{"id":253633,"date":"2025-07-12T06:23:15","date_gmt":"2025-07-12T06:23:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=253633"},"modified":"2025-07-12T06:23:18","modified_gmt":"2025-07-12T06:23:18","slug":"determine-the-molecular-formula-of-a-compound-that-has-a-molar-mass-of-122-11-g-mol-and-an-empirical-formula-of-c2h5o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/12\/determine-the-molecular-formula-of-a-compound-that-has-a-molar-mass-of-122-11-g-mol-and-an-empirical-formula-of-c2h5o2\/","title":{"rendered":". Determine the molecular formula of a compound that has a molar mass of 122.11 g\/mol and an empirical formula of C2H5O2."},"content":{"rendered":"\n

122.11 g\/mol and an empirical formula of C2H5O2. Determine the molecular formula of a compound that has a molar mass of 122.11 g\/mol and an empirical formula of C2H5O2. C2H5O2 C8H20O8 C6H15O6 C4H10O4 C3H7O3<\/p>\n\n\n\n

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

To determine the molecular formula of a compound, we need to use the empirical formula and the molar mass of the compound.<\/p>\n\n\n\n

    \n
  1. Step 1: Find the molar mass of the empirical formula.<\/strong><\/li>\n<\/ol>\n\n\n\n

    The empirical formula is C2H5O2C_2H_5O_2C2\u200bH5\u200bO2\u200b. We will calculate the molar mass of this empirical formula by adding the atomic masses of each element:<\/p>\n\n\n\n

      \n
    • Carbon (C): 12.01\u2009g\/mol12.01 \\, \\text{g\/mol}12.01g\/mol<\/li>\n\n\n\n
    • Hydrogen (H): 1.008\u2009g\/mol1.008 \\, \\text{g\/mol}1.008g\/mol<\/li>\n\n\n\n
    • Oxygen (O): 16.00\u2009g\/mol16.00 \\, \\text{g\/mol}16.00g\/mol<\/li>\n<\/ul>\n\n\n\n

      So, the molar mass of the empirical formula C2H5O2C_2H_5O_2C2\u200bH5\u200bO2\u200b is: Molar mass of empirical formula=(2\u00d712.01)+(5\u00d71.008)+(2\u00d716.00)\\text{Molar mass of empirical formula} = (2 \\times 12.01) + (5 \\times 1.008) + (2 \\times 16.00)Molar mass of empirical formula=(2\u00d712.01)+(5\u00d71.008)+(2\u00d716.00) =24.02+5.04+32.00=61.06\u2009g\/mol= 24.02 + 5.04 + 32.00 = 61.06 \\, \\text{g\/mol}=24.02+5.04+32.00=61.06g\/mol<\/p>\n\n\n\n

        \n
      1. Step 2: Compare the molar mass of the empirical formula to the given molar mass.<\/strong><\/li>\n<\/ol>\n\n\n\n

        The molar mass of the compound is given as 122.11\u2009g\/mol122.11 \\, \\text{g\/mol}122.11g\/mol.<\/p>\n\n\n\n

        Now, divide the molar mass of the compound by the molar mass of the empirical formula to find how many times the empirical formula is repeated in the molecular formula: Molar mass of the compoundMolar mass of empirical formula=122.1161.06\u22482\\frac{\\text{Molar mass of the compound}}{\\text{Molar mass of empirical formula}} = \\frac{122.11}{61.06} \\approx 2Molar mass of empirical formulaMolar mass of the compound\u200b=61.06122.11\u200b\u22482<\/p>\n\n\n\n

          \n
        1. Step 3: Multiply the empirical formula by this factor.<\/strong><\/li>\n<\/ol>\n\n\n\n

          Since the ratio is approximately 2, we multiply the empirical formula C2H5O2C_2H_5O_2C2\u200bH5\u200bO2\u200b by 2: Molecular formula=(C2H5O2)\u00d72=C4H10O4\\text{Molecular formula} = (C_2H_5O_2) \\times 2 = C_4H_{10}O_4Molecular formula=(C2\u200bH5\u200bO2\u200b)\u00d72=C4\u200bH10\u200bO4\u200b<\/p>\n\n\n\n

            \n
          1. Step 4: Identify the correct molecular formula.<\/strong><\/li>\n<\/ol>\n\n\n\n

            From the options provided:<\/p>\n\n\n\n

              \n
            • C2H5O2C_2H_5O_2C2\u200bH5\u200bO2\u200b is the empirical formula, not the molecular formula.<\/li>\n\n\n\n
            • C8H20O8C_8H_{20}O_8C8\u200bH20\u200bO8\u200b, C6H15O6C_6H_{15}O_6C6\u200bH15\u200bO6\u200b, and C3H7O3C_3H_7O_3C3\u200bH7\u200bO3\u200b are not correct since they do not match the factor of 2.<\/li>\n\n\n\n
            • The correct molecular formula is C4H10O4C_4H_{10}O_4C4\u200bH10\u200bO4\u200b.<\/li>\n<\/ul>\n\n\n\n

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

              The molecular formula of the compound is C4H10O4<\/strong>.<\/p>\n\n\n\n

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

              122.11 g\/mol and an empirical formula of C2H5O2. Determine the molecular formula of a compound that has a molar mass of 122.11 g\/mol and an empirical formula of C2H5O2. C2H5O2 C8H20O8 C6H15O6 C4H10O4 C3H7O3 The Correct Answer and Explanation is: To determine the molecular formula of a compound, we need to use the empirical formula […]<\/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-253633","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253633","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=253633"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253633\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253633"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253633"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253633"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}