 1.75<\/td>\n                <td data-label=\"What it suggests\">Corresponds to a 5:4 or 7:4 relationship.<\/td>\n                <td data-label=\"Action\">Multiply all ratios by 4.<\/td>\n              <\/tr>\n            <\/tbody>\n          <\/table>\n        <\/div>\n\n        <div class=\"ols-definition-box\">\n          <p><strong>Worked example with Step 3:<\/strong> a compound gives mole ratios of N : O = 1 : 1.5 after Step 2. Multiply both by 2 to give N : O = 2 : 3. The empirical formula is <strong>N<sub>2<\/sub>O<sub>3<\/sub><\/strong>.<\/p>\n        <\/div>\n        <div class=\"ols-key-box\">\n          <p><strong>Exam tip:<\/strong> small rounding differences are expected. If a ratio comes out as 1.98 or 2.03, round it to 2. Only multiply up when the value is clearly not close to a whole number (for example, 1.5 or 1.33).<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Card 6: Combustion Analysis -->\n      <article class=\"ols-note-card purple\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">6<\/div><h2>Empirical Formulae from Combustion Data<\/h2><\/div>\n        <p>When an organic compound is burned completely in excess oxygen, the products are <strong>carbon dioxide<\/strong> and <strong>water<\/strong>. These products can be collected and weighed to find the empirical formula of the original compound.<\/p>\n        <p>The method for using combustion data is:<\/p>\n        <ol>\n          <li>Find the mass of carbon from the mass of CO<sub>2<\/sub> produced: mass of C = mass of CO<sub>2<\/sub> &times; (12 \/ 44).<\/li>\n          <li>Find the mass of hydrogen from the mass of H<sub>2<\/sub>O produced: mass of H = mass of H<sub>2<\/sub>O &times; (2 \/ 18).<\/li>\n          <li>If the compound contains oxygen, find the mass of O by subtracting the masses of C and H from the original sample mass.<\/li>\n          <li>Apply the standard three-step method to the masses of C, H, and O.<\/li>\n        <\/ol>\n        <div class=\"ols-key-box\">\n          <p><strong>Key assumption:<\/strong> all the carbon in the original compound ends up as CO<sub>2<\/sub>, and all the hydrogen ends up as H<sub>2<\/sub>O. This is only valid when combustion is complete.<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Image 3: formula explained \/ moles connection -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Finding-a-formula-Empirical-formula-explained.webp\" alt=\"Finding a formula: empirical formula explained with atomic masses and mole ratios\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Combustion analysis links the masses of CO<sub>2<\/sub> and H<sub>2<\/sub>O back to the carbon and hydrogen content of the original compound.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 7: Empirical Formula and Molecular Formula -->\n      <article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">7<\/div><h2>Linking the Empirical Formula to the Molecular Formula<\/h2><\/div>\n        <p>Once you have the empirical formula, you can find the <strong>molecular formula<\/strong> if you are also given the relative molecular mass (M<sub>r<\/sub>) of the compound.<\/p>\n        <p>The relationship is:<\/p>\n        <div class=\"ols-definition-box\">\n          <p><strong>Molecular formula<\/strong> = (empirical formula) &times; n, where <strong>n = M<sub>r<\/sub> of compound &divide; M<sub>r<\/sub> of empirical formula unit<\/strong><\/p>\n        <\/div>\n        <p>For example, if the empirical formula is CH<sub>2<\/sub>O (M<sub>r<\/sub> = 30) and the compound has M<sub>r<\/sub> = 180, then n = 180 &divide; 30 = 6, giving the molecular formula <strong>C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/strong>.<\/p>\n\n        <div class=\"ols-table-wrap\">\n          <table class=\"ols-table\">\n            <thead>\n              <tr>\n                <th>Compound<\/th>\n                <th>Empirical formula<\/th>\n                <th>M<sub>r<\/sub> of compound<\/th>\n                <th>n<\/th>\n                <th>Molecular formula<\/th>\n              <\/tr>\n            <\/thead>\n            <tbody>\n              <tr>\n                <td data-label=\"Compound\">Glucose<\/td>\n                <td data-label=\"Empirical formula\">CH<sub>2<\/sub>O<\/td>\n                <td data-label=\"Mr of compound\">180<\/td>\n                <td data-label=\"n\">6<\/td>\n                <td data-label=\"Molecular formula\">C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Compound\">Ethene<\/td>\n                <td data-label=\"Empirical formula\">CH<sub>2<\/sub><\/td>\n                <td data-label=\"Mr of compound\">28<\/td>\n                <td data-label=\"n\">2<\/td>\n                <td data-label=\"Molecular formula\">C<sub>2<\/sub>H<sub>4<\/sub><\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Compound\">Hydrogen peroxide<\/td>\n                <td data-label=\"Empirical formula\">HO<\/td>\n                <td data-label=\"Mr of compound\">34<\/td>\n                <td data-label=\"n\">2<\/td>\n                <td data-label=\"Molecular formula\">H<sub>2<\/sub>O<sub>2<\/sub><\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Compound\">Water<\/td>\n                <td data-label=\"Empirical formula\">H<sub>2<\/sub>O<\/td>\n                <td data-label=\"Mr of compound\">18<\/td>\n                <td data-label=\"n\">1<\/td>\n                <td data-label=\"Molecular formula\">H<sub>2<\/sub>O<\/td>\n              <\/tr>\n            <\/tbody>\n          <\/table>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Note:<\/strong> when the empirical formula and the molecular formula are the same for a molecular substance, n = 1. Ionic compounds do not have molecular formulae because they consist of giant lattices; their formulae show the simplest whole-number ratio of ions.