{"id":232447,"date":"2025-06-12T01:03:24","date_gmt":"2025-06-12T01:03:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232447"},"modified":"2025-06-12T01:03:26","modified_gmt":"2025-06-12T01:03:26","slug":"which-is-the-percent-of-phosphorus-in-zn3po42-if-the-molar-mass-of-zn3po42-is-386-17-grams","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/12\/which-is-the-percent-of-phosphorus-in-zn3po42-if-the-molar-mass-of-zn3po42-is-386-17-grams\/","title":{"rendered":"Which is the percent of phosphorus in Zn3(PO4)2 if the molar mass of Zn3(PO4)2 is 386.17 grams"},"content":{"rendered":"\n

Which is the percent of phosphorus in Zn3(PO4)2 if the molar mass of Zn3(PO4)2 is 386.17 grams<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the percent of phosphorus<\/strong> in zinc phosphate<\/strong>, Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b, we use the formula:Percent of element=(Total mass of the element in 1 mole of compoundMolar mass of the compound)\u00d7100\\text{Percent of element} = \\left( \\frac{\\text{Total mass of the element in 1 mole of compound}}{\\text{Molar mass of the compound}} \\right) \\times 100Percent of element=(Molar mass of the compoundTotal mass of the element in 1 mole of compound\u200b)\u00d7100<\/p>\n\n\n\n

Step 1: Determine the number of phosphorus atoms<\/h3>\n\n\n\n

The chemical formula Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b contains two phosphate ions<\/strong>, PO43\u2212\\text{PO}_4^{3-}PO43\u2212\u200b. Each phosphate ion contains one phosphorus atom<\/strong>, so there are:2 P atoms per formula unit2 \\text{ P atoms per formula unit}2 P atoms per formula unit<\/p>\n\n\n\n

Step 2: Molar mass of phosphorus<\/h3>\n\n\n\n

The atomic mass of phosphorus (P) is approximately 30.97 g\/mol<\/strong>.<\/p>\n\n\n\n

So, total mass of phosphorus in one mole of Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b:2\u00d730.97=61.94 g2 \\times 30.97 = 61.94 \\text{ g}2\u00d730.97=61.94 g<\/p>\n\n\n\n

Step 3: Molar mass of the compound<\/h3>\n\n\n\n

You are given the molar mass of Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b as 386.17 g\/mol<\/strong>.<\/p>\n\n\n\n

Step 4: Calculate percent of phosphorus<\/h3>\n\n\n\n

Percent of P=(61.94386.17)\u00d7100\u224816.04%\\text{Percent of P} = \\left( \\frac{61.94}{386.17} \\right) \\times 100 \\approx 16.04\\%Percent of P=(386.1761.94\u200b)\u00d7100\u224816.04%<\/p>\n\n\n\n

\n\n\n\n

Final Answer:<\/h3>\n\n\n\n

The percent of phosphorus in Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b is approximately 16.04%.<\/strong><\/p>\n\n\n\n

\n\n\n\n

Explanation):<\/h3>\n\n\n\n

To determine the percentage composition of an element in a compound, we use the concept of molar mass and the proportion of the element’s mass relative to the total molar mass of the compound. In this case, we are interested in finding the percent composition of phosphorus in the compound zinc phosphate, which has the formula Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b.<\/p>\n\n\n\n

This formula indicates that each formula unit of zinc phosphate contains three zinc (Zn) atoms and two phosphate (PO\u2084\u00b3\u207b) ions. Since each phosphate ion contains one phosphorus atom, the compound contains a total of two phosphorus atoms. The atomic mass of phosphorus is approximately 30.97 g\/mol, so the total mass of phosphorus in one mole of zinc phosphate is:2\u00d730.97=61.94 g2 \\times 30.97 = 61.94 \\text{ g}2\u00d730.97=61.94 g<\/p>\n\n\n\n

The given molar mass of zinc phosphate is 386.17 g\/mol. To calculate the percent composition, we divide the total mass of phosphorus in the compound by the molar mass of the compound, then multiply by 100 to convert to a percentage:(61.94386.17)\u00d7100\u224816.04%\\left( \\frac{61.94}{386.17} \\right) \\times 100 \\approx 16.04\\%(386.1761.94\u200b)\u00d7100\u224816.04%<\/p>\n\n\n\n

This result tells us that in one mole (or 386.17 grams) of zinc phosphate, about 16.04% of the mass is due to phosphorus. This kind of calculation is important in chemistry for understanding the elemental makeup of compounds, especially in contexts like nutritional labeling, industrial formulations, and stoichiometric calculations.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Which is the percent of phosphorus in Zn3(PO4)2 if the molar mass of Zn3(PO4)2 is 386.17 grams The Correct Answer and Explanation is: To find the percent of phosphorus in zinc phosphate, Zn3(PO4)2\\text{Zn}_3(\\text{PO}_4)_2Zn3\u200b(PO4\u200b)2\u200b, we use the formula:Percent of element=(Total mass of the element in 1 mole of compoundMolar mass of the compound)\u00d7100\\text{Percent of element} = \\left( \\frac{\\text{Total mass of the element in 1 mole of compound}}{\\text{Molar mass of the compound}} […]<\/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-232447","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232447","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=232447"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232447\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232447"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232447"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232447"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}