{"id":183514,"date":"2025-01-16T14:01:58","date_gmt":"2025-01-16T14:01:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183514"},"modified":"2025-01-16T14:02:00","modified_gmt":"2025-01-16T14:02:00","slug":"the-vanillin-molecule-is-the-primary-molecule-present-in-vanilla-extract","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/the-vanillin-molecule-is-the-primary-molecule-present-in-vanilla-extract\/","title":{"rendered":"The vanillin molecule is the primary molecule present in vanilla extract"},"content":{"rendered":"\n<p>1) The vanillin molecule is the primary molecule present in vanilla extract. The molecular mass of vanillin is 152.08 grams per mole and contains 63.18% carbon, 5.26% hydrogen, and 31.56% oxygen. What is the molecular formula of vanillin?2) A sample of fuel is found to contain 87.4% nitrogen and 12.6% hydrogen. If the molecular mass of the fuel is 32.05 grams\/mole, what is the molecular formula of the fuel?3) A compound is analyzed and found to have a mass of 289.9 grams\/mole and contain 49.67% carbon, 48.92% chlorine and 1.39% hydrogen. What is the molecular formula of the compound?4) A compound is analyzed and found to contain 35.66% carbon, 16.24% hydrogen and 45.10% nitrogen. What is the empirical formula of the compound?<\/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>Let&#8217;s break down each problem and calculate the molecular formulas step by step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1) Molecular formula of vanillin:<\/h3>\n\n\n\n<p>Vanillin&#8217;s molecular mass is given as 152.08 g\/mol, and it is made up of 63.18% carbon (C), 5.26% hydrogen (H), and 31.56% oxygen (O). We will use the percentages to find the empirical formula and then the molecular formula.<\/p>\n\n\n\n<p><strong>Step 1: Find the mass of each element in one mole of vanillin.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: ( 63.18\\% \\times 152.08 \\, \\text{g} = 95.98 \\, \\text{g of C} )<\/li>\n\n\n\n<li>Hydrogen: ( 5.26\\% \\times 152.08 \\, \\text{g} = 8.00 \\, \\text{g of H} )<\/li>\n\n\n\n<li>Oxygen: ( 31.56\\% \\times 152.08 \\, \\text{g} = 48.04 \\, \\text{g of O} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Convert grams to moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of carbon: ( \\frac{95.98 \\, \\text{g}}{12.01 \\, \\text{g\/mol}} = 7.99 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of hydrogen: ( \\frac{8.00 \\, \\text{g}}{1.008 \\, \\text{g\/mol}} = 7.94 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of oxygen: ( \\frac{48.04 \\, \\text{g}}{16.00 \\, \\text{g\/mol}} = 3.00 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Determine the ratio of moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: ( \\frac{7.99}{3.00} = 2.66 )<\/li>\n\n\n\n<li>Hydrogen: ( \\frac{7.94}{3.00} = 2.65 )<\/li>\n\n\n\n<li>Oxygen: ( \\frac{3.00}{3.00} = 1 )<\/li>\n<\/ul>\n\n\n\n<p>The ratio of C:H:O is approximately 2.66:2.65:1, which simplifies to 3:3:1. The empirical formula is ( \\text{C}_8 \\text{H}_8 \\text{O} ).<\/p>\n\n\n\n<p><strong>Step 4: Compare the empirical formula mass with the molecular mass.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Empirical formula mass = ( 8 \\times 12.01 + 8 \\times 1.008 + 16.00 = 152.08 \\, \\text{g\/mol} )<\/li>\n<\/ul>\n\n\n\n<p>Since the empirical formula mass is equal to the molecular mass, the molecular formula of vanillin is ( \\text{C}_8 \\text{H}_8 \\text{O} ).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2) Molecular formula of the fuel:<\/h3>\n\n\n\n<p>The fuel contains 87.4% nitrogen (N) and 12.6% hydrogen (H), and its molecular mass is 32.05 g\/mol.<\/p>\n\n\n\n<p><strong>Step 1: Find the mass of each element in one mole of the fuel.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen: ( 87.4\\% \\times 32.05 \\, \\text{g} = 27.99 \\, \\text{g of N} )<\/li>\n\n\n\n<li>Hydrogen: ( 12.6\\% \\times 32.05 \\, \\text{g} = 4.03 \\, \\text{g of H} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Convert grams to moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of nitrogen: ( \\frac{27.99 \\, \\text{g}}{14.01 \\, \\text{g\/mol}} = 2.00 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of hydrogen: ( \\frac{4.03 \\, \\text{g}}{1.008 \\, \\text{g\/mol}} = 4.00 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Determine the ratio of moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen: ( \\frac{2.00}{2.00} = 1 )<\/li>\n\n\n\n<li>Hydrogen: ( \\frac{4.00}{2.00} = 2 )<\/li>\n<\/ul>\n\n\n\n<p>The ratio of N:H is 1:2. Thus, the empirical formula is ( \\text{NH}_2 ).