To determine the empirical and molecular formulas from the given percent compositions and molar mass, follow these steps:<\/p>\n\n\n\n
Step 1: Convert the percent composition to grams<\/h3>\n\n\n\n
Since the percent composition is given, assume you have 100 grams of the compound. This makes the calculations easier because the percentages directly translate into grams.<\/p>\n\n\n\n
- \n
- Carbon (C): 26.4% \u2192 26.4 grams<\/li>\n\n\n\n
- Hydrogen (H): 3.3% \u2192 3.3 grams<\/li>\n\n\n\n
- Oxygen (O): 70.3% \u2192 70.3 grams<\/li>\n<\/ul>\n\n\n\n
Step 2: Convert grams to moles<\/h3>\n\n\n\n
Next, convert the mass of each element into moles using the atomic masses:<\/p>\n\n\n\n
- \n
- Carbon (C) has an atomic mass of 12.01 g\/mol.<\/li>\n\n\n\n
- Hydrogen (H) has an atomic mass of 1.008 g\/mol.<\/li>\n\n\n\n
- Oxygen (O) has an atomic mass of 16.00 g\/mol.<\/li>\n<\/ul>\n\n\n\n
Using the formula: Moles of element=mass of elementatomic mass of element\\text{Moles of element} = \\frac{\\text{mass of element}}{\\text{atomic mass of element}}<\/p>\n\n\n\n
- \n
- Moles of C: 26.412.01=2.20\\frac{26.4}{12.01} = 2.20 moles<\/li>\n\n\n\n
- Moles of H: 3.31.008=3.28\\frac{3.3}{1.008} = 3.28 moles<\/li>\n\n\n\n
- Moles of O: 70.316.00=4.39\\frac{70.3}{16.00} = 4.39 moles<\/li>\n<\/ul>\n\n\n\n
Step 3: Find the simplest mole ratio<\/h3>\n\n\n\n
Now, divide the number of moles of each element by the smallest number of moles calculated (in this case, 2.20 moles of C):<\/p>\n\n\n\n
- \n
- Carbon: 2.202.20=1\\frac{2.20}{2.20} = 1<\/li>\n\n\n\n
- Hydrogen: 3.282.20=1.49\u22481.5\\frac{3.28}{2.20} = 1.49 \\approx 1.5<\/li>\n\n\n\n
- Oxygen: 4.392.20=2.00\\frac{4.39}{2.20} = 2.00<\/li>\n<\/ul>\n\n\n\n
Since the ratio of hydrogen is approximately 1.5, multiply all the ratios by 2 to get whole numbers:<\/p>\n\n\n\n
- \n
- Carbon: 1\u00d72=21 \\times 2 = 2<\/li>\n\n\n\n
- Hydrogen: 1.5\u00d72=31.5 \\times 2 = 3<\/li>\n\n\n\n
- Oxygen: 2\u00d72=42 \\times 2 = 4<\/li>\n<\/ul>\n\n\n\n
So, the empirical formula is C\u2082H\u2083O\u2084<\/strong>.<\/p>\n\n\n\n
Step 4: Determine the molecular formula<\/h3>\n\n\n\n
Now, use the molar mass to find the molecular formula. The empirical formula mass is the sum of the atomic masses of each element in the empirical formula:<\/p>\n\n\n\n
- \n
- Empirical formula mass = (2\u00d712.01)+(3\u00d71.008)+(4\u00d716.00)=24.02+3.024+64.00=91.044(2 \\times 12.01) + (3 \\times 1.008) + (4 \\times 16.00) = 24.02 + 3.024 + 64.00 = 91.044 g\/mol<\/li>\n<\/ul>\n\n\n\n
Since the given molar mass is 91.0 g\/mol and the empirical formula mass is approximately 91.0 g\/mol, the empirical formula and the molecular formula are the same.<\/p>\n\n\n\n
Thus, the molecular formula is C\u2082H\u2083O\u2084<\/strong>.<\/p>\n\n\n\n
\n\n\n\nIn summary, the empirical and molecular formula for the compound with the given percent compositions and molar mass is C\u2082H\u2083O\u2084<\/strong>.<\/p>\n\n\n\n
Now, I’ll generate an image to visually present this solution.<\/p>\n\n\n\n
Here is the educational diagram that outlines the process of calculating both the empirical and molecular formulas. It visually represents the steps involved, from converting percent compositions to moles, finding the mole ratio, and determining the empirical and molecular formulas. This should help in understanding how to approach similar problems. Let me know if you need further clarification!<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-587.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Empirical\/Molecular Formula Practice WorksheetDirections: Find the empirical AND molecular formulas given the percent compositions or masses. SHOW YOUR WORK to receive full credit.1)26.4% Carbon3.3% Hydrogen70.3% OxygenMolar Mass: 91.0 g\/mol The Correct Answer and Explanation is : To determine the empirical and molecular formulas from the given percent compositions and molar mass, follow these steps: Step […]<\/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-191545","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191545","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=191545"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191545\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191545"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191545"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191545"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Empirical formula mass = (2\u00d712.01)+(3\u00d71.008)+(4\u00d716.00)=24.02+3.024+64.00=91.044(2 \\times 12.01) + (3 \\times 1.008) + (4 \\times 16.00) = 24.02 + 3.024 + 64.00 = 91.044 g\/mol<\/li>\n<\/ul>\n\n\n\n