{"id":2925,"date":"2016-08-24T19:49:09","date_gmt":"2016-08-24T19:49:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2925"},"modified":"2017-08-31T19:09:37","modified_gmt":"2017-08-31T19:09:37","slug":"calculating-molar-mass","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/calculating-molar-mass\/","title":{"raw":"Calculating Molar Mass","rendered":"Calculating Molar Mass"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YzY2MGY4ZWUyZjQ1ZWY2ZDQ3MmRhZTA3NTFmMDRkYzc.-ub1\">\r\n \t<li>Perform calculations for the determination of molar mass from changes in boiling or freezing points.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>How much antifreeze is needed?<\/h3>\r\n[caption id=\"\" align=\"aligncenter\" width=\"200\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212517\/20140811155609739434.jpeg\" alt=\"It is possible to calculate quantities like molecular weight from freezing point depression\" width=\"200\" height=\"500\" \/> Pouring antifreeze. From <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:USMC-110329-M-PE262-004.jpg\">Wikimedia<\/a>.[\/caption]\r\n<p id=\"x-ck12-MGU2OWYxOTYxZjkzOTljN2ZjMDk4YjQ5MWUyM2Y5NDM.-uec\">We know that we can put antifreeze into a radiator and keep an engine from freezing up. By knowing how cold it will get and how much water is in the radiator, we can determine how much antifreeze to add to achieve our desired freezing point depression. We can do this because we know what the antifreeze is. Can we switch things around and get some information about the properties of the antifreeze (such as its molecular weight) from the freezing point decrease? It turns out that we can do this fairly easily and accurately.<\/p>\r\n\r\n<\/div>\r\n<div id=\"x-ck12-MmY0YmEzNTIxY2RlYzAyYjZkNTdmYWM5NmM4OGRjYmI.-dvd\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2NDg5NDU4MS0wMS02LUltYWdlLS0tMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212522\/20140811155609916214.jpeg\" alt=\"Changes in melting point and boiling point can be used to determine molecular weight\" width=\"500\" height=\"341\" longdesc=\"Changes%20in%20temperature.\" \/> Figure 1. Changes in temperature. Figure by Laura Guerin, CK-12 Foundation.[\/caption]\r\n\r\n<\/div>\r\n<h2>Calculating Molar Mass<\/h2>\r\n<p id=\"x-ck12-ZGRiOGYyNWNkNTllMjgyZTkzYThkMTA4OTg1NTIyZjM.-ezv\">In the laboratory, freezing point or boiling point data can be used to determine the molar mass of an unknown solute. Since we know the relationship between a decrease in freezing point and the concentration of solute, if we dissolve a known mass of our unknown solute into a known amount of solvent, we can calculate the molar mass of the solute. The\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/d358324009cfdaad1a755c105a2819f8.png\" alt=\"K_f\" width=\"22\" height=\"18\" \/> or\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of the solvent must be known. We also need to know if the solute is an electrolyte or a nonelectrolyte. If the solvent is an electrolyte, you would need to know the number of ions is produced when it dissociates.<\/p>\r\n\r\n<h3>Sample Problem: Molar Mass from Freezing Point Depression<\/h3>\r\n<p id=\"x-ck12-NGM5ZDU3MDFkYmQwODQ2NGIxMjNjY2ZjZWVmMTIxMmI.-pzk\">38.7 g of a nonelectrolyte is dissolved into 218 g of water. The freezing point of the solution is measured to be -5.53\u00b0C. Calculate the molar mass of the solute.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-wqb\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-lo3\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MjA2MzczNjU5YTlkMjY2NzUxYjQ2MWU1OTAzMDRjMGY.-ky2\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212524\/335f7628c3c8b6039277bf9105e8d918.png\" alt=\"Delta T_f=-5.53^circ text{C}\" width=\"129\" height=\"19\" \/><\/em><\/li>\r\n \t<li>mass H <sub> 2 <\/sub> O 218 g = 0.218 kg<\/li>\r\n \t<li>mass solute = 38.7 g<\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212524\/9d645bc14534e15572a0a5e24ae22f71.png\" alt=\"K_f(text{H}_2text{O})=-1.86^circ text{C}\/m\" width=\"193\" height=\"20\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bwd\">Unknown<\/p>\r\n\r\n<ul id=\"x-ck12-MGVlMWNlZWNmNTg4OWVhNzQxYTM5ODQ5M2NiNjJkZGI.