Composition of Chemical Compounds

A chemical formula conveys considerable quantitative information about a compound and its constituent elements. We have already learned how to determine the molar mass of a compound, and, in this section, we consider some other types of calculations based on the chemical formula.

The colorless, volatile liquid halothane has been used as a fire extinguisher and also as an inhalation anesthetic. Its empirical and molecular formulas are C₂HBrClF₃, its molecular mass is 197.382 u, and its molar mass is 197.382 g/mol, as calculated below:

MC₂HB_rClF₃ = 2M_C + M_H + M_Br + M_Cl + 3M_F

= [(2 x 12.0107) + 1.00794 + 79.904 + 35.453 + (3 x 18.9984) L G / mol

= 197.382 g/mol 

The molecular formula of C₂HBrClF₃ tells us that per mole of halothane there are two moles of C atoms, one mole each of H, Br, and Cl atoms, and three moles of F atoms. This factual statement can be turned into conversion factors to answer such questions as, "How many C atoms are present per mole of halo thane?" In this case, the factor needed is 2 mol C/mol C₂HBrClF₃. That is, 

? C atoms = 1.000 mol C₂HB_RClF₃ x 2 mol C / 1 molC₂HBrClF₃ x 6.022 x 10²³ C atoms / 1 mol c

= 1.204 x 1024 C atoms  

In Example 3-3, we use another conversion factor derived from the formula for halothane. This factor is shown in blue in the setup, which includes other familiar factors to make the conversion pathway:

mL à g à mol C₂HBrClF₃ à mol F

Example 3-3 | Using Relationship Derived From A Chemical Formula

How many moles of F atoms are in a 75.0 mL sample of halothane (d = 1.871 g/mL)?

Analyze

The conversion pathway for this problem us given above. First, convert the volume of the sample to mass; this requires density as a conversion factor. Next, convert the mass of halothane to its amount in moles; this requires the inverse of the molar mass as a conversation factor. The final conversion factor is based on the formula of halothane.

 

? mol F  = 75.0 mL C₂HBrClF3 x 1.871 g C₂HBrClF₃ / 1 mL C₂HB_RClF₃

x 1 mol C₂HB_RClF₃/197.4 C₂HB_RClF₃ x 3 mol F/1 mol C₂HB_RClF₃

 

Assess

If we had been asked for the number of moles of C instead, the final conversion factor in the calculation above would have been (2 mol C/1 mol C₂HB_RClF₃). 

Calculating Percent Composition from a Chemical Formula

When chemists believe that they have synthesized a new compound, a sample is generally sent to an analytical laboratory where its percent composition is determined. This experimentally determined percent composition is then compared with the percent composition calculated from the formula of the expected compound. In this way, chemists can see if the compound obtained could be the one expected.

Equation (3.1) establishes how the mass percent of an element in a compound is calculated. In applying the equation, as in Example 3-4, think in terms of the following steps.

1. Determine the molar mass of the compound. This is the denominator in equation (3.1).

2. Determine the contribution of the given element to the molar mass. This product of the formula subscript and the molar mass of the element appears in the numerator of equation (3.1).

3. Formulate the ratio of the mass of the given element to the mass of the compound as a whole. This is the ratio of the numerator from step 2 to the denominator from step 1.

4. Multiply this ratio by 100% to obtain the mass percent of the element.

The mass composition of a compound is the collection of mass percentages of the individual elements in the compound.

The percentages of the elements in a compound should add up to 100.00%, and we can use this fact in two ways.

1.    Check the accuracy of the computations by ensuring that the percentages total 100.00%. As applied to the results of Example 3-4:

           12.17% + 0.51% + 40.48% + 17.96% + 28.88% = 100.00%

     2. Determine the percentages of all the elements but one. Obtain that one            difference (subtraction). From Example 3-4:                                         

% H = 100.00% - % C - % Br - % Cl - % F

= 100.00% - 12.17% - 40.48% - 17.96% - 28.88%

= 0.51%

Establishing Formulas from the Experimentally Determined Percent Composition of Compounds

At times, a chemist isolates a chemical compound—say, from an exotic tropical plant—and has no idea what it is. A report from an analytical laboratory on the percent composition of the compound yields data needed to determine its formula.

Percent composition establishes the relative proportions of the elements in a compound on a mass basis. A chemical formula requires these proportions to be on a mole basis, that is, in terms of numbers of atoms. Consider the following five-step approach to determining a formula from the experimentally determined percent composition of the compound 2-deoxyribose, a sugar that is a basic constituent of DNA (deoxyribonucleic acid). The mass percent composition of 2-deoxyribose is 44.77% C, 7.52% H, and 47.71% O.

1. Although we could choose any sample size, if we take one of exactly 100 g, the masses of the elements are numerically equal to their percentages, that is, 44.77 g C, 7.52 g H, and 47.71 g O.

2. Convert the masses of the elements in the 100.00 g sample to amounts in moles.

3. Write a tentative formula based on the numbers of moles just determined.

C_3.727H_7.46O_2.982

4. Attempt to convert the subscripts in the tentative formula to small whole numbers. This requires dividing each of the subscripts by the smallest one (2.982).

If all subscripts at this point differ only slightly from whole numbers— which is not the case here—round them off to whole numbers, concluding the calculation at this point.

5. If one or more subscripts is still not a whole number—which is the case here—multiply all subscripts by a small whole number that will make them all integral. Thus, multiply by 4 here.

The formula that we get by the method just outlined, C₅H₁₀O₄, is the simplest possible formula—the empirical formula. The actual molecular formula may be equal to, or some multiple of, the empirical formula, such as C₁₀H₂₀O₈, C₁₅H₃₀O₁₂, C₂₀H₄₀O₁₆, and so on. To find the multiplying factor, we must compare the formula mass based on the empirical formula with the true molecular mass of the compound. We can establish the molecular mass from a separate experiment (by methods introduced in next blog posts). The experimentally determined molecular mass of 2-deoxyribose is 134 u. The formula mass based on the empirical formula, C₅HK₃O₄, is 134.1 u. The measured molecular mass is the same as the empirical formula mass. The molecular formula is also C₅H₁₀O₄.

We outline the five-step approach described above in the flow diagram below and then apply the approach to Example 3-5, where we will find that the empirical formula and the molecular formula are not the same.

Combustion Analysis

Figure 3-6 illustrates an experimental method for establishing an empirical formula for compounds that are easily burned, such as compounds containing carbon and hydrogen with oxygen, nitrogen, and a few other elements. In combustion analysis, a weighed sample of a compound is burned in a stream of oxygen gas. The water vapor and carbon dioxide gas produced in the combustion are absorbed by appropriate substances. The increases in mass of these absorbers correspond to the masses of water and carbon dioxide. We can think of the matter as shown below. (The subscripts x, y, and z are integers whose values we do not know initially.)

After combustion, all the carbon atoms in the sample are found in the C02. All the H atoms are in the H20. Moreover, the only source of the carbon and hydrogen atoms was the sample being analyzed. Oxygen atoms in the C02 and H20 could have come partly from the sample and partly from the oxygen gas consumed in the combustion. Thus, the quantity of oxygen in the sample has to be determined indirectly. These ideas are applied in Example 3-6.

We have just seen how combustion reactions can be used to analyze chemical substances, but not all samples can be easily burned. Fortunately, several other types of reactions can be used for chemical analyses. Also, modern methods in chemistry rely much more on physical measurements with instruments than on chemical reactions. We will cite some of these methods later in other posts (use search box above to look for the relative content).