{"id":28984,"date":"2025-06-20T18:36:52","date_gmt":"2025-06-20T18:36:52","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=28984"},"modified":"2025-06-20T18:36:53","modified_gmt":"2025-06-20T18:36:53","slug":"the-empirical-formula-of-a-compound-was-determined-to-be-p2o3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-empirical-formula-of-a-compound-was-determined-to-be-p2o3\/","title":{"rendered":"The empirical formula of a compound was determined to be P2O3."},"content":{"rendered":"\n<p>The empirical formula of a compound was determined to be P2O3. The molecular weight of the compound is 220 g\/mol. What is the molecular formula of the compound?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-green-cyan-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the molecular formula of a compound from its empirical formula and molecular weight, follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 1: Calculate the molar mass of the empirical formula P\u2082O\u2083.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Atomic mass of phosphorus (P) = 30.97 g\/mol<\/li>\n\n\n\n<li>Atomic mass of oxygen (O) = 16.00 g\/mol<\/li>\n\n\n\n<li>Molar mass of P\u2082O\u2083 = (2 \u00d7 30.97) + (3 \u00d7 16.00)<\/li>\n\n\n\n<li>Molar mass = 61.94 + 48.00 = <strong>109.94 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 2: Divide the molecular weight by the empirical formula weight.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Given molecular weight = 220 g\/mol<\/li>\n\n\n\n<li>Empirical formula weight = 109.94 g\/mol<\/li>\n\n\n\n<li>Ratio = 220 \/ 109.94 \u2248 <strong>2<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 3: Multiply the subscripts in the empirical formula by this ratio.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Empirical formula = P\u2082O\u2083<\/li>\n\n\n\n<li>Multiply each subscript by 2:<br>(2 \u00d7 2) = 4 for P<br>(3 \u00d7 2) = 6 for O<\/li>\n\n\n\n<li><strong>Molecular formula = P\u2084O\u2086<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer: <strong>P\u2084O\u2086<\/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:<\/h3>\n\n\n\n<p>The empirical formula represents the simplest whole number ratio of atoms in a compound. In this case, the empirical formula P\u2082O\u2083 tells us that for every 2 phosphorus atoms, there are 3 oxygen atoms. However, this does not always represent the actual number of atoms in a molecule.<\/p>\n\n\n\n<p>To find the molecular formula, which shows the actual number of each type of atom in a molecule, we need to know the molecular weight of the compound. Once we know both the empirical formula mass and the molecular weight, we divide the molecular weight by the empirical formula mass. This gives us a multiplier. We apply this multiplier to each subscript in the empirical formula to get the molecular formula.<\/p>\n\n\n\n<p>Since the molecular weight is about twice the mass of the empirical formula, it means the molecular formula must have twice as many atoms of each element. That results in the formula P\u2084O\u2086.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-282.jpeg\" alt=\"\" class=\"wp-image-28985\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-282.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-282-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-282-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The empirical formula of a compound was determined to be P2O3. The molecular weight of the compound is 220 g\/mol. What is the molecular formula of the compound? The Correct Answer and Explanation is: To determine the molecular formula of a compound from its empirical formula and molecular weight, follow these steps: Step 1: Calculate [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-28984","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28984","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=28984"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28984\/revisions"}],"predecessor-version":[{"id":28986,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28984\/revisions\/28986"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28984"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28984"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28984"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}