-tfx\">\r\n \t<li>molar mass solute = ? g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZTY1MWU5YTY5ZGM1NThlMWYzZjNhMjcyNTQ2NDUxMzI.-jfl\">Use the freeing point depression <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212525\/a8438cf659fbbfcd6f776ebee1f7356a.png\" alt=\"(Delta T_f)\" width=\"46\" height=\"20\" \/> \u00a0to calculate the molality of the solution. Then use the molality equation to calculate the moles of solute. Then divide the grams of solute by the moles to determine the molar mass.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-yrl\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-h9u\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212525\/799827c2aecfdda518f2b4b77e357767.png\" alt=\"m=frac{Delta T_f}{K_f} &amp;=frac{-5.53^circ text{C}}{-1.86^circ text{C}\/m}=2.97 m\\text{mol solute} &amp;= m times text{kg H}_2text{O}=2.97 m times 0.218 text{ kg} = 0.648 text{ mol}\\frac{38.7 text{ g}}{0.648 text{ mol}} &amp;= 59.7 text{ g\/mol}\" width=\"466\" height=\"115\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-bdl\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NzEwYmE1Y2U0NmY1ZTM5OTUyMjEyM2VhMDE2MTdkMDI.-arx\">The molar mass of the unknown solute is 59.7 g\/mol. Knowing the molar mass is an important step in determining the identity of an unknown. A similar problem could be done with the change in boiling point.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-NDIwODhmZmZiODY1MTIxMGFiYjUzZmEwYWFmM2EzOTg.-vru\">\r\n \t<li>Determination of molar mass by measuring freezing point depression is described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZDJhNWU1Y2FmMTUzMzdhOTQ4N2YxZTMyODlhOTE3NTk.-4xg\">\r\n \t<li>What do we need to know about the solvent to use this technique?<\/li>\r\n \t<li>Will it work with ionizable compounds?<\/li>\r\n \t<li>Can we use boiling point elevation to determine molar mass?<\/li>\r\n<\/ol>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YzY2MGY4ZWUyZjQ1ZWY2ZDQ3MmRhZTA3NTFmMDRkYzc.-ub1\">\n<li>Perform calculations for the determination of molar mass from changes in boiling or freezing points.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>How much antifreeze is needed?<\/h3>\n<div style=\"width: 210px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212517\/20140811155609739434.jpeg\" alt=\"It is possible to calculate quantities like molecular weight from freezing point depression\" width=\"200\" height=\"500\" \/><\/p>\n<p class=\"wp-caption-text\">Pouring antifreeze. From <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:USMC-110329-M-PE262-004.jpg\">Wikimedia<\/a>.<\/p>\n<\/div>\n<p id=\"x-ck12-MGU2OWYxOTYxZjkzOTljN2ZjMDk4YjQ5MWUyM2Y5NDM.-uec\">We know that we can put antifreeze into a radiator and keep an engine from freezing up. By knowing how cold it will get and how much water is in the radiator, we can determine how much antifreeze to add to achieve our desired freezing point depression. We can do this because we know what the antifreeze is. Can we switch things around and get some information about the properties of the antifreeze (such as its molecular weight) from the freezing point decrease? It turns out that we can do this fairly easily and accurately.<\/p>\n<\/div>\n<div id=\"x-ck12-MmY0YmEzNTIxY2RlYzAyYjZkNTdmYWM5NmM4OGRjYmI.-dvd\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2NDg5NDU4MS0wMS02LUltYWdlLS0tMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212522\/20140811155609916214.jpeg\" alt=\"Changes in melting point and boiling point can be used to determine molecular weight\" width=\"500\" height=\"341\" longdesc=\"Changes%20in%20temperature.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Changes in temperature. Figure by Laura Guerin, CK-12 Foundation.<\/p>\n<\/div>\n<\/div>\n<h2>Calculating Molar Mass<\/h2>\n<p id=\"x-ck12-ZGRiOGYyNWNkNTllMjgyZTkzYThkMTA4OTg1NTIyZjM.-ezv\">In the laboratory, freezing point or boiling point data can be used to determine the molar mass of an unknown solute. Since we know the relationship between a decrease in freezing point and the concentration of solute, if we dissolve a known mass of our unknown solute into a known amount of solvent, we can calculate the molar mass of the solute. The\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/d358324009cfdaad1a755c105a2819f8.png\" alt=\"K_f\" width=\"22\" height=\"18\" \/> or\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of the solvent must be known. We also need to know if the solute is an electrolyte or a nonelectrolyte. If the solvent is an electrolyte, you would need to know the number of ions is produced when it dissociates.