{"id":190657,"date":"2025-02-13T06:32:42","date_gmt":"2025-02-13T06:32:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190657"},"modified":"2025-02-13T06:32:45","modified_gmt":"2025-02-13T06:32:45","slug":"name-each-compound-and-calculate-its-formula-mass","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/13\/name-each-compound-and-calculate-its-formula-mass\/","title":{"rendered":"Name each compound and calculate its formula mass"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Name each compound and calculate its formula mass.<br>(a) PtO\u2082<br>(b) N\u2082O\u2083<br>(c) Al(CIO3)3<br>(d) PBrs<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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<h3 class=\"wp-block-heading\">Compound Names and Formula Mass Calculations<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">(a) <strong>PtO\u2082<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Name<\/strong>: Platinum(IV) oxide<\/li>\n\n\n\n<li><strong>Formula Mass Calculation<\/strong>:<\/li>\n\n\n\n<li>Pt (Platinum): 195.08 amu<\/li>\n\n\n\n<li>O (Oxygen): 16.00 amu \u00d7 2 = 32.00 amu<\/li>\n\n\n\n<li><strong>Total<\/strong>: 195.08 amu + 32.00 amu = 227.08 amu<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">(b) <strong>N\u2082O\u2083<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Name<\/strong>: Dinitrogen trioxide<\/li>\n\n\n\n<li><strong>Formula Mass Calculation<\/strong>:<\/li>\n\n\n\n<li>N (Nitrogen): 14.01 amu \u00d7 2 = 28.02 amu<\/li>\n\n\n\n<li>O (Oxygen): 16.00 amu \u00d7 3 = 48.00 amu<\/li>\n\n\n\n<li><strong>Total<\/strong>: 28.02 amu + 48.00 amu = 76.02 amu<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">(c) <strong>Al(ClO\u2083)\u2083<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Name<\/strong>: Aluminum chlorate<\/li>\n\n\n\n<li><strong>Formula Mass Calculation<\/strong>:<\/li>\n\n\n\n<li>Al (Aluminum): 26.98 amu<\/li>\n\n\n\n<li>ClO\u2083 (Chlorate): 35.45 amu (Cl) + 16.00 amu \u00d7 3 (O) = 83.45 amu per ClO\u2083; 83.45 amu \u00d7 3 = 250.35 amu<\/li>\n\n\n\n<li><strong>Total<\/strong>: 26.98 amu + 250.35 amu = 277.33 amu<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">(d) <strong>PBr\u2085<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Name<\/strong>: Phosphorus pentabromide<\/li>\n\n\n\n<li><strong>Formula Mass Calculation<\/strong>:<\/li>\n\n\n\n<li>P (Phosphorus): 30.97 amu<\/li>\n\n\n\n<li>Br (Bromine): 79.90 amu \u00d7 5 = 399.50 amu<\/li>\n\n\n\n<li><strong>Total<\/strong>: 30.97 amu + 399.50 amu = 430.47 amu<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The <strong>formula mass<\/strong> of a compound is calculated by adding the atomic masses (amu) of each element in the compound, multiplied by the number of atoms of that element present in the formula. The atomic masses are taken from the periodic table and typically rounded to two decimal places for simplicity.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determining Names<\/strong>: The name of a compound is determined based on its chemical formula. For simple binary compounds (like PtO\u2082), the name reflects the elements involved and their oxidation states or ratios (e.g., Platinum(IV) oxide). For compounds with polyatomic ions (like Al(ClO\u2083)\u2083), the name includes the cation followed by the name of the polyatomic anion (e.g., Aluminum chlorate).<\/li>\n\n\n\n<li><strong>Calculating Formula Mass<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Identify the elements<\/strong> and their respective quantities in the compound.<\/li>\n\n\n\n<li><strong>Multiply the atomic mass<\/strong> of each element by the number of atoms of that element in the compound.<\/li>\n\n\n\n<li><strong>Sum these values<\/strong> to get the total formula mass of the compound.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">These calculations are crucial in various chemical contexts, such as stoichiometry, preparing solutions, and conducting chemical reactions, where precise measurements are necessary to ensure the desired outcomes. Understanding how to calculate and interpret formula masses is fundamental in chemistry, reinforcing the relationship between the molecular structure of compounds and their physical properties.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Name each compound and calculate its formula mass.(a) PtO\u2082(b) N\u2082O\u2083(c) Al(CIO3)3(d) PBrs The Correct Answer and Explanation is : Compound Names and Formula Mass Calculations (a) PtO\u2082 (b) N\u2082O\u2083 (c) Al(ClO\u2083)\u2083 (d) PBr\u2085 Explanation The formula mass of a compound is calculated by adding the atomic masses (amu) of each element in the compound, multiplied [&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-190657","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190657","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=190657"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190657\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190657"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190657"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190657"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}