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<!DOCTYPE html>
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<title>Determining Empirical and Molecular Formulas</title>
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<div id="book-content">
<div class="section" id="averill_1.0-ch03_s02" condition="start-of-chunk" version="5.0" lang="en">
<h2 class="title editable block">
<span class="title-prefix">3.2</span> Determining Empirical and Molecular Formulas</h2>
<div class="learning_objectives editable block" id="averill_1.0-ch03_s02_n01">
<h3 class="title">Learning Objectives</h3>
<ol class="orderedlist" id="averill_1.0-ch03_s02_l01">
<li>To determine the empirical formula of a compound from its composition by mass.</li>
<li>To derive the molecular formula of a compound from its empirical formula.</li>
</ol>
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_p01">When a new chemical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, chemists determine its elemental composition, its empirical formula, and its structure to understand its properties. In this section, we focus on how to determine the empirical formula of a compound and then use it to determine the molecular formula if the molar mass of the compound is known.</p>
<div class="section" id="averill_1.0-ch03_s02_s01">
<h2 class="title editable block">Calculating Mass Percentages</h2>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p01">The law of definite proportions states that a chemical compound always contains the same proportion of elements by mass; that is, the <span class="margin_term"><a class="glossterm">percent composition</a><span class="glossdef">The percentage of each element present in a pure substance. With few exceptions, the percent composition of a chemical compound is constant (see law of definite proportions).</span></span>—the percentage of each element present in a pure substance—is constant (although we now know there are exceptions to this law). For example, sucrose (cane sugar) is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. First we will use the molecular formula of sucrose (C<sub class="subscript">12</sub>H<sub class="subscript">22</sub>O<sub class="subscript">11</sub>) to calculate the mass percentage of the component elements; then we will show how mass percentages can be used to determine an empirical formula.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p02">According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. We can use this information to calculate the mass of each element in 1 mol of sucrose, which will give us the molar mass of sucrose. We can then use these masses to calculate the percent composition of sucrose. To three decimal places, the calculations are the following:</p>
<div class="equation block" id="averill_1.0-ch03_s02_s01_eq01">
<p class="title editable"><span class="title-prefix">Equation 3.3</span> </p>
<math xml:id="averill_1.0-ch03_m020" display="block">
<semantics>
<mtable columnalign="left">
<mtr>
<mtd>
<mtext>mass of C/mol of sucrose </mtext>
<mo>=</mo>
<mtext> 12 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol C</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mn>12.011</mn>
<mtext> g C</mtext>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol C</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>144.132</mn>
<mtext> g C</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass of H/mol of sucrose </mtext>
<mo>=</mo>
<mtext> 22 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol H</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mn>1.008</mn>
<mtext> g H</mtext>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol H</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>22.176</mn>
<mtext> g H</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass of O/mol of sucrose </mtext>
<mo>=</mo>
<mtext> 11 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol O</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mn>15.999</mn>
<mtext> g O</mtext>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol O</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>175.989</mn>
<mtext> g O</mtext>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p03">Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p04">The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places:</p>
<span class="informalequation block">
<math xml:id="averill_1.0-ch03_m021" display="block">
<semantics>
<mtable columnalign="left">
<mtr>
<mtd>
<mtext>mass % C in sucrose </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>mass of C/mol sucrose</mtext>
</mrow>
<mrow>
<mtext>molar mass of sucrose</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mn>100</mn>
<mo>=</mo>
<mfrac>
<mrow>
<mn>144.132</mn>
<mtext> g C</mtext>
</mrow>
<mrow>
<mtext>342</mtext>
<mtext>.297 g/mol</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mn>100</mn>
<mo>=</mo>
<mn>42.12</mn>
