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# % Composition, Empirical Formulas

A molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be 6.022 × 1023, a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. The chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound’s percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound’s empirical formula. The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula.

## 33.1 Molecular and Empirical Formulas

### Learning Objectives

By the end of this section, you will be able to:

• Symbolize the composition of molecules using molecular formulas and empirical formulas
• Represent the bonding arrangement of atoms within molecules using structural formulas
• Define the amount unit mole and the related quantity Avogadro's number
• Explain the relation between mass, moles, and numbers of atoms or molecules and perform calculations deriving these quantities from one another

A molecular formula is a representation of a molecule that uses chemical symbols to indicate the types of atoms followed by subscripts to show the number of atoms of each type in the molecule. (A subscript is used only when more than one atom of a given type is present.) Molecular formulas are also used as abbreviations for the names of compounds.

The structural formula for a compound gives the same information as its molecular formula (the types and numbers of atoms in the molecule) but also shows how the atoms are connected in the molecule. The structural formula for methane contains symbols for one C atom and four H atoms, indicating the number of atoms in the molecule (Figure 33.1). The lines represent bonds that hold the atoms together. (A chemical bond is an attraction between atoms or ions that holds them together in a molecule or a crystal.) We will discuss chemical bonds and see how to predict the arrangement of atoms in a molecule later. For now, simply know that the lines are an indication of how the atoms are connected in a molecule. A ball-and-stick model shows the geometric arrangement of the atoms with atomic sizes not to scale, and a space-filling model shows the relative sizes of the atoms.

Figure 33.1

A methane molecule can be represented as (a) a molecular formula, (b) a structural formula, (c) a ball-and-stick model, and (d) a space-filling model. Carbon and hydrogen atoms are represented by black and white spheres, respectively.

Although many elements consist of discrete, individual atoms, some exist as molecules made up of two or more atoms of the element chemically bonded together. For example, most samples of the elements hydrogen, oxygen, and nitrogen are composed of molecules that contain two atoms each (called diatomic molecules) and thus have the molecular formulas H2, O2, and N2, respectively. Other elements commonly found as diatomic molecules are fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2). The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is S8 (Figure 33.2).

Figure 33.2

A molecule of sulfur is composed of eight sulfur atoms and is therefore written as S8. It can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. Sulfur atoms are represented by yellow spheres.

It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, H2 and 2H represent distinctly different species. H2 is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2H, on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression 2H2 represents two molecules of diatomic hydrogen (Figure 33.3).

Figure 33.3

The symbols H, 2H, H2, and 2H2 represent very different entities.

Compounds are formed when two or more elements chemically combine, resulting in the formation of bonds. For example, hydrogen and oxygen can react to form water, and sodium and chlorine can react to form table salt. We sometimes describe the composition of these compounds with an empirical formula, which indicates the types of atoms present and the simplest whole-number ratio of the number of atoms (or ions) in the compound. For example, titanium dioxide (used as pigment in white paint and in the thick, white, blocking type of sunscreen) has an empirical formula of TiO2. This identifies the elements titanium (Ti) and oxygen (O) as the constituents of titanium dioxide, and indicates the presence of twice as many atoms of the element oxygen as atoms of the element titanium (Figure 33.4).

Figure 33.4

(a) The white compound titanium dioxide provides effective protection from the sun. (b) A crystal of titanium dioxide, TiO2, contains titanium and oxygen in a ratio of 1 to 2. The titanium atoms are gray and the oxygen atoms are red. (credit a: modification of work by “osseous”/Flickr)

As discussed previously, we can describe a compound with a molecular formula, in which the subscripts indicate the actual numbers of atoms of each element in a molecule of the compound. In many cases, the molecular formula of a substance is derived from experimental determination of both its empirical formula and its molecular mass (the sum of atomic masses for all atoms composing the molecule). For example, it can be determined experimentally that benzene contains two elements, carbon (C) and hydrogen (H), and that for every carbon atom in benzene, there is one hydrogen atom. Thus, the empirical formula is CH. An experimental determination of the molecular mass reveals that a molecule of benzene contains six carbon atoms and six hydrogen atoms, so the molecular formula for benzene is C6H6 (Figure 33.5).

