{"id":45191,"date":"2025-07-01T03:47:24","date_gmt":"2025-07-01T03:47:24","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=45191"},"modified":"2025-07-01T03:47:25","modified_gmt":"2025-07-01T03:47:25","slug":"dibenzalacetone-c17h14o-is-a-common-ingredient-in-sunscreens-that-is-prepared-by-the-following-chemical-reaction","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/dibenzalacetone-c17h14o-is-a-common-ingredient-in-sunscreens-that-is-prepared-by-the-following-chemical-reaction\/","title":{"rendered":"Dibenzalacetone, C17H14O, is a common ingredient in sunscreens that is prepared by the following chemical reaction"},"content":{"rendered":"\n<p>Dibenzalacetone, C17H14O, is a common ingredient in sunscreens that is prepared by the following chemical reaction: 2 C6H6O + C6H5CHO -> C17H14O + 2 H2O. In a lab experiment, 3.95 mL of C6H6O (molar mass 106.12 g mol-1; density = 1.094 g mL-1) and an excess of C6H5CHO were used to synthesize C17H14O (molar mass = 234.29 g mol-1). The target product was produced as a yellow solid with a mass of 3.85 g. What was the percent yield of C17H14O? Choose the closest answer. a) 56.6% b) 80.7% c) 96.6% d) 40.4% e) 20.4%<\/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<p>To find the percent yield of C17H14O, we need to follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the moles of C6H6O used<\/h3>\n\n\n\n<p>The first step is to convert the volume of C6H6O into mass and then into moles. We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of C6H6O = 3.95 mL<\/li>\n\n\n\n<li>Density of C6H6O = 1.094 g\/mL<\/li>\n\n\n\n<li>Molar mass of C6H6O = 106.12 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>First, calculate the mass of C6H6O:Mass&nbsp;of&nbsp;C6H6O=Volume\u00d7Density=3.95\u2009mL\u00d71.094\u2009g\/mL=4.32\u2009g\\text{Mass of C6H6O} = \\text{Volume} \\times \\text{Density} = 3.95 \\, \\text{mL} \\times 1.094 \\, \\text{g\/mL} = 4.32 \\, \\text{g}Mass&nbsp;of&nbsp;C6H6O=Volume\u00d7Density=3.95mL\u00d71.094g\/mL=4.32g<\/p>\n\n\n\n<p>Now, convert the mass into moles:Moles&nbsp;of&nbsp;C6H6O=MassMolar&nbsp;mass=4.32\u2009g106.12\u2009g\/mol=0.0407\u2009mol\\text{Moles of C6H6O} = \\frac{\\text{Mass}}{\\text{Molar mass}} = \\frac{4.32 \\, \\text{g}}{106.12 \\, \\text{g\/mol}} = 0.0407 \\, \\text{mol}Moles&nbsp;of&nbsp;C6H6O=Molar&nbsp;massMass\u200b=106.12g\/mol4.32g\u200b=0.0407mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the theoretical moles of C17H14O produced<\/h3>\n\n\n\n<p>From the balanced chemical equation:2\u2009C6H6O+C6H5CHO\u2192C17H14O+2\u2009H2O2 \\, \\text{C6H6O} + \\text{C6H5CHO} \\rightarrow \\text{C17H14O} + 2 \\, \\text{H2O}2C6H6O+C6H5CHO\u2192C17H14O+2H2O<\/p>\n\n\n\n<p>The molar ratio of C6H6O to C17H14O is 2:1. This means that for every 2 moles of C6H6O, 1 mole of C17H14O is produced.<\/p>\n\n\n\n<p>So, the moles of C17H14O produced will be half the moles of C6H6O:Moles&nbsp;of&nbsp;C17H14O=0.0407\u2009mol2=0.02035\u2009mol\\text{Moles of C17H14O} = \\frac{0.0407 \\, \\text{mol}}{2} = 0.02035 \\, \\text{mol}Moles&nbsp;of&nbsp;C17H14O=20.0407mol\u200b=0.02035mol<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the theoretical mass of C17H14O<\/h3>\n\n\n\n<p>Now, convert the moles of C17H14O into mass using its molar mass:Mass&nbsp;of&nbsp;C17H14O=Moles\u00d7Molar&nbsp;mass=0.02035\u2009mol\u00d7234.29\u2009g\/mol=4.77\u2009g\\text{Mass of C17H14O} = \\text{Moles} \\times \\text{Molar mass} = 0.02035 \\, \\text{mol} \\times 234.29 \\, \\text{g\/mol} = 4.77 \\, \\text{g}Mass&nbsp;of&nbsp;C17H14O=Moles\u00d7Molar&nbsp;mass=0.02035mol\u00d7234.29g\/mol=4.77g<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate the percent yield<\/h3>\n\n\n\n<p>The percent yield is calculated by comparing the actual yield to the theoretical yield:Percent&nbsp;yield=(Actual&nbsp;yieldTheoretical&nbsp;yield)\u00d7100\\text{Percent yield} = \\left( \\frac{\\text{Actual yield}}{\\text{Theoretical yield}} \\right) \\times 100Percent&nbsp;yield=(Theoretical&nbsp;yieldActual&nbsp;yield\u200b)\u00d7100<\/p>\n\n\n\n<p>The actual yield is 3.85 g (given), and the theoretical yield is 4.77 g (calculated above):Percent&nbsp;yield=(3.85\u2009g4.77\u2009g)\u00d7100=80.7%\\text{Percent yield} = \\left( \\frac{3.85 \\, \\text{g}}{4.77 \\, \\text{g}} \\right) \\times 100 = 80.7\\%Percent&nbsp;yield=(4.77g3.85g\u200b)\u00d7100=80.7%<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The closest answer is <strong>b) 80.7%<\/strong>.<\/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\/07\/learnexams-banner8-4.jpeg\" alt=\"\" class=\"wp-image-45192\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-4.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-4-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-4-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Dibenzalacetone, C17H14O, is a common ingredient in sunscreens that is prepared by the following chemical reaction: 2 C6H6O + C6H5CHO -> C17H14O + 2 H2O. In a lab experiment, 3.95 mL of C6H6O (molar mass 106.12 g mol-1; density = 1.094 g mL-1) and an excess of C6H5CHO were used to synthesize C17H14O (molar mass [&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-45191","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45191","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=45191"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45191\/revisions"}],"predecessor-version":[{"id":45193,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45191\/revisions\/45193"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45191"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45191"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45191"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}