{"id":257625,"date":"2025-07-17T15:25:09","date_gmt":"2025-07-17T15:25:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257625"},"modified":"2025-07-17T15:25:14","modified_gmt":"2025-07-17T15:25:14","slug":"a-metal-m-forms-a-metal-carbonate-m2co3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/a-metal-m-forms-a-metal-carbonate-m2co3\/","title":{"rendered":"A metal M forms a metal carbonate M2CO3"},"content":{"rendered":"\n

A metal M forms a metal carbonate M2CO3. If the carbonate contains 48% oxygen by mass, then determine the atomic weight of the metal.<\/p>\n\n\n\n

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

To find the atomic weight of the metal MMM in the metal carbonate M2CO3M_2CO_3M2\u200bCO3\u200b, we can use the information about the percentage of oxygen in the compound and apply the concept of mass balance.<\/p>\n\n\n\n

Step 1: Write the molecular formula<\/h3>\n\n\n\n

We are given that the metal carbonate is M2CO3M_2CO_3M2\u200bCO3\u200b, meaning it contains 2 atoms of the metal MMM, 1 atom of carbon, and 3 atoms of oxygen.<\/p>\n\n\n\n

Step 2: Calculate the molar mass of M2CO3M_2CO_3M2\u200bCO3\u200b<\/h3>\n\n\n\n

To calculate the molar mass of M2CO3M_2CO_3M2\u200bCO3\u200b, we need to sum the molar masses of each component:<\/p>\n\n\n\n

    \n
  • The molar mass of oxygen (OOO) is 16 g\/mol.<\/li>\n\n\n\n
  • The molar mass of carbon (CCC) is 12 g\/mol.<\/li>\n<\/ul>\n\n\n\n

    So, the mass of the oxygen atoms in M2CO3M_2CO_3M2\u200bCO3\u200b is: Mass of oxygen=3\u00d716=48\u2009g\/mol\\text{Mass of oxygen} = 3 \\times 16 = 48 \\, \\text{g\/mol}Mass of oxygen=3\u00d716=48g\/mol<\/p>\n\n\n\n

    Now, let the atomic mass of the metal MMM be xxx g\/mol. The total mass of 2 atoms of MMM is: Mass of metal=2x\u2009g\/mol\\text{Mass of metal} = 2x \\, \\text{g\/mol}Mass of metal=2xg\/mol<\/p>\n\n\n\n

    Thus, the total molar mass of M2CO3M_2CO_3M2\u200bCO3\u200b is: Total molar mass=2x+12+48=2x+60\u2009g\/mol\\text{Total molar mass} = 2x + 12 + 48 = 2x + 60 \\, \\text{g\/mol}Total molar mass=2x+12+48=2x+60g\/mol<\/p>\n\n\n\n

    Step 3: Use the mass percent of oxygen<\/h3>\n\n\n\n

    We are told that oxygen makes up 48% by mass of M2CO3M_2CO_3M2\u200bCO3\u200b. This means that the mass of oxygen in the compound, which is 48 g\/mol, accounts for 48% of the total molar mass of M2CO3M_2CO_3M2\u200bCO3\u200b.<\/p>\n\n\n\n

    Thus, we can set up the equation: Mass of oxygenTotal molar mass=482x+60=0.48\\frac{\\text{Mass of oxygen}}{\\text{Total molar mass}} = \\frac{48}{2x + 60} = 0.48Total molar massMass of oxygen\u200b=2x+6048\u200b=0.48<\/p>\n\n\n\n

    Step 4: Solve for xxx<\/h3>\n\n\n\n

    Now, solve for xxx: 48=0.48\u00d7(2x+60)48 = 0.48 \\times (2x + 60)48=0.48\u00d7(2x+60) 48=0.48\u00d72x+0.48\u00d76048 = 0.48 \\times 2x + 0.48 \\times 6048=0.48\u00d72x+0.48\u00d760 48=0.96x+28.848 = 0.96x + 28.848=0.96x+28.8 48\u221228.8=0.96×48 – 28.8 = 0.96×48\u221228.8=0.96x 19.2=0.96×19.2 = 0.96×19.2=0.96x x=19.20.96=20x = \\frac{19.2}{0.96} = 20x=0.9619.2\u200b=20<\/p>\n\n\n\n

    Step 5: Conclusion<\/h3>\n\n\n\n

    The atomic weight of the metal MMM is 20\u2009g\/mol20 \\, \\text{g\/mol}20g\/mol.<\/p>\n\n\n\n

    So, the metal MMM is calcium (Ca), as its atomic weight is 20 g\/mol.<\/p>\n\n\n\n

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

    A metal M forms a metal carbonate M2CO3. If the carbonate contains 48% oxygen by mass, then determine the atomic weight of the metal. The Correct Answer and Explanation is: To find the atomic weight of the metal MMM in the metal carbonate M2CO3M_2CO_3M2\u200bCO3\u200b, we can use the information about the percentage of oxygen in […]<\/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-257625","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257625","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=257625"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257625\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257625"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257625"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}