{"id":8080,"date":"2026-09-15T08:24:26","date_gmt":"2026-09-15T07:24:26","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-using-reacting-masses\/"},"modified":"2026-09-19T20:33:28","modified_gmt":"2026-09-19T19:33:28","slug":"calculations-using-reacting-masses","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-using-reacting-masses\/","title":{"rendered":"Calculations using Reacting Masses"},"content":{"rendered":"\n<!--\n===============================================================================\nONLINE LEARNING SYSTEM COPYRIGHT NOTICE\n\u00a9 Online Learning System. 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.ols-figure-placeholder .ols-figure-image { background: transparent; }\n    .ols-figure-placeholder { background: #ffffff; border: 1px solid rgba(28, 36, 75, 0.12); border-radius: 22px; padding: 14px; margin: 18px 0 6px; }\n    .ols-figure-placeholder .ols-figure-caption p { margin: 10px 0 0; font-size: 14px; color: #667085; text-align: center; }\n<\/style>\n\n  <!--\n    OLS protected content block.\n    Page: Calculations Using Reacting Masses.\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\">\n  <div class=\"ols-sidebar-header\">\n    <h3>Revision Notes<\/h3>\n    <p>Cambridge International AS Level Chemistry<\/p>\n  <\/div>\n  <div class=\"ols-topic-group\">\n    <h4>Topic 2 Atoms, Molecules and Stoichiometry<\/h4>\n    <ul class=\"ols-topic-list\">\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic overview<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/atoms-elements-and-molecules\/\">2.1 \/ 2.2 Atoms, Elements and Molecules<\/a><\/li>\n<li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/ions-and-ionic-formulae\/\">2.3 Ions and Ionic Formulae<\/a><\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/\">2.3 Equations and Reaction Types<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/writing-chemical-equations\/\">Writing Chemical Equations<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/typical-reactions-of-acids\/\">Typical Reactions of Acids<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/ionic-and-full-equations\/\">Ionic and Full Equations<\/a><\/li>\n        <\/ul>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/\">2.3 Empirical and Molecular Formulae<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/empirical-formulae\/\">Empirical Formulae<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/molecular-formulae\/\">Molecular Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/hydrated-salts-and-water-of-crystallisation\/\">2.3 Hydrated Salts and Water of Crystallisation<\/a><\/li>\n        <\/ul>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/\">2.4 Mole Calculations<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/comparing-masses-of-substances\/\">Comparing Masses of Substances<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-involving-moles\/\">Calculations involving Moles<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-using-reacting-masses\/\">Calculations using Reacting Masses<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/the-yield-of-a-reaction\/\">The Yield of a Reaction<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/atom-economy\/\">Atom Economy<\/a><\/li>\n        <\/ul>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/\">2.4 Solutions and Gases<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/molar-volume-calculations\/\">Molar Volume Calculations<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/concentrations-of-solutions\/\">Concentrations of Solutions<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/titration-calculations\/\">Titration Calculations<\/a><\/li>\n        <\/ul>\n      <\/li>\n    <\/ul>\n  <\/div>\n  <div class=\"ols-topic-group\">\n    <h4>Other Topics<\/h4>\n    <ul class=\"ols-topic-list\">\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-1-atomic-structure\/\">Topic 1 Atomic Structure<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-3-chemical-bonding\/\">Topic 3 Chemical Bonding<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-4-states-of-matter\/\">Topic 4 States of Matter<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-9-the-periodic-table-chemical-periodicity\/\">Topic 9 Chemical Periodicity<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-13-an-introduction-to-as-level-organic-chemistry\/\">Topic 13 Introduction to Organic Chemistry<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-14-hydrocarbons\/\">Topic 14 Hydrocarbons<\/a><\/li>\n    <\/ul>\n  <\/div>\n<\/aside>\n<script>\r\n(function() {\r\n  function normalisePath(path) {\r\n    return String(path || '')\r\n      .split('?')