{"id":224362,"date":"2025-06-03T04:41:48","date_gmt":"2025-06-03T04:41:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224362"},"modified":"2025-06-03T04:41:50","modified_gmt":"2025-06-03T04:41:50","slug":"use-the-ar-values-below-to-calculate","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/03\/use-the-ar-values-below-to-calculate\/","title":{"rendered":"Use the Ar values below to calculate"},"content":{"rendered":"\n<p>Use the Ar values below to calculate: (Ar values: Fe = 55.80, B = 10.80, O = 16.00) 1-the relative molecular mass of iron(III) borate, Fe(BO2 )3 . 2- the number of moles in 92.1g of iron(III) borate<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s solve each part step by step using the given <strong>Ar values<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Fe = 55.80<\/li>\n\n\n\n<li>B = 10.80<\/li>\n\n\n\n<li>O = 16.00<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Relative Molecular Mass of Iron(III) Borate, Fe(BO\u2082)\u2083<\/strong><\/h3>\n\n\n\n<p>Iron(III) borate has the chemical formula <strong>Fe(BO\u2082)\u2083<\/strong>.<\/p>\n\n\n\n<p>We break this compound down:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 Fe atom<\/li>\n\n\n\n<li>3 BO\u2082 groups\n<ul class=\"wp-block-list\">\n<li>Each BO\u2082 group has 1 B and 2 O atoms<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>So, in total:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 Fe<\/li>\n\n\n\n<li>3 \u00d7 1 = 3 B<\/li>\n\n\n\n<li>3 \u00d7 2 = 6 O<\/li>\n<\/ul>\n\n\n\n<p>Now multiply the Ar (atomic relative mass) by the number of atoms:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Fe: 1 \u00d7 55.80 = <strong>55.80<\/strong><\/li>\n\n\n\n<li>B: 3 \u00d7 10.80 = <strong>32.40<\/strong><\/li>\n\n\n\n<li>O: 6 \u00d7 16.00 = <strong>96.00<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Add them all together:<br><strong>Relative molecular mass = 55.80 + 32.40 + 96.00 = 184.20 g\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Number of Moles in 92.1 g of Iron(III) Borate<\/strong><\/h3>\n\n\n\n<p>Use the formula: Moles=MassRelative&nbsp;Molecular&nbsp;Mass\\text{Moles} = \\frac{\\text{Mass}}{\\text{Relative Molecular Mass}}<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass = 92.1 g<\/li>\n\n\n\n<li>Relative Molecular Mass = 184.20 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Moles=92.1184.20\u22480.500&nbsp;mol\\text{Moles} = \\frac{92.1}{184.20} \u2248 0.500 \\text{ mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>To find the <strong>relative molecular mass<\/strong> of a compound, we sum up the relative atomic masses (Ar) of each element, considering how many atoms of each are present. For <strong>iron(III) borate<\/strong>, the chemical formula is <strong>Fe(BO\u2082)\u2083<\/strong>. This means the compound consists of one iron (Fe) atom and three borate units (BO\u2082). Each BO\u2082 group contains one boron (B) and two oxygen (O) atoms.<\/p>\n\n\n\n<p>By multiplying the Ar values with the number of respective atoms\u2014Fe (55.80), B (10.80), and O (16.00)\u2014we find the total relative molecular mass is <strong>184.20 g\/mol<\/strong>.<\/p>\n\n\n\n<p>In part two, we apply a basic concept from chemistry: the <strong>mole<\/strong>. A mole is a standard unit for amount of substance, and one mole of a compound has a mass equal to its molecular mass in grams. To calculate how many moles are in a given mass (here, 92.1 g), we divide the mass by the relative molecular mass. Moles=Given&nbsp;MassMolar&nbsp;Mass=92.1184.20\u22480.500&nbsp;mol\\text{Moles} = \\frac{\\text{Given Mass}}{\\text{Molar Mass}} = \\frac{92.1}{184.20} \u2248 0.500 \\text{ mol}<\/p>\n\n\n\n<p>This tells us that <strong>92.1 g of Fe(BO\u2082)\u2083<\/strong> contains <strong>0.500 moles<\/strong> of the compound. These calculations are foundational in chemistry, especially when converting between mass and amount in reactions, determining reagent quantities, or calculating yield.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner6-24.jpeg\" alt=\"\" class=\"wp-image-224363\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Use the Ar values below to calculate: (Ar values: Fe = 55.80, B = 10.80, O = 16.00) 1-the relative molecular mass of iron(III) borate, Fe(BO2 )3 . 2- the number of moles in 92.1g of iron(III) borate The Correct Answer and Explanation is: Let&#8217;s solve each part step by step using the given Ar [&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-224362","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224362","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=224362"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224362\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224362"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224362"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}