{"id":220764,"date":"2025-05-28T18:31:23","date_gmt":"2025-05-28T18:31:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220764"},"modified":"2025-05-28T18:31:26","modified_gmt":"2025-05-28T18:31:26","slug":"determine-the-empirical-and-molecular-formula-of-each-of-the-following-substances","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/determine-the-empirical-and-molecular-formula-of-each-of-the-following-substances\/","title":{"rendered":"\u00a0Determine the empirical and molecular formula of each of the following substances"},"content":{"rendered":"\n<figure class=\"wp-block-image size-full is-resized\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-493.png\" alt=\"\" class=\"wp-image-220765\" style=\"width:706px;height:auto\"\/><\/figure>\n\n\n\n<p>What is the 5. A compound whose empirical formula is XF3 consists of 65% F by mass. atomic mass of X?:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) Caffeine, a stimulant found in coffee, contains 49.5% C, 5.15% H, 28.9% N, and 16.5% O by mass and has a molar mass of 195 g mol-1<\/li>\n\n\n\n<li>(b) Monosodium glutamate (MSG), a flavour enhancer in certain foods, contains 35.51% C, 4.77% H, 37.85% O, 8.29% N, and 13.60% Na, and has a molar mass of 169 g mol<\/li>\n<\/ul>\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<h3 class=\"wp-block-heading\"><strong>Question 5 Answer:<\/strong><\/h3>\n\n\n\n<p>We are told the empirical formula is XF\u2083 and the compound is 65% fluorine by mass. Let\u2019s find the atomic mass of element X.<\/p>\n\n\n\n<p>Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of 1 mole of XF\u2083 = M<\/li>\n\n\n\n<li>Mass of 3 fluorine atoms = 3 \u00d7 19.0 = 57.0 g<\/li>\n\n\n\n<li>Since fluorine makes up 65% of the compound by mass:<\/li>\n<\/ul>\n\n\n\n<p>57.0M=0.65\u21d2M=57.00.65=87.69\u2009g\/mol\\frac{57.0}{M} = 0.65 \\Rightarrow M = \\frac{57.0}{0.65} = 87.69 \\, \\text{g\/mol}<\/p>\n\n\n\n<p>Atomic mass of X \u2248 <strong>87.7 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>Question 6(a): Caffeine<\/strong><\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C = 49.5%, H = 5.15%, N = 28.9%, O = 16.5%<\/li>\n\n\n\n<li>Molar mass = 195 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Assume 100 g of caffeine:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C: 49.5 g \u2192 49.5 \/ 12.01 = 4.12 mol<\/li>\n\n\n\n<li>H: 5.15 g \u2192 5.15 \/ 1.008 = 5.11 mol<\/li>\n\n\n\n<li>N: 28.9 g \u2192 28.9 \/ 14.01 = 2.06 mol<\/li>\n\n\n\n<li>O: 16.5 g \u2192 16.5 \/ 16.00 = 1.03 mol<\/li>\n<\/ul>\n\n\n\n<p>Divide all by the smallest (1.03):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C: 4.12 \/ 1.03 \u2248 4<\/li>\n\n\n\n<li>H: 5.11 \/ 1.03 \u2248 5<\/li>\n\n\n\n<li>N: 2.06 \/ 1.03 \u2248 2<\/li>\n\n\n\n<li>O: 1.03 \/ 1.03 = 1<\/li>\n<\/ul>\n\n\n\n<p><strong>Empirical formula<\/strong> = C\u2084H\u2085N\u2082O<br>Empirical mass = (4\u00d712.01) + (5\u00d71.008) + (2\u00d714.01) + (1\u00d716.00) = 97.1 g\/mol<br>195 \/ 97.1 \u2248 2 \u2192 Multiply subscripts by 2<\/p>\n\n\n\n<p><strong>Molecular formula<\/strong> = <strong>C\u2088H\u2081\u2080N\u2084O\u2082<\/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>Question 6(b): MSG<\/strong><\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C = 35.51%, H = 4.77%, O = 37.85%, N = 8.29%, Na = 13.60%<\/li>\n\n\n\n<li>Molar mass = 169 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Assume 100 g of MSG:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C: 35.51 g \u2192 2.96 mol<\/li>\n\n\n\n<li>H: 4.77 g \u2192 4.73 mol<\/li>\n\n\n\n<li>O: 37.85 g \u2192 2.37 mol<\/li>\n\n\n\n<li>N: 8.29 g \u2192 0.592 mol<\/li>\n\n\n\n<li>Na: 13.60 g \u2192 0.592 mol<\/li>\n<\/ul>\n\n\n\n<p>Divide all by 0.592:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C: 2.96 \/ 0.592 \u2248 5<\/li>\n\n\n\n<li>H: 4.73 \/ 0.592 \u2248 8<\/li>\n\n\n\n<li>O: 2.37 \/ 0.592 \u2248 4<\/li>\n\n\n\n<li>N: 0.592 \/ 0.592 = 1<\/li>\n\n\n\n<li>Na: 0.592 \/ 0.592 = 1<\/li>\n<\/ul>\n\n\n\n<p><strong>Empirical formula<\/strong> = C\u2085H\u2088NO\u2084Na<br>Empirical mass = 169 g\/mol<\/p>\n\n\n\n<p><strong>Molecular formula<\/strong> = <strong>C\u2085H\u2088NO\u2084Na<\/strong> (same as empirical)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Q5 Answer:<\/strong> Atomic mass of X \u2248 <strong>87.7 g\/mol<\/strong><\/li>\n\n\n\n<li><strong>Q6(a):<\/strong> Empirical = <strong>C\u2084H\u2085N\u2082O<\/strong>, Molecular = <strong>C\u2088H\u2081\u2080N\u2084O\u2082<\/strong><\/li>\n\n\n\n<li><strong>Q6(b):<\/strong> Empirical &amp; Molecular = <strong>C\u2085H\u2088NO\u2084Na<\/strong><\/li>\n<\/ul>\n\n\n\n<p>This analysis uses percent-by-mass conversions to moles and ratio simplifications to find empirical formulas. For molecular formulas, the molar mass confirms the multiplier of empirical units.<\/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\/learnexams-banner4-76.jpeg\" alt=\"\" class=\"wp-image-220766\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the 5. A compound whose empirical formula is XF3 consists of 65% F by mass. atomic mass of X?: The Correct Answer and Explanation is: Question 5 Answer: We are told the empirical formula is XF\u2083 and the compound is 65% fluorine by mass. Let\u2019s find the atomic mass of element X. Let: [&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-220764","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220764","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=220764"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220764\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220764"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220764"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}