{"id":218810,"date":"2025-05-24T13:29:08","date_gmt":"2025-05-24T13:29:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218810"},"modified":"2025-05-24T13:29:10","modified_gmt":"2025-05-24T13:29:10","slug":"the-active-compound-in-pepto-bismol-contains-c-h-o-and-bi","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/the-active-compound-in-pepto-bismol-contains-c-h-o-and-bi\/","title":{"rendered":"The active compound in Pepto-Bismol contains C, H, O, and Bi."},"content":{"rendered":"\n<p>None<br>The active compound in Pepto-Bismol contains C, H, O, and Bi.<br>(a) When 0.22105 g of it was burned in excess O2, 0.1422 g of bismuth(III) oxide, 0.1880 g of carbon dioxide, and 0.02750 g of water were formed. What is the empirical formula of this compound?<br>(b) Given a molar mass of 1086 g\/mol, determine the molecular formula. (c) Complete and balance the acid-base reaction between bismuth(III) hydroxide and salicylic acid (HC7H5O3), which is used to form this compound.<br>(d) A dose of Pepto-Bismol contains 0.600 mg of active ingredient. If the yield of the reaction in part (c) is 88.0%, what mass (in mg) of bismuth(III) hydroxide is required to prepare one dose?<\/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<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given Data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of compound burned: 0.22105 g<\/li>\n\n\n\n<li>Products formed:\n<ul class=\"wp-block-list\">\n<li>Bi\u2082O\u2083 (bismuth(III) oxide): 0.1422 g<\/li>\n\n\n\n<li>CO\u2082: 0.1880 g<\/li>\n\n\n\n<li>H\u2082O: 0.02750 g<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">(a) Find the empirical formula of the compound<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine the moles of each element from the products<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Bismuth (Bi):<\/strong><br>From Bi\u2082O\u2083:<br>Molar mass Bi\u2082O\u2083 = (2 \u00d7 208.98) + (3 \u00d7 16.00) = 417.96 + 48 = 465.96 g\/mol<br>Moles of Bi\u2082O\u2083 = 0.1422 g \/ 465.96 g\/mol = 3.051 \u00d7 10\u207b\u2074 mol<br>Moles of Bi atoms = 2 \u00d7 3.051 \u00d7 10\u207b\u2074 = 6.102 \u00d7 10\u207b\u2074 mol<\/li>\n<\/ol>\n\n\n\n<p>Mass of Bi = moles \u00d7 atomic mass = 6.102 \u00d7 10\u207b\u2074 mol \u00d7 208.98 g\/mol = 0.1275 g<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Carbon (C):<\/strong><br>From CO\u2082:<br>Molar mass CO\u2082 = 44.01 g\/mol<br>Moles of CO\u2082 = 0.1880 g \/ 44.01 g\/mol = 4.273 \u00d7 10\u207b\u00b3 mol<br>Moles of C = 4.273 \u00d7 10\u207b\u00b3 mol<br>Mass of C = 4.273 \u00d7 10\u207b\u00b3 mol \u00d7 12.01 g\/mol = 0.0513 g<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>Hydrogen (H):<\/strong><br>From H\u2082O:<br>Molar mass H\u2082O = 18.02 g\/mol<br>Moles of H\u2082O = 0.02750 g \/ 18.02 g\/mol = 1.526 \u00d7 10\u207b\u00b3 mol<br>Moles of H = 2 \u00d7 1.526 \u00d7 10\u207b\u00b3 mol = 3.052 \u00d7 10\u207b\u00b3 mol<br>Mass of H = 3.052 \u00d7 10\u207b\u00b3 mol \u00d7 1.008 g\/mol = 0.00308 g<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li><strong>Oxygen (O):<\/strong><br>Total mass of sample = 0.22105 g<br>Sum of masses of Bi, C, and H = 0.1275 + 0.0513 + 0.00308 = 0.1819 g<br>Mass of O = 0.22105 g &#8211; 0.1819 g = 0.03915 g<br>Moles of O = 0.03915 g \/ 16.00 g\/mol = 2.447 \u00d7 10\u207b\u00b3 mol<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find the mole ratio of elements<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Element<\/th><th>Moles<\/th><th>Divide by smallest<\/th><th>Ratio (approx.)