Figure 33.5

Benzene, C6H6, is produced during oil refining and has many industrial uses. A benzene molecule can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. (d) Benzene is a clear liquid. (credit d: modification of work by Sahar Atwa)

If we know a compound’s formula, we can easily determine the empirical formula. (This is somewhat of an academic exercise; the reverse chronology is generally followed in actual practice.) For example, the molecular formula for acetic acid, the component that gives vinegar its sharp taste, is C2H4O2. This formula indicates that a molecule of acetic acid (Figure 33.6) contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. The ratio of atoms is 2:4:2. Dividing by the lowest common denominator (2) gives the simplest, whole-number ratio of atoms, 1:2:1, so the empirical formula is CH2O. Note that a molecular formula is always a whole-number multiple of an empirical formula.

Figure 33.6

(a) Vinegar contains acetic acid, C2H4O2, which has an empirical formula of CH2O. It can be represented as (b) a structural formula and (c) as a ball-and-stick model. (credit a: modification of work by “HomeSpot HQ”/Flickr)

### Example 33.1

#### Empirical and Molecular Formulas

Molecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose?

#### Solution

The molecular formula is C6H12O6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH2O.

A molecule of metaldehyde (a pesticide used for snails and slugs) contains 8 carbon atoms, 16 hydrogen atoms, and 4 oxygen atoms. What are the molecular and empirical formulas of metaldehyde?

Molecular formula, C8H16O4; empirical formula, C2H4O

### Portrait of a Chemist

#### Lee Cronin

What is it that chemists do? According to Lee Cronin (Figure 33.7), chemists make very complicated molecules by “chopping up” small molecules and “reverse engineering” them. He wonders if we could “make a really cool universal chemistry set” by what he calls “app-ing” chemistry. Could we “app” chemistry?

In a 2012 TED talk, Lee describes one fascinating possibility: combining a collection of chemical “inks” with a 3D printer capable of fabricating a reaction apparatus (tiny test tubes, beakers, and the like) to fashion a “universal toolkit of chemistry.” This toolkit could be used to create custom-tailored drugs to fight a new superbug or to “print” medicine personally configured to your genetic makeup, environment, and health situation. Says Cronin, “What Apple did for music, I’d like to do for the discovery and distribution of prescription drugs.”1

Figure 33.7

Chemist Lee Cronin has been named one of the UK’s 10 most inspirational scientists. The youngest chair at the University of Glasgow, Lee runs a large research group, collaborates with many scientists worldwide, has published over 250 papers in top scientific journals, and has given more than 150 invited talks. His research focuses on complex chemical systems and their potential to transform technology, but also branches into nanoscience, solar fuels, synthetic biology, and even artificial life and evolution. (credit: image courtesy of Lee Cronin)

It is important to be aware that it may be possible for the same atoms to be arranged in different ways: Compounds with the same molecular formula may have different atom-to-atom bonding and therefore different structures. For example, could there be another compound with the same formula as acetic acid, C2H4O2? And if so, what would be the structure of its molecules?

If you predict that another compound with the formula C2H4O2 could exist, then you demonstrated good chemical insight and are correct. Two C atoms, four H atoms, and two O atoms can also be arranged to form a methyl formate, which is used in manufacturing, as an insecticide, and for quick-drying finishes. Methyl formate molecules have one of the oxygen atoms between the two carbon atoms, differing from the arrangement in acetic acid molecules. Acetic acid and methyl formate are examples of isomers—compounds with the same chemical formula but different molecular structures (Figure 33.8). Note that this small difference in the arrangement of the atoms has a major effect on their respective chemical properties. You would certainly not want to use a solution of methyl formate as a substitute for a solution of acetic acid (vinegar) when you make salad dressing.

Figure 33.8

Molecules of (a) acetic acid and methyl formate (b) are structural isomers; they have the same formula (C2H4O2) but different structures (and therefore different chemical properties).

Many types of isomers exist (Figure 33.9). Acetic acid and methyl formate are structural isomers, compounds in which the molecules differ in how the atoms are connected to each other. There are also various types of spatial isomers, in which the relative orientations of the atoms in space can be different. For example, the compound carvone (found in caraway seeds, spearmint, and mandarin orange peels) consists of two isomers that are mirror images of each other. S-(+)-carvone smells like caraway, and R-(−)-carvone smells like spearmint.

