Calculating Mass PercentagesDetermining the Empirical Formula of PenicillinCombustion AnalysisFrom Empirical Formula to Molecular Formula

Learning Objectives

To recognize the empirical formula of a compound from its complace by mass. To derive the molecular formula of a compound from its empirical formula.

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When a new slrfc.orgical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural resource, slrfc.orgists recognize its elemental composition, its empirical formula, and also its structure to understand also its properties. In this area, we emphasis on just how to recognize the empirical formula of a compound and also then use it to determine the molecular formula if the molar mass of the compound is well-known.


Calculating Mass Percentages

The regulation of definite proparts states that a slrfc.orgical compound constantly includes the same propercent of elements by mass; that is, the percent complace (the portion of each facet present in a pure substance). With few exceptions, the percent composition of a slrfc.orgical compound is consistent (see regulation of definite proportions).—the percent of each aspect existing in a pure substance—is constant (although we now understand tbelow are exceptions to this law). For example, succlimbed (cane sugar) is 42.11% carbon, 6.48% hydrogen, and also 51.41% oxygen by mass. This indicates that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and also 51.41 g of oxygen. First we will certainly use the molecular formula of succlimbed (C12H22O11) to calculate the mass percentage of the component elements; then we will present how mass percentages deserve to be used to recognize an empirical formula.

According to its molecular formula, each molecule of sucincreased consists of 12 carbon atoms, 22 hydrogen atoms, and also 11 oxygen atoms. A mole of succlimbed molecules therefore consists of 12 mol of carbon atoms, 22 mol of hydrogen atoms, and also 11 mol of oxygen atoms. We deserve to usage this indevelopment to calculate the mass of each facet in 1 mol of sucincreased, which will offer us the molar mass of succlimbed. We have the right to then usage these masses to calculate the percent complace of sucrose. To 3 decimal areas, the calculations are the following:


( eginmatrixmass; of; C/mol; of; succlimbed & =12cancelmolleft ( dfrac12.011; g; Ccancel1; mol; C appropriate ) & =144.132; g ; C\& & \mass; of; H/mol; of; sucrose & =22cancelmolleft ( dfrac1.008; g; Hcancel1; mol; H ight ) & =22.176; g ; H\& & \mass; of; O/mol; of; sucincreased & =11cancelmolleft ( dfrac15.999; g; Ocancel1; mol; O appropriate ) & =175.989; g ; Oendmatrix) ag7.2.1 )


Hence 1 mol of succlimbed has a mass of 342.297 g; note that more than fifty percent of the mass (175.989 g) is oxygen, and nearly fifty percent of the mass (144.132 g) is carbon.

The mass portion of each element in sucrose is the mass of the facet existing in 1 mol of sucrose divided by the molar mass of sucincreased, multiplied by 100 to provide a portion. The result is displayed to 2 decimal places:

( eginmatrixmass; \%; C; in; succlimbed & =dfracmass; of; C/mol; sucrosemolar; mass; of; sucrose imes 100 & =dfrac144.132; g; C342.297; g/mol imes 100 & =42.12 \%\& & \mass; \%; H; in; sucrose & =dfracmass; of; H/mol; sucrosemolar; mass; of; sucrose imes 100 & =dfrac22.176; g; H342.297; g/mol imes 100 & =6.48 \%\& & \mass; \%; O; in; sucrose & =fracmass; of; O/mol; sucrosemolar; mass; of; sucrose imes 100 & =dfrac175.989; g; O342.297; g/mol imes 100 & =51.41 \%endmatrix )​

You have the right to inspect your occupational by verifying that the amount of the percentperiods of all the facets in the compound is 100%:

( 42.12 \% + 6.48 \% + 51.41 \% = 100.01 \% )​

If the sum is not 100%, you have actually made an error in your calculations. (Rounding to the correct number of decimal locations can, but, cause the total to be slightly various from 100%.) Hence 100.00 g of sucrose has 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to 2 decimal places, the percent composition of succlimbed is indeed 42.12% carbon, 6.48% hydrogen, and also 51.41% oxygen.


Figure (PageIndex1) Percent Composition

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(a) Penicillium mold is thriving in a culture dish; the photo reflects its impact on bacterial growth. (b) In this photomicrograph of Penicillium, its rod- and pencil-shaped branches are visible. The name originates from the Latin penicillus, definition “paintbrush.”


Although Fleming was unable to isolate penicillin in pure form, the medical prominence of his discovery engendered researchers in various other laboratories. Finally, in 1940, two slrfc.orgists at Oxford University, Howard Florey (1898–1968) and also Ernst Chain (1906–1979), were able to isolate an active product, which they called penicillin G. Within 3 years, penicillin G remained in widespcheck out usage for dealing with pneumonia, gangrene, gonorrhea, and also other conditions, and also its usage considerably raised the survival price of wounded soldiers in World War II. As a result of their work-related, Fleming, Florey, and Chain shared the Nobel Prize in Medicine in 1945.