<\/th><\/tr><\/thead><tbody><tr><td>Bi<\/td><td>6.102 \u00d7 10\u207b\u2074<\/td><td>6.102 \u00d7 10\u207b\u2074<\/td><td>1<\/td><\/tr><tr><td>C<\/td><td>4.273 \u00d7 10\u207b\u00b3<\/td><td>6.102 \u00d7 10\u207b\u2074<\/td><td>7.0<\/td><\/tr><tr><td>H<\/td><td>3.052 \u00d7 10\u207b\u00b3<\/td><td>6.102 \u00d7 10\u207b\u2074<\/td><td>5.0<\/td><\/tr><tr><td>O<\/td><td>2.447 \u00d7 10\u207b\u00b3<\/td><td>6.102 \u00d7 10\u207b\u2074<\/td><td>4.0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Write empirical formula<\/h3>\n\n\n\n<p>The mole ratio approximates to:<\/p>\n\n\n\n<p><strong>Bi\u2081 C\u2087 H\u2085 O\u2084<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">(b) Molecular formula<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Given molar mass = 1086 g\/mol<\/p>\n\n\n\n<p>Calculate molar mass of empirical formula:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bi: 208.98 \u00d7 1 = 208.98 g\/mol<\/li>\n\n\n\n<li>C: 12.01 \u00d7 7 = 84.07 g\/mol<\/li>\n\n\n\n<li>H: 1.008 \u00d7 5 = 5.04 g\/mol<\/li>\n\n\n\n<li>O: 16.00 \u00d7 4 = 64.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Sum = 208.98 + 84.07 + 5.04 + 64.00 = 362.09 g\/mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Calculate multiplier:<\/p>\n\n\n\n<p>n=1086362.09\u22483n = \\frac{1086}{362.09} \\approx 3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>So molecular formula is:<\/p>\n\n\n\n<p><strong>Bi\u2083 C\u2082\u2081 H\u2081\u2085 O\u2081\u2082<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">(c) Balanced acid-base reaction between bismuth(III) hydroxide and salicylic acid (HC\u2087H\u2085O\u2083)<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bismuth(III) hydroxide: Bi(OH)\u2083<\/li>\n\n\n\n<li>Salicylic acid: HC\u2087H\u2085O\u2083 (one acidic H per molecule)<\/li>\n<\/ul>\n\n\n\n<p><strong>Reaction:<\/strong><\/p>\n\n\n\n<p>Bi(OH)3+3&nbsp;HC7H5O3\u2192Bi(C7H5O3)3+3&nbsp;H2O\\mathrm{Bi(OH)_3 + 3 \\ HC_7H_5O_3 \\rightarrow Bi(C_7H_5O_3)_3 + 3 \\ H_2O}<\/p>\n\n\n\n<p>Bismuth(III) hydroxide reacts with 3 moles of salicylic acid to form bismuth(III) salicylate and water.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">(d) Mass of Bi(OH)\u2083 required to prepare one dose (0.600 mg active ingredient) at 88.0% yield<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find mass of Bi(OH)\u2083 required theoretically for 0.600 mg of product<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molecular mass of Bi(OH)\u2083:<br>Bi = 208.98 g\/mol<br>O = 16.00 \u00d7 3 = 48.00 g\/mol<br>H = 1.008 \u00d7 3 = 3.02 g\/mol<br>Total = 208.98 + 48 + 3.02 = 260.0 g\/mol<\/li>\n\n\n\n<li>Molecular mass of active ingredient (bismuth salicylate) = 1086 g\/mol<\/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\">Step 2: Moles of active ingredient in 0.600 mg dose<\/h3>\n\n\n\n<p>Moles=0.600\u00d710\u221231086=5.53\u00d710\u22127&nbsp;mol\\text{Moles} = \\frac{0.600 \\times 10^{-3}}{1086} = 5.53 \\times 10^{-7} \\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\">Step 3: Moles of Bi(OH)\u2083 needed (1:1 mole ratio)<\/h3>\n\n\n\n<p>Same moles of Bi(OH)\u2083 required = 5.53 \u00d7 10\u207b\u2077 mol<\/p>\n\n\n\n<p>Mass of Bi(OH)\u2083 theoretically required:<\/p>\n\n\n\n<p>5.53\u00d710\u22127\u00d7260.0=1.44\u00d710\u22124&nbsp;g=0.144&nbsp;mg5.53 \\times 10^{-7} \\times 260.0 = 1.44 \\times 10^{-4} \\text{ g} = 0.144 \\text{ mg}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Adjust for 88.0% yield<\/h3>\n\n\n\n<p>Mass of Bi(OH)\u2083 required