Examples empirical formula

For our example (empirical formula CH2O), let s assume that the molecular weight is determined to be about 60. The weight of one CH2O unit is 30, so our unknown compound must contain twice this many atoms. The molecular formula must be C2H4O2. The compound might be acetic acid. 

As an example, the empirical formula of phenylalanine may be split into a more extended form that shows the presence of a phenyl ring, as well as an amino and a carboxylic acid group (the condensed form in Figure 2-4). 

The earlier sections have only considered the way atoms are bonded to each other in a molecule (topology) and how this is translated into a computer-readable form. Chemists define this arrangement of the bonds as the constitution of a molecule. The example in Figure 2-39, Section 2.5.2.1, shows that molecules with a given empirical formula, e.g., C H O, can have several different structures, which are called isomers [lOOj. Isomeric structures can be divided into constitutional isomers and stereoisomers (see Figure 2-67). 

There are also many empirical formulas used for calculatiag the friction head loss in piping systems. These must be used carefuUy because many are based on the properties of specific fluids and are not appHcable over a broad range of fluids, temperatures, and pressures. For example, the Ha2en and Wdhams formula widely used for water flow ... 

Referring to Examples 12.2 and 12.3 above, the heat produced according to the empirical formula is as follows ... 

From Cg/ZgNO (problem 4), for example, the empirical formula C Hjo is derived and compared with the alkane formula Cc>H2o, a hydrogen deficit of ten and thus of five double-bond equivalents is deduced. If the NMR spectra have too few signals in the shift range appropriate for multiple bonds, then the double-bond equivalents indicate rings (see, for example, a-pinene. Fig. 2.4). 

The empirical formula tells only the relative number of atoms of each element in a molecule. For example, consider ethane. Analysis shows that this is a compound of carbon and hydrogen and that there are three hydrogen atoms for... 

To find the empirical formula of vitamin C from the data in Example F.l we must express the ratios of numbers of atoms as the simplest whole numbers. First, we divide each number by the smallest value (3.41), which gives a ratio of 1.00 1.33 1.00. Molecules contain only whole numbers of atoms, and one of these numbers is still not a whole number. Hence, we must multiply each number by the correct factor so that all numbers can be rounded off to whole numbers. Because 1.33 is Vi (within experimental error), we multiply all three numbers by 3 to obtain 3.00 3.99 3.00, or approximately 3 4 3. Now we know that the empirical formula of vitamin C is C3H403. 

EXAMPLE F.2 Determining the empirical formula from mass percentage composition... 

EXAMPLE F.3 Sample exercise Determining the molecular formula from the empirical formula... 

J 3 Determine the molecular formula of a compound from its empirical formula and its molar mass (Example F.3). 

EXAMPLE M.4 Determining an empirical formula by combustion analysis... 

Determine the empirical formula of an organic compound containing carbon, hydrogen, and oxygen by combustion analysis (Example M.4). 

The situation is confused, however, by the case of certain chemicals. Styrene, for example, was known from the mid-nineteenth century as a clear organic liquid of characteristic pungent odour. It was also known to convert itself under certain circumstances into a clear resinous solid that was almost odour-free, this resin then being called metastyrene. The formation of metastyrene from styrene was described as a polymerisation and metastyrene was held to be a polymer of styrene. However, these terms applied only in the sense that there was no change in empirical formula despite the very profound alteration in chemical and physical properties. There was no understanding of the cause of this change and certainly the chemists of the time had no idea of what had happened to the styrene that was remotely akin to the modem view of polymerisation. 

The index is augmented by successively permuted versions of all empirical formulas. As an example, C3H3AIO9 will appear as such and, at the appropriate positions in the alphanumeric sequence, as H3A109 C3, AIO/C3H3 and 09 C3H3A1. The asterisk identifies a permuted formula and allows the original formula to be reconstructed by shifting to the front the elements that follow the asterisk. 

Proceed exactly as in Example. The flowchart in Figure 3-15 outlines the process. From masses of products, determine masses of elements. Then convert masses of elements to moles of elements. From moles of the elements, find the empirical formula. Finally, use information about the molar mass to obtain the molecular formula. 

EXAMPLE 4.12. What is the empirical formula of a compound which contains 40.0% sulfur and 60.0% oxygen, by mass ... 

EXAMPLE 4.14. Determine the empirical formula of a sample of a compound which contains 79.59g Fc and 30.40 g O. 

Although the compounds [Pt(en)2Cl2]Br2 and [Pt(en)2Br2]Cl2 have the same empirical formulas, they are quite different compounds. For example, the first gives Br- when dissolved in water, whereas the second gives Cl-. This occurs because in the first case, the Cl- ions are coordinated to the Pt4+, whereas in the second case the Br- ions are coordinated to the metal ion. The isomerism in cases such as these is known as ionization isomerism. It is easy to see that many pairs of compounds could be considered as ionization isomers, among them the following examples ... 

The formula of an ionic compound is always an empirical formula, for example the empirical formula of aluminium oxide is AI2O3. 

Example Calculate the empirical formula of the compound containing 6 g of carbon and 1 g of hydrogen (relative atomic masses C = 12,... 

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