{"id":13628,"date":"2014-02-13T23:26:11","date_gmt":"2014-02-13T23:26:11","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=13628"},"modified":"2026-03-25T11:46:08","modified_gmt":"2026-03-25T16:46:08","slug":"how-to-balance-a-chemical-equation","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/how-to-balance-a-chemical-equation\/","title":{"rendered":"How Do You Balance Chemical Equations?"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_bx9eSF8tAco\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_bx9eSF8tAco\" data-source-videoID=\"bx9eSF8tAco\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"How Do You Balance Chemical Equations? 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In other words, when reactants recombine or decompose to form products, we can\u2019t lose or create atoms- we have to end with the same number we started with. <\/p>\n<p>Chemists represent this in a balanced chemical equation, in which the number of each type of <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/structure-of-atoms\/\">atom<\/a> on the reactant side equals the number on the product side. Conversely, an unbalanced reaction has unequal numbers of each type of atom on the reactant and product side of the equation.  <\/p>\n<h2><span id=\"Balancing_Equations_Examples\" class=\"m-toc-anchor\"><\/span>Balancing Equations Examples<\/h2>\n<p>\nFor example, here\u2019s the unbalanced reaction of the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/combustion\/\">combustion<\/a> of methane. <\/p>\n<div class=\"examplesentence\">\\(\\text{CH}_4 + \\text{O}_2 \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nMethane reacts with oxygen to produce carbon dioxide and water. While this equation accurately represents the chemical identity of the reactants and products, it\u2019s unbalanced. If we count the atoms on each side of the reaction, we\u2019ll discover that they are not equal! <\/p>\n<p>It\u2019s often helpful to keep track of our atom count in a table. <\/p>\n<p>While counting, remember that the subscripts (the 4 in methane, 2 in oxygen, 2 in carbon dioxide, and 2 in water) they denote the number of atoms of the element preceding the subscript within the compound. <\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 70%;\">\n<thead>\n<tr>\n<th>Element<\/th>\n<th>Reactant Side<\/th>\n<th>Product Side<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>C<\/td>\n<td>1<\/td>\n<td>1<\/td>\n<td>Yes<\/td>\n<\/tr>\n<tr>\n<td>H<\/td>\n<td>4<\/td>\n<td>2<\/td>\n<td>No<\/td>\n<\/tr>\n<tr>\n<td>O<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>No<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nOn the reactant side, we have 1 carbon, 4 hydrogen, and 2 oxygen atoms. On the product side, we have 1 carbon, 2 hydrogen, and 3 oxygen atoms. So, we need to balance this reaction. <\/p>\n<p>Note that to balance equations, we have to honor the chemistry, meaning that you can\u2019t simply add in extra compounds or atoms to achieve balance. This is because, of course, the equation must still reflect the chemistry that actually takes place. So for example, while this equation is technically balanced in that there is the same number of each element on either side of the reaction, it\u2019s chemically inaccurate. <\/p>\n<div class=\"examplesentence\">\\(\\text{CH}_4 + \\text{O}_2 + \\text{O} \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O} + \\text{H}_2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis reaction doesn\u2019t produce \\(\\text{H}_2\\) gas and certainly doesn\u2019t use a lone oxygen atom as a reactant. <\/p>\n<p>We also can\u2019t adjust the subscripts of compounds to achieve balance, as this would change the chemical identity of the species. <\/p>\n<div class=\"examplesentence\">\\(\\text{CH}_4 + \\text{O}_3 \\rightarrow \\text{CO}_2 + \\text{H}_4\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo again, while this is a balanced equation, we are no longer