<\/p>\n        <\/div>\n      <\/article>\n\n      <!-- Image 4: moles made simple -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Moles-made-simple-understanding-formulae.webp\" alt=\"Moles made simple: understanding how mole calculations connect to chemical formulae\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Understanding moles is the foundation of the empirical formula method: every step converts mass data into a mole ratio before simplifying.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 8: Common Exam Points -->\n      <article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">8<\/div><h2>Common Exam Points<\/h2><\/div>\n        <ul>\n          <li>The empirical formula shows the <strong>simplest whole-number ratio<\/strong> of atoms, not the actual number of atoms per molecule.<\/li>\n          <li>When percentage data is given, you can treat the percentages directly as masses in grams because the method only depends on the ratio.<\/li>\n          <li>Always divide by the <strong>relative atomic mass<\/strong> (A<sub>r<\/sub>), not the molecular mass, in Step 1.<\/li>\n          <li>After dividing by the smallest, look carefully at whether the ratios are already whole numbers before deciding whether Step 3 is needed.<\/li>\n          <li>A ratio of 1.5 means you must multiply by 2; a ratio of 1.33 means you must multiply by 3. Do not round 1.5 to 2.<\/li>\n          <li>Ionic compounds only have empirical formulae, never molecular formulae, because they consist of a lattice rather than discrete molecules.<\/li>\n          <li>To find the molecular formula, you need both the empirical formula and the relative molecular mass (M<sub>r<\/sub>).<\/li>\n          <li>Show all working clearly: marks are awarded for each correct step, even if the final formula contains an error.<\/li>\n        <\/ul>\n      <\/article>\n\n      <!-- Check Your Understanding -->\n      <section class=\"ols-h5p-card\">\n        <h2>Check Your Understanding<\/h2>\n        <p>Use this activity to practise finding the empirical formula from mass and percentage composition data.<\/p>\n        <div class=\"ols-h5p-grid\">\n          <div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-290\" class=\"h5p-iframe\" data-content-id=\"290\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Relative Formula Mass Practice\"><\/iframe><\/div><\/div>\n        <\/div>\n      <\/section>\n\n      <!-- QuickSnap Summary -->\n      <section class=\"ols-quicksnap-card\">\n        <h2>QuickSnap<\/h2>\n        <p>This text summary condenses the page into the essential exam ideas.<\/p>\n        <ul class=\"ols-quicksnap-list\">\n          <li><strong>Empirical formula:<\/strong> the simplest whole-number ratio of atoms of each element in a compound.<\/li>\n          <li><strong>Step 1:<\/strong> divide the mass (or percentage) of each element by its relative atomic mass to get moles.<\/li>\n          <li><strong>Step 2:<\/strong> divide all amounts in moles by the smallest mole value.<\/li>\n          <li><strong>Step 3:<\/strong> multiply up to whole numbers if any ratio is not already a whole number (for example, multiply by 2 if a ratio is 1.5).<\/li>\n          <li><strong>Percentage data:<\/strong> treat percentage values exactly like masses in grams; the method is identical.<\/li>\n          <li><strong>Combustion analysis:<\/strong> use masses of CO<sub>2<\/sub> and H<sub>2<\/sub>O to find masses of C and H, then apply the method.<\/li>\n          <li><strong>Molecular formula:<\/strong> multiply the empirical formula by n, where n = M<sub>r<\/sub> of compound divided by M<sub>r<\/sub> of one empirical unit.<\/li>\n          <li><strong>Ionic compounds:<\/strong> their formulae represent the simplest whole-number ratio of ions in the lattice, not discrete molecules.<\/li>\n        <\/ul>\n      <\/section>\n\n      <!