{"id":246232,"date":"2025-07-07T03:47:57","date_gmt":"2025-07-07T03:47:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246232"},"modified":"2025-07-07T03:47:59","modified_gmt":"2025-07-07T03:47:59","slug":"cobalt-forms-two-different-oxides","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/cobalt-forms-two-different-oxides\/","title":{"rendered":"Cobalt forms two different oxides"},"content":{"rendered":"\n

Cobalt forms two different oxides. The mass percentages of cobalt in the pure oxides are measured as being 78.65% and 71.06%. A. Calculate the formula of each oxide using the percent by mass, then name them. B. Calculate the mole fraction of cobalt and mass percentage of cobalt in a mixture containing 8.000 kg of the first oxide and 4.550 kg of the second oxide. C. A sample of mixed cobalt oxides with a mass of 5.283 kg is completely reduced until the total mass is 4.025 kg of pure cobalt metal. What percentage of cobalt ions in the original sample had an oxidation state of +2? (by mole)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

A. Calculate the formula of each oxide using percent by mass, then name them<\/strong><\/h3>\n\n\n\n
    \n
  1. Oxide 1 (78.65% Co)<\/strong>:\n
      \n
    • Let’s assume we have 100 grams of the oxide. The mass of cobalt is 78.65 grams, and the mass of oxygen is 100 – 78.65 = 21.35 grams.<\/li>\n\n\n\n
    • Moles of cobalt (Co) = 78.65 g \/ 58.93 g\/mol = 1.335 mol<\/li>\n\n\n\n
    • Moles of oxygen (O) = 21.35 g \/ 16.00 g\/mol = 1.335 mol<\/li>\n\n\n\n
    • The ratio of moles of Co to O is 1:1, so the formula of this oxide is CoO<\/strong>.<\/li>\n\n\n\n
    • Name<\/strong>: Cobalt(I) oxide.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Oxide 2 (71.06% Co)<\/strong>:\n
        \n
      • Assuming 100 grams of the oxide, the mass of cobalt is 71.06 grams, and the mass of oxygen is 100 – 71.06 = 28.94 grams.<\/li>\n\n\n\n
      • Moles of cobalt (Co) = 71.06 g \/ 58.93 g\/mol = 1.206 mol<\/li>\n\n\n\n
      • Moles of oxygen (O) = 28.94 g \/ 16.00 g\/mol = 1.809 mol<\/li>\n\n\n\n
      • The ratio of moles of Co to O is 1:1.5, so we multiply both by 2 to obtain whole numbers: 2:3<\/strong>. Therefore, the formula is Co\u2082O\u2083<\/strong>.<\/li>\n\n\n\n
      • Name<\/strong>: Cobalt(III) oxide.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
        \n\n\n\n

        B. Mole fraction of cobalt and mass percentage of cobalt in a mixture<\/strong><\/h3>\n\n\n\n

        Given:<\/p>\n\n\n\n

          \n
        • First oxide: 8.000 kg of CoO (molar mass = 58.93 g\/mol + 16.00 g\/mol = 74.93 g\/mol)<\/li>\n\n\n\n
        • Second oxide: 4.550 kg of Co\u2082O\u2083 (molar mass = 2(58.93 g\/mol) + 3(16.00 g\/mol) = 174.86 g\/mol)<\/li>\n<\/ul>\n\n\n\n

          Step 1: Calculate the moles of cobalt in each oxide:<\/strong><\/p>\n\n\n\n

            \n
          • Moles of Co in 8.000 kg of CoO: Moles\u00a0of\u00a0Co=8000\u2009g74.93\u2009g\/mol=106.76\u2009mol\\text{Moles of Co} = \\frac{8000 \\, \\text{g}}{74.93 \\, \\text{g\/mol}} = 106.76 \\, \\text{mol}Moles\u00a0of\u00a0Co=74.93g\/mol8000g\u200b=106.76mol<\/li>\n\n\n\n
          • Moles of Co in 4.550 kg of Co\u2082O\u2083: Moles\u00a0of\u00a0Co=4550\u2009g174.86\u2009g\/mol=26.05\u2009mol\\text{Moles of Co} = \\frac{4550 \\, \\text{g}}{174.86 \\, \\text{g\/mol}} = 26.05 \\, \\text{mol}Moles\u00a0of\u00a0Co=174.86g\/mol4550g\u200b=26.05mol<\/li>\n<\/ul>\n\n\n\n

            Step 2: Calculate total moles of cobalt:<\/strong>Total moles of Co=106.76\u2009mol+26.05\u2009mol=132.81\u2009mol\\text{Total moles of Co} = 106.76 \\, \\text{mol} + 26.05 \\, \\text{mol} = 132.81 \\, \\text{mol}Total moles of Co=106.76mol+26.05mol=132.81mol<\/p>\n\n\n\n

            Step 3: Calculate the total mass of the mixture:<\/strong>Total mass=8.000\u2009kg+4.550\u2009kg=12.550\u2009kg=12550\u2009g\\text{Total mass} = 8.000 \\, \\text{kg} + 4.550 \\, \\text{kg} = 12.550 \\, \\text{kg} = 12550 \\, \\text{g}Total mass=8.000kg+4.550kg=12.550kg=12550g<\/p>\n\n\n\n