<mi>%</mi>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass % H in sucrose </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>mass of H/mol sucrose</mtext>
</mrow>
<mrow>
<mtext>molar mass of sucrose</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mn>100</mn>
<mo>=</mo>
<mfrac>
<mrow>
<mn>22.176</mn>
<mtext> g H</mtext>
</mrow>
<mrow>
<mtext>342</mtext>
<mtext>.297 g/mol</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mn>100</mn>
<mo>=</mo>
<mn>6.48</mn>
<mi>%</mi>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass % O in sucrose </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>mass of O/mol sucrose</mtext>
</mrow>
<mrow>
<mtext>molar mass of sucrose</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mn>100</mn>
<mo>=</mo>
<mfrac>
<mrow>
<mn>175.989</mn>
<mtext> g O</mtext>
</mrow>
<mrow>
<mtext>342</mtext>
<mtext>.297 g/mol</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mn>100</mn>
<mo>=</mo>
<mn>51.41</mn>
<mi>%</mi>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</span>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p05">You can check your work by verifying that the sum of the percentages of all the elements in the compound is 100%:</p>
<span class="informalequation block">
<span class="mathphrase">42.12% + 6.48% + 51.41% = 100.01%</span>
</span>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p06">If the sum is not 100%, you have made an error in your calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.</p>
<div class="informalfigure large block">
<img src="section_07/d59e0a50a4d588e2e662c9694ade7155.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_s01_p07">We could also calculate the mass percentages using atomic masses and molecular masses, with atomic mass units. Because the answer we are seeking is a ratio, expressed as a percentage, the units of mass cancel whether they are grams (using molar masses) or atomic mass units (using atomic and molecular masses).</p>
<div class="exercises block" id="averill_1.0-ch03_s02_s01_n01">
<h3 class="title">Example 5</h3>
<p class="para" id="averill_1.0-ch03_s02_s01_p08">Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C<sub class="subscript">14</sub>H<sub class="subscript">18</sub>N<sub class="subscript">2</sub>O<sub class="subscript">5</sub>.</p>
<div class="informalfigure medium">
<img src="section_07/ddf03d4083033d5552e6dfc4b05c359e.jpg">
</div>
<ol class="orderedlist" id="averill_1.0-ch03_s02_s01_l01" numeration="loweralpha">
<li>Calculate the mass percentage of each element in aspartame.</li>
<li>Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.</li>
</ol>
<p class="para" id="averill_1.0-ch03_s02_s01_p09"><strong class="emphasis bold">Given: </strong>molecular formula and mass of sample</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p10"><strong class="emphasis bold">Asked for: </strong>mass percentage of all elements and mass of one element in sample</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p11">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p12"><strong class="emphasis bold">A</strong> Use atomic masses from the periodic table to calculate the molar mass of aspartame.</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p13"><strong class="emphasis bold">B</strong> Divide the mass of each element by the molar mass of aspartame; then multiply by 100 to obtain percentages.</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p14"><strong class="emphasis bold">C</strong> To find the mass of an element contained in a given mass of aspartame, multiply the mass of aspartame by the mass percentage of that element, expressed as a decimal.</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p15">
<strong class="emphasis bold">Solution:</strong>
</p>
<ol class="orderedlist" id="averill_1.0-ch03_s02_s01_l02" numeration="loweralpha">
<li>
<p class="para"><strong class="emphasis bold">A</strong> We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m022" display="block">
<semantics>
<mrow>
<mtable rowlines="none none none solid" columnalign="right">
<mtr columnalign="right">
<mtd columnalign="right">
<mrow>
<mtext>14C</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>(14 mol C)(12</mtext>
<mtext>.011 g/mol C)</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mo>=</mo>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext> 168</mtext>
<mtext>.154 g</mtext>
</mrow>
</mtd>
</mtr>
<mtr columnalign="right">
<mtd columnalign="right">
<mrow>
<mtext>18H</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>(18 mol H)(1</mtext>
<mtext>.008 g/mol H)</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mo>=</mo>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>18</mtext>
<mtext>.114 g</mtext>
</mrow>
</mtd>
</mtr>
<mtr columnalign="right">
<mtd columnalign="right">
<mrow>
<mtext>2N</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>(2 mol N)(14</mtext>
<mtext>.007 g/mol N)</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mo>=</mo>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext> 28</mtext>
