{"id":220722,"date":"2025-05-28T17:09:01","date_gmt":"2025-05-28T17:09:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220722"},"modified":"2025-05-28T17:09:05","modified_gmt":"2025-05-28T17:09:05","slug":"an-analysis-of-an-oxide-of-nitrogen-with-a-molecular-weight-of-92-02-amu-has-a-percent-composition-of-69-57-oxygen-and-30-43-nitrogen-what-are-the-empirical-and-molecular-formulas-for-this-nitrogen-o","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/28\/an-analysis-of-an-oxide-of-nitrogen-with-a-molecular-weight-of-92-02-amu-has-a-percent-composition-of-69-57-oxygen-and-30-43-nitrogen-what-are-the-empirical-and-molecular-formulas-for-this-nitrogen-o\/","title":{"rendered":"An analysis of an oxide of nitrogen with a molecular weight of 92.02 amu has a percent composition of 69.57% oxygen and 30.43% nitrogen, what are the empirical and molecular formulas for this nitrogen oxide"},"content":{"rendered":"\n<p>An analysis of an oxide of nitrogen with a molecular weight of 92.02 amu has a percent composition of 69.57% oxygen and 30.43% nitrogen, what are the empirical and molecular formulas for this nitrogen oxide? Complete and balance the equation for its formation from the elements nitrogen and oxygen.<\/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><strong>Correct Answer:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Empirical Formula:<\/strong> NO\u2082<\/li>\n\n\n\n<li><strong>Molecular Formula:<\/strong> N\u2082O\u2084<\/li>\n\n\n\n<li><strong>Balanced Formation Equation:<\/strong> N2(g)+2O2(g)\u2192N2O4(g)N_2(g) + 2O_2(g) \\rightarrow N_2O_4(g)<\/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>Explanation <\/strong><\/h3>\n\n\n\n<p>To determine the <strong>empirical formula<\/strong>, we use the percent composition data provided:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>30.43% nitrogen (N)<\/li>\n\n\n\n<li>69.57% oxygen (O)<\/li>\n<\/ul>\n\n\n\n<p>Assume we have 100 grams of the compound. This gives us:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>30.43 g N \u2192 30.43\u00a0g14.01\u00a0g\/mol\u22482.17\u00a0mol\u00a0N\\frac{30.43 \\text{ g}}{14.01 \\text{ g\/mol}} \u2248 2.17 \\text{ mol N}<\/li>\n\n\n\n<li>69.57 g O \u2192 69.57\u00a0g16.00\u00a0g\/mol\u22484.35\u00a0mol\u00a0O\\frac{69.57 \\text{ g}}{16.00 \\text{ g\/mol}} \u2248 4.35 \\text{ mol O}<\/li>\n<\/ul>\n\n\n\n<p>Now, find the simplest whole-number ratio: 2.172.17:4.352.17=1:2\\frac{2.17}{2.17} : \\frac{4.35}{2.17} = 1 : 2<\/p>\n\n\n\n<p>So, the <strong>empirical formula<\/strong> is <strong>NO\u2082<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Next, we calculate the molar mass of the empirical formula NO\u2082: 14.01+(2\u00d716.00)=46.01&nbsp;g\/mol14.01 + (2 \u00d7 16.00) = 46.01 \\text{ g\/mol}<\/p>\n\n\n\n<p>Now compare this with the given <strong>molecular weight<\/strong> of the compound: 92.02 amu 92.0246.01\u22482\\frac{92.02}{46.01} \u2248 2<\/p>\n\n\n\n<p>So the molecular formula is <strong>2 \u00d7 (NO\u2082) = N\u2082O\u2084<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>To write the <strong>balanced formation reaction<\/strong> of N\u2082O\u2084 from its elements:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen gas: N\u2082(g)<\/li>\n\n\n\n<li>Oxygen gas: O\u2082(g)<\/li>\n<\/ul>\n\n\n\n<p>In N\u2082O\u2084, there are 2 nitrogen atoms and 4 oxygen atoms per molecule. To form one molecule of N\u2082O\u2084: N2(g)+2O2(g)\u2192N2O4(g)N_2(g) + 2O_2(g) \\rightarrow N_2O_4(g)<\/p>\n\n\n\n<p>This is already balanced with respect to both nitrogen and oxygen atoms.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Empirical Formula: <strong>NO\u2082<\/strong><\/li>\n\n\n\n<li>Molecular Formula: <strong>N\u2082O\u2084<\/strong><\/li>\n\n\n\n<li>Balanced Formation Equation: N2(g)+2O2(g)\u2192N2O4(g)N_2(g) + 2O_2(g) \\rightarrow N_2O_4(g)<\/li>\n<\/ul>\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-banner4-74.jpeg\" alt=\"\" class=\"wp-image-220723\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>An analysis of an oxide of nitrogen with a molecular weight of 92.02 amu has a percent composition of 69.57% oxygen and 30.43% nitrogen, what are the empirical and molecular formulas for this nitrogen oxide? Complete and balance the equation for its formation from the elements nitrogen and oxygen. The Correct Answer and Explanation is: [&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-220722","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220722","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=220722"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220722\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}