{"id":386,"date":"2017-08-21T06:46:37","date_gmt":"2017-08-21T06:46:37","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=386"},"modified":"2017-10-20T07:42:33","modified_gmt":"2017-10-20T07:42:33","slug":"chemical-calculations-and-yields","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/chemical-calculations-and-yields\/","title":{"rendered":"Chemical Calculations and Yields"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/the-mole\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/kinetics\">Next Lesson\u00a0\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Chemical Calculations and Yields<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review how to balance chemical reactions and use mole ratios and molar masses to determine theoretical yields.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>The mole concept is used in chemistry as a way to count atoms and molecules by weighing them.<\/li>\n<li>The number of moles of a substance can be determined from the amount and the molar mass.<\/li>\n<li>If the number of moles of a substance is known, the number of grams present can be determined<\/li>\n<\/ul>\n<section>\n<h3>Chemical Calculations<\/h3>\n<p>The\u00a0coefficients in a balanced equation represent the number\u00a0of molecules or atoms that are reacting and are produced. For example, in the\u00a0formation of\u00a0water, 2 molecules of hydrogen gas react with 1 molecule of oxygen\u00a0producing 2\u00a0molecules of water. If 4 molecules of hydrogen gas are present, then 2\u00a0molecules of oxygen gas will be needed to produce 4 molecules of water.<\/p>\n<p class=\"center\">2H<sub>2<\/sub>(<em>g<\/em>)\u00a0+ O<sub>2<\/sub>(<em>g<\/em>) \u2192 2H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p>But\u00a0what if there are 400 molecules of hydrogen gas? How much\u00a0oxygen gas would be required to use up all the hydrogen gas?<\/p>\n<p>It\u00a0can be seen that the ratio of hydrogen\u00a0molecules\u00a0to oxygen\u00a0molecules\u00a0required\u00a0is always 2:1. But in the laboratory, when a measurable amount of\u00a0reactants are\u00a0necessary, it is advantageous to use moles to count the molecules. So 2\u00a0moles\u00a0of H<sub>2<\/sub> (representing 1.20 \u00d7 10<sup>24\u00a0<\/sup>molecules) would react\u00a0with 1 mole of O<sub>2<\/sub> (representing 6.02\u00a0\u00d710<sup>23<\/sup> molecules) and\u00a0produce 2 moles of H<sub>2<\/sub>O (representing 1.20\u00a0\u00d7 10<sup>24\u00a0<\/sup>molecules).<\/p>\n<p>In\u00a0other words, all coefficients in a balanced equation\u00a0represent the number of moles of substances as well as the number of molecules, and\u00a0can give\u00a0a ratio between the compounds and elements in a reaction. In the case\u00a0of the\u00a0reaction of sodium with chlorine to produce sodium chloride:<\/p>\n<p class=\"center\">2Na(<em>s<\/em>)\u00a0+ Cl<sub>2<\/sub> (<em>g<\/em>) \u2192 2NaCl(<em>s<\/em>)<\/p>\n<p>The ratio\u00a0of moles of Na to moles of Cl<sub>2<\/sub> is 2:1, but the\u00a0ratio of moles of Na to moles of NaCl is 2:2, and the ratio of moles of Cl<sub>2<\/sub> to moles of NaCl is 1:2.\u00a0These mole ratios can be developed for any balanced equation and can be\u00a0used to\u00a0determine how much product is made from a given amount of starting\u00a0material.<\/p>\n<h4>Sample Problem<\/h4>\n<p>How\u00a0many moles of Cl<sub>2<\/sub> would be required if 5.0\u00a0mol of Na was completely reacted?<\/p>\n<p>The mole ratio of Na:Cl<sub>2 <\/sub>is\u00a02:1,\u00a0so we multiply the 5.0 reacted moles of Na by the mole ratio:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_001-1.gif\" width=\"398\" height=\"42\" \/><\/center>Please note that the mole\u00a0ratios used have an infinite\u00a0number of significant figures and the number of significant figures in\u00a0the\u00a0answer will be determined by the number in the starting figure.<\/p>\n<p>A chemist measures out molecules or moles not by counting, but by weighing. The molar mass\u00a0of a<br \/>\ncompound provides the connection between the mass\u00a0of a sample of a substance and the number of moles\u00a0of that substance in the sample.<\/p>\n<p>The molar mass can be used\u00a0to\u00a0set up more realistic calculations from equations.\u00a0For\u00a0example, what if 46.0 g of sodium metal were put into a\u00a0container of chlorine gas and allowed to completely react? How many\u00a0grams of\u00a0sodium chloride would be produced? Using a periodic table, the molar\u00a0mass of\u00a0sodium is found to be 23.0 g\/mol. The molar mass of sodium chloride is\u00a058.5 g\/mol. This additional information is used to complete the\u00a0calculation as\u00a0before.