{"id":18584,"date":"2025-06-13T07:40:18","date_gmt":"2025-06-13T07:40:18","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=18584"},"modified":"2025-06-13T07:40:20","modified_gmt":"2025-06-13T07:40:20","slug":"determine-the-percent-composition-of-chromium-in-chromiumii-dichromate","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-percent-composition-of-chromium-in-chromiumii-dichromate\/","title":{"rendered":"Determine the percent composition of chromium in chromium(II) dichromate"},"content":{"rendered":"\n<p>Determine the percent composition of chromium in chromium(II) dichromate.<\/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>percent composition of chromium (Cr)<\/strong> in <strong>chromium(II) dichromate<\/strong>, we must:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine the chemical formula<\/h3>\n\n\n\n<p>Chromium(II) = Cr\u00b2\u207a<br>Dichromate = Cr\u2082O\u2087\u00b2\u207b<br>To balance charges:<br>1 Cr\u00b2\u207a ion balances with 1 Cr\u2082O\u2087\u00b2\u207b ion<br>So the formula is: <strong>Cr(Cr\u2082O\u2087)<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the molar mass of Cr(Cr\u2082O\u2087)<\/h3>\n\n\n\n<p>Break it down:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chromium (Cr): 3 atoms<br>Atomic mass \u2248 <strong>51.996 g\/mol<\/strong><br>\u2192 3 \u00d7 51.996 = <strong>155.988 g\/mol<\/strong><\/li>\n\n\n\n<li>Oxygen (O): 7 atoms<br>Atomic mass \u2248 <strong>16.00 g\/mol<\/strong><br>\u2192 7 \u00d7 16.00 = <strong>112.00 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Total molar mass = 155.988 + 112.00 = 267.988 g\/mol<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the percent composition of chromium<\/h3>\n\n\n\n<p>Percent&nbsp;Cr=(Mass&nbsp;of&nbsp;CrMolar&nbsp;Mass&nbsp;of&nbsp;Cr(Cr2O7))\u00d7100\\text{Percent Cr} = \\left( \\frac{\\text{Mass of Cr}}{\\text{Molar Mass of Cr(Cr}_2\\text{O}_7)} \\right) \\times 100 =(155.988267.988)\u00d7100\u224858.22%= \\left( \\frac{155.988}{267.988} \\right) \\times 100 \\approx \\boxed{58.22\\%}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (Approx. 300 words):<\/h3>\n\n\n\n<p>Percent composition tells us what portion (by mass) of a compound is made up of a specific element. In this case, we are asked to find the percent composition of chromium in <strong>chromium(II) dichromate<\/strong>.<\/p>\n\n\n\n<p>First, we identify the chemical formula. Chromium(II) means the chromium ion has a +2 charge (Cr\u00b2\u207a), and dichromate is a polyatomic ion with the formula Cr\u2082O\u2087\u00b2\u207b. To create a neutral compound, one Cr\u00b2\u207a ion must pair with one Cr\u2082O\u2087\u00b2\u207b ion, since their charges cancel out. This gives the formula <strong>Cr(Cr\u2082O\u2087)<\/strong>. This compound contains <strong>3 chromium atoms<\/strong>\u2014one from Cr\u00b2\u207a and two from Cr\u2082O\u2087\u00b2\u207b.<\/p>\n\n\n\n<p>Next, we calculate the molar mass (the mass of one mole of the compound). Each chromium atom has an atomic mass of about 51.996 g\/mol, and there are three chromium atoms total, so their combined mass is about 155.988 g\/mol. There are 7 oxygen atoms in Cr\u2082O\u2087, each about 16.00 g\/mol, so the oxygen total is 112.00 g\/mol. Together, the molar mass of the compound is 267.988 g\/mol.<\/p>\n\n\n\n<p>Finally, to find the percent composition of chromium, we divide the mass of chromium (155.988 g\/mol) by the total molar mass (267.988 g\/mol) and multiply by 100. The result is <strong>58.22%<\/strong>, meaning chromium makes up about 58.22% of the mass of chromium(II) dichromate.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Determine the percent composition of chromium in chromium(II) dichromate. The correct answer and explanation is: To determine the percent composition of chromium (Cr) in chromium(II) dichromate, we must: Step 1: Determine the chemical formula Chromium(II) = Cr\u00b2\u207aDichromate = Cr\u2082O\u2087\u00b2\u207bTo balance charges:1 Cr\u00b2\u207a ion balances with 1 Cr\u2082O\u2087\u00b2\u207b ionSo the formula is: Cr(Cr\u2082O\u2087) Step 2: Determine [&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-18584","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18584","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=18584"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18584\/revisions"}],"predecessor-version":[{"id":18585,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18584\/revisions\/18585"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18584"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18584"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}