{"id":257765,"date":"2025-07-17T17:22:39","date_gmt":"2025-07-17T17:22:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257765"},"modified":"2025-07-17T17:22:41","modified_gmt":"2025-07-17T17:22:41","slug":"help-with-calculations-of-greener-bromination-of-trans-stilbene-synthesizing-trans-12-dibromo-12-diphenylethane-my-initial-mass-of-trans-stilbene-0-2033-grams","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/help-with-calculations-of-greener-bromination-of-trans-stilbene-synthesizing-trans-12-dibromo-12-diphenylethane-my-initial-mass-of-trans-stilbene-0-2033-grams\/","title":{"rendered":"Help with calculations of Greener Bromination of Trans-stilbene Synthesizing trans-1,2-dibromo-1,2-diphenylethane My initial mass of Trans-stilbene = 0.2033 grams"},"content":{"rendered":"\n

Help with calculations of Greener Bromination of Trans-stilbene Synthesizing trans-1,2-dibromo-1,2-diphenylethane My initial mass of Trans-stilbene = 0.2033 grams At the end of the experiment, the mass of the crystal yielded 0.2211 grams Can you please help me fill in the ??? blanks? Calculations: Theoretical Yield = 0.2 grams E-Stilbene 1 mol E-Stilbene 180.25 grams E-Stilbene \u00c3\u2014 1 mol C14H12Br2 1 mol C14H12Br2 = 0.0011 moles C14H12Br2 Actual Yield = grams C14H12Br21 \u00c3\u2014 1 mol C14H12Br2 340.058 grams C14H12Br2 = 0.000159 moles C14H12Br2 Chemical Percent % Yield = Actual Yield\/Theoretical Yield \u00c3\u2014 100% = 0.000159 moles\/0.0011 moles \u00c3\u2014 100% = ???% Atom Economy: Atom economy = MW of products\/MW of reactants \u00c3\u2014 100% = 340 grams C14H12Br2 \/ (180.25 grams E-Stilbene + 159.808 grams Bromine) \u00c3\u2014 100% = 100% Experimental atom economy = ????? grams C14H12Br2 \/ 0.2 grams E-stilbene + ?????? grams Bromine \u00c3\u2014 100% = ???% ???? \u00c2\u00b4 ???? = ????% \u00e2\u2030\u02c6<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s break down the calculations for your experiment step by step. You already have some values provided, so I’ll walk you through completing the missing pieces.<\/p>\n\n\n\n

Theoretical Yield Calculation:<\/strong><\/h3>\n\n\n\n
    \n
  1. Starting with Trans-stilbene (C14H12)<\/strong>:
    • Given mass = 0.2033 grams.<\/li>
    • Molar mass of Trans-stilbene = 180.25 g\/mol.<\/li><\/ul>Using the formula: Moles\u00a0of\u00a0Trans-stilbene=Mass\u00a0of\u00a0Trans-stilbeneMolar\u00a0mass\u00a0of\u00a0Trans-stilbene=0.2033\u2009g180.25\u2009g\/mol=0.00113\u2009mol.\\text{Moles of Trans-stilbene} = \\frac{\\text{Mass of Trans-stilbene}}{\\text{Molar mass of Trans-stilbene}} = \\frac{0.2033 \\, \\text{g}}{180.25 \\, \\text{g\/mol}} = 0.00113 \\, \\text{mol}.Moles\u00a0of\u00a0Trans-stilbene=Molar\u00a0mass\u00a0of\u00a0Trans-stilbeneMass\u00a0of\u00a0Trans-stilbene\u200b=180.25g\/mol0.2033g\u200b=0.00113mol.<\/li>\n\n\n\n
    • Moles of Product (C14H12Br2)<\/strong>:\n
        \n
      • From the stoichiometry of the reaction, 1 mol of Trans-stilbene reacts with 1 mol of Bromine to produce 1 mol of C14H12Br2.<\/li>\n\n\n\n
      • Thus, 0.00113 mol of Trans-stilbene will produce 0.00113 mol of C14H12Br2 (theoretical yield).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Theoretical Yield in grams<\/strong>:
        • Molar mass of C14H12Br2 = 340.058 g\/mol.<\/li><\/ul>Theoretical\u00a0yield\u00a0(grams)=Moles\u00a0of\u00a0product\u00d7Molar\u00a0mass\u00a0of\u00a0product=0.00113\u2009mol\u00d7340.058\u2009g\/mol=0.384\u2009g.\\text{Theoretical yield (grams)} = \\text{Moles of product} \\times \\text{Molar mass of product} = 0.00113 \\, \\text{mol} \\times 340.058 \\, \\text{g\/mol} = 0.384 \\, \\text{g}.Theoretical\u00a0yield\u00a0(grams)=Moles\u00a0of\u00a0product\u00d7Molar\u00a0mass\u00a0of\u00a0product=0.00113mol\u00d7340.058g\/mol=0.384g. (You mentioned a theoretical yield of 0.2 g in your question, so double-check if this is the number you’d like to use, but this is the result based on stoichiometry).<\/li>\n<\/ol>\n\n\n\n