-- Lightbox element -->\n      <div class=\"ols-image-lightbox\" id=\"olsImageLightboxEmpiricalFormulaAQA001\" aria-hidden=\"true\" role=\"dialog\" aria-modal=\"true\" aria-label=\"Expanded revision image\">\n        <div class=\"ols-image-lightbox-inner\">\n          <button class=\"ols-image-lightbox-close\" type=\"button\" aria-label=\"Close enlarged image\">&times;<\/button>\n          <img decoding=\"async\" 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<\/style>\r\n\r\n  <div class=\"ols-course-cta-card\">\r\n\r\n    <div class=\"ols-course-cta-top\">\r\n\r\n      <a\r\n        class=\"ols-course-cta-image-link\"\r\n        href=\"[insert here]\"\r\n        aria-label=\"Open the AQA A Level Chemistry 3.1.2 Amount of Substance course page\">\r\n\r\n        <div class=\"ols-course-cta-image\">\r\n          <img decoding=\"async\"\r\n            src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/06\/3-1-2-Amount-of-Substance.jpg\"\r\n            alt=\"AQA A Level Chemistry 3.1.2 Amount of Substance interactive course banner\"\r\n          >\r\n        <\/div>\r\n\r\n      <\/a>\r\n\r\n      <div class=\"ols-course-cta-content\">\r\n\r\n        <div class=\"ols-course-cta-header-row\">\r\n\r\n          <div class=\"ols-course-cta-kicker\">AQA 7405<\/div>\r\n          <div class=\"ols-course-cta-kicker\">Paper 1 &amp; Paper 2<\/div>\r\n          <div class=\"ols-course-cta-kicker\">3.1.2 Amount of Substance<\/div>\r\n\r\n        <\/div>\r\n\r\n        <h2>\r\n          Master Amount of Substance for AQA A Level Chemistry\r\n        <\/h2>\r\n\r\n        <div class=\"ols-course-cta-heading-button\">\r\n          <a\r\n            class=\"ols-course-button ols-course-button-top\"\r\n            href=\"[insert here]\">\r\n            View Course\r\n          <\/a>\r\n        <\/div>\r\n\r\n      <\/div>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-cta-intro-stats\">\r\n\r\n      <p class=\"ols-course-cta-intro\">\r\n        Continue from these free revision notes into the full 3.1.2 Amount of Substance course, covering moles, Avogadro constant, empirical and molecular formulae, reacting masses, concentration, titrations, gas volumes, percentage yield and atom economy with guided video teaching, diagnostic MCQ practice, teacher-marked short-answer questions and a personalised progress report.\r\n      <\/p>\r\n\r\n      <div class=\"ols-course-cta-stats\">\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>Guided learning<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>Video lessons<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>MCQ practice<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>SAQ practice<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n      <\/div>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-cta-features\">\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Guided video teaching<\/h3>\r\n        <p>\r\n          Learn the chemistry and exam technique through structured video lessons with worked examples and walkthroughs.\r\n        <\/p>\r\n      <\/div>\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Instant MCQ feedback<\/h3>\r\n        <p>\r\n          Auto-marked MCQ quizzes provide immediate diagnostic feedback for every answer choice.\r\n        <\/p>\r\n      <\/div>\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Teacher-marked SAQs<\/h3>\r\n        <p>\r\n          Submit written exam responses and receive chemistry specialist feedback with improvement guidance.\r\n        <\/p>\r\n      <\/div>\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Progress tracking<\/h3>\r\n        <p>\r\n          Identify strengths and weaknesses across the full 3.1.2 Amount of Substance specification, including mole calculations, formulae, solution calculations, gas calculations, yield and atom economy.\r\n        <\/p>\r\n      <\/div>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-animation-wrap\">\r\n\r\n      <h3 class=\"ols-course-animation-title\">\r\n        See how the course works\r\n      <\/h3>\r\n\r\n      <div class=\"ols-animation-frame-shell\">\r\n        <iframe\r\n          id=\"olsCourseAnimationFrame312\"\r\n          title=\"OLS course preview animation\"\r\n          loading=\"lazy\"\r\n          referrerpolicy=\"no-referrer\">\r\n        <\/iframe>\r\n\r\n        <button\r\n          class=\"ols-animation-start-overlay\"\r\n          id=\"olsCourseAnimationStart312\"\r\n          type=\"button\"\r\n          aria-label=\"Play OLS course preview animation\">\r\n          <span class=\"ols-animation-start-content\">\r\n            <span class=\"ols-animation-play-circle\" aria-hidden=\"true\">\r\n              <span class=\"ols-animation-play-icon\"><\/span>\r\n            <\/span>\r\n            <span class=\"ols-animation-start-text\">\r\n              Play course preview animation\r\n            <\/span>\r\n          <\/span>\r\n        <\/button>\r\n\r\n        <button\r\n          class=\"ols-animation-pause-button\"\r\n          id=\"olsCourseAnimationPause312\"\r\n          type=\"button\"\r\n          aria-label=\"Pause course preview animation\">\r\n          Pause\r\n        <\/button>\r\n      <\/div>\r\n\r\n      <p class=\"ols-course-video-status\">\r\n        Click play to start the course preview animation.