{"id":5129,"date":"2019-02-17T15:15:15","date_gmt":"2019-02-17T09:45:15","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=5129"},"modified":"2022-11-04T12:06:30","modified_gmt":"2022-11-04T06:36:30","slug":"weight-percent-to-mole-fraction","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/weight-percent-to-mole-fraction\/","title":{"rendered":"Weight Percent Formula | How to convert weight percent to mole fraction"},"content":{"rendered":"<p>In this post , we will be checking out what is weight percent or mass percent,what is Weight percent or mass percent formula, how to convert <strong>weight percent to mole fraction<\/strong><\/p>\n<p><strong>Weight percent or mass percent<\/strong> is defined as the weight of the solute in 100 gm of the solution<\/p>\n<p>$Weight \\; percent= \\frac {mass \\; of \\; the \\; solute}{mass \\; of \\; the \\; solution} \\times 100$<\/p>\n<p>So , if m is the mass of Solute and M is the mass of solvent, then Weight percent formula will be<\/p>\n<p>$Weight \\; percent= \\frac {m}{m+M} \\times 100$<\/p>\n<p>It is generally given as 30% NaCL by weight solution or 50% HCL by weight solution<\/p>\n<p><strong>Example on Weight percent<\/strong><\/p>\n<p>1. A solution is prepared by adding 3 g of a substance A to 17 g of water. Calculate the mass per cent of the solute.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>$Weight \\; percent= \\frac {mass \\; of \\; the \\; solute}{mass \\; of \\; the \\; solution} \\times 100$<\/p>\n<p>$=\\frac {3}{3+17} \\times 100$<\/p>\n<p>=15%<\/p>\n<p>2. How many grams of 5.0% by weight NaCL solutions are required to 6.4 g NaCL?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>A 5% by weight NaCL solutions contains 5 g of NaCL in 100 g of solution<\/p>\n<p>So 1 g of NaCL will be contained in = 100\/5 =20 g of solution<\/p>\n<p>Therefore 6.4 g of NaCL will be present in = 20 X 6.4 = 128 g of solution<\/p>\n<p>Alternate Method<\/p>\n<p>$Weight \\; percent= \\frac {mass \\; of \\; the \\; solute}{mass \\; of \\; the \\; solution} \\times 100$<\/p>\n<p>$5= \\frac {6.4}{x}\u00a0 \\times 100$<\/p>\n<p>or x = 128 g<\/p>\n<p><strong>How to convert Weight percent to mole fraction<\/strong><\/p>\n<p>Now if weight percent of the solution is known, we can calculate the <a href=\"https:\/\/physicscatalyst.com\/chemistry\/mole-fraction.php\"  rel=\"noopener noreferrer\">mole fraction<\/a> of the solution or the solvent easily as explained with below example<\/p>\n<p><strong>Example<\/strong><\/p>\n<p>Calculate the mole fraction of HCl and H<sub>2<\/sub>O in a solution of HCL acid in water, containing 20% HCl by weight.<br \/>\n<strong>Solution<\/strong><br \/>\nThe solution contains 20 grams of HCL acid and 80 grams of water. Also,<br \/>\nMolar mass of HCl is 36.5 grams\/mole.<br \/>\nMolar mass of water is 18 grams\/mole.<br \/>\nMoles of HCl = 20\/36.5 = .547 moles of HCl<br \/>\nMoles of water = 80\/18 = 4.44 moles of H2O.<\/p>\n<p>$ Mole \\; Fraction \\; of \\; solute = \\frac {no \\; of \\; moles \\; of \\; solute}{no\u00a0\\; of \\; moles \\; of \\; solute + no \\; of \\; moles \\; of \\; solvent}$<br \/>\n$Mole \\; fraction \\; of \\; HCl = \\frac {.547}{.547 + 4.44}= .109$<br \/>\n$Mole fraction of H2O = \\frac {4.44}{.547 + 4.44} = .891$<\/p>\n<p>So in short, following steps need to be executed<\/p>\n<p><a href=\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2019\/02\/weight-percent-to-mole-fraction.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-5130\" src=\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2019\/02\/weight-percent-to-mole-fraction.png\" alt=\"\" width=\"601\" height=\"107\" srcset=\"https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2019\/02\/weight-percent-to-mole-fraction.png 601w, https:\/\/physicscatalyst.com\/article\/wp-content\/uploads\/2019\/02\/weight-percent-to-mole-fraction-300x53.png 300w\" sizes=\"auto, (max-width: 601px) 100vw, 601px\" \/><\/a><\/p>\n<ol>\n<li>Find the weight of Solute and solvent from weight percent or mass percent<\/li>\n<li>Find the molecules mass of the solute and solvent<\/li>\n<li>Calculate the moles of solute and solvent using weight and molecular mass<\/li>\n<li>Apply the mole fraction formula to obtain the mole fraction<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Practice Questions<\/strong><\/p>\n<ol>\n<li>\u00a0Find the mass percent if\u00a0 10 g of X is dissolved in\u00a0 80 g of Y?<\/li>\n<li>Calculate the mole fraction of HCl and H<sub>2<\/sub>O in a solution of HCL acid in water, containing 10% HCl by weight.<\/li>\n<li>\u00a0A solution is prepared by adding 2 g of substance to 18 g of water. Calculate the mass percent of solute<\/li>\n<li>20% (w\/W) NaOH solution . Calculate the Mole fraction of NaOH and water<\/li>\n<\/ol>\n<p><strong>Related Articles<\/strong><\/p>\n<p><a href=\"https:\/\/physicscatalyst.com\/chemistry\/molarity-formula.php\"  rel=\"noopener noreferrer\">Molarity Formula<\/a><\/p>\n<p><a href=\"https:\/\/physicscatalyst.com\/chemistry\/molality-formula.php\"  rel=\"noopener noreferrer\">Molality Formula<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this post , we will be checking out what is weight percent or mass percent,what is Weight percent or mass percent formula, how to convert weight percent to mole fraction Weight percent or mass percent is defined as the weight of the solute in 100 gm of the solution $Weight \\; percent= \\frac {mass [&hellip;]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","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":"","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 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{mass&hellip;","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/5129","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=5129"}],"version-history":[{"count":1,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/5129\/revisions"}],"predecessor-version":[{"id":7292,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/5129\/revisions\/7292"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=5129"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=5129"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=5129"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}