To solve this problem, we need to calculate the mass of C17H14O that was isolated from the reaction, considering the yield percentage. Here\u2019s the step-by-step process:<\/p>\n\n\n\n
Step 1: Write the reaction<\/h3>\n\n\n\n
The balanced chemical reaction is:2\u2009C7H6O+C3H6O\u27f6C17H14O+2\u2009H2O2 \\, C_7H_6O + C_3H_6O \\longrightarrow C_{17}H_{14}O + 2 \\, H_2O2C7\u200bH6\u200bO+C3\u200bH6\u200bO\u27f6C17\u200bH14\u200bO+2H2\u200bO<\/p>\n\n\n\n
Step 2: Calculate the moles of reactants<\/h3>\n\n\n\n
We are given the masses of C7H6O (4.25 g) and C3H6O (1.10 g). To find the moles of each reactant, we use the formula:moles=massmolar mass\\text{moles} = \\frac{\\text{mass}}{\\text{molar mass}}moles=molar massmass\u200b<\/p>\n\n\n\n
- \n
- Molar mass of C7H6O = 106.12 g\/mol<\/li>\n\n\n\n
- Molar mass of C3H6O = 58.08 g\/mol<\/li>\n<\/ul>\n\n\n\n
For C7H6O:<\/strong>moles of C7H6O=4.25\u2009g106.12\u2009g\/mol=0.0400\u2009mol\\text{moles of C7H6O} = \\frac{4.25 \\, \\text{g}}{106.12 \\, \\text{g\/mol}} = 0.0400 \\, \\text{mol}moles of C7H6O=106.12g\/mol4.25g\u200b=0.0400mol<\/p>\n\n\n\n
For C3H6O:<\/strong>moles of C3H6O=1.10\u2009g58.08\u2009g\/mol=0.0189\u2009mol\\text{moles of C3H6O} = \\frac{1.10 \\, \\text{g}}{58.08 \\, \\text{g\/mol}} = 0.0189 \\, \\text{mol}moles of C3H6O=58.08g\/mol1.10g\u200b=0.0189mol<\/p>\n\n\n\n
Step 3: Determine the limiting reagent<\/h3>\n\n\n\n
From the balanced equation, we see that 2 moles of C7H6O react with 1 mole of C3H6O. Therefore, the molar ratio of C7H6O to C3H6O is 2:1.<\/p>\n\n\n\n
- \n
- For 0.0400 mol of C7H6O, we would need:<\/li>\n<\/ul>\n\n\n\n
0.0400\u2009mol2=0.0200\u2009mol of C3H6O\\frac{0.0400 \\, \\text{mol}}{2} = 0.0200 \\, \\text{mol of C3H6O}20.0400mol\u200b=0.0200mol of C3H6O<\/p>\n\n\n\n
- \n
- But we only have 0.0189 mol of C3H6O, so C3H6O is the limiting reagent.<\/li>\n<\/ul>\n\n\n\n
Step 4: Calculate the theoretical yield of C17H14O<\/h3>\n\n\n\n
From the balanced equation, 1 mole of C3H6O produces 1 mole of C17H14O. Therefore, the moles of C17H14O produced will be equal to the moles of C3H6O used, which is 0.0189 mol.<\/p>\n\n\n\n
To find the theoretical mass of C17H14O:mass of C17H14O=moles\u00d7molar mass of C17H14O\\text{mass of C17H14O} = \\text{moles} \\times \\text{molar mass of C17H14O}mass of C17H14O=moles\u00d7molar mass of C17H14Omass of C17H14O=0.0189\u2009mol\u00d7234.29\u2009g\/mol=4.43\u2009g\\text{mass of C17H14O} = 0.0189 \\, \\text{mol} \\times 234.29 \\, \\text{g\/mol} = 4.43 \\, \\text{g}mass of C17H14O=0.0189mol\u00d7234.29g\/mol=4.43g<\/p>\n\n\n\n
Step 5: Account for the percent yield<\/h3>\n\n\n\n
The reaction yields 78.4% of the theoretical product. So, the actual mass of C17H14O isolated is:actual mass=4.43\u2009g\u00d778.4100=3.47\u2009g\\text{actual mass} = 4.43 \\, \\text{g} \\times \\frac{78.4}{100} = 3.47 \\, \\text{g}actual mass=4.43g\u00d710078.4\u200b=3.47g<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The mass of C17H14O isolated is 3.47 g<\/strong>.<\/p>\n\n\n\n
![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-11.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Dibenzalacetone, C17H14O, is a common ingredient in sunscreens that can be prepared by the following chemical reaction.2 C7H6O + C3H6O ? C17H14O + 2 H2OIn a lab experiment, 4.25 g of C7H6O and 1.10 g of C3H6O were used to synthesize C17H14O. The target product was produced as a yellow solid with a 78.4% yield.What […]<\/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-45223","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45223","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=45223"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45223\/revisions"}],"predecessor-version":[{"id":45225,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45223\/revisions\/45225"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- But we only have 0.0189 mol of C3H6O, so C3H6O is the limiting reagent.<\/li>\n<\/ul>\n\n\n\n
- For 0.0400 mol of C7H6O, we would need:<\/li>\n<\/ul>\n\n\n\n