{"id":33366,"date":"2025-06-23T08:15:13","date_gmt":"2025-06-23T08:15:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=33366"},"modified":"2025-06-23T08:15:15","modified_gmt":"2025-06-23T08:15:15","slug":"given-the-following-information-calculate-the-mass-in-grams-of-2-methyl-1-butene-produced-from-6-28-g-of-2-methyl-2-butanol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/given-the-following-information-calculate-the-mass-in-grams-of-2-methyl-1-butene-produced-from-6-28-g-of-2-methyl-2-butanol\/","title":{"rendered":"Given the following information, calculate the mass in grams of 2-methyl-1-butene produced from 6.28 g of 2-methyl-2-butanol"},"content":{"rendered":"\n<p>Given the following information, calculate the mass in grams of 2-methyl-1-butene produced from 6.28 g of 2-methyl-2-butanol: 8 OH H2SO4 H2O 90% Yield C5H12O C5H10 C5H10<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Answer:<\/h3>\n\n\n\n<p><strong>Step 1: Calculate Molar Masses<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of <strong>2-methyl-2-butanol (C\u2085H\u2081\u2082O)<\/strong><br>= (5 \u00d7 12.01) + (12 \u00d7 1.008) + (16.00)<br>= 60.05 + 12.096 + 16.00<br>= <strong>88.15 g\/mol<\/strong><\/li>\n\n\n\n<li>Molar mass of <strong>2-methyl-1-butene (C\u2085H\u2081\u2080)<\/strong><br>= (5 \u00d7 12.01) + (10 \u00d7 1.008)<br>= 60.05 + 10.08<br>= <strong>70.13 g\/mol<\/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\">Step 2: Calculate Moles of 2-methyl-2-butanol<\/h3>\n\n\n\n<p>Given mass = 6.28 g<br>Molar mass = 88.15 g\/mol<\/p>\n\n\n\n<p>Moles = Mass \/ Molar mass = 6.28 g \u00f7 88.15 g\/mol = <strong>0.07126 moles<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate Theoretical Mass of 2-methyl-1-butene<\/h3>\n\n\n\n<p>The reaction is:<br><strong>C\u2085H\u2081\u2082O \u2192 C\u2085H\u2081\u2080 + H\u2082O<\/strong><\/p>\n\n\n\n<p>The mole ratio is 1:1, so 0.07126 moles of alcohol produce 0.07126 moles of alkene.<\/p>\n\n\n\n<p>Mass of alkene = Moles \u00d7 Molar mass<br>= 0.07126 moles \u00d7 70.13 g\/mol<br>= <strong>4.997 g<\/strong> (Theoretical mass)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Apply the 90 Percent Yield<\/h3>\n\n\n\n<p>Actual mass produced = Theoretical mass \u00d7 Percent yield<br>= 4.997 g \u00d7 0.90<br>= <strong>4.497 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>4.50 grams of 2-methyl-1-butene produced<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"> Explanation:<\/h2>\n\n\n\n<p>This question requires calculating the mass of product formed from a known mass of reactant, considering the percent yield. The reaction involves dehydration of <strong>2-methyl-2-butanol<\/strong> using sulfuric acid to produce <strong>2-methyl-1-butene<\/strong> and water.<\/p>\n\n\n\n<p>First, the molar masses of both the reactant and the product are calculated using atomic weights of carbon, hydrogen, and oxygen. The molar mass of <strong>2-methyl-2-butanol<\/strong> is 88.15 g\/mol, and the molar mass of <strong>2-methyl-1-butene<\/strong> is 70.13 g\/mol.<\/p>\n\n\n\n<p>Next, using the given mass of the alcohol (6.28 grams), the number of moles is found by dividing the mass by the molar mass. Since the mole ratio between the alcohol and the alkene is 1:1, the same number of moles of alkene is expected.<\/p>\n\n\n\n<p>To find the theoretical mass of <strong>2-methyl-1-butene<\/strong>, multiply the number of moles by its molar mass. This gives 4.997 grams, which represents the maximum amount that can be formed if the reaction were 100 percent efficient.<\/p>\n\n\n\n<p>However, the reaction proceeds with a 90 percent yield, meaning only 90 percent of the theoretical mass is actually obtained. Therefore, multiplying the theoretical mass by 0.90 gives the actual mass of product formed, which is <strong>4.50 grams<\/strong>.<\/p>\n\n\n\n<p>This method is common in stoichiometry to predict real-life chemical production outcomes where reactions are rarely perfectly efficient. Percent yield reflects practical limitations like side reactions, incomplete conversions, or experimental losses.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-505.jpeg\" alt=\"\" class=\"wp-image-33374\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-505.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-505-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-505-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Given the following information, calculate the mass in grams of 2-methyl-1-butene produced from 6.28 g of 2-methyl-2-butanol: 8 OH H2SO4 H2O 90% Yield C5H12O C5H10 C5H10 The Correct Answer and Explanation is: Correct Answer: Step 1: Calculate Molar Masses Step 2: Calculate Moles of 2-methyl-2-butanol Given mass = 6.28 gMolar mass = 88.15 g\/mol Moles [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-33366","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/33366","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=33366"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/33366\/revisions"}],"predecessor-version":[{"id":33380,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/33366\/revisions\/33380"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=33366"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=33366"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=33366"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}