representing the combustion of methane. \\(\\text{O}_3\\) is ozone, and is chemically very different from \\(\\text{O}_2\\), and \\(\\text{H}_4\\text{O}\\) is hydrogen hydrate. By changing the subscripts, we change the chemical identity of the species and wind up with an entirely different and probably unviable reaction. <\/p>\n<p>Instead, to balance a reaction, we adjust the number of each species in the chemical reaction. This is represented in the chemical equation by adding coefficients in front of the chemical species. <\/p>\n<div class=\"examplesentence\">__\\(\\text{CH}_4 +\\) __\\(\\text{O}_2 ->\\) __\\(\\text{CO}_2 +\\) __\\(\\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor example, if we put a 2 in front of water, this means we have two water molecules, or 4 hydrogen and 2 oxygen atoms. <\/p>\n<div class=\"examplesentence\">__\\(\\text{CH}_4 +\\) __\\(\\text{O}_2 \\rightarrow \\) __\\(\\text{CO}_2 + 2\\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNotice that by doubling the water, hydrogen (in addition to carbon) is now balanced. <\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 70%;\">\n<thead>\n<tr>\n<th>Element<\/th>\n<th>Reactant Side<\/th>\n<th>Product Side<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>C<\/td>\n<td>1<\/td>\n<td>1<\/td>\n<td>Yes<\/td>\n<\/tr>\n<tr>\n<td>H<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<td><strong>Yes<\/strong><\/td>\n<\/tr>\n<tr>\n<td>O<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>No<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nNow to finish we just need to balance oxygen. We have 2 on the reactant side and 4 on the product side, so by simply adding a second molecule of oxygen to the reactant side, we have ourselves a balanced reaction! <\/p>\n<div class=\"examplesentence\">\\(1\\text{CH}_4 +2\\text{O}_2 \\rightarrow 1\\text{CO}_2 + 2\\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNote that it\u2019s standard to drop the coefficient when it\u2019s 1.<\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 70%;\">\n<thead>\n<tr>\n<th>Element<\/th>\n<th>Reactant Side<\/th>\n<th>Product Side<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>C<\/td>\n<td>1<\/td>\n<td>1<\/td>\n<td>Yes<\/td>\n<\/tr>\n<tr>\n<td>H<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<td>Yes<\/td>\n<\/tr>\n<tr>\n<td>O<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<td>Yes<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<br \/>\nWe can now read this equation as 1 molecule of methane reacts with 2 molecules of oxygen to produce 1 molecule of carbon dioxide and 2 molecules of water.<\/p>\n<p>Notice that the equation is also balanced with these coefficients:<\/p>\n<div class=\"examplesentence\">\\(2\\text{CH}_4 + 4\\text{O}_2 \\rightarrow 2\\text{CO}_2 + 4\\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe\u2019ve simply multiplied each coefficient by 2, which does give us a balanced chemical equation; however, we typically want the lowest whole number coefficients. So, similarly, while this<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{2}\\text{CH}_4 + \\text{O}_2 \\rightarrow \\frac{1}{2}\\text{CO}_2 + \\text{H}_2\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nis also technically balanced, it uses fractions rather than whole numbers as coefficients. So once you\u2019re finished balancing your reaction, make sure there\u2019s no common denominator between coefficients or any fractional coefficients. If there are, either divide out by the common denominator or multiply to get the lowest whole number coefficients.