{"id":268,"date":"2012-04-10T17:15:41","date_gmt":"2012-04-10T17:15:41","guid":{"rendered":"https:\/\/wordpress.clarku.edu\/debrobertson\/?page_id=268"},"modified":"2012-05-23T17:20:16","modified_gmt":"2012-05-23T17:20:16","slug":"introduction-to-chemical-concepts","status":"publish","type":"page","link":"https:\/\/wordpress.clarku.edu\/debrobertson\/laboratory-protocols\/introduction-to-chemical-concepts\/","title":{"rendered":"Introduction to Chemical Concepts"},"content":{"rendered":"<p><center><\/center><em>EXPRESSIONS OF SOLUTES:<\/em><\/p>\n<ul>\n<li><strong><span style=\"color: #0080ff;\">MOLARITY<\/span><\/strong>\u00a0(M):\n<ul>\n<li>by definition, 1 mole per liter.<\/li>\n<li>Solvent is added and brought up to 1 L<\/li>\n<\/ul>\n<\/li>\n<li><strong><span style=\"color: #0080ff;\">MOLALITY<\/span><\/strong>\u00a0(m):\n<ul>\n<li>by definition, 1 mole per kg of solvent<\/li>\n<li>for example, an aqueous solution with water, 1 kg of H<sub>2<\/sub>0 is approximately 1 L<\/li>\n<li>1mole is added to 1 kg of solvent<\/li>\n<\/ul>\n<\/li>\n<li><strong><span style=\"color: #0080ff;\">NORMALITY<\/span><\/strong>\u00a0(N):\n<ul>\n<li>by definition, one equivalent per liter<\/li>\n<li>What is an equivalent?\n<ul>\n<li>Molecular weight divided by the number of replaceable protons. For example, HCl there is one replaceable proton and therefore normality = molarity<\/li>\n<li>In contrast, H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0or H<sub>3<\/sub>PO<sub>4<\/sub>\u00a0have two and three replaceable protons, respectively. In the case of H<sub>2<\/sub>SO<sub>4<\/sub>, a 1 N solution would be equal to a 0.5 M solution.<\/li>\n<\/ul>\n<\/li>\n<li>The advantage of normality is in titration, 1 equivalent of acid will completely normalize one equivalent of base.<\/li>\n<\/ul>\n<\/li>\n<li><strong><span style=\"color: #0080ff;\">EQUILAVENTS<\/span><\/strong>\u00a0(Eq):\n<ul>\n<li>Electrolyte concentrations are expressed in milliequivalents (mEq)<\/li>\n<li>For single elements, an equivalent is the molecular weight divided by the absolute value of the oxidative state.\n<ul>\n<li>Na<sup>+1<\/sup>: 23 gm = 1 Eq and 23 mg = 1 mEq<\/li>\n<li>Cl<sup>-1<\/sup>: 35.5 gm = 1 Eq and 35.5 mg= 1 mEq<\/li>\n<li>Ca<sup>+2<\/sup>: 20.0 gm = 1 Eq and 20 mg = 1 mEq<\/li>\n<\/ul>\n<\/li>\n<li>For compounds, equivalents are calculated as the molecular weight divided by the absolute value of the highest oxidative state\n<ul>\n<li>NaCl: 58.5 gm = 1 Eq NaCl<\/li>\n<li>CaCl<sub>2<\/sub>: 55.5 gm = 1 Eq CaCl<sub>2<\/sub><\/li>\n<li>AlCl3: 44 gm = 1 Eq AlCl<sub>2<\/sub><\/li>\n<\/ul>\n<\/li>\n<li>Equivalent weights do not always remain the same since they result from the net electron change<\/li>\n<\/ul>\n<\/li>\n<li><strong><span style=\"color: #0080ff;\">PERCENT COMPOSITION SOLUTIONS<\/span><\/strong>\n<ul>\n<li>weight\/weight (w\/w)\n<ul>\n<li>gm of solute per gm of solvent<\/li>\n<li>Example:\n<ul>\n<li>10% (w\/w) NaOH: 10 gm of NaOH and 90 gm H<sub>2<\/sub>0<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>weight\/volume (w\/v)\n<ul>\n<li>Example:\n<ul>\n<li>10% (w\/v) NaOH: 10 gm of NaOH dissolved in a small amount of solvent then brought up to 100 mL<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>volume\/volume (v\/v)\n<ul>\n<li>Example:\n<ul>\n<li>30% Ethanol: 30 mL of 100% Ethanol and 70 mL H<sub>2<\/sub>0<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li>CLINICAL EXAMPLES\n<ul>\n<li>mg % = mg\/dL = mg\/100 mL<\/li>\n<\/ul>\n<\/li>\n<li>TRACE METALS\n<ul>\n<li>ppm (parts per million): mg \/ L<\/li>\n<li>ppb (parts per billion) \u00b5g \/ L<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>EXPRESSIONS OF SOLUTES: MOLARITY\u00a0(M): by definition, 1 mole per liter. Solvent is added and brought up to 1 L MOLALITY\u00a0(m): by definition, 1 mole per kg of solvent for example, an aqueous solution with water, 1 kg of H20 is &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/wordpress.clarku.edu\/debrobertson\/laboratory-protocols\/introduction-to-chemical-concepts\/\"> <span class=\"screen-reader-text\">Introduction to Chemical Concepts<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":72,"featured_media":0,"parent":8,"menu_order":80,"comment_status":"closed","ping_status":"closed","template":"onecolumn-page.php","meta":{"ngg_post_thumbnail":0,"footnotes":""},"class_list":["post-268","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/pages\/268","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/users\/72"}],"replies":[{"embeddable":true,"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/comments?post=268"}],"version-history":[{"count":0,"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/pages\/268\/revisions"}],"up":[{"embeddable":true,"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/pages\/8"}],"wp:attachment":[{"href":"https:\/\/wordpress.clarku.edu\/debrobertson\/wp-json\/wp\/v2\/media?parent=268"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}