\r\n      <\/p>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-cta-bottom\">\r\n\r\n      <a\r\n        class=\"ols-course-button\"\r\n        href=\"[insert here]\">\r\n        View Course\r\n      <\/a>\r\n\r\n    <\/div>\r\n\r\n  <\/div>\r\n\r\n  <template id=\"olsCourseAnimationTemplate312\">\r\n<!DOCTYPE html>\r\n<html lang=\"en\">\r\n<head>\r\n<meta charset=\"UTF-8\">\r\n<meta name=\"viewport\" content=\"width=device-width, initial-scale=1.0\">\r\n<title>OLS Amount of Substance Promo Animation<\/title>\r\n\r\n<style>\r\n*{box-sizing:border-box;}\r\n\r\nhtml.ols-animation-paused *,\r\nhtml.ols-animation-paused *::before,\r\nhtml.ols-animation-paused *::after{\r\n  animation-play-state:paused!important;\r\n}\r\n\r\n:root{\r\n  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pointer-events:auto;\r\n}\r\n\r\n.ols-scene.fade-out{\r\n  animation:olsSceneFadeOut 0.85s ease forwards;\r\n}\r\n\r\n@keyframes olsSceneFadeOut{\r\n  to{\r\n    opacity:0;\r\n    visibility:hidden;\r\n  }\r\n}\r\n\r\n.ols-title-cursor{\r\n  display:inline-block;\r\n  width:5px;\r\n  height:0.85em;\r\n  background:#1C244B;\r\n  margin-left:8px;\r\n  transform:translateY(8px);\r\n  animation:olsBlink 0.8s infinite;\r\n}\r\n\r\n.cursor{\r\n  display:inline-block;\r\n  width:3px;\r\n  height:28px;\r\n  background:#1c244b;\r\n  margin-left:3px;\r\n  transform:translateY(5px);\r\n  animation:olsBlink 0.8s infinite;\r\n}\r\n\r\n@keyframes olsBlink{\r\n  0%,45%{opacity:1;}\r\n  46%,100%{opacity:0;}\r\n}\r\n\r\n.ols-video-stage,\r\n.ols-mcq-stage,\r\n.ols-saq-stage{\r\n  width:100vw;\r\n  height:100vh;\r\n  display:flex;\r\n  justify-content:center;\r\n  align-items:center;\r\n  padding:0;\r\n  overflow:hidden;\r\n  background:\r\n    radial-gradient(circle at top left,rgba(70,127,247,0.14),transparent 34%),\r\n    linear-gradient(135deg,#f8fbff 0%,#e9efff 100%);\r\n}\r\n\r\n.ols-video-title-screen,\r\n.ols-mcq-intro-screen,\r\n.ols-saq-intro{\r\n  position:absolute;\r\n  inset:0;\r\n  z-index:50;\r\n  display:flex;\r\n  justify-content:center;\r\n  align-items:center;\r\n  background:\r\n    radial-gradient(circle at top left,rgba(70,127,247,0.14),transparent 34%),\r\n    linear-gradient(135deg,#f8fbff 0%,#e9efff 100%);\r\n  opacity:1;\r\n  visibility:visible;\r\n  pointer-events:none;\r\n}\r\n\r\n.ols-video-title-screen.hide{animation:olsVideoTitleFadeOut 0.85s ease forwards;}\r\n.ols-mcq-intro-screen.hide{animation:olsMcqIntroFadeOut 0.85s ease forwards;}\r\n.ols-saq-intro.hide{animation:olsSaqIntroFadeOut 0.85s ease forwards;}\r\n\r\n@keyframes olsVideoTitleFadeOut{to{opacity:0;visibility:hidden;}}\r\n@keyframes olsMcqIntroFadeOut{to{opacity:0;visibility:hidden;}}\r\n@keyframes olsSaqIntroFadeOut{to{opacity:0;visibility:hidden;}}\r\n\r\n.ols-video-title-wrap,\r\n.ols-mcq-intro-title-wrap,\r\n.ols-saq-intro-title-wrap{\r\n  max-width:1250px;\r\n  padding:0 34px;\r\n  text-align:center;\r\n}\r\n\r\n.ols-video-title,\r\n.ols-mcq-intro-title,\r\n.ols-saq-intro-title{\r\n  margin:0;\r\n  font-family:Poppins,Arial,sans-serif;\r\n  font-size:clamp(52px,8vw,124px);\r\n  font-weight:600;\r\n  line-height:1.08;\r\n  color:#1C244B;\r\n  text-shadow:-9px 0 9px rgba(28,36,75,0.16);\r\n  letter-spacing:-0.045em;\r\n}\r\n\r\n#videoTitleText,\r\n#mcqIntroTitleText,\r\n#saqIntroTitleText{\r\n  white-space:pre-wrap;\r\n}\r\n\r\n.ols-video-scene{\r\n  width:100%;\r\n  height:100%;\r\n  display:flex;\r\n  justify-content:center;\r\n  align-items:center;\r\n  opacity:0;\r\n  visibility:hidden;\r\n}\r\n\r\n.ols-video-scene.show{\r\n  visibility:visible;\r\n  animation:olsVideoSceneIn 1s ease forwards;\r\n}\r\n\r\n@keyframes olsVideoSceneIn{to{opacity:1;}}\r\n\r\n.ols-video-scale-wrap{\r\n  width:calc(var(--ols-video-screen-w) * 1px);\r\n  height:calc(var(--ols-video-screen-h) * 1px);\r\n  transform:scale(var(--ols-video-screen-scale));\r\n  transform-origin:center center;\r\n  display:flex;\r\n  justify-content:center;\r\n  align-items:center;\r\n}\r\n\r\n.ols-video-card{\r\n  width:1180px;\r\n  height:664px;\r\n  border-radius:28px;\r\n  overflow:hidden;\r\n  background:#1C244B;\r\n  box-shadow:0 28px 80px rgba(28,36,75,0.28);\r\n  transform:translateY(24px) scale(0.96);\r\n  opacity:0;\r\n}\r\n\r\n.ols-video-scene.show .ols-video-card{\r\n  animation:olsVideoCardIn 1s cubic-bezier(.18,.89,.32,1.12) forwards;\r\n  animation-delay:0.2s;\r\n}\r\n\r\n@keyframes olsVideoCardIn{\r\n  to{\r\n    transform:translateY(0) scale(1);\r\n    opacity:1;\r\n  }\r\n}\r\n\r\n.ols-video-card video{\r\n  display:block;\r\n  width:100%;\r\n  height:100%;\r\n  aspect-ratio:16 \/ 9;\r\n  object-fit:cover;\r\n  border:0;\r\n}\r\n\r\n.ols-mcq-scale-wrap,\r\n.ols-saq-scale-wrap{\r\n  width:calc(var(--ols-mcq-screen-w) * 1px);\r\n  height:calc(var(--ols-mcq-screen-h) * 1px);\r\n  transform:scale(var(--ols-mcq-screen-scale));\r\n  transform-origin:center center;\r\n  display:flex;\r\n  justify-content:center;\r\n  align-items:flex-start;\r\n}\r\n\r\n.ols-saq-scale-wrap{\r\n  width:calc(var(--ols-saq-screen-w) * 1px);\r\n  height:calc(var(--ols-saq-screen-h) * 1px);\r\n  transform:scale(var(--ols-saq-screen-scale));\r\n}\r\n\r\n.ols-mcq-screen{\r\n  width:1360px;\r\n  height:1130px;\r\n  border-radius:22px;\r\n  overflow:hidden;\r\n  background:#eaf0ff;\r\n  box-shadow:0 26px 70px rgba(28,36,75,0.22);\r\n  opacity:0;\r\n  transform:translateY(24px) scale(0.96);\r\n}\r\n\r\n.ols-mcq-screen.show{\r\n  animation:olsMcqScreenEnter 1s ease forwards;\r\n}\r\n\r\n@keyframes olsMcqScreenEnter{\r\n  