<\/p>\n\n\n\n<p><strong>Step 4: Compare the empirical formula mass with the molecular mass.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Empirical formula mass = ( 14.01 + 2 \\times 1.008 = 16.03 \\, \\text{g\/mol} )<\/li>\n<\/ul>\n\n\n\n<p>The ratio of the molecular mass to the empirical formula mass is ( \\frac{32.05}{16.03} = 2 ). Thus, the molecular formula is ( \\text{N}_2 \\text{H}_4 ).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3) Molecular formula of the compound:<\/h3>\n\n\n\n<p>The compound has a molecular mass of 289.9 g\/mol and contains 49.67% carbon, 48.92% chlorine, and 1.39% hydrogen.<\/p>\n\n\n\n<p><strong>Step 1: Find the mass of each element in one mole of the compound.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: ( 49.67\\% \\times 289.9 \\, \\text{g} = 144.92 \\, \\text{g of C} )<\/li>\n\n\n\n<li>Chlorine: ( 48.92\\% \\times 289.9 \\, \\text{g} = 141.89 \\, \\text{g of Cl} )<\/li>\n\n\n\n<li>Hydrogen: ( 1.39\\% \\times 289.9 \\, \\text{g} = 4.03 \\, \\text{g of H} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Convert grams to moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of carbon: ( \\frac{144.92 \\, \\text{g}}{12.01 \\, \\text{g\/mol}} = 12.08 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of chlorine: ( \\frac{141.89 \\, \\text{g}}{35.45 \\, \\text{g\/mol}} = 4.00 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of hydrogen: ( \\frac{4.03 \\, \\text{g}}{1.008 \\, \\text{g\/mol}} = 4.00 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Determine the ratio of moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: ( \\frac{12.08}{4.00} = 3.02 )<\/li>\n\n\n\n<li>Chlorine: ( \\frac{4.00}{4.00} = 1 )<\/li>\n\n\n\n<li>Hydrogen: ( \\frac{4.00}{4.00} = 1 )<\/li>\n<\/ul>\n\n\n\n<p>The ratio of C:Cl:H is approximately 3:1:1. The empirical formula is ( \\text{C}_3 \\text{H}_3 \\text{Cl} ).<\/p>\n\n\n\n<p><strong>Step 4: Compare the empirical formula mass with the molecular mass.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Empirical formula mass = ( 3 \\times 12.01 + 3 \\times 1.008 + 35.45 = 106.29 \\, \\text{g\/mol} )<\/li>\n<\/ul>\n\n\n\n<p>The ratio of the molecular mass to the empirical formula mass is ( \\frac{289.9}{106.29} \\approx 2.73 ). This suggests that the molecular formula is approximately ( \\text{C}_9 \\text{H}_9 \\text{Cl}_3 ).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">4) Empirical formula of the compound:<\/h3>\n\n\n\n<p>The compound contains 35.66% carbon, 16.24% hydrogen, and 45.10% nitrogen.<\/p>\n\n\n\n<p><strong>Step 1: Find the mass of each element in one mole of the compound.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: ( 35.66\\% \\times 100 \\, \\text{g} = 35.66 \\, \\text{g of C} )<\/li>\n\n\n\n<li>Hydrogen: ( 16.24\\% \\times 100 \\, \\text{g} = 16.24 \\, \\text{g of H} )<\/li>\n\n\n\n<li>Nitrogen: ( 45.10\\% \\times 100 \\, \\text{g} = 45.10 \\, \\text{g of N} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Convert grams to moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of carbon: ( \\frac{35.66 \\, \\text{g}}{12.01 \\, \\text{g\/mol}} = 2.97 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of hydrogen: ( \\frac{16.24 \\, \\text{g}}{1.008 \\, \\text{g\/mol}} = 16.12 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of nitrogen: ( \\frac{45.10 \\, \\text{g}}{14.01 \\, \\text{g\/mol}} = 3.22 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Determine the ratio of moles.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon: ( \\frac{2.97}{2.97} = 1 )<\/li>\n\n\n\n<li>Hydrogen: ( \\frac{16.12}{2.97} = 5.43 )<\/li>\n\n\n\n<li>Nitrogen: ( \\frac{3.22}{2.97} = 1.08 )<\/li>\n<\/ul>\n\n\n\n<p>The ratio of C:H:N is approximately 1:5:1, which rounds to an empirical formula of ( \\text{CH}_5 \\text{N} ). However, this seems unusual as ( \\text{H}_5 ) is rare, so further analysis of empirical formulas may be needed.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>1) The vanillin molecule is the primary molecule present in vanilla extract. The molecular mass of vanillin is 152.08 grams per mole and contains 63.18% carbon, 5.26% hydrogen, and 31.56% oxygen. What is the molecular formula of vanillin?2) A sample of fuel is found to contain 87.4% nitrogen and 12.6% hydrogen. If the molecular mass [&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-183514","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183514","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=183514"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183514\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}