{"id":207265,"date":"2025-04-21T07:41:04","date_gmt":"2025-04-21T07:41:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=207265"},"modified":"2025-04-21T07:41:06","modified_gmt":"2025-04-21T07:41:06","slug":"calculate-the-molar-mass-of-propane","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/21\/calculate-the-molar-mass-of-propane\/","title":{"rendered":"Calculate the molar mass of propane"},"content":{"rendered":"\n<p>Calculate the molar mass of propane, C3H8.<br>Select one:<\/p>\n\n\n\n<p>44.1 g\/mol<br>36.0 g\/mol<br>24.0 g\/mol<br>60.1 g\/mol<br>26.0 g\/mol<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>The <strong>correct answer is: 44.1 g\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>To calculate the <strong>molar mass<\/strong> of a compound, we need to sum the atomic masses of all atoms in the molecule. Let&#8217;s break it down for <strong>propane (C\u2083H\u2088)<\/strong>:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. Identify the Elements in Propane (C\u2083H\u2088):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C)<\/li>\n\n\n\n<li>Hydrogen (H)<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>2. Determine the Number of Each Atom:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 Carbon atoms<\/li>\n\n\n\n<li>8 Hydrogen atoms<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>3. Use the Periodic Table for Atomic Masses:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Atomic mass of <strong>Carbon (C)<\/strong> = <strong>12.01 g\/mol<\/strong><\/li>\n\n\n\n<li>Atomic mass of <strong>Hydrogen (H)<\/strong> = <strong>1.008 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>4. Calculate Total Mass:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: 3 \u00d7 12.01 = <strong>36.03 g\/mol<\/strong><\/li>\n\n\n\n<li>Hydrogen: 8 \u00d7 1.008 = <strong>8.064 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>5. Add Them Together:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total molar mass = 36.03 + 8.064 = <strong>44.094 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Rounded to <strong>one decimal place<\/strong>, the molar mass of propane is:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>\u2705 44.1 g\/mol<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Why This Matters:<\/strong><\/h3>\n\n\n\n<p>Molar mass is a critical concept in <strong>chemistry<\/strong> because it allows scientists and students to convert between <strong>grams and moles<\/strong>, which are essential for calculations in stoichiometry, gas laws, and reaction yields.<\/p>\n\n\n\n<p>For example, if you were to combust propane for energy, knowing the molar mass would help you determine how much CO\u2082 is produced per mole or per gram of propane used.<\/p>\n\n\n\n<p>Understanding how to calculate molar mass also reinforces a basic skill in interpreting chemical formulas and using the periodic table \u2014 fundamental for all chemistry students and professionals.<\/p>\n\n\n\n<p>In short, the correct molar mass of propane is <strong>44.1 g\/mol<\/strong>, and it serves as the bridge between microscopic particles and macroscopic measurements.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/04\/image-199.png\" alt=\"\" class=\"wp-image-207266\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the molar mass of propane, C3H8.Select one: 44.1 g\/mol36.0 g\/mol24.0 g\/mol60.1 g\/mol26.0 g\/mol The correct answer and explanation is : The correct answer is: 44.1 g\/mol Explanation: To calculate the molar mass of a compound, we need to sum the atomic masses of all atoms in the molecule. Let&#8217;s break it down for propane [&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-207265","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207265","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=207265"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207265\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207265"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207265"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207265"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}