from{opacity:0;transform:translateY(24px) scale(0.96);}\r\n  to{opacity:1;transform:translateY(0) scale(1);}\r\n}\r\n\r\n.mcq-question-area{\r\n  background:#eaf0ff;\r\n  padding:28px 30px 30px;\r\n  position:relative;\r\n  height:610px;\r\n}\r\n\r\n.mcq-question-lead{\r\n  margin:0 0 14px;\r\n  font-size:24px;\r\n  line-height:1.35;\r\n  color:#111827;\r\n}\r\n\r\n.mcq-ionic-table{\r\n  border-collapse:collapse;\r\n  width:500px;\r\n  margin-bottom:8px;\r\n  font-size:21px;\r\n  color:#111827;\r\n}\r\n\r\n.mcq-ionic-table th,\r\n.mcq-ionic-table td{\r\n  border:1.5px solid #c9ced8;\r\n  padding:12px 18px;\r\n  text-align:center;\r\n}\r\n\r\n.mcq-ionic-table th{\r\n  background:#f2f2f2;\r\n  font-weight:700;\r\n}\r\n\r\n.mcq-ionic-table td{\r\n  background:#eaf0ff;\r\n}\r\n\r\n.mcq-question-main{\r\n  margin:6px 0 28px;\r\n  font-size:24px;\r\n  line-height:1.35;\r\n  color:#111827;\r\n}\r\n\r\n.mcq-options-wrap{\r\n  display:flex;\r\n  flex-direction:column;\r\n  gap:17px;\r\n  max-width:1200px;\r\n}\r\n\r\n.mcq-option-row{\r\n  display:flex;\r\n  align-items:center;\r\n  min-height:34px;\r\n  position:relative;\r\n}\r\n\r\n.mcq-radio{\r\n  width:28px;\r\n  height:28px;\r\n  border-radius:50%;\r\n  border:2px solid #6b7280;\r\n  margin-right:16px;\r\n  background:transparent;\r\n  position:relative;\r\n  flex:0 0 auto;\r\n}\r\n\r\n.mcq-radio::after{\r\n  content:\"\";\r\n  position:absolute;\r\n  inset:5px;\r\n  border-radius:50%;\r\n  background:#6b7280;\r\n  opacity:0;\r\n  transform:scale(0.4);\r\n  transition:opacity 0.25s ease,transform 0.25s ease;\r\n}\r\n\r\n.mcq-option-row.selected .mcq-radio::after{\r\n  opacity:1;\r\n  transform:scale(1);\r\n}\r\n\r\n.mcq-radio-ripple{\r\n  position:absolute;\r\n  left:14px;\r\n  top:50%;\r\n  width:28px;\r\n  height:28px;\r\n  border-radius:50%;\r\n  border:2px solid rgba(28,36,75,0.28);\r\n  transform:translate(-50%,-50%) scale(1);\r\n  opacity:0;\r\n  pointer-events:none;\r\n}\r\n\r\n.mcq-option-row.clicking .mcq-radio-ripple{\r\n  animation:mcqRipple 0.75s ease forwards;\r\n}\r\n\r\n@keyframes mcqRipple{\r\n  0%{opacity:0.65;transform:translate(-50%,-50%) scale(1);}\r\n  100%{opacity:0;transform:translate(-50%,-50%) scale(2.5);}\r\n}\r\n\r\n.mcq-option-label{\r\n  font-size:23px;\r\n  color:#111827;\r\n}\r\n\r\n.mcq-specific-feedback{\r\n  display:inline-flex;\r\n  align-items:center;\r\n  margin-left:18px;\r\n  min-height:38px;\r\n  opacity:0;\r\n  transform:translateX(-8px);\r\n}\r\n\r\n.mcq-specific-feedback.show{\r\n  opacity:1;\r\n  transform:translateX(0);\r\n  transition:opacity 0.35s ease,transform 0.35s ease;\r\n}\r\n\r\n.mcq-cross-icon{\r\n  width:26px;\r\n  height:26px;\r\n  border-radius:50%;\r\n  border:2px solid #d83255;\r\n  color:#d83255;\r\n  display:inline-flex;\r\n  justify-content:center;\r\n  align-items:center;\r\n  font-size:18px;\r\n  font-weight:700;\r\n  margin-right:12px;\r\n  opacity:0;\r\n  transform:scale(0.4);\r\n}\r\n\r\n.mcq-cross-icon.show{\r\n  animation:mcqCrossPop 0.35s ease forwards;\r\n}\r\n\r\n@keyframes mcqCrossPop{\r\n  70%{opacity:1;transform:scale(1.18);}\r\n  100%{opacity:1;transform:scale(1);}\r\n}\r\n\r\n.mcq-specific-feedback-text{\r\n  display:inline-block;\r\n  background:#fff4b8;\r\n  color:#111827;\r\n  padding:8px 16px;\r\n  font-size:22px;\r\n  line-height:1.35;\r\n  min-width:850px;\r\n  min-height:44px;\r\n}\r\n\r\n.mcq-submit-button{\r\n  position:absolute;\r\n  right:34px;\r\n  bottom:30px;\r\n  border:0;\r\n  border-radius:999px;\r\n  background:#1C244B;\r\n  color:#ffffff;\r\n  padding:14px 28px;\r\n  font-size:18px;\r\n  font-weight:600;\r\n  box-shadow:0 14px 32px rgba(28,36,75,0.25);\r\n  cursor:default;\r\n  transition:transform 0.2s ease,box-shadow 0.2s ease,background 0.2s ease;\r\n}\r\n\r\n.mcq-submit-button.clicked{\r\n  transform:translateY(2px) scale(0.97);\r\n  background:#26315f;\r\n  box-shadow:0 7px 18px rgba(28,36,75,0.22);\r\n}\r\n\r\n.mcq-general-feedback{\r\n  height:520px;\r\n  background:#fff3c9;\r\n  color:#8a6a00;\r\n  padding:26px 30px 28px;\r\n  font-size:23px;\r\n  line-height:1.34;\r\n  opacity:0;\r\n  transform:translateY(20px);\r\n  transition:opacity 0.7s ease,transform 0.7s ease;\r\n  overflow:hidden;\r\n}\r\n\r\n.mcq-general-feedback.show{\r\n  opacity:1;\r\n  transform:translateY(0);\r\n}\r\n\r\n#mcqGeneralText{\r\n  white-space:pre-wrap;\r\n}\r\n\r\n.mcq-specific-cursor{background:#111827;}\r\n.mcq-general-cursor{background:#8a6a00;}\r\n\r\n.ols-saq-screen{\r\n  width:1360px;\r\n  height:965px;\r\n  border-radius:22px;\r\n  overflow:hidden;\r\n  background:#eaf0ff;\r\n  box-shadow:0 26px 70px rgba(28,36,75,0.22);\r\n  opacity:0;\r\n  transform:translateY(24px) scale(0.96);\r\n}\r\n\r\n.ols-saq-screen.show{\r\n  animation:olsSaqScreenEnter 1s ease forwards;\r\n}\r\n\r\n@keyframes olsSaqScreenEnter{\r\n  from{opacity:0;transform:translateY(24px) scale(0.96);}\r\n  to{opacity:1;transform:translateY(0) scale(1);}\r\n}\r\n\r\n.saq-question-area{\r\n  height:335px;\r\n  background:#eaf0ff;\r\n  padding:30px 40px 32px;\r\n}\r\n\r\n.saq-question-text{\r\n  font-size:25px;\r\n  line-height:1.42;\r\n  color:#111827;\r\n  