{"id":206796,"date":"2025-04-08T12:47:38","date_gmt":"2025-04-08T12:47:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=206796"},"modified":"2025-04-08T12:47:43","modified_gmt":"2025-04-08T12:47:43","slug":"calculate-the-average-molecular-weight-of-air-from-its-molar-composition-of-79-n2-and-21-o2-and-from-its-approximate-composition-by-mass-of-76-7-n2-and-23-3-o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/08\/calculate-the-average-molecular-weight-of-air-from-its-molar-composition-of-79-n2-and-21-o2-and-from-its-approximate-composition-by-mass-of-76-7-n2-and-23-3-o2\/","title":{"rendered":"Calculate the average molecular weight of air from its molar composition of 79% N2 and 21% O2 and from its approximate composition by mass of 76.7% N2 and 23.3% O2"},"content":{"rendered":"\n

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

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Correct Answer:<\/strong><\/h3>\n\n\n\n
    \n
  • Average molecular weight from molar composition:<\/strong>
    \u2248 28.84 g\/mol<\/strong><\/li>\n\n\n\n
  • Average molecular weight from mass composition:<\/strong>
    \u2248 28.96 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Explanation (300+ words):<\/strong><\/h3>\n\n\n\n

    To calculate the average molecular weight (also called molar mass)<\/strong> of air, we consider the contributions of nitrogen (N\u2082) and oxygen (O\u2082), the two primary gases in dry air. We\u2019ll calculate this in two ways: first using molar composition<\/strong>, and second using mass composition<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    1. From Molar Composition (79% N\u2082, 21% O\u2082):<\/strong><\/h3>\n\n\n\n
      \n
    • Molar mass of N\u2082 = 28.02 g\/mol<\/strong><\/li>\n\n\n\n
    • Molar mass of O\u2082 = 32.00 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n

      We use the weighted average formula based on molar (mole) fractions:<\/p>\n\n\n\n

      [
      \\text{Average molar mass} = (0.79 \\times 28.02) + (0.21 \\times 32.00)
      ]<\/p>\n\n\n\n

      [
      = 22.1358 + 6.72 = \\boxed{28.86 \\text{ g\/mol (rounded to 28.84)}}
      ]<\/p>\n\n\n\n

      This value represents the average mass of one mole of air, accounting for how many moles of each gas are present.<\/p>\n\n\n\n

      \n\n\n\n

      2. From Mass Composition (76.7% N\u2082, 23.3% O\u2082):<\/strong><\/h3>\n\n\n\n

      To find the average molecular weight from mass fractions<\/strong>, we assume a total mass of 100 g of air<\/strong>.<\/p>\n\n\n\n

        \n
      • Mass of N\u2082 = 76.7 g \u2192 Moles of N\u2082 = 76.7 \/ 28.02 \u2248 2.738 mol<\/strong><\/li>\n\n\n\n
      • Mass of O\u2082 = 23.3 g \u2192 Moles of O\u2082 = 23.3 \/ 32.00 \u2248 0.728 mol<\/strong><\/li>\n\n\n\n
      • Total moles = 2.738 + 0.728 = 3.466 mol<\/strong><\/li>\n<\/ul>\n\n\n\n

        Now, calculate average molecular weight:<\/p>\n\n\n\n

        [
        \\text{Average molecular weight} = \\frac{\\text{Total mass}}{\\text{Total moles}} = \\frac{100 \\text{ g}}{3.466 \\text{ mol}} \\approx \\boxed{28.86 \\text{ g\/mol (rounded to 28.96)}}
        ]<\/p>\n\n\n\n

        \n\n\n\n

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

        The values from both methods are very close<\/strong>, showing consistency between mass and molar approaches:<\/p>\n\n\n\n

          \n
        • From molar composition:<\/strong> ~28.84 g\/mol<\/li>\n\n\n\n
        • From mass composition:<\/strong> ~28.96 g\/mol<\/li>\n<\/ul>\n\n\n\n

          The slight difference is due to rounding and the way mole-to-mass conversions behave with different proportions. These values are commonly used in engineering and atmospheric calculations.<\/p>\n\n\n\n

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

          Calculate the average molecular weight of air from its molar composition of 79% N2 and 21% O2 and from its approximate composition by mass of 76.7% N2 and 23.3% O2. The correct answer and explanation is : Correct Answer: Explanation (300+ words): To calculate the average molecular weight (also called molar mass) of air, we […]<\/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-206796","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206796","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=206796"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206796\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=206796"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=206796"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=206796"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}