{"id":190,"date":"2017-12-14T21:28:45","date_gmt":"2017-12-14T21:28:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/yields\/"},"modified":"2017-12-19T15:17:39","modified_gmt":"2017-12-19T15:17:39","slug":"yields","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/yields\/","title":{"raw":"Yields","rendered":"Yields"},"content":{"raw":"<div id=\"ball-ch05_s05\" class=\"section\" lang=\"en\">\r\n<div id=\"ball-ch05_s05_n01\" class=\"learning_objectives editable block\">\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n1. Define and determine theoretical yields, actual yields, and percent yields.\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"ball-ch05_s05_p01\" class=\"para editable block\">In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions: (1) the reaction goes exactly as written, and (2) the reaction proceeds completely. In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO<sub class=\"subscript\">2<\/sub>. However, we will continue to ignore side reactions, unless otherwise noted.<\/p>\r\n<p id=\"ball-ch05_s05_p02\" class=\"para editable block\">The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons (some of which we will consider in <a class=\"xref\" href=\"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/introduction-to-chemical-equilibrium\/\">Chapter 13 \"Chemical Equilibrium\"<\/a>). When we calculate an amount of product assuming that all the reactant reacts, we calculate the <span class=\"margin_term\"><a class=\"glossterm\">theoretical yield<\/a><\/span>, an amount that is theoretically produced as calculated using the balanced chemical reaction.<\/p>\r\n<p id=\"ball-ch05_s05_p03\" class=\"para editable block\">In many cases, however, this is not what really happens. In many cases, less\u2014sometimes much less\u2014of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the <span class=\"margin_term\"><a class=\"glossterm\">actual yield<\/a><\/span>. By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made.<\/p>\r\n<p id=\"ball-ch05_s05_p04\" class=\"para editable block\">Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The <span class=\"margin_term\"><a class=\"glossterm\">percent yield<\/a><\/span>\u00a0is a comparison between the actual yield and the theoretical yield and is defined as<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-22-at-12.54.26-PM.png\"><img class=\"aligncenter wp-image-3742\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212838\/Screen-Shot-2014-07-22-at-12.54.26-PM-1.png\" alt=\"Screen Shot 2014-07-22 at 12.54.26 PM\" width=\"373\" height=\"73\" \/><\/a>\r\n<p id=\"ball-ch05_s05_p05\" class=\"para editable block\">It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%\u2014again, if percent yield is greater than 100%, an error has been made.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 13<\/h3>\r\n<p id=\"ball-ch05_s05_p06\" class=\"para\">A worker reacts 30.5 g of Zn with nitric acid and evapourates the remaining water to obtain 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. What are the theoretical yield, the actual yield, and the percent yield?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 HNO<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span>\r\n<p class=\"simpara\">Solution<\/p>\r\n<p id=\"ball-ch05_s05_p07\" class=\"para\">A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g\/mol) and Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> (189.41 g\/mol). In three steps, the mass-mass calculation is<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/305GZN.png\"><img class=\"aligncenter wp-image-3743\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212841\/305GZN-1024x100-1.png\" alt=\"305GZN\" width=\"600\" height=\"59\" \/><\/a>\r\n<p id=\"ball-ch05_s05_p08\" class=\"para\">Thus, the theoretical yield is 88.3 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/642GZN.png\"><img class=\"aligncenter wp-image-3744\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212844\/642GZN-1.png\" alt=\"642GZN\" width=\"282\" height=\"70\" \/><\/a>\r\n<p id=\"ball-ch05_s05_p09\" class=\"para\">The worker achieved almost three-fourths of the possible yield.<\/p>\r\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\r\n<p id=\"ball-ch05_s05_p10\" class=\"para\">A synthesis produced 2.05 g of NH<sub class=\"subscript\">3<\/sub> from 16.5 g of N<sub class=\"subscript\">2<\/sub>. What is the theoretical yield and the percent yield?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span>\r\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\r\n<p id=\"ball-ch05_s05_p11\" class=\"para\">theoretical yield = 20.1 g; percent yield = 10.2%<\/p>\r\n\r\n<\/div>\r\n<div id=\"ball-ch05_s05_n03\" class=\"callout block\">\r\n<h3 class=\"title\">Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification<\/h3>\r\n<p id=\"ball-ch05_s05_p12\" class=\"para\">Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.