[0]\r\n      .split('#')[0]\r\n      .replace(\/\\\/+$\/, '')\r\n      .toLowerCase();\r\n  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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\/cie\/\">Cambridge International (CIE)<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic 2 Atoms, Molecules and Stoichiometry<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/\">Mole Calculations<\/a> \/\n<span>Calculations using Reacting Masses<\/span>\n<\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Calculations Using Reacting Masses<\/h1>\n        <p class=\"ols-page-intro\">A focused revision guide to using balanced equations, mole ratios and molar masses to calculate the mass of a reactant or product in a chemical reaction.<\/p>\n\n        <div class=\"ols-badges\">\n<div class=\"ols-badge\">AS Level<\/div>\n<div class=\"ols-badge\">Topic 2: Atoms, Molecules and Stoichiometry<\/div>\n<div class=\"ols-badge\">9701 Papers 1 and 2<\/div>\n<\/div>\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      <section class=\"ols-h5p-card ols-h5p-inline ols-h5p-recap\">\n<span class=\"ols-h5p-kicker\">Before you start<\/span>\n<h2>GCSE Recap: Balanced Equations and Conservation of Mass<\/h2>\n<p>Before you start, check that you can balance an equation and use conservation of mass.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"617\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">1<\/div><h2>Why Equations Can Be Used for Mass Calculations<\/h2><\/div>\n        <p>A balanced chemical equation tells you the <strong>mole ratio<\/strong> between reactants and products. Once you know how many moles of one substance are involved, the equation lets you work out how many moles of another substance react or form.<\/p>\n        <p>Reacting mass calculations connect three ideas: <strong>mass<\/strong>, <strong>amount in moles<\/strong> and the <strong>balanced equation<\/strong>.<\/p>\n        <div class=\"ols-definition-box\"><p><strong>Reacting mass calculation:<\/strong> a calculation that uses a balanced equation to find the mass of a reactant used or a product formed.<\/p><\/div>\n        <div class=\"ols-key-box\"><p><strong>Key idea:<\/strong> the balancing numbers in the equation are mole ratios, not mass ratios.<\/p><\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: What the Balancing Numbers Mean<\/h2>\n<p>Decide whether the statement about this equation is true or false.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-618\" class=\"h5p-iframe\" data-content-id=\"618\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reacting Masses True or False: Reading the Balancing Numbers\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <article class=\"ols-note-card\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">2<\/div><h2>The Core Formula<\/h2><\/div>\n        <p>The most important formula for reacting mass calculations links mass, amount and molar mass.<\/p>\n        <div class=\"ols-equation-box\"><strong>amount in mol = mass \u00f7 M<sub>r<\/sub><\/strong><br><strong>mass = amount in mol \u00d7 M<sub>r<\/sub><\/strong><\/div>\n        <p>You usually use the first form when converting the given mass into moles, and the second form when converting the final amount in moles back into a mass.