Ordinary table sugar is primarily sucrose, which has the chemical formula C 12 H 22 O 11. What is the mass percentages of the following in sucrose?: a.) carbon b.) hydrogen c.) oxygen<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the mass percentages of carbon, hydrogen, and oxygen in sucrose (C\u2081\u2082H\u2082\u2082O\u2081\u2081), we first need to determine the molar mass of sucrose and then the individual contributions of carbon, hydrogen, and oxygen.<\/p>\n\n\n\n The molecular formula for sucrose is C\u2081\u2082H\u2082\u2082O\u2081\u2081, so we need the atomic masses of carbon (C), hydrogen (H), and oxygen (O):<\/p>\n\n\n\n Now, calculate the molar mass of sucrose: Now, to find the mass percentages of carbon, hydrogen, and oxygen in sucrose, divide the mass contribution of each element by the molar mass of sucrose and multiply by 100:<\/p>\n\n\n\n The mass percentages give us an understanding of how much each element contributes to the total mass of sucrose. Since carbon has the highest percentage (42.09%), it plays the most significant role in determining the overall structure of the molecule, with oxygen being the second largest contributor (51.43%), and hydrogen making the smallest contribution (6.48%). These percentages are derived from the elemental composition of the molecule, which is critical in understanding its chemical properties and behavior, especially in biological and industrial contexts like metabolism and food production.<\/p>\n","protected":false},"excerpt":{"rendered":" Ordinary table sugar is primarily sucrose, which has the chemical formula C 12 H 22 O 11. What is the mass percentages of the following in sucrose?: a.) carbon b.) hydrogen c.) oxygen The Correct Answer and Explanation is : To calculate the mass percentages of carbon, hydrogen, and oxygen in sucrose (C\u2081\u2082H\u2082\u2082O\u2081\u2081), we first […]<\/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-189576","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189576","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=189576"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189576\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189576"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189576"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189576"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate the molar mass of sucrose<\/h3>\n\n\n\n
\n
[
\\text{Molar mass of sucrose} = (12 \\times 12.01) + (22 \\times 1.008) + (11 \\times 16.00)
]
[
= 144.12 + 22.176 + 176.00 = 342.296 \\, \\text{g\/mol}
]<\/p>\n\n\n\nStep 2: Determine the mass contributions of carbon, hydrogen, and oxygen<\/h3>\n\n\n\n
\n
[
\\text{Mass of carbon} = 12 \\times 12.01 = 144.12 \\, \\text{g\/mol}
]<\/li>\n\n\n\n
[
\\text{Mass of hydrogen} = 22 \\times 1.008 = 22.176 \\, \\text{g\/mol}
]<\/li>\n\n\n\n
[
\\text{Mass of oxygen} = 11 \\times 16.00 = 176.00 \\, \\text{g\/mol}
]<\/li>\n<\/ul>\n\n\n\nStep 3: Calculate the mass percentages<\/h3>\n\n\n\n
\n
[
\\left(\\frac{144.12}{342.296}\\right) \\times 100 = 42.09\\%
]<\/li>\n\n\n\n
[
\\left(\\frac{22.176}{342.296}\\right) \\times 100 = 6.48\\%
]<\/li>\n\n\n\n
[
\\left(\\frac{176.00}{342.296}\\right) \\times 100 = 51.43\\%
]<\/li>\n<\/ul>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n
Explanation:<\/h3>\n\n\n\n