{"id":234353,"date":"2025-06-14T04:53:17","date_gmt":"2025-06-14T04:53:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234353"},"modified":"2025-06-14T04:53:19","modified_gmt":"2025-06-14T04:53:19","slug":"the-calculated-masses-from-iii-of-anhydrous-salt-and-water-and-the-identity-of-the-hydrated-use-compound-on-the-data-slip-supplied-with-the-experiment-determine-the-number-of-hydrated-waters-in","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/14\/the-calculated-masses-from-iii-of-anhydrous-salt-and-water-and-the-identity-of-the-hydrated-use-compound-on-the-data-slip-supplied-with-the-experiment-determine-the-number-of-hydrated-waters-in\/","title":{"rendered":"The calculated masses (from III) of anhydrous salt and water, and the identity of the hydrated Use- compound (on the data slip supplied with the experiment), determine the number of hydrated waters in the hydrated compound of anhydrous salt."},"content":{"rendered":"\n<p>The calculated masses (from III) of anhydrous salt and water, and the identity of the hydrated Use- compound (on the data slip supplied with the experiment), determine the number of hydrated waters in the hydrated compound of anhydrous salt. Formula for anhydrous salt mass = water lost. The moles of anhydrous salt and the moles of water are calculated. Determine the simplest mole &#8220;hydrated water (per mole) of compound and get the moles. Formula of hydrated compound calculation determines the formula for hydrated compound [I on data slip]. Slip number Data for hydrate (supplied of hydrated compound before heating 3ZS99 mass upon heating water 41260623 loss IeDS of residue after heating: anhydrous salt. Calculate the moles of anhydrous salt moles ullei. Determine the empirical formula of hydrated compound. Formula of hydrated compound.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the formula of a hydrated compound, follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given Data:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mass of hydrated compound (before heating):<\/strong> 3.2599 g<\/li>\n\n\n\n<li><strong>Mass of anhydrous salt (after heating):<\/strong> 2.60623 g<\/li>\n\n\n\n<li><strong>Mass of water lost =<\/strong> 3.2599 g \u2013 2.60623 g = <strong>0.65367 g<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Calculate the moles of anhydrous salt<\/strong><\/h3>\n\n\n\n<p>To calculate moles of the anhydrous salt, we need its molar mass. Suppose from the data slip the <strong>anhydrous salt is CuSO\u2084<\/strong>, whose molar mass is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cu = 63.55 g\/mol<\/li>\n\n\n\n<li>S = 32.07 g\/mol<\/li>\n\n\n\n<li>O\u2084 = 4 \u00d7 16.00 = 64.00 g\/mol<\/li>\n\n\n\n<li><strong>Total = 63.55 + 32.07 + 64.00 = 159.62 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;CuSO\u2084=2.60623&nbsp;g159.62&nbsp;g\/mol=0.01633&nbsp;mol\\text{Moles of CuSO\u2084} = \\frac{2.60623\\text{ g}}{159.62\\text{ g\/mol}} = 0.01633 \\text{ mol}Moles&nbsp;of&nbsp;CuSO\u2084=159.62&nbsp;g\/mol2.60623&nbsp;g\u200b=0.01633&nbsp;mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Calculate the moles of water lost<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of H\u2082O = 18.02 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;H\u2082O=0.65367&nbsp;g18.02&nbsp;g\/mol=0.03627&nbsp;mol\\text{Moles of H\u2082O} = \\frac{0.65367\\text{ g}}{18.02\\text{ g\/mol}} = 0.03627 \\text{ mol}Moles&nbsp;of&nbsp;H\u2082O=18.02&nbsp;g\/mol0.65367&nbsp;g\u200b=0.03627&nbsp;mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Determine the mole ratio of water to salt<\/strong><\/h3>\n\n\n\n<p>Mole&nbsp;ratio=0.036270.01633\u22482.22\\text{Mole ratio} = \\frac{0.03627}{0.01633} \u2248 2.22Mole&nbsp;ratio=0.016330.03627\u200b\u22482.22<\/p>\n\n\n\n<p>This ratio is approximately <strong>2.22 : 1<\/strong>, which is close to <strong>2:1<\/strong>. Since empirical formulas require whole numbers, round to the <strong>nearest whole number<\/strong>.<\/p>\n\n\n\n<p>So, <strong>2 moles of water per mole of salt<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Write the formula of the hydrated compound<\/strong><\/h3>\n\n\n\n<p>If the anhydrous salt is <strong>CuSO\u2084<\/strong>, and there are <strong>2 water molecules<\/strong> per formula unit, the formula of the hydrated compound is: CuSO\u2084\\cdotp2H\u2082O\\boxed{\\text{CuSO\u2084\u00b72H\u2082O}}CuSO\u2084\\cdotp2H\u2082O\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p><strong>Empirical formula of hydrated compound: CuSO\u2084\u00b72H\u2082O<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300 words):<\/strong><\/h3>\n\n\n\n<p>The empirical formula of a hydrated compound tells us the simplest whole-number ratio of the salt to water molecules in the compound. In this experiment, the mass of the hydrate before heating was measured to be 3.2599 g, and after heating, the mass was reduced to 2.60623 g. This mass difference is due to the loss of water during heating. The mass of the water lost is found by subtracting the mass after heating from the initial mass, resulting in 0.65367 g of water.<\/p>\n\n\n\n<p>To determine how many moles of each component are present, divide the mass of each substance by its molar mass. The molar mass of copper(II) sulfate (CuSO\u2084) is 159.62 g\/mol, and water\u2019s molar mass is 18.02 g\/mol. This gives us 0.01633 mol of CuSO\u2084 and 0.03627 mol of H\u2082O. To find the number of water molecules per salt formula unit, divide the moles of water by the moles of anhydrous salt. The ratio is approximately 2.22:1, which we round to 2:1 for simplicity and accuracy in chemistry.<\/p>\n\n\n\n<p>Thus, for every mole of CuSO\u2084, there are 2 moles of water, leading to the hydrated formula CuSO\u2084\u00b72H\u2082O. This empirical formula provides essential insight into the composition of the hydrated compound and is crucial in both laboratory and industrial settings, where knowing the exact hydration state affects reactivity, storage, and usage.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-531.jpeg\" alt=\"\" class=\"wp-image-234355\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The calculated masses (from III) of anhydrous salt and water, and the identity of the hydrated Use- compound (on the data slip supplied with the experiment), determine the number of hydrated waters in the hydrated compound of anhydrous salt. Formula for anhydrous salt mass = water lost. The moles of anhydrous salt and the moles [&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-234353","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234353","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=234353"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234353\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234353"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234353"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234353"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}