          Actual Yield Calculation:<\/strong><\/h3>\n\n\n\n

          You have the actual yield of the product, which is 0.2211 grams. To find the moles of C14H12Br2<\/strong>, we can use the molar mass of C14H12Br2. Moles of C14H12Br2=Mass of C14H12Br2Molar mass of C14H12Br2=0.2211\u2009g340.058\u2009g\/mol=0.00065\u2009mol.\\text{Moles of C14H12Br2} = \\frac{\\text{Mass of C14H12Br2}}{\\text{Molar mass of C14H12Br2}} = \\frac{0.2211 \\, \\text{g}}{340.058 \\, \\text{g\/mol}} = 0.00065 \\, \\text{mol}.Moles of C14H12Br2=Molar mass of C14H12Br2Mass of C14H12Br2\u200b=340.058g\/mol0.2211g\u200b=0.00065mol.<\/p>\n\n\n\n

          Percent Yield Calculation:<\/strong><\/h3>\n\n\n\n

          Percent Yield=Actual YieldTheoretical Yield\u00d7100=0.00065\u2009mol0.00113\u2009mol\u00d7100=57.5%.\\text{Percent Yield} = \\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}} \\times 100 = \\frac{0.00065 \\, \\text{mol}}{0.00113 \\, \\text{mol}} \\times 100 = 57.5\\%.Percent Yield=Theoretical YieldActual Yield\u200b\u00d7100=0.00113mol0.00065mol\u200b\u00d7100=57.5%.<\/p>\n\n\n\n

          Atom Economy Calculation:<\/strong><\/h3>\n\n\n\n
            \n
          1. Molar mass of reactants<\/strong>:\n
              \n
            • Molar mass of Trans-stilbene (C14H12) = 180.25 g\/mol.<\/li>\n\n\n\n
            • Molar mass of Bromine (Br2) = 159.808 g\/mol.<\/li>\n\n\n\n
            • Total mass of reactants = 180.25 + 159.808 = 340.058 g\/mol.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Atom economy formula<\/strong>: Atom\u00a0Economy=Molar\u00a0mass\u00a0of\u00a0productMolar\u00a0mass\u00a0of\u00a0reactants\u00d7100=340.058\u2009g\/mol340.058\u2009g\/mol\u00d7100=100%.\\text{Atom Economy} = \\frac{\\text{Molar mass of product}}{\\text{Molar mass of reactants}} \\times 100 = \\frac{340.058 \\, \\text{g\/mol}}{340.058 \\, \\text{g\/mol}} \\times 100 = 100\\%.Atom\u00a0Economy=Molar\u00a0mass\u00a0of\u00a0reactantsMolar\u00a0mass\u00a0of\u00a0product\u200b\u00d7100=340.058g\/mol340.058g\/mol\u200b\u00d7100=100%.<\/li>\n<\/ol>\n\n\n\n

              Experimental Atom Economy:<\/strong><\/h3>\n\n\n\n

              For this part, let’s complete the experimental atom economy.<\/p>\n\n\n\n

                \n
              • The experimental mass of C14H12Br2 = 0.2211 g (which you already have).<\/li>\n\n\n\n
              • The mass of Bromine used is: Mass\u00a0of\u00a0Bromine=Mass\u00a0of\u00a0C14H12Br2\u2212Mass\u00a0of\u00a0Trans-stilbene=0.2211\u2009g\u22120.2033\u2009g=0.0178\u2009g.\\text{Mass of Bromine} = \\text{Mass of C14H12Br2} – \\text{Mass of Trans-stilbene} = 0.2211 \\, \\text{g} – 0.2033 \\, \\text{g} = 0.0178 \\, \\text{g}.Mass\u00a0of\u00a0Bromine=Mass\u00a0of\u00a0C14H12Br2\u2212Mass\u00a0of\u00a0Trans-stilbene=0.2211g\u22120.2033g=0.0178g.<\/li>\n<\/ul>\n\n\n\n

                Thus, the Experimental Atom Economy<\/strong> is: Experimental Atom Economy=0.2211\u2009g0.2033\u2009g+0.0178\u2009g\u00d7100=0.22110.2211\u00d7100=100%.\\text{Experimental Atom Economy} = \\frac{0.2211 \\, \\text{g}}{0.2033 \\, \\text{g} + 0.0178 \\, \\text{g}} \\times 100 = \\frac{0.2211}{0.2211} \\times 100 = 100\\%.Experimental Atom Economy=0.2033g+0.0178g0.2211g\u200b\u00d7100=0.22110.2211\u200b\u00d7100=100%.<\/p>\n\n\n\n

                Final Summary:<\/strong><\/h3>\n\n\n\n
                  \n
                • Theoretical Yield<\/strong> = 0.384 grams.<\/li>\n\n\n\n
                • Actual Yield<\/strong> = 0.2211 grams.<\/li>\n\n\n\n
                • Percent Yield<\/strong> = 57.5%.<\/li>\n\n\n\n
                • Atom Economy<\/strong> = 100%.<\/li>\n\n\n\n
                • Experimental Atom Economy<\/strong> = 100%.<\/li>\n<\/ul>\n\n\n\n
                  \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                  Help with calculations of Greener Bromination of Trans-stilbene Synthesizing trans-1,2-dibromo-1,2-diphenylethane My initial mass of Trans-stilbene = 0.2033 grams At the end of the experiment, the mass of the crystal yielded 0.2211 grams Can you please help me fill in the ??? blanks? Calculations: Theoretical Yield = 0.2 grams E-Stilbene 1 mol E-Stilbene 180.25 grams E-Stilbene […]<\/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-257765","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257765","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=257765"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257765\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257765"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257765"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257765"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}