{"id":251908,"date":"2025-07-10T21:09:08","date_gmt":"2025-07-10T21:09:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=251908"},"modified":"2025-07-10T21:09:15","modified_gmt":"2025-07-10T21:09:15","slug":"what-is-the-molecular-mass-of-baking-soda-nahco3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/what-is-the-molecular-mass-of-baking-soda-nahco3\/","title":{"rendered":"What is the molecular mass of baking soda, NaHCO3"},"content":{"rendered":"\n<p>What is the molecular mass of baking soda, NaHCO3? (specify units) b) how many grams of oxygen are in 0.25 mol of baking soda? c) how many hydrogen atoms are present in 0.25 mol of baking soda? Please write out all the steps<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s break it down step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">a) Molecular Mass of Baking Soda (NaHCO\u2083)<\/h3>\n\n\n\n<p>First, to find the molecular mass of NaHCO\u2083 (baking soda), we add up the atomic masses of each element in the compound. Here are the atomic masses of each element:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sodium (Na):<\/strong> 22.99 g\/mol<\/li>\n\n\n\n<li><strong>Hydrogen (H):<\/strong> 1.008 g\/mol<\/li>\n\n\n\n<li><strong>Carbon (C):<\/strong> 12.01 g\/mol<\/li>\n\n\n\n<li><strong>Oxygen (O):<\/strong> 16.00 g\/mol (and there are three oxygen atoms)<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the total molecular mass:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sodium: 1 \u00d7 22.99 g\/mol = 22.99 g\/mol<\/li>\n\n\n\n<li>Hydrogen: 1 \u00d7 1.008 g\/mol = 1.008 g\/mol<\/li>\n\n\n\n<li>Carbon: 1 \u00d7 12.01 g\/mol = 12.01 g\/mol<\/li>\n\n\n\n<li>Oxygen: 3 \u00d7 16.00 g\/mol = 48.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>So, the molecular mass of NaHCO\u2083 is: 22.99+1.008+12.01+48.00=84.008\u2009g\/mol22.99 + 1.008 + 12.01 + 48.00 = 84.008 \\, \\text{g\/mol}22.99+1.008+12.01+48.00=84.008g\/mol<\/p>\n\n\n\n<p>Thus, the molecular mass of NaHCO\u2083 is <strong>84.01 g\/mol<\/strong> (rounded to 2 decimal places).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b) Grams of Oxygen in 0.25 mol of Baking Soda<\/h3>\n\n\n\n<p>To find how many grams of oxygen are in 0.25 mol of NaHCO\u2083, we use the molecular mass of oxygen in NaHCO\u2083.<\/p>\n\n\n\n<p>From the formula, there are 3 oxygen atoms per molecule of NaHCO\u2083. The molar mass of oxygen in NaHCO\u2083 is: 3\u00d716.00\u2009g\/mol=48.00\u2009g\/mol3 \\times 16.00 \\, \\text{g\/mol} = 48.00 \\, \\text{g\/mol}3\u00d716.00g\/mol=48.00g\/mol<\/p>\n\n\n\n<p>Now, calculate how many grams of oxygen are in 0.25 mol of NaHCO\u2083: Grams&nbsp;of&nbsp;oxygen=0.25\u2009mol\u00d748.00\u2009g\/mol=12.00\u2009grams&nbsp;of&nbsp;oxygen\\text{Grams of oxygen} = 0.25 \\, \\text{mol} \\times 48.00 \\, \\text{g\/mol} = 12.00 \\, \\text{grams of oxygen}Grams&nbsp;of&nbsp;oxygen=0.25mol\u00d748.00g\/mol=12.00grams&nbsp;of&nbsp;oxygen<\/p>\n\n\n\n<p>Thus, there are <strong>12.00 grams<\/strong> of oxygen in 0.25 mol of baking soda.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c) Number of Hydrogen Atoms in 0.25 mol of Baking Soda<\/h3>\n\n\n\n<p>Next, we need to find how many hydrogen atoms are present in 0.25 mol of NaHCO\u2083.<\/p>\n\n\n\n<p>From the chemical formula, NaHCO\u2083 contains 1 hydrogen atom per molecule. The number of molecules in 1 mole is <strong>Avogadro\u2019s number<\/strong>, which is 6.022\u00d710236.022 \\times 10^{23}6.022\u00d71023 molecules per mole. Therefore, in 0.25 mol of NaHCO\u2083, there are: 0.25\u2009mol\u00d76.022\u00d71023\u2009atoms\/mol=1.5055\u00d71023\u2009hydrogen&nbsp;atoms0.25 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{atoms\/mol} = 1.5055 \\times 10^{23} \\, \\text{hydrogen atoms}0.25mol\u00d76.022\u00d71023atoms\/mol=1.5055\u00d71023hydrogen&nbsp;atoms<\/p>\n\n\n\n<p>Thus, there are <strong>1.5055 \u00d7 10\u00b2\u00b3 hydrogen atoms<\/strong> in 0.25 mol of baking soda.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a) Molecular mass of NaHCO\u2083:<\/strong> 84.01 g\/mol<\/li>\n\n\n\n<li><strong>b) Grams of oxygen in 0.25 mol of NaHCO\u2083:<\/strong> 12.00 g<\/li>\n\n\n\n<li><strong>c) Number of hydrogen atoms in 0.25 mol of NaHCO\u2083:<\/strong> 1.5055 \u00d7 10\u00b2\u00b3 atoms<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-189.jpeg\" alt=\"\" class=\"wp-image-251909\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the molecular mass of baking soda, NaHCO3? (specify units) b) how many grams of oxygen are in 0.25 mol of baking soda? c) how many hydrogen atoms are present in 0.25 mol of baking soda? Please write out all the steps The Correct Answer and Explanation is: Let&#8217;s break it down step by [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-251908","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251908","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=251908"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251908\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=251908"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=251908"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=251908"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}