            Step 4: Calculate the mole fraction of cobalt:<\/strong>Mole fraction of Co=132.81\u2009molTotal moles of all substances=132.81\u2009mol12550\u2009g74.93\u2009g\/mol+12550\u2009g174.86\u2009g\/mol\\text{Mole fraction of Co} = \\frac{132.81 \\, \\text{mol}}{\\text{Total moles of all substances}} = \\frac{132.81 \\, \\text{mol}}{\\frac{12550 \\, \\text{g}}{74.93 \\, \\text{g\/mol}} + \\frac{12550 \\, \\text{g}}{174.86 \\, \\text{g\/mol}}}Mole fraction of Co=Total moles of all substances132.81mol\u200b=74.93g\/mol12550g\u200b+174.86g\/mol12550g\u200b132.81mol\u200bTotal moles of the mixture=12550\u2009g74.93\u2009g\/mol+12550\u2009g174.86\u2009g\/mol=167.89\u2009mol\\text{Total moles of the mixture} = \\frac{12550 \\, \\text{g}}{74.93 \\, \\text{g\/mol}} + \\frac{12550 \\, \\text{g}}{174.86 \\, \\text{g\/mol}} = 167.89 \\, \\text{mol}Total moles of the mixture=74.93g\/mol12550g\u200b+174.86g\/mol12550g\u200b=167.89molMole fraction of Co=132.81\u2009mol167.89\u2009mol=0.79\\text{Mole fraction of Co} = \\frac{132.81 \\, \\text{mol}}{167.89 \\, \\text{mol}} = 0.79Mole fraction of Co=167.89mol132.81mol\u200b=0.79<\/p>\n\n\n\n

            Step 5: Calculate the mass percentage of cobalt:<\/strong>Mass percentage of Co=132.81\u2009mol\u00d758.93\u2009g\/mol12550\u2009g\u00d7100=7835.81\u2009g12550\u2009g\u00d7100=62.5%\\text{Mass percentage of Co} = \\frac{132.81 \\, \\text{mol} \\times 58.93 \\, \\text{g\/mol}}{12550 \\, \\text{g}} \\times 100 = \\frac{7835.81 \\, \\text{g}}{12550 \\, \\text{g}} \\times 100 = 62.5\\%Mass percentage of Co=12550g132.81mol\u00d758.93g\/mol\u200b\u00d7100=12550g7835.81g\u200b\u00d7100=62.5%<\/p>\n\n\n\n

            \n\n\n\n

            C. Percentage of cobalt ions with oxidation state of +2<\/strong><\/h3>\n\n\n\n

            Given:<\/p>\n\n\n\n

              \n
            • Mass of the sample before reduction = 5.283 kg = 5283 g<\/li>\n\n\n\n
            • Mass after reduction (pure cobalt) = 4.025 kg = 4025 g<\/li>\n\n\n\n
            • The remaining mass after reduction is pure cobalt, which means the rest of the mass was in the form of cobalt oxides.<\/li>\n<\/ul>\n\n\n\n

              Step 1: Determine the mass of cobalt oxides:<\/strong>Mass of cobalt oxides=5283\u2009g\u22124025\u2009g=1258\u2009g\\text{Mass of cobalt oxides} = 5283 \\, \\text{g} – 4025 \\, \\text{g} = 1258 \\, \\text{g}Mass of cobalt oxides=5283g\u22124025g=1258g<\/p>\n\n\n\n

              Step 2: Calculate the moles of cobalt in the reduced metal:<\/strong>Moles of Co=4025\u2009g58.93\u2009g\/mol=68.3\u2009mol\\text{Moles of Co} = \\frac{4025 \\, \\text{g}}{58.93 \\, \\text{g\/mol}} = 68.3 \\, \\text{mol}Moles of Co=58.93g\/mol4025g\u200b=68.3mol<\/p>\n\n\n\n

              Step 3: Determine the percentage of Co with oxidation state +2:<\/strong><\/p>\n\n\n\n

              To find the percentage of cobalt ions with oxidation state +2, we need to assume that cobalt in CoO has an oxidation state of +2 and cobalt in Co\u2082O\u2083 has an oxidation state of +3. From the initial sample, we know that the cobalt comes from a mixture of these two oxides. If we focus on CoO, where cobalt has an oxidation state of +2, and calculate the moles of Co in CoO:<\/p>\n\n\n\n

              Step 4: Use stoichiometry to calculate the mole fraction of Co\u00b2\u207a:<\/strong><\/p>\n\n\n\n

              Let\u2019s assume that 100% of the initial sample came from CoO. Since the mass of cobalt is reduced to pure cobalt metal, the percentage of Co with oxidation state +2 will be directly proportional to the amount of Co from CoO:Moles of Co in CoO=1258\u2009g74.93\u2009g\/mol=16.79\u2009mol\\text{Moles of Co in CoO} = \\frac{1258 \\, \\text{g}}{74.93 \\, \\text{g\/mol}} = 16.79 \\, \\text{mol}Moles of Co in CoO=74.93g\/mol1258g\u200b=16.79mol<\/p>\n\n\n\n

              Thus, the percentage of cobalt with oxidation state +2 is given by:Percentage of Co\u00b2\u207a=16.79\u2009mol68.3\u2009mol\u00d7100=24.6%\\text{Percentage of Co\u00b2\u207a} = \\frac{16.79 \\, \\text{mol}}{68.3 \\, \\text{mol}} \\times 100 = 24.6\\%Percentage of Co\u00b2\u207a=68.3mol16.79mol\u200b\u00d7100=24.6%<\/p>\n\n\n\n

              So, about 24.6%<\/strong> of the cobalt ions had an oxidation state of +2 in the original sample.<\/p>\n\n\n\n

              \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

              Cobalt forms two different oxides. The mass percentages of cobalt in the pure oxides are measured as being 78.65% and 71.06%. A. Calculate the formula of each oxide using the percent by mass, then name them. B. Calculate the mole fraction of cobalt and mass percentage of cobalt in a mixture containing 8.000 kg of […]<\/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-246232","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246232","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=246232"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246232\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246232"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246232"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246232"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}