practically:<\/p>\n\n\n\n<p>0.1440.88=0.164&nbsp;mg\\frac{0.144}{0.88} = 0.164 \\text{ mg}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\"><strong>Final Answers:<\/strong><\/h1>\n\n\n\n<p><strong>(a)<\/strong> Empirical formula: <strong>BiC\u2087H\u2085O\u2084<\/strong><br><strong>(b)<\/strong> Molecular formula: <strong>Bi\u2083C\u2082\u2081H\u2081\u2085O\u2081\u2082<\/strong><br><strong>(c)<\/strong> Balanced reaction:<br>Bi(OH)3+3&nbsp;HC7H5O3\u2192Bi(C7H5O3)3+3&nbsp;H2O\\mathrm{Bi(OH)_3 + 3 \\ HC_7H_5O_3 \\rightarrow Bi(C_7H_5O_3)_3 + 3 \\ H_2O}<br><strong>(d)<\/strong> Mass of Bi(OH)\u2083 required = <strong>0.164 mg<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\">Explanation<\/h1>\n\n\n\n<p>The problem involves analyzing combustion data to determine the empirical formula of the active compound in Pepto-Bismol, followed by finding the molecular formula using the molar mass, balancing a key acid-base reaction, and calculating reagent mass for synthesis.<\/p>\n\n\n\n<p>In part (a), the compound is combusted completely to produce bismuth(III) oxide, carbon dioxide, and water. By quantifying the amounts of these products, the moles of each element in the original compound can be calculated. Bismuth is determined from Bi\u2082O\u2083, carbon from CO\u2082, and hydrogen from H\u2082O. The remaining mass is attributed to oxygen. Dividing each elemental mole amount by the smallest mole quantity yields the mole ratio that forms the empirical formula BiC\u2087H\u2085O\u2084.<\/p>\n\n\n\n<p>Part (b) compares the empirical formula mass (~362 g\/mol) with the experimentally given molar mass (1086 g\/mol). The molecular formula is a multiple (approximately 3) of the empirical formula, yielding Bi\u2083C\u2082\u2081H\u2081\u2085O\u2081\u2082.<\/p>\n\n\n\n<p>Part (c) involves the acid-base reaction between bismuth(III) hydroxide and salicylic acid. Bismuth(III) hydroxide acts as a base, neutralizing three moles of salicylic acid to form bismuth(III) salicylate and water. The balanced equation shows the 1:3 molar ratio.<\/p>\n\n\n\n<p>Finally, part (d) calculates the mass of Bi(OH)\u2083 needed to prepare one 0.600 mg dose of the active ingredient assuming 88% reaction yield. The moles of active compound are converted to the equivalent moles of Bi(OH)\u2083 (1:1 ratio), and then adjusted for the reaction efficiency. The result is 0.164 mg of Bi(OH)\u2083 required.<\/p>\n\n\n\n<p>This detailed analysis combines stoichiometric calculations, empirical data interpretation, and reaction chemistry to characterize the compound and its synthesis accurately.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-351.png\" alt=\"\" class=\"wp-image-218811\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>NoneThe active compound in Pepto-Bismol contains C, H, O, and Bi.(a) When 0.22105 g of it was burned in excess O2, 0.1422 g of bismuth(III) oxide, 0.1880 g of carbon dioxide, and 0.02750 g of water were formed. What is the empirical formula of this compound?(b) Given a molar mass of 1086 g\/mol, determine the [&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-218810","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218810","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=218810"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218810\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218810"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218810"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218810"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}