{"id":184153,"date":"2025-01-19T11:09:38","date_gmt":"2025-01-19T11:09:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=184153"},"modified":"2025-01-19T11:09:40","modified_gmt":"2025-01-19T11:09:40","slug":"calculate-the-molar-mass-of-phosphoric-acid-h3po4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/19\/calculate-the-molar-mass-of-phosphoric-acid-h3po4\/","title":{"rendered":"Calculate the molar mass of phosphoric acid H3PO4"},"content":{"rendered":"\n<p>Calculate the molar mass of phosphoric acid H3PO4.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-321.png\" alt=\"\" class=\"wp-image-184154\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-322.png\" alt=\"\" class=\"wp-image-184155\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Molar Mass of Phosphoric Acid (H\u2083PO\u2084)<\/h3>\n\n\n\n<p>The molar mass of phosphoric acid is calculated by summing the atomic masses of its constituent atoms. The chemical formula H3PO4H_3PO_4 indicates it contains:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hydrogen (H):<\/strong> 3 atoms<\/li>\n\n\n\n<li><strong>Phosphorus (P):<\/strong> 1 atom<\/li>\n\n\n\n<li><strong>Oxygen (O):<\/strong> 4 atoms<\/li>\n<\/ul>\n\n\n\n<p>Using the periodic table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Atomic mass of H = 1.01 g\/mol<\/li>\n\n\n\n<li>Atomic mass of P = 30.97 g\/mol<\/li>\n\n\n\n<li>Atomic mass of O = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Calculation:<\/h4>\n\n\n\n<p>Molar&nbsp;mass=(3\u00d71.01)+(1\u00d730.97)+(4\u00d716.00)\\text{Molar mass} = (3 \\times 1.01) + (1 \\times 30.97) + (4 \\times 16.00) Molar&nbsp;mass=3.03+30.97+64.00=98.00\u2009g\/mol\\text{Molar mass} = 3.03 + 30.97 + 64.00 = 98.00 \\, \\text{g\/mol}<\/p>\n\n\n\n<p>Thus, the molar mass of phosphoric acid (H3PO4H_3PO_4) is <strong>98.00 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\">Explanation:<\/h3>\n\n\n\n<p>Molar mass is the mass of one mole of a chemical compound, expressed in grams per mole (g\/molg\/mol). It is a crucial property in chemistry for understanding the quantities of substances involved in chemical reactions.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Steps:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Understand the Formula<\/strong>: The formula H3PO4H_3PO_4 shows the ratio of atoms in the compound: 3 hydrogens, 1 phosphorus, and 4 oxygens.<\/li>\n\n\n\n<li><strong>Find Atomic Masses<\/strong>: Atomic masses of individual elements are obtained from the periodic table. These values reflect the average mass of an atom, accounting for isotopes.<\/li>\n\n\n\n<li><strong>Multiply by Number of Atoms<\/strong>: For each element, multiply its atomic mass by the number of atoms in the compound.\n<ul class=\"wp-block-list\">\n<li>For hydrogen: 3\u00d71.01\u2009g\/mol3 \\times 1.01 \\, \\text{g\/mol}<\/li>\n\n\n\n<li>For phosphorus: 1\u00d730.97\u2009g\/mol1 \\times 30.97 \\, \\text{g\/mol}<\/li>\n\n\n\n<li>For oxygen: 4\u00d716.00\u2009g\/mol4 \\times 16.00 \\, \\text{g\/mol}.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Sum the Results<\/strong>: Adding these contributions gives the total molar mass.<\/li>\n<\/ol>\n\n\n\n<p>Phosphoric acid is a triprotic acid widely used in fertilizers, food flavoring, and rust removal. Understanding its molar mass allows chemists to measure precise quantities for reactions, ensuring accurate stoichiometric calculations.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the molar mass of phosphoric acid H3PO4. The correct answer and explanation is: Molar Mass of Phosphoric Acid (H\u2083PO\u2084) The molar mass of phosphoric acid is calculated by summing the atomic masses of its constituent atoms. The chemical formula H3PO4H_3PO_4 indicates it contains: Using the periodic table: Calculation: Molar&nbsp;mass=(3\u00d71.01)+(1\u00d730.97)+(4\u00d716.00)\\text{Molar mass} = (3 \\times 1.01) [&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-184153","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184153","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=184153"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184153\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184153"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184153"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184153"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}