{"id":6941,"date":"2014-12-09T19:57:24","date_gmt":"2014-12-09T19:57:24","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=6941"},"modified":"2017-09-07T18:23:57","modified_gmt":"2017-09-07T18:23:57","slug":"chemistry-how-to-find-the-empirical-formula","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/chemistry-how-to-find-the-empirical-formula\/","title":{"rendered":"Chemistry:  how to find the empirical formula"},"content":{"rendered":"<h1>Tutoring high school chemistry, you&#8217;re asked this every year. \u00a0The tutor shows a simple example of how to find the empirical formula.<\/h1>\n<p>Back on September 12, I wrote a post on <a href=\"?p=5485\">percent composition<\/a>. Finding the empirical formula can be the reverse process: you know the percent composition, but need the chemical formula of the compound.<\/p>\n<p><strong>Example 1<\/strong>: \u00a0A compound consists of 60.00% carbon, 13.33% hydrogen, and 26.67% oxygen. Find its emprical formula.<\/p>\n<p>Solution:<\/p>\n<p>Step 1: Assume 100.00g sample.<\/p>\n<p>Step 2: Calculate the mass in grams of each atom present:<\/p>\n<table>\n<tbody>\n<tr>\n<th>atom<\/th>\n<th>percent composition<\/th>\n<th>mass present (g)<\/th>\n<\/tr>\n<tr>\n<td>carbon<\/td>\n<td>60.00%<\/td>\n<td>0.6000 x 100.00=60.00<\/td>\n<\/tr>\n<tr>\n<td>hydrogen<\/td>\n<td>13.33%<\/td>\n<td>0.1333 x 100.00=13.33<\/td>\n<\/tr>\n<tr>\n<td>oxygen<\/td>\n<td>26.67%<\/td>\n<td>0.2667 x 100=26.67<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Step 3: From the mass of each atom present, calculate its moles present. For instance, for carbon:<\/p>\n<p>60.00g x 1mol\/12.0g = 5.00 mol<\/p>\n<p>Similarly, for the hydrogen:<\/p>\n<p>13.33g x 1mol\/1.01g = 13.2 mol<\/p>\n<p>Finally, for oxygen:<\/p>\n<p>26.67g x 1mol\/16.0g = 1.67 mol<\/p>\n<p>We tabulate our results so far:<\/p>\n<table>\n<tbody>\n<tr>\n<th>atom<\/th>\n<th>moles present<\/th>\n<\/tr>\n<tr>\n<td>carbon<\/td>\n<td>5.00<\/td>\n<\/tr>\n<tr>\n<td>hydrogen<\/td>\n<td>13.2<\/td>\n<\/tr>\n<tr>\n<td>oxygen<\/td>\n<td>1.67<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Step 4: Find the atom with the least moles present. Divide the other atoms&#8217; moles by that one.<\/p>\n<p>In our case, oxygen has the least moles, with 1.67. We divide each of the moles present by 1.67, then round each answer to the nearest whole number. (For more about this, see the comments below):<\/p>\n<table>\n<tbody>\n<tr>\n<th>atom<\/th>\n<th>moles present<\/th>\n<th>moles present \u00f7 1.67<\/th>\n<th>rounded answer<\/th>\n<\/tr>\n<tr>\n<td>carbon<\/td>\n<td>5.00<\/td>\n<td>2.99<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>hydrogen<\/td>\n<td>13.2<\/td>\n<td>7.90<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>oxygen<\/td>\n<td>1.67<\/td>\n<td>1.00<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Step 5:<\/p>\n<p>From the rounded answers in the table in Step 4, we have our number of each atom:<\/p>\n<p>atomnumber in empirical formula<\/p>\n<table>\n<tbody>\n<tr>\n<td>carbon<\/td>\n<td>3<\/td>\n<td>hydrogen<\/td>\n<td>8<\/td>\n<td>oxygen<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now we know our empirical formula: C<sub>3<\/sub>H<sub>8<\/sub>O<\/p>\n<p>Comments:<\/p>\n<ol>\n<li>It&#8217;s very important to keep at least 4 digits until the end in order to preserve the ratios.<\/li>\n<li>Sometimes, after dividing out the lowest number of moles as done in Step 4, you get a number like 0.67 or 2.5, which is not very near the next whole number. I&#8217;ll treat such a case in a coming post.<\/li>\n<\/ol>\n<p>Jack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Tutoring high school chemistry, you&#8217;re asked this every year. \u00a0The tutor shows a simple example of how to find the empirical formula. Back on September 12, I wrote a post on percent composition. Finding the empirical formula can be the &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/chemistry-how-to-find-the-empirical-formula\/\"> <span class=\"screen-reader-text\">Chemistry:  how to find the empirical formula<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[341],"class_list":["post-6941","post","type-post","status-publish","format-standard","hentry","category-chemistry","tag-empirical-formula"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/6941","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=6941"}],"version-history":[{"count":27,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/6941\/revisions"}],"predecessor-version":[{"id":23448,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/6941\/revisions\/23448"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=6941"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=6941"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=6941"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}