<\/p>\n<h3>Sample Problem: Molar Mass from Freezing Point Depression<\/h3>\n<p id=\"x-ck12-NGM5ZDU3MDFkYmQwODQ2NGIxMjNjY2ZjZWVmMTIxMmI.-pzk\">38.7 g of a nonelectrolyte is dissolved into 218 g of water. The freezing point of the solution is measured to be -5.53\u00b0C. Calculate the molar mass of the solute.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-wqb\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-lo3\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-MjA2MzczNjU5YTlkMjY2NzUxYjQ2MWU1OTAzMDRjMGY.-ky2\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212524\/335f7628c3c8b6039277bf9105e8d918.png\" alt=\"Delta T_f=-5.53^circ text{C}\" width=\"129\" height=\"19\" \/><\/em><\/li>\n<li>mass H <sub> 2 <\/sub> O 218 g = 0.218 kg<\/li>\n<li>mass solute = 38.7 g<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212524\/9d645bc14534e15572a0a5e24ae22f71.png\" alt=\"K_f(text{H}_2text{O})=-1.86^circ text{C}\/m\" width=\"193\" height=\"20\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bwd\">Unknown<\/p>\n<ul id=\"x-ck12-MGVlMWNlZWNmNTg4OWVhNzQxYTM5ODQ5M2NiNjJkZGI.-tfx\">\n<li>molar mass solute = ? g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ZTY1MWU5YTY5ZGM1NThlMWYzZjNhMjcyNTQ2NDUxMzI.-jfl\">Use the freeing point depression <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212525\/a8438cf659fbbfcd6f776ebee1f7356a.png\" alt=\"(Delta T_f)\" width=\"46\" height=\"20\" \/> \u00a0to calculate the molality of the solution. Then use the molality equation to calculate the moles of solute. Then divide the grams of solute by the moles to determine the molar mass.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-yrl\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-h9u\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212525\/799827c2aecfdda518f2b4b77e357767.png\" alt=\"m=frac{Delta T_f}{K_f} &amp;=frac{-5.53^circ text{C}}{-1.86^circ text{C}\/m}=2.97 m\\text{mol solute} &amp;= m times text{kg H}_2text{O}=2.97 m times 0.218 text{ kg} = 0.648 text{ mol}\\frac{38.7 text{ g}}{0.648 text{ mol}} &amp;= 59.7 text{ g\/mol}\" width=\"466\" height=\"115\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-bdl\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NzEwYmE1Y2U0NmY1ZTM5OTUyMjEyM2VhMDE2MTdkMDI.-arx\">The molar mass of the unknown solute is 59.7 g\/mol. Knowing the molar mass is an important step in determining the identity of an unknown. A similar problem could be done with the change in boiling point.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-NDIwODhmZmZiODY1MTIxMGFiYjUzZmEwYWFmM2EzOTg.-vru\">\n<li>Determination of molar mass by measuring freezing point depression is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZDJhNWU1Y2FmMTUzMzdhOTQ4N2YxZTMyODlhOTE3NTk.-4xg\">\n<li>What do we need to know about the solvent to use this technique?<\/li>\n<li>Will it work with ionizable compounds?<\/li>\n<li>Can we use boiling point elevation to determine molar mass?<\/li>\n<\/ol>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2925\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":29,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2925","chapter","type-chapter","status-publish","hentry"],"part":2337,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2925","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/29"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2925\/revisions"}],"predecessor-version":[{"id":3690,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2925\/revisions\/3690"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2337"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2925\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2925"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2925"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2925"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2925"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}