<\/p>\n<h2><span id=\"The_Importance_of_Balancing_Equations\" class=\"m-toc-anchor\"><\/span>The Importance of Balancing Equations<\/h2>\n<p>\nNow that we understand how to balance equations, let\u2019s take a step back and think about why they are important. These coefficients not only ensure we follow the law of conservation of mass, but also dictate the calculation of reactants and products, or the stoichiometry, of the reaction (you\u2019ll sometimes see the coefficients referred to as the stoichiometric coefficients). <\/p>\n<p>For example, in the combustion of methane, to produce 1 mole of carbon dioxide, we need 2 moles of oxygen but only 1 mole of methane. Put another way, the mole ratio is 2:1 for oxygen and carbon dioxide but 1 to 1 for methane and carbon dioxide. Thus, these coefficients are necessary for calculating limiting reagents, yields, molarity, or other calculations that requires converting between moles of species within a reaction. Basically, it\u2019s pretty important to have your chemical equation balanced correctly. <\/p>\n<p>Though there is no one sure-fire way to balance chemical equations, there are a few useful techniques to apply. To summarize- it\u2019s usually best to balance the elements in the more complex compounds (or multiple compounds) first and leave simple species for the latter steps. Second, for elements in only one species on either side, try using the subscript of the opposite side as a coefficient. Third, while neither of our examples dealt with charged ions, if you do encounter a chemical equation with ions, make sure that the charge is balanced as well! After all, you don\u2019t want to lose electrons! <\/p>\n<p>And lastly, some equations are simply going to be more challenging than others. It\u2019s important not to get frustrated. Simply jump in and start by balancing one element and evaluate how that changed the entire chemical equation. Then move on to a second element, add a coefficient, and re-evaluate. Keep running through this process until you\u2019ve got yourself a balanced equation- but remember to check that you\u2019ve found the set of lowest whole-number coefficients. <\/p>\n<h2><span id=\"Practice_Problems\" class=\"m-toc-anchor\"><\/span>Practice Problems<\/h2>\n<p>\nNow that we\u2019ve pointed out a few tips for balancing reactions, let\u2019s practice some more. Pause the video and balance the following chemical equations:<\/p>\n<ul>\n<li style=\"margin-bottom: 12px;\">Oxidation of nitric oxide by oxygen: \\(\\text{NO} + \\text{O}_2 \\rightarrow \\text{NO}_2\\)<\/li>\n<li style=\"margin-bottom: 12px;\">Roasting of the sulfide ore pyrite: \\(\\text{FeS}_2 + \\text{O}_2 \\rightarrow \\text{Fe}_2\\text{O}_3 + \\text{SO}_2\\)<\/li>\n<li style=\"margin-bottom: 12px;\">Combustion of ethane: \\(\\text{C}_2\\text{H}_6 + \\text{O}_2 \\rightarrow \\text{CO}_2 + \\text{H}_2\\text{O}\\)<\/li>\n<li>Decomposition of lead nitrate: \\(\\text{Pb(NO}_3\\text{)}_2 \\rightarrow \\text{PbO} + \\text{NO}_2 + \\text{O}_2\\)<\/li>\n<\/ul>\n<div style=\"text-align: center; margin-bottom: 20px;\"><button class=\"buttontranscript\" onClick=\"toggle('Answer1')\">Show Answers<\/button><\/div>\n<div id=\"Answer1\" style=\"display:none; box-shadow: 1.5px 1.5px 5px grey; background-color:#E0E0E0; padding: 30px; padding-bottom: 15px; width: 60%; margin: auto; text-align: center;\">\n<strong>Oxidation of nitric oxide by oxygen<\/strong><br \/>\n\\(2\\text{NO} + \\text{O}_2 \\rightarrow 2\\text{NO}_2\\)<\/p>\n<hr>\n<p>\n<strong>Roasting of the sulfide ore pyrite<\/strong><br \/>\n\\(4\\text{FeS}_2 + 11\\text{O}_2\\rightarrow 2\\text{Fe}_2\\text{O}_3 + 8\\text{SO}_2\\)<\/p>\n<hr>\n<p>\n<strong>Combustion of ethane<\/strong><br \/>\n\\(2\\text{C}_2\\text{H}_6 + 7\\text{O}_2 \\rightarrow 4\\text{CO}_2 + 6\\text{H}_2\\text{O}\\)<\/p>\n<hr>\n<p>\n<strong>Decomposition of lead nitrate <\/strong><br \/>\n\\(2\\text{Pb(NO}_3\\text{)}_2 \\rightarrow 2\\text{PbO} + 4\\text{NO}_2 + \\text{O}_2\\)\n<\/div>\n<p>\n&nbsp;<br \/>\nThe last three were pretty challenging, so don\u2019t feel discouraged if they took you a while or you weren\u2019t able to solve them. The best thing to do is to keep practicing and you\u2019ll start to feel more comfortable with the process. <\/p>\n<p>That\u2019s it for today\u2019s video. 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