{"id":244764,"date":"2025-07-05T10:51:23","date_gmt":"2025-07-05T10:51:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244764"},"modified":"2025-07-05T10:51:25","modified_gmt":"2025-07-05T10:51:25","slug":"the-carbohydrate-fructose-found-in-honey-and-fruits-has-an-empirical-formula-ofch2o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/the-carbohydrate-fructose-found-in-honey-and-fruits-has-an-empirical-formula-ofch2o\/","title":{"rendered":"The carbohydrate fructose found in honey and fruits has an empirical formula ofCH2O"},"content":{"rendered":"\n

The carbohydrate fructose found in honey and fruits has an empirical formula ofCH2O
. If the experimental molar mass of fructose is 180g
, what is its molecular formula?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The empirical formula of fructose is CH2OCH_2OCH2\u200bO, which means that the ratio of carbon, hydrogen, and oxygen atoms in the simplest form is 1:2:1. The next step is to determine the molecular formula, which tells us how many of each type of atom are in a single molecule of fructose.<\/p>\n\n\n\n

Step-by-Step Solution:<\/h3>\n\n\n\n
    \n
  1. Calculate the molar mass of the empirical formula<\/strong>:
    The empirical formula for fructose is CH2OCH_2OCH2\u200bO, so the molar mass of one molecule of CH2OCH_2OCH2\u200bO is:
    • Carbon (C) has an atomic mass of 12.01 g\/mol.<\/li>
    • Hydrogen (H) has an atomic mass of 1.008 g\/mol.<\/li>
    • Oxygen (O) has an atomic mass of 16.00 g\/mol.<\/li><\/ul>Therefore, the molar mass of CH2OCH_2OCH2\u200bO is: 12.01+(2\u00d71.008)+16.00=30.03\u2009g\/mol12.01 + (2 \\times 1.008) + 16.00 = 30.03 \\, \\text{g\/mol}12.01+(2\u00d71.008)+16.00=30.03g\/mol<\/li>\n\n\n\n
    • Divide the experimental molar mass by the empirical formula mass<\/strong>:
      The experimental molar mass of fructose is given as 180 g\/mol. To find how many times the empirical formula fits into the molecular formula, divide the molar mass of fructose by the empirical formula’s molar mass: 18030.03\u22486\\frac{180}{30.03} \\approx 630.03180\u200b\u22486<\/li>\n\n\n\n
    • Determine the molecular formula<\/strong>:
      Since the empirical formula mass is 30.03 g\/mol, and the ratio is 6, we multiply the empirical formula by 6 to find the molecular formula: (CH2O)6=C6H12O6(CH_2O)_6 = C_6H_{12}O_6(CH2\u200bO)6\u200b=C6\u200bH12\u200bO6\u200b So, the molecular formula of fructose is C6H12O6C_6H_{12}O_6C6\u200bH12\u200bO6\u200b<\/strong>.<\/li>\n<\/ol>\n\n\n\n

      Conclusion:<\/h3>\n\n\n\n

      The molecular formula of fructose is C6H12O6C_6H_{12}O_6C6\u200bH12\u200bO6\u200b. This shows that fructose is composed of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms in each molecule, and it is commonly referred to as a simple sugar or monosaccharide. The empirical formula CH2OCH_2OCH2\u200bO represents the simplest ratio of the elements, while the molecular formula C6H12O6C_6H_{12}O_6C6\u200bH12\u200bO6\u200b describes the actual number of atoms in a molecule of fructose.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      The carbohydrate fructose found in honey and fruits has an empirical formula ofCH2O. If the experimental molar mass of fructose is 180g, what is its molecular formula? The Correct Answer and Explanation is: The empirical formula of fructose is CH2OCH_2OCH2\u200bO, which means that the ratio of carbon, hydrogen, and oxygen atoms in the simplest form […]<\/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-244764","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244764","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=244764"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244764\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244764"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244764"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}