{"id":270427,"date":"2025-07-25T07:20:20","date_gmt":"2025-07-25T07:20:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=270427"},"modified":"2025-07-25T07:20:22","modified_gmt":"2025-07-25T07:20:22","slug":"calculate-the-percent-composition-of-gallium-sulfite","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/25\/calculate-the-percent-composition-of-gallium-sulfite\/","title":{"rendered":"Calculate the percent composition of gallium sulfite."},"content":{"rendered":"\n
Calculate the percent composition of gallium sulfite. 2) A compound contains 92.3% C and 7.7% H and has a molar mass of 78.0 g\/mol. Determine the molecular formula. (3) Sorbitol, used as a sweetener in some sugar free foods, has a molecular formula of 182 g\/mol and percent composition of 39.56% carbon, 7.74% hydrogen, and 52.70% of oxygen. What is the empirical formula and molecular formula of sorbitol.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
) Percent Composition of Gallium Sulfite (Ga2(SO3)3)<\/h3>\n\n\n\n
To find the percent composition of gallium sulfite, we need to calculate the molar mass of Ga2(SO3)3 and the contribution of each element.<\/p>\n\n\n\n
Step 1: Calculate the Molar Mass of Ga2(SO3)3<\/h4>\n\n\n\n
\n
Gallium (Ga) has a molar mass of approximately 69.72 g\/mol.<\/li>\n\n\n\n
Sulfur (S) has a molar mass of approximately 32.07 g\/mol.<\/li>\n\n\n\n
Oxygen (O) has a molar mass of approximately 16.00 g\/mol.<\/li>\n<\/ul>\n\n\n\n
Step 2: Calculate the Percent Composition of Each Element<\/h4>\n\n\n\n
\n
Percent composition of Ga = (139.44 \/ 379.65) \u00d7 100 = 36.74%<\/strong><\/li>\n\n\n\n
Percent composition of S = (96.21 \/ 379.65) \u00d7 100 = 25.35%<\/strong><\/li>\n\n\n\n
Percent composition of O = (240.21 \/ 379.65) \u00d7 100 = 63.26%<\/strong><\/li>\n<\/ul>\n\n\n\n
2) Molecular Formula of a Compound with 92.3% C and 7.7% H (Molar Mass = 78.0 g\/mol)<\/h3>\n\n\n\n
Step 1: Find the Empirical Formula<\/h4>\n\n\n\n
Given:<\/p>\n\n\n\n
\n
92.3% C, 7.7% H<\/li>\n\n\n\n
Molar mass = 78.0 g\/mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Convert Percentages to Grams (Assume 100 g of the compound)<\/h4>\n\n\n\n
\n
92.3 g of carbon<\/li>\n\n\n\n
7.7 g of hydrogen<\/li>\n<\/ul>\n\n\n\n
Step 3: Convert Grams to Moles<\/h4>\n\n\n\n
\n
Moles of C = 92.3 g \/ 12.01 g\/mol = 7.69 mol<\/li>\n\n\n\n
Moles of H = 7.7 g \/ 1.008 g\/mol = 7.64 mol<\/li>\n<\/ul>\n\n\n\n
Step 4: Determine the Ratio of Elements<\/h4>\n\n\n\n
\n
The ratio of moles of C to H is approximately 1:1.<\/li>\n\n\n\n
So, the empirical formula is CH<\/strong>.<\/li>\n<\/ul>\n\n\n\n
Step 5: Determine the Molecular Formula<\/h4>\n\n\n\n
The empirical formula mass of CH is 12.01 + 1.008 = 13.018 g\/mol.<\/p>\n\n\n\n
\n
Molar mass of the compound is 78.0 g\/mol.<\/li>\n\n\n\n
Ratio of molar mass to empirical formula mass = 78.0 \/ 13.018 = 6. Thus, the molecular formula is C6H6<\/strong> (benzene).<\/li>\n<\/ul>\n\n\n\n
3) Empirical and Molecular Formulas of Sorbitol<\/h3>\n\n\n\n
Given:<\/p>\n\n\n\n
\n
Molecular formula = 182 g\/mol<\/li>\n\n\n\n
Percent composition:\n
\n
39.56% C<\/li>\n\n\n\n
7.74% H<\/li>\n\n\n\n
52.70% O<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
Step 1: Convert Percentages to Grams (Assume 100 g of sorbitol)<\/h4>\n\n\n\n
\n
39.56 g of carbon<\/li>\n\n\n\n
7.74 g of hydrogen<\/li>\n\n\n\n
52.70 g of oxygen<\/li>\n<\/ul>\n\n\n\n
Step 2: Convert Grams to Moles<\/h4>\n\n\n\n
\n
Moles of C = 39.56 g \/ 12.01 g\/mol = 3.296 mol<\/li>\n\n\n\n
Moles of H = 7.74 g \/ 1.008 g\/mol = 7.67 mol<\/li>\n\n\n\n
Moles of O = 52.70 g \/ 16.00 g\/mol = 3.294 mol<\/li>\n<\/ul>\n\n\n\n
Step 3: Find the Empirical Formula<\/h4>\n\n\n\n
\n
The mole ratio of C:H:O is approximately 3:8:3.<\/li>\n\n\n\n
The empirical formula is C3H8O3<\/strong>.<\/li>\n<\/ul>\n\n\n\n
Step 4: Determine the Molecular Formula<\/h4>\n\n\n\n
The empirical formula mass of C3H8O3 is: (3 \u00d7 12.01) + (8 \u00d7 1.008) + (3 \u00d7 16.00) = 36.03 + 8.064 + 48.00 = 92.094 g\/mol.<\/p>\n\n\n\n
\n
The molar mass of sorbitol is 182 g\/mol.<\/li>\n\n\n\n
Ratio of molar mass to empirical formula mass = 182 \/ 92.094 = 2. Thus, the molecular formula is C6H14O6<\/strong>.<\/li>\n<\/ul>\n\n\n\n\n\n\n\n
Summary:<\/h3>\n\n\n\n
\n
Gallium sulfite (Ga2(SO3)3) has a percent composition of 36.74% Ga<\/strong>, 25.35% S<\/strong>, and 63.26% O<\/strong>.<\/li>\n\n\n\n
The molecular formula of the compound with 92.3% C and 7.7% H and molar mass of 78.0 g\/mol is C6H6<\/strong> (benzene).<\/li>\n\n\n\n
The empirical formula of sorbitol is C3H8O3<\/strong>, and its molecular formula is C6H14O6<\/strong>.<\/li>\n<\/ul>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Calculate the percent composition of gallium sulfite. 2) A compound contains 92.3% C and 7.7% H and has a molar mass of 78.0 g\/mol. Determine the molecular formula. (3) Sorbitol, used as a sweetener in some sugar free foods, has a molecular formula of 182 g\/mol and percent composition of 39.56% carbon, 7.74% hydrogen, and […]<\/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-270427","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/270427","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=270427"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/270427\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=270427"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=270427"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=270427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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