<\/p>\n        <div class=\"ols-table-wrap\">\n          <table class=\"ols-table\">\n            <thead>\n              <tr><th>Quantity<\/th><th>Meaning<\/th><th>Typical unit<\/th><\/tr>\n            <\/thead>\n            <tbody>\n              <tr><td data-label=\"Quantity\">mass<\/td><td data-label=\"Meaning\">The mass of the substance in the question<\/td><td data-label=\"Typical unit\">g, kg or tonnes<\/td><\/tr>\n              <tr><td data-label=\"Quantity\">amount<\/td><td data-label=\"Meaning\">The number of moles of particles<\/td><td data-label=\"Typical unit\">mol<\/td><\/tr>\n              <tr><td data-label=\"Quantity\">M<sub>r<\/sub><\/td><td data-label=\"Meaning\">Relative formula mass or relative molecular mass<\/td><td data-label=\"Typical unit\">g mol<sup>-1<\/sup> when used as molar mass<\/td><\/tr>\n            <\/tbody>\n          <\/table>\n        <\/div>\n      <\/article>\n\n      <article class=\"ols-note-card purple\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">3<\/div><h2>The Standard Method<\/h2><\/div>\n        <p>Most reacting mass questions can be solved using the same sequence. The method is especially reliable because it keeps the mole ratio separate from the mass calculation.<\/p>\n        <div class=\"ols-rule-list\">\n          <div class=\"ols-rule-item\"><h3>Step 1: Write or check the balanced equation<\/h3><p>The coefficients in the equation give the mole ratio between substances.<\/p><\/div>\n          <div class=\"ols-rule-item\"><h3>Step 2: Convert the given mass into moles<\/h3><p>Use amount = mass \u00f7 M<sub>r<\/sub>.<\/p><\/div>\n          <div class=\"ols-rule-item\"><h3>Step 3: Use the mole ratio<\/h3><p>Use the balanced equation to convert from moles of the known substance to moles of the required substance.<\/p><\/div>\n          <div class=\"ols-rule-item\"><h3>Step 4: Convert moles into the required mass<\/h3><p>Use mass = amount \u00d7 M<sub>r<\/sub>.<\/p><\/div>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Order the Standard Method<\/h2>\n<p>Drag the steps of this reacting mass calculation into the right order.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-619\" class=\"h5p-iframe\" data-content-id=\"619\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reacting Masses Put in Order: The Standard Method for Aluminium Oxide\"><\/iframe><\/div><\/div>\n<\/section>\n\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\/Using-moles-to-calculate-mass.webp\" alt=\"Using moles and a balanced equation to calculate reacting masses\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>The calculation pathway is mass \u2192 moles \u2192 equation ratio \u2192 moles \u2192 mass.<\/p>\n        <\/div>\n      <\/div>\n\n      <article class=\"ols-note-card orange\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">4<\/div><h2>Worked Example: Sodium Hydrogencarbonate<\/h2><\/div>\n        <p>Calculate the mass of carbon dioxide produced when <strong>5.50 g<\/strong> of sodium hydrogencarbonate is heated.<\/p>\n        <div class=\"ols-equation-box\">2NaHCO<sub>3<\/sub> \u2192 Na<sub>2<\/sub>CO<sub>3<\/sub> + CO<sub>2<\/sub> + H<sub>2<\/sub>O<\/div>\n        <div class=\"ols-calculation-stack\">\n          <div class=\"ols-calculation-step\"><h3>Step 1: Find moles of NaHCO<sub>3<\/sub><\/h3><p>M<sub>r<\/sub> of NaHCO<sub>3<\/sub> = 84<br>amount = 5.50 \u00f7 84 = 0.0655 mol<\/p><\/div>\n          <div class=\"ols-calculation-step\"><h3>Step 2: Use the equation ratio<\/h3><p>2 mol NaHCO<sub>3<\/sub> gives 1 mol CO<sub>2<\/sub><br>0.0655 mol NaHCO<sub>3<\/sub> gives 0.0328 mol CO<sub>2<\/sub><\/p><\/div>\n          <div class=\"ols-calculation-step\"><h3>Step 3: Find the mass of CO<sub>2<\/sub><\/h3><p>M<sub>r<\/sub> of CO<sub>2<\/sub> = 44.0<br>mass = 0.0328 \u00d7 44.0 = <strong>1.44 g<\/strong><\/p><\/div>\n        <\/div>\n        <div class=\"ols-key-box\"><p><strong>Exam focus:<\/strong> the mass of CO<sub>2<\/sub> is not found by directly comparing 84 and 44. The equation shows that 2 mol NaHCO<sub>3<\/sub> forms only 1 mol CO<sub>2<\/sub>.<\/p><\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Reacting Masses in Grams<\/h2>\n<p>Work each calculation out on paper before you turn the card.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-308\" class=\"h5p-iframe\" data-content-id=\"308\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reacting Masses Flip Cards: Grams of Reactant and Product\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">5<\/div><h2>Using Direct Proportion<\/h2><\/div>\n        <p>Some reacting mass questions can be solved by treating one formula unit or one mole ratio as a direct proportion. This is useful when the units are large, such as kilograms or tonnes.