{"id":28988,"date":"2025-06-20T18:39:16","date_gmt":"2025-06-20T18:39:16","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=28988"},"modified":"2025-06-20T18:39:17","modified_gmt":"2025-06-20T18:39:17","slug":"determine-the-molecular-formula-for-the-compound-with-the-empirical-formula-p2o3-if-the-molar-mass-is-219-90-g-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-molecular-formula-for-the-compound-with-the-empirical-formula-p2o3-if-the-molar-mass-is-219-90-g-mol\/","title":{"rendered":"Determine the molecular formula for the compound with the empirical formula P2O3 if the molar mass is 219.90 g\/mol."},"content":{"rendered":"\n<p>Determine the molecular formula for the compound with the empirical formula P2O3 if the molar mass is 219.90 g\/mol.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-purple-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the <strong>molecular formula<\/strong> of a compound, you need to compare the <strong>empirical formula mass<\/strong> to the <strong>given molar mass<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Determine the empirical formula mass of P\u2082O\u2083<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Phosphorus (P) atomic mass = 30.97 g\/mol<\/li>\n\n\n\n<li>Oxygen (O) atomic mass = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Empirical formula: <strong>P\u2082O\u2083<\/strong><\/p>\n\n\n\n<p>Empirical formula mass = (2 \u00d7 30.97) + (3 \u00d7 16.00)<br>= 61.94 + 48.00<br>= <strong>109.94 g\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Compare empirical formula mass to molar mass<\/strong><\/h3>\n\n\n\n<p>Given <strong>molar mass<\/strong> = 219.90 g\/mol<br>Empirical formula mass = 109.94 g\/mol<\/p>\n\n\n\n<p>Now calculate the ratio:<\/p>\n\n\n\n<p>219.90 \u00f7 109.94 \u2248 <strong>2.00<\/strong><\/p>\n\n\n\n<p>This means the molecular formula contains <strong>two empirical units<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Multiply the subscripts in the empirical formula<\/strong><\/h3>\n\n\n\n<p>Empirical formula: <strong>P\u2082O\u2083<\/strong><br>Multiply each subscript by 2:<\/p>\n\n\n\n<p>Molecular formula = <strong>P\u2084O\u2086<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 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\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The empirical formula of a compound represents the simplest whole-number ratio of atoms in the molecule. In this problem, the empirical formula is P\u2082O\u2083, which means the compound contains two atoms of phosphorus for every three atoms of oxygen. However, this may not reflect the actual number of atoms in a single molecule of the substance. To find the true molecular formula, you must compare the empirical formula mass with the actual molar mass of the compound.<\/p>\n\n\n\n<p>First, calculate the mass of the empirical formula. Two phosphorus atoms contribute a total mass of 61.94 g\/mol and three oxygen atoms add up to 48.00 g\/mol, giving a total empirical formula mass of 109.94 g\/mol. When you divide the given molar mass of 219.90 g\/mol by the empirical formula mass, you get a factor of approximately 2.00. This factor tells you how many times the empirical formula must be multiplied to reach the true molecular formula.<\/p>\n\n\n\n<p>By multiplying each subscript in P\u2082O\u2083 by 2, the molecular formula becomes P\u2084O\u2086. This new formula reflects the actual number of atoms in one molecule, and its mass matches the given molar mass. P\u2084O\u2086 is also known as tetraphosphorus hexoxide and is a well-known phosphorus oxide. This step-by-step process demonstrates how understanding the relationship between empirical and molecular formulas allows chemists to determine molecular composition based on measurable properties like molar mass.<\/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-banner8-283.jpeg\" alt=\"\" class=\"wp-image-28989\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-283.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-283-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-283-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the molecular formula for the compound with the empirical formula P2O3 if the molar mass is 219.90 g\/mol. The Correct Answer and Explanation is: To determine the molecular formula of a compound, you need to compare the empirical formula mass to the given molar mass. Step 1: Determine the empirical formula mass of P\u2082O\u2083 [&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-28988","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28988","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=28988"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28988\/revisions"}],"predecessor-version":[{"id":28990,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28988\/revisions\/28990"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28988"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28988"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28988"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}