<mtext>.014 g</mtext>
</mrow>
</mtd>
</mtr>
<mtr columnalign="right">
<mtd columnalign="right">
<mrow>
<mo>+</mo>
<mtext>5O </mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>(5 mol O)(15</mtext>
<mtext>.999 g/mol O)</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mo>=</mo>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>79</mtext>
<mtext>.995 g</mtext>
</mrow>
</mtd>
</mtr>
<mtr columnalign="right">
<mtd columnalign="right">
<mrow>
<msub>
<mtext>C</mtext>
<mrow>
<mtext>14</mtext>
</mrow>
</msub>
<msub>
<mtext>H</mtext>
<mrow>
<mtext>18</mtext>
</mrow>
</msub>
<msub>
<mtext>N</mtext>
<mtext>2</mtext>
</msub>
<msub>
<mtext>O</mtext>
<mtext>5</mtext>
</msub>
</mrow>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>molar mass of aspartame</mtext>
</mrow>
</mtd>
<mtd columnalign="right">
<mo>=</mo>
</mtd>
<mtd columnalign="right">
<mrow>
<mtext>294</mtext>
<mtext>.277 g/mol</mtext>
</mrow>
</mtd>
</mtr>
</mtable>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch03_s02_s01_p16">Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p17"><strong class="emphasis bold">B</strong> To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m023" display="block">
<semantics>
<mtable columnalign="left">
<mtr>
<mtd>
<mtext>mass % C </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>168</mtext>
<mtext>.154 g C</mtext>
</mrow>
<mrow>
<mtext>294</mtext>
<mtext>.277 g aspartame</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mtext>100 </mtext>
<mo>=</mo>
<mtext> 57</mtext>
<mtext>.14% C</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass % H </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>18</mtext>
<mtext>.114 g H</mtext>
</mrow>
<mrow>
<mtext>294</mtext>
<mtext>.277 g aspartame</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mtext>100 </mtext>
<mo>=</mo>
<mtext> 6</mtext>
<mtext>.16% H</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass % N </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>28</mtext>
<mtext>.014 g N</mtext>
</mrow>
<mrow>
<mtext>294</mtext>
<mtext>.277 g aspartame</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mtext>100 </mtext>
<mo>=</mo>
<mtext> 9</mtext>
<mtext>.52% N</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>mass % O </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>79</mtext>
<mtext>.995 g O</mtext>
</mrow>
<mrow>
<mtext>294</mtext>
<mtext>.277 g aspartame</mtext>
</mrow>
</mfrac>
<mo>×</mo>
<mtext>100 </mtext>
<mo>=</mo>
<mtext> 27</mtext>
<mtext>.18% O</mtext>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch03_s02_s01_p18">As a check, we can add the percentages together:</p>
<span class="informalequation">
<span class="mathphrase">57.14% + 6.16% + 9.52% + 27.18% = 100.00%</span>
</span>
<p class="para" id="averill_1.0-ch03_s02_s01_p19">If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation.</p>
</li>
<li>
<p class="para"><strong class="emphasis bold">C</strong> The mass of carbon in 1.00 g of aspartame is calculated as follows:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m024" display="block">
<semantics>
<mrow>
<mtext>mass of C </mtext>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.00 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g aspartame</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mn>57.14</mn>
<mtext> g C</mtext>
</mrow>
<mrow>
<mtext>100 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g aspartame</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>0.571</mn>
<mtext> g C</mtext>
</mrow>
</semantics>
</math>
</span>
</li>
</ol>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p20">Calculate the mass percentage of each element in aluminum oxide (Al<sub class="subscript">2</sub>O<sub class="subscript">3</sub>). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide.</p>
<p class="para" id="averill_1.0-ch03_s02_s01_p21"><strong class="emphasis bold">Answer: </strong>52.93% aluminum; 47.08% oxygen; 1.92 g Al</p>
</div>
</div>
<div class="section" id="averill_1.0-ch03_s02_s02">
<h2 class="title editable block">Determining the Empirical Formula of Penicillin</h2>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p01">Just as we can use the empirical formula of a substance to determine its percent composition, we can use the percent composition of a sample to determine its empirical formula, which can then be used to determine its molecular formula. Such a procedure was actually used to determine the empirical and molecular formulas of the first antibiotic to be discovered: penicillin.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p02"><em class="emphasis">Antibiotics</em> are chemical compounds that selectively kill microorganisms, many of which cause diseases. Although we may take antibiotics for granted today, penicillin was discovered only about 80 years ago. The subsequent development of a wide array of other antibiotics for treating many common diseases has contributed greatly to the substantial increase in life expectancy over the past 50 years. The discovery of penicillin is a historical detective story in which the use of mass percentages to determine empirical formulas played a key role.