margin-bottom:22px;\r\n}\r\n\r\n.saq-student-box{\r\n  height:195px;\r\n  background:#f3f4f6;\r\n  border:2px solid #cfd6e3;\r\n  border-radius:8px;\r\n  padding:23px 26px;\r\n  font-size:24px;\r\n  line-height:1.55;\r\n  color:#4b5563;\r\n  position:relative;\r\n  overflow:hidden;\r\n}\r\n\r\n.saq-teacher-feedback{\r\n  height:630px;\r\n  background:#dff3e6;\r\n  color:#075f3b;\r\n  padding:28px 40px 30px;\r\n  font-size:23px;\r\n  line-height:1.47;\r\n  opacity:0;\r\n  transform:translateY(20px);\r\n  transition:opacity 0.7s ease,transform 0.7s ease;\r\n  overflow:hidden;\r\n}\r\n\r\n.saq-teacher-feedback.show{\r\n  opacity:1;\r\n  transform:translateY(0);\r\n}\r\n\r\n#saqTeacherText{\r\n  white-space:pre-wrap;\r\n}\r\n\r\n.saq-teacher-cursor{\r\n  background:#075f3b;\r\n}\r\n\r\n@media(max-width:900px){\r\n  .ols-video-title,\r\n  .ols-mcq-intro-title,\r\n  .ols-saq-intro-title{\r\n    font-size:clamp(42px,11vw,78px);\r\n    line-height:1.12;\r\n  }\r\n\r\n  .ols-title-cursor{\r\n    width:4px;\r\n    margin-left:6px;\r\n  }\r\n\r\n  .ols-video-card{\r\n    border-radius:20px;\r\n  }\r\n}\r\n<\/style>\r\n<\/head>\r\n\r\n<body>\r\n<div class=\"ols-combined-stage\">\r\n\r\n  <section id=\"sceneVideo\" class=\"ols-scene active\">\r\n    <div class=\"ols-video-stage\">\r\n      <div id=\"videoTitleScreen\" class=\"ols-video-title-screen\">\r\n        <div class=\"ols-video-title-wrap\">\r\n          <h1 class=\"ols-video-title\">\r\n            <span id=\"videoTitleText\"><\/span><span id=\"videoTitleCursor\" class=\"ols-title-cursor\"><\/span>\r\n          <\/h1>\r\n        <\/div>\r\n      <\/div>\r\n\r\n      <section id=\"videoScene\" class=\"ols-video-scene\">\r\n        <div class=\"ols-video-scale-wrap\">\r\n          <div class=\"ols-video-card\">\r\n            <video\r\n              id=\"lessonVideo\"\r\n              src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Virtual-Lesson.mp4\"\r\n              muted\r\n              playsinline\r\n              preload=\"auto\">\r\n            <\/video>\r\n          <\/div>\r\n        <\/div>\r\n      <\/section>\r\n    <\/div>\r\n  <\/section>\r\n\r\n  <section id=\"sceneMcq\" class=\"ols-scene\">\r\n    <div class=\"ols-mcq-stage\">\r\n      <div id=\"mcqIntroScreen\" class=\"ols-mcq-intro-screen\">\r\n        <div class=\"ols-mcq-intro-title-wrap\">\r\n          <h1 class=\"ols-mcq-intro-title\">\r\n            <span id=\"mcqIntroTitleText\"><\/span><span id=\"mcqIntroTitleCursor\" class=\"ols-title-cursor\"><\/span>\r\n          <\/h1>\r\n        <\/div>\r\n      <\/div>\r\n\r\n      <div class=\"ols-mcq-scale-wrap\">\r\n        <main id=\"mcqScreen\" class=\"ols-mcq-screen\">\r\n          <section class=\"mcq-question-area\">\r\n            <p class=\"mcq-question-lead\">A student prepares a sodium hydroxide solution.<\/p>\r\n\r\n            <table class=\"mcq-ionic-table\">\r\n              <thead>\r\n                <tr>\r\n                  <th>Quantity<\/th>\r\n                  <th>Value<\/th>\r\n                <\/tr>\r\n              <\/thead>\r\n              <tbody>\r\n                <tr><td>Concentration of NaOH<\/td><td>0.200 mol dm<sup>\u22123<\/sup><\/td><\/tr>\r\n                <tr><td>Volume used<\/td><td>25.0 cm<sup>3<\/sup><\/td><\/tr>\r\n                <tr><td>Volume in dm<sup>3<\/sup><\/td><td>0.0250 dm<sup>3<\/sup><\/td><\/tr>\r\n              <\/tbody>\r\n            <\/table>\r\n\r\n            <p class=\"mcq-question-main\">What amount of NaOH is present in the 25.0 cm<sup>3<\/sup> sample?<\/p>\r\n\r\n            <div class=\"mcq-options-wrap\">\r\n              <div id=\"mcqWrongOption\" class=\"mcq-option-row\">\r\n                <span class=\"mcq-radio\"><\/span>\r\n                <span class=\"mcq-radio-ripple\"><\/span>\r\n                <span class=\"mcq-option-label\">5.00 mol<\/span>\r\n\r\n                <span id=\"mcqSpecificFeedback\" class=\"mcq-specific-feedback\">\r\n                  <span id=\"mcqCrossIcon\" class=\"mcq-cross-icon\">\u00d7<\/span>\r\n                  <span class=\"mcq-specific-feedback-text\">\r\n                    <span id=\"mcqSpecificText\"><\/span><span id=\"mcqSpecificCursor\" class=\"cursor mcq-specific-cursor\" style=\"display:none;\"><\/span>\r\n                  <\/span>\r\n                <\/span>\r\n              <\/div>\r\n\r\n              <div class=\"mcq-option-row\"><span class=\"mcq-radio\"><\/span><span class=\"mcq-radio-ripple\"><\/span><span class=\"mcq-option-label\">0.500 mol<\/span><\/div>\r\n              <div class=\"mcq-option-row\"><span class=\"mcq-radio\"><\/span><span class=\"mcq-radio-ripple\"><\/span><span class=\"mcq-option-label\">0.00500 mol<\/span><\/div>\r\n              <div class=\"mcq-option-row\"><span class=\"mcq-radio\"><\/span><span class=\"mcq-radio-ripple\"><\/span><span class=\"mcq-option-label\">0.000500 mol<\/span><\/div>\r\n            <\/div>\r\n\r\n            <button id=\"mcqSubmitButton\" class=\"mcq-submit-button\">Submit answer<\/button>\r\n          <\/section>\r\n\r\n          <section id=\"mcqGeneralFeedback\" class=\"mcq-general-feedback\">\r\n            <span