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1.07:_Molecular_Weight_and_the_Mole
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<span id="Molecular_Weight"></span><h2 style="background-color: unset;" class="lt-bio-3747">Molecular Weight</h2> <p class="lt-bio-3747" style="background-color: unset;">The weight of a molecule is the sum of the weights of the atoms of which it is made. The unit of weight is the <strong>dalton</strong>, one-twelfth the weight of an atom of <sup><font size="2">12</font></sup>C. Thus the <strong>molecular weight</strong> (<strong>MW</strong>) of water is 18 daltons. (We shall ignore the tiny error introduced by the presence of traces of other isotopes - <sup><font size="2">17</font></sup>O, <sup><font size="2">18</font></sup>O, and <sup><font size="2">2</font></sup>H among the predominant <sup><font size="2">1</font></sup>H and <sup><font size="2">16</font></sup>O atoms.)</p> <p class="lt-bio-3747" style="background-color: unset;">Why is it important to know the molecular weight of a compound?</p> <p class="lt-bio-3747" style="background-color: unset;">For an example, let us assume that you want to study the response of honeybees to solutions of various kinds of sugars. One way to do this would be to make up several different solutions and see which one the bees prefer to harvest.</p> <p class="lt-bio-3747" style="background-color: unset;">You might offer the bees the choice between, say, a 35% solution of sucrose (common table sugar) and a 35% solution of glucose (a natural component of honey). This would involve, in each case, dissolving 350 parts by weight (e.g., grams) of sugar in 650 parts (g) of water, thus producing 1000 g of each solution. But there is a problem with this approach. The willingness of the honeybee to respond to the presence of sugar dissolved in water is dependent on the <strong>number</strong> of sugar molecules in a given volume of the solution.</p> <p class="lt-bio-3747" style="background-color: unset;">The sucrose molecule (MW = 342) is almost twice as heavy as the glucose molecule (MW = 180). So a 35% solution of glucose would contain almost twice as many molecules as a 35% solution of sucrose. To correct the problem, you should make the solution with the weights of sucrose and glucose in a ratio of 342:180. Then you would have the same concentration of molecules in each; that is, drop for drop, each solution would contain the same number of molecules.</p> <span id="Mole"></span><h2 style="background-color: unset;" class="lt-bio-3747">Mole</h2> <p class="lt-bio-3747" style="background-color: unset;">A mole is the quantity of a substance whose weight in grams is equal to the molecular weight of the substance. If you weight out exactly 342 grams (g) of sucrose, you will have weighed out <strong>1 mole</strong> of it. Thus 1 mole of glucose weighs 180 g. Furthermore, if you dissolve 1 mole of a substance in enough water to make 1 <strong>liter</strong> (<strong>L</strong>) of solution, you have made a <strong>1-molar</strong> (<strong>1 <em>M</em></strong>) solution.</p> <p class="lt-bio-3747" style="background-color: unset;">A <strong>1 <em>M</em></strong> solution of these sugars would probably be too strong for the experiment with the bees. It might be better to make up a liter of each solution containing 34.2 g and 18.0 g respectively. Such solutions would be designated <strong>one-tenth molar</strong> (<strong>0.1 <em>M</em></strong>) solutions. Drop for drop, these two solutions would still contain exactly the same number of molecules because they are of the same <strong>molarity</strong>.</p> <div class="box-note"><span id="Avogadro's_Number"></span><h5 class="box-legend"><span class="lt-icon-default">Avogadro's Number</span></h5> <p class="lt-bio-3747" style="background-color: unset;">How many molecules are there in a mole?</p> <p class="lt-bio-3747" style="background-color: unset;"><strong>Solution</strong></p> <p class="lt-bio-3747" style="background-color: unset;">The number is approximately 6 x 10<sup><font size="2">23</font></sup>. This number is called Avogadro's number after the chemist who first attempted to determine it.</p> <p class="lt-bio-3747" style="background-color: unset;">Avogadro's number applies to a mole of any substance: molecule or ion. Thus we can properly refer to a mole of hydrogen ions (1 g).