{"id":195597,"date":"2025-02-28T13:29:55","date_gmt":"2025-02-28T13:29:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195597"},"modified":"2025-02-28T13:29:58","modified_gmt":"2025-02-28T13:29:58","slug":"calculate-the-formula-weights-of-the-following-substances","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/calculate-the-formula-weights-of-the-following-substances\/","title":{"rendered":"Calculate the formula weights of the following substances"},"content":{"rendered":"\n<ol class=\"wp-block-list\">\n<li>Calculate the formula weights of the following substances: (a) BaCl2 \u00b7 2H2O, (b) KHC2O4 \u00b7 H2C2O4, (c) Ag2Cr2O7, (d) Ca3(PO4)2.<\/li>\n\n\n\n<li>Calculate the number of millimoles contained in 500 mg of each of the following substances: (a) BaCrO4, (b) CHCl3, (c) KIO3 \u00b7 HIO3, (d) MgNH4PO4, (e) Mg2P2O7, (f) FeSO4 \u00b7 C2H4(NH3)2SO4 \u00b7 4H2O<\/li>\n<\/ol>\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 address your queries, we&#8217;ll first calculate the formula weights (molar masses) of the specified substances and then determine the number of millimoles in 500 mg of each compound.<\/p>\n\n\n\n<p><strong>1. Formula Weights:<\/strong><\/p>\n\n\n\n<p>The formula weight (molar mass) of a compound is the sum of the atomic masses of all atoms present in its formula.<\/p>\n\n\n\n<p><strong>(a) BaCl\u2082\u00b72H\u2082O (Barium chloride dihydrate):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Barium (Ba):<\/strong> 1 atom \u00d7 137.33 g\/mol = 137.33 g\/mol<\/li>\n\n\n\n<li><strong>Chlorine (Cl):<\/strong> 2 atoms \u00d7 35.45 g\/mol = 70.90 g\/mol<\/li>\n\n\n\n<li><strong>Water (H\u2082O):<\/strong> 2 molecules \u00d7 (2 \u00d7 1.01 g\/mol + 16.00 g\/mol) = 36.04 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Molar Mass:<\/strong> 137.33 + 70.90 + 36.04 = 244.27 g\/mol<\/p>\n\n\n\n<p><strong>(b) KHC\u2082O\u2084\u00b7H\u2082C\u2082O\u2084 (Potassium hydrogen oxalate and oxalic acid):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Potassium (K):<\/strong> 1 atom \u00d7 39.10 g\/mol = 39.10 g\/mol<\/li>\n\n\n\n<li><strong>Hydrogen (H):<\/strong> 3 atoms \u00d7 1.01 g\/mol = 3.03 g\/mol<\/li>\n\n\n\n<li><strong>Carbon (C):<\/strong> 4 atoms \u00d7 12.01 g\/mol = 48.04 g\/mol<\/li>\n\n\n\n<li><strong>Oxygen (O):<\/strong> 8 atoms \u00d7 16.00 g\/mol = 128.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Molar Mass:<\/strong> 39.10 + 3.03 + 48.04 + 128.00 = 218.17 g\/mol<\/p>\n\n\n\n<p><strong>(c) Ag\u2082Cr\u2082O\u2087 (Silver dichromate):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Silver (Ag):<\/strong> 2 atoms \u00d7 107.87 g\/mol = 215.74 g\/mol<\/li>\n\n\n\n<li><strong>Chromium (Cr):<\/strong> 2 atoms \u00d7 52.00 g\/mol = 104.00 g\/mol<\/li>\n\n\n\n<li><strong>Oxygen (O):<\/strong> 7 atoms \u00d7 16.00 g\/mol = 112.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Molar Mass:<\/strong> 215.74 + 104.00 + 112.00 = 431.74 g\/mol<\/p>\n\n\n\n<p><strong>(d) Ca\u2083(PO\u2084)\u2082 (Calcium phosphate):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Calcium (Ca):<\/strong> 3 atoms \u00d7 40.08 g\/mol = 120.24 g\/mol<\/li>\n\n\n\n<li><strong>Phosphorus (P):<\/strong> 2 atoms \u00d7 30.97 g\/mol = 61.94 g\/mol<\/li>\n\n\n\n<li><strong>Oxygen (O):<\/strong> 8 atoms \u00d7 16.00 g\/mol = 128.