{"id":88362,"date":"2025-06-01T07:08:46","date_gmt":"2025-06-01T07:08:46","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88362"},"modified":"2025-06-01T07:08:46","modified_gmt":"2025-06-01T07:08:46","slug":"the-compound-c-6-h-12-o-4-contains","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/","title":{"rendered":"The compound C 6 H 12 O 4 contains"},"content":{"rendered":"

The compound C6<\/sub>H12<\/sub>O4<\/sub> contains<\/p>\n

[amp_mcq option1=”22 atoms per mole” option2=”twice the mass percent of H as compared to the mass percent of C” option3=”six times the mass percent of C as compared to the mass percent of H” option4=”thrice the mass percent of H as compared to the mass percent of O” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2017<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nWe need to analyze the composition of the compound C\u2086H\u2081\u2082O\u2084 in terms of the relative masses or mass percentages of its constituent elements.
\n<\/section>\n
\nThe compound is C\u2086H\u2081\u2082O\u2084. Using approximate atomic masses (C=12, H=1, O=16):
\nMass of C in one molecule = 6 * 12 = 72
\nMass of H in one molecule = 12 * 1 = 12
\nMass of O in one molecule = 4 * 16 = 64
\nTotal mass of one molecule (or molar mass) = 72 + 12 + 64 = 148.<\/p>\n

Let’s evaluate the options:
\nA) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro’s number of molecules, so one mole contains 22 * Avogadro’s number of atoms, not just 22 atoms. Incorrect.
\nB) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12\/148)*100% vs 2 * (72\/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect.
\nC) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let’s check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element \/ total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true.
\nD) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect.<\/p>\n

Option C is the only statement that accurately reflects the mass composition of the compound, interpreting “six times the mass percent of C as compared to the mass percent of H” as the mass percent of C being six times the mass percent of H.
\n<\/section>\n

\nCalculating mass percentages:
\n%C = (72\/148) * 100% \u2248 48.65%
\n%H = (12\/148) * 100% \u2248 8.11%
\n%O = (64\/148) * 100% \u2248 43.24%
\nCheck C: 6 * %H = 6 * 8.11% \u2248 48.66%, which is very close to %C (48.65%).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The compound C6H12O4 contains [amp_mcq option1=”22 atoms per mole” option2=”twice the mass percent of H as compared to the mass percent of C” option3=”six times the mass percent of C as compared to the mass percent of H” option4=”thrice the mass percent of H as compared to the mass percent of O” correct=”option3″] This question … <\/p>\n