As soon as they had thrived in isolating pure penicillin G, Florey and Chain subjected the compound to a procedure dubbed combustion evaluation (explained later in this section) to recognize what aspects were existing and also in what amounts. The outcomes of such analyses are generally reported as mass percentperiods. They found that a typical sample of penicillin G contains 53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur, and also 6.5% sodium by mass. The sum of these numbers is just 82.1%, rather than 100.0%, which means that tright here have to be one or even more added facets. A reasonable candiday is oxygen, which is a widespread component of compounds that contain carbon and also hydrogen;Do not assume that the “missing” mass is always because of oxygen. It could be any type of various other facet. for technical factors, however, it is challenging to analyze for oxygen directly. If we assume that all the lacking mass is due to oxygen, then penicillin G includes (100.0% − 82.1%) = 17.9% oxygen. From these mass percentperiods, the empirical formula and also inevitably the molecular formula of the compound can be figured out.

To identify the empirical formula from the mass percentages of the elements in a compound such as penicillin G, we need to transform the mass percentperiods 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 supplied for analyses are mainly much smaller sized, commonly in milligrams. This assumption simplifies the arithmetic because a 53.9% mass percentage of carbon coincides to 53.9 g of carbon in a 100.0 g sample of penicillin G; also, 4.8% hydrogen corresponds to 4.8 g of hydrogen in 100.0 g of penicillin G; and so forth for the various other facets. We have the right to then divide each mass by the molar mass of the aspect to identify exactly how many moles of each aspect are existing in the 100.0 g sample:


( fracmassleft ( g ight )molar;massleft ( g/mol ideal )=left ( cancelg est )left ( fracmolcancelg appropriate )=mol ag7.2.2 )

( eginmatrix53.9; cancelg; Cdfrac1; mol; C12.011; cancelg; C &=4.49; mol; C \& \4.8; cancelg; Hdfrac1; mol; H1.008; cancelg; H &=4.49; mol; H \& \7.9; cancelg; Ndfrac1; mol; N14.007; cancelg; N &=0.56; mol; N \& \9.0; cancelg; Sdfrac1; mol; S32.065; cancelg; S &=0.28; mol; S \& \6.5; cancelg; Nadfrac1; mol; Na22.990; cancelg; Na &=0.28; mol; Na \& \17.9; cancelg; Odfrac1; mol; O15.999; cancelg; O &=1.12; mol; Oendmatrix )


( eginmatrixC:dfrac4.490.28 = 16 & H:dfrac4.80.28 = 17 & N:dfrac0.560.28 = 2.0 \& & \S:dfrac0.280.28 = 1.0 & Na:dfrac0.280.28 = 1.0 & O:dfrac1.120.28 = 1.0endmatrix ag7.2.3 )


Figure (PageIndex3) Structural Formulaand Ball-and-Stick Model of theAnion of Penicillin G

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From Empirical Formula to Molecular Formula

The empirical formula offers just the relative numbers of atoms in a substance in the smallest feasible ratio. For a covalent substance, we are usually even more interested in the molecular formula, which gives the actual variety of atoms of each type current per molecule. Without additional indevelopment, yet, it is impossible to know whether the formula of penicillin G, for instance, is C16H17N2NaO4S or an integral multiple, such as C32H34N4Na2O8S2, C48H51N6Na3O12S3, or (C16H17N2NaO4S)n, wbelow n is an integer. (The actual framework of penicillin G is presented in Figure 7.2.3).

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Consider glucose, the sugar that circulates in our blood to provide fuel for our bodies and specifically for our brains. Results from combustion evaluation of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Due to the fact that combustion occurs in the presence of oxygen, it is impossible to straight determine the percent of oxygen in a compound by making use of burning analysis; other even more complex methods are vital. If we assume that the continuing to be portion is due to oxygen, then glucose would certainly contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and also 53.79 g of oxygen. To calculate the variety of moles of each facet in the 100.0 g sample, we divide the mass of each facet by its molar mass:


( eginmatrixmoles; C &=39.68; cancelg; Cdfrac1; mol; C12.011; cancelgC & = 3.304; mol; C \& & \moles; H & =6.58; cancelg; Hdfrac1; mol; H1.0079; cancelgH & = 6.53; mol; H \& & \moles; O & =53.79; cancelg; Odfrac1; mol; C15.9994; cancelgO & = 3.362; mol; Oendmatrix ag7.2.4 )


Once again, we discover the subscripts of the aspects in the empirical formula by separating the variety of moles of each aspect by the variety of moles of the element current in the smallest amount:

( eginmatrixC:dfrac3.3043.304=1.000 & H:dfrac6.533.304=1.98 & O:dfrac3.3623.304=1.018endmatrix )

The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is roughly 2. The empirical formula of glucose is therefore CH2O, however what is its molecular formula?

Many known compounds have the empirical formula CH2O, consisting of formaldehyde, which is provided to maintain organic specimens and has properties that are incredibly different from the sugar circulating in our blood. At this point, we cannot know whether glucose is CH2O, C2H4O2, or any other (CH2O)n. We can, yet, usage the experimentally determined molar mass of glucose (180 g/mol) to deal with this dilemma.

First, we calculate the formula mass, the molar mass of the formula unit, which is the amount of the atomic masses of the facets in the empirical formula multiplied by their corresponding subscripts. For glucose,


( n=dfrac180; g30.026; g; CH_2O=5.99approx 6 CH^_2O; formula; units ag7.2.5 )