{"id":217345,"date":"2025-05-21T18:33:43","date_gmt":"2025-05-21T18:33:43","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217345"},"modified":"2025-05-21T18:33:46","modified_gmt":"2025-05-21T18:33:46","slug":"the-molar-mass-of-butane-c4h10-should-be-58-g-mo","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/21\/the-molar-mass-of-butane-c4h10-should-be-58-g-mo\/","title":{"rendered":"The molar mass of butane (C4H10) should be 58 g\/mo"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The molar mass of butane (C4H10) should be 58 g\/mol (as calculated from periodic table). If a student measured the molar mass of butane to be 65, what is the % error of the measurement? Use no decimal place, such as &#8220;5&#8221;, or &#8220;13&#8221;.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the <strong>percentage error<\/strong>, we use the following formula: Percentage&nbsp;Error=(\u2223Measured&nbsp;Value\u2212Actual&nbsp;Value\u2223Actual&nbsp;Value)\u00d7100\\text{Percentage Error} = \\left( \\frac{|\\text{Measured Value} &#8211; \\text{Actual Value}|}{\\text{Actual Value}} \\right) \\times 100<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In this problem:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Measured Value<\/strong> = 65 g\/mol<\/li>\n\n\n\n<li><strong>Actual Value<\/strong> = 58 g\/mol (the correct molar mass of butane)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Now substitute into the formula: Percentage&nbsp;Error=(\u222365\u221258\u222358)\u00d7100=(758)\u00d7100\u224812.07\\text{Percentage Error} = \\left( \\frac{|65 &#8211; 58|}{58} \\right) \\times 100 = \\left( \\frac{7}{58} \\right) \\times 100 \\approx 12.07<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Rounded to <strong>no decimal place<\/strong>, the percentage error is: 12\\boxed{12}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (Approximately 300 Words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Percentage error is a common way to express the accuracy of a measurement by comparing the measured value to the actual or accepted value. It helps identify how far off a result is from the truth, which is useful in experimental and laboratory settings.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In this case, we are comparing the <strong>measured molar mass of butane<\/strong>, 65 g\/mol, to the <strong>actual molar mass<\/strong>, which is <strong>58 g\/mol<\/strong>. The actual value is determined using atomic masses from the periodic table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) has an atomic mass of about 12.01 g\/mol.<\/li>\n\n\n\n<li>Hydrogen (H) has an atomic mass of about 1.008 g\/mol.<\/li>\n\n\n\n<li>Butane (C\u2084H\u2081\u2080) has:<br>4\u00d712.01+10\u00d71.008=48.04+10.08=58.12\u2009g\/mol4 \\times 12.01 + 10 \\times 1.008 = 48.04 + 10.08 = 58.12 \\, \\text{g\/mol},<br>which is commonly rounded to <strong>58 g\/mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">The student\u2019s value of 65 g\/mol is higher than the actual value. The absolute difference is 7 g\/mol. To express how significant this difference is relative to the actual value, we divide by 58 and multiply by 100. This gives a <strong>percentage error of about 12.07%<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By rounding to the nearest whole number (as instructed), we get a final answer of <strong>12%<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Percentage error helps evaluate the reliability and precision of experimental procedures. A smaller percentage indicates a more accurate measurement, while larger errors suggest the need to improve methods, calibration, or technique.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The molar mass of butane (C4H10) should be 58 g\/mol (as calculated from periodic table). If a student measured the molar mass of butane to be 65, what is the % error of the measurement? Use no decimal place, such as &#8220;5&#8221;, or &#8220;13&#8221;. The Correct Answer and Explanation is: To find the percentage error, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-217345","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217345","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=217345"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217345\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217345"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217345"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217345"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}