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Molar Mass:<\/strong> 120.24 + 61.94 + 128.00 = 310.18 g\/mol<\/p>\n\n\n\n<p><strong>2. Millimoles in 500 mg:<\/strong><\/p>\n\n\n\n<p>To find the number of millimoles in 500 mg (0.500 g) of each substance, use the formula:<\/p>\n\n\n\n<p><strong>Millimoles (mmol) = (mass in grams \/ molar mass in g\/mol) \u00d7 1000<\/strong><\/p>\n\n\n\n<p><strong>(a) BaCrO\u2084 (Barium chromate):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar Mass:<\/strong> 137.33 (Ba) + 51.996 (Cr) + 4 \u00d7 16.00 (O) = 253.326 g\/mol<\/li>\n\n\n\n<li><strong>Millimoles:<\/strong> (0.500 g \/ 253.326 g\/mol) \u00d7 1000 \u2248 1.97 mmol<\/li>\n<\/ul>\n\n\n\n<p><strong>(b) CHCl\u2083 (Chloroform):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar Mass:<\/strong> 12.01 (C) + 1.008 (H) + 3 \u00d7 35.45 (Cl) = 119.37 g\/mol<\/li>\n\n\n\n<li><strong>Millimoles:<\/strong> (0.500 g \/ 119.37 g\/mol) \u00d7 1000 \u2248 4.19 mmol<\/li>\n<\/ul>\n\n\n\n<p><strong>(c) KIO\u2083\u00b7HIO\u2083 (Potassium iodate and iodic acid):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar Mass:<\/strong> 39.10 (K) + 126.90 (I) + 3 \u00d7 16.00 (O) + 1.01 (H) + 126.90 (I) + 3 \u00d7 16.00 (O) = 389.01 g\/mol<\/li>\n\n\n\n<li><strong>Millimoles:<\/strong> (0.500 g \/ 389.01 g\/mol) \u00d7 1000 \u2248 1.29 mmol<\/li>\n<\/ul>\n\n\n\n<p><strong>(d) MgNH\u2084PO\u2084 (Magnesium ammonium phosphate):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar Mass:<\/strong> 24.31 (Mg) + 14.01 (N) + 4 \u00d7 1.01 (H) + 30.97 (P) + 4 \u00d7 16.00 (O) = 137.33 g\/mol<\/li>\n\n\n\n<li><strong>Millimoles:<\/strong> (0.500 g \/ 137.33 g\/mol) \u00d7 1000 \u2248 3.64 mmol<\/li>\n<\/ul>\n\n\n\n<p><strong>(e) Mg\u2082P\u2082O\u2087 (Magnesium pyrophosphate):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar Mass:<\/strong> 2 \u00d7 24.31 (Mg) + 2 \u00d7 30.97 (P) + 7 \u00d7 16.00 (O) = 222.55 g\/mol<\/li>\n\n\n\n<li><strong>Millimoles:<\/strong> (0.500 g \/ 222.55 g\/mol) \u00d7 1000 \u2248 2.25 mmol<\/li>\n<\/ul>\n\n\n\n<p><strong>(f) FeSO\u2084\u00b7C\u2082H\u2084(NH\u2083)\u2082SO\u2084\u00b74H\u2082O (Iron(II) sulfate complex):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar Mass:<\/strong> 55.85 (Fe) + 2 \u00d7 32.07 (S) + 12 \u00d7 16.00 (O) + 2 \u00d7 12.01 (C)<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is : To address your queries, we&#8217;ll first calculate the formula weights (molar masses) of the specified substances and then determine the number of millimoles in 500 mg of each compound. 1. Formula Weights: The formula weight (molar mass) of a compound is the sum of the atomic masses of [&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-195597","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195597","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=195597"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195597\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195597"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195597"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195597"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}