{"id":234461,"date":"2025-06-14T06:39:04","date_gmt":"2025-06-14T06:39:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234461"},"modified":"2025-06-14T06:39:07","modified_gmt":"2025-06-14T06:39:07","slug":"the-natural-abundances-of-chlorines-two-major-isotopes-are-35cl-75-5-and-37cl-24-5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/14\/the-natural-abundances-of-chlorines-two-major-isotopes-are-35cl-75-5-and-37cl-24-5\/","title":{"rendered":"The natural abundances of chlorine’s two major isotopes are 35Cl (75.5%) and 37Cl (24.5%)"},"content":{"rendered":"\n

The natural abundances of chlorine’s two major isotopes are 35Cl (75.5%) and 37Cl (24.5%). If the mass spectrum of chlorobenzene contains a peak at m\/z 114 with an abundance of 5.0%, calculate the percent abundance of the peak at m\/z 12.<\/p>\n\n\n\n

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

To calculate the percent abundance of the peak at m\/z = 112<\/strong> (which you likely meant instead of “12”, as 12 is too low for chlorobenzene), we consider the mass spectrum of chlorobenzene (C\u2086H\u2085Cl)<\/strong>, which arises from combinations of chlorine isotopes:<\/p>\n\n\n\n

Step 1: Isotopes and Molar Mass<\/h3>\n\n\n\n
    \n
  • Chlorine has two main isotopes:\n
      \n
    • \u00b3\u2075Cl<\/strong> with a mass of ~34.97 amu and abundance 75.5% (0.755)<\/strong><\/li>\n\n\n\n
    • \u00b3\u2077Cl<\/strong> with a mass of ~36.97 amu and abundance 24.5% (0.245)<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Benzene (C\u2086H\u2085) has a mass of approximately 77 amu.<\/li>\n<\/ul>\n\n\n\n

      So:<\/p>\n\n\n\n

        \n
      • Chlorobenzene with \u00b3\u2075Cl<\/strong>: m\/z \u2248 77 + 35 = 112<\/strong><\/li>\n\n\n\n
      • Chlorobenzene with \u00b3\u2077Cl<\/strong>: m\/z \u2248 77 + 37 = 114<\/strong><\/li>\n<\/ul>\n\n\n\n

        Given:<\/p>\n\n\n\n

          \n
        • m\/z 114 peak<\/strong> (\u00b3\u2077Cl) has relative abundance = 5.0%<\/strong><\/li>\n\n\n\n
        • Need to find the abundance of m\/z 112<\/strong> (\u00b3\u2075Cl)<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Step 2: Use Isotopic Ratio<\/h3>\n\n\n\n

          The abundance ratio of the two peaks is directly proportional to the isotopic abundances of \u00b3\u2075Cl and \u00b3\u2077Cl:Ratio=Abundance of 37ClAbundance of 35Cl=0.2450.755=5.0%x\\text{Ratio} = \\frac{\\text{Abundance of }^{37}\\text{Cl}}{\\text{Abundance of }^{35}\\text{Cl}} = \\frac{0.245}{0.755} = \\frac{5.0\\%}{x}Ratio=Abundance of 35ClAbundance of 37Cl\u200b=0.7550.245\u200b=x5.0%\u200b<\/p>\n\n\n\n

          Solving:0.2450.755=5.0x\\frac{0.245}{0.755} = \\frac{5.0}{x}0.7550.245\u200b=x5.0\u200bx=5.0\u00d70.7550.245=3.7750.245\u224815.4%x = \\frac{5.0 \\times 0.755}{0.245} = \\frac{3.775}{0.245} \\approx 15.4\\%x=0.2455.0\u00d70.755\u200b=0.2453.775\u200b\u224815.4%<\/p>\n\n\n\n

          \n\n\n\n

          \u2705 Final Answer:<\/h3>\n\n\n\n

          The percent abundance of the peak at m\/z 112 is approximately 15.4%.<\/strong><\/p>\n\n\n\n

          \n\n\n\n

          \ud83d\udcd8 Textbook-Style Explanation<\/h3>\n\n\n\n

          The mass spectrum of chlorobenzene (C\u2086H\u2085Cl) shows distinct peaks due to the presence of chlorine isotopes. Chlorine naturally exists primarily as two isotopes: \u00b3\u2075Cl and \u00b3\u2077Cl. Their respective abundances are approximately 75.5% and 24.5%. In mass spectrometry, these isotopic variations cause the molecular ion peak of chlorobenzene to appear at different m\/z values depending on the isotope present.<\/p>\n\n\n\n

          The molecular mass of the C\u2086H\u2085 (phenyl) group is about 77 amu. When bonded with \u00b3\u2075Cl, the molecular ion has a mass of roughly 112 amu (77 + 35). When bonded with \u00b3\u2077Cl, the ion has a mass of 114 amu (77 + 37). Therefore, peaks in the mass spectrum are expected at m\/z = 112 and m\/z = 114, corresponding to \u00b3\u2075Cl and \u00b3\u2077Cl variants, respectively.<\/p>\n\n\n\n

          Since isotopic abundance affects peak intensity, the relative peak heights mirror the abundance ratio. The ratio of \u00b3\u2077Cl to \u00b3\u2075Cl is:24.575.5\u22480.3245\\frac{24.5}{75.5} \\approx 0.324575.524.5\u200b\u22480.3245<\/p>\n\n\n\n

          If the less abundant isotope peak at m\/z 114 (\u00b3\u2077Cl) has a reported abundance of 5.0%, we apply this ratio to determine the peak at m\/z 112:5.0%x=0.2450.755\u21d2x\u224815.4%\\frac{5.0\\%}{x} = \\frac{0.245}{0.755} \\Rightarrow x \\approx 15.4\\%x5.0%\u200b=0.7550.245\u200b\u21d2x\u224815.4%<\/p>\n\n\n\n

          This reflects the greater natural abundance of \u00b3\u2075Cl. Hence, the m\/z 112 peak is expected to be more intense, consistent with the isotope’s prevalence in nature.<\/p>\n\n\n\n

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

          The natural abundances of chlorine’s two major isotopes are 35Cl (75.5%) and 37Cl (24.5%). If the mass spectrum of chlorobenzene contains a peak at m\/z 114 with an abundance of 5.0%, calculate the percent abundance of the peak at m\/z 12. The Correct Answer and Explanation is: To calculate the percent abundance of the peak […]<\/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-234461","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234461","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=234461"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234461\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234461"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234461"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234461"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}