<\/p>\r\n<p id=\"ball-ch05_s05_p13\" class=\"para\">Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C<sub class=\"subscript\">13<\/sub>H<sub class=\"subscript\">21<\/sub>NO<sub class=\"subscript\">2<\/sub>; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:<\/p>\r\n\r\n<div class=\"informaltable\">\r\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td>impure albuterol \u2192\u00a0intermediate A<\/td>\r\n<td>percent yield = 70%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate A \u2192\u00a0intermediate B<\/td>\r\n<td>percent yield = 100%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate B \u2192\u00a0intermediate C<\/td>\r\n<td>percent yield = 40%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate C \u2192\u00a0intermediate D<\/td>\r\n<td>percent yield = 72%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>intermediate D \u2192\u00a0purified albuterol<\/td>\r\n<td>percent yield = 35%<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">overall percent yield = 70% \u00d7 100% \u00d7 40% \u00d7 72% \u00d7 35% = 7.5%<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<p id=\"ball-ch05_s05_p14\" class=\"para\">That is, only about <em class=\"emphasis\">one-fourteenth<\/em> of the original material was turned into the purified drug. This gives you one reason why some drugs are so expensive; a lot of material is lost in making a high-purity pharmaceutical.<\/p>\r\n\r\n<\/div>\r\n<div id=\"ball-ch05_s05_n04\" class=\"key_takeaways editable block\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Takeaways<\/h3>\r\n<ul id=\"ball-ch05_s05_l02\" class=\"itemizedlist\">\r\n \t<li>Theoretical yield is what you calculate the yield will be using the balanced chemical reaction.<\/li>\r\n \t<li>Actual yield is what you actually get in a chemical reaction.<\/li>\r\n \t<li>Percent yield is a comparison of the actual yield with the theoretical yield.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div id=\"ball-ch05_s05_qs01\" class=\"qandaset block\">\r\n<ol id=\"ball-ch05_s05_qs01_qd01\" class=\"qandadiv\">\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p1\" class=\"para\">What is the difference between the theoretical yield and the actual yield?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p3\" class=\"para\">What is the difference between the actual yield and the percent yield?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa03\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p5\" class=\"para\">A worker isolates 2.675 g of SiF<sub class=\"subscript\">4<\/sub> after reacting 2.339 g of SiO<sub class=\"subscript\">2<\/sub> with HF. What are the theoretical yield and the actual yield?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">SiO<sub class=\"subscript\">2<\/sub>(s) +\u00a04 HF(g) \u2192\u00a0SiF<sub class=\"subscript\">4<\/sub>(g) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa04\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p7\" class=\"para\">A worker synthesizes aspirin, C<sub class=\"subscript\">9<\/sub>H<sub class=\"subscript\">8<\/sub>O<sub class=\"subscript\">4<\/sub>, according to this chemical equation. If 12.66 g of C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">3<\/sub> are reacted and 12.03 g of aspirin are isolated, what are the theoretical yield and the actual yield?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">3<\/sub> \u2192\u00a0C<sub class=\"subscript\">9<\/sub>H<sub class=\"subscript\">8<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a0HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa05\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p9\" class=\"para\">A chemist decomposes 1.006 g of NaHCO<sub class=\"subscript\">3<\/sub> and obtains 0.0334 g of Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>. What are the theoretical yield and the actual yield?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 NaHCO<sub class=\"subscript\">3<\/sub>(s) \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0CO<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa06\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p11\" class=\"para\">A chemist combusts a 3.009 g sample of C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">12<\/sub> and obtains 3.774 g of H<sub class=\"subscript\">2<\/sub>O. What are the theoretical yield and the actual yield?<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">12<\/sub>(\u2113) +\u00a08 O<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a05 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa07\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p13\" class=\"para\">What is the percent yield in Exercise 3?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa08\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p15\" class=\"para\">What is the percent yield in Exercise 4?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa09\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p17\" class=\"para\">What is the percent yield in Exercise 5?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch05_s05_qs01_qd01_qa10\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch05_s05_qs01_p19\" class=\"para\">What is the percent yield in Exercise 6?