{"id":188757,"date":"2025-02-07T09:12:54","date_gmt":"2025-02-07T09:12:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188757"},"modified":"2025-02-07T09:12:56","modified_gmt":"2025-02-07T09:12:56","slug":"iron-has-a-reported-average-atomic-mass-of-55-85-amu","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/iron-has-a-reported-average-atomic-mass-of-55-85-amu\/","title":{"rendered":"Iron has a reported average atomic mass of 55.85 amu"},"content":{"rendered":"\n<p>Iron has a reported average atomic mass of 55.85 amu. If iron has two isotopes, one of which has 29 neutrons and an isotopic abundance of 57.50%, what is the mass of the other isotope?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To find the mass of the other isotope of iron, we can use the weighted average formula for atomic mass. The atomic mass of an element is a weighted average of the masses of its isotopes, where the abundance of each isotope is considered.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Understand the given data<\/h3>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The average atomic mass of iron (Fe) = 55.85 amu.<\/li>\n\n\n\n<li>One isotope has 29 neutrons, with an isotopic abundance of 57.50%.<\/li>\n\n\n\n<li>We need to determine the mass of the other isotope.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the number of protons and neutrons in the first isotope<\/h3>\n\n\n\n<p>The atomic number of iron (Fe) is 26, meaning it has 26 protons. The first isotope has 29 neutrons, so its mass number (total number of protons and neutrons) is:<\/p>\n\n\n\n<p>[<br>\\text{Mass number} = 26 + 29 = 55 \\, \\text{amu}.<br>]<\/p>\n\n\n\n<p>Thus, the first isotope has a mass of approximately 55 amu.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Use the weighted average formula for atomic mass<\/h3>\n\n\n\n<p>The atomic mass is a weighted average of the two isotopes:<\/p>\n\n\n\n<p>[<br>\\text{Average atomic mass} = (f_1 \\times m_1) + (f_2 \\times m_2),<br>]<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( f_1 ) and ( f_2 ) are the fractional abundances of the isotopes (expressed as decimals),<\/li>\n\n\n\n<li>( m_1 ) and ( m_2 ) are the masses of the isotopes.<\/li>\n<\/ul>\n\n\n\n<p>We know:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( f_1 = 57.50\\% = 0.575 ),<\/li>\n\n\n\n<li>( m_1 = 55 \\, \\text{amu} ),<\/li>\n\n\n\n<li>The average atomic mass = 55.85 amu.<\/li>\n<\/ul>\n\n\n\n<p>Since the total abundance is 100%, the abundance of the second isotope, ( f_2 ), is:<\/p>\n\n\n\n<p>[<br>f_2 = 100\\% &#8211; 57.50\\% = 42.50\\% = 0.425.<br>]<\/p>\n\n\n\n<p>Now, we can use the weighted average formula to solve for the mass of the second isotope, ( m_2 ):<\/p>\n\n\n\n<p>[<br>55.85 = (0.575 \\times 55) + (0.425 \\times m_2),<br>]<\/p>\n\n\n\n<p>[<br>55.85 = 31.625 + (0.425 \\times m_2),<br>]<\/p>\n\n\n\n<p>[<br>55.85 &#8211; 31.625 = 0.425 \\times m_2,<br>]<\/p>\n\n\n\n<p>[<br>24.225 = 0.425 \\times m_2,<br>]<\/p>\n\n\n\n<p>[<br>m_2 = \\frac{24.225}{0.425} = 57.<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Conclusion<\/h3>\n\n\n\n<p>The mass of the second isotope is approximately <strong>57 amu<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>Isotopes are atoms of the same element with the same number of protons but a different number of neutrons, which leads to a different mass. In this case, the average atomic mass of iron (55.85 amu) is a weighted average of the masses of two isotopes, one with a mass of 55 amu and the other with a mass of 57 amu. By using the known isotopic abundance and atomic mass, we can calculate the mass of the second isotope.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Iron has a reported average atomic mass of 55.85 amu. If iron has two isotopes, one of which has 29 neutrons and an isotopic abundance of 57.50%, what is the mass of the other isotope? The Correct Answer and Explanation is : To find the mass of the other isotope of iron, we can use [&hellip;]<\/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-188757","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188757","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=188757"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188757\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188757"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188757"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}