{"id":249,"date":"2017-08-21T04:50:03","date_gmt":"2017-08-21T04:50:03","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=249"},"modified":"2017-09-08T10:43:56","modified_gmt":"2017-09-08T10:43:56","slug":"percent-composition-and-formulas","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/percent-composition-and-formulas\/","title":{"rendered":"Percent Composition and Formulas"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/acids-and-oxides\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/chemical-naming-and-structure\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-3\">Review \u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Percent Composition and Formulas<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review percent composition, empirical formulas, and molecular formulas.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Acids are formed from the combination of hydrogen ion(s) and an anion.<\/li>\n<li>Acid names are generated from the anion name as follows:\n<ul>\n<li>For ions that end with \u201c\u2013ide,\u201d place \u201chydro\u201d in front of the anion name and replace \u201c\u2013ide\u201d with \u201c\u2013ic acid.\u201d<\/li>\n<li>For ions that end with \u201c\u2013ite,\u201d replace \u201c\u2013ite\u201d with \u201c\u2013ous acid.\u201d<\/li>\n<li>For ions that end with \u201c\u2013ate,\u201d replace \u201c\u2013ate\u201d with \u201c\u2013ic acid.\u201d<\/li>\n<\/ul>\n<\/li>\n<li>Metal oxides react with water to form metal hydroxides, which are basic (alkaline).<\/li>\n<li>Nonmetal oxides react with water to form oxyacids.<\/li>\n<li>When nonmetal oxides react with water, the oxidation number of the nonmetal does not change.<\/li>\n<\/ul>\n<section>\n<h3>Percent Composition<\/h3>\n<p>Chemical formulas are extremely useful, in that they show us the ratio of atoms in a compound. And because a mole is simply a whole number multiple of atoms, a formula also shows us the ratio of moles. Thus, water\u2019s formula, H<sub>2<\/sub>O, lets us know that the compound contains two atoms of hydrogen for every atom of oxygen or two moles of hydrogen for every mole of oxygen.<\/p>\n<p>But in the chemistry laboratory, we often measure mass. And because atoms of different elements can have widely different masses, the ratio of atoms or moles is often radically different from the ratio of masses. For this reason, we sometimes use percent composition.<\/p>\n<p>Percent composition is nothing more than the percent by mass of each element in a compound. When a compound\u2019s formula is known, it is most easily calculated by determining the compound\u2019s molar mass, element by element. We then divide the mass of each element by the total mass of the compound and multiply by 100%.<\/p>\n<p>Consider calcium sulfate, CaSO<sub>4<\/sub>. A mole of this compound contains 40.1 g Ca (1 mole), 32.1 g S (1 mole) and 64.0 g O (4 moles), for a total mass of 136.2 g. The percent composition of CaSO<sub>4<\/sub> would be calculated as follows:<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_001.gif\" \/><\/center>The percent composition shows us that while oxygen comprises two-thirds of the atoms in this compound, it contributes less than half of the mass.<\/p>\n<h3>Empirical Formula<\/h3>\n<p>When we analyze an unknown substance in the laboratory, we usually obtain mass data first. From that data, we would like to determine the compound\u2019s chemical formula. The first step in this process is to determine the empirical formula.<\/p>\n<p>To find an empirical formula, it is helpful to remember two things. First, the ratio of atoms is the same as the ratio of moles; therefore, any data that we have must be converted to moles. Second, since compounds have definite composition, we may choose any sample size we wish when calculating.<\/p>\n<p>Consider a compound whose composition is 74.5% Pb and 25.5% Cl. Again, any sample size will work, but if we choose 100 grams, our calculation will start off more easily, because the sample will contain 74.5 g Pb and 25.5 g Cl. Our first step is to convert these masses into moles using molar mass, as shown below:<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_003.gif\" \/><\/center>Now we have a ratio of moles, but it is not a whole number ratio. The whole number ratio may be obvious to you by inspection, but if it is not, the next step is to divide each number of moles by the smaller number. This reduces the ratio so that it contains \u201c1\u201d as its smallest number. This step is shown below:<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_004.gif\" \/><\/center>Now we can clearly see that we have a 1:2 ratio, which is an empirical formula of PbCl<sub>2<\/sub>.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Which of the following is the correct empirical formula for a compound that is 30.4% N and 69.6% O by mass?<\/p>\n<ol>\n<li>NO<\/li>\n<li>NO<sub>2<\/sub><\/li>\n<li>N<sub>2<\/sub>O<\/li>\n<li>N<sub>2<\/sub>O<sub>4<\/sub><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is B. We may eliminate choice D immediately as its ratio can be reduced, so it is not an empirical formula. The calculation below shows how the empirical formula of NO<sub>2<\/sub> is found:<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_005.gif\" \/>\n<\/p>\n<\/section>\n<h3>Molecular Formula<\/h3>\n<p>The empirical formula is a valuable first step, but does\u00a0not provide enough\u00a0information to be of great use. After all, glucose (C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>),\u00a0acetic acid (CH<sub>3<\/sub>COOH) and formaldehyde (CH<sub>2<\/sub>O)\u00a0all have\u00a0the same 1:2:1 ratio of C:H:O, but they have vastly different\u00a0properties. What\u00a0we really want is the <abbr title=\" A formula that shows the actual number of atoms in a compound\">molecular\u00a0formula<\/abbr>.