<\/p>\n        <p>For example, in a blast furnace, haematite is reduced to iron:<\/p>\n        <div class=\"ols-equation-box\">Fe<sub>2<\/sub>O<sub>3<\/sub> + 3CO \u2192 2Fe + 3CO<sub>2<\/sub><\/div>\n        <p>From the equation, <strong>1 mol Fe<sub>2<\/sub>O<sub>3<\/sub><\/strong> gives <strong>2 mol Fe<\/strong>. Using A<sub>r<\/sub> values Fe = 56 and O = 16:<\/p>\n        <div class=\"ols-equation-box\">160 g Fe<sub>2<\/sub>O<sub>3<\/sub> gives 112 g Fe<br>160 tonnes Fe<sub>2<\/sub>O<sub>3<\/sub> gives 112 tonnes Fe<br>16 tonnes Fe<sub>2<\/sub>O<sub>3<\/sub> gives 11.2 tonnes Fe<\/div>\n        <div class=\"ols-key-box\"><p><strong>Key idea:<\/strong> if the same mass unit is used throughout, the ratio still works. Do not mix grams and tonnes in the same line unless you deliberately convert units.<\/p><\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Tonnes and Kilograms<\/h2>\n<p>Use direct proportion and type each answer to 3 significant figures.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-620\" class=\"h5p-iframe\" data-content-id=\"620\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reacting Masses Fill in the Blanks: Direct Proportion in Tonnes and Kilograms\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <article class=\"ols-note-card purple\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">6<\/div><h2>Multi-Step Equation Chains<\/h2><\/div>\n        <p>Sometimes the product of one equation is used as the reactant in the next equation. The safest approach is to follow the mole ratio through each equation before converting back to mass.<\/p>\n        <p>For example, nitrogen can be converted through several steps to nitric acid:<\/p>\n        <div class=\"ols-equation-box\">N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub><br>4NH<sub>3<\/sub> + 5O<sub>2<\/sub> \u2192 4NO + 6H<sub>2<\/sub>O<br>2NO + O<sub>2<\/sub> \u2192 2NO<sub>2<\/sub><br>2H<sub>2<\/sub>O + 4NO<sub>2<\/sub> + O<sub>2<\/sub> \u2192 4HNO<sub>3<\/sub><\/div>\n        <p>Tracing the ratios shows that <strong>1 mol N<sub>2<\/sub><\/strong> eventually gives <strong>2 mol HNO<sub>3<\/sub><\/strong>. Since M<sub>r<\/sub> of N<sub>2<\/sub> = 28 and M<sub>r<\/sub> of HNO<sub>3<\/sub> = 63:<\/p>\n        <div class=\"ols-equation-box\">28 tonnes N<sub>2<\/sub> gives 2 \u00d7 63 tonnes HNO<sub>3<\/sub><br>1 tonne N<sub>2<\/sub> gives (2 \u00d7 63) \u00f7 28 = <strong>4.5 tonnes HNO<sub>3<\/sub><\/strong><\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Follow the Chain<\/h2>\n<p>Trace the mole ratio through all three equations before you calculate a mass.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-309\" class=\"h5p-iframe\" data-content-id=\"309\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reacting Masses MCQ: Sulfur to Sulfuric Acid in Three Steps\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <article class=\"ols-note-card\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">7<\/div><h2>Units and Significant Figures<\/h2><\/div>\n        <p>Reacting mass calculations often lose marks through unit errors rather than chemistry errors. Keep the unit consistent and present the final answer to a sensible number of significant figures.<\/p>\n        <div class=\"ols-table-wrap\">\n          <table class=\"ols-table\">\n            <thead>\n              <tr><th>Issue<\/th><th>What to do<\/th><th>Example<\/th><\/tr>\n            <\/thead>\n            <tbody>\n              <tr><td data-label=\"Issue\">Mixed units<\/td><td data-label=\"What to do\">Convert before calculating, or keep the same unit throughout a proportion.<\/td><td data-label=\"Example\">Use all grams, all kilograms or all tonnes.<\/td><\/tr>\n              <tr><td data-label=\"Issue\">Premature rounding<\/td><td data-label=\"What to do\">Keep extra figures in intermediate steps.<\/td><td data-label=\"Example\">Use 0.03275 mol before rounding the final answer.<\/td><\/tr>\n              <tr><td data-label=\"Issue\">Missing equation ratio<\/td><td data-label=\"What to do\">Always use the coefficients from the balanced equation.<\/td><td data-label=\"Example\">2NaHCO<sub>3<\/sub> : 1CO<sub>2<\/sub>, not 1 : 1.<\/td><\/tr>\n            <\/tbody>\n          <\/table>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Sound Working<\/h2>\n<p>In each round, choose the statement that shows sound working.