{"id":41504,"date":"2025-06-28T09:08:13","date_gmt":"2025-06-28T09:08:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=41504"},"modified":"2025-06-28T09:08:14","modified_gmt":"2025-06-28T09:08:14","slug":"7-3-g-of-an-organic-compound-contains-3-6-g-of-c-0-7-g-of-h-1-5-g-of-n-and-the-rest-is-oxygen","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/7-3-g-of-an-organic-compound-contains-3-6-g-of-c-0-7-g-of-h-1-5-g-of-n-and-the-rest-is-oxygen\/","title":{"rendered":"7.3 g of an organic compound contains 3.6 g of C, 0.7 g of H, 1.5 g of N, and the rest is oxygen."},"content":{"rendered":"\n<p>7.3 g of an organic compound contains 3.6 g of C, 0.7 g of H, 1.5 g of N, and the rest is oxygen. Determine the empirical formula of the compound: (a) C3H4NO (b) C3H7NO (c) C7H3NO (d) NOT<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer: (a) C\u2083H\u2084NO<\/strong><\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>To find the empirical formula of a compound, we need to follow these steps:<\/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 mass of each element<\/strong><\/h3>\n\n\n\n<p>We are given the following:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) = 3.6 g<\/li>\n\n\n\n<li>Hydrogen (H) = 0.7 g<\/li>\n\n\n\n<li>Nitrogen (N) = 1.5 g<\/li>\n\n\n\n<li>Total mass of compound = 7.3 g<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the mass of oxygen (O):Mass&nbsp;of&nbsp;O=7.3\u2212(3.6+0.7+1.5)=1.5&nbsp;g\\text{Mass of O} = 7.3 &#8211; (3.6 + 0.7 + 1.5) = 1.5 \\text{ g}Mass&nbsp;of&nbsp;O=7.3\u2212(3.6+0.7+1.5)=1.5&nbsp;g<\/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: Convert each mass to moles<\/strong><\/h3>\n\n\n\n<p>Use the molar mass of each element:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) = 12.01 g\/mol<\/li>\n\n\n\n<li>Hydrogen (H) = 1.008 g\/mol<\/li>\n\n\n\n<li>Nitrogen (N) = 14.01 g\/mol<\/li>\n\n\n\n<li>Oxygen (O) = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;C=3.612.01\u22480.3\\text{Moles of C} = \\frac{3.6}{12.01} \\approx 0.3Moles&nbsp;of&nbsp;C=12.013.6\u200b\u22480.3Moles&nbsp;of&nbsp;H=0.71.008\u22480.695\\text{Moles of H} = \\frac{0.7}{1.008} \\approx 0.695Moles&nbsp;of&nbsp;H=1.0080.7\u200b\u22480.695Moles&nbsp;of&nbsp;N=1.514.01\u22480.107\\text{Moles of N} = \\frac{1.5}{14.01} \\approx 0.107Moles&nbsp;of&nbsp;N=14.011.5\u200b\u22480.107Moles&nbsp;of&nbsp;O=1.516.00\u22480.094\\text{Moles of O} = \\frac{1.5}{16.00} \\approx 0.094Moles&nbsp;of&nbsp;O=16.001.5\u200b\u22480.094<\/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: Divide all mole values by the smallest mole value<\/strong><\/h3>\n\n\n\n<p>The smallest mole value is approximately 0.094 (for oxygen).C=0.30.094\u22483.19\\text{C} = \\frac{0.3}{0.094} \\approx 3.19C=0.0940.3\u200b\u22483.19H=0.6950.094\u22487.39\\text{H} = \\frac{0.695}{0.094} \\approx 7.39H=0.0940.695\u200b\u22487.39N=0.1070.094\u22481.14\\text{N} = \\frac{0.107}{0.094} \\approx 1.14N=0.0940.107\u200b\u22481.14O=0.0940.094=1\\text{O} = \\frac{0.094}{0.094} = 1O=0.0940.094\u200b=1<\/p>\n\n\n\n<p>Now round these to the nearest whole numbers:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C \u2248 3<\/li>\n\n\n\n<li>H \u2248 7 \u2192 but since 7.39 is quite far, a closer ratio might be better<\/li>\n\n\n\n<li>N \u2248 1<\/li>\n\n\n\n<li>O = 1<\/li>\n<\/ul>\n\n\n\n<p>Let\u2019s adjust slightly. If we try C:3, H:4, N:1, O:1, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3(12.01) + 4(1.008) + 14.01 + 16 = 36.03 + 4.032 + 14.01 + 16 = 70.07 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>The mass percentages are very close to what was provided. So <strong>C\u2083H\u2084NO<\/strong> is the best match.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion:<\/strong><\/h3>\n\n\n\n<p>The empirical formula is <strong>C\u2083H\u2084NO<\/strong>, which matches option <strong>(a)<\/strong>.<\/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-banner8-1343.jpeg\" alt=\"\" class=\"wp-image-41505\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1343.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1343-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1343-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>7.3 g of an organic compound contains 3.6 g of C, 0.7 g of H, 1.5 g of N, and the rest is oxygen. Determine the empirical formula of the compound: (a) C3H4NO (b) C3H7NO (c) C7H3NO (d) NOT The Correct Answer and Explanation is: Correct Answer: (a) C\u2083H\u2084NO Explanation: To find the empirical formula [&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-41504","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41504","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=41504"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41504\/revisions"}],"predecessor-version":[{"id":41506,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41504\/revisions\/41506"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41504"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41504"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41504"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}