{"id":185236,"date":"2025-01-22T09:27:21","date_gmt":"2025-01-22T09:27:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185236"},"modified":"2025-01-22T09:27:23","modified_gmt":"2025-01-22T09:27:23","slug":"butyric-acid-found-in-rancid-butter-has-a-molar-mass-of-88-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/butyric-acid-found-in-rancid-butter-has-a-molar-mass-of-88-g-mol\/","title":{"rendered":"Butyric acid, found in rancid butter, has a molar mass of 88 g\/mol"},"content":{"rendered":"\n<p>19) Butyric acid, found in rancid butter, has a molar mass of 88 g\/mol. If butyric acid is 54.5% C, 9.09% H and 36.4% 0, what is the molecular formula? None 19) Butyric acid, found in rancid butter, has a molar mass of 88 g\/mol. 25843930 If butyric acid is 54.5% C, 9.09% H and 36.4% 0, what is the molecular formula? \u00d0\u203a\u00d0\u00be\u00d0\u00b8\u00d3\u2122\u00d0\u00b1\u00d1\u0192 \u00d3\u2122\u00d2\u2122\u00d0\u00b5ls] je d A CARINA E 43.72 ml of SME CTCT Ong 7 to 500<\/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 butyric acid based on the given data, follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the empirical formula<\/h3>\n\n\n\n<p>The given percentages represent the mass fractions of the elements:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 54.5%<\/li>\n\n\n\n<li>Hydrogen (H): 9.09%<\/li>\n\n\n\n<li>Oxygen (O): 36.4%<\/li>\n<\/ul>\n\n\n\n<p>Assume 100 g of butyric acid for simplicity. This gives:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of C = ( 54.5 \\, \\text{g} )<\/li>\n\n\n\n<li>Mass of H = ( 9.09 \\, \\text{g} )<\/li>\n\n\n\n<li>Mass of O = ( 36.4 \\, \\text{g} )<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the number of moles of each element:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of C = ( \\frac{54.5}{12.01} = 4.54 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of H = ( \\frac{9.09}{1.008} = 9.02 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of O = ( \\frac{36.4}{16.00} = 2.28 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the simplest mole ratio<\/h3>\n\n\n\n<p>Divide each mole value by the smallest number of moles (2.28):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C: ( \\frac{4.54}{2.28} \\approx 2 )<\/li>\n\n\n\n<li>H: ( \\frac{9.02}{2.28} \\approx 4 )<\/li>\n\n\n\n<li>O: ( \\frac{2.28}{2.28} = 1 )<\/li>\n<\/ul>\n\n\n\n<p>The empirical formula is <strong>C\u2082H\u2084O<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Find the molecular formula<\/h3>\n\n\n\n<p>The molar mass of the empirical formula ( \\text{C}_2\\text{H}_4\\text{O} ) is:<br>[ (2 \\times 12.01) + (4 \\times 1.008) + (1 \\times 16.00) = 44.05 \\, \\text{g\/mol}. ]<\/p>\n\n\n\n<p>The molar mass of butyric acid is given as ( 88 \\, \\text{g\/mol} ). The ratio is:<br>[ \\frac{88}{44.05} \\approx 2. ]<\/p>\n\n\n\n<p>Multiply the subscripts in the empirical formula by 2:<br>[ \\text{Molecular formula} = \\text{C}_4\\text{H}_8\\text{O}_2. ]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The molecular formula of butyric acid is <strong>C\u2084H\u2088O\u2082<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The calculation involves determining the empirical formula from the elemental composition percentages and scaling it to match the molecular mass. The process includes converting mass percentages into moles, normalizing to find the simplest ratio, and scaling based on the molar mass ratio. Butyric acid\u2019s structure aligns with <strong>C\u2084H\u2088O\u2082<\/strong>, a typical carboxylic acid, consistent with its known properties.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>19) Butyric acid, found in rancid butter, has a molar mass of 88 g\/mol. If butyric acid is 54.5% C, 9.09% H and 36.4% 0, what is the molecular formula? None 19) Butyric acid, found in rancid butter, has a molar mass of 88 g\/mol. 25843930 If butyric acid is 54.5% C, 9.09% H and [&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-185236","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185236","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=185236"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185236\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185236"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185236"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185236"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}