{"id":5864,"date":"2025-05-21T10:25:02","date_gmt":"2025-05-21T10:25:02","guid":{"rendered":"https:\/\/yaveni.com\/blog\/?p=5864"},"modified":"2025-05-21T10:25:03","modified_gmt":"2025-05-21T10:25:03","slug":"a-compound-is-54-5-carbon-9-1-hydrogen-and-36-4-oxygen","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-compound-is-54-5-carbon-9-1-hydrogen-and-36-4-oxygen\/","title":{"rendered":"\u00a0A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen"},"content":{"rendered":"\n<p>\u00a0A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molar mass is 88.0 g\/mol. What is its molecular formula?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the <strong>molecular formula<\/strong> of the compound, we follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine the empirical formula<\/h3>\n\n\n\n<p>Assume a 100 g sample of the compound. This allows us to treat the percentages as grams:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 54.5 g<\/li>\n\n\n\n<li>Hydrogen (H): 9.1 g<\/li>\n\n\n\n<li>Oxygen (O): 36.4 g<\/li>\n<\/ul>\n\n\n\n<p>Convert grams to moles:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of C = 54.5 g \u00f7 12.01 g\/mol = 4.54 mol<\/li>\n\n\n\n<li>Moles of H = 9.1 g \u00f7 1.008 g\/mol = 9.03 mol<\/li>\n\n\n\n<li>Moles of O = 36.4 g \u00f7 16.00 g\/mol = 2.28 mol<\/li>\n<\/ul>\n\n\n\n<p>Next, divide each mole value by the <strong>smallest number of moles<\/strong> to find the simplest whole number ratio:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C: 4.54 \u00f7 2.28 \u2248 1.99 \u2248 2<\/li>\n\n\n\n<li>H: 9.03 \u00f7 2.28 \u2248 3.96 \u2248 4<\/li>\n\n\n\n<li>O: 2.28 \u00f7 2.28 = 1<\/li>\n<\/ul>\n\n\n\n<p>So, the <strong>empirical formula<\/strong> is <strong>C\u2082H\u2084O<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the molecular formula<\/h3>\n\n\n\n<p>Calculate the <strong>empirical formula mass<\/strong> of C\u2082H\u2084O:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(2 \u00d7 12.01) + (4 \u00d7 1.008) + (1 \u00d7 16.00) = 24.02 + 4.032 + 16.00 = <strong>44.05 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Now, divide the <strong>given molar mass<\/strong> by the empirical formula mass:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>88.0 g\/mol \u00f7 44.05 g\/mol \u2248 2.00<\/li>\n<\/ul>\n\n\n\n<p>This means the molecular formula is <strong>2 times<\/strong> the empirical formula:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer: <strong>C\u2084H\u2088O\u2082<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 words):<\/h3>\n\n\n\n<p>The process of determining a molecular formula begins with analyzing the compound&#8217;s percent composition by mass. These percentages indicate the relative amount of each element present in a compound. By assuming a 100 g sample, we simplify the math so that the mass percentages become actual gram values, which are then converted into moles using the atomic masses from the periodic table.<\/p>\n\n\n\n<p>After obtaining the moles of each element, we find the simplest whole-number mole ratio by dividing each by the smallest mole quantity among the elements. These ratios give the empirical formula, representing the simplest integer ratio of atoms in the compound. In our case, the empirical formula is C\u2082H\u2084O.<\/p>\n\n\n\n<p>However, the empirical formula does not necessarily reflect the true number of atoms in a molecule. The molecular formula is a multiple of the empirical formula and is determined using the compound&#8217;s given molar mass. By dividing the molar mass (88.0 g\/mol) by the empirical formula mass (44.05 g\/mol), we find the multiple to be 2. Multiplying the subscripts in the empirical formula by this multiple gives the molecular formula: <strong>C\u2084H\u2088O\u2082<\/strong>.<\/p>\n\n\n\n<p>This molecular formula provides the actual number of carbon, hydrogen, and oxygen atoms in a molecule of the compound and is essential for understanding the compound\u2019s structure, properties, and reactivity.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molar mass is 88.0 g\/mol. What is its molecular formula? The correct answer and explanation is : To determine the molecular formula of the compound, we follow these steps: Step 1: Determine the empirical formula Assume a 100 g sample of the compound. This [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-5864","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/5864","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=5864"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/5864\/revisions"}],"predecessor-version":[{"id":5865,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/5864\/revisions\/5865"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=5864"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=5864"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=5864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}