The formula for copper(II) phosphate is Cu\u2083(PO\u2084)\u2082<\/strong>.<\/p>\n\n\n\n To find the number of phosphorus (P) atoms in 3.70 mol of copper(II) phosphate, follow these steps:<\/p>\n\n\n\n Finally, to get the number of phosphorus atoms in terms of mass or moles, we use these basic relations, where each mole of P atoms equals Avogadro’s number. In conclusion, 3.70 mol of Cu\u2083(PO\u2084)\u2082 contains 7.40 mol of phosphorus atoms<\/strong>.<\/p>\n\n\n\n What is the formula for copper(II) phosphate? Capitalization and punctuation count. Formula: Cu3(PO4)2. How many atoms of phosphorus are in 3.70 mol of copper(II) phosphate? Number of P atoms: 229.4. The Correct Answer and Explanation is: The formula for copper(II) phosphate is Cu\u2083(PO\u2084)\u2082. To find the number of phosphorus (P) atoms in 3.70 mol of […]<\/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-248464","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248464","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=248464"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248464\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
The chemical formula Cu\u2083(PO\u2084)\u2082 indicates that each molecule of copper(II) phosphate contains 2 phosphate ions (PO\u2084\u00b2\u207b), and each phosphate ion contains 1 phosphorus (P) atom. Therefore, for each formula unit of copper(II) phosphate, there are 2 phosphorus atoms.<\/li>\n\n\n\n
One mole of any substance contains Avogadro’s number of molecules, which is approximately 6.022\u00d710236.022 \\times 10^{23}6.022\u00d71023 molecules. So, 3.70 mol of copper(II) phosphate will contain: 3.70\u2009mol\u00d7(6.022\u00d71023\u2009molecules\/mol)=2.23\u00d71024\u2009molecules.3.70 \\, \\text{mol} \\times (6.022 \\times 10^{23} \\, \\text{molecules\/mol}) = 2.23 \\times 10^{24} \\, \\text{molecules}.3.70mol\u00d7(6.022\u00d71023molecules\/mol)=2.23\u00d71024molecules.<\/li>\n\n\n\n
Since each molecule of copper(II) phosphate contains 2 phosphorus atoms, the total number of phosphorus atoms in 3.70 mol is: 2.23\u00d71024\u2009molecules\u00d72\u2009P\u00a0atoms\/molecule=4.46\u00d71024\u2009P\u00a0atoms.2.23 \\times 10^{24} \\, \\text{molecules} \\times 2 \\, \\text{P atoms\/molecule} = 4.46 \\times 10^{24} \\, \\text{P atoms}.2.23\u00d71024molecules\u00d72P\u00a0atoms\/molecule=4.46\u00d71024P\u00a0atoms.<\/li>\n\n\n\n
In terms of moles, the number of phosphorus atoms in 3.70 mol of copper(II) phosphate is 2 times the moles of the compound, as there are 2 phosphorus atoms per formula unit of Cu\u2083(PO\u2084)\u2082: 3.70\u2009mol\u00d72\u2009P\u00a0atoms\/mol=7.40\u2009mol\u00a0of\u00a0P\u00a0atoms.3.70 \\, \\text{mol} \\times 2 \\, \\text{P atoms\/mol} = 7.40 \\, \\text{mol of P atoms}.3.70mol\u00d72P\u00a0atoms\/mol=7.40mol\u00a0of\u00a0P\u00a0atoms.<\/li>\n<\/ol>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-239.jpeg\")
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