<\/p>\n<p class=\"center\">2Na(<em>s<\/em>) + Cl<sub>2<\/sub>(<em>g<\/em>) \u2192 2NaCl(<em>s<\/em>)<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_002.gif\" width=\"508\" height=\"50\" \/><\/center>Furthermore, the amount of\u00a0chlorine gas needed could be\u00a0determined using a similar calculation. Calculation of volumes of gases\u00a0will be\u00a0addressed in the chapter on gases, but a similar calculation is used.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>How many grams of chlorine gas are needed to completely react with 46.0 g Na?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Calculate the molar mass of Cl<sub>2<\/sub> as 71.0 g\/mol. Then set up the problem\u00a0as before:<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_003-1.gif\" width=\"518\" height=\"50\" \/><\/p>\n<\/section>\n<h3>Percent Yield<\/h3>\n<p>When calculating the\u00a0expected amount of product from a\u00a0given amount of one of the starting materials, the calculated value is\u00a0called\u00a0the <abbr title=\"The amount of product that would be expected to be produced if all conditions in the reaction were to be, in theory, perfect\">theoretical\u00a0yield<\/abbr>. That is the amount that\u00a0would be expected to be\u00a0produced\u00a0as calculated from the balanced equation.\u00a0However, in reality, many conditions may be present to prevent\u00a0obtaining the\u00a0theoretical yield. For example the starting materials\u00a0or product may\u00a0have been\u00a0impure,\u00a0the\u00a0recovery of the product may have been incomplete, or the reaction may have reached\u00a0equilibrium before it was complete. When\u00a0a reaction is finished, and the actual yield of product is determined, that amount\u00a0is\u00a0called the <abbr title=\"The actual amount of product that is made during a chemical reaction in the laboratory\">experimental\u00a0yield<\/abbr>.\u00a0In order to see how\u00a0effective the experimental conditions\u00a0were in obtaining the desired amount of product, chemists calculate the\u00a0percentage yield, or <abbr title=\"The percent of the expected amount (theoretical yield) that is obtained\">percent yield<\/abbr>. The percent\u00a0yield is\u00a0the\u00a0percent of the expected amount that is obtained. Thus the percentage yield is calculated by\u00a0taking the actual yield and dividing by the theoretical yield, then\u00a0multiplying by 100 to convert to percentage.<\/p>\n<p>Often,\u00a0the\u00a0percent yield is less than 100%, but sometimes a\u00a0perplexing percent yield of over 100% is obtained. Closer examination usually reveals a product that\u00a0is contaminated with impurities or that is not completely dry.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>In the production of lime,\u00a0CaO, from the heating of\u00a0limestone, CaCO<sub>3<\/sub>, carbon dioxide gas is also\u00a0produced: CaCO<sub>3<\/sub>(<em>s<\/em>) \u2192 CaO(<em>s<\/em>)+ CO<sub>2<\/sub>(<em>g<\/em>).<\/p>\n<p>In an experiment\u00a0using 2.00 kg of limestone,\u00a00.982 kg of lime was produced.\u00a0What is the percentage yield?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The first\u00a0step to getting the\u00a0solution consists of calculating the theoretical yield using the\u00a0information given, 2.00 kg of CaCO<sub>3<\/sub> and the correct mole ratio and\u00a0molar masses.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_004-1.gif\" width=\"613\" height=\"50\" \/><\/p>\n<p>But only 0.982 kg was produced, so the percentage yield is calculated by\u00a0taking the actual yield and dividing by the theoretical yield, then\u00a0multiplying by 100 to convert to percentage.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_005.gif\" width=\"260\" height=\"48\" \/><\/p>\n<\/div>\n<\/section>\n<h3>Limiting and Excess Reagents<\/h3>\n<p>So far the discussion of calculations of yields has been based on a given quantity of one starting material, assuming that there is enough of the other reactant(s) to consume the starting material in question. However, that is not always the case. Many times the amounts of both reactants are given, and they are not <abbr title=\"The exact number of moles of each reactant to complete the reaction according to mole ratio from the balanced equation\">stoichiometrically<\/abbr> equivalent\u2014that is, present in the correct mole ratio for a complete reaction. In such a case, one is called the <abbr title=\"Reactant which is totally consumed during a chemical reaction; It limits the amount of product and the extent of the reaction.