{"id":218082,"date":"2025-05-23T07:37:46","date_gmt":"2025-05-23T07:37:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218082"},"modified":"2025-05-23T07:37:49","modified_gmt":"2025-05-23T07:37:49","slug":"calculate-the-average-atomic-mass-of-element","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/calculate-the-average-atomic-mass-of-element\/","title":{"rendered":"Calculate the average atomic mass of element"},"content":{"rendered":"\n

Calculate the average atomic mass of element. What is the identity of the element?
The atomic masses of and are 6.0151 u and 7.0160 u, respectively. Calculate the natural percentage abundances of these two isotopes. The average atomic mass of Li is 6.941 u.
How many atoms of Au are in an ounce of gold? (Use the atomic mass unit (u) approach<\/p>\n\n\n\n

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

Let\u2019s solve these problems step-by-step using atomic mass unit (u) concepts and the definition of average atomic mass.<\/p>\n\n\n\n

\n\n\n\n

Part 1: Identity and Average Atomic Mass of Lithium<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Mass of isotope 1 (\u2076Li): 6.0151 u<\/li>\n\n\n\n
  • Mass of isotope 2 (\u2077Li): 7.0160 u<\/li>\n\n\n\n
  • Average atomic mass of Li: 6.941 u<\/li>\n<\/ul>\n\n\n\n

    Let x<\/strong> be the fractional abundance of \u2076Li, so the fractional abundance of \u2077Li is (1 \u2212 x).<\/p>\n\n\n\n

    Using the formula for average atomic mass:<\/p>\n\n\n\n

    $$
    6.941 = (6.0151)(x) + (7.0160)(1 – x)
    $$<\/p>\n\n\n\n

    Expanding and simplifying:<\/p>\n\n\n\n

    $$
    6.941 = 6.0151x + 7.0160 – 7.0160x
    $$<\/p>\n\n\n\n

    $$
    6.941 = 7.0160 – (7.0160 – 6.0151)x
    $$<\/p>\n\n\n\n

    $$
    6.941 = 7.0160 – 1.0009x
    $$<\/p>\n\n\n\n

    $$
    1.0009x = 7.0160 – 6.941 = 0.0750
    $$<\/p>\n\n\n\n

    $$
    x = \\frac{0.0750}{1.0009} \u2248 0.0749
    $$<\/p>\n\n\n\n

    So, \u2076Li is about 7.49%<\/strong>, and \u2077Li is about 92.51%<\/strong> abundant.<\/p>\n\n\n\n

    \n\n\n\n

    Part 2: Identity of the Element<\/strong><\/h3>\n\n\n\n

    The element described is Lithium (Li)<\/strong>, since its average atomic mass is 6.941 u<\/strong>, which is listed on the periodic table as the atomic mass of lithium. This confirms the isotope data pertains to lithium.<\/p>\n\n\n\n

    \n\n\n\n

    Part 3: Number of Atoms of Gold (Au) in 1 Ounce<\/strong><\/h3>\n\n\n\n

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

      \n
    • Mass of 1 ounce = 28.3495 grams<\/li>\n\n\n\n
    • Atomic mass of Au (gold) = 196.97 u<\/li>\n\n\n\n
    • 1 mole of atoms = 6.022 \u00d7 10\u00b2\u00b3 atoms<\/li>\n\n\n\n
    • 1 mole of Au weighs 196.97 grams<\/li>\n<\/ul>\n\n\n\n

      Now, calculate the number of moles in 1 ounce of gold:<\/p>\n\n\n\n

      $$
      \\text{Moles of Au} = \\frac{28.3495 \\text{ g}}{196.97 \\text{ g\/mol}} \u2248 0.1438 \\text{ mol}
      $$<\/p>\n\n\n\n

      Now, multiply by Avogadro’s number to find the number of atoms:<\/p>\n\n\n\n

      $$
      \\text{Atoms of Au} = 0.1438 \\text{ mol} \u00d7 6.022 \u00d7 10^{23} \\text{ atoms\/mol} \u2248 8.66 \u00d7 10^{22} \\text{ atoms}
      <\/p>\n\n\n\n

      <\/h3>\n\n\n\n

      The average atomic mass of an element reflects the weighted average of all naturally occurring isotopes based on their relative abundances. In the case of lithium (Li), it has two stable isotopes: lithium-6 and lithium-7. Their respective atomic masses are 6.0151 u and 7.0160 u. The weighted average formula allows us to solve for the natural abundances using the known average atomic mass of 6.941 u. Solving the equation yields about 7.49% for \u2076Li and 92.51% for \u2077Li, showing \u2077Li is the dominant isotope in nature.<\/p>\n\n\n\n

      Next, we explored the number of atoms in an ounce of gold (Au). One mole of gold has a mass of 196.97 grams and contains Avogadro’s number of atoms, 6.022 \u00d7 10\u00b2\u00b3. By converting ounces to grams (1 oz = 28.3495 g) and using dimensional analysis, we calculated the number of moles of gold in an ounce to be approximately 0.1438. Multiplying this by Avogadro\u2019s number gives about 8.66 \u00d7 10\u00b2\u00b2 atoms in one ounce of gold.<\/p>\n\n\n\n

      These calculations illustrate how atomic mass units (u), isotopic abundances, and mole-based conversions interconnect. They are essential in chemistry for understanding atomic structure, isotopic distribution, and real-world applications, such as determining the atomic quantity in a given sample.<\/p>\n\n\n\n

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

      Calculate the average atomic mass of element. What is the identity of the element?The atomic masses of and are 6.0151 u and 7.0160 u, respectively. Calculate the natural percentage abundances of these two isotopes. The average atomic mass of Li is 6.941 u.How many atoms of Au are in an ounce of gold? (Use the […]<\/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-218082","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218082","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=218082"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218082\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218082"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218082"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218082"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}