To determine the mass of 4.00 moles of glucose (C\u2086H\u2081\u2082O\u2086), we need to use the concept of molar mass. The molar mass is the mass of one mole of a substance, and it is usually given in grams per mole (g\/mol). To calculate the molar mass of glucose, we first need to add up the atomic masses of all the atoms in its formula.<\/p>\n\n\n\n
The molecular formula for glucose is C\u2086H\u2081\u2082O\u2086, meaning it consists of:<\/p>\n\n\n\n
- \n
- 6 carbon (C) atoms<\/li>\n\n\n\n
- 12 hydrogen (H) atoms<\/li>\n\n\n\n
- 6 oxygen (O) atoms<\/li>\n<\/ul>\n\n\n\n
Now, let’s find the atomic masses:<\/p>\n\n\n\n
- \n
- Carbon (C) has an atomic mass of approximately 12.01 g\/mol<\/strong>.<\/li>\n\n\n\n
- Hydrogen (H) has an atomic mass of approximately 1.008 g\/mol<\/strong>.<\/li>\n\n\n\n
- Oxygen (O) has an atomic mass of approximately 16.00 g\/mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n
Using this information, we can calculate the molar mass of glucose:<\/p>\n\n\n\n
[
\\text{Molar Mass of C\u2086H\u2081\u2082O\u2086} = (6 \\times 12.01 \\, \\text{g\/mol}) + (12 \\times 1.008 \\, \\text{g\/mol}) + (6 \\times 16.00 \\, \\text{g\/mol})
]<\/p>\n\n\n\n[
\\text{Molar Mass of C\u2086H\u2081\u2082O\u2086} = 72.06 \\, \\text{g\/mol} + 12.096 \\, \\text{g\/mol} + 96.00 \\, \\text{g\/mol} = 180.156 \\, \\text{g\/mol}
]<\/p>\n\n\n\nSo, the molar mass of glucose is 180.156 g\/mol<\/strong>.<\/p>\n\n\n\n
Next, to calculate the mass of 4.00 moles of glucose, we multiply the number of moles by the molar mass:<\/p>\n\n\n\n
[
\\text{Mass of 4.00 moles of glucose} = 4.00 \\, \\text{moles} \\times 180.156 \\, \\text{g\/mol}
]<\/p>\n\n\n\n[
\\text{Mass of 4.00 moles of glucose} = 720.624 \\, \\text{g}
]<\/p>\n\n\n\nThus, the mass of 4.00 moles of glucose is 720.624 grams<\/strong>.<\/p>\n\n\n\n
Explanation Summary:<\/h3>\n\n\n\n
The calculation involves determining the molar mass of glucose by adding up the atomic masses of all its constituent elements. Once the molar mass is known, the mass of 4.00 moles is calculated by multiplying the molar mass by the number of moles. The final result is 720.624 grams. This process is fundamental in chemistry for converting between moles and mass.<\/p>\n\n\n\n
Let me generate an image showing this calculation for you.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-161.png\")
<\/figure>\n\n\n\nHere is the diagram showing the calculation for the mass of 4.00 moles of glucose (C\u2086H\u2081\u2082O\u2086). It includes the molecular formula, atomic masses of carbon, hydrogen, and oxygen, as well as the step-by-step process leading to the final result of 720.624 grams.<\/p>\n","protected":false},"excerpt":{"rendered":"
What is the mass of 4.00 moles of glucose CsH12O6? The Correct Answer and Explanation is : To determine the mass of 4.00 moles of glucose (C\u2086H\u2081\u2082O\u2086), we need to use the concept of molar mass. The molar mass is the mass of one mole of a substance, and it is usually given in grams […]<\/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-196327","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196327","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=196327"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196327\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196327"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196327"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196327"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Hydrogen (H) has an atomic mass of approximately 1.008 g\/mol<\/strong>.<\/li>\n\n\n\n
- Carbon (C) has an atomic mass of approximately 12.01 g\/mol<\/strong>.<\/li>\n\n\n\n