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<b>Answers<\/b>\r\n\r\n<strong>1.<\/strong>\r\n\r\nTheoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.\r\n\r\n<strong>3.<\/strong>\r\n\r\ntheoretical yield = 4.052 g; actual yield = 2.675 g\r\n\r\n<strong>5.<\/strong>\r\n\r\ntheoretical yield = 0.635 g; actual yield = 0.0334 g\r\n\r\n<strong>7.<\/strong>\r\n\r\n66.02%\r\n\r\n<strong>9.<\/strong>\r\n\r\n5.26%\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"ball-ch05_s05\" class=\"section\" lang=\"en\">\n<div id=\"ball-ch05_s05_n01\" class=\"learning_objectives editable block\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>1. Define and determine theoretical yields, actual yields, and percent yields.<\/p>\n<\/div>\n<\/div>\n<p id=\"ball-ch05_s05_p01\" class=\"para editable block\">In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions: (1) the reaction goes exactly as written, and (2) the reaction proceeds completely. In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO<sub class=\"subscript\">2<\/sub>. However, we will continue to ignore side reactions, unless otherwise noted.<\/p>\n<p id=\"ball-ch05_s05_p02\" class=\"para editable block\">The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons (some of which we will consider in <a class=\"xref\" href=\"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/introduction-to-chemical-equilibrium\/\">Chapter 13 &#8220;Chemical Equilibrium&#8221;<\/a>). When we calculate an amount of product assuming that all the reactant reacts, we calculate the <span class=\"margin_term\"><a class=\"glossterm\">theoretical yield<\/a><\/span>, an amount that is theoretically produced as calculated using the balanced chemical reaction.<\/p>\n<p id=\"ball-ch05_s05_p03\" class=\"para editable block\">In many cases, however, this is not what really happens. In many cases, less\u2014sometimes much less\u2014of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the <span class=\"margin_term\"><a class=\"glossterm\">actual yield<\/a><\/span>. By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made.<\/p>\n<p id=\"ball-ch05_s05_p04\" class=\"para editable block\">Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The <span class=\"margin_term\"><a class=\"glossterm\">percent yield<\/a><\/span>\u00a0is a comparison between the actual yield and the theoretical yield and is defined as<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-22-at-12.54.26-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3742\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212838\/Screen-Shot-2014-07-22-at-12.54.26-PM-1.png\" alt=\"Screen Shot 2014-07-22 at 12.54.26 PM\" width=\"373\" height=\"73\" \/><\/a><\/p>\n<p id=\"ball-ch05_s05_p05\" class=\"para editable block\">It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%\u2014again, if percent yield is greater than 100%, an error has been made.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 13<\/h3>\n<p id=\"ball-ch05_s05_p06\" class=\"para\">A worker reacts 30.5 g of Zn with nitric acid and evapourates the remaining water to obtain 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. What are the theoretical yield, the actual yield, and the percent yield?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 HNO<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch05_s05_p07\" class=\"para\">A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g\/mol) and Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> (189.41 g\/mol). In three steps, the mass-mass calculation is<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/305GZN.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3743\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212841\/305GZN-1024x100-1.png\" alt=\"305GZN\" width=\"600\" height=\"59\" \/><\/a><\/p>\n<p id=\"ball-ch05_s05_p08\" class=\"para\">Thus, the theoretical yield is 88.3 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/642GZN.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3744\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14212844\/642GZN-1.png\" alt=\"642GZN\" width=\"282\" height=\"70\" \/><\/a><\/p>\n<p id=\"ball-ch05_s05_p09\" class=\"para\">The worker achieved almost three-fourths of the possible yield.<\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch05_s05_p10\" class=\"para\">A synthesis produced 2.05 g of NH<sub class=\"subscript\">3<\/sub> from 16.5 g of N<sub class=\"subscript\">2<\/sub>. What is the theoretical yield and the percent yield?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch05_s05_p11\" class=\"para\">theoretical yield = 20.1 g; percent yield = 10.2%<\/p>\n<\/div>\n<div id=\"ball-ch05_s05_n03\" class=\"callout block\">\n<h3 class=\"title\">Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification<\/h3>\n<p id=\"ball-ch05_s05_p12\" class=\"para\">Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.