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p03">In 1928, Alexander Fleming, a young microbiologist at the University of London, was working with a common bacterium that causes boils and other infections such as blood poisoning. For laboratory study, bacteria are commonly grown on the surface of a nutrient-containing gel in small, flat culture dishes. One day Fleming noticed that one of his cultures was contaminated by a bluish-green mold similar to the mold found on spoiled bread or fruit. Such accidents are rather common, and most laboratory workers would have simply thrown the cultures away. Fleming noticed, however, that the bacteria were growing everywhere on the gel <em class="emphasis">except</em> near the contaminating mold (part (a) in <a class="xref" href="#averill_1.0-ch03_s02_s02_f01">Figure 3.3 "
"</a>), and he hypothesized that the mold must be producing a substance that either killed the bacteria or prevented their growth. To test this hypothesis, he grew the mold in a liquid and then filtered the liquid and added it to various bacteria cultures. The liquid killed not only the bacteria Fleming had originally been studying but also a wide range of other disease-causing bacteria. Because the mold was a member of the <em class="emphasis">Penicillium</em> family (named for their pencil-shaped branches under the microscope) (part (b) in <a class="xref" href="#averill_1.0-ch03_s02_s02_f01">Figure 3.3 "
"</a>), Fleming called the active ingredient in the broth <em class="emphasis">penicillin</em>.</p>
<div class="figure large medium-height editable block" id="averill_1.0-ch03_s02_s02_f01">
<p class="title"><span class="title-prefix">Figure 3.3</span>
<em class="emphasis">Penicillium</em>
</p>
<img src="section_07/5ab6694188604c7969842f4a14045c42.jpg">
<p class="para">(a) <em class="emphasis">Penicillium</em> mold is growing in a culture dish; the photo shows its effect on bacterial growth. (b) In this photomicrograph of <em class="emphasis">Penicillium</em>, its rod- and pencil-shaped branches are visible. The name comes from the Latin <em class="emphasis">penicillus</em>, meaning “paintbrush.”</p>
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p04">Although Fleming was unable to isolate penicillin in pure form, the medical importance of his discovery stimulated researchers in other laboratories. Finally, in 1940, two chemists at Oxford University, Howard Florey (1898–1968) and Ernst Chain (1906–1979), were able to isolate an active product, which they called penicillin G. Within three years, penicillin G was in widespread use for treating pneumonia, gangrene, gonorrhea, and other diseases, and its use greatly increased the survival rate of wounded soldiers in World War II. As a result of their work, Fleming, Florey, and Chain shared the Nobel Prize in Medicine in 1945.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p05">As soon as they had succeeded in isolating pure penicillin G, Florey and Chain subjected the compound to a procedure called combustion analysis (described later in this section) to determine what elements were present and in what quantities. The results of such analyses are usually reported as mass percentages. They discovered that a typical sample of penicillin G contains 53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur, and 6.5% sodium by mass. The sum of these numbers is only 82.1%, rather than 100.0%, which implies that there must be one or more additional elements. A reasonable candidate is oxygen, which is a common component of compounds that contain carbon and hydrogen;<span class="footnote" id="averill_1.0-fn03_001">Do not assume that the “missing” mass is always due to oxygen. It could be any other element.</span> for technical reasons, however, it is difficult to analyze for oxygen directly. If we assume that all the missing mass is due to oxygen, then penicillin G contains (100.0% − 82.1%) = 17.9% oxygen. From these mass percentages, the empirical formula and eventually the molecular formula of the compound can be determined.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p06">To determine the empirical formula from the mass percentages of the elements in a compound such as penicillin G, we need to convert the mass percentages to relative numbers of atoms. For convenience, we assume that we are dealing with a 100.0 g sample of the compound, even though the sizes of samples used for analyses are generally much smaller, usually in milligrams. This assumption simplifies the arithmetic because a 53.9% mass percentage of carbon corresponds to 53.9 g of carbon in a 100.0 g sample of penicillin G; likewise, 4.8% hydrogen corresponds to 4.8 g of hydrogen in 100.0 g of penicillin G; and so forth for the other elements. We can then divide each mass by the molar mass of the element to determine how many moles of each element are present in the 100.0 g sample:</p>