id=\"mcqGeneralText\"><\/span><span id=\"mcqGeneralCursor\" class=\"cursor mcq-general-cursor\" style=\"display:none;\"><\/span>\r\n          <\/section>\r\n        <\/main>\r\n      <\/div>\r\n    <\/div>\r\n  <\/section>\r\n\r\n  <section id=\"sceneSaq\" class=\"ols-scene\">\r\n    <div class=\"ols-saq-stage\">\r\n      <div id=\"saqIntro\" class=\"ols-saq-intro\">\r\n        <div class=\"ols-saq-intro-title-wrap\">\r\n          <h1 class=\"ols-saq-intro-title\">\r\n            <span id=\"saqIntroTitleText\"><\/span><span id=\"saqIntroTitleCursor\" class=\"ols-title-cursor\"><\/span>\r\n          <\/h1>\r\n        <\/div>\r\n      <\/div>\r\n\r\n      <div class=\"ols-saq-scale-wrap\">\r\n        <main id=\"saqScreen\" class=\"ols-saq-screen\">\r\n          <section class=\"saq-question-area\">\r\n            <div class=\"saq-question-text\">\r\n              <strong>(ii)<\/strong>&nbsp;&nbsp; Calcium carbonate reacts with hydrochloric acid.<br>\r\n              Calculate the mass of CaCO<sub>3<\/sub> that reacts with 25.0 cm<sup>3<\/sup> of 0.200 mol dm<sup>\u22123<\/sup> HCl.<br>\r\n              CaCO<sub>3<\/sub> + 2HCl \u2192 CaCl<sub>2<\/sub> + H<sub>2<\/sub>O + CO<sub>2<\/sub>\r\n            <\/div>\r\n\r\n            <div class=\"saq-student-box\">\r\n              <span id=\"saqStudentText\"><\/span><span id=\"saqStudentCursor\" class=\"cursor\"><\/span>\r\n            <\/div>\r\n          <\/section>\r\n\r\n          <section id=\"saqTeacherFeedback\" class=\"saq-teacher-feedback\">\r\n            <span id=\"saqTeacherText\"><\/span><span id=\"saqTeacherCursor\" class=\"cursor saq-teacher-cursor\" 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Convert cm\u00b3 to dm\u00b3 before multiplying by concentration.\";\r\nconst mcqGeneralFeedbackText=\r\n\"Your answer is incorrect.\\n\\n\"+\r\n\"For solution concentration calculations, always convert the volume into dm\u00b3 before using n = cV.\\n\\n\"+\r\n\"\u2022 Volume = 25.0 cm\u00b3 = 25.0 \u00f7 1000 = 0.0250 dm\u00b3.\\n\"+\r\n\"\u2022 Concentration = 0.200 mol dm\u207b\u00b3.\\n\"+\r\n\"\u2022 Amount = concentration \u00d7 volume.\\n\"+\r\n\"\u2022 n = 0.200 \u00d7 0.0250 = 0.00500 mol.\\n\\n\"+\r\n\"The correct answer is: 0.00500 mol\";\r\n\r\nconst mcqIntroScreen=document.getElementById(\"mcqIntroScreen\");\r\nconst mcqIntroTitleText=document.getElementById(\"mcqIntroTitleText\");\r\nconst mcqIntroTitleCursor=document.getElementById(\"mcqIntroTitleCursor\");\r\nconst mcqScreen=document.getElementById(\"mcqScreen\");\r\nconst mcqWrongOption=document.getElementById(\"mcqWrongOption\");\r\nconst mcqSubmitButton=document.getElementById(\"mcqSubmitButton\");\r\nconst mcqSpecificFeedback=document.getElementById(\"mcqSpecificFeedback\");\r\nconst mcqCrossIcon=document.getElementById(\"mcqCrossIcon\");\r\nconst mcqSpecificText=document.getElementById(\"mcqSpecificText\");\r\nconst mcqSpecificCursor=document.getElementById(\"mcqSpecificCursor\");\r\nconst mcqGeneralFeedback=document.getElementById(\"mcqGeneralFeedback\");\r\nconst mcqGeneralText=document.getElementById(\"mcqGeneralText\");\r\nconst mcqGeneralCursor=document.getElementById(\"mcqGeneralCursor\");\r\n\r\nfunction selectMcqWrongOption(){\r\n  mcqWrongOption.classList.add(\"clicking\");\r\n\r\n  setTimeout(function(){\r\n    mcqWrongOption.classList.add(\"selected\");\r\n  },220);\r\n\r\n  setTimeout(function(){\r\n    mcqWrongOption.classList.remove(\"clicking\");\r\n  },850);\r\n}\r\n\r\nfunction clickMcqSubmit(){\r\n  mcqSubmitButton.classList.add(\"clicked\");\r\n\r\n  setTimeout(function(){\r\n    mcqSubmitButton.classList.remove(\"clicked\");\r\n  },240);\r\n}\r\n\r\nfunction showMcqSpecificFeedback(){\r\n  mcqSpecificFeedback.classList.add(\"show\");\r\n\r\n  setTimeout(function(){\r\n    mcqCrossIcon.classList.add(\"show\");\r\n  },180);\r\n\r\n  setTimeout(function(){\r\n    mcqSpecificCursor.style.display=\"inline-block\";\r\n\r\n    typeWriter(mcqSpecificText,mcqSpecificFeedbackText,12,function(){\r\n      mcqSpecificCursor.style.display=\"none\";\r\n\r\n      setTimeout(function(){\r\n        showMcqGeneralFeedback();\r\n      },650);\r\n    });\r\n  },560);\r\n}\r\n\r\nfunction showMcqGeneralFeedback(){\r\n  mcqGeneralFeedback.classList.add(\"show\");\r\n\r\n  setTimeout(function(){\r\n    mcqGeneralCursor.style.display=\"inline-block\";\r\n\r\n    typeWriter(mcqGeneralText,mcqGeneralFeedbackText,8,function(){\r\n      mcqGeneralCursor.style.display=\"none\";\r\n\r\n      setTimeout(function(){\r\n        switchScene(sceneMcq,sceneSaq,function(){\r\n          startSaqIntro();\r\n        });\r\n      },2600);\r\n    });\r\n  },600);\r\n}\r\n\r\nfunction startMcqAnimation(){\r\n  fitMcqScreen();\r\n  mcqScreen.classList.add(\"show\");\r\n\r\n  setTimeout(function(){\r\n    selectMcqWrongOption();\r\n\r\n    setTimeout(function(){\r\n      clickMcqSubmit();\r\n\r\n      setTimeout(function(){\r\n        showMcqSpecificFeedback();\r\n      },620);\r\n    },1150);\r\n  },1300);\r\n}\r\n\r\nfunction startMcqIntro(){\r\n  fitMcqScreen();\r\n\r\n  setTimeout(function(){\r\n    typeWriter(mcqIntroTitleText,mcqIntroTitle,44,function(){\r\n      setTimeout(function(){\r\n        mcqIntroTitleCursor.style.display=\"none\";\r\n\r\n        setTimeout(function(){\r\n          mcqIntroScreen.classList.add(\"hide\");\r\n\r\n          setTimeout(function(){\r\n            startMcqAnimation();\r\n          },850);\r\n        },3000);\r\n      },250);\r\n    });\r\n  },700);\r\n}\r\n\r\nconst saqIntroTitle=\"Chemistry Specialist\\nMarked Exam Feedback\";\r\nconst saqStudentAnswer=\"Moles of HCl = 0.200 \u00d7 0.0250 = 0.00500 mol. The mass of calcium carbonate is 0.00500 \u00d7 100 = 0.500 g.