Determining Empirical and Molecular Formulas (6.2)

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH₃), ammonium nitrate (NH₄NO₃), and urea (CH₄N₂O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH₃ contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

$$\begin{array}{c}\\ \%\text{N}=\frac{14.01\text{amu N}}{17.03\text{amu}{\text{NH}}_3}\times 100\%=82.27\%\\ \%\text{H}=\frac{3.024\text{amu H}}{17.03\text{amu}{\text{NH}}_3}\times 100\%=17.76\%\end{array}$$

This same approach may be taken considering a pair of molecules, a dozen molecules, a mole of molecules, and so on. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated in Example 6.4. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound’s elements.

Determination of Empirical Formulas

As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of a compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

$$\begin{array}{l}\\ 1.71\text{g C}\times \frac{1\text{mol C}}{12.01\text{g C}}=0.142\text{mol C}\\ 0.287\text{g H}\times \frac{1\text{mol H}}{1.008\text{g H}}=0.284\text{mol H}\end{array}$$Thus, this compound may be represented by the formula C_0.142H_0.248. Per convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

$${\text{C}}_{\frac{0.142}{0.142}}{\text{H}}_{\frac{0.284}{0.142}}\text{or}{\text{CH}}_2$$

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

The empirical formula for this compound is thus CH₂. This may or not be the compound’s molecular formula as well; however, additional information is needed to make that determination (as discussed later in this section).

Consider as another example a sample of a compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

$${\text{Cl}}_{0.150}{\text{O}}_{0.525}={\text{Cl}}_{\frac{0.150}{0.150}}{\text{O}}_{\frac{0.525}{0.150}}={\text{ClO}}_{3.5}$$

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl₂O₇ as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

1. Deriving the number of moles of each element from its mass
2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Figure 6.7 outlines this procedure in flowchart fashion for a substance containing elements A and X.

EXAMPLE 6.5

Determining a Compound’s Empirical Formula from the Masses of Its Elements

A sample of the black mineral hematite (Figure 6.8), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

Solution

This problem provides the mass in grams of each element. Begin by finding the moles of each:

$$\begin{array}{l}\\ 34.97\text{g Fe}\left(\frac{\text{mol Fe}}{55.85\text{g}}\right)& =& 0.6261\text{mol Fe}\hfill \\ 15.03\text{g O}\left(\frac{\text{mol O}}{16.00\text{g}}\right)& =& 0.9394\text{mol O}\hfill \end{array}$$

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

$$\begin{array}{}\\ \\ \frac{0.6261}{0.6261}=1.000\text{mol Fe}\\ \frac{0.9394}{0.6261}=1.500\text{mol O}\end{array}$$

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe₁O_1.5). Finally, multiply the ratio by two to get the smallest possible whole-number subscripts while still maintaining the correct iron-to-oxygen ratio:

$$2\left({\text{Fe}}_1{\text{O}}_{1.5}\right)={\text{Fe}}_2{\text{O}}_3$$The empirical formula is Fe₂O₃.

Check Your Learning

What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

ANSWER:

N₂O₅

LINK TO LEARNING

For additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip.

Deriving Empirical Formulas from Percent Composition

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.

EXAMPLE 6.6

Determining an Empirical Formula from Percent Composition

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O (Figure 6.9). What is the empirical formula for this gas?

Solution

Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum, meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

$$\begin{array}{l}\\ 27.29\%\text{C}& =\hfill & \frac{27.29\text{g C}}{100\text{g compound}}\hfill \\ 72.71\%\text{O}& =\hfill & \frac{72.71\text{g O}}{100\text{g compound}}\hfill \end{array}$$The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

$$\begin{array}{l}\\ 27.29\text{g C}\left(\frac{\text{mol C}}{12.01\text{g}}\right)& =\hfill & 2.272\text{mol C}\hfill \\ 72.71\text{g O}\left(\frac{\text{mol O}}{16.00\text{g}}\right)& =\hfill & 4.544\text{mol O}\hfill \end{array}$$

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

$$\begin{array}{}\\ \\ \frac{2.272\text{mol C}}{2.272}=1\\ \frac{4.544\text{mol O}}{2.272}=2\end{array}$$

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO₂.

Check Your Learning

What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?

Answer: CH₂O

 

Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in a previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (n):

$$\frac{\text{molecular or molar mass}\left(\text{amu or}\frac{\text{g}}{\text{mol}}\right)}{\text{empirical formula mass}\left(\text{amu or}\frac{\text{g}}{\text{mol}}\right)}=n\text{formula units/molecule}$$

The molecular formula is then obtained by multiplying each subscript in the empirical formula by n, as shown by the generic empirical formula A_xB_y:

$${({\text{A}}_{\text{x}}{\text{B}}_{\text{y}})}_{\text{n}}={\text{A}}_{\text{nx}}{\text{B}}_{\text{ny}}$$

For example, consider a covalent compound whose empirical formula is determined to be CH₂O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

$$\frac{180\text{amu/molecule}}{30\frac{\text{amu}}{\text{formula unit}}}=6\text{formula units/molecule}$$

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

$${\text{(CH}}_{\text{2}}{\text{O)}}_{\text{6}}={\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$$

Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, one mole of empirical formula units and molecules is considered, as opposed to single units and molecules.

EXAMPLE 6.7

Determination of the Molecular Formula for Nicotine

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

Solution

Determining the molecular formula from the provided data will require the comparison of the compound’s empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound’s empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:

$$\begin{array}{l}\\ \left(74.02\text{g C}\right)\left(\frac{1\text{mol C}}{12.01\text{g C}}\right)& =\hfill & 6.163\text{mol C}\hfill \\ \left(8.710\text{g H}\right)\left(\frac{1\text{mol H}}{1.008\text{g H}}\right)& =\hfill & 8.641\text{mol H}\hfill \\ \left(17.27\text{g N}\right)\left(\frac{1\text{mol N}}{14.01\text{g N}}\right)& =\hfill & 1.233\text{mol N}\hfill \end{array}$$Next, calculate the molar ratios of these elements relative to the least abundant element, N.

$$6.163\text{mol}C/1.233\text{mol}\mathrm{N}=5$$ 

$$8.641\text{mol}H/1.233\text{mol}\mathrm{N}=7$$

 

$$1.233\text{mol}N/1.233\text{mol}\mathrm{N}=1$$

$$\begin{array}{lll}\frac{1.233}{1.233}& =\hfill & 1.000\text{mol N}\hfill \\ \frac{6.163}{1.233}& =\hfill & 4.998\text{mol C}\hfill \\ \frac{8.624}{1.233}& =\hfill & 6.994\text{mol H}\hfill \end{array}$$

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C₅H₇N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/ mol formula unit.

Calculate the molar mass for nicotine from the given mass and molar amount of the compound:

$$\frac{40.57\text{g nicotine}}{0.2500\text{mol nicotine}}=\frac{162.3\text{g}}{\text{mol}}$$

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

$$\frac{162.3\text{g/mol}}{81.13\frac{\text{g}}{\text{formula unit}}}=2\text{formula units/molecule}$$

Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

$${\text{(C}}_{\text{5}}{\text{H}}_{\text{7}}{\text{N)}}_{\text{2}}={\text{C}}_{\text{10}}{\text{H}}_{\text{14}}{\text{N}}_{\text{2}}$$

Check Your Learning

What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O and a molecular mass of 194.2 amu?

Answer:C₈H₁₀N₄O₂