A sample containing the amino acid alanine, CH3CH(NH2)COOH, plus inert matter is analyzed by the Kjeldahl method. A 2.00-g sample is digested, the NH3 is distilled and collected in 50.0 mL of 0.150 M H2SO4, and a volume of 9.0 mL of 0.100 M NaOH is required for back-titration. Calculate the percent alanine in the sample.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the percent alanine<\/strong> in the sample using the Kjeldahl method<\/strong>, we need to calculate how much nitrogen (N) originated from alanine and then relate it to the molecular weight of alanine to find its mass fraction in the sample.<\/p>\n\n\n\n $$ $$ H\u2082SO\u2084 reacts with NH\u2083 in a 1:2 ratio (1 mol H\u2082SO\u2084 reacts with 2 mol NH\u2083). But for titration purposes, we use:<\/p>\n\n\n\n $$ $$ Since 1 mol H\u2082SO\u2084 traps 2 mol NH\u2083:<\/p>\n\n\n\n $$ This is the amount of nitrogen in moles (1 mol NH\u2083 = 1 mol N).<\/p>\n\n\n\n $$ Alanine (CH\u2083CH(NH\u2082)COOH) has one N atom per molecule<\/strong>. Molar mass = 89.09 g\/mol.<\/p>\n\n\n\n From nitrogen content:<\/p>\n\n\n\n $$ $$ $$ The Kjeldahl method<\/strong> is a classical technique for determining nitrogen content in organic compounds. Since amino acids like alanine<\/strong> contain nitrogen (from the amino group, \u2013NH\u2082), the amount of nitrogen in a sample can be used to determine the quantity of alanine present.<\/p>\n\n\n\n In the procedure described, a 2.00 g sample<\/strong> containing alanine and inert (non-nitrogen-containing) matter is digested with sulfuric acid, converting nitrogen into ammonium ions (NH\u2084\u207a). These are then released as ammonia (NH\u2083) by making the solution basic and are trapped in a known volume and concentration of sulfuric acid (H\u2082SO\u2084). The remaining (unreacted) acid is then back-titrated<\/strong> with NaOH to determine how much acid was actually used to trap NH\u2083.<\/p>\n\n\n\n The amount of H\u2082SO\u2084 that reacted with NH\u2083 gives the number of moles of ammonia, and therefore nitrogen. Since alanine contains one nitrogen atom per molecule<\/strong>, the number of moles of nitrogen is equal to the number of moles of alanine. Using alanine’s molar mass (89.09 g\/mol)<\/strong>, we calculate how much alanine corresponds to that nitrogen amount.<\/p>\n\n\n\n Finally, the mass of alanine<\/strong> is divided by the total sample mass to calculate the percentage of alanine<\/strong>. This allows us to determine how much of the original 2.00 g sample was actually alanine. The final result is that 58.8%<\/strong> of the sample is alanine.<\/p>\n","protected":false},"excerpt":{"rendered":" A sample containing the amino acid alanine, CH3CH(NH2)COOH, plus inert matter is analyzed by the Kjeldahl method. A 2.00-g sample is digested, the NH3 is distilled and collected in 50.0 mL of 0.150 M H2SO4, and a volume of 9.0 mL of 0.100 M NaOH is required for back-titration. Calculate the percent alanine in the […]<\/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-211857","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211857","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=211857"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211857\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211857"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211857"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211857"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep-by-step Solution:<\/strong><\/h3>\n\n\n\n
1. Acid amount (H\u2082SO\u2084) added to trap NH\u2083:<\/strong><\/h4>\n\n\n\n
\\text{mol H}_2\\text{SO}_4 = M \\times V = 0.150 \\, \\text{mol\/L} \\times 0.0500 \\, \\text{L} = 0.00750 \\, \\text{mol}
$$<\/p>\n\n\n\n2. Base (NaOH) used in back titration:<\/strong><\/h4>\n\n\n\n
\\text{mol NaOH} = 0.100 \\, \\text{mol\/L} \\times 0.0090 \\, \\text{L} = 0.00090 \\, \\text{mol}
$$<\/p>\n\n\n\n
\\text{mol H}_2\\text{SO}_4 \\text{ neutralized by NH}_3 = \\text{total mol H}_2\\text{SO}_4 – \\text{mol NaOH}
$$<\/p>\n\n\n\n
= 0.00750 – 0.00090 = 0.00660 \\, \\text{mol H}_2\\text{SO}_4}
$$<\/p>\n\n\n\n
\\text{mol NH}_3 = 2 \\times 0.00660 = 0.01320 \\, \\text{mol}
$$<\/p>\n\n\n\n3. Mass of nitrogen:<\/strong><\/h4>\n\n\n\n
\\text{mass N} = 0.01320 \\, \\text{mol} \\times 14.01 \\, \\text{g\/mol} = 0.1849 \\, \\text{g}
$$<\/p>\n\n\n\n4. Mass of alanine:<\/strong><\/h4>\n\n\n\n
\\text{mol alanine} = \\text{mol N} = 0.01320 \\, \\text{mol}
$$<\/p>\n\n\n\n
\\text{mass alanine} = 0.01320 \\times 89.09 = 1.176 \\, \\text{g}
$$<\/p>\n\n\n\n5. Percent alanine:<\/strong><\/h4>\n\n\n\n
\\% \\text{alanine} = \\frac{1.176}{2.00} \\times 100\\% = \\boxed{58.8\\%}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n