{"id":185888,"date":"2025-01-23T16:46:25","date_gmt":"2025-01-23T16:46:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185888"},"modified":"2025-01-23T16:46:27","modified_gmt":"2025-01-23T16:46:27","slug":"menthol-molar-mass-156-30g-mol-contains-carbon-hydrogen-and-oxygen","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/23\/menthol-molar-mass-156-30g-mol-contains-carbon-hydrogen-and-oxygen\/","title":{"rendered":"Menthol (molar mass = 156.30g\/mol), contains carbon, hydrogen, and oxygen"},"content":{"rendered":"\n<p>Menthol (molar mass = 156.30g\/mol), contains carbon, hydrogen, and oxygen.\u00c3\u201a\u00c2 When 0.1595g of menthol is combusted, 0.449g of CO2 and 0.184g of H2O are formed.\u00c3\u201a\u00c2 Determine the molecular formula of menthol.<\/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>To determine the molecular formula of menthol (C\u2081\u2080H\u2082\u2080O) from combustion data, we analyze the masses of carbon dioxide (CO\u2082) and water (H\u2082O) produced during combustion.<\/p>\n\n\n\n<p><strong>Step 1: Calculate the moles of CO\u2082 and H\u2082O produced.<\/strong><\/p>\n\n\n\n<p>Given masses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of CO\u2082 = 0.449 g<\/li>\n\n\n\n<li>Mass of H\u2082O = 0.184 g<\/li>\n<\/ul>\n\n\n\n<p>Molar masses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of CO\u2082 = 44.01 g\/mol<\/li>\n\n\n\n<li>Molar mass of H\u2082O = 18.015 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Moles of CO\u2082 = 0.449 g \/ 44.01 g\/mol \u2248 0.0102 mol<br>Moles of H\u2082O = 0.184 g \/ 18.015 g\/mol \u2248 0.0102 mol<\/p>\n\n\n\n<p><strong>Step 2: Determine the moles of carbon and hydrogen in the sample.<\/strong><\/p>\n\n\n\n<p>Each mole of CO\u2082 contains 1 mole of carbon atoms, and each mole of H\u2082O contains 2 moles of hydrogen atoms.<\/p>\n\n\n\n<p>Moles of carbon = 0.0102 mol<br>Moles of hydrogen = 0.0102 mol \u00d7 2 = 0.0204 mol<\/p>\n\n\n\n<p><strong>Step 3: Calculate the masses of carbon and hydrogen.<\/strong><\/p>\n\n\n\n<p>Mass of carbon = 0.0102 mol \u00d7 12.011 g\/mol \u2248 0.122 g<br>Mass of hydrogen = 0.0204 mol \u00d7 1.008 g\/mol \u2248 0.0206 g<\/p>\n\n\n\n<p><strong>Step 4: Determine the mass of oxygen in the sample.<\/strong><\/p>\n\n\n\n<p>Mass of oxygen = Total mass of sample \u2013 (Mass of carbon + Mass of hydrogen)<br>Mass of oxygen = 0.1595 g \u2013 (0.122 g + 0.0206 g) = 0.0169 g<\/p>\n\n\n\n<p><strong>Step 5: Calculate the moles of oxygen.<\/strong><\/p>\n\n\n\n<p>Moles of oxygen = 0.0169 g \/ 16.00 g\/mol \u2248 0.00106 mol<\/p>\n\n\n\n<p><strong>Step 6: Determine the empirical formula.<\/strong><\/p>\n\n\n\n<p>The mole ratio of C:H:O is approximately 10:20:1.<br>Empirical formula: C\u2081\u2080H\u2082\u2080O<\/p>\n\n\n\n<p><strong>Step 7: Determine the molecular formula.<\/strong><\/p>\n\n\n\n<p>Given molar mass of menthol = 156.30 g\/mol<br>Empirical formula mass = (10 \u00d7 12.011) + (20 \u00d7 1.008) + (1 \u00d7 16.00) = 156.30 g\/mol<\/p>\n\n\n\n<p>Since the empirical formula mass equals the molar mass, the empirical formula is also the molecular formula.<br>Molecular formula: C\u2081\u2080H\u2082\u2080O<\/p>\n\n\n\n<p>Therefore, the molecular formula of menthol is C\u2081\u2080H\u2082\u2080O.<\/p>\n\n\n\n<p>This analysis demonstrates how combustion data can be used to determine the molecular formula of a compound by calculating the moles of carbon, hydrogen, and oxygen, and comparing the empirical formula mass to the known molar mass.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Menthol (molar mass = 156.30g\/mol), contains carbon, hydrogen, and oxygen.\u00c3\u201a\u00c2 When 0.1595g of menthol is combusted, 0.449g of CO2 and 0.184g of H2O are formed.\u00c3\u201a\u00c2 Determine the molecular formula of menthol. The Correct Answer and Explanation is : To determine the molecular formula of menthol (C\u2081\u2080H\u2082\u2080O) from combustion data, we analyze the masses of carbon [&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-185888","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185888","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=185888"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185888\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185888"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185888"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185888"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}