Detailed SolutionThe compound C 6 H 12 O 4 contains<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1101,1096,1239],"class_list":["post-88362","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1101","tag-chemistry","tag-inorganic-chemistry","no-featured-image-padding"],"yoast_head":"\nThe compound C 6 H 12 O 4 contains<\/title>\n<meta name=\"description\" content=\"We need to analyze the composition of the compound C\u2086H\u2081\u2082O\u2084 in terms of the relative masses or mass percentages of its constituent elements. The compound is C\u2086H\u2081\u2082O\u2084. Using approximate atomic masses (C=12, H=1, O=16): Mass of C in one molecule = 6 * 12 = 72 Mass of H in one molecule = 12 * 1 = 12 Mass of O in one molecule = 4 * 16 = 64 Total mass of one molecule (or molar mass) = 72 + 12 + 64 = 148. Let's evaluate the options: A) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro's number of molecules, so one mole contains 22 * Avogadro's number of atoms, not just 22 atoms. Incorrect. B) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12\/148)*100% vs 2 * (72\/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect. C) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let's check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element \/ total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true. D) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect. Option C is the only statement that accurately reflects the mass composition of the compound, interpreting "six times the mass percent of C as compared to the mass percent of H" as the mass percent of C being six times the mass percent of H.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The compound C 6 H 12 O 4 contains\" \/>\n<meta property=\"og:description\" content=\"We need to analyze the composition of the compound C\u2086H\u2081\u2082O\u2084 in terms of the relative masses or mass percentages of its constituent elements. The compound is C\u2086H\u2081\u2082O\u2084. Using approximate atomic masses (C=12, H=1, O=16): Mass of C in one molecule = 6 * 12 = 72 Mass of H in one molecule = 12 * 1 = 12 Mass of O in one molecule = 4 * 16 = 64 Total mass of one molecule (or molar mass) = 72 + 12 + 64 = 148. Let's evaluate the options: A) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro's number of molecules, so one mole contains 22 * Avogadro's number of atoms, not just 22 atoms. Incorrect. B) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12\/148)*100% vs 2 * (72\/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect. C) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let's check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element \/ total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true. D) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect. Option C is the only statement that accurately reflects the mass composition of the compound, interpreting "six times the mass percent of C as compared to the mass percent of H" as the mass percent of C being six times the mass percent of H.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:08:46+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"The compound C 6 H 12 O 4 contains","description":"We need to analyze the composition of the compound C\u2086H\u2081\u2082O\u2084 in terms of the relative masses or mass percentages of its constituent elements. The compound is C\u2086H\u2081\u2082O\u2084. Using approximate atomic masses (C=12, H=1, O=16): Mass of C in one molecule = 6 * 12 = 72 Mass of H in one molecule = 12 * 1 = 12 Mass of O in one molecule = 4 * 16 = 64 Total mass of one molecule (or molar mass) = 72 + 12 + 64 = 148. Let's evaluate the options: A) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro's number of molecules, so one mole contains 22 * Avogadro's number of atoms, not just 22 atoms. Incorrect. B) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12\/148)*100% vs 2 * (72\/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect. C) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let's check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element \/ total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true. D) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect. Option C is the only statement that accurately reflects the mass composition of the compound, interpreting \"six times the mass percent of C as compared to the mass percent of H\" as the mass percent of C being six times the mass percent of H.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/","og_locale":"en_US","og_type":"article","og_title":"The compound C 6 H 12 O 4 contains","og_description":"We need to analyze the composition of the compound C\u2086H\u2081\u2082O\u2084 in terms of the relative masses or mass percentages of its constituent elements. The compound is C\u2086H\u2081\u2082O\u2084. Using approximate atomic masses (C=12, H=1, O=16): Mass of C in one molecule = 6 * 12 = 72 Mass of H in one molecule = 12 * 1 = 12 Mass of O in one molecule = 4 * 16 = 64 Total mass of one molecule (or molar mass) = 72 + 12 + 64 = 148. Let's evaluate the options: A) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro's number of molecules, so one mole contains 22 * Avogadro's number of atoms, not just 22 atoms. Incorrect. B) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12\/148)*100% vs 2 * (72\/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect. C) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let's check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element \/ total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true. D) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect. Option C is the only statement that accurately reflects the mass composition of the compound, interpreting \"six times the mass percent of C as compared to the mass percent of H\" as the mass percent of C being six times the mass percent of H.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:08:46+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/","name":"The compound C 6 H 12 O 4 contains","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:08:46+00:00","dateModified":"2025-06-01T07:08:46+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"We need to analyze the composition of the compound C\u2086H\u2081\u2082O\u2084 in terms of the relative masses or mass percentages of its constituent elements. The compound is C\u2086H\u2081\u2082O\u2084. Using approximate atomic masses (C=12, H=1, O=16): Mass of C in one molecule = 6 * 12 = 72 Mass of H in one molecule = 12 * 1 = 12 Mass of O in one molecule = 4 * 16 = 64 Total mass of one molecule (or molar mass) = 72 + 12 + 64 = 148. Let's evaluate the options: A) 22 atoms per mole: One molecule contains 6 + 12 + 4 = 22 atoms. One mole contains Avogadro's number of molecules, so one mole contains 22 * Avogadro's number of atoms, not just 22 atoms. Incorrect. B) twice the mass percent of H as compared to the mass percent of C: Mass of C is 72, Mass of H is 12. Is %H = 2 * %C? (12\/148)*100% vs 2 * (72\/148)*100%. This simplifies to 12 vs 2*72=144. Clearly incorrect. Is %C = 2 * %H? 72 vs 2*12=24. Incorrect. C) six times the mass percent of C as compared to the mass percent of H: This is poorly phrased but implies a ratio. Let's check if %C = 6 * %H. Mass of C is 72, Mass of H is 12. 72 = 6 * 12. Yes, the mass of Carbon in the compound is six times the mass of Hydrogen. Since the mass percentages are (mass of element \/ total mass) * 100%, if Mass of C = 6 * Mass of H, then %C = 6 * %H. This holds true. D) thrice the mass percent of H as compared to the mass percent of O: Mass of H is 12, Mass of O is 64. Is %O = 3 * %H? 64 vs 3*12=36. Incorrect. Is %H = 3 * %O? 12 vs 3*64. Incorrect. Option C is the only statement that accurately reflects the mass composition of the compound, interpreting \"six times the mass percent of C as compared to the mass percent of H\" as the mass percent of C being six times the mass percent of H.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-compound-c-6-h-12-o-4-contains\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"The compound C 6 H 12 O 4 contains"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88362","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88362"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88362\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88362"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88362"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}