{"id":268792,"date":"2025-07-24T08:20:56","date_gmt":"2025-07-24T08:20:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=268792"},"modified":"2025-07-24T08:21:01","modified_gmt":"2025-07-24T08:21:01","slug":"calculate-the-average-molecular-weight-of-air-from-its-approximate-molar-composition-of-79-n2-21-ao2-and-2-from-its-approximate-composition-by-mass-of-76-7-n2-23-3-o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/24\/calculate-the-average-molecular-weight-of-air-from-its-approximate-molar-composition-of-79-n2-21-ao2-and-2-from-its-approximate-composition-by-mass-of-76-7-n2-23-3-o2\/","title":{"rendered":"Calculate the average molecular weight of air from its approximate molar composition of 79% N2, 21% aO2 and 2 from its approximate composition by mass of 76.7% N2 23.3% O2"},"content":{"rendered":"\n

Calculate the average molecular weight of air from its approximate molar composition of 79% N2, 21% aO2 and 2 from its approximate composition by mass of 76.7% N2 23.3% O2<\/p>\n\n\n\n

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

To calculate the average molecular weight of air, we will use the concept of a weighted average, which takes into account the proportions of nitrogen (N2) and oxygen (O2) in air.<\/p>\n\n\n\n

We will approach this in two steps based on two different sets of compositions:<\/p>\n\n\n\n

1. By molar composition (79% N2, 21% O2)<\/strong><\/h3>\n\n\n\n
    \n
  • Molar mass of N2<\/strong>: 28.0 g\/mol<\/li>\n\n\n\n
  • Molar mass of O2<\/strong>: 32.0 g\/mol<\/li>\n<\/ul>\n\n\n\n

    To find the average molecular weight of air based on the molar composition, we multiply each gas’s molar mass by its proportion in the air and then add the results:Average molecular weight=(0.79\u00d728.0)+(0.21\u00d732.0)\\text{Average molecular weight} = (0.79 \\times 28.0) + (0.21 \\times 32.0)Average molecular weight=(0.79\u00d728.0)+(0.21\u00d732.0)Average molecular weight=22.12+6.72=28.84 g\/mol\\text{Average molecular weight} = 22.12 + 6.72 = 28.84 \\text{ g\/mol}Average molecular weight=22.12+6.72=28.84 g\/mol<\/p>\n\n\n\n

    So, the average molecular weight of air based on molar composition is 28.84 g\/mol<\/strong>.<\/p>\n\n\n\n

    2. By mass composition (76.7% N2, 23.3% O2)<\/strong><\/h3>\n\n\n\n
      \n
    • Molar mass of N2<\/strong>: 28.0 g\/mol<\/li>\n\n\n\n
    • Molar mass of O2<\/strong>: 32.0 g\/mol<\/li>\n<\/ul>\n\n\n\n

      We need to convert the mass percentages into mole fractions before we can calculate the molecular weight. First, we assume we have 100 grams of air, so:<\/p>\n\n\n\n

        \n
      • Mass of N2 = 76.7 grams<\/li>\n\n\n\n
      • Mass of O2 = 23.3 grams<\/li>\n<\/ul>\n\n\n\n

        Now, calculate the number of moles of each component:<\/p>\n\n\n\n

          \n
        • Moles of N2 = 76.7 g \/ 28.0 g\/mol = 2.74 mol<\/li>\n\n\n\n
        • Moles of O2 = 23.3 g \/ 32.0 g\/mol = 0.73 mol<\/li>\n<\/ul>\n\n\n\n

          Now, calculate the total moles of air:Total moles of air=2.74+0.73=3.47 mol\\text{Total moles of air} = 2.74 + 0.73 = 3.47 \\text{ mol}Total moles of air=2.74+0.73=3.47 mol<\/p>\n\n\n\n

          Next, calculate the mole fractions:<\/p>\n\n\n\n

            \n
          • Mole fraction of N2 = 2.74 mol \/ 3.47 mol = 0.79<\/li>\n\n\n\n
          • Mole fraction of O2 = 0.73 mol \/ 3.47 mol = 0.21<\/li>\n<\/ul>\n\n\n\n

            Finally, calculate the average molecular weight using the mole fractions and molar masses:Average molecular weight=(0.79\u00d728.0)+(0.21\u00d732.0)\\text{Average molecular weight} = (0.79 \\times 28.0) + (0.21 \\times 32.0)Average molecular weight=(0.79\u00d728.0)+(0.21\u00d732.0)Average molecular weight=22.12+6.72=28.84 g\/mol\\text{Average molecular weight} = 22.12 + 6.72 = 28.84 \\text{ g\/mol}Average molecular weight=22.12+6.72=28.84 g\/mol<\/p>\n\n\n\n

            In this case, the average molecular weight of air based on mass composition is also 28.84 g\/mol<\/strong>.<\/p>\n\n\n\n

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

            The average molecular weight of air is approximately 28.84 g\/mol<\/strong>, whether you use the molar composition (79% N2, 21% O2) or the mass composition (76.7% N2, 23.3% O2).<\/p>\n\n\n\n

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

            Calculate the average molecular weight of air from its approximate molar composition of 79% N2, 21% aO2 and 2 from its approximate composition by mass of 76.7% N2 23.3% O2 The Correct Answer and Explanation is: To calculate the average molecular weight of air, we will use the concept of a weighted average, which takes […]<\/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-268792","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268792","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=268792"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/268792\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=268792"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=268792"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=268792"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}