\";\r\nconst saqTeacherFeedbackText=\r\n\"Comment:\\n\"+\r\n\"Your first step is correct: you converted 25.0 cm\u00b3 to 0.0250 dm\u00b3 and calculated 0.00500 mol of HCl. The error is that you did not use the mole ratio in the balanced equation.\\n\\n\"+\r\n\"The equation shows that 1 mol of CaCO\u2083 reacts with 2 mol of HCl. Therefore the amount of CaCO\u2083 is half the amount of HCl.\\n\\n\"+\r\n\"Correct method:\\n\"+\r\n\"\u2022 n(HCl) = cV = 0.200 \u00d7 0.0250 = 0.00500 mol \u2713\\n\"+\r\n\"\u2022 n(CaCO\u2083) = 0.00500 \u00f7 2 = 0.00250 mol \u2713\\n\"+\r\n\"\u2022 M\u1d63(CaCO\u2083) = 40.1 + 12.0 + (3 \u00d7 16.0) = 100.1 \u2713\\n\"+\r\n\"\u2022 mass = n \u00d7 M\u1d63 = 0.00250 \u00d7 100.1 = 0.250 g \u2713\\n\\n\"+\r\n\"Next time, always use the balanced equation after calculating moles. 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The molecular formula shows the actual number of atoms in one molecule. They can be the same (for example, H<sub>2<\/sub>O) or different (for example, glucose has the empirical formula CH<sub>2<\/sub>O but the molecular formula C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>).<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>Can I use percentage composition directly in the calculation?<\/h3>\n            <p>Yes. You treat the percentage values exactly as if they were masses in grams and apply the same three steps. The calculation works because empirical formulae depend only on the ratio of amounts, not on the actual sample size.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>What do I do if the mole ratio comes out as 1.5?<\/h3>\n            <p>A ratio of 1.5 cannot be rounded to 2. You must multiply all the ratios by 2 to convert them to whole numbers. So a ratio of 1 : 1.5 becomes 2 : 3 after multiplying by 2.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>Do ionic compounds have molecular formulae?<\/h3>\n            <p>No. Ionic compounds consist of a giant lattice rather than discrete molecules, so their formulae represent the simplest whole-number ratio of ions. For example, sodium chloride is written as NaCl, which represents the simplest ratio of Na<sup>+<\/sup> ions to Cl<sup>&#8211;<\/sup> ions.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>How do I find the empirical formula from combustion data?<\/h3>\n            <p>Convert the mass of CO<sub>2<\/sub> to mass of C using the fraction 12\/44, and the mass of H<sub>2<\/sub>O to mass of H using 2\/18. If the compound contains oxygen, subtract the masses of C and H from the original sample mass to find the mass of O. Then apply the standard three-step method.<\/p>\n          <\/div>\n        <\/div>\n      <\/section>\n\n      <!-- Related Topics -->\n      <section class=\"ols-related-card\">\n        <h2>Related Topics<\/h2>\n        <p>Build the surrounding AQA 3.1.2 skills needed to work confidently with empirical formula and mole calculations.<\/p>\n        <div class=\"ols-related-grid\">\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/\">Empirical and Molecular Formulae<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/molecular-formula\/\">Molecular Formula<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/relative-masses\/\">Relative Masses<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/the-mole-and-avogadro-constant\/\">The Mole and Avogadro Constant<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/calculations-using-reacting-masses\/\">Calculations Using Reacting Masses<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/\">3.1.2 Amount of Substance<\/a>\n        <\/div>\n      <\/section>\n\n      <!-- Copyright attribution -->\n      <section class=\"ols-attribution-card\">\n        <p><strong>Copyright notice:<\/strong> This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.<\/p>\n      <\/section>\n\n      <!-- JSON-LD Schema -->\n      <script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@graph\": [\n    {\n      \"@type\": \"WebPage\",\n      \"@id\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/empirical-formula\/#webpage\",\n      \"url\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/empirical-formula\/\",\n      \"name\": \"Empirical Formula - AQA A Level Chemistry Revision Notes\",\n      \"description\": \"AQA A Level Chemistry revision notes explaining empirical formula, mole-ratio calculations, percentage composition, combustion analysis and the link to molecular formula.\",\n      \"inLanguage\": \"en-GB\",\n      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