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-621\" class=\"h5p-iframe\" data-content-id=\"621\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reacting Masses Summary: Units, Rounding and Ratios\"><\/iframe><\/div><\/div>\n<\/section>\n\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 balancing numbers in an equation represent a mole ratio.<\/li>\n          <li>Do not compare masses directly unless you have first converted through moles or built a valid proportion.<\/li>\n          <li>Use amount = mass \u00f7 M<sub>r<\/sub> to find moles from mass.<\/li>\n          <li>Use mass = amount \u00d7 M<sub>r<\/sub> to convert moles back into mass.<\/li>\n          <li>Use the balanced equation to move between the known substance and the required substance.<\/li>\n          <li>Check whether the question asks for a reactant mass or a product mass.<\/li>\n          <li>Keep units consistent, especially when questions use kg or tonnes.<\/li>\n          <li>Show enough working so the mole ratio and formula mass steps are visible.<\/li>\n        <\/ul>\n      <\/article>\n\n      <!-- supp-insert LR1 -->\n<article class=\"ols-note-card orange\">\n<div class=\"ols-note-title\">\n<div class=\"ols-note-icon\">9<\/div>\n<h2>Limiting and Excess Reagents<\/h2>\n<\/div>\n<p>Reactants are rarely mixed in exactly the ratio the equation demands. The reactant that runs out first is the <strong>limiting reagent<\/strong>; it decides how much product can form. The other reactant is in <strong>excess<\/strong> and some of it is left over.<\/p>\n<p><strong>Worked example.<\/strong> 5.6 g of iron is heated with 4.0 g of sulfur: <strong>Fe + S \u2192 FeS<\/strong>.<\/p>\n<ol>\n<li>n(Fe) = 5.6 \/ 55.8 = 0.100 mol; n(S) = 4.0 \/ 32.1 = 0.125 mol<\/li>\n<li>The equation needs 1 : 1, so 0.100 mol of iron needs 0.100 mol of sulfur. There is 0.125 mol of sulfur, so <strong>sulfur is in excess and iron is limiting<\/strong>.<\/li>\n<li>Product from the limiting reagent: n(FeS) = 0.100 mol, mass = 0.100 \u00d7 87.9 = <strong>8.79 g<\/strong><\/li>\n<li>Sulfur left over: 0.125 \u2212 0.100 = 0.025 mol = 0.80 g<\/li>\n<\/ol>\n<p>Always divide each reactant&#8217;s moles by its coefficient in the equation before comparing; the smaller result identifies the limiting reagent.<\/p>\n<div class=\"ols-key-box\">\n<p><strong>Exam sentence:<\/strong> The limiting reagent is the reactant that is completely used up; the amount of product is calculated from its moles, not from the reactant in excess.<\/p>\n<\/div>\n<\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Which Reactant Runs Out?<\/h2>\n<p>Find the limiting reagent, then use it to calculate the mass of product.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-521\" class=\"h5p-iframe\" data-content-id=\"521\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Limiting Reagent MCQ: Magnesium Burning in Oxygen\"><\/iframe><\/div><\/div>\n<\/section>\n\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>Balanced equations:<\/strong> show mole ratios between reactants and products.<\/li>\n          <li><strong>First conversion:<\/strong> use amount = mass \u00f7 M<sub>r<\/sub>.<\/li>\n          <li><strong>Equation step:<\/strong> use the coefficients to convert from one substance to another.<\/li>\n          <li><strong>Final conversion:<\/strong> use mass = amount \u00d7 M<sub>r<\/sub>.<\/li>\n          <li><strong>Direct proportion:<\/strong> works when the equation ratio and units are handled consistently.<\/li>\n          <li><strong>Common mistake:<\/strong> treating the equation numbers as mass ratios instead of mole ratios.<\/li>\n        <\/ul>\n      <\/section>\n\n      <div class=\"ols-image-lightbox\" id=\"olsImageLightboxReactingMasses001\" 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\">\u00d7<\/button>\n          <img decoding=\"async\" class=\"ols-image-lightbox-img\" src=\"\" alt=\"\">\n        <\/div>\n      <\/div>\n\n      <script>\n        (function(){\n          var page = document.querySelector(\".ols-reacting-masses-page\");\n          if (!page) { return; }\n          var lightbox = page.querySelector(\"#olsImageLightboxReactingMasses001\");\n          if (!lightbox) { return; }\n          var lightboxImage = lightbox.querySelector(\".ols-image-lightbox-img\");\n          var closeButton = lightbox.querySelector(\".ols-image-lightbox-close\");\n          var clickableImages = page.querySelectorAll(\"img.ols-lightbox-target\");\n\n          function openLightbox(image) {\n            lightboxImage.src = image.currentSrc || image.src;\n            lightboxImage.alt = image.alt || \"Expanded revision image\";\n            lightbox.classList.add(\"is-open\");\n            lightbox.setAttribute(\"aria-hidden\", \"false\");\n            if (!document.body.dataset.olsPreviousOverflow) {\n              document.body.dataset.olsPreviousOverflow = document.body.style.overflow || \"default\";\n            }\n            document.body.style.overflow = \"hidden\";\n          }\n\n          function closeLightbox() {\n            lightbox.classList.remove(\"is-open\");\n            lightbox.setAttribute(\"aria-hidden\", \"true\");\n            lightboxImage.src = \"\";\n            if (document.body.dataset.olsPreviousOverflow) {\n              document.body.style.overflow = document.body.dataset.olsPreviousOverflow === \"default\" ? \"\" : document.body.dataset.olsPreviousOverflow;\n              delete document.body.dataset.olsPreviousOverflow;\n            }\n          }\n\n          clickableImages.forEach(function(image){\n            image.addEventListener(\"click\", function(event){\n              event.preventDefault();\n              event.stopPropagation();\n              openLightbox(image);\n            });\n          });\n          closeButton.addEventListener(\"click\", closeLightbox);\n          lightbox.addEventListener(\"click\", function(event){ if (event.target === lightbox) { closeLightbox(); } });\n          document.addEventListener(\"keydown\", function(event){ if (event.key === \"Escape\" && lightbox.classList.contains(\"is-open\")) { closeLightbox(); } });\n        })();\n      <\/script>\n\n      <section class=\"ols-faq-card\">\n        <h2>FAQs<\/h2>\n        <p>These questions address the most common points students confuse when calculating reacting masses.<\/p>\n        <div class=\"ols-faq-list\">\n          <div class=\"ols-faq-item\"><h3>What is a reacting mass calculation?<\/h3><p>It is a calculation that uses a balanced chemical equation to work out the mass of a reactant used or a product formed.<\/p><\/div>\n          <div class=\"ols-faq-item\"><h3>Why do I need the balanced equation?<\/h3><p>The balanced equation gives the mole ratio between substances. Without this ratio, the calculation may use the wrong number of moles.<\/p><\/div>\n          <div class=\"ols-faq-item\"><h3>Can I compare masses directly?<\/h3><p>Not usually. The equation gives mole ratios, so you should convert through moles unless you have built a valid direct proportion from molar masses.<\/p><\/div>\n          <div class=\"ols-faq-item\"><h3>Does the method work with tonnes?<\/h3><p>Yes. The same ratio works with grams, kilograms or tonnes as long as the unit is used consistently throughout the proportion.<\/p><\/div>\n          <div class=\"ols-faq-item\"><h3>What is the most common mistake?<\/h3><p>The most common mistake is forgetting to use the coefficient ratio from the balanced equation, such as treating 2NaHCO<sub>3<\/sub> \u2192 CO<sub>2<\/sub> as a 1 : 1 mole relationship.<\/p><\/div>\n        <\/div>\n      <\/section>\n\n      <section class=\"ols-related-card\">\n        <h2>Related Topics<\/h2>\n        <p>Use these topics to strengthen the equation writing and mole calculation skills that support reacting mass questions.<\/p>\n        <div class=\"ols-related-grid\">\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/writing-chemical-equations\/\">Writing Chemical Equations<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/\">Equations and Reaction Types<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/ionic-and-full-equations\/\">Ionic Equations<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-involving-moles\/\">Calculations Involving Moles<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-1-atomic-structure\/mass-spectrometry\/ram-calculations\/\">RAM Calculations<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic 2 Overview<\/a>\n        <\/div>\n      <\/section>\n\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. 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