{"id":188333,"date":"2025-02-06T18:21:17","date_gmt":"2025-02-06T18:21:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188333"},"modified":"2025-02-06T18:21:20","modified_gmt":"2025-02-06T18:21:20","slug":"fun-sheet-to-accompany-chapter-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/fun-sheet-to-accompany-chapter-3\/","title":{"rendered":"Fun Sheet to Accompany Chapter 3"},"content":{"rendered":"\n<p>Fun Sheet to Accompany Chapter 3<br>31 Atomic Mass<br>1) Chlorine exists as a mixture of 2 major isotopes, Chlorine-35 and Chlorine-37. Calculate the average atontic mass for Chlorine. See Problem 3.5<br>Isotope<br>35<br>17<br>37C1<br>Explain why the atomic masses on the periodic table are decimals.<br>3.2 Molar Mass of an Element and Avogadro&#8217;s Number<br>2) Setting up using the conversion factor between the number of atoms in a mole &amp; Avogadro&#8217;s number, convert 6.00 x 10&#8243; atoms of Cobalt (Co) to moles of cobalt atom. See Problem 3.14<br>3) Setting up using the conversion factarx of molar mass and\/or Avogadro&#8217;s number:<br>a) A modern penny weighs 2.5 g, but contains only 0.63 g of copper(Cu). How many copper atoms are present in a modern penny? See Problem 3.20<br>b) Determine the number of atoms of sulfur in 5.10 moles of sulfur? See Problem 3.13<br>4) Which sample has more atoms: 1.10 g of hydrogen atoma, or 14.7 g of chromium atoms7See Problem 3:21<br>33 Molecular Mass<br>5) Calculate the molecular mass (in amu) and molar mass(in grams) of each of the following CHA<br>substances: See Problem 3.23-24<br>b) K-50<br>c)<br>Cas(PO)<br>Abundance(%)<br>75.53<br>24.47<br>34.968<br>36.956<\/p>\n\n\n\n<p><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\">Question 1: Atomic Mass of Chlorine (Cl)<\/h3>\n\n\n\n<p>Chlorine exists as a mixture of two isotopes, <strong>Chlorine-35<\/strong> and <strong>Chlorine-37<\/strong>, with the following natural abundances:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chlorine-35 has an abundance of <strong>75.53%<\/strong>.<\/li>\n\n\n\n<li>Chlorine-37 has an abundance of <strong>24.47%<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>To calculate the average atomic mass, we use the formula for the weighted average: Average&nbsp;Atomic&nbsp;Mass=(m1\u00d7f1)+(m2\u00d7f2)\\text{Average Atomic Mass} = (m_1 \\times f_1) + (m_2 \\times f_2)<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>m1m_1 and m2m_2 are the atomic masses of Chlorine-35 (34.968 amu) and Chlorine-37 (36.956 amu), respectively.<\/li>\n\n\n\n<li>f1f_1 and f2f_2 are the fractional abundances of Chlorine-35 and Chlorine-37.<\/li>\n<\/ul>\n\n\n\n<p>Substituting the values: Average&nbsp;Atomic&nbsp;Mass=(34.968\u2009amu\u00d70.7553)+(36.956\u2009amu\u00d70.2447)\\text{Average Atomic Mass} = (34.968 \\, \\text{amu} \\times 0.7553) + (36.956 \\, \\text{amu} \\times 0.2447) Average&nbsp;Atomic&nbsp;Mass=26.409\u2009amu+9.038\u2009amu=35.447\u2009amu\\text{Average Atomic Mass} = 26.409 \\, \\text{amu} + 9.038 \\, \\text{amu} = 35.447 \\, \\text{amu}<\/p>\n\n\n\n<p>So, the average atomic mass of Chlorine is <strong>35.447 amu<\/strong>.<\/p>\n\n\n\n<p><strong>Explanation for Decimals on the Periodic Table:<\/strong> Atomic masses on the periodic table are often decimal numbers because they represent weighted averages of the atomic masses of all naturally occurring isotopes of an element. Since elements can have multiple isotopes with different abundances, the atomic mass reflects these variations.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Question 2: Moles of Cobalt (Co)<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Atoms of Cobalt (Co)<\/strong> = 6.00\u00d710246.00 \\times 10^{24}<\/li>\n\n\n\n<li><strong>Avogadro&#8217;s Number<\/strong> = 6.022\u00d710236.022 \\times 10^{23} atoms per mole.<\/li>\n<\/ul>\n\n\n\n<p>To convert atoms to moles, we use the conversion factor: Moles&nbsp;of&nbsp;Co=6.00\u00d71024\u2009atoms6.022\u00d71023\u2009atoms\/mol=9.97\u2009moles\\text{Moles of Co} = \\frac{6.00 \\times 10^{24} \\, \\text{atoms}}{6.022 \\times 10^{23} \\, \\text{atoms\/mol}} = 9.97 \\, \\text{moles}<\/p>\n\n\n\n<p>So, the number of moles of cobalt is <strong>9.97 moles<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Question 3a: Copper Atoms in a Penny<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of copper in the penny = <strong>0.63 g<\/strong><\/li>\n\n\n\n<li>Molar mass of copper (Cu) = <strong>63.55 g\/mol<\/strong><\/li>\n\n\n\n<li>Avogadro\u2019s Number = 6.022\u00d710236.022 \\times 10^{23} atoms\/mol.