<\/p>\n<p>Let\u2019s consider a compound that is 43.7% P and\u00a056.3% O. We would begin by\u00a0calculating its empirical formula as follows:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_006.gif\" alt=\"placeholder.png\" width=\"364\" height=\"90\" \/><\/center>This example brings up an interesting point. The\u00a0empirical formula is the\u00a0lowest whole number ratio, but it is not required to have\u00a0\u201c1\u201d in it. Our\u00a0calculations, however, will always reduce the lowest number in the\u00a0ratio to\u00a0\u201c1\u201d. Seeing 2.50 moles of O should tell us that we\u00a0need to multiply the entire\u00a0ratio by two, which would be an empirical formula of P<sub>2<\/sub>O<sub>5<\/sub>.\u00a0To go further, we require additional information. The\u00a0molecular formula will\u00a0always be a whole number multiple of the empirical formula. If we know<br \/>\nthe\u00a0molar mass of the compound, we may divide it by the mass of the\u00a0empirical\u00a0formula to obtain this whole number. For the purpose of the example in\u00a0progress, experiment has shown the compound\u2019s molar mass\u00a0to be\u00a0284 grams per mole. The mass of the empirical\u00a0formula is therefore\u00a0162 grams per mole [(2\u00d731.0)+(5\u00d716.0)]. The\u00a0multiple is calculated by dividing, as shown below:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_008.gif\" alt=\"placeholder.png\" width=\"300\" height=\"80\" \/><\/center>We need to multiply the empirical formula of P<sub>2<\/sub>O<sub>5<\/sub> by two to obtain the correct molecular formula of P<sub>4<\/sub>O<sub>10<\/sub>.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Which of the following is the correct molecular formula\u00a0for a compound that is 26.7% C, 2.24% H and 71.1% O, and has a molar\u00a0mass of\u00a090.0 grams per mole?<\/p>\n<ol>\n<li>C<sub>2<\/sub>H<sub>2<\/sub>O<sub>4<\/sub><\/li>\n<li>C<sub>4<\/sub>H<sub>6<\/sub>O<sub>4<\/sub><\/li>\n<li>C<sub>2<\/sub>H<sub>4<\/sub>O<sub>2<\/sub><\/li>\n<li>C<sub>3<\/sub>H<sub>6<\/sub>O<sub>3<\/sub><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is A.\u00a0We must first calculate the empirical formula of CHO<sub>2<\/sub> as follows:<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_009.gif\" \/><\/p>\n<p class=\"q-reveal\">Now we find the whole number multiple by dividing the given molar mass\u00a0by the mass of the empirical formula that we just calculated, as shown\u00a0below:<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_010.gif\" \/><br \/>\nMultiplying the empirical formula through by two gives us C<sub>2<\/sub>H<sub>2<\/sub>O<sub>4<\/sub>.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Which of the following is the correct empirical formula for a compound that is 26.6% K, 35.4% Cr, and 38.1% O?<\/p>\n<ol>\n<li>KCrO<sub>4<\/sub><\/li>\n<li>K<sub>2<\/sub>CrO<sub>4<\/sub><\/li>\n<li>K<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub><\/li>\n<li>KCrO<sub>7<\/sub><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Choice C is correct. The formula of K<sub>2<\/sub>Cr<sub>2<\/sub>O<sub>7<\/sub> is found by multiplying by two the numbers from the calculation shown below:<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_012.gif\" \/><\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>What is the correct molecular formula for a compound that is 57.1% C, 4.81% H, and 38.1% O, and that has a molar mass of 126 grams per mole?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The answer is C<sub>6<\/sub>H<sub>6<\/sub>O<sub>3<\/sub>. The empirical formula is calculated as follows:<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_013.gif\" \/><br \/>\nOur empirical formula is C<sub>2<\/sub>H<sub>2<\/sub>O. Taking the given molar mass and dividing by the mass of the empirical formula will give us the multiple, as follows:<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s9_014.gif\" \/><br \/>\nMultiplying the empirical formula by three gives the result, C<sub>6<\/sub>H<sub>6<\/sub>O<sub>3<\/sub>.<\/p>\n<\/section>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/acids-and-oxides\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/chemical-naming-and-structure\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-3\">Review \u27a1<\/a><\/p>\n<\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Review \u27a1 Percent Composition and Formulas Objective In this lesson we will review percent composition, empirical formulas, and molecular formulas. Previously we covered&#8230; Acids are formed from the combination of hydrogen ion(s) and an anion. Acid names are generated from the anion name as follows: For ions that end with \u201c\u2013ide,\u201d place [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-249","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/249","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=249"}],"version-history":[{"count":11,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/249\/revisions"}],"predecessor-version":[{"id":699,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/249\/revisions\/699"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=249"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}