<\/p>\n<p id=\"ball-ch05_s05_p13\" class=\"para\">Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C<sub class=\"subscript\">13<\/sub>H<sub class=\"subscript\">21<\/sub>NO<sub class=\"subscript\">2<\/sub>; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:<\/p>\n<div class=\"informaltable\">\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td>impure albuterol \u2192\u00a0intermediate A<\/td>\n<td>percent yield = 70%<\/td>\n<\/tr>\n<tr>\n<td>intermediate A \u2192\u00a0intermediate B<\/td>\n<td>percent yield = 100%<\/td>\n<\/tr>\n<tr>\n<td>intermediate B \u2192\u00a0intermediate C<\/td>\n<td>percent yield = 40%<\/td>\n<\/tr>\n<tr>\n<td>intermediate C \u2192\u00a0intermediate D<\/td>\n<td>percent yield = 72%<\/td>\n<\/tr>\n<tr>\n<td>intermediate D \u2192\u00a0purified albuterol<\/td>\n<td>percent yield = 35%<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">overall percent yield = 70% \u00d7 100% \u00d7 40% \u00d7 72% \u00d7 35% = 7.5%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p id=\"ball-ch05_s05_p14\" class=\"para\">That is, only about <em class=\"emphasis\">one-fourteenth<\/em> of the original material was turned into the purified drug. This gives you one reason why some drugs are so expensive; a lot of material is lost in making a high-purity pharmaceutical.<\/p>\n<\/div>\n<div id=\"ball-ch05_s05_n04\" class=\"key_takeaways editable block\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch05_s05_l02\" class=\"itemizedlist\">\n<li>Theoretical yield is what you calculate the yield will be using the balanced chemical reaction.<\/li>\n<li>Actual yield is what you actually get in a chemical reaction.<\/li>\n<li>Percent yield is a comparison of the actual yield with the theoretical yield.<\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div id=\"ball-ch05_s05_qs01\" class=\"qandaset block\">\n<ol id=\"ball-ch05_s05_qs01_qd01\" class=\"qandadiv\">\n<li id=\"ball-ch05_s05_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p1\" class=\"para\">What is the difference between the theoretical yield and the actual yield?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p3\" class=\"para\">What is the difference between the actual yield and the percent yield?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p5\" class=\"para\">A worker isolates 2.675 g of SiF<sub class=\"subscript\">4<\/sub> after reacting 2.339 g of SiO<sub class=\"subscript\">2<\/sub> with HF. What are the theoretical yield and the actual yield?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">SiO<sub class=\"subscript\">2<\/sub>(s) +\u00a04 HF(g) \u2192\u00a0SiF<sub class=\"subscript\">4<\/sub>(g) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p7\" class=\"para\">A worker synthesizes aspirin, C<sub class=\"subscript\">9<\/sub>H<sub class=\"subscript\">8<\/sub>O<sub class=\"subscript\">4<\/sub>, according to this chemical equation. If 12.66 g of C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">3<\/sub> are reacted and 12.03 g of aspirin are isolated, what are the theoretical yield and the actual yield?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">6<\/sub>O<sub class=\"subscript\">3<\/sub> \u2192\u00a0C<sub class=\"subscript\">9<\/sub>H<sub class=\"subscript\">8<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a0HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p9\" class=\"para\">A chemist decomposes 1.006 g of NaHCO<sub class=\"subscript\">3<\/sub> and obtains 0.0334 g of Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>. What are the theoretical yield and the actual yield?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 NaHCO<sub class=\"subscript\">3<\/sub>(s) \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0CO<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p11\" class=\"para\">A chemist combusts a 3.009 g sample of C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">12<\/sub> and obtains 3.774 g of H<sub class=\"subscript\">2<\/sub>O. What are the theoretical yield and the actual yield?<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">12<\/sub>(\u2113) +\u00a08 O<sub class=\"subscript\">2<\/sub>(g) \u2192\u00a05 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p13\" class=\"para\">What is the percent yield in Exercise 3?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p15\" class=\"para\">What is the percent yield in Exercise 4?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p17\" class=\"para\">What is the percent yield in Exercise 5?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch05_s05_qs01_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch05_s05_qs01_p19\" class=\"para\">What is the percent yield in Exercise 6?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p><b>Answers<\/b><\/p>\n<p><strong>1.<\/strong><\/p>\n<p>Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>theoretical yield = 4.052 g; actual yield = 2.675 g<\/p>\n<p><strong>5.<\/strong><\/p>\n<p>theoretical yield = 0.635 g; actual yield = 0.0334 g<\/p>\n<p><strong>7.<\/strong><\/p>\n<p>66.02%<\/p>\n<p><strong>9.<\/strong><\/p>\n<p>5.26%<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-190\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":23485,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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