<div class="equation block" id="averill_1.0-ch03_s02_s02_eq01">
<p class="title editable"><span class="title-prefix">Equation 3.4</span> </p>
<math xml:id="averill_1.0-ch03_m025" display="block">
<semantics>
<mtable>
<mtr>
<mtd>
<mfrac>
<mrow>
<mtext>mass (g)</mtext>
</mrow>
<mrow>
<mtext>molar mass (g/mol)</mtext>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> (</mtext>
<menclose notation="updiagonalstrike">
<mtext>g</mtext>
</menclose>
<mtext>)</mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>mol</mtext>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mtext>g</mtext>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> mol</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>53</mtext>
<mtext>.9 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g C</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol C</mtext>
</mrow>
<mrow>
<mtext>12</mtext>
<mtext>.011 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g C</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 4</mtext>
<mtext>.49 mol C</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>4</mtext>
<mtext>.8 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g H</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol H</mtext>
</mrow>
<mrow>
<mtext>1</mtext>
<mtext>.008 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g H</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 4</mtext>
<mtext>.8 mol H</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>7</mtext>
<mtext>.9 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g N</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol N</mtext>
</mrow>
<mrow>
<mtext>14</mtext>
<mtext>.007 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g N</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.56 mol N</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>9</mtext>
<mtext>.0 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g S</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol S</mtext>
</mrow>
<mrow>
<mtext>32</mtext>
<mtext>.065 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g S</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.28 mol S</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>6</mtext>
<mtext>.5 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g Na</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol Na</mtext>
</mrow>
<mrow>
<mtext>22</mtext>
<mtext>.990 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g Na</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.28 mol Na</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>17</mtext>
<mtext>.9 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g O</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol O</mtext>
</mrow>
<mrow>
<mtext>15</mtext>
<mtext>.999 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g O</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.12 mol O</mtext>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p07">Thus 100.0 g of penicillin G contains 4.49 mol of carbon, 4.8 mol of hydrogen, 0.56 mol of nitrogen, 0.28 mol of sulfur, 0.28 mol of sodium, and 1.12 mol of oxygen (assuming that all the missing mass was oxygen). The number of significant figures in the numbers of moles of elements varies between two and three because some of the analytical data were reported to only two significant figures.</p>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p08">These results tell us the ratios of the moles of the various elements in the sample (4.49 mol of carbon to 4.8 mol of hydrogen to 0.56 mol of nitrogen, and so forth), but they are not the whole-number ratios we need for the empirical formula—the empirical formula expresses the <em class="emphasis">relative</em> numbers of atoms in the <em class="emphasis">smallest whole numbers possible</em>. To obtain whole numbers, we divide the numbers of moles of all the elements in the sample by the number of moles of the element present in the lowest relative amount, which in this example is sulfur or sodium. The results will be the subscripts of the elements in the empirical formula. To two significant figures, the results are</p>
<div class="equation block" id="averill_1.0-ch03_s02_s02_eq02">
<p class="title editable"><span class="title-prefix">Equation 3.5</span> </p>
<math xml:id="averill_1.0-ch03_m026" display="block">
<semantics>
<mrow>
<mtable columnalign="left">
<mtr columnalign="left">
<mtd columnalign="left">
<mrow>
<mtext>C: </mtext>
<mfrac>
<mrow>
<mn>4.49</mn>
</mrow>
<mrow>
<mn>0.28</mn>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 16</mtext>
</mrow>
</mtd>
<mtd columnalign="left">
<mrow>
<mtext>H: </mtext>
<mfrac>
<mrow>
<mn>4.8</mn>
</mrow>
<mrow>
<mn>0.28</mn>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 17</mtext>
</mrow>
</mtd>
<mtd columnalign="left">
<mrow>
<mtext>N: </mtext>
<mfrac>
<mrow>
<mn>0.56</mn>
</mrow>
<mrow>