Figure 33.9

Molecules of carvone are spatial isomers; they only differ in the relative orientations of the atoms in space. (credit bottom left: modification of work by “Miansari66”/Wikimedia Commons; credit bottom right: modification of work by Forest & Kim Starr)

### Footnotes

1. Lee Cronin, “Print Your Own Medicine,” Talk presented at TED Global 2012, Edinburgh, Scotland, June 2012.

Supplemental exercises are available if you would like more practice with these concepts.

## 33.2 Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

$%\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass H}}{\text{mass compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$
$%\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass C}}{\text{mass compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

$%\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{2.5\phantom{\rule{0.2em}{0ex}}\text{g H}}{10.0\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=25%$
$%\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{7.5\phantom{\rule{0.2em}{0ex}}\text{g C}}{10.0\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=75%$

### Example 33.2

#### Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

#### Solution

To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
$\begin{array}{c}\\ %\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{7.34\phantom{\rule{0.2em}{0ex}}\text{g C}}{12.04\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=61.0%\\ %\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{1.85\phantom{\rule{0.2em}{0ex}}\text{g H}}{12.04\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=15.4%\\ %\phantom{\rule{0.2em}{0ex}}\text{N}=\phantom{\rule{0.2em}{0ex}}\frac{2.85\phantom{\rule{0.2em}{0ex}}\text{g N}}{12.04\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=23.7%\end{array}$

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

12.1% C, 16.1% O, 71.79% Cl

### Determining Percent Composition from Molecular or Empirical Formulas

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:

$\begin{array}{c}\\ %\phantom{\rule{0.2em}{0ex}}\text{N}=\phantom{\rule{0.2em}{0ex}}\frac{14.01\phantom{\rule{0.2em}{0ex}}\text{amu N}}{17.03\phantom{\rule{0.2em}{0ex}}\text{amu}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=82.27%\\ %\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{3.024\phantom{\rule{0.2em}{0ex}}\text{amu H}}{17.03\phantom{\rule{0.2em}{0ex}}\text{amu}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=17.76%\end{array}$

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. 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 33.3. 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.

### Example 33.3

#### Determining Percent Composition from a Molecular Formula

Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?

#### Solution

To calculate the percent composition, the masses of C, H, and O in a known mass of C9H8O4 are needed. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:
$\begin{array}{c}\\ %\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{9\phantom{\rule{0.2em}{0ex}}\text{mol C}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{molar mass C}}{\text{molar mass}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{9}{\text{H}}_{8}{\text{O}}_{4}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100=\phantom{\rule{0.2em}{0ex}}\frac{9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}12.01\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{180.159\phantom{\rule{0.2em}{0ex}}\text{g/mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100=\phantom{\rule{0.2em}{0ex}}\frac{108.09\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{180.159\phantom{\rule{0.2em}{0ex}}\text{g/mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100\\ %\phantom{\rule{0.2em}{0ex}}\text{C}=60.00%\phantom{\rule{0.2em}{0ex}}\text{C}\hfill \end{array}$
$\begin{array}{c}\\ %\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{8\phantom{\rule{0.2em}{0ex}}\text{mol H}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{molar mass H}}{\text{molar mass}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{9}{\text{H}}_{8}{\text{O}}_{4}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100=\phantom{\rule{0.2em}{0ex}}\frac{8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1.008\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{180.159\phantom{\rule{0.2em}{0ex}}\text{g/mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100=\phantom{\rule{0.2em}{0ex}}\frac{8.064\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{180.159\phantom{\rule{0.2em}{0ex}}\text{g/mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100\\ %\phantom{\rule{0.2em}{0ex}}\text{H}=4.476%\phantom{\rule{0.2em}{0ex}}\text{H}\hfill \end{array}$
$\begin{array}{c}\\ %\phantom{\rule{0.2em}{0ex}}\text{O}=\phantom{\rule{0.2em}{0ex}}\frac{4\phantom{\rule{0.2em}{0ex}}\text{mol O}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{molar mass O}}{\text{molar mass}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{9}{\text{H}}_{8}{\text{O}}_{4}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100=\phantom{\rule{0.2em}{0ex}}\frac{4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}16.00\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{180.159\phantom{\rule{0.2em}{0ex}}\text{g/mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100=\phantom{\rule{0.2em}{0ex}}\frac{64.00\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{180.159\phantom{\rule{0.2em}{0ex}}\text{g/mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100\\ %\phantom{\rule{0.2em}{0ex}}\text{O}=35.52%\hfill \end{array}$

Note that these percentages sum to equal 100.00% when appropriately rounded.