\">limiting reactant<\/abbr> and the other is called the <abbr title=\"The reactant which remains after a reaction is completed\">excess reactant<\/abbr>. The limiting reactant is the one which is used up completely and thus limits the maximum amount of product that is formed. The reactant left over is called the excess reactant.<\/p>\n<p>To determine the maximum amount of product, a chemical calculation must be performed as before. One of the easiest ways to determine the limiting reactant is to perform two calculations, beginning each with the amount of given reactant, and comparing the amounts of the desired product. The reactant which produced the least amount of product is the limiting reactant and this amount is the <em>maximum<\/em> of the product that can be produced.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Magnesium ribbon will react with aqueous hydrogen chloride to produce\u00a0hydrogen\u00a0gas and magnesium chloride.<\/p>\n<p>If 15.0 g Mg(<i>s<\/i>) reacts with 0.25 mol of HCl(<em>aq<\/em>),\u00a0how many moles of hydrogen gas are produced?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>This is a single replacement\u00a0reaction which can be written as shown:<\/p>\n<p class=\"center\">Mg(<em>s<\/em>) + 2HCl(<em>aq<\/em>) \u2192 H<sub>2<\/sub>(<em>g<\/em>) +\u00a0MgCl<sub>2<\/sub>(<em>aq<\/em>)<\/p>\n<p>Comparing the amounts produced in each case indicates that no\u00a0more than 0.125 mol of hydrogen gas can be produced because there is\u00a0not enough HCl present to produce any more.<\/p>\n<p>The hydrogen chloride is\u00a0the limiting reactant and since the Mg will be in excess not all of it\u00a0will react. The amount left over can also be calculated.<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_006.gif\" \/><\/p>\n<p>Moles Mg in\u00a0excess would be found by calculating the moles present (0.617) and\u00a0subtracting the amount reacted (0.125mol); 0.492 mole Mg will be left\u00a0which corresponds to 12.0g (rounded to three significant digits from\u00a011.9556)<\/p>\n<\/div>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>What is the limiting reactant when 3.0 g of Al reacts\u00a0with 5.6 g of Cl<sub>2<\/sub>(<em>g<\/em>)\u00a0producing AlCl<sub>3<\/sub>? And how\u00a0much of the excess reactant will be left?<\/p>\n<ol>\n<li>LR Cl<sub>2<\/sub>, 0.057 mol Al in excess<\/li>\n<li>LR Cl<sub>2<\/sub>, 0.11 mol Al in excess<\/li>\n<li>LR Al, 0.052 mol Cl<sub>2<\/sub> in excess<\/li>\n<li>LR Al, 0.165 mol Cl<sub>2<\/sub> in excess<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct answer is A.\u00a0That\u2019s because the moles of Al needed to react with 5.6 g of\u00a0chlorine is 0.053 mol:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_012.gif\" width=\"512\" height=\"50\" \/><\/p>\n<p>The amount of aluminum available to react is 0.11 mol:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s4_013.gif\" width=\"226\" height=\"48\" \/><\/p>\n<p>Since aluminum\u00a0is present in a greater amount than needed, the chlorine must be the\u00a0limiting reactant.<\/p>\n<p>To determine the amount of excess aluminum present,\u00a0the amount of aluminum reacted must be calculated and subtracted from\u00a0the starting amount. The number of moles of aluminum present is 0.11\u00a0moles. The amount required is 0.053 moles. The difference is 0.057\u00a0moles which is 1.5 g.<\/p>\n<\/div>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/the-mole\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/kinetics\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Chemical Calculations and Yields Objective In this lesson we will review how to balance chemical reactions and use mole ratios and molar masses to determine theoretical yields. Previously we covered&#8230; The mole concept is used in chemistry as a way to count atoms and molecules by weighing them. The number [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-386","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/386","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=386"}],"version-history":[{"count":20,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/386\/revisions"}],"predecessor-version":[{"id":971,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/386\/revisions\/971"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=386"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}