</p> </div> <footer class="mt-content-footer"> <style>/*<![CDATA[*/#mt-toc-container {display: none !important;}/*]]>*/</style><script type="text/javascript">/*<![CDATA[*/ $(function() { if(!window['autoDefinitionList']){ window['autoDefinitionList'] = true; $('dl').find('dt').on('click', function() { $(this).next().toggle('350'); }); } });/*]]>*/</script> <script defer="true" src="https://static.cloudflareinsights.com/beacon.min.js" data-cf-beacon="{"token": "483ec2414e274209a7e93c253192df0b"}"></script><script src="https://cdn.libretexts.net/github/LibreTextsMain/Miscellaneous/h5p-resizer.js"></script><script src="https://cdnjs.cloudflare.com/ajax/libs/iframe-resizer/4.2.11/iframeResizer.contentWindow.min.js" integrity="sha512-FOf4suFgz7OrWmBiyyWW48u/+6GaaAFSDHagh2EBu/GH/1+OQSYc0NFGeGeZK0gZ3vuU1ovmzVzD6bxmT4vayg==" crossorigin="anonymous"></script><script src="https://cdnjs.cloudflare.com/ajax/libs/iframe-resizer/4.2.11/iframeResizer.min.js" integrity="sha512-HY1lApSG7xxx8mYzs/lxRs+c5AaDThRaa3pvQB6puiswvf2lWqMJVf+8qSGiL4ZXfHQoPIqbd1TlpqfycPo3cQ==" crossorigin="anonymous"></script><script>/*<![CDATA[*/window.addEventListener('load', function(){$('iframe').iFrameResize({warningTimeout:0, scrolling: 'omit'});})/*]]>*/</script><script>/*<![CDATA[*/ window.PageNum = "auto"; window.InitialOffset = "false"; window.PageName = "1.7: Molecular Weight and the Mole"; /*]]>*/</script> <script type="text/javascript">/*<![CDATA[*/ // var front = window.PageNum.trim(); if(front=="auto"){ front = window.PageName.replace('\"', '\\\"').trim(); //front = "'..string.matchreplace(PageName,'\"','\\\"')..'".trim(); if(front.includes(":")){ front = front.split(":")[0].trim(); if(front.includes(".")){ front = front.split("."); front = front.map((int)=>int.includes("0")?parseInt(int,10):int).join("."); } front+="."; } else { front = ""; } } front = front.trim(); function loadMathJaxScript() { try { const script = document.createElement('script'); script.id = "mathjax-script"; script.src = "https://cdn.jsdelivr.net/npm/mathjax@4/tex-mml-svg.js"; script.type = "text/javascript"; script.defer = true; document.head.appendChild(script); } catch (err) { console.error(err); } } document.addEventListener('DOMContentLoaded', (e) => { loadMathJaxScript(); }); if (window.PageName !== 'Realtime MathJax'){ MathJax = { options: { ignoreHtmlClass: "tex2jax_ignore", processHtmlClass: "tex2jax_process", menuOptions: { settings: { zscale: "150%", zoom: "Double-Click", assistiveMml: true, // true to enable assitive MathML collapsible: false, // true to enable collapsible math }, }, }, output: { scale: 0.85, mtextInheritFont: false, displayOverflow: "linebreak", linebreaks: { width: "100%", }, }, startup: { pageReady: () => { if (window.activateBeeLine) { window.activateBeeLine(); } return MathJax.startup.defaultPageReady(); }, }, chtml: { matchFontHeight: true, }, tex: { tags: "all", tagformat: { number: (n) => { if (window.InitialOffset) { const offset = Number(window.InitialOffset); if(!offset) { return front + n; // If offset is falsy (nan, undefined, etc.) } const added = Number(n) + offset; return front + added; } else { return front + n; } }, }, macros: { eatSpaces: ['#1', 2, ['', ' ', '\\endSpaces']], PageIndex: ['{' + front.replace(/\./g, '{.}') + '\\eatSpaces#1 \\endSpaces}', 1], test: ["{" + front + "#1}", 1], mhchemrightleftharpoons: "{\\unicode{x21CC}\\,}", xrightleftharpoons: ['\\mhchemxrightleftharpoons[#1]{#2}', 2, ''] }, packages: { "[+]": [ "mhchem", "color", "cancel", "ams", "tagformat" ], }, }, loader: { '[tex]/mhchem': { ready() { const {MapHandler} = MathJax._.input.tex.MapHandler; const mhchem = MapHandler.getMap('mhchem-chars'); mhchem.lookup('mhchemrightarrow')._char = '\uE42D'; mhchem.lookup('mhchemleftarrow')._char = '\uE42C'; } }, load: [ "[tex]/mhchem", "[tex]/color", "[tex]/cancel", "[tex]/tagformat", ], }, }; }; ///*]]>*/</script> <hr class="autoattribution-divider" /><div class="autoattribution"><p>This page titled <a target="_blank" class="internal mt-self-link" href="/Sandboxes/johnnyphung/biology/01:_The_Chemical_Basis_of_Life/1.07:_Molecular_Weight_and_the_Mole">1.7: Molecular Weight and the Mole</a> is shared under a <a rel="nofollow" href="https://creativecommons.org/licenses/by/3.0" target="_blank">CC BY 3.0</a> license and was authored, remixed, and/or curated by <a rel="nofollow" target="_blank" href="http://www.biology-pages.info/">John W. 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