To three significant digits, what is the mass percentage of iron in the compound Fe2O3?

69.9% Fe

### 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 compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

$\begin{array}{l}\\ 1.71\phantom{\rule{0.2em}{0ex}}\text{g C}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol C}}{12.01\phantom{\rule{0.2em}{0ex}}\text{g C}}\phantom{\rule{0.2em}{0ex}}=0.142\phantom{\rule{0.2em}{0ex}}\text{mol C}\\ 0.287\phantom{\rule{0.2em}{0ex}}\text{g H}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol H}}{1.008\phantom{\rule{0.2em}{0ex}}\text{g H}}\phantom{\rule{0.2em}{0ex}}=0.284\phantom{\rule{0.2em}{0ex}}\text{mol H}\end{array}$

Thus, this compound may be represented by the formula C0.142H0.284. Per convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

${\text{C}}_{\phantom{\rule{0.2em}{0ex}}\frac{0.142}{0.142}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{\phantom{\rule{0.2em}{0ex}}\frac{0.284}{0.142}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{2}$

(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 may 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 compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

${\text{Cl}}_{0.150}{\text{O}}_{0.525}={\text{Cl}}_{\phantom{\rule{0.2em}{0ex}}\frac{0.150}{0.150}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{\phantom{\rule{0.2em}{0ex}}\frac{0.525}{0.150}\phantom{\rule{0.2em}{0ex}}}={\text{ClO}}_{3.5}$

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 33.10 outlines this procedure in flow chart fashion for a substance containing elements A and X.

Figure 33.10

The empirical formula of a compound can be derived from the masses of all elements in the sample.

### Example 33.4

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

A sample of the black mineral hematite (Figure 33.11), 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?

Figure 33.11

Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)

#### Solution

This problem provides the mass in grams of each element. Begin by finding the moles of each:
$\begin{array}{l}\\ 34.97\phantom{\rule{0.2em}{0ex}}\text{g Fe}\left(\frac{\text{mol Fe}}{55.85\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =& 0.6261\phantom{\rule{0.2em}{0ex}}\text{mol Fe}\hfill \\ 15.03\phantom{\rule{0.2em}{0ex}}\text{g O}\left(\frac{\text{mol O}}{16.00\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =& 0.9394\phantom{\rule{0.2em}{0ex}}\text{mol O}\hfill \end{array}$

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

$\begin{array}{}\\ \\ \phantom{\rule{0.2em}{0ex}}\frac{0.6261}{0.6261}\phantom{\rule{0.2em}{0ex}}=1.000\phantom{\rule{0.2em}{0ex}}\text{mol Fe}\\ \phantom{\rule{0.2em}{0ex}}\frac{0.9394}{0.6261}\phantom{\rule{0.2em}{0ex}}=1.500\phantom{\rule{0.2em}{0ex}}\text{mol O}\end{array}$

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

$2\left({\text{Fe}}_{1}{\text{O}}_{1.5}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{2}{\text{O}}_{3}$

The empirical formula is Fe2O3.

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

N2O5

### 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 33.5

#### 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 33.12). What is the empirical formula for this gas?

Figure 33.12

An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: “Dual Freq”/Wikimedia Commons)

#### 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:
$\begin{array}{l}\\ 27.29%\phantom{\rule{0.2em}{0ex}}\text{C}& =\hfill & \phantom{\rule{0.2em}{0ex}}\frac{27.29\phantom{\rule{0.2em}{0ex}}\text{g C}}{100\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}\hfill \\ 72.71%\phantom{\rule{0.2em}{0ex}}\text{O}& =\hfill & \phantom{\rule{0.2em}{0ex}}\frac{72.71\phantom{\rule{0.2em}{0ex}}\text{g O}}{100\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}\hfill \end{array}$

The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

$\begin{array}{l}\\ 27.29\phantom{\rule{0.2em}{0ex}}\text{g C}\left(\frac{\text{mol C}}{12.01\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =\hfill & 2.272\phantom{\rule{0.2em}{0ex}}\text{mol C}\hfill \\ 72.71\phantom{\rule{0.2em}{0ex}}\text{g O}\left(\frac{\text{mol O}}{16.00\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =\hfill & 4.544\phantom{\rule{0.2em}{0ex}}\text{mol O}\hfill \end{array}$