<mn>0.28</mn>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 2</mtext>
<mtext>.0</mtext>
</mrow>
</mtd>
</mtr>
<mtr columnalign="left">
<mtd columnalign="left">
<mrow>
<mtext>S: </mtext>
<mfrac>
<mrow>
<mn>0.28</mn>
</mrow>
<mrow>
<mn>0.28</mn>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.0</mtext>
</mrow>
</mtd>
<mtd columnalign="left">
<mrow>
<mtext>Na: </mtext>
<mfrac>
<mrow>
<mn>0.28</mn>
</mrow>
<mrow>
<mn>0.28</mn>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.0</mtext>
</mrow>
</mtd>
<mtd columnalign="left">
<mrow>
<mtext>O: </mtext>
<mfrac>
<mrow>
<mn>1.12</mn>
</mrow>
<mrow>
<mn>0.28</mn>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 4</mtext>
<mtext>.0</mtext>
</mrow>
</mtd>
</mtr>
</mtable>
</mrow>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p09">The empirical formula of penicillin G is therefore C<sub class="subscript">16</sub>H<sub class="subscript">17</sub>N<sub class="subscript">2</sub>NaO<sub class="subscript">4</sub>S. Other experiments have shown that penicillin G is actually an ionic compound that contains Na<sup class="superscript">+</sup> cations and [C<sub class="subscript">16</sub>H<sub class="subscript">17</sub>N<sub class="subscript">2</sub>O<sub class="subscript">4</sub>S]<sup class="superscript">−</sup> anions in a 1:1 ratio. The complex structure of penicillin G (<a class="xref" href="#averill_1.0-ch03_s02_s02_f02">Figure 3.4 "Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G"</a>) was not determined until 1948.</p>
<div class="figure large medium-height editable block" id="averill_1.0-ch03_s02_s02_f02">
<p class="title"><span class="title-prefix">Figure 3.4</span> Structural Formula and Ball-and-Stick Model of the Anion of Penicillin G</p>
<img src="section_07/48da8115076217f5ce75e6636bd427fc.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch03_s02_s02_p10">In some cases, one or more of the subscripts in a formula calculated using this procedure may not be integers. Does this mean that the compound of interest contains a nonintegral number of atoms? No; rounding errors in the calculations as well as experimental errors in the data can result in nonintegral ratios. When this happens, you must exercise some judgment in interpreting the results, as illustrated in Example 6. In particular, ratios of 1.50, 1.33, or 1.25 suggest that you should multiply <em class="emphasis">all</em> subscripts in the formula by 2, 3, or 4, respectively. Only if the ratio is within 5% of an integral value should you consider rounding to the nearest integer.</p>
<div class="exercises block" id="averill_1.0-ch03_s02_s02_n01">
<h3 class="title">Example 6</h3>
<p class="para" id="averill_1.0-ch03_s02_s02_p11">Calculate the empirical formula of the ionic compound calcium phosphate, a major component of fertilizer and a polishing agent in toothpastes. Elemental analysis indicates that it contains 38.77% calcium, 19.97% phosphorus, and 41.27% oxygen.</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p12"><strong class="emphasis bold">Given: </strong>percent composition</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p13"><strong class="emphasis bold">Asked for: </strong>empirical formula</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p14">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p15"><strong class="emphasis bold">A</strong> Assume a 100 g sample and calculate the number of moles of each element in that sample.</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p16"><strong class="emphasis bold">B</strong> Obtain the relative numbers of atoms of each element in the compound by dividing the number of moles of each element in the 100 g sample by the number of moles of the element present in the smallest amount.</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p17"><strong class="emphasis bold">C</strong> If the ratios are not integers, multiply all subscripts by the same number to give integral values.</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p18"><strong class="emphasis bold">D</strong> Because this is an ionic compound, identify the anion and cation and write the formula so that the charges balance.</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p19">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch03_s02_s02_p20"><strong class="emphasis bold">A</strong> A 100 g sample of calcium phosphate contains 38.77 g of calcium, 19.97 g of phosphorus, and 41.27 g of oxygen. Dividing the mass of each element in the 100 g sample by its molar mass gives the number of moles of each element in the sample:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m027" display="block">
<semantics>
<mtable columnalign="left">
<mtr>
<mtd>
<mtext>moles Ca </mtext>
<mo>=</mo>
<mtext> 38</mtext>
<mtext>.77 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g Ca</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>1 mol Ca</mtext>
</mrow>
<mrow>
<mtext>40</mtext>
<mtext>.078 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g Ca</mtext>
</mrow>