Citric acid, the compound responsible for the sour taste of lemons, has an elemental composition of 37.51%\u00c3\u201a\u00c2 C, 4.20%\u00c3\u201a\u00c2 H, and 58.29%\u00c3\u201a\u00c2 O\u00c3\u201a\u00c2 Calculate the empirical formula of citric acid. Express your answer as a chemical formula. \u00c3\u201a\u00c2<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the empirical formula of citric acid, we need to convert the given percentages of carbon (C), hydrogen (H), and oxygen (O) into moles, then find the simplest whole-number ratio of these elements. Here’s a step-by-step breakdown of the process:<\/p>\n\n\n\n Since the percentages are given as mass percentages, assume you have a 100 g sample of citric acid. This simplifies the calculations because the mass of each element will be the same as its percentage in grams.<\/p>\n\n\n\n To find the number of moles of each element, use the molar mass of each element.<\/p>\n\n\n\n Now, divide the number of moles of each element by the smallest number of moles obtained in the previous step (in this case, 3.12 mol of carbon).<\/p>\n\n\n\n These ratios suggest that the numbers of atoms of each element are approximately 1:1.34:1.17. To obtain whole numbers, multiply all of the ratios by 3 to eliminate the decimals:<\/p>\n\n\n\n So, the simplest whole-number ratio is 3:4:3.<\/p>\n\n\n\n The empirical formula, based on the ratio of 3:4:3 for C, H, and O, is:<\/p>\n\n\n\n [ Thus, the empirical formula of citric acid is C\u2086H\u2088O\u2086<\/strong>.<\/p>\n\n\n\n Citric acid is a compound that is naturally found in citrus fruits, and its sour taste is due to the presence of carboxyl groups (-COOH). By calculating the empirical formula, we determine the simplest ratio of atoms in the molecule, which helps chemists understand the basic composition of the compound without knowing its molecular structure. The empirical formula is important because it serves as a starting point for further investigations into the molecular formula and structure of the compound.<\/p>\n","protected":false},"excerpt":{"rendered":" Citric acid, the compound responsible for the sour taste of lemons, has an elemental composition of 37.51%\u00c3\u201a\u00c2 C, 4.20%\u00c3\u201a\u00c2 H, and 58.29%\u00c3\u201a\u00c2 O\u00c3\u201a\u00c2 Calculate the empirical formula of citric acid. Express your answer as a chemical formula. \u00c3\u201a\u00c2 The Correct Answer and Explanation is : To determine the empirical formula of citric acid, we need […]<\/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-189215","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189215","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=189215"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189215\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189215"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189215"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Assume a 100 g sample<\/h3>\n\n\n\n
\n
Step 2: Convert the masses of each element to moles<\/h3>\n\n\n\n
\n
[
\\frac{37.51 \\, \\text{g}}{12.01 \\, \\text{g\/mol}} = 3.12 \\, \\text{mol C}
]<\/li>\n\n\n\n
[
\\frac{4.20 \\, \\text{g}}{1.008 \\, \\text{g\/mol}} = 4.17 \\, \\text{mol H}
]<\/li>\n\n\n\n
[
\\frac{58.29 \\, \\text{g}}{16.00 \\, \\text{g\/mol}} = 3.64 \\, \\text{mol O}
]<\/li>\n<\/ul>\n\n\n\nStep 3: Determine the simplest ratio<\/h3>\n\n\n\n
\n
\n
Step 4: Write the empirical formula<\/h3>\n\n\n\n
\\text{C}_6\\text{H}_8\\text{O}_6
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n