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

$\begin{array}{}\\ \\ \phantom{\rule{0.2em}{0ex}}\frac{2.272\phantom{\rule{0.2em}{0ex}}\text{mol C}}{2.272}\phantom{\rule{0.2em}{0ex}}=1\\ \phantom{\rule{0.2em}{0ex}}\frac{4.544\phantom{\rule{0.2em}{0ex}}\text{mol O}}{2.272}\phantom{\rule{0.2em}{0ex}}=2\end{array}$

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

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

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):

$\phantom{\rule{0.2em}{0ex}}\frac{\text{molecular or molar mass}\phantom{\rule{0.2em}{0ex}}\left(\text{amu or}\phantom{\rule{0.2em}{0ex}}\frac{\text{g}}{\text{mol}}\right)}{\text{empirical formula mass}\phantom{\rule{0.2em}{0ex}}\left(\text{amu or}\phantom{\rule{0.2em}{0ex}}\frac{\text{g}}{\text{mol}}\right)}\phantom{\rule{0.2em}{0ex}}=n\phantom{\rule{0.2em}{0ex}}\text{formula units/molecule}$

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

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

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:

$\phantom{\rule{0.2em}{0ex}}\frac{180\phantom{\rule{0.2em}{0ex}}\text{amu/molecule}}{30\phantom{\rule{0.2em}{0ex}}\frac{\text{amu}}{\text{formula unit}}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}=6\phantom{\rule{0.2em}{0ex}}\text{formula units/molecule}$

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 33.6

#### 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 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:
$\begin{array}{l}\\ \left(74.02\phantom{\rule{0.2em}{0ex}}\text{g C}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol C}}{12.01\phantom{\rule{0.2em}{0ex}}\text{g C}}\right)& =\hfill & 6.163\phantom{\rule{0.2em}{0ex}}\text{mol C}\hfill \\ \left(8.710\phantom{\rule{0.2em}{0ex}}\text{g H}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol H}}{1.01\phantom{\rule{0.2em}{0ex}}\text{g H}}\right)& =\hfill & 8.624\phantom{\rule{0.2em}{0ex}}\text{mol H}\hfill \\ \left(17.27\phantom{\rule{0.2em}{0ex}}\text{g N}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol N}}{14.01\phantom{\rule{0.2em}{0ex}}\text{g N}}\right)& =\hfill & 1.233\phantom{\rule{0.2em}{0ex}}\text{mol N}\hfill \end{array}$

Next, calculate the molar ratios of these elements relative to the least abundant element, N.

$6.163\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}C/\phantom{\rule{0.2em}{0ex}}1.233\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}N=5$
$8.264\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}H/\phantom{\rule{0.2em}{0ex}}1.233\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}N=7$
$1.233\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}N/\phantom{\rule{0.2em}{0ex}}1.233\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}N=1$
$\begin{array}{lll}\frac{1.233}{1.233}& =\hfill & 1.000\phantom{\rule{0.2em}{0ex}}\text{mol N}\hfill \\ \frac{6.163}{1.233}& =\hfill & 4.998\phantom{\rule{0.2em}{0ex}}\text{mol C}\hfill \\ \frac{8.624}{1.233}& =\hfill & 6.994\phantom{\rule{0.2em}{0ex}}\text{mol H}\hfill \end{array}$

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 compound:

$\phantom{\rule{0.2em}{0ex}}\frac{40.57\phantom{\rule{0.2em}{0ex}}\text{g nicotine}}{0.2500\phantom{\rule{0.2em}{0ex}}\text{mol nicotine}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{162.3\phantom{\rule{0.2em}{0ex}}\text{g}}{\text{mol}}\phantom{\rule{0.2em}{0ex}}$

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

$\phantom{\rule{0.2em}{0ex}}\frac{162.3\phantom{\rule{0.2em}{0ex}}\text{g/mol}}{81.13\phantom{\rule{0.2em}{0ex}}\frac{\text{g}}{\text{formula unit}}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.2em}{0ex}}\text{formula units/molecule}$

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

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

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?

C8H10N4O2

Supplemental exercises are available if you would like more practice with these concepts.

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### Suggested Citation

General College Chemistry. https://edtechbooks.org/general_college_chemistry

Previous Version(s)

Flowers, P., Neth, E. J., Robinson, W. R., Theopold, K., & Langley, R. (2019). Chemistry in Context. In Chemistry: Atoms First 2e. OpenStax. https://openstax.org/details/books/chemistry-atoms-first-2e

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