<\/li>\n<\/ul>\n\n\n\n<p>First, convert grams of copper to moles using molar mass: Moles&nbsp;of&nbsp;Cu=0.63\u2009g63.55\u2009g\/mol=0.00991\u2009mol\\text{Moles of Cu} = \\frac{0.63 \\, \\text{g}}{63.55 \\, \\text{g\/mol}} = 0.00991 \\, \\text{mol}<\/p>\n\n\n\n<p>Next, convert moles to atoms: Atoms&nbsp;of&nbsp;Cu=0.00991\u2009mol\u00d76.022\u00d71023\u2009atoms\/mol=5.96\u00d71021\u2009atoms\\text{Atoms of Cu} = 0.00991 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{atoms\/mol} = 5.96 \\times 10^{21} \\, \\text{atoms}<\/p>\n\n\n\n<p>So, there are approximately <strong>5.96 \u00d7 10\u00b2\u00b9 atoms of copper<\/strong> in the penny.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Question 3b: Atoms of Sulfur in 5.10 Moles<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of sulfur = <strong>5.10 mol<\/strong><\/li>\n\n\n\n<li>Avogadro\u2019s Number = 6.022\u00d710236.022 \\times 10^{23} atoms\/mol.<\/li>\n<\/ul>\n\n\n\n<p>To find the number of atoms: Atoms&nbsp;of&nbsp;S=5.10\u2009mol\u00d76.022\u00d71023\u2009atoms\/mol=3.07\u00d71024\u2009atoms\\text{Atoms of S} = 5.10 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{atoms\/mol} = 3.07 \\times 10^{24} \\, \\text{atoms}<\/p>\n\n\n\n<p>So, there are <strong>3.07 \u00d7 10\u00b2\u2074 atoms of sulfur<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Question 4: More Atoms of Hydrogen or Chromium?<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1.10 g of hydrogen<\/li>\n\n\n\n<li>14.7 g of chromium<\/li>\n<\/ul>\n\n\n\n<p>For hydrogen (H):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of hydrogen = <strong>1.008 g\/mol<\/strong><\/li>\n\n\n\n<li>Moles of hydrogen = 1.10\u2009g1.008\u2009g\/mol=1.092\u2009mol\\frac{1.10 \\, \\text{g}}{1.008 \\, \\text{g\/mol}} = 1.092 \\, \\text{mol}<\/li>\n\n\n\n<li>Atoms of hydrogen = 1.092\u2009mol\u00d76.022\u00d71023\u2009atoms\/mol=6.57\u00d71023\u2009atoms1.092 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{atoms\/mol} = 6.57 \\times 10^{23} \\, \\text{atoms}<\/li>\n<\/ul>\n\n\n\n<p>For chromium (Cr):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of chromium = <strong>52.00 g\/mol<\/strong><\/li>\n\n\n\n<li>Moles of chromium = 14.7\u2009g52.00\u2009g\/mol=0.2827\u2009mol\\frac{14.7 \\, \\text{g}}{52.00 \\, \\text{g\/mol}} = 0.2827 \\, \\text{mol}<\/li>\n\n\n\n<li>Atoms of chromium = 0.2827\u2009mol\u00d76.022\u00d71023\u2009atoms\/mol=1.70\u00d71023\u2009atoms0.2827 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{atoms\/mol} = 1.70 \\times 10^{23} \\, \\text{atoms}<\/li>\n<\/ul>\n\n\n\n<p><strong>Conclusion:<\/strong> There are more atoms in 1.10 g of hydrogen (6.57\u00d710236.57 \\times 10^{23}) than in 14.7 g of chromium (1.70\u00d710231.70 \\times 10^{23}).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Question 5: Molecular Mass<\/h3>\n\n\n\n<p>For molecular mass calculations, sum the atomic masses of all atoms in the formula. For example:<\/p>\n\n\n\n<p>For C2H6\\text{C}_2\\text{H}_6 (ethane):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon = 12.01 g\/mol, Hydrogen = 1.008 g\/mol.<\/li>\n\n\n\n<li>Molecular mass = 2\u00d712.01+6\u00d71.008=24.02+6.048=30.07\u2009g\/mol2 \\times 12.01 + 6 \\times 1.008 = 24.02 + 6.048 = 30.07 \\, \\text{g\/mol}.<\/li>\n<\/ul>\n\n\n\n<p>This is the process for determining molecular mass.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Fun Sheet to Accompany Chapter 331 Atomic Mass1) Chlorine exists as a mixture of 2 major isotopes, Chlorine-35 and Chlorine-37. Calculate the average atontic mass for Chlorine. See Problem 3.5Isotope351737C1Explain why the atomic masses on the periodic table are decimals.3.2 Molar Mass of an Element and Avogadro&#8217;s Number2) Setting up using the conversion factor between [&